e^c-A^-t- A R CORNELL UNIVERSITY LIBRARY 924 103 882 26 The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924103882126 ELEMENTARY ALGEBRA REVISED BY FREDERICK H. SOMERVILLE, B.S. HEADMASTER, HILLMAN ACADEMY, WILKES-BARKE, PA. FORMERLY OF LAWRENCEVILLE SCHOOL AMERICAN BOOK COMPANY NEW YORK CINCINNATI CHICAGO Copyright, 1908, 1913, bt FEEDEEICK H. 80MEEVILLE. Entbbed at Stationers' Hall, London, EL. ALG. — REV. ■w. p. 3 PREFACE In arrangement this text follows the plan of the older stand- ard works, supplementing their strong points by the addition of the newer features demanded by modem conditions. A teachable algebra must provide a practical and well-graded general course, and, in addition, must meet the varied require- ments of the college entrance standards. In recognition of these needs this text provides certain features, among which may be mentioned the following : 1. Negative numbers are introduced as a dii-ect and natural extension of the arithmetical numbers already familiar. 2. Kew definitions and symbols are introduced slowly, so that the beginner may not be confused at the outset by a mass of new and strange terms. 3. The difficulties of the beginner are kept constantly in mind, and his progress is assisted by frequent illustrations that are sound and free from technicalities. 4. Oral drills, which form an important part of the early practice, occur wherever practicable. 5. Easy and well-graded exercises, provided in abundance, give constant opportunity for drill work on all important topics. 6. The drill work, which begins with simple exercises, in- creases in difficulty by gradual steps until it adequately meets the average college I'equirements. 7. Review exercises, covering both elementary practice and college preparation, are frequent and comprehensive. 3 4 PREFACE Special attention is directed to the thorough treatment of the graph, and to the exercises applying the more common formulas used in physics. In both of these subjects the treat- ment is non-technical, and is easily within the grasp of the young student. In several of the subjects where more than one good method might be given, it has been thought best to suggest a single practical process. One practical method thoroughly mastered is usually best for the young student, who has no need for comparative discussions that iuterest only the mature mind. In response to the suggestions of several prominent teachers, the drill-work of the text has been increased by more than fourteen hundred new examples. For further practice of an elementary character there has been included an exercise con- sisting of nine hundred examples so classified as to supple- ment those of the earlier chapters ; and for classes whose aim is college preparation a large group based upon recent exam- ination questions has been provided. With these additional drill features, the five thousand examples in the text include more than twelve hundred equations and problems. The author gratefully acknowledges his indebtedness to those friends whose suggestions and encouragment have been of material aid in the preparation of this text. Particular acknowledgment is made to those who have so generously responded with suggestions for the present revision. FREDERICK H. SOMERVILLE. The William Penn Charter School, Philadelphia. CONTENTS CHAPTER PAGE 1. Introduction, Symbols, Negative Numbers ... 9 Symbols of Operation . . .... 11 Symbols of Quality 13 Negative Numbers 14 Algebraic Expressions 20 II. Addition, Parentheses 22 Addition of Monomials 22 Addition of Polynomials .... . . 24 Parentheses . 26 III. Subtraction, Review 32 Subtraction of Polynomials 34 General Review 35 IV. Multiplication 38 Multiplication of a Monomial by a Monomial . . 42 Multiplication of a Polynomial by a Monomial . . 43 Multiplication of a Polynomial by a Polynomial . . 45 General Review 50 V. Division, Review 52 Division of a Monomial by a Monomial . . 54 Division of a Polynomial by a Monomial . . .55 Division of a Polynomial by a Polynomial . 56 General Review . . . . , . . . .62 VI. The Linear Equation, The Problem 64 General Solution of tlie Linear Equation . . . .68 The Solution of Problems 70 VII. Substitution ... 85 The Use of Formulas . 87 VIII. Special Cases in Multiplication and Division 89 Multiplication . . .... '89 Division 96 5 CONTENTS !X. X. XI. Factoring Expressions having the Same Monomial factor Trinomial Expressions Binomial Expressions Expressions of Four or More Terms factored by Grouping Supplementary Factoring ... . . Highest Common Factor Highest Common Factor of Monomials Highest Common Factor of Polynomials by Factoring Fractions, Transformations . Changes in the Form of a Fraction Transformations of Fractions XII. Fractions (Continued), Lowest Common Multiple, Lowest Common Denominator, Addition of Fractions Lowest Common Multiple .... Lowest Common Multiple of Monomials . Lowest Common Multiple of Polynomials by Factoring Lowest Common Denominator Addition and Subtraction of Fractions xra. XIV. XV. XVI. Fractions (Continued), Multiplication, Division, The Com plex Form .... Multiplication of Fractions .... Division of Fractions .... The Complex Form Fractional and Literal Linear Equations, Problems Special Fornis .... Problems leading to Fractional Linear Equations Applications of General Symbols, Review The General Statement of a Problem Use of Physical Formulas Transformation of Formulas . General Review .... Simultaneous Linear Equations, Problems Elimination by Substitution CONTENTS 7 OHATTER PAGE Elimination by Comparison 176 Elimination by Addition or Subtraction .... 176 Systems involving Three or More Unknown Quantities . 178 Fractional Forms 181 Literal Forms 185 Problems producing Simultaneous Linear Equations . 187 Problems from College Examinations .... 193 XVII. Graphical Representation of Linear Equations . .197 Graph of a Point 197 Graph of a Linear Equation in Two Unknown Numbers . 200 Graphs of Simultaneous Linear Equations . . . 201 XVIII. Involution and Evolution 205 Involution 205 Evolution . . 209 Square Root of Polynomials 213 Square Root of Arithmetical Numbers .... 217 XIX. Theory of Exponents 221 The Zero Exponent .... ... 221 The Negative Exponent 222 The Fractional Form of the Exponent .... 223 Applications ... 227 XX. Radicals, Imaginary Numbers, Review .... 236 Transformation of Radicals 237 Operations with Radicals 242 Equations involving Irrational Expressions . . . 262 Imaginary and Complex Numbers 264 Operations with Imaginary Numbers .... 255 General Review .... . . 261 XXI. Quadratic Equations . . 266 Pure Quadratic Equations .... . 266 Affected Quadratic Equations 268 Discussion of Afiected Quadratic Equations . . 277 Graphs of Affected Quadratic Equations .... 282 XXII. The Quadratic Form, Higher Equations, Irrational Equa- tions • • 286 The Quadratic Form . . .... 286 CONTENTS CHAPTER XXIII. XXIV. XXV. XXVI. XXVII. XXVIII. XXIX. Higher Equations solved by Quadratic Methods . Problems Producing Affected Quadratic Equations Irrational Equations . . • . . Simultaneous Quadratic Equations, Problems Solution by Substitution . .... Solution by Comparison and Factoring Solution of Symmetrical Types Solution of Miscellaneous Types ... Graphs of Quadratic Equations in Two Variables Problems producing Simultaneous Quadratic Equations Ratio, Proportion, Variation . Eatio . . . . Proportion . Variation .... The Progressions Arithmetical Progression Geometrical Progression The Binomial Theorem — Positive Integral Exponent Proof of the Binomial Formula .... Applications Logarithms .... The Parts of a Logarithm The Use of the Four-place Table The Properties of Logarithms The Cologarithm .... Use of Logarithms in Computations Miscellaneous Applications of Logarithms Supplementary Exercises Supplementary Topics, Review The Factor Theorem .... The Theory of Divisors of Binomials . H. C. F. of Expressions not readily Factorable L. C. M. of Expressions not readily Factorable Cube Root General Review .... Questions from College Examinations INDEX PAGE 286 200 2I»2 204 2il4 295 296 298 302 .310 315 315 317 330 338 338 348 358 359 361 364 365 367 374 376 377 379 385 417 417 419 421 424 426 430 437 445 ELEMENTARY ALGEBRA CHAPTER I INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS 1. The definite number symbols of arithmetic, 1, 2, 3, 4, 5, etc., are symbols that express in each case a number with one definite value. Thus, 3 units, 5 units, 7 units, etc., represent particular groups, the symbols, 3, 5, and 7 having each a particular name and value in the number system that we have learned to use. 2. The general number symbols of algebra are symbols that may represent not one alone but many values, and for these general symbols the letters of the alphabet are in common use. THE ADVANTAGE OF THE GENERAL NUMBER SYMBOL 3. Many of the familiar principles of arithmetic may be stated much more briefly, and usually with greater clearness, if general or literal number symbols are employed. To illustrate : (a) The area of a rectangle is equal to the product of its height, or altitude, by its length, or base. Or, arith- metically, Area = altitude x base. Using only the first letters of each word, we may write. Area = a x b, and this latter expression, while equally clear in meaning, serves as a general expression for obtaining the area of any rectangle with any values of a and 6. 9 10 INTRODUCTION. SYMBOLS. NEGATIVE NUMBEKS (6) The familiar problem of simple interest gives arithmetically, Interest = principal x rate x time. A much simpler form of expressing the same principle is obtained here as above, using only the first letters of each element for the general expression. Thus: I = p x. r xt. (c) Two principles relating to the operation of division in arithmetic may be clearly and briefly expressed by the use of literal symbols. Thus : Dividend -=- divisor = quotient. That is, — = ?• d divisor x quotient = Dividend. Or, d x q = D. 4. In a more extended manner the literal symbol permits a breadth and power of expression not hitherto possible with the number symbols of arithmetic. Many problems involving unknown quantities are readily stated and solved by means of general symbols, and clearness of expression in such problems is invariably gained by their use. To illustrate a common use of literal symbols, consider the following problem : Two brothers, John and William, possess together 200 books, and John has 20 books more than William. Write expressions that clearly state these conditions. (Compare carefully the two methods of expression.) The Arithmetical Expression (a) The number of John's books + the number of William's books=200 (6) The number of John's books — the number of William's books= 20 The Algebraic Expression Let us assume that x = the number of books that John has, and that y = the number of books that William has. Then from the conditions given in the problem : (a) x + y = 200. (6) x-y = 20. THE SYMBOLS OF OPERATION 11 By a simple process the values of x and y are readily determined. From this parallel between arithmetical and algebraic forms of ex- pression the brevity and the advantage of the literal or general symbol for number is clearly manifest. The later processes of algebra will con- stantly furnish the means wherewith we may broaden our power of expression, and the meaning of algebra will be interpreted as merely an extension of our processes with number. THE SYMBOLS OF OPERATION 5. The principal signs for operations in algebra are identical with those of the corresponding operations in arithmetic. 6. Addition is indicated by the "phis" sign, +. Thus, « 4- & is the indicated sum of the quantity a and the quantity 6. The expression is read " a plus 6." 7. Subtraction is indicated by the "minus" sign, — . Thus, a — 6 is the indicated difference between the quantity b and the quantity a. The expression is read " a minus 6." 8. Multiplication is usually indicated by an absence of sign between the quantities to be multiplied. Thus, ab is the indicated product of the quantities a and &. abx is the indicated product of the quantities a, b, and x. Sometimes a dot is used to indicate a multiplication. Thus, a • 6 is the product of a and 6. The ordinary symbol, " X ," is occasionally used in algebraic expression. An indicated product may be read by the use of the word " times " or by reading the literal symbols only. Thus, ab may be read "a times 6," or simply " ab." 9. Division is indicated by the sign " -=-," or by writing in the fractional form. 12 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS Thus, a H- 6 is the indicated quotient of the quantity a divided by the quantity 6. - is the fractional form for the same indicated quotient. o Both forms are read " a divided by 6." 10. Indicated operations are of constant occurrence in alge- braic processes, for the literal symbols do not permit the combining of two or more into a single symbol as in the case of numerals. Thus: Arithmetically, 5 + 3 + 7 may be written " 15," for the symbol 15 is the symbol for the group made up of the three groups, 5, 3, and 7. Algebraically, a + b + c cannot be rewritten unless particular values are assigned to the symbols a, h, and c. The sum is an indicated result. Algebraic expression, therefore, confines us to a constant use of indicated operations, and we must clearly understand the meaning of : I. An Indicated Addition, a + 6. II. An Indicated Subtraction, a — b. III. An Indicated Multiplication, db. IV. An Indicated Division, 2 . ' 6 11. Equality of quantities or expressions in algebra is indi- cated by the sign of equality, =, read "equals," or "is equal to." Thus, a + 6 = c + d is an indicated equality between two quantities, a + 6 and c + d. 12. If two or more numbers are multiplied together, each of them, or the product of two or more of them, is a factor of the product ; and any factor of a product may be considered the coefficient of the product of the other factors. Thus : In 5 o, 5 is the coefficient of a. In acmx, a is the coefficient of cmx, In ax, a is the coefficient of x, or ae may be the coefficient of or X is the coefficient of o. mx, etc. THE SyjIBOLS OF gUALlTY 16 Coefficients are the direct results of additions, for o (I is merely an abbreviation ofa + n + n+a + a 4 xy is au abbreviation of .nj + .rij + .)•;/ + xy. If the coefficient of a quantity is "unity ' or "1,"' it is not usaally written or read. Tlius, a is the same as 1 a. xy is the same as 1 xy. 13. The parenthesis is used to indicate that two or inore quantities are to be treated as a single quantity. The ordinary form. ( ), is most common. For clearness in the discussion of elementary principles, the parenthesis will frequently be made use of to inclose single quantities. 14. An axiom is a statement of a truth so simple as to he accepted without proof. Two of the axioms necessary iu early discussions are : Axiom 1. If equals are added to equals, the sumfi are equal. Axiti-M 2. If equals are subtracted from equals, the re, mahnhrs arc equal. THE SYMBOLS OF QUALITY 15. In scientific and in many everyday discussions greater clearness and convenience have resulted from a definite method of indicating opposition of quality. For example: Temperature above and below the zero point. Latitude north and south of the equator, Assets, or possessions, and liabilities, or debts, in business, etc., represent cases in which direct opposites of quality or kind are under discussion; hence, a need exists for a form of ex- pression that shall indicate kind as well as amount, quality as well as quantity. To supply this need the plus and minus signs are in genei-al use, their direct opposition making them useful as signs of 14 INTRODUCTION, SYMBOLS. NEGATIVE NUMBERS quality as well as of operation; and we will now consider their use as THE + AND - SIGHS OF QUALITY 16. A few selected examples of common occurrence clearly illustrate the application of quality signs to opposites in kind In the laboratory : Temperature above 0° is considered as +. Temperature below 0° is considered as — . In navigation : Latitude north of the equator is considered as +. Latitude south of the equator is considered as — . lu business administration : Assets, or possessions, are considered as + . Liabilities, or debts, are considered as — . The following illustration emphasizes the advantage of in- dicating opposites in kind by the use of the + and — signs of quality. A thermometer registers 10° above at 8 a.m., 15" above at 11 a.m., 5° below at 4 p.m., and 10° below at 10 p.m. At the right we have tabu- lated the conditions in a concise form made possible only through the use of quality signs. By applying this idea of opposition to arithmetical numbers we may establish NEGATIVE NUMBERS 17. It is first necessary to show that there exists a need for a definite method of indicating opposition of quality in number. Consider the subtractions : (1) 5-4= (2) 5-5= (3) 5-6 = There are three defijiite cases included. In (1) a subtrahend less than the minuend. In (2) a subtrahend equal to the minuend. In (.3) a subtrahend greater than the minuend. 8 a.m. + 10° 11 A.M. + 15° 4 P.M. - 5° 10 P.M. -10° NEGATIVE NUMBERS 15 The first two cases are faniiliai' in aritliiiietical processes, but the tliird raises a new question, and we ask, 5 — 6 = what number ? 18. On any convenient straight line denote the middle point by " 0," and mark off equal points of division both to the right and to the left of 0. Tlie two directions from are clearly defined cases ofop^sition, and this opposition may be indicated by marking the suc- cessive division points at the right of with numerals having plus signs, -5 -4 -3 -2 -1 +1 +2 +3 +4 +5 '''■■■ and, similarly, the division points Ze/lt of with numerals hav- ing minus signs. The result is a series of positive and negative numbers established from a given point which we may call " zero." With this extended number system we may at once obtain a clear and logical answer for the question raised above. The minuend remaining the same in each case, the result for each subtraction is established by merely counting off the subtrahend from the minuend, the direction of counting being toward zero. Therefore, For (1) TJie subtraction of a positive number from a greater positive number gives a positive result. Or, 5-4 = 1. For (2) The subtraction of a positive number from an equal positive number gives a zero result. Or, 5-5 = 0. For (3) The subtraction of a positive number from a less posi- tive number gives a negative result. Or, 5 - 6 = - 1. 16 INTEODUCTION. SYMBOLS NEGATIVE NUMBERS We may conclude, therefore, that : 19. The need of a negative number system is that subtraction may be always possible. On the principle of opposition the idea of negative number is as firmly established as that of positive number, and the definition of algebra as an extension of number is even further warranted. 20. The extension of arithmetical number to include nega- tive as ■well as positive number establishes algebraic number. 21. It is often necessary to refer to the magnitude of a num- ber regardless of its quality, the number of units in the group being the only consideration. Thus, + 5 and — 5 have the same magnitude, or absolute value, for each stands for the same group idea in tlie numeral classification. They differ in their qualities, however, being exact opposites. In like manner, + a and — a, + xy and — xy, are of the same absolute value in each respective case, no matter what values are represented by a or by xy. POSITIVE AND NEGATIVE NUMBERS COMBINED The simplest combination of algebraic numbers is addition, and since two groups of opposite kinds result from the positive and negative qualities, we must consider an elementary dis- cussion of algebraic addition under three heads. I. Positive Units + Positive Units. If a rise in temperature is 15° + 15° and a further rise of 10° occurs, -1- 10° we have a final reading of ^. 25° II. Negative Units + Negative Units If a fall in temperature is 12° _ 12° and a further fall of 16° occurs, _ 16'^ we have a final reading of . ^T^"" POSITIVE AND NEGATIVE NUMBERS COMBINED 17 III. Positive Units + Negative Units. If a rise in temperature is 20° and j. 20° an immediate fall of 15° occurs, _ 15° we have finally . .... , y If a fall in temperature is 80° and _ go° an immediate rise of 24° occurs, j. 24° we have finally . . .... "^ — 0° The student should verify these illustrations by making a sketch of a thermometer and applying each case given above. In general, addition of algebraic numbers results as follows : (1) If a and 6 are positive numbers : (+ a) + (+ 6) = + a + 6. 9 ....... ■ Numerical Illustration : V +5 +3 y (+5) + (+3) = + 5 + 3=+8. (2) If a and 6 are negative numbers : (-o) + (-6) = -a-6 ■.■■..-- O Kiimerical Illustration : V2Z2__ _p ~^ ^ (-5) + (-3) = -5-3 = -8. (8) If a is positive and & negative : (+o) + (-6)= +a-fc. o +5. Numerical Illustration : »-+2— ' -3 (+5)+.(-3)=+5-3=+2. -5. (4) If o is negative and 6 positive : (-«!) + (+ 6) = - a + 6. Numerical Illustration: +3 (-5)+(+3)=-5 + 3 = -2. From the four cases we may state the general principles for combining by addition any given groups of positive quantitiess, negative quantities, or positive and negative quantities : SOM. EI,. ALC. — 2 18 INTRODUCTION, SYMBOLS. NEGATIVE NUMBERS 22. The siini of tivo groups ofxilus, or po^itifc, miits is a posi- tive quantity. 23. The sum of two groups of minus, or negative, units is a negative quantity. 24. The sum of two groups of units of opposite quality is posi- tive if the number of units in the positive group ix the greater, but negative if the number of units in the negative group is the greater. From Art. 24 we derive the following important principle : 25. The sum of two units of the same absolute value but oppo- site in sign is 0. In general: (+ o) + (- a) = + n - a = 0. Numerical Illustration : (+ 5) + ( -.5) = + 5 - 5 = 0. Exercise 1 1. If distances to the right are to be considered as posi- tive in a discussion, what shall we consider distances to the left? 2. If the year 20 a.d. is considered as the year " + 20," how shall we express the year 20 b.c. ? 3. Draw a sketch of a thermometer, and indicate upon it the temperature points + 35°, - 18°, + 12°, and — 12°. 4. On your sketch determine the number of degrees passed through if the temperature rises from 8° to 35° ; * from to 15°; from - 15° to 10°. 5. On your sketch determine the number of degrees passed through in a fall of temperature from 35° to 10° ; from 20° to -10°. * Note that the quality of a unit is considered plus when no sign of quality is given. POSITIVE AND NEGATIVE NUMBERS COMBINED 19 6. Show by your sketch that a rise of 25° followed by a fall of 15° results in an actual change of 10° from the starting point. 7. In example 6 is the condition true for the single case when you begin at 0°, or is the result the same no matter at what point you begin ? Illustrate. 8. Show that a fall of 18° from the zero point succeeded by a rise of 30° results in a, final reading of + 12°. 9. Show that a rise of 10° from the zero point succeeded by a further rise of 18°, and later by a fall of 40°, results in a final reading of 12° below zero. Express " 12° below zero " in a simpler form. 10. From your sketch determine the final reading when, after a rise of 40° from 0, there occurs a fall of 15°, a succeed- ing rise of 7°, and another fall of 36°. Express these changes with proper signs. 11. C is a point on the line A C B AB. A traveler starts at C, goes +io ~, 10 miles toward B, turns back 7 "^ ~+9 ., miles toward A, and returns 9 miles toward B. Determine his final distance from C, and also his position at either the right or the left of C. (Assume distances to the right of C as +•) Would the information given be sufficient to determine the result without using the sketch ? 12. Determine the result of a journey 8 miles from C toward B, returning 6 miles toward A. Give the total distance trav- eled and the final position. 13. Determine the result of a journey 9 miles from C toward B, 16 miles back toward A, and then 8 miles toward B. Make a drawing similar to that illustrating example 11, and prove your answer by reference to the drawing. 20 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS 14. Determine the result of a journey 17 miles toward B, 10 miles back toward A, o miles more toward A, back 19 miles toward B, the starting point being at C. Make a drawing illustrating the entire journey. 15. With the same distances and directions as in problem 14, determine the result if the starting point had been at A, giv- ing for the answer the final distance of the traveler from C as well as from A. (Assume that from G to A is 30 miles.) ALGEBRAIC EXPRESSIONS 26. An algebraic expression is an algebraic symbol, or group of algebraic symbols, representing some quantity. An ex- pression is numerical when made up wholly of numerical symbols, and is literal when made up wholly or in part of literal symbols. 20 -I- 10 — 13 X 2 is a numerical expression. ah + mil — xy and 5 a — 6 ex — 12 are literal expressions. 27. The parts of an expression connected by the -|- or — signs are the terms. In the expression the terms are ^ b -y 2{d + k) ab, + (G-d),- mn, + ^5-±^, and - ?'• + « - ^ . ^ b-y 2{d + k) A parenthesis, or a sign of the same significance, may inclose a group as one term. A fraction as an indicated quotient is also a single term. The sign between two terms is the sign of the term following. The sign of the first term of an expression is not usually written if it is +. In the term ab of the given expression both the sign, +, and the coefficient, 1, are understood. ALGEBRAIC EXPRESSIONS 21 28. Terms not differing excepting as to their numerical coefficients are like or similar terms. 5 a% and — 3 ax are similar terms. 3 aft and 14 mn are dissimilar terms. 29. Tlie common expressions of algebra are frequently named in accordance with the number of terms composing them. The following names are generally used : A Monomial. An algebraic expression of one term. 4 a, 5OTn, — ?> xy, and 17 xyz are monomials. A Binomial. An algebraic expression of two terms. a + b, 3 m — X, 10 — T ay, — 4 mnx + 11 .? are binomials. A Trinomial. An algebraic expression of three terms. 3 a; — 7 m + 8, 4 a& — 11 ao — 10 mny are trinomials. A Pol3momial. Any expression having two or more terms. While the binomial and the trinomial both come under this head they occur so frequently that common practice gives each a distinct name. Expressions having four or more terms are ordinarily named polynomials. Oral Drill Kead the following algebraic expressions : 1. a + 3 a; — 4 mn + cdn — 3 xy. 2. 2mx — 3acd + (a + x) — (m + n). 3. (a — x) — {c—d) + (m — y + z). 4. 5 ayz — 2(2 m — w) + a{a — x}+3a(2x — 3y). 5. — mnx + 3 a(c — 2 d + 1) — (a — m + n)x + 12(4 — a;). 6. (a-x)(c + y)-(x + 2)(:y-3)-(x + l){x + 2)(x + 3). b^y 3{c-d) ^ ^ i'; V -r j^_y CHAPTER 11 ADDITION. PARENTHESES 30. Addition in algebra, as in arithmetic, is the process of combining two or more expressions into an equivalent ex- pression or sum. The given expressions to be added are the addends. THE NUMBER PRINCIPLES OF ADDITION 31. The Law of Order. Algebraic members may be added in any order. In general : a + 6 = 6 + a. Numerical Illustration : 5 + 3 = 3 + 5. 32. The Law of Grouping. The sum of three or more alge- braic numbers is the same in whatever manner the numbers are grouped. In general: a+6 + c=a + (6+c) =(a+6)+c=(o+c) + 6. Numerical Illustration: 2+3+4=2+ (3+4) = (2+3)+4=(2 + 4)+.S. A rigid proof of these laws is not necessary at this point', but may be reserved for later work in elementary algebra. The law of order is frequently called the commutative law, and the law of grouping is called the associative law. ADDITION OF MONOMIALS The principles underlying the addition of the simplest forms of algebraic expressiors have already been developed, and they are readily applied in the more difficult forms of later work. 22 ADDITION OF MONOMIALS 23 (1) The sum of like quantities having the same sign, all + or all-. By Articles 22 and 23 : + 7 + 12o -7 -12a + 3 + 5a -3 - 5a + 10 + 17a -10 -17 a In general : 33. Tlie coefficient of the sum of similar terms having like signs is the sum of the coefficients of the given terms with the common sign. (2) The sum of like quantities having different signs, some + and some — . By Article 24 : +7 + 12 a - 7 - 12 o - 3 - 5a + o + 5a + 4 +7a — 4 — 7a In general : 34. The coefficient of the sum of similar terms having unlike sii/iis is the arithmetical difference between the sum of the + co- efficients and the sum of the — coefficients, with the sign of the greater. Oral Drill (See also page 385) Add orally : 1. 5a 3 a 2. 6a 9a 3. ix 7x 4. 9mti 8mn 5. 3 bed 11 bed 6. 7 amx 19 amx 7. 17 cmy 19 cmy 8. -A ah -Sab 9. -Sen — 5 en 10. -8xz —xz 11. -15 cz -11 cz 12. — 3 CXI/ — licxy 13. — 12a6a; -19 a6a; 14. — i"> diiy -dny 15. 10 ac — 6ac 16. — 19 >n.x 9 m.v 17. -8-7/ 15 cy 18. 11' ax -&ax 19. -15 « 15 cz 20. 27 axy — 16 axy 21. — 7 cdmn 45 cdmn 24 ADDITION. PARENTHESES If the sum of three or more tei'ms is required, we apply the law of grouping (Art. 32), and separately add the + and the — terms. Thus, 5-8 + 11 -16 + 3 = 5 + 11 + 3-8- 16 = 19-24 = — 5. Result. Let the student apply this principle in the following : Oral Drill Add orally : 1. 2 — 7 + 6-9. 7. —8x + 9x-6x + 5x. 2. —8 + 3 — 12 + 7. 8. — 11 ac + 14: ao — ac — 2 ac. 3. —9 + 8 — 15 — 11. 9. 3 mux — 8 mnx — 9 m7ix + 13 mnx. 4. 7 — 14 — 3 + 10. 10. — 1 cy — Gey + 3 cy — 11 cy. 5. 13 — 18 + 7 - 21. 11. 3 a - T) a + 8 a — 11 a - 3 a + a. 6. 3a — 4a + 6a — 3a. 12. — -i xy+Sxy — T xy+xy—xy+8xy. 13. 5 am — 8 am — 24 am + 13 am — am + 6 am — 11 am. 14. 6 m — 7 — 4 m + 11 — 5 m — 17 — m + 5 m + 13. 15. — 4: ex + 8 ex — 3 ex + 2 ex — OCX + ex — 15 ex — lA ex. ADDITION OF POLYNOMIALS The principles established for the addition of monomials apply directly to the addition of polynomials. Illustrations : 1. Find the sum of 5a + 76 — 2 c, 2a — 36 + 8 c, and -3a + 26-10c. In the customary form : 5a + 76— 2c 2a-36+ 8c -8a + 26-10c 4a + 66— 4c Result. For the sum : the coefficient ofa= 5 + 2— 3= 4, the coefficient of 6= 7-3+2= 6, the coefficient ofo=-2+8-10=-4. AUDITION OF POLYNOMIALS 25 It frequently happens that not all of the terms considered are found in each of the given expressions, in whiqh case we arrange the work so that space will be given to such terms as an examination shows need for. 2. Add ia + Sb+Sm, 2b + 3c — d, 2a + 3d + 2m-x, and 5 6 — 5 m — 3 a;. 4a+36 +3m + 26 + 3c - d 2a +'id + 2in— X + 5 6 — 5 m — 3a: 6a + 10 6 + 3c + 2 m + 7 a + b — c 9. From 7 a + 3 m — 9 take 2 « — 5 m + 8. 10. From 16« + Gx-;i)/ take «-7a;-15 2/. (r«-H 5X-''/'^-^ 11. From 4a + 9rf — 18 take — 3 « + 9 cZ — 15. ' 12. From 5x—7y + 3z take .'! ;/ — 7 z. 13. From — 7 a + 6 c — 3 d + 5 take 2 a — 5c + 3. 14. From 11 a + 3 m — a; take 2 a + 7 x — m. 15. From 2 b + 5 r — ?, n take -2u ~b +.4 c — 3 n. 16. Subtract lo a.v — 3ay — 19 from 17 a.r — 5 ((y — 11. 17. Subtract lQ~3x — ly-z from z + 11 - 2 x - 3 y. '^ 18. Subtract 5c + 6d — 4m + » from 3 c — m . 19. Subtract 5 a — 11 & — 2 c from ."> o — 8 c + 2 m. 20. From the sum of 3 /;( — ;/. + 2 j} and .'5 //( — 4 ?i — 5p take the sum of m + 3 « — ji and 3 'in — 7 /i + G yj. 21. From the sum of 5 .t - 3 ?/ + 2 ;: + 3 and 3 .r + 7 subtract the sum oi 3x — y + z and 7 — 2 x — y + 11 z. 22. Subtract the sum of 4 a — 11 r + d and 3 6 — c + 10 from the sum of 3 & + cZ — 8 c and 4 <( — 4 c + 9. 23. Take the sum of 3?)i — ?/4-2;/, 2?; — 3m— 4?/, and vi+p—z from the sum of m — 3 «, 3 n -\--p, and j) + ») — 2 ?/ + » - z. GENERAL REVIEW 35 50. Addition and Subtraction with Dissimilar Coefficients. When tlie coefficients of similar terms are themselves dis- similar, the processes of addition and subtraction result in expressions with compound coefficients, i.e. coefficients having two or more terms. The principle is easily understood from the following illustrations : Addition Subtuaction ax bx (a + 6) a; cm am cdz — acm cm - c) m — 9 amdz (1- - a) cm (o - (c + 9 am) dz Exercise 6 Find the sum of : 1. am + dm. 4. —ll'nuc.ij + 2pxy. 7. hcdx+\2cdx. 2. cdx + 6 ;/.(■. 5. cdn — 7 dn. ^ 8. 5 cdx — 12 bdx. 3. 'dbcd~'2 )ii.rd. 6. ucx — 19b.v. ^9. — mxy — 11 cxy. 10. bx + aij and iiu + ny. 13. 5 ab — '1 cd and Sb + 7 d. 11. ax + nz and hx + pz. 14. 6 az + 5 by and nz — o y. 12. arm + Z»i and dm + c?«. 15. 11 )nxi/ + "i a;» and axy — cxz. 16. 2 Z//)i + 3 .^• + ?/ and 5 lii + c.b — my. 17. 3 a& + 7 ac + 11, mh — 11 c — 7, and 5 & — nc + 3. 18. (3 + <')x+C2c — 9)x+((} — 3c)x+(a + G+l)x. ^ (Examples l-]2 inclusive will serve as an exercise for subtraction ; the tirst expression ni each being the given minuend, the second expression, the subtrahend.) GENERAL REVIEW Exercise 7 1. Find the sum of 3 a + 2 ?; - 6 m, 2 b - r> c + x, —ia-T b + c — m, and a — 3 c + x. 2. Simplify 1 — J 1 — [1 ^ (1 - 1 + m) ] - m ( - »?i. 3. Add "a+2b-c, 2a-3b + 3c, ia + 3b — 7c, and — 8 a - 4 6 + .1 c. 36 SUBTRACTION. REVIEW" 4. Subtract ix+ 3ij — 2 from 5 x, and add x — oy + 2 to the result. ^ 5. Subtract 2x-3y from 4.'); + 2;/, siibtract this result from 7 x — 3y, and add x — y to the tinal remainder. 6. What expression must be added to 4 a — 3 ?» + « to pro- duce 7 a — om + 3 x? 7. What is the value of (-3) - (+ 2) - (-3) + (- 5) -(-2)? - 8. Collect —3a — b + ic,2c — 3d^-x,a — 4:X — d,5c + 2a + 3 a;, and id + b — 7 c. 9. Simplify a; — 1 — [a; — 1 — [a; — 1 — a; — 1 — .r] | . 10. Subtract ix—3y + ll from unity, and add 5 a; — 3 y+ 12 to the result. 11. Combine the m-terms, the n-terms, and the avterms in the following, inclosing the resulting coefficients in parentheses: am + 3 bn + X + m + n — ax -\- 2 n — ex. 12. A given minuend is 7 x + 12 y — 7, and the corresponding difference, ix — y + 2. Find the subtrahend. 13. To what expression must you add 2a — 3c+wito pro- duce 5a + 7c — 9m? 14. Subtract 2 a — 7 a; + 3 from the sum of 3 a; + 2 a, -ia — 10 a;, 5 a — 7, and — 4 a; — 11 a. 15. From (a-\-c)y + (m + n)z take (a — c)y — (m — 7i)z. 16. Subtract x + 17 y — z from 12 a; + 3 « and add the result to3.r+(2/ + ll2)-42/. 17. Add the sum of 4a; + 7c and 2 a; + 3c to the remainder that results when a; + 4 c is subtracted from 5 c — 11 a;. 18. From (a + 4).r + (o + 3)^/ subtract {a + l)x-\-{a + 2)y. 19. Add 3.« + 2m — 1 and 2x—3m-\-7, and subtract the sum from Q>x — m + %. GENERAL REVIEW 37 20. Simplify a- 1 — a + l+[a — 1 — a — a-1— (a- 1) — (1-a)]. ^ 21. From 7m + 3a; — 12 take the sum of 3 a; + 7 and 2m-3y-3. \ 22. What expression must be added to a + b + c to pro- duce ? ^ 23. What expression must be subtracted from to give a+b+c? ' 24. From what expression must 4 a; — 7 m + 10 be sub- tracted to give a remainder of 3x + 6m — 4:n + 2? 25. Inclose the last four terms oia — Sb + im~5n + 7x in a parenthesis preceded by a minus sign. Y^ 26. Simplify and collect a+[— 2m — 4a-|-a; — (— 2 a; — m -f- a) — 3 a;]. f 27. If a; 4-7 2^ — 9 is subtracted from 0, what expression results ? 28. Isthesumof [-7-f(-2)-(-3)] + [-{-2-K-l)|] positive or negative ? 29. Prove that [3-(-2)-(-l)] + [(-2)-|-(-l)-(-3)] + [_2_(_l) + (-5)] = 0. 30. Simplify and collect 10-[9-8-(7-6-J5-4-3^^S-l)]. M. 31. Show that 2-(-3-|-a-l)-3 + (-2-a-|-l)=-2a. \1 32. Collect the coefficients of x, y, and z in abx — acy + abz — mnx + mpy — vpz + x — y~z. -^ 33. Simplify and collect l_[_5_(_l + n^)_l|-l]_a. ^ 34. What expression must be subtracted from a — a; -j- 3 to ^ give -(a + [-a;-(-2a-^+l)-3])? CHAPTER IV MULTIPLICATION 51. Multiplication is an abbreviated form of addition. Thus, 3x4 = 4 + 4 + 4. 5a = a + a + a + a + a. For the purpose of arithmetic multiplication has been defined as the process of taking one quantity (the multiplicand) as many times as there are units in another quantity (the multiplier). This definition will not hold true when the mul- tiplier is negative or fractional. Hence, the need for the following definition : 52. Multiplication is the process of performing on one factor (the multiplicand) the same operation that was per- formed upon unity to produce the other factor (the multiplier). The result of a multiplication is a product. Illustration : The Integral Multiplier. 3x5 means that 5 is taken three times in asum. Or, 3x5=5 + 5+5. By the same process the multiplier was obtained from unity, for 3 = 1 + 1 + 1. The Fractional Multiplier. Obtained from unity a multiplier, 2f = l + l + J + J + J. For unity was taken twice as an addend and J of unity taken three times as an addend. In like manner, 2| x 5 = 5 + 5 + J + J + | = 10 + Y" = 13|. THE NUMBER PRINCIPLES OF MULTIPLICATION 53. The Law of Order. Algebraic numbers may be multiplied in any order. In general : ab = ba. Numerical Illustration : .'1x5 = 5 x 3. 38 SIGNS IN MULTIPLICATION 39 54. The Law of Grouping. Tlie product of three or more algebraic numbers is the same in whatever manner the numbers are grouped. In general: a6c = a(6c) = (a6)c = (ac)6. Numerical Illustration : 2 . 3 • 5 = 2(3 • 5) = (2 • 3)5 = (2 . 5)3. 55. The Law of Distribution. The product of a polynomial by a monomial equals the sum of the joroducts obtained by multi- plying each term of the polynomial by the monomial. In general : a{x -{■ y + z)= ax + ay + az. Numerical Illustration : 2(3 + 4 + 5) = 6 + 8 + 10. As in the case of addition, no rigid proof of these laws is ordinarily required until the later practice of elementary algebra. Here, as in addition, the law of order is frequently called the commutative law, and the law of grouping is called the associative law. SIGNS IN MULTIPLICATION Upon the definition of multiplication we may establish the results for all possible cases in which the multiplicand or multiplier, or both, are iiegative numbers. (1) A positive multiplier indicates a product to be added. (2) A negative multiplier indicates a product to be sub- tracted. (1) The Positive Multiplier. Expressed with all signs (mul- tiplier = + 3) : (+3) X (-1-5) = +5+ 5 -1-5 = + 15. (+.3) X (-5) = -5- 5-5= -15. (2) The Negative Multiplier. Expressed with all signs (mul- tiplier = — 3) : (- 3) X (-f 5) = - (+ 5 -K 5 + 5) = - (-1- 15) = - 15. (- 3) X (- 5) = - (- 5 - 6 - 5) = - (- 15) = -I- 15. 40 MULTIPLICATION Comparing the four cases in the ordinary form of multiplica- tion, we have + 5 - 5 + 5 - 5 + 3 + -3 - 3 - 3 + 15 -15 -15 +15 general : (+«)(+?,) = + nb. (-(j)( + 6) = -ab. ( + «)(- -6) = - ah. (-a) (-6)= +ab. 56. Like signs in multiplication give a positive result. 57. Unlike signs in multiplication give a negative result. Oral Drill (See also page 388) Give the products in the following, each with its proper sign: 1. (3) (-6). 7., (-6) (-7). 13. (2)(-6)(0). 2. (4)(-o). 8. (-9)(0). 14. (-5) (3) (-8). 3. (-3) (6). 9. (-3) (-11). 15. -(2) (3) (-4). 4. (-4) (5). 10. (-4) (-3) (2). 16. -(3) (-5) (-2). 5. (-5) (-4). 11. (2) (-6) (3). 17. -(-4)(-l)(-2). 6. (5)(0). 12. (_2)(5)(-3). 18. -(-3) (-5) (-8). COEFFICIENTS IN MULTIPLICATION 58. The coefficient of a term in a prodioct of two algebraic expressions is the product of the coefficients in the given multiplier and multiplicand. The principle is established by means of the Law of Grouping. For the coefBcient of mn in the product of am times en: By Art. 54 : am x en = (a x c)(m x n) = (ac)(mn) = aenin. And the required coefficient is ac. EXPONENTS IN MULTIPLICATION 41 EXPONENTS IN MULTIPLICATION 59. An exponent is a symbol, numerical or literal, written above and to the right of a given quantity, to indicate how many times that quantity occurs as a factor. Thus, if three a's occur as factors of a number, we write a', and avoid the otherwise cumbersome form of a y. ay. a. In like manner, axaxax6x6 = 0^6'', and is read " a cube, 6 square." 60. The product of two or more equal factors is a power. Any one of the equal factors of a power is a root. In common practice, literal and other factors having ex- ponents greater than 3 are read as powers. Thus, a° is read "a sixth power," or merely " a sixth." ahpz'^ is read " a cube, y seventh, 2 square." The exponent " 1 " is neither written nor read. That is, a is the same as a}. The difference between coefficients and exponents must be clearly understood. A numerical illustrar tiou emphasizes that difference. Thus : If 5 is a coefficient, 6x2 = 2 + 2 + 2 + 2 + 2 = 10. If 6 is an exponent, 2= =2x2x2x2x2 = 32. The General Law for Exponents in Multiplication By definition, a^ = a y. a x a, a* = axayaxa. Therefore, a^ya'^ = ayaxayayaxaxa = a'. Similarly, a''xa* = a^+i = a^, »»2 X to' X m = m^+'+i = m^. In general, therefore, we have the following : If TO and n are any positive integers : a'" = ax axfl!---toTO factors, a" = a X a X a ••• to n factors. Hence, a™ x a» = (a x a x a •■• to m factors) (a x a x o ••■ to w factors) = Ca X a X a ■•• to m + Ji factors) = ra"'+». In the same manner, a"' x «" x o^ = 0"'+"+'', and so on, indefluitely. 42 MULTIPLICATION This principle establishes the first index law, m and n be- ing positive and integral. The general statement of this important law follows : 61. The po'odiict of tivo or more powers of a given factor is a power whose exponent is the sum of the given exponents of that factor. Oral Drill (See also page 388) Give orally the products of the following : 1. a^ X a'. 5. x^ X :^. 9. c'' X c^ X c'. 2. m* X m^ 6. X^ X Cf23. 10. 2/» X y' X f. 3. d' X d\ 7. m X m- X m'. 11. a^ X a' X a^ X a'. 4. %" X a'. 8. (K' X «■ X a. 12. »^ X «^ X m" X nl MULTIPLICATION OF A MONOMIAL BY A MONOMIAL By application of the law of order and the principles for signs and exponents, we obtain a process for the multiplication of a monomial by a monomial. Illustrations : 1. Multiply 3 «%' by 12 a%\ By the Law of Order, 3 a?-V x 12 a^tfi = 3 x 12 x a^ x a* x 6' X 6^. By the Law of Grouping, = (3 x 12) (a'^ x a«) (6* x V^). By Arts. 58 and 61, = 36 a^fts. Result. 2. Multiply -7 aVx^z by 5 d'Wy. -7 a266a^2 x 5 a'ft^j, = -7 x 5 x o'^ X a' x 6^ x 52 x x^ x 2 x j/ (53) = ( - 7 X 5) (a2 X a') (fi^ X 62) (x^) (z) (?,) (64) = -35a''6'x3y2. Result. (67) (01) Therefore, to multiply a monomial by a monomial : 62. Observing the law of signs, obtain the product of the numerical coefficients. Tlie exponent of each literal factor in the product is the sum of the exponents of that factor in the multipili- cand and multiplier. MULTIPLICATION OF POLYNOMIAL BY MONOMIAL iS Oral Drill (See also page 388) Give orally the products of the following : 1. 2. 3. 4. 5. 6. 7. 5a 3x - -4 a 5 m — r>x —3x — 5y -3 — 7x -r,a -8m — 11 a; Id x -13y 8. 9. 10. 11. 12. 13. -3ab' inv'n 5 a^ — 3 xy' — 6 vi^ny — 3 7nnx 2ab — 3 7nn^ -7 afy"- — Tafiy — 11 mnPy 10 m^a; 14. 4 abc by 3 acd. 20. - c'd'm'' by — 6 mhi. 15. 4 aa^ by — 7 a«/2. 21. (fdxy by — 11 c*dy. 16. a'¥c* by a=62(;^. 22. - 10 x'f-z by a^y''. 17. 4 a;^?/z by — aPyzK 23. — 11 cmn^?/ by — 5 mV. 18. 3 a' by - 4 amn. 24. 13 cWaf by - 2 cW . 19. - 12 a^w" by — 2 m?n-z. 25. 15 a^i^a;^ by — 8 ^I'mrj. MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL The process of multiplying a polynomial by a monomial results directly from the number principle for multiplication assumed in Art. 55. That is : a(x + y + z) =az+ay + ag. In common practice the multiplicand and multiplier are written as in arithmetic, excepting that the multiplier is usually written at the extreme left. Illustration : Multiply 3 m'— 5 m^n + 7 miv' — 2 w^ by — 2 mn. .S ms - 5 mH + 7 mre^ -2n^ Each term of the prod- ~^™" uct is obtained by the -6 m*n + 10 m^n^ - 14 mH^ + 4 mn* Result, principles of Art. 62, for the operation is made up of successive nmltipllcations of a monomial by a. monomial. 44 MULTIPLICATION Hence, to multiply a polynomial by a monomial : 63. Multiply separately each term of the multiplicand by the multiplier, and connect the terms of the resulting polynomial by the proper signs. Exercise 8 Multiply : ^ 1. 2. 3. 4. Sa+7x 2a2 — 10a llax — 15ay a^ — lOx — 11 2 a 3 a' — axy a^ { 5. 6. ^ 7. 2m^— 10 mn + 15 n^ —a? — x^y + xy^ a*^ -S a^x + Zaoi?—x* 3 71 — 35* ax 8. 10a(7a&-8ac + 116c). 11. 3 a(a' — a^6 + aft^). 9. -3a;(.i;^ — a; + ll). 12. - 4 6c (6a; - 6m + 3 fen) . 10. — 2 m^(m'?i — mV + m?i'). 13. 11 a^a;(«^ + 9 a; — 15). ^ 64. The degree of a term is. determined by the number of literal factors in that term. 7 a'x^ is a term of the 5th degree, for 3 + 2 = 5. 65. The degree of an algebraic expression is determined by the term of highest degree in that expression. 5 rtfin + mn + n^ is an expression of the 3d degree. 66. An algebraic expression is arranged in order when its terms are written in accordance with the powers of some letter in the expression. If the powers of the selected letter increase from left to right, the expression is arranged in ascending 'order. Thus, x - 2 a;2 + 5 a;8 _ 7 X* + 10 x^. If the powers of the selected letter decrease from left to right, the expression is arranged in descending order. Thus, 4 a;9 - - 5 »7 + 3 x5 - 2 x3 - 3 X. MULTIPLICATION OF POLYNOMIAL BY POLYNOMIAL 45 67. Tlie degree of a product is equal to the sum of the degrees of its factors. 68. A polynomial is called homogeneous when its terms are all of the same degree. Thus, K* — 4 x^y + 6 xhf^ — 4 xy^ + «/* is a homogeneous polynomial. MULTIPLICATION OF A POLYNOMIAL BY A POLYNOMIAL A further application of the law of distribution for multipli- cation (Art. 55) establishes the principle for multiplying a polynomial by a polynomial. By Art. 65 : (a + 6) (a: + i/) = a (a; + y) + 6 (x + y) = ax + ay + hx-\- by. The polynomial multiplicand, (x + y), is multiplied by each separate term of the polynomial, (a + 6), and the resulting products are added. The process will be clearly understood from the following coinparison : Numerical Illustration : Algebraic Illustration : 12 10+2 a + 6 Multiplicand. 13 10+ 3 a + 7 Multiplier. 36 100 + 20 a^+ 6a 12 30 + 6 + 7 a + 35 156 = 100 + 50 + 6 a^ + 12 a + 35 Product. Explanation ; Explanation : 10 ao + 2) = 100 + 20 a (a + 5) = a2 + 5 (K 3 (10 + 2) = 30 + 6 7 (a + 5) = 7 a + :V) 100 + (20 + 30) + 6 = 156 a2 + (5 a + 7 a) + 35 = a2 + 12 a + 35 We have, therefore, the following general process for multi- plying a polynomial by a polynomial : 69. Arrange the terms of each polynomial according to the ascending order or the descending order of the same letter. Multiply all the terms of the multiplicand by each term of the multiplier. Add the partial products thus formed. 46 MULTIPLICATION Illustrations : 1. Multiply 2a + 7by3a-8. „ . „ Explanation : 6 a2 + 21 a 3 a (2 a + 7) = 6 a^ + 21 a -16a -56 -8(2a + 7) = -16a-56 6a2+ 50-56 Result. 60^ + (21 a - 16a) - 56 = 6 a^ + 5a - 56. 2. Multiply a^-2a2+3a-2 by a2 + 3a-2. a3-2 a2+3 a-2 Explanation : g3.f 3a-2 a2 (a^-2 a^ + S a - 2) =fli5_2 a* + S a»-2 a^ a6-2a*+3a'- 2a= 3a (a3-2a- + 3n-2)=3 a«-6 a'+9a2-6a + 3a*-6o'+ 9a'^- 6 a -2 (a3-2 aH"- a-2) = -2 a3 + 4 a2-6a+4. -2a°+ 4ff^- 6a+4 Adding tlie partial products, we have 06+ a*-5a3+lla-'-12a + 4 a^ + a* - Sa^ + 11 a^ - 12 a + 4. Expressions given with their terms not arranged should both be arranged in the same order before multiplication. 3. Multiply 1 — 7x^ + 0^ + 5x hy ~4,x-l + 2x\ Arrange both multiplicand and multiplier in the descending order, x' — 7a;2 + 5z + l (Let the student complete this multiplication, 2 a:'' - 4 a: — 1 writing out a complete explanation in the same form as those accompanying examples 1 and 2.) 70. Cheeking. A convenient check for work in multipli- cation can usually be made by the substitution of a small num- ber as shown in addition. Thus, in Ex. 2, if a = 2 : Multiplicand a'-2a2+3a-2= 8-8-1-0-2=4 Multiplier a^ + 3 a -2= 4-l-()-2=_8 Product a5 ^. o4 _ 5 a3 +11 o2 _ 12 a 4- 4 = 32 -M6 - 40 H- 44 -24 -f4=32 It is well to remember that this check will not always serve as a test for both coefficients and exponents. If the value, 1, had been used above, only the coefficients would have been tested, for any power of 1 is 1. MULTIl'LICATION OF MISCELLANEOUS TYPES 47 Exercise 9 ( Sec also page 38!») Multiply : 1. 4a + 7 by 3« + 5. 7. 7 ac- 3 by 5 ac + 1. 2. :;x + 4: by 7. 1- -3. 8. 11 abc- 3 by 5abc + 2. 3. r. m — 9 by 4 m +7. 9. da--^— 7 by 3a;- + l. 4. 2c — r. 2/ by 3 f + 11?/. 10. 7 m;* -.t= by 3 m» + 2a^. > 5. 4 m — 11 by 4m + ll. 11. 16.r~10xtj by 5a-'' + ll.w/. 6. 5 cd — 7 by 6 cd + 5. 12. 4 ab'' — W.vy by ab- — 3 i^a^. 13. 0^-3 0^ + 3 0-1 by c2-2c + l. 14. m^ — 2»rt+l by m^ — 2 m. + 1. 15. a= + 2a--;!a + 4 by a--3a-l. 16. 3(:/--5d + 2 by (J?-d-l. 17. 4/r-7/-3.y + 2 by ?/=-5?/-4. 18. x^ — .i"(/ + .17/- — ]f by )~ + «•.?/ + ?/-. 19. 27 - 18 m + 12 m- - 8 m-'' by 3 + 2 m. 20. 1-2 ac + 4 a^c- - 8 a'c^ + 16 a'^'' by 1 + 2 ac. '21. .V — 7 — ;> .1" + a^ by a; — 3 + 2 a^. 22. m- — 2 »i^ + 7 — 2 m^ — m by 2 m- — 9 — 3 tm? — 2 })i. 23. 12-7.i- + 5.r'-2.i- by - 8 a.-^ + a; - 3 .r" + a-^. 24. - 11 « - 7 ((•' + 17 + ((^ - 3 a- by 3 a - 10 - 7a- + 2 iv MDLTIPLICATION OF MISCELLANEOUS TYPES 71. Illustrations: 1. Multiply rt + 6 + 2 by o + & -2. « + 6 + 2 a + 6-2 a?- + a6 + 2 a + a6 + 62 + 2 6 -2a -26-4 «= + 2 a6 +6'' - 4 Result. 48 MULTIPLICATION 2. Multiply a' + b^ + c^ + 2 ab — ac — be hy a + b +c. Arranging in the descending powers of a : a^ + 2ab - ac+ 6^ - be + c^ a + b + c 0,^ + 2 a^b — a^c + ab'^ — abc + ac" + a^b +2ab^- abc +b''-b^c + bc^ + g% +2abe- ac^ + b^c — bc^ + c^ a^ + 3a^b +3 ab^ + b^ + c^ Result. 3. Multiply (a - 2)'. (a - 2)5 = (a - 2)2(a - 2) = (a2_4« + 4)(a-2) = a^ - 6 ai2 + 12 a - 8. Result. 4. (a + 5)(a + 6)(a-3). (a + 5)(a + 6)(a - 3) = [(a +5)(a + 6)](a - 3) = (a2-+ lla! + 30)(a-3) 5. (a + a;)(a + 2^). o + a; a + y = a3 + 8 a2- -3a- 90. Result. 6. a- a- a;)(a -a; -2^ -y> a2 — OK -ay + xy a2 — ax -ay + xy a^ + ax + ay + xy a:^ + ax + ay + xy Or, a^ + (x + y)a + xy. Result. Or, a^ — (x + y)a + xy. Result Iizercise 10 Perform the following indicated operations : — 1. (c + a; + 3)(c + a;-3). 2. (a+-'tn + y)(a + m — y). 3. (a + c + m + x)(a + c—'m~x). 4. (m^ + 2 m7i + m- + 2/)(m2 + 2 mn + m' — y). 5. (c^ + x^ + z^ + 2cx~cz—xz)(c + x + z). 6. (m + x + n + y){m — n—M — y). MULTIPLICATION OF MISCELLANEOUS TYPES 49 7. (a + 3bxy. 9. (3m-2ny. 11. (5u^-2ayy. 8. {3mn-2nyy. 10. (Aa^-Tte^f., ' 12. (3c m^ir:r + 5 mnoi? — 2 a;* \ by 2 wi.-»i- — 3 mux -\-2a?. 16. 8 aV - 8 aVd - 4aVd2 + 11 a^cW - 10 acd* + 3 d= by 2 aV — 3 ace? + d\ 17. 9a2-a*-16a-a= + 6a= + 3by5a-2a2-l + 3a^ 18. 16 a;«+ 1 - 4 x^- 4 a;^ by 2 a;2- 1 + 4 af'- 2 «. 19. 5m2-6-llm^ + 6m«-6m« + 2m«'by 2 + 2m*-ml 20. 6 m^n^ — 13 mS + 6 7?i'n* + 3 mV + 6 mif + 6 m^ + 8 fl' - 15 wi*n3 by 3 m* - mn^ - 2 mh< - 2 n* + mV. 21. m" — n' hj m — n. 24. m* — ?i^ by m + n. 22. m'-n-hj m + n. 25. 8 a^-27 by 2 a - 3. ^23. m^—n^hjm — n. 26. m' — 32 m° by m — 2 w. 27. 81 — m'' by 3 + m. ..^28. 27 + x'' by 3 + a;l 29. a'^ — 27 2/" by X — 3 2/1 30. 27.^■^ + 64^/''by 3a; + 4 2/^ 31. 16 nl*n* - 81 aY" by 2 mw + 3 a?y*. 32. m^ + 2 mn + n^ — a:^ by m + »i + »• 33. c^ — 4c + 4 — d^'by c — 2 + rt. 34. m- + 6m + 9 — 25 a;* by ??i + 3 — 5 a;l 35. 16 mV + 40 mw?/ + 25 2/^^ — 81 by imn + ^y + d. 36. a^ + m" + a^ + 2 am + 2 ax-\-2 mx by a + m + x. 37. 9 m^n^ + ix'z^ + 25- 12 mnas3 - 30 mn + 20 a;z by 3 mn — 2xz — 5. 38. m' + w'+p'' — 3m«p by m + w+p. 39. a'^b" - 2 a^b^ + 1 hy a'b^- 2 ab + 1. 2. 9 + 15 - 30 + 2. 8, 3. -9+ "' + 1--17. 9, 4. - 15 - -'4 - 8 + 34. 10, 5. lli - 3 + 1 - 19. 11, 6. -16 + IT -11 -34. 12 ORAL AND WRITTEN llEVIEW 61 ORAL AND WRITTEN REVIEW Exercise 14 Give orally the results of the following : 1. 5-7 + 11-8. 7. 17-10 + 19-24. -16 + 11 + 19-14. 6 a — 5 a + 8 « — 12 a + a. — 6tt+13((— a+lla— 9«. 4a;-10+3a!-5a; + 19. 5 a; - 16 + 3 a; - 8 X + 29. 13. — 8 am + 12 a?7i — 16 am — am. 14. 12 ac — 5 uc + 11 ac— 21 ac. 15. -4c + ll-6c-18 + 7c. 16. 18 0!- 14 + 19 a; +3- 20 .r. 17. (-4) + (-3)-(-l). 25. (-2)(-3) + (-3)(4). 18. (_ 8) -(-.-,) +(-3). 26. (-2) (-5) -(3) (-2). 19. (_2a) + (_3a)-(7a). 27. - (3)(- 7) + (-2)(- 10). 20. _ (3 a) -(-2 a) -(a). 28. - (2) (3) - (3)(-4). 21. 4a-(-6a) + (3a-l). 29. a(2) -a(3) + a(6). 22. 5a!+(-2a!)-(-a;+l). 30. «(- 3)-a;(- 1) + a;(- 2). 23. -3a;-(-5a;)-(a;+2). 31. (- 3) (2) (x) -(*•)(- 2). ' 24. _(5a;-l)-(-a; + 3). 32. _ (2)(- .v) + (- a;)(- 4). 33. a(a + 6 + C). 42. a-(a'' + a + l). 34. a{x + y — z). 43. x-{a?-x-2). 35. m{a—h — c). 44. ax{aV + 2 a£-^2i). 36. — m (m + ?i — 1) . 45. c- (1 — c + c^ — (?). 37. 2x{x+y-z). 46. - a''(a' + 3 a^ - 2 a + 1). 38. — m(2m — n — 3). 47. 2 aa;(a^- ax + a;^). 39. 3 a;(- a; -4 ?/'+«). 48. xij {3?f — oi?y^ + xy — V). 40. —x{-x-\-y — z). 49. -3x?/(x'-ar'2/ + 29^2_32^2) 41. ,_2c(-c + 3d-l). 50. 4a2(2a'-3a=+4a-2). 62 DIVISION. REVIEW 51. Simplify 2 a -[3 a- 2f a- 3(a-l)- 2\ + 5a]. 52. Subtract (x — a){c — y) from (a — x)(c — y). 53. Simplify {a - c + xf +{g — x + af +(x — « + cf. 54. Collect (a + \)x + (a — l)y + (a + 2) z +(a - l)x + (a + l)2/+(a-2)2. 55. Given a minuend, Sai' + Sa;^ — 4a;+2, and a subtrahend, 2a!' + 5a;'' — 4a; — 3, find the difference. 56. Given a multiplicand, a^ + 4 a^ — a + 3, and a multi- plier, a'' + 2 a — 1, find the product. 57. Given a dividend, a* — 4 a&' + &^ — 4 a'& + 6 a%^, and a divisor, a^ + &" — 2 ah, find the quotient. 58. Given a product, x"* — 4a;'-|-6a;'' — 4a; + l, and a multi- plicand, 1 — 2 X + x^, find the multiplier. GENERAL REVIEW Exercise 15 1. Show that (a; - 1)=^ - (.v - 2) = 1 -|- {x -l){x - 2). 2. Simplify (a+ !/-(« + l)(a - l)-[a(2-a)-(2a-l)]. 3. Prove that (a + ?)i)(« — iii)-\-{m + l)(m — 1) + (l_a)(l-f a) = 0. 4. Divide Ax^ -2x^ - x" -2x - l\)y 2o? - x^ - x- 1. 5. Add the quotient of (af" — 1) t- (a; — 1) to that of {x^-2x + l)^{x-l). 6. What is the coeflBcient of «c in the simplified form of (ac + 3y-3ac(ac-l)? 7. Simplify (in + l)(TO-t- 3)(m + 5) — (m— l)(m — 3)(?7i— 5). 8. Show that {x + y -\-z — l){x + y — z + l) — i{xy + z) + (x-y + z + 1)(1 -x + y + z)=0. " 9. Divide 2 a;'' — 3 a;.y — 5 a;z — 2y- —5yz — 3z'' by 2x -)- y + z. 10. A certain product is C a^ -)- 4 a'y — 9 ay — 3 ay^ + 2y*, and the multiplier 2 a'^ -|- 2 ay — y\ Find the multiplicand. ■ 11. Simplify 3[a; - 2]x - 3(2 x - o x + 7)J]. GENERAL REVIE\V C3 -18. Show that (l-;;.r + .r^=+.r(1 — .r)(2 - .r)(.S -x) — 1 =0. 13. Simplify -' jr -.",-(,•) .r + 3 ar)-a;(,r -3)-(a; + 1X2 — x^) and subtract the result from 5 — 2x. { 14. Simplify (a + 4) (a + 3) (a + 2) - (a + 3) (a + 2) (a + 1) -(a+2)(a+l)-(a + l). \ 15. Prove that (1 + c'^ (1 + a)^ - 2 (1 - ac) (a - c) = (l + cy(l+a^. 16. Subtract a + 3 from the square of a + 2, and multiply the result by the quotient obtained when «^— 1 is divided by a* + a^+ a' + a + l. < 17. Simplify a-a--[3a-(a;-a)] + [(2a;-3a)-(a;-2a)]. \ 18. Divide c'-Scd + d' + l by c^-cd-c + d^-d + l. \ 19. Find the continued product of a^ — ab + b% a' + ab + 6', and a* - a'b" + b\ ^ 20. Simplify (aV-aV + CKB-l) (ax + 1) - (aV + 1) (ax + 1) (ax - 1). 21. Multiply 8a^ — 27 by a + 2 and divide the product by 2a-3. 22. Divide c + 36 c»- 18 0^-73 0' + 12 by -5e + 4-6c*. 23. Simplify [(a;24.3x + 2)(ar'-9)]-- [(x + 3)(a^-x-6)]. 24. Show that 3^ + f+l—3xy—(l — xy)- y(f — «) — x(x'-y)=0. 25. Divide 82 mV + 40 - 67mV + 18 j/iV -45 wW by 3 m*n* — 4 mV + 5. 26. What must be the value of m in order that ar' + lSx f m may be exactly divisible by a; + 4 ? 27. Show that 2(4 + x* + a= -ax-2a-2 x) = (2-a;)2 + (a-2/+(x-a)^ 28. What must be the value of m + n in order that x* + 3 x' -f- 2 0^ + mx + n may be exactly divisible by x- + 2 x + 1 ? 29. By how much does (a'x* + 3 ax + 2)^ exceed 2(3 a V -f 2 a^a;* + 6 ax) ? CHAPTER VI THE LINEAR EQUATION. THE PROBLEM 87. An equation is a statement that two numbers or two expressions are equal. Thus, 3a;+5 = a;+7. 88. The expression at the left of the sign of equality is the left member (or first member), and the expression at the right, the right member (or secoud member) of the equation. 89. An equation is a conditional equation if its members are equal for particular values of the unknown quantity. Thus, 3x + 5 = x + 7isaii equation only when the value of x is 1. 90. An equation is an identical equation when its members are equal for any and all values of the unknown quantity. Thus, x^—l = (x+l)(x—l) is an equation for any value of x whatsoever. A conditional equation is usually referred to as an equation ; an iden- tical equation, as an identity. 91. To solve an equation is to obtain the value of the un- known number that will, when substituted for that unknown number, make the members of the equation equal. 93. The value found to make the members of an equation equal, or to satisfy the equation, is a root of the equation. A root of an equation when substituted for the unknown quan- tity reduces the original equation to an identity. 64 AXIOMS 65 93. A linear or simple equation is an equation wMch, when reduced to its simplest form, has no power of the unknown quantity higher than the iirst power. Thus : 5 a: = 15 is a linear or simple equation in x. 7 !/ = 35 is a linear or simple equation in y. 3a; + 2 = 2a; + 7isa linear equation in a, but is not reduced in form. (x + 5)2 = X- + V .r + 6 is a linear equation in x, for, when simplified, the resulting equation will have only the first power of x. While the final letters, x, y, and z, are most commonly used for representing unknown quantities in equations, any other letters may and will be used in later practice. 94. The solution of equations is based upon the truths known as AXIOMS 1. If equals are added to equals, the sums are equal. 2. If equals are subtracted from equals, the remainders are equal. 3. If equals are multiplied by equals, the products are equal. 4. If equals are divided by equals, the quotients are equal. 5. If two quantities are equal to the same quantity, they are equal to each other. In general, these axioms may be illustrated as follows. Given the equation A = B. By Axiom 1 A = B By Axiom 2 A = B Add 0, C= C Subtract O, C= C A+ C = B+ C A- C = B- C Or, briefly : 95. The same number may be added to, or s^ibtra<:ted from, both members of an equation. By Axiom 3 A = B By Axiom 4 A = B Multiply by C, C= C Divide by C, C= AC=BC A^B C O SOM. KL. ALG. 5 66 THE LINEAR EQUATION. THE PHOBLEM Or, briefly : 96. Both members of an equation may be multiplied, or divided, by the same number. By Axiom 5 It A = B and B = D ; we have, A = D. THE TRANSPOSITION OF TERMS 97. Most equations are given in such a form that the known and the unknown terms occur together in both members. Transposition is the process of changing the form of an equar tion so that the unknown terms shall all be in one member, usually the left, and the known terms all in the other. The process is based on Art. 95. Given the general equation, ax — c = bx + d (1) • By Axiom 1 , adding c = e ax =bx + c + d (2) By Axiom 2, subtracting bx = bx ax — bx = c + d (3) Compare carefully (1) and (3). In (3) we find c in the right member with its sign changed from — to +. In (3) we find bx in the left member with its sign changed from + to — . In general: 98. Any term in an equation may be transposed from one member to the other member if its sign is changed. As a direct consequence of the use of the axioms, we have : (1) The same term with the same sign in both members of an equation may be discarded. Given the equation, 3x + a — n = 2x + a + m. "Whence, Zx — n = 2x + ni. THE TRANSPOSITION OF TERMS 67 (2) The sign of every term in an equation may he changed without destroying the equality. Given the equation, — 5x+m = 6 — a. Multiplying by — 1, 6 a; — m = a — 6. Tlie slijn of a root in a solution depends upon the law of signs for division. Thus: 5a; = 20 5 x = - 20 -5a; = 20 -5a;=-20 x=4 x=— 4 x=— 4 x=4 In general : 99. WJien both members of an equation are reduced to simplest form, like signs in both members give a positive root, unlike signs a negative root. If the coefficient of the unknown quantity in a simplified equation is not exactly contained in the known quantity, the root is a fraction ; and if, in a simplified equation, the member containing the known quantities reduces to zero, the root is zero. Thatis:If 3x = 5 -4x = 7 And if 5a! = -9x = x = | x=-J 05 = a; = Oral DriU Give orally the roots of the following : 1. 5.r = 30. 6. 6a; = -36. 11. 8a; = 7. 2. 7.r = 42. 7. -7a;=21. 12. 7y = 13. 3. 4. 1- = 28. 8. -8.1= -66. 13. -5a; = 16. 4. 3^v = -18. 9. -3.r = -39. 14. 7 .v =0. 5. —3 2/ =18. 10. 5z=3. 15. — 3a; = 0. THE VERIFICATION OF LINEAR EQUATIONS 100. To substitute a root in an equation is to replace the unknown literal factor in each term by the value of the root obtained. 68 THE LINEAR EQUATION. THE WIOBLEM 101. To verify a root is to show that, by the substitution of this value, the given equation reduces to an identity. The verification of a root, as illustrated in the solutions following, should always be made in the original equation. THE GENERAL SOLUTION OF THE LINEAR EQUATION Illustrations : 1. Solve 5a;-4=3a! + 12. 5a;-4=3a; + 12. Transposing 3 a; to the left member and — 4 to the right member, 5a;- 3a; = 4 + 12. Uniting terms, 2 a; = 16. Dividing both members by the coefficient of x, X = 8, the root. Verification : In the original equation, 5x — 4 = 3a;+12. Substitute 8 for x, 5 (8) - 4 = 3 (8) + 12. 40 - 4 = 24 + 12. 36 = 36. Therefore, 8 is the correct value of the root, for, by substituting 8 for « in the original equation, we obtain an identity. 2. Solve 6 a;- [3 - (a;- 2 a;- 1)]=- 10. 5 a: - [3 - (x - 2 a; - 1)] = - 10. Eemovlng parentheses, 5a; — 3 + x- 2a; + l = -10. Transposing, 5a; + x- 2x = + 3 — 1 — 10. Uniting, 4 a; = — 8. Dividing both members by 4, a; = — 2, the root. Verification : In the original equation, 5 a; — [3 — (a; — 2 a; — 1)] = — 10. Substitute - 2 for x, 5( -2) - [3 - ( - 2 - 2(-2)-l)] = - 10. Simplifying, -10 -8-2 + 4 + 1 = - 10. - 10 = - 10. GENERAL SOLUTION OF THE LINEAR EQUATION 69 3. Solve (x + 3) (2 .r - 5)= !>(.« - 2y - 2(x + 1). {x + S){-2.r-b)=2{x-2y--2(x + l). Multiplying, 2 r- + x - 15 = 2(a;2 _ 4 ;r , + 4) _ (2 a: + 2). Removing parentheses, 2 .1- + ,»■ — 16 = 2 a;^ — 8 a; + 8 — 2 a; — 2. Discarding .r--terms and transposing, .<• + 8 a- + 2 3- = 15 + 8 - 2. .r = {\, the root. Verification : Substituting J} for r in tlie original equation, m + 3) Hi - 5) = 2(1^ - 2)-^ - 2(^ + 1) . (M)(-H)=2(-«^-2(Jf). From the foregoing, we may state the general method : 102. Perfomn all indicated midtiplicMtiuns und remove all parentheses. Transpose the teiins containing the unknown quantity to one member, and all knoion terms to the other member of the equation. Collect the tertns in each member. Divide both members by the coefficient of the unknown quantity. Exercise 16 (See also page 390. ) Find and verify the roots of the following ; 1. 4 .r 4- 5 = 3 .V + 9. 9. - 7 .v - 5 = — 5+ 2 .r. 2. 6 .V + 7 = 5 X — 11. 10. 12 .^■ - 7 - 14 .r — 12 .r. 3. 7.r-2 -5.i' = 12. 11. oa; - 3 + 19 - 7 ,v = 0. 4. 8.1- -12 = 3.1- + 8. 12. 14;r + 7-(.');+20)=0. 5. 4.V+ r> = .r- 28. 13. T) -(3 a- + 11) = -9 a-. ^6. 7-2.i-=9.v-92. 14. 5. (,--(2. 1- + 3) =12. 7. o.'B- 1 = 3.1- + 1. 15. 7a!-2(a; + l)=13. 8. 3.r+ 7-a;=-3. 16. 3(.^ + 1)- 8 - 6 :i- = 4. 70 THE LINEAR EQUATION. THE PROBLEM 17. 2a;-l+(4a;-2)=l. 18. P,x + 7 -2(x + l) = Cy 19. 3 x~{2x + l)-{Ax- 3) = 0. 20. 4a;-(5a; + l)=8-(6a; + 9). 21. -(2a; + ll) + 13 = 5-0r + l). 22. 12 .«-(« + 13) =11 a; -(a; + 15). 23. 16-(a; + 3)-(5a;-4)=8. 24. {x + 7) — {3x + l)=5x + -l. 25. 2(a;+l) + 3(a;-2)=4(a;r|-3). 26. 3(a;-3)-2(a;-2)-(a;-l)=x. ~27. 4(a;-l) = 6(a; + ll)-7(a;-13). 28. 6(a;4-l)-7 = 7-[a;-(a;+l)]. 29. 12-2(a;-5) + [3a;-(2-a;)]=l. 30. ll-(2a--3)-5 = -3a;-(5-7a!). 31. (a; + 1) (a; + 3) = (a; -2) (a; -5). 32. (2a;-3)(a;-7) = (a;-l)(a; + 4) + a;2. 33. (a;-3)'' + 2(a;-4)2-3(a;-6)(a; + 5)=7. 34. 5(a;-l)2-3(a;-2)2 = (2a;-l)(3 + .T)-6. 35. 4[3a;-2(a;24-l)] = 7-4a;(2a;-16). ^36. -[2(a;-3)(a;-5)-(a; + 7)(3-a;)]=-3(a;2 + 3). 37. (a; + 2)3-(a;-l)'-(3a; + l)(3a;-4) = 0. 38. 2.7 a; - (11 - 1.3 a;) - 6.7 a; = .62 + .4 a; - 11. 39. 0.007 X - 2 (.0035 a; + .07)= .017 -(.14 - .85 x). THE SOLUTION OF PROBLEMS 103. A problem is a question to be solved. In general, a problem is a statement of conditions involving . an unknown number or numbers. We seek the value of that unknown number, and by assuming a literal symbol for the unknown we are able to state the given conditions in terms of that unknown. LITERAL SYMBOLS KOR UNKNOWN QUAMTITIES 71 104. The solution of a problem is (1) a translation of the language into the symbol-expression of algebra by means of an assumed value for the unknown ; (2) the translation of the given conditions into equations ; and (3) the finding of the root of the derived equation. The result of a solution should always be verified in the given conditions. LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 105. A few simple statements will illustrate the ease with which the conditions of a problem are translated, or expressed, in algebraic symbols. It will be seen that in each case we first assume a value for an unknown quantity upon which the statements seem to depend, and then write expressions for the different statements. Illustrations : 1. If a; denotes a certain number, write an expression for 10 more than x. Since we may assume x = the given number, By adding 10, » + 10 = the required number. And, fulfilling the given condition, we have written a number greater than the given number by 10 ; the words "more than" being translated by the symbol of addition, + . 2. In a certain exercise John solved twice as many examples as William. Write an expression for the total number of ex- amples both together solved. We assume that x = the number that William solved. Then 2 a; = the number that John solved. By addition, Sx = the number both together solved. Here we assumed a literal symbol for that particular number upon which the problem seemed to depend ; that is, the number of examples solved by William. A simple application of multiplication and addition completes tlic translation of the conditions. 72 THE LINEAR EQUATION. THE PROBLEM 3. A man is y years of age. (a) How old was he 10 years ago ? (&) How old will he be in z years '! (a) Since y = the number of years in his present age, By subtraction, i/ - 10 = his age (in years) 10 years ago. And "years ago," a decrease, is translated by the sign — . (6) Since y = the number of years in his present age. By addition, ?/ + a = his age (in years) z years from now. And "in" referring to the future is an increase translated by the sign +. 4. m yards of cloth cost $ 3 per yard, and n yards of silk, f 5 per yard. Write an expression for the total cost of both in dollars. At % 3 per yard m yards of cloth cost (3 x m) dollars = 3 m dollars. At f 5 per yard n yards of silk cost (5 x n) dollars = 5 re dollars. Therefore, adding, (3 m + 5 ii) dollars equals the total cost of both. 5. Write three consecutive numbers, the least of the three being x. The difference between any two consecutive numbers is 1. Therefore, when x = the first and least number, a; + 1 = the second number, a; + 2 = the third number. 6. Write three consecutive even numbers, the least being m. The difference between consecutive even numbers is 2. Therefore, as above, m, m + 2, and m + 4, are the required numbers. 7. Write three consecutive odd numbers, the greatest being a ; and write an expression for their sum. The difference between consecutive odd numbers is 2. Therefore, a = the first and greatest number, a — 2 = the second number, a — i = the third number. By addition, 3 a — 6 = the sum of the three numbers. LITERAL SYMBOLS TOR UNKNOWN QUANTITIES 73 8. A boy has x dimes and y nickels and spends 10 cents. Write an expression for the amount he has remaining. Problems involving value must be so stated that different denominations are all expressed in one and the same denomination. Hence, the value of x dimes = (10 • x) cents = 10 a; cents. Hence, the value of y nickels = (5 • y) cents = ty cents. Therefore, he had at first, (10 a; + 5 j/) cents. Subtracting the 10 cents he spent, we have for an expression of his final amount, (10 a; + 5?/) — 10 cents. Result. Oral DriU 1. If a; denotes a certain number, give an expression for 16 less than x. 2. If y denotes a certain number, give an expression for 12 less than the double of y. 3. If X denotes the number of square inches in a certain surface, how many square inches are there in m similar surfaces ? 4. John has x marbles, William y marbles, and Charles z marbles. What is the expression for the total number all together have ? 5. A boy had m marbles and lost 1 of them. How many had he left ? 6. A boy earned x cents, found three times as many, and spent c cents. How many cents had he finally ? 7. A boy solved n examples and his sister solved 6 more than twice as many. How many examples did the sister solve ? 8. John caught m trout and his brother caught 3 less than three times as many. How many did both together catch ? 9. If William solves x examples, how many examples must John solve so that both together shall solve y examples ? 74 THE LINEAR EQUATION. THE PROBLEM 10. Three men together buy a field. B pays twice as much as A, C four times as much as B. If A pays d dollars, what is the cost of the field ? 11. A horse cost y dollars, a harness x dollars, and a wagon as much as the combined cost of a horse and two harnesses. What did all three together cost ? 12. If a line 10 inches in length is increased by n inches, what is the length of the new line ? 13. How much remains of a line m inches long if n inches are cut from one end and p inches are cut from the other end ? 14. A rectangle is x inches long and y inches wide, and a strip 2 inches wide is cut from one end. What is the area of the part cut off and also of the part remaining ? 15. The sum of two numbers is 10 and the smaller number is n. What is the larger number ? 16. The sum of three numbers is 60; one is a;, and another 35. What is the expression for the third number ? 17. Name three consecutive numbers, the least of the three being n. 18. Name three consecutive even numbers, the least number being n. 19. Name three consecutive odd numbers, the least of the three being m. 20. Name five consecutive numbers, the middle one being x. 21. If X is an even number, what is the next odd number above x ? 22. If a; is an odd number, what is the next even numbei above x ? 23. Name the three consecutive odd numbers below n, n being an odd number. 24. Name the three consecutive odd numbers below n, n being an even number. LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 75 25. What is the sum of the three consecutive even numbers below a; + 1, a; + 1 being odd '! 26. If X is the middle one of three consecutive odd numbers, what is the expression for their indicated product ? 27. If a man is x years old, how old will he be in a years ? 28. If a man is x years old now, how many years will pass before he is a years old ? 29. If a man is y years old now, how old was he z years ago? '30. What is the sum of the ages of a father and son if the son is X years old and the father m times as old as the son ? 31. A man is m times as old as his son and n times as old as his daughter. If the daughter is x years old, what is the sum of the ages of all three ? 32. In a period of c years a man earns d dollars. If he spends n dollars each one of the c years, what is the expression for his final saving ? 33. How many cents are represented by x dimes ? 34. Express the condition that x dimes shall equal y nickels. 35. One dollar is lost from a purse that had contained x dollars and y quarters. Write the expression for the amount remaining in cents. 36. A man travels a miles an hour for h hours. What is the total distance he travels ? 37. A boy rides x miles in a train, then y miles by boat, and, finally, by automobile twice as far as he has already traveled. Write the expression for the total number of miles in his journey. 38. What is the expression for the statement that the square of the difference of two nvimbers, a and b, is 2 less than the sum of the squares of two other numbers, m and n ? 76 THE LINEAK EQUATION. THE PROBLEM 39. A room is x yards long and y yards wide. What is the expression for its area in square yards ? in square feet ? 40. If a man is now a; + 1 years old, how many years ago was he 30 years old? How many years must pass before he will be 60 years old ? When was he a; — 1 years old ? When will he be 2 a; years old ? PROBLEMS LEADING TO LINEAR EQUATIONS 106. From the oral work just covered the student will see that no general directions can be given for the statement of problems. The following suggestions will be of assistance in the state- ment of all problems, and will serve as a general outline of the method by which we attack the different types. 107. In stating a problem : 1. Study the problem to find that number whose vcUue is re- quired. 2. Represent this unknown number or quantity by any conven- ient literal symbol. 3. The problem will state certain existing conditions or relations. Express those conditions in terms of your literal symbol. 4. Some statement in the problem will furnish a verbal equation. Translate this verbal equation into algebraic expression by means of your stated conditions. The following illustrations will show the ease with which certain common words and phrases may be translated into the common operations of algebra. Illustrations : 1. The greater of two numbers is 3 more than the less, and four times the less number exceeds twice the greater number by 8. Find the two numbers. PROBLEMS LEADING TO LINEAR EQUATIONS 77 Let X = the smaller number. Then a; + 3 = the larger number. Now the word " exceeds," in this case, may be translated by the sign " — ," and the word " by " may be translated by the sign " =." Hence, we select the conditional sentence : "four times the less exceeds twice the greater by 8," and translate :4 a; — 2(x + 3) =8. Solving the equation 4 a; — 2 (a; + 3) = 8, a; = 7, the smaller number. Consequently, adding 3, a: + 3 = 10, the larger number. The results verify in the original condition, hence in the equation. 2. Find the number which, multiplied by 4, exceeds 40 by as much as the number itself is less than 40. Let X = the number. Then 4 a; = four times the number. Translating the conditional sentence : " number . . . multiplied by 4 exceeds 40 by as much as 40 exceeds the number," X X 4-40 = 40 - X. Solving the equation, 4 a; — 40 = 40 — a;, X = 16, the number. Verification : 4(16) - 40 = 40 - 16. 64 - 40 = 40 - 16. 24 = 24. 3. A father is four times as old as his son, but in 24 years the father will be only twice as old as the son. What is the present age of each ? Let X = the number of years in the son's present age. Then 4 a; = the number of years in the father's present age. Therefore, x + 24 = the son's age after 24 years. 4 a; + 24 = the father's age after 24 years. Now, "... in 24 years the father will be only twice as old as the son." Or, 24 + 4 a; = 2(a; + 24). Solving the equation, 24 + 4 a; = 2(a; + 24). X = 12, the son's age now. 4 a; = 48, the father's age now. Verification : In 24 years the father will be (48 + 24) = 72 years of age. In 24 years the son will be (12 + 24) = 36 years of age. 78 THE LINEAR EQUATION. THE I'ROBLEM rhe problems in the following exercise are at first classi- fied in four groups involving only the simplest of commonly occurring conditions. Exercise 17 (See also page 391) (I) Problems involving One Number. 1. Four times a certain number is 36. Find the number. Let X = the required number. From the problem 4 a; = four times that number. But the problem states that 36 = four times that number. Hence, from fiur assumed condition and from the given condition, we have two expressions, 4 x and 36, representing the same quantity. Therefore, 4 a; = 36. a; = 9, the required number. Verification : 4(9) = 36. 36 = 36. 2. What is that number which, when decreased by 5, gives a remainder of 19 ? 3. Three times a certain number is diminished by 7 and the remainder is 11. What is the number 'i 4. William has three times as many books as John, and both together have 32 volumes. How many has each ? 5. If four times a certain number is added to five times the same number, the sum is 36. What is the number? 6k Four times a certain number is siditracted from eleven times the same number, and the remainder is 42. Find the number. 7. I double a certain number and subtract 7 from the result, and my remainder is 1 more than the original number. What was the number ? e\ 8. If a certain number is increased by 6, the sum is 8 less than twice the original number. Find the number. TROBLEMS LEADING TO LINEAR EQUATIONS 71) 9. Twelve times a certain mimbev is decreased hy 5, and the remainder is 15 more than seven times the oi-iginal num- ber. What was the number ? 10. Find that number -which, if doubled, exceeds 60 by as much as the number itself is less than 75. 11. What number is that which, if doubled and subtracted from 60, gives a remainder 5 less than three times the origi- nal number ? (II) Problems involving Two or More Numbers. 12. The sum of two numbers is 24, and the greater numbei is 3 more than twice the smaller number. Find the numbers. Let X = the smaller number. Then 24 — a; = the larger number. Now " the greater number is 3 more than twice the smaller," Hence, 24 - a; = 3 + 2 a;. Solving, x — 1, the smaller number. Whence, 24 — 7 = 17, the larger number Verification : 24 - 7 = 3 + 2(7). 17 = 3 + 14. 17 = 17. 13. One number exceeds another number by 5, and their sum is 49. Find the numbers. 14. One number is four times as large as a second number, and their sum is 21 more than twice the smaller number. Find the numbers. 15. The sum of three numbers is 108. The second number is twice the first number, and the third is equal to twice the sum of the first and second. Find the three numbers. 16. Find the three consecutive numbers whose sum is 54. 17. Find the three consecutive odd numbers whose sum is 39. 18. Find the five consecutive odd numbers whose sum shall be equal to nine times the smallest of the numbers. 80 THE LINEAR EQUATION. THE PROBLEM 19. Find four consecutive odd numbers such that twice the sum of the three smallest shall be 16 more than three times the greatest one. 20. Divide 17 into two parts such that the smaller part plus four times the larger part shall be 50. (Hint : Let % = the smaller part ; 17 — a; = the larger part.) 21. Divide 64 into two parts such that three times the smaller part added to twice the larger part shall be 168. 22. Divide 100 into two parts such that twice the larger part shall be 60 more than three times the smaller part. 23. Divide 75 into two parts such that three times the larger part decreased by 6 shall equal four times the smaller part increased by 9. (Ill) Problems involving the Element of Time. 24. A man is twice as old as his brother, but 5 years ago he was three times as old. Find the present age of each. Let X = the number of years in the brother's present age. Then 2 k = the number of years in the present age of the man. Now a; — 5 = the brother's age 5 years ago. And 2a; — 5 = the man's age 5 years ago. From the statement in the problem : 2a;-5 = 3(a;-5). Solving, X = 10, the brother's age now. 2 a t= 20, the man's age now. 25. A boy is 5 years older than his sister, and in 4 years the sum of their ages will be 29 years. Find the present age of each. 26. A man is twice as old as his son, but 10 years hence the sum of their ages will be 83 years. What is the present age of each ? 27. Five years ago the sum of the ages of A and B was 50 years, but at the present time B is four times as old as A. How old is each now ? PROBLEMS LEADING TO LINEAR EQUATIONS 81 28. In 7 years the sum jjf the ages of A and B will be 26 years less than three times A's present age. If A is now three times as old as B, find the age of each after 7 years. 29. In 7 years a boy will be twice as old as his brother, and at the present time the sum of their ages is 13 years. Find the present age of each. 30. A boy is three times as old as his sister, but in 4 years he will be only twice as old. What is the present age of each ? 31. A young man is 23 years of age and his brother is 11 years old. Ho'w many years ago was the older brother three times as old as the younger ? 32. A man 60 years old has a boy of 9 years. In how many years will the father be three times as old as, the son? 33. The sum of the present ages of a man and his son is 60 years, and in 2 years the man will be three times as old as the son. What will be the age of each when the sum of their ages is 100 years ? (IV) Problems involving the Element of Value. 34. A man pays a bill of $49 with five-dollar and two-dol- lar bills, using the same number of each kind. How many bills of each kind are used ? Let X = the number of bills of each kind. Then 5x = the value of the fives in dollars, and 2x = the value of the twos in dollars. Therefore, 7 a; = 49. From which a; = 7, the number of bills of each kind. \ 35. Divide $100 among A, B, and C, so that B shall receive three times as much as A, and C $20 more than A and B together. 36. A has $ 16 less than B, and C has as many dollars as A and B together. All three have $ 60. How many dollars has each ? SOM. EL. ALG. — 6 82 THE LINEAR EQUATION. 'IMIK PHOBLEM ^ 37. A number of yards of cloth cost $3 per yard, and the same number of yards of silk, $7 per yard. The cost of both pieces was $ 100. How many yards were there iu eacli piece '.' 38. $ 41 was paid to 16 men for a day's work, a part of the men receiving $2 per day and the other part $3 per day. How many men worked at each rate ? 39. A boy has $42 in two-dollar bills and half dollars, and there are three times as many coins as bills. How many has he of each kind ? 40. $ 2.10 was paid for 8 dozen oranges, part costing 20 cents a dozen and part costing 30 cents a dozen. How many dozen were there in each lot ? 41. A merchant bought 50 postage-stamps, the lot being made up of the five-cent and the two-cent denominations. Twice the cost of the two-cent stamjis was 48 cents more than three times the cost of the five-cent stamps. How many of each kind were bought, and what was the total amount paid for them ? MISCELLANEOUS PROBLEMS Exercise 18 1. What number increased by 15 is equal to 49 ? 2. AVhat number diminished by 18 is equal to 47 ? 3. What number increased by twice itself is equal to 69? 4. Divide 1 00 into two parts whose difference is 36. 5. Divide 120 into two parts whose difference is 24. 6. Find three consecutive numbers whose sum is 96. 7. Find three consecutive numbers whose sum is 159. .8. Find five consecutive numbers whose sum is 250. 9. Find three consecutive odd numbers whose s'im ^is 111. 10. Find five consecutive even numbers whose sum is 100. MISCKLLANEUUS PROBLEMS 83 11. Find two numbers whose difference is 14 and whose sum is 48. 12. Find two numbers whose sum is 50, such that twice the larger number equals 3 times the smaller. 13. Divide 60 into two parts, one of which shall be 24 more than the other. 14. Divide 84 into two parts, so that four times the smaller part shall equal three times the larger. 15. Find two numbers whose difference is 8, such that 4 times the smaller shall exceed the larger by 28. ^ . 16. A spent twice as much as B, and B twice as much as C. If all together spent $ 105, how much did each spend ? 17. Three times a certain number diminished by 4 equals the number increased by 34. Find the number. 18. Divide 70 into two parts such that 3 times one plus 4 times the other shall be equal to 230. 19. A is 5 times as old as B, but in 5 years he will be only 3 times as old. How old is each now ? 20. A house and lot together cost $6300, and the house cost 6 times as much as the lot. Find the cost of each. 21. Divide S 3000 among A, B, and C, so that A shall get $ 200 more than B, and $ 200 less than C. 22. A has $10 more than B, B has $40 more than C, and all together liave $ 700. AVhat amount has each ? 23. Four times the greater of two numbers equals five times the smaller, and their sum is 108. What are the numbers ? 24. A man is 30 years older than his son, and 12 years ago he was 4 times as old. Find the present age Of each. 84 THE LINEAR EQUATION. THE PROBLEM 25. A man has the same number of dimes as quarters, and . in all he has $ 8.40. How many coins has he in all ? 26. Three men invested $4800 in a business. A put in $400 more than B, and C, $200 less than A. What amount did each invest ? 27. A had three times as many dollars as B. A spent $ 50 while B added $ 125 to his money, and B then had $ 7 more than A. How much had each at first ? 28. Two men worked the same number of days for a con- tractor, one receiving S4, the other $3, per day. Both to- gether were paid $ 106. How many days did they work ? 29. Mr. Smith weighs 15 pounds more than double the weight of his boy, and both together weigh 255 pounds. What is the weight of each ? 30. Two kinds of coal cost $5 and $7 each per" ton, and a school paid $288 for a supply made up of the same number of tons of each. How many tons were bought? 31. Mr. Smith's farm is 5 times as large as Mr. Brown's. After Smith sells 120 acres to Brown, the Smith farm is still twice as large as the other. How many acres are there in each? 32. The sum of the ages of a man and his son is 50 years. If the boy were 7 years older, the father would still be twice his age. How old is each ? 33. Twelve books were bought for $4.50. For a part of them the price paid was 30 cents each, and for the other part 40 cents. What number were bought at each price ? 34. Two automobiles start from the same point at the same time and go in opposite directions. One goes 20 miles, the other 25 miles per hour. After how many hours will they be 135 miles apart ? CHAPTER VII SUBSTITUTION 108. Substitution is the process of replacing literal factors in algebraic terms by numerical or by other literal values. Illustration : Find the value of 10 {x + 1)^ — 3 (a; - 5)^ +2x^ when a; = 3. 10 (s- + 1)2 _ 3 (x - 5)2 + 2 :f2 = 10 (3 + 1)2 - ■?. (3 - 5)2 + 2 (3)2. = 10 (4)2 -3 (-2)2 + 2. 9 = 10 ■ 16 - . 4 + 18 = 160 -12 + 18 = 166. Result. From the foregoing process we may state the general method : 109. Replace the literal factors in the terms of the given ex- pression by their respective given values. Simplify the result. Exercise 19 Find the numerical values of the following when a=4, 6=3, c = 4, and d = l: 1. a + b + c. 7. ab — 3 be + 5 ad. 2. a-2b + 5c. 8. 2ab-3cd + 2bd. 3. 4 & — 3 c + 2 d. 9.5ad — 3bd+8cd. 4. 7a — ri + 9c — 6. 10. ab—{hc — cd). 5. 10 a + 6 -(3 c — d). 11. S abc - 5 bed + 2 acd. 6. 4 J) _ [3 c — (2 a + d)]. 12. acd — {abc — bed ^ acd). 85 St; SUBSTITUTION Find the value of : 13. a(a + c) — c(c — a) when a = 4 and c = 2. 14. (a + 1)2 - (a + l)(a - 1) - (a - 1)^ when a = 5. 15. (a+b)~(a + b){a -b)—(a — by whon a = 4, and b = 3. 16. 10(a + ly- (3 a- 2)^- (u - 1)^ when a = - 2. 17. 4c(x + yy— 8(a; + y)(x — y) when a; = 2 and y=0- 18. 3(m — a;)''— 2(m+x)(m — x) + x when m = — 5 and a;=L'. 19. d — 2(c + d) (c — 2 d) — 3(c — c?) when c = 2 and d = — 2. 20. a{a +b) - b(a — by—(a +by when a = - 3 and b=-^2. 21. 7a(a + l) — (2 + a)(3a— 5)— 3a(a — 1) when a = 0. 22. (3 m — 2)2 + (m — n+ ly — (m — ny when m = n = — 3. 23. (x + n)(x + w — 1) when x = m and n = — m. 24. (2 a — 3 « + a;)(a; — 2 a) — 2 acx when a = a; and c = 0. 25. 5 ac — (a + c)(a — 4 c) + (a — c)^ when c = 0. 26. (2 x + iy— 3 x' — i x(x— 5) when a;= — 2a. 27. (3 X - ly - (3 X + 1)(2 a; - 6) when x = ab. 28. (3 a-2 - 2 a; + 1) - (3 a- - 2y when a- = - mn. 29. (m + «)(7)i, — a;) — (m — a;)" + 2 m when 7;i = — x. 30. 2(m — a;) — 2(?)i + 3 x){m — 2 a;) — 3 ma; when m = a and a; = — 2 a. 31. (a - 2 6 + 8)2 _ a;(a; + 2 a) - (a - 2 6)' - 1 when a = 2i and a; = 0. 32. {6x'' — llxy— 10 ^z'') — 6 (a;" — 2 a;?/) + 11 y^ when x = m and ;'/ = — m. 33. (a; — a) (a; + a — y) — (« — 2/) (a; + «) = 2 R.) 9. If the diameter (JD) of a sphere is 2 feet, what is the volume (F) of the sphere from the formula, V= ? (7r = 3.1416 -.) ^ 10. Find the last term (J) in a series of numerical terms of which the first term (a) is 3, the number of terms (n) is 8, and the difference between the terms (d) is 2, the formula be- ing l=ia+ {n — l)cl. 11. If o = 3, 7" = 2, and n = 5, what is the value of I in the expression I = ar"~^ ? 12. If r —5, s = 4, and n = 6, find the value of a in the ex- pression a = (/• — l)s -f- ?•""'. CHAPTER VIII SPECIAL CASES IN MULTIPLICATION AND DIVISION MULTIPLICATION The product of simple forms of binomials may often be written without actual multiplication, the result being obtained by observing certain laws seen to exist in the process of actual multiplication. Three common cases are : The square of the sum of two quan- tities. a + b o + 6 II The square of the difference of two a — a — ab ab + h^ •■- ab - ab + b"- III The product of the sum and difference of two quantities, a + b a — b a^ + ab -ab-V (I), (II), and (III), a^+2ab + b^ a^-2ab + b^ Therefore, we may state, from respectively : 111. The square of the sum of two quantities equals the square of the first, plus twice the product of the first by the second, plus the square of the second. 112. The square of the difference of two quantities equals the square of the first, minus twice the product of the first by the second, plus the square of the second. 113. The product of the sum and difference of two quantities equals the difference of their squares. Ability to apply these principles rapidly is essential in all later practice. 89 yU SPECIAL CASES IN ML'LTll'LlCATlON AND DIVISION Oral Drill (See also page 396) Give orally the products of : 1. {a + mf. 4. (a; + 4)1 7. {Sa + 2)\ 2. {x + zf. 5. (a + 3 my. 8. {4.a + Sf. 3. (a; + 3)1 6. (c + 5 d)l 9. (5 c +7)1 10. {c -xf. 13. (c-8)l 16. (cd-4)2. 11. (&-4)2. 14. (3 a -5)2. 17. («%2-6)l 12. {m-Bf. 15. (7 a -3)2. 18. (cd-3ca;)2. 19. (a + x)(a — x). 27. (3 ?ft?i + 5 ma;)^. 20. (m + y)Cm — y). 28. (4c-V72 — 5)^. 21. (a; + 4j(a;-4). 29. (2 m' + 9)(2 m^- 9). 22. (2a + l)(2a-l). 30. (5 xyz + 7 yV)- . 23. (3a + 5)(3a-5). 31. (3 x*- 13)(.3 a;* + ]3). 24. (7a + 10)(7a — 10). 32. (am'' + xyz){am^ — xyz). 25. (5x-8y){5x + 8y). 33. (3 c^ + ll)^. 26. (3 a^ + 5)(3 a^ - 5). 34. (7m'-9xy. IV. THE DIFFERENCE OF TWO SQUARES OBTAINED FROM TRINOMIALS 114. Many products of two trinomials may be so written as to come under the binomial case of Art. 113. In such multiplications we group two of the three terms in each quantity so an to produce the same binomial expressions in each, the pa- renthesis being treated as a single term,. Three different cases may occur : (1) (2) (3) (a+6 + f)(a + 6-c) (a-hb + c)(a-b + c) (a+b- c)(a -b + c) Tlie terms inclosed in parentheses must form the same binomial in each exprestiion. THE DIFFERENCE OF TWO SQUARES 91 From 1. (a + 6 + c)(a + 6 - c) = [(a + 6) + c][(o + 6) - '] = (a + b)' - C = a' + 2ab + b- — c-. Result. From 2, (a + 6 + c) (a - 6 + c) = [(a + c) + 6] [(a + c) - h] = (a + c)2 - 6- = a^ + 2 ac + c- — 6^. Result. From 3. In this case only one teiin has the same sign in each ex- pression, the o-term. Hence the last two terras of each expression are inclosed in parentheses, the parentliesis in one case being preceded by the minus sign. (o + 6 - c)ia - 6 + c) = [a + (6 - c)][a - (6 - c)] = a- - (6 - e)2 = a2 _ (hi _ 2 6c + c2) = a- — b- + 2bc — (■■-. Result. Exercise 21 Write by inspection the following products : 1. [(a + a;) + 4][(a + s-)-4]. 8. {m + x + 2)(m + x-2). 2. [(m + a;) + 2][(jn. + a-) — 2]. 9. (m — ii + c)(m— n—c). 3. [(c-2)+m][(e-2)-m]. 10. (r + c + l)((r + c- 1). 4. [(a^ + 1) + a] [(a^ +!)-«]. H. (c-d + 2)(c-cZ-2). 5. [a + (d + x)]la-(d + x)]. 12. (m + w -4)(m- « +4). 6. [d+(3/-2)][f«-(2/-^)]- 13. (.v'-2-x)(x'-2 + x). 7. (a+a; + 2/)(a + a^-y). 14. (x" + x-2){ic? -x + 2). 15. (a;' + a^+n(.»--.r=+l). 16. (m*-2m-' + l)(m<+2m- + l). 17. (x*-x'-(^){x' + x' — 6). 18. [(c + d) + (m + l)][(c + d)-(m + l)]. 19. [(n + a;) - (?/ - z)][(« + x) + (y - 2)]. 20. (c^-c^i-f-lXcS+c^ + c-l). 92 SPECIAL CASES IN JIULTIPLICATION AND DIVISION V. THE PRODUCT OF TWO BINOMIALS HAVING A COMMON LITERAL TERM By actual multiplication : a + 7 a + 5 a^ + 7 a + 5 g + 35 a' + 12 a + 35 Hence, in the product : a^ = a X a, the product of the given first terms. + 12a=(+7 + 5')a the product of the common literal term by the sum of given last terms. + 35 = (+ 5)(+ 7) the product of the given last terms. In like manner : (1 ) (k - 4) (a; - 9) = a;2 -13 x + 36. x^ = x-x. -13a;=(-4-9)a:. + 36= (-4)(- 9). (2) (x + 9)(x - 3) = x-^ + 6 a; - 27. x^ = x-x. +6x= C + 9-3)x. -27=(+9)(-3). (3) (m-15)(m + 7) = m2-8m- 105. m^ = m-7n. -8 TO = (-15 + 7)m. -105=(- 15)( + 7) In general : {x + d)(x + b) = x'' + (a + b)x + ab. Therefore, in the product of two binomials having a com- mon literal term: 115. Tlie first term is the product of the given first terms. Hie second term is the product of the common literal term by the algebraic sum of the given second terms. Tlie third term is the product of the given second terms. Oral Drill (See also page 396) Give orally the following products : 1. (x + 3)(x + 4). 6. (c + 12)(c + l). 2. (a;-t-4)(a; + 5). 7. (w + ll)(w + 12). 3. '(x+5)(x + 7). 8. (m + 3)(m + 20). 4. (m + 6)(m + 4). 9. (d + 12)(d + 15). 5- (2/ + 9)0/ + 4). 10. (c + 15)(c + 16). THE PRODUCT OF ANV TWO BINOMIALS 93 11. (6-4)(6-7). 15. (a58-9)(a;z-l). 12. (?i - ll)(n - 10). 16. (c2_3)(c^-4). 13. {ax — 3)(ax-5). n. (a? - xy){ai' - 2 xy). 14. (cd-7)(cd-3). 18. (mn-3n){mn-7n). 19. (.r-3)(a; + 6). 24. (a-13)(a+10). 20. (c-3)(c+8). 25. (a'^+5)(a=-3). 21. (a;-9)(a; + 7). 26. (m' + 8)(m'- 3). 22. (a; + 9)(a; - 10). 27. (a&-9)(a6 + 12). 23. (m + ll)(m-12). 28. {ca?-l){c3? + S). VI. THE PRODUCT OF AKT TWO BINOKUALS By actual multiplication : 2 o + 5 3a +4 f 6 a2 + 15 a + 8a + 20 6 «« + 23 a + 20 The two multiplications resulting in the terms + 15 a and + 8a are cross products. It will greatly assist the beginner to imagine that the terms entering the cross product have this connection : I P=i — \ (2a+5)(3a+4) And, by inspection, the middle term results from (+2a)(+4)+ (+3a)(+5) = + 8a + 15a = + 23 o. Considering other possible cases : (1) In (4 a — -7) (3 a — 5) the middle term in the product will be (+4a)(-5) + (+3a)(-7) = - 20 a -21 a = -41 a. (2) In (2 a + 5) (3 a — 4) the middle term in the product will be (+2a)(-4) + (+3a)(+5)=-8a + 15a = + 7 a. (3) In (3 a — 2) (9 a + 4) the middle term in the product will be (+3n)(-4) + (+9a)(-2) = + 12a-18a= -6 a. 94 SPECIAL CASES IN MULTIPLICATION AND DIVISION In general : (ax + b) (ex + (l)= acx^ + (ad + hc.)x + hd. Therefore, in the product of any two binomials : 116. The first term is the product of the given first terms. The second term is the algebraic sum of the cross products of the given terms. The third term is the product of the given second terms. Exercise 22 (See also page 397) Write by inspection the following products : 1. (3a; + 2)(a; + l). 13. 2. (2 a + !)(•'! a + 2). 14. 3. (3m+2)(2m + 3). 15. 4. (4 a; + !)(»; + 3). 16. 5.' (6c + 4)(2c + 3). 17. 6. (6b + 5)(2b + 3). 18. 7. (6« + 7)(3 2 + 4). 19. 8. (7m + 2)(2m + 9). 20. 9. (3a-7)(2a-5). 21. 10. (3m-l)(2m-3). 22. 11. (5c-2)(3c-l). 23. 12. i6y-l){2y-3). 24. 7.v-2)(3a;-7). 10c-ll)(3c-7). 7 n-9)(8n-5). 5 mn — 1)(3 mn — 4). 4 a + 7 m) (3 a — 2 m) 7c-3d)(8c-|-3d). 2ac — 3 x){ac->r 11 x) 14m — 5na;)(2m + «■») • llx + iac- 12 be. Result. In genera] : 117. The sc/KdVP of any polynomial Is the sum of the. squares of the several teriiis loijcthi'i- ivith twice the product of each term by each of the terms that folloiv it. Exercise 23 (See also page 397) Write by inspection the following products : 1. (a + m + n)'-'. 7. (a + c + m + xy. 2. (,_• + d + .!•)-' 8. {m -2n + 3x- 2)1 3. (,f + ,. _ my. 9. (3 a - 2 6 - c - 1)^ 4. (a + 2b + 3 cy. 10. *(1 - c - c' - cy. 5. , 2 „,, _ 3 »( + 4)^ 11. *(m^ -m' + m- ly. 6. (3m-4n-5y. 12. *{d' - 3cP + 4.d - 2y. DIVISION In certain cases where both dividends and divisors are bi- nomials we are able to write the quotients without actual divi- sion. Such divisions are limited to the case.s where the terms of the dividend are like powers ; that is, both terms must be squares, both cubes, both fourth powers, etc. The powers of the binomial divisors are also like. I. THE DIFFERENCE OF TWO SQUARES By Art. 113, (a + 6)(a - 6) = a^ - b'K Therefore, by division : «i^6! = a_6, and ^^ = a+b. a+b a -6 * After squaring, the terms should be collected. 96 SPECIAL CASES IX MULTIPLICATION AND DIVISION Hence the general principle may be stated as follows : 118. The difference of the squares of two quantities may he divided by either the sum or the difference of the quantities.- If the divisor is the sum of the quantities, the quotient will he the difference of the quantities ; and if the divisor is the difference of the quantities, the quotient will he the sum. Oral Drill (See also page 397) Give orally the quotients of : 1. (a2-m2)-=-(a-m). 8. (16-9m2)-i-(4-3TO). (26a;2-492/^--(5a!-7?/). 10. (49c2d2-9)-=-(7cd + 3). (100-81a^)-f-(10-9a;=). (ar=/ - 121) -(a^ + 11). (81 mV-l&Qc") -=- (9 mn - 13 c^. 7. (9d2-25)^(3d+5). 14. (196c«-81d^)-f-(14c» + 9d2). II. THE DIFFERENCE OF TWO CUBES yj3 _ 7j8 By actual division : — —a^^ah + h^. a — b From the form of the quotient and its relation to the divisor we state: 119. The difference of the cubes of two quantities may be divided by the difference of the quantities. TJie quotient is the square of the first quantity, plus the product of the two quantities, plus the square of the second quantity. Oral Drill (See also page 398) Give orally the quotients of : 1. (a''-c5)^-(a-c). 3. (a? - ^) -i- (x - z). 2. (?M'-r')-=-(m-.r). 4. (ar' - 1) -^ (a; - 1). 2. (m=-a;^)-^(m-a;). 9. 3. (a=_4)-(a + 2). 10. 4. (af-9)^(x + 3). 11. 5. (c2-36)--(c-6). 12, 6. (4m2-l)-=-(2m-l). 13, THE SUM OF TWO CUBES 97 5. (c'_8)-(c-2). 9. (r\P-12r,)^(rd~5). 6. (d3-27)-(rf-3). 10. (8.r^-27«r)-=-(2.r~3m). 7. (64-ar')H-(4-a!). n, (512cW-729) --(8ed-9). 8. (3^f - 1) -i- (xy - 1). 12. {2Way'f-1000z'')-i-(6xy-10z). in. THE SUM OF TWO CUBES By actual division : " "^ =a^ — ab + b^ a + h From the form of the quotient and its relation to the divisor, we state: 120. Tlie sum of the cubes of two quantities may be divided by the sum of the quantities. The quotient is the square of the first quantity, minus the product of the two quantities, plus the square of the second quantity. Oral Drill (See also page 398) Give orally the quotients of : 1. {^a?j^c'')^{a + c). 7. {125 + d') ~ {5 + d) . 2. (m?+o^)^{m + x). 8. (27 + d^) -=- (3 + d). 3. (c= + /)-(c + 2/). 9. (my + l)-=-(m?/ + l). 4. (a^ + 1) -=- (a + 1). 10. (mV + 125) -- (ma; + 5). 5. (a? + ?,)^{x + 2). 11. (8c' + 272=)H-(2c + 3z). 6. («= + 27)H-(n+3). 12. (125 &^ + 64) -f- (5 & + 4). 13. (729 mV + 1000 x^) h- (9 mn' + 10 a"). IV. THE SUM OR DIFFERENCE OF ANY TWO LIKE POWERS (a) The Dipferenck Even Powers Odd Powers a2 - 62 a3 - 63 a* — 6* may each be divided by a^ — h^ may each be divided a^ — 6^ (a + 6) or (a — 6). a' — 6' by (a — 6) only. o8 - 6* a? - V etc. etc. SOM. EL. AT.G. 7 98 SPECIAL CASES IN MULTIPLICATION AND DI\'ISION (6) The Sum Even Powers Odd Powers a-2 + 62 a^+V a +b* are not divisible by n^ + b^ may each be divided a' + 66 either (a + 6) or (a — 6). a- + 6' by (a + 6) only. 2r ~ 30 p + 9. TRINOMIAL EXPKESSIUNS 103 19. 36 c= + 25 m^- 60 cm. 22. Gix^ -SOx'y^ + 25y*. 20. 72 cd + 16 c2 + 81 cP. 23. 49 a^ft^ + 140 abc f 100 c^ 21. 36a;< + 84a^ + 49. 24. SlaV-lQSa^a;?/^ 121 j/l (ft) The Trinomial whose Highest Power has the Cokffi- ciENT Unity Type Form ••• x^ + (c '+ d)x -\- cd. The Origin : If two binomials, (;« + c) and (x + d), are multi- plied, we have (Art. 115) : (0;+ c)(« + d)= x{x + d) + c(x + (?) = a;2 + dx + ex + cd = a;2 + (c + d) X + cd. In the resulting product : The first term x^ = the product of the given first terms. The second term (c+ d)x = the product of the given first term hy the alge- hraic sum of the given second terms. The third term cd = the product of the given second terms. Hence, the coefficient of tJie second term of the product is the sum of the gii'en second terms. Illustrations : 1. x^ + 8.« + 15 = (a; + ?) (a; + ?). We require the two factors of + 1-") whose sum is + 8 : +3 and + 5. Therefore, x^ + 8 x + 15 = (x + .3) (x + 5). Result. 2. ar'-8a; + 1.5 = (,v-?)(a;- ?). We require the two factors of + 15 whose sum is — 8 : — 3 and — 5. Therefore, x^ - 8 x + 15 = (x - 3) (:c - 5). Result. 3. ar" + 2a; -15 = (a; + ?)(»;-?). In this expression the sign of 15 is -, hence the signs of its factors are unlike. The sign of 2 is +, hence the greater factor of — 15 is + We require the two factors of — 15 whose sum is + 2 : +5 and — 3. Therefore, x^ + 2 x - 15 = (x + 5) (x - 0). Result. 104 FACTORING 4. oi?-2x-15 = (x-?)(x+?). In this expression the sign of 15 is — , hence the signs of its factors are unlike. The sign of 2 is — , hence the greater factor of — 15 is —. Therefore, a;^ — 2 a; - 15 = (a; - 5) (a; + 3). Result. From the illustrations we make the following important conclusions : 131. (1) The first aid to -^factoring such expressions is the sign of the third term. (2) If that sign is +, the signs of the second terms in the required factors are like; but if that sign is —, the signs of the second terms of the factors are unlike. (3) The sign of the given second term is the same as that of the greater factor of the given third term. Exercise 27 Give orally the factors of the three groups, a, b, and c. (a) The third term +. The second term +. The sign of the last term of each factor + . Illustration : x^ + 10 a; + 24 = (s + 4) (a; + 6). 1. x' + Tx + U. 5. d' + 20d + 36. 2. m^ + 8m + 16. 6. 2/' +19 2/ + 48. 3. c' + 12c + 20. 7. c2 + 31c + 68. 4. a:^+12a + S2. 8. p^ + 37p+70. (b) The third term +. The second term — . The sign of the last term of each factor — . Illustration : x^ - 10 a; + 24 = (a; - 4) (x - 6). 9. a--8a + 12. 13. a;^ — 18a; + 32. 10. a^-9a; + 18. 14. c2-16c + 39. 11. c2-9c + 14. 15. n^-llw + 24. 12. m^- 10m + 21. 16. f-l'iy + 2i. TRINOMIAL EXPRESSIONS 105 (c) The third term — . The second term either + or — . The signs of the last terms of the factors unlike, the greater last term having the same sign as the second term of the given trinomial. Illustrations : a;^ + 2 a; - 24 = (x + 6) (x - 4). a;2 - 2 a; - 24 = (X - 6) (X + 4). - 17. a;2 + a;-20. 21. 0^-15 0-34 18. m^-3m-10. 22. a2-9a-70. 19. c2-5c-14. 23. z= + 13«-48. 20. a^ + 8a;-20. 24. x^-%x-12. (d) Write the factors of: ~ 25. m^ —15m- 54. 34. mV - 8 mn - 84. 26. c''-c-132. 35. a^y^ - 72 -xy. 27. y'-lly-26. ^36. 28 - 16 xa + a^g^. 28. llO-SSc-c^. 37. aV -15 az- 76. 29. ar'/-25a;2/ + 46. 38. 96 + 28 xyz + xYz'. 30. n'-9n'-112. 39. 0^ + 33 0^-70. 31. 2a;-120 + a^. 40. 63 a=6^+a<6« - 130. 32. 6= + 13&-140. 41. mV-mn- 210. 33. cP-6d-135. 42. a*bV-7a^b^c-144:. 43. aVx= + 5aca;-36. 44. 45 + 4 mn- — m^n*. 45. 33 + 8 c^d-c'dl 46. 54-15af'-a;«. 47. ac''d-aVd2 + 12. 48. 2 mna;-mVa;' + 143. 49. 380-cdV-c2dV. 106 FACTORING (c) The Trinomial whose Highest Power has a Coefficient Greater than Unity , Type Form • • • acx^ + iad + bc)x + bd. The Origin: If two binomials, (aa! + 6) and (ccB + d), are multiplied, we have (Art. 116) : (ax + 6) (ex + d) = ax (ex + d) + & (ex + S) = acx^ + adx + 6cx + bd = acx'-* + (ad + 6c) x + bd. Now the coefBcient of x in the result (ad + be) is made up of the coefficients that, multiplied, would produce abed. Therefore, 132. We require the two factors of abed that, added, will produce ad + be. The application of the principle in practice will be readily understood from the following illustrations : 1. Factor 6 a;2 + 25 a; + 14. 6 X 14 = 84. Required the factors of 84 that, added, equal 25. By trial we find them to be 4 and 21. Therefore, 6x2 + 25x + 14 = 6x2 + 4x + 21 x + 14 = (6x2 + 4x) + (21x + ^4) = 2x(3x + 2)+7(3x+ 2) = (2 X + 7) (3 X + 2) . Result. 2. Factor 14 a^ + 31 a - 10. 14 X -10= -140. Required the factors of —140 that, added, equal 31. By trial they are found to be +35 and —4. Therefore, 14 a^ + 31 a - 10 = 14 a^ + 35 <, _ 4 „ _ 10 = (]4a2 + 35a)-(4fi + 10) = 7a(2a + 5)-2(2a + 5) = (7a -2)(2a+-5). Result. Several excellent methods for factoring a trinomial of this type form might be given, but to understand and apply accurately one method is a better plan for the beginner. Y^x BINOMIAL EXPRESSIONS 107 Exercise 28 rite the factors of : 1. 4m2 + 8m + 3. 14. 9m=' + 23m + 10. 2. 2c2 + 3c + l. 15. 322 + 5«-22. 3. 622_72 + 2. • 16. (;a2 + 7a6-56^ 4. 6d2-nd + 4. 17. 12 m- - 23 mn + 10 w'. 5. 9w2-9?/+2. 18. 10c2-3cd-]8#. 6. 6?/^-i;!2/ + 6. 19. Qx'-Slxz + SSz'. 7. 6c2 + 17c + 12. 20. 16c2 + 18cd-9d2. 8. 8a2-2a-3. 21. 10 a^ + aa; - 24 a;^. 9. 9m= + 21m + 10. 22. . 9 w^ - 15 va; - 50 a.-^. 10. Gz''-Jz-5. ^^,^.^ 23. 7c'-50cz + 7z\ 11. 10a;2-^17a; + 3. ■ 24. 20 a' -27 am- Um^ 12. 10a^ + a;-3. 25. 10 x^- 2m- 21 m-. 13. 6z''-z-12. 26. 8)r-10na;-25a!2. 27. 20 aF - 37 az- IS z^ BINOMIAL EXPRESSIONS (a) The Dikference of Two Squares Type Form ••• jr^ —/I The Origin: If two binomials, (x + y) and (x—y), are mul- tiplied, we have (Art. 113): (a; + 2/) (a: - 2/) = a;2 - j^a. Therefore, to factor the difference of two squares : 133. Extract the square roots of the given square terms. One factor is the sum of these square roots ; the other factor, their difference. 108 FACTORING Illustrations : 1. 4o2_26a;2 = (2o + 5c<;)(2a-5a;). 2. 25 o* - 9a;'' = (Sa^ + 3a;3) (5 a^ - 3a;»). 3. 16 a* -81 = (4a2 + 9) (4ai'-9) =:(4a2 + 9)(2a + 3)(2a-3). Note that the second factor of (3) can be refactored into two other factors. 16. 4a^-25y«. 17. UmP-^dn". 18. ma?-25%\ 19. 49n2-1002/l 20. 121a%2-144c^ 29. m< — 9»«. 30. ci2-64m*. 31. 36aV-252/«. 32. 49m"'-4»j^^. (6) The Difference of Two Cubes Type Form ••• x^—^. The Origin : Since the product of a divisor by a quotient equals the corresponding dividend, we have, from Art. 119, an expression (ar* — if') equal to the product of the expressions (x — y) and {x^ + xy + y^). Therefore, for the factors of the difference of two cubes : 134. One factor is the difference of the cube roots of the quantities. The other factor is the sum of the squares of the cube roots of the quantities plus their product. Factor ( arally : Exercise 29 1. a'-f. 6. d'-W. , 11. 9a;^-25. 2. aP-z". 7. a;2-25. 12. 16 m^- 9. 3. c' -1. 8. m''-49. , 13. 4n^-26. 4. x'-i. 9. n^-Sl. 14. 9a2-64. 5. a^-9. 10. 4c^-9. 15. x''-'dy\ Write the factors of : 21. a'- -x\ 25. a?-a\ 22. c*- -16d*. 26. c'^-ie. 23. 64: ^*-y\ 27. C8-266. 24. 81 i m*-l&n\ 28. a;i«-l. BINOMIAL EXPRESSIONS 109 Illustrations : 1. e3_8 = (c-2)(c2 + 2c + 4). 2. 27 ms - 64 = (3 m - 4)(9 m^ + 12 m + 16). 3. 64 «6 _ 125 y8«8 = (4 x^ _ 5 y^) (16 a:* + 20 x^e + 26 yS^'). Exercise 30 Factor orally : 1. c'-d». 5. m'-8. 9. 27 -a?. 2. a»-m*. 6. a" -27. 10. 64-al\^^a 3. n^-st?. 7. ?/-64. 11. cW-64. 4. c'-l. 8. c=-125.. 12. aW-125. AVrite the factors of: 13. 8 0^-27. 17. 8 m' -343. 21. 125m»-l. 14. 27 -125 m'. 18. 8c^-27. 22. 27mV-64.'t'. 15. 12oaf'-64?/'. 19. 6i2m'-a;«. 23. 125 x^-T29z^'-. 16. 216 a;' -ay. 20. 1253?f-&^z\ 24. 729 m^^ - 1000 ny. (c) The Sum of Two Cubes Type Form ••■ jr'+^. The Origin : As in the preceding case, we refer to the principle of division by which (a? + ff) is shown to contain {x + y), the quotient being (x' — xy + if) (Art. 120). Therefore, for the factors of the sum of two cubes : 135. One factor is the sian of the cube roots of the quantities. The other factor is the sum of the sqiiares of the cube roots of the quantities minus their product. Illustrations : 1. cS + 27=(c+8)(c2-3c+9). 2. 8+125c3 = (2 + 5c)(4-10c + 25c2). 3. 27 a;6 + 64 n^ = (3 a:- + 4 re') (9 x* - 12 x-u' + 16 n^). 110 FACTORING Exercise 31 Factor orally : l.a^+c^. 4. m' + l. 7. m? + 125. 10. 64: + T>f. 2. m^ + x'. 5. x> + 8. 8. f + 216. 11. mV + 343. 3.cP+f. 6. 25 + 27. 9. 27 + aV. 12. 2/-V + 512. Write the factors of : 13. So' +27 b\ 17. 125a!« + l. 21. 512 + 126 a^^ 14. 64 a^ + 125. 18. 64 o^ + 27 y\ 22. 729 + 64 a?f. 15. 27 m^ + 6471=. 19. xY + V2nz^. 23. 1000 a-» + 27 y". 16. 125 71^ + 216 of. 20. aY + 2162'. 24. 1728 a;'^ + 1331?/^. EXPRESSIONS OF POUR OR MORE TERMS FACTORED BY GROUPING (n) The Grouping of Tekjis to show a Common Polynomiai Factor Type Form • • • ax + ay + bx + by. The Origin: If any binomial, (x + j/), is multiplied by a binomial having dissimilar terms, (a + &), we have (Art. 55) : (a + 6)(x + 2/) = a(a;+ y)+b(x + y) = ax + ay + bx + by. In the resulting product note that a is common to the first two terms, and that 6 is common to the last two terras. Note, also, that dividing the first two terms by a, and the last two terms by 6, gives the same quotient, {x + y). Therefore : (x +y) is n common polynomial factor. And from a(x+ y) + b{x + y) we obtain hy adding the coefficients of the common factor : (a + 6) (a; + ?/) , the factors. Illustrations : 1. ac + be + ad + bd= (ac + be) + (ad + bd) = e{a + b) +d(_a + b) = (,c + d)Ca + b). EXPRESSIONS OF FOUR OR MORE TERMS 111 2. -' c- - 6 (• + cd - 3 (i! = (2 c2 - 6 c) + (cd -3d) = 2p(c-3) +d{c-3) = {2c + d)(c -3). 3. a^b-U-1 a'^ + 2ab =a^b + 2ab-1 a^- 14 = ia^b + 2ab) - (7 a^ + 14) = a6(a2+2)-7(a2 + 2) = (a6- 7)(a2 + 2). 136. Note that the proper arrangement of an expression is first necessary. We are assisted in grouping the terms by noting that terms hearing to each other the same relation are grouped together. With expressions of more than four terms the principle is unchanged. Exercise 32 Write the factors of : 1. am + an+mx + nx. 10. abxy—cxy—cz + abz. 2. am, -{■ an ~ am -— en. 11. x- -\-hx-\- ax-'r ah. 3. ac — ad—hc + hd. 12. ■if — my + 2y — 2m. 4. ax + ay + x + y. 13. a^+aa^ + cc + a. 5. asB-az — x + z. ,14. z^+ 2z — 3z- — 6. 6. ax + 3a + 2x + 6. 15. a^ + 5a^ + 10 + 2a\ 7. xy — 4:X + 5y — 20. 16. a-^ + M — 2 a; — 7 a^. 8 afta; — 2 a?* + c.r — 2 c. 17. — adx + 3d — 3 cx + aca^. 9. 2mwa; — 5 a;— 6m«+15. 18. 3 d — 10 d' — 15 + 2 dl (i) The Grouping op Terms to form the Difference of Two Squares Type Form ••• je^ + 2x/ +y—z^. The Origin ; By multiplication, (x + y + z)(x + y - z) = l{x + y) + z]l(x + y) - «] = (a + yy - *' = a;2 + 2 a;(/ + 2/2 - z^. \ 112 FACTORING The product is the difference of a trinomial square and a monomial square. Therefore, an expression of this type is factored by grouping three of the terms that will together form a perfect trinomial square ; the fourth term being a per- fect monomial square. The result is a difference of two squares. Illustrations : 1. 0= + 2 a6 + 62 - c2 = (_a^ + 2ab + b^- c^ = (a + by - c2 = (fl! + 6 + c)(o + 6-c). 2. i^-i/^~2yz-z^ = x^- (y^ + 2yz + z^) = x^-(y + zV' = [x+(2/-|-z)][s-(2/-l-«)] = (X + y + z)(x - y - z). 3. a2-6aaj + 9a;2_4TO2_i2TO_9 = (a2-6aa; + 9a;2') _ (4 m^ -(- 12 to + 9) = (a-3a;)2- (2 m + 3)2 = [(a -3 a;) + (2 m 4- 3)1 [(a -3 a;) - (2 to +3)] = (a-3a; + 2m-r3)(a-3a;-2TO-3). The process consists mainly in finding three terms that, when grouped, form a perfect trinomial square. The key to the group- ing is the given term that is not a perfect monomial square. Or, considering the given square terms, we may state : 137. (1) When only one given square term is plus, it is ivritten first, and the other three terms are inclosed in a negative paren- thesis. (2) When only one given square term is minus, it is written last, and the other three terms are written first in a positive parenthesis. Szercise 33 Write the factors of : 1. a^ + 2ax-\-oi?-m\ 3. c^ + d'-y'--2 cd. 2. y'-2yz + z'-4:. 4. l-c^-2 cd-d!'. EEPEATEIJ FACTOKING 113 5. a?-m?-l + 2m. 10. (r - 10 c.c + 'jr, x' - 49 m\ 6. ie' — cP + l — ic. 11. .^;--4x//-9;r//- + 4 2/2. '' 7. 9-m'-2my-y\ 12. f- + 16 z" - li; + 8 tjz. 8. 4c2 + c*-4c^-4. 13. a^2/' - a^!^^ + 16 a6 - 64. 9. 12 a; +■ 4 a!^- 9 2/2 + 9. 14. 49cW + 10 cW-c^d*-25. 15. 81 - 100 a:^y* + 60 xYz - 9 «2. 16. 4:X^-Ax+l-9m' + 6mn-n\ 17. c^ — a^ + a^ — 2/2 — 2 ccc — 2 aj/. 18. c'-d^-x^ + m^-2cm + 2dx. 19. 36c<+l-49a;''-92/2-12c2-42arV. REPEATED FACTORING 138. Any process in factoring may result in factors that may still be resolved into other factors. The review examples following frequently combine two or more types already con- sidered. The following hints on factoring in general will assist. 1. Remove monoynial factors common to all terms. 2. What is the number of terms in the expression to be fac- tored ? (a) If two terms, -which of the three types ? (b) If three terms, which of the three types ? (c) lifoiir terms, how shall they be grouped f 3. Continue the processes until the resulting factors are prime. Illustrations : 1 3a'-3aa^ = 3a(a6-a6) = 3a(a8+s8)(a8-a;') = 3 a (a + a;) (a2 - ax + a;2) (a - «) (a^ + ax + a;8) . 2. 12 x5 - 75 x' + 108 K = 3 a; (4 x* - 25 x^ + 36) = 3x(4x2-9)(x2-4) ='3x(2x + 3)(2x-3)(x-l-2)(x-2). SOM. EL. ALG. — 8 114 FACTORING 3. 8a'x-2ax + 8 aH - 2 a*x= 8 a'x + 8 a^x -1aH-1a% = 2aa:(Ja'+4a<-«-''- 1) = 2 ax [(4 a' + 4 «<) - (a' + 1)] = 2 ax [4 a' («»+])- (a-'' + 1 J] = 2 ax (4 a* -!)(«■>+ 1) = 2na:(2a2 + l)(2a-2-])(a + l)(a2-a+l). MISCELLANEOUS FACTORING Exercise 34 (See also pages 399, 400) Write the factors of : 1. 5a!2 + 25a; + 20. 21. 75 a'' - 90 a^ + 27 a. 2. 5a^-45af. 22. Ca^aj+Qato-Sa^j/— 12a62/, 3. 8 a^ - 24 a + 18. 23. 5 m^ + 135. 4. a*-8a. 24. 15c3+33c2 + 6c. ^5. 2am + 2mx-Aa-4:X. 25. 242af'-98.r. 6. 2-2c2-4cd-2dl 26. 12 m* + 10m=-8 ml 7. 3 m< 4- 81 m. 27. 160 x^ + 20 a.-^. ~ 8. Sa'^- 9a-84. 28. 3 a^ - 15 a^ -|- 12 a;. 9. 2 c^ - 128 cl 29. 98 ar^ + 18 xtf - 84 x^y. 1 0. 16 TO V + 8 mnx + a^. 30. m'' — 2 m'' + 8 m - 16. 11. 15 - 60 ajy + 60 a;^/. 31. 16 a" - 25 o-.t- - 10 aa,-^. 12. 7 ci"a^ — 175 ax. 32. 81 ai^y* - 3 a;?/. 13. c^+c'-ScP-Sc. 33. 4cd + 2eW + 2-2a,-2. 14. 14af-82a,-2-12a;. 34. 250 c^- 16 c. 15. 49 m^ — 84 m" 4- 36 m. 35. 8 mV+4mV— 112tom. 16. 4 a;' -36 0)2 + 56 a;. 36. 5 c= - 10 e'' - 315 c. 17. 7 a;^ - 7 a;. 37. 21 af* + 77 x' - 140 a;. 18. 4 o^ + 16 a" 4- 15 a. 38. 512 - 32 mV. 19. 8a^-18c2 + 24a& + 186l 39. 27 a;" + 215 ar'- 8. 20. 24 a?! - 55 «2 - 24 a-. 40. 3 a? - 51 x' + 48 x. SUl'l'LEMKNTARY FACTORING 116 41. 686 jc* - 2 iuy'. 45. 242 a^ - 748 a^ + 578. 42. 8irt« + 8m2-18m^-18. 46. 2»*+642/2-8a;'t/''-16.T;^ 43. 8a3 + 2a'^-8a*-2a. 47. 64a«+729a2. 44. 32 a* - 2 a?y\ 48. 9,a^-VQa?- 432 x. SUPPLEMENTARY FACTORING ('0 Compound Expressions in Binomial and in Trinomial Forms (Jomparative illustrations with corresponding processes : 1. a:2 + 12 a; + 35 = (jc + 5)(a; + 7). Similarly, (a - 6)2 + 12 (a - 6) + 35 == (a - 6 + 5)(a -6 + 7). 2. 3 x" + 10 xy + 3 j/2 = (3 a; + j^)(x + 3 y). Similarly, 3 (a - 6)-2 + 10 (a - 6) (c - d) + 3 (c - (i)^ = [3 (o - 6) + (c - d)] [(a - 6) + 3 (c - d) ] =:(3a-36+c-d)(a-6 + 3c-3d). 3. a2-9a;2 = (a + 3a;)(a-3a;). Similarly, (« + 2)2 - 9 (X - 1)2 = [(X + 2) + 3 (X - 1)] [(X + 2) - 3 (X - 1)] = (x + 2 + 3x-3)(x + 2-3x + 3) = (4x-l)(-2x + 5) = -(4x-l)(2x-5). 4. a^ f 8 = (a + 2)(«(2-2a + 4). Similarly, (a - 6)8 + 8 = [(a - 6) + 2][(a - 6)2 - 2 (a - 6) + 4] = (a-6 + 2)(a2_'2o6 + 62-2a + 26 + 4). 139. If similar monomial terms occur in the different-compound terms of an expression, the factors can usually be simplified by collecting like terms. 116 FACTORING Exercise 35 rind and simplify the factors of : 1. 16(a+l)^-9a^. 6. (a^ - 1)"* - 9 (a; + 2)^ 2. 49(c-l)2-4. 7. 27 (a- 1)8 + a^. 3. (a + xf + 10 (a + x) + 24. 8. 2 (a; + 1)^ + ll(a;+l) + 12. 4. (m + 2y - 15 (m+2) + 56. 9. (2 m - 1)< - 16 (2 m + 1)^ 5. (2/ _ 3)2 -7(2/ -3) -30. 10. 8 (a - 2)= - 27 (a + I)''. 11. 6 a^ - 12 aa; + 6 a.-2 - 5 a + 5 a; + 1. 12. 3(a;+l)2 + 7(a.-2-l)-6(a;-l)2. (6) The Difference of Two Squares obtained by Addition AND Subtraction of a Monomial Perfect Square 140. No general statement of this process can be given in a simple form. Illustration : Factor 9a^+6a^ + 49. The first and third terms are positive perfect squares. Hence, if the middle term were twice the product of their square roots, the expression would be a perfect trinomial square. For a perfect trinomial square the middle term should be 2(3af^(7) =42 .^•^. Adding + 36 .r^ to the expression, we obtain the perfect trinomial square required, and subtracting the same square, + 36 .i?, the expression is unchanged in value hut is now the difference of two perfect squares. Therefore: 9a:* + 6a;2 + 49= 9a^+ 6a;2+49 + 36 x"- - 36 a;' = 9a^ + 42a;2 + 49_36x2 = (3x2 + 7)2-36a;2 = (3 x2 + 7 + 6 a;)(3 x2 + 7 - 6 x). It is important to note that a positive perfect square only can be added. SUPPLEMENTARY FACTORING 117 Exercise 36 Write the factors of: 1. x* + r + l. 7. 25 of -51 a? + 25. 2. a* + 3 0^ + 4. 8. 81 m^ + 45 m- + 49. 3. m*-7n2 + l. 9. 36 c* - 61 c^m^ + 26 m*. 4. c^-28c2 + 16. 10. 64 ffl< + 79 a-.)r + 100 a;^ 5. 9c^ + 5c2ar' + 25a^. 11. 64m* + 76 mV4-49 n^ 6. 9d*-55dV + 25a;*. 12. 81 x^ - 169 o^z^ + 64 »♦. (c) The Sum and the Differexce of Equal Odd Powers By actual division : a + x a^ — o^ a* ~ aH + ah? — ax^ + x*. = «» + aH + ah? + ax' + x*. u — u> Hence, for the factors : a^ + x^={a + x) (a* - a'a; + a^^ - aifi + x«). a5 - x' = (a - X) (a* + a^x + aV + ax** + x*). By Art. 121, the factors of any similar cases may be found. In general, therefore : 141. (1) One of the factors of the sum of equal odd powers is the sum of the quantities. Tlie other factor is the quotient obtained by using the binomial as a divisor. (2) One of the factors of the difference of equal odd powers is the difference of the quantities. TJie other factor is the quotient obtained by using the binomial as a divisor. Write the factors of : Exercise 37 1. m'-l. 5. m^ + »i'. 9. 32c«+243. 2. c' + ar'. 6. 32 - n». 10. a>V - 32. 3. m' — nl 7. 128 +c^ 11. «i" + n'\ 4. a^ + xl 8. ary - 243. 12. a'V-oifif. 118 FAC'l'OUING BUSCBLLANEOnS FACTORIITG Exercise 38 1. 3a!»-54a^ + 243a;. 14. 72m« + 94m< + 128. 2. (c3 + 8) + 2c(c + 2). 15. 2a\a+2)-5cv'-Sa + 4:. 3. 4a2 + 9ar'-(12ax+16). 16. (-•-' + a;^)-' - 4 cV. 4.' 5(c»+l)-15c-15. 17. (4ar'+32)-5a;2-17a;-14. 5. (m^- 12)2 -ml ig. 98 a^ - 16 x^z^ + 8 k*. 6. (a + 2)2-7(a + 2) + 12. 19. m«-3m* + 3m^- 1. 7. 4 + 2a; — (am + a;)am. 20. (a;- + a;— 1)^— (ck=— a;— 1)-. 8. 3(a;+l)'-19(a;+l)+6. 21. 2(3^-1) + 7 (ar'- 1). 9. (m^ - 12)2- (m^- 6)2. 22. m< - 9 a;^ + m-' + 3 .rl 10. 2 a;« + 38 ar'- 432. 23. 12ni'-m(n~l)-{n-l)^ 11. (a2 + 4)^-125al 24. 2di"-1024d 12. (7n+a!)2— 7— 3(m+a;+l). 25. cx+my—mx—cz—cy+mz. 13. (c2 + 4)=-16c*-64c=. 26. 4 m V - (m^ + n^ - 1)^. 27. (c4-d)(m'-l)-(m + l)(c2-d^). 28. 12 62_l8ma; + 12a6-3a!2 + 3a2-27m2. 29. 7 ?)i2 _ 7 a^ + 7 w2 — 14 {xz — mn) — 7 z^. 30. (m — l)(a2_aa;)-)-(l — 7ft)(aa; — a^). 31. (a;-3)3-2(a;-3)2-15(a;-3). 32. (a2 + 4a + 3)2-23(a2 + 4a + 3) + 120. 33. (a; -2) (a; -3) (a; -4) -(a; -2) + (a; -2) (a; -3). 34. a?h + b'c + ac' - ah - a¥ - bc\ 35. (3a; + 2)(9a^ + 2a; + 12)-(27ar' + 8). 36. 8(a-a;)2 + 5a2_5a'2-3(a + a;)2. 37. (c - 1) (c2 - 9)-(c + 5) (c - 3)- 3 c2 + 9 c. 38. (a; + 22/)2-3(a; + 22/ + l)-15. 39. m2(a; — 1)— 2ma; — m + a52(m — 1) — 03. 40. a^ (a; - 5)2 + 2 a; (a^- a; -20) + (a; + 4)2. CHAPTER X HIGHEST COMMON FACTOR 142. A common factor of two or more algebraic expressions is an expression tliat divides each of them without a re- mainder. Thus : aV, a'6*, and a^fi^ may each be divided by ab. Therefore, ab is a common factor of a'6', a^b*, and a'6'. In this definition the algebraic expressions are understood to include only rational and integral expressions. 143. Expressions having no common factor except 1 are said to be prime to each other. Thus : 3 abx and 7 cmn are prime to each other. 144. The highest common factor of two or more algebraic ex- pressions is the expression of highest degree that divides each of them without a remainder. Thus, g^ a'^V is the expression of highest degree that will divide each of the three without a remainder. That is, ifib^ is the highest common factor of a^6', a'6*, and a^6'. 145. The highest common factor of two or more expressions is the product of the lowest powers of the factors common to the given expressions. The abbreviation " H. C. F." is commonly used in practice. 119 120 HIGHEST COMMON FACTOR THE H. C. F. OF MONOMIALS 146. The H. C . F. of monomials is readily found by inspection. Oral DrUl Give orally the H. C. F. of : 1. a%* and a^b^. 8. 16 mn^ and 20 m/'x. 2. m-nd' and mn'x. 9. 12 mV and 15 m*n'. 3. 2 mW and 4 m^a;^. 10. 8 cWm and 12 c^d*w. 4. 3c»d»and6cW. 11. 36 a^i/^a and 42 my. 5. 3 a^ni'if and 9 am^y\ 12. 51 a^y'i^ and 17 ayz'. 6. 10aWn2andl5m-». 13. 12 of yz, 16 a^y'z-, said 20 xfz. 7. 16 afy and 24 y''z. 14. 18 cdm, 24 cmn, 30 cdw, and 36 dmn. 15. 6c%, lOccZ'?/, 15cdy% and20c^dy 16. 33 mrtx, 44 mnt/, 55 ma;?/, and 66 mnscy. THE H, C.F. OF POLYNOMIALS BY FACTORING 147. The H. C. F. of factorable polynomials is readily found by inspection of the factors. Illustrations : 1. Find the H. C. F. of a'-Sa^-lOft, a='-8a2 + 16a, and a'-25a. Factoring, a^ - 3 a'^ - 10 a = a(a - 5) (a + 2) a3_8a2+ 15a=a(a-3)(a-5) a^-25a = a(a + 5)(a- 5) Therefore, H. C. F. = a(a - 5) Result. 2. Find the H. C. F. of m= - 27, 9 - m% and m^ - 6 m^ + 9 m. Factoring, m' - 27 = (m - 3) (m^ _|_ 3 „ + 9) 9 - m2 = _ (m + 3) (^ - 3) m^ - 6 m" + 9m = ■m(m - Sy Therefore, H. C. F. = (m - 3) Result. The student will recall that (9-m^) = (iS+ )M)(3-m) = (m + 3)(-m + 3)= -(TO + 3)(m-3). H.C.P. Oi- POLYNOMIALS BY FACTORING 121 Exercise 39 By factoring obtain the H. C. P. of : 1. am + miax + x. 8. a^ — 1,(1 — a)". 2. am + m, mx + m. 9. a? + 1, (1 + a)l 3. cx — dx,cy—dy. 10. c' — l, 2 0^ + 20 + 2. 4. mx + m, mx — m. 11. c' — cd-, l — 1 W^ — w + 3 a" — 3a— 2 _ aM- a^ + a- O. — * !■ a - 5 a= + 2 «* + 4 g + 1 - 3 m* — m^ — m +1 a^ + a — 1 m^ — 1 CHAPTER XII FRACTIONS (Continued) — LOWEST COMMON MULTIPLE. LOWEST COMMON DENOMINATOR. ADDITION THE LOWEST COMMON MULTIPLE 157. A common multiple of two or more algebraic expres- sions is an expression that may be divided by each of them without a remainder. Thus, a^ft* will contain a'6', a'6*, and a^b^. Tlierefore, a''b'' is a common multiple of rt^ft', a'6*, and a'^b^. In this definition the algebraic expressions are understood to include only rational and integral expressions. 158. The lowest common multiple of two or more algebraic expressions is the expression of lowest degree and least numeri- cal coefficient that will contain each of them without a re- mainder. Thus : 13 a^b^ 30 a'b^ is the expression of lowest degree and least 2 a'6* numerical coefficient that will contain each of them 5 a^fts ■without a remainder. That is, 30 a'fi^ jg the lowest common multiple of 3 a'h', 2 a^b*, and 5 a-'bK 159. 77*6 lowest common multiple of two or more expressions is the product of the highest powers of all the factors that occur in the given expressions. The abbreviation "L. CM." is commonly used in practice. 127 128 FRACTIONS THE L. CM. OF MONOMIALS 160. The L. C. M. of monomials is readily found by in- spection. Oral Drill Give orally the L. CM. of: 1. (?m, and rni?. 7. 16 a^c and 24 c?c. 2. my and my. 8. 9 a^/ and 18 a^2/«a^ 3. 2 m^n and 3 m'n. 9. 10 c^x and 8 c(^y. 4. 3 QKf and 6 aaj. 10. 16 mv?y and 12 mV?/'. 5. 8 tmj and 12 ny. 11. 15 a^z and 20 c^zl 6. 10 mVa; and 6 mnV. 12. 26 a^mV and 20 aWc*. THE L. CM. OF POLYNOMIALS BY FACTORING 161. The L. C. M. of factorable polynomials is readily found by inspection of the factors. Illustrations : 1. Find the L. C. M. of o" - ab and ah — 6*. Factoring, o^ — a6 = a (a — 6) a6 - 6^ = 6 (g - 6) Therefore, L. C. M. = ab (a- 6) Result. 2. FindtheL.C. M. ofa^+a;-2, K^-a; — 6, and a!'-4a! + 3. Factoring, x^ + x-2= (x + 2)(a5 - 1), a;2 - a; _ 6 = (a; - .3) (k + 2), x2-4x + S = (x-3)(x-l), Tlierefore, L. CM. = (x + 2)(a; - l)(a; .;- 3). Result. 3. Find the L. C. M. of 2 a' + 8 a^ -f 8 a, 4 a; - a'a;, and 5o2-20a + 20. Factoring, 2 a' + 8 a^ + 8 a = 2 o (a + 2)(a + 2) i X - a'^x = - X (a + 2)(a - 2) 5 g'' - 20 g + 20 = 5 (a- 2)(a - 2) Therefore, L. C. M. = 10 ax (a + 2)2(a - 2)2. Result. THE LOWEST COMMON DENOMINATOR 129 Exerciae 42 By factoring obtain the L. C. I\] . of : 1. x' + x,xij + ij. 5. 2a(a4-l), 3a2 + 3a. 2. am — a, m- — m. 6. lu'ir — 9, mhv — G nin + 9. 3. c^ + c, -;- - c. 7. c'W _ 1, c2^i2 _ i . 4. /-I, (2/_l)2. 8. a?~'La' + 4=a,9,a-a\ 9. ,;3 _ 6 c^ + 9 c, & - 81 c. la m? + 3 7it- -f- 2 ))i, 7)!.^ + 4 m^ + 3 m. 11. 3 .V (.r' + .r + 1), 4.r^ + 4a;. 12. 7 16-' — irn a, 3 a''' — 30 u- + 75 a. 13. (a; + 2/)^0/_a;)^a:2-7/l 14. y^y-z), y{y^-z'),y + z. 15. 3 (a;= + xy), 8 (xy - y-'), 12 («^ - J/^- 16. 2c2-c^-c, 2r'-3c^-2c. 17. 5a;^-5.r-^, 4.x'''-8a-'2 + 4a;. 18. 7a-7a^ 4 a (a - 1)V3 a^ + 6 a- + 3 a. 19. jft" + 1 , 3 (2 + 3 m + m=), i m — 4 771^ + 4: m*. 20. 18 - 2 a;^ 3 af'- 81, 4x^-324. 21. am + a+ m + 1, run' + cmi — m" — m. 22. 5c(c-2y-, 4c.,; + 12c-24-8«, 24c2-3c». 23. (a + xf - 1, a- - (a; - 1)^, (a - If - x\ 24. 4.r2-14a;+12, 8.r2-32a; + 30, 18 a;^ - 69 a; + 63. 25. 6 a^x + 6 a^ — 6 a;^ — 6 a;, 4 d^m — irr — imx + A x, 2d'x-2ti?-2a^ + 2x. THE LOWEST COMMON DENOMINATOR 162. The lowest common denominator of two or more algebraic fractions is the lowest common multiple of their denominators. The abbreviation "L. C. D." is commonly used in practice. SOM. EL. ALO. 9 130 FRACTIONS Illustrations : 1. Change -— , — , and - to equivalent fractions having a common denominator. The L. C. M. of the denominators 3, 4, and 6 Is 12. L. C. D. = 12. Dividing each denominator into tlie common denominator, we have : 12 = 4,12^3,12^2 3 '4 '6 Multiplying both nvunerator and denominator of each fraction by the respective quotients from the division, we have : 3 4 12 §Jx-= — 4 4 3 12 x_ X x2 = 2_? 6 6 2 12 Sx^3x^S^9x_ Result. x_x ^l^2x^ 3^g^^,^_ 2. Change — and ^„ to equivalent fractions having X- + X ar — a; a common denominator. The L. C. M. of x^ + x and x^ - x is x(x + l)(x — 1). Hence L. C. D. =x(a; + l) (a;- 1). Dividing each denominator into the L. C. D., we have : x(x + l)(x-l) _^ ^ . a: ( a: + 1) Ca: - 1) = ^ I i_ a;^ + X ' a;2 — X Multiplying both numerator and denominator of each fraction by the corresponding quotients obtained : X - 1 „ X - 1 _ Cx - ly y^£SZjL=A±^ZJJ^. Result. x^ + X X — I X (x'^ — 1) X+i^X+l^ix+m. R^^^jt, x^-x x + 1 x(x2-l) In general, to change two or more fractions to equivalent fractions having a common denominator : THE LOWEST COMMON DENOMIXATOR 131 163. If necessary, reduce the fractions to their loicest terms. Find the lowest common multiple of the given denominators, for the common denominator. Divide each given denominator into the common denominator. Multiply both numerator and denominator of each given fractiov by the respective quotients obtained. EzerciBe 43 Change to equivalent fractions having a common denominator : 1. om 2 m om 4 ' 3 ' 2 7. 2 3 c+1' c-1 2. 6 a 3 a. a 6 ' 4 ' 8' 8. m m m+n n—m 3. 5 3 4 2a' 4a'3a' 9. c c <^-r (c-iy 4. 2 A 1. ax' bx' ab 10. 2 5 y(j/-^y(y-^r 5. 3 2 4 a'b' aV ab^ 11. a a 3a + 4' 9a^-16 6. a b c • 12. X x m'ny' mny ' mny"^ a^ — 1' a^ + x' + x 13 ^ 2 m' — m -6' m^ + 2 m — 15 11 ^ 2 *'• 2c^-c -1' &(?-C :-2 IB " 'a' a — a 1 *"■ d' + 'La + Z'- -12' a 16. \, x+r 2 Vf' 3 (a; + (x + lf c 3c 3(c + l)'4(c-l)'2(c= + 2c + l) 132 FRACTIONS jg m m m 19. 20. wi^ + in m? — m + 1' m* + m 3 a; 4 a; 5a; 2 a! -2' 3 a; + 3' 4 a;2 - 4' 1 1 1 3y-15' 5y^- 125y'7?/ + 35 ADDITION AND SUBTRACTION OF FRACTIONS By Art. 75, -+- + -= ^ • ■' a; a; a; x That is, two or more fractions having the same denominator may be added by adding their numerators, and writing the sum over that common denominator. By Art. 163, any two or more fractions may be changed to equivalent fractions having the same or a common denominator. From these two statements it is clear that any given frac- tions may be added. The term " simplify " is used to include both operations of addition and subtraction. Illustrations : X X 1. Simplify x—1 x+l The L. C. D. is x^ - 1. Dividinn; each given denominator into the L. C. D. and multiplying the corresponding numerators by the respective quotients, we obtain : X X _ (x + 1) X — (x — 1) X x — 1 x + l~ x^-1 x'^ + x — x^ + x 2x Eesult. ADDITION AND SUBTRACTION OF FRACTIONS 133 2. Simplify -^ — --2- m^ — 1 m^ + m + 1 Every integer may be considered as having a denominator 1. There- fore, in adding fractions and integers the integers are multiplied by the L. C. D. The L.C.D. = m3-l. Dividing each denominator into the L. C. D., and multiplying as be- fore, we have : 2m3 2 m + 1 ^ 2m^ -2(m^ -1) ~ (m + 1) (m-1) m^ — 1 1 ni' -t- 7)1 + 1 m' — 1 ^ 2 m'i - (2 m" -2) -(m^-1) )yl-i - 1 = ^-"^\ Result. m^ — 1 In general, to add algebraic fractions : 164. Reduce the given fractions to lowest terms. Divide each denominator into the lowest common denominator, multiply the corresponding numerators by the quotients obtained, and turite the sum of the residtirig products for the numerator of the result. The sign of each fraction becomes the sign of its numerator in the addition. Oral DrUl Simplify orally : 1. '*+^ a; X 6. XXX 11. A + A + 1, ax ay xy a b 2. m m 7. a b c m m m' 12. X y z 3 ^ + -. ^- 3^4 8. 3_2_o XXX 13. ^ + 1 + ^- b a X , X a 2 a 9. 3x 2x 14. m .. _ « n m 2a 3a ^- X 2x 10. 5 3 8 a; 7 x 15. ^-4 + 2 + ^. XT y^ 134 FRACTIONS Exercise 44 (See also page 401) Simplify : x—1 x+1 « + ^-3 l-g + g;, 3. ^±1-1 _^+^. 14. 2 6 . 2m — 1 , wi — 1 Tm-f-l ,_ 4. ; 1 — -! — 15. a -1-1 ^■^a-1 4 3 d' ■\- a a' -a' 3c 2c -2 4c 3c -f 3 12 5 3 10 7£c2-7 3a;^-f-3 5 c-5 ., c-h3 ^g m+_3 m+J. 12 " 6 ■ m-l-4 m + 2' a^-l-2 g-l a-(-l ^ m -j- 9 w — 9 4 a- 3 a 2 a ' m — 9 m-|-9 a — b , b — c , c — a ,„ 5aj-t-7 5a; — 7 7 1 r • lo* • ab be ca 6 a; — 7 5 a; -f 7 8. U^-. 19. ^-1 ^^+1 X X -\- y c^ — c-|-lc'*'-l-c-|-l 9. -^+-^. 20. ^^-1 -F 2^ + 1 ■ 2m m -1-1 (2 c -I- 1)^ (2 - 1)2 ^ ,-5 3 „, m — 5a; lOma; lU. -• /SI. ■ — - — I — - 8xy 5xy — 1 m + 5x m^ — 253? n. L I^. 22 *-2 ^^+2 aa; aa,-^ — 1 a;'' — 2 a; -|- 4 a? + 2x + 4: 1 23. '-|-7r'-f-12c c' -I- 8 c' -1- 15 c 24- ^^+2 . ^ -, a; — 1 a;-|-l ar— 1 „_ aa; a.v 2 aa;^ /&0. 1 — «-!-;(; « — a; a-* — a;^ 26. ■'^-l 3 x + 1 ^ (3x+7)x x + 1 X ^ 1 a-- — 1 ADDITION AND SUBTRACTION OF FKACTIONS 136 „^ a* + m* _ a* m* 28. 29. a — 1 cd+l a^ + 11 2a + 2 3a -3 6a- -6 3x4-2 13 .»; - 38 2 a; - 3 X — 5 X- — ix — 5 X + 1 30 4c' 2o 8c^ + 2c + l • 4c^+2c+l 2c-l Sc'-l „., a — 1 a + 1 a — 7 '*■'■• "":; ;: r~7; :; r. r~?i + 32. a' — 5a + 6 a--^oa + 2 a^ — 4a + 3 1 1 , 3n; a;- + 2a; + 4 x' — 2x + 4: .r* + 4a;^ + 16 165. The process of addition of certain forms of fractions is simplified by application of the principles governing the signs of fractions (Art. 162). 3 2 r,a 1. Simplify a + 1 1 — a a'- — 1 Changing the form of the second fraction, in order that the terms of the factors in all the denominators may be in the same order, we have (Art. 152) : 2 _ 2 _ 2 ^ 2 1 — a — a+1 — (a — 1) a — 1 Therefore, _A^ + ^ _ -^ = 3(. - 1) + 2(. + 1) - ,5« a + 1 a - 1 ((■' — 1 a^— 1 - ~ ^ ^ Result. 2. Simplify a^ - 1 1 - cfi 1 1 (a-l)(a-c) (l-a)(l-c) (e-a)(c-l) Changing the signs of both factors of the denominator of the second fraction, we have changed the signs of an even number of factors, and the sign of the fraction is not changed. Changing the sign of one factor in the denominator of the third fraction changes the sign of the fraction. 136 i-KACTlONS Therefore, (a - 1) (a - c) (1 - a) (1 - c) (c - a) (,c - 1) 1 1 + - (0-l)(a-c) (a-l)(c-l) (fl!-c)(c-l) L. C. D. = (« - 1) (a - c) (c - 1) _ Cc - 1) - Ca - C-) + (a - 1) (a - 1) (a - c) (c - 1) 2c-2 Simplify : (a - 1) (a -<:)( >-!) 2 Result. (a-l)(a. -c) Exercise 45 3c 2 2 l-c"" c -1 c + 1 1 , 1 1 10 a; 5a; + l 5 a;- 1 'l. -25a!2 m^ + 3 m + 9 m^- om + 9 3. m + 3 3 — m . c — 2d 2c — d . 3(cy — dy) c + 2/ 2/— c c^ — 2/^ _ m + ?i , a + ?i (m — a) (c — m) (a — m)(G — a) (a;-l)(aj-2) (a; - 2) (3 - a;) (1 - a;) (a; - 3) a; + a 2(c — a) x + c (as — a) (a; — b) (c — a;) (a; — a) (6 —a;) (c — a;) 8. , 1 ,+ i 1 (6 — c) (6 — a) (c — a) (c — &) (a —l))(c — a) _ (1 — g — a;) x(x — 1 — a) a{a —1 — x) (x — 1) (1 — a) (a; — a) {x — 1) (a — 1) (a — a;) CHAPTER XIII FRACTIONS (Continued) — MULTIPLICATION. DIVISION. THE COMPLEX FORM MULTIPLICATION OF FRACTIONS (a) A Fraction multiplied by a Fraction The product of two fractions is obtained as follows : Given two fractions, - and -, and let X y TO = 2 (1) andn=- (2). Multiplying (1) by (2), mn = - .—• X y (3) Multiplying (1) by x, Multiplying (2) by y, Therefore, multiplying. mx = a. ny = b. mnxy = ab. (Ax. 3) (Ax. 3) Dividing by xy, mn = — • xy (Ax. 4) But from (3), a b mn=- .-• X y Therefore, a b_ab x' y xy (Ax. 6) In general : 166. 27*6 product of two fractions is a fraction wliose numerator is the product of the given numerators, and whose denominator is the product of the given denominators. (6) A Fraction multiplied by an Integer Since any integral expression, b, has a denominator, 1 : TT \ 4. laa a ^, f. a b ab Fri im Art. 166, - x o = - • - = — X XIX 137 138 FRACTIONS That is : 167. A fraction is multiplied by an integral expression if its numerator is multiplied by that expression. The process of multiplication of fractions is simplified if factors common to numerators and denominators of the given fractions are canceled before multiplication. Illustrations : 1 T\/r 14.- 1 8 ah'^ , 25 mV 1. Multiply — by -— 15 amV l&cx 8 a^d^ ^ 25 rrfix'^ _ 8 • 25 gV-iroV _ 6afc _ jjgg^jt 15 amW 16 ex 15 ■ 16 acm'a;* 6 m The cancellations are : 8 in 16, 5 in 25 and 15, a in a', c in c', m? in m', and a;'' in a;*. « -ii/r li.- 1 m, + l , (m — T)' . m^ — 4 2. Multiply —31- X } > ., X — — - • m — \ m" — 6m-\- z mr — 1 Writing each fraction with numerators and denominators factored, ffl + 1 ^ (to-1)^ ^ m— 4 ^ m + 1 ^ Ora-l)(m-l) ^ (m + 2)(TO-2) TO — 1 m'^ — 3to + 2 m^-l m-l (m-l)(m— 2) (m+l)(m-l) Canceling oommon factors = "' "*" • Result. m — 1 In applying cancellation select factors for divisors from the numerator only. Begin at the left of the numerator and seek a possible cancellation for each new factor considered. Oral Drill Multiply orally: J axy ^^m 3 m ax 5. 17 c^ 2 n^ 6 n^ 51 c' „ 3a;/.,10c=. 5c ^ 9y 6. Sax' 39 y'z 13 xh 10 x'y 2 mn ,, 9 <^ 3cd ''7 cm 7. 42oi?f'^^ 10 2 15 xy 28 afiyz ^ 5 m^ ^ 21 c'd 3cd'^ 10 am 8. 32 xSi ^^ 27 cSn' 9 c^'a; ' 16 my MULTIPLICATION OF FRACTIONS 139 Exercise 46 Simplify : cd a + 1 a^_4 ^c + ir (c-iy a-2 d'-l ' c c ' {c'-lf X — 1 tf° + l X — 2 12 ^> \A\ •n^—mn m^— 4mn4-3n^ „ a'-4-2a; + 4 a.'^ + S ^' b. -; : X (m—nf m' — ii^ ay' — 4: x' — S 8. a!=-9 a;^— 3 a;— 18 c^-Tc +12.^ c2-c-2 X c^ — 3 c— 4 r— 5 c+6 . 3 — 4 g + 0' y a^ — 16 a 11. _g^9 x(a^ + 2a-15)x ^'-^%+^^ a' + 125 ^ ^ a + 3 m^ + 2 w?t + W' — a;^ m^—2 mx + a;^ — w' (m + 11 -f a;)^ 9?i^ — 2 wmi + w^ — a;* c + 3 c^-81 c^-9 (3 - c)^ (c- 3)2 ■ (c + 3)2 " c2 + 9 ■ (c - .■!)-' 12. 13. ar'— 9 am + 2 g gift — 2 mj — 2 g + 4 m^ — 4 &a; — 3 6 go; + 3 a — 2 a; — 6 a^-8 a^-2a; + 4 a; + 2 '■^- ar' + S ar^ + 2a; + 4 a;-2 3a;2_a;_2 3a;^-3a; 10-lOa; 16. •6x' + 2x '5v?-lQx + Z Q3? + Ga? + &x I-IO FRACTIONS DIVISION OF FRACTIONS (o) A Fraction divided by a Fraction The quotient of a fraction divided by a fraction is obtained' as follows: We will assume that - h- - = m. (1) % y Now a dividend equals the product of the corresponding divisor and quotient. Therefore, - = - x m. X- y Multiplying by^, -x^ = -xmx^. '^•' ° •' 6' X b y b Whence, - x f = m. (2) X -^- = - x^- X y X h Hence, from (1) and (2), ^^ * = « x ^ In general : 168. Tlie quotient of a fraction divided by a fraction is the product of the dividend by the inverted divisor. 169. The reciprocal of a quantity is the quotient obtained by dividing 1 by that quantity. Thus : - is the reciprocal of a. a 2 Q 9 1 -!-- = -, the reciprocal of - • 3 2 3 (J) A Fraction divided by an Integer From Art. 168, « h- ?/ = ^ h- ^ X X 1 = «xl X y — A. xy DIVISION OF FUACTIONS 141 In general : 170. The quotient of a fraction by an integer is the product of the given fraction by the reciprocal of the integer. The first step in a division of any fraction is, therefore, the inversion of the divisor ; wlience the process becomes a multi- plication of fractions. Illustrations : 1. Divide l^ by ?1^. 32 cy ^ 40 cy 7 g'a: . 21 n^z ^ 7 a-x ^ 40 ey ^ 5 a; ^g^^j^ 32 cy ' 'iOoy 32 cy 21 a'z 12 z 2. Divide f-"-^ by -"^I^3a-4 a^-5a+6 aJ'-8a + 15 a'^ - a - 2 . ga _ ,3 g - 4 _ (g - 2)(o! + 1) . (g - 4) (a + 1) g'^ - 5 g + 6 ' g'^ - 8 g + 15 (g - 2) (g - 3) ' (g - 8) (g - 5) ^ (g + l) ^ Jg-.3)(g-5) (g - 3) (a - 4) {a + 1) g — 5 a-4 Result. 3. Divide , «'^ + 27 by a^ - 3 a; + 9. »^' + 27 .^(^._3^ + 9) = . x'+27 „ x^ + 6x- + 9x a;' + 6 u;-^ + 9 X x^ -3x + 9 x(x + 3)(x + 3) x:^~3x + 9 1 X (a; + 3) Exercise 47 Result. Simplify : 2. a^-a;^ a — x (r + -t^"' r--9 c-3 (.t:+2)- . .^'-4 c- + 2c ■ c^ + 3c + 2 " a;-5 ■x'-126 14:2 FRACTIONS , (?—cx—6x^ . c + Sx _ 0^ — 4a; + 3 . x—1 c=-9ca;^ ""0+20;' ' 3-x ' 2-x' g a^ + Qx-l . jg'-Sx+g g a^-1 . 1 - a; ar'-4a;-21 ■ ar^ + a;-6 " ' 1-x^ ' ai' + l+x' x^-Sx + lS . 7a;-10-ar^ 9. 10. 6-5x + x' ' x^-ix+i (a;-l)^-a^ . x'-jl-af (x-af-l ' ar^-(a-l)2' ~, a'-3a ^ a^ + 3a + 9 . a'-27 12. a?+3a" a + 3 (a + 3/ a^ - a; - 12 a^-2a;-3 . x- + x 9^_9 ^a^_2a;-8'' a; -2' .„ a? — ax — a + x a? — ?> lo. — -- — z — X 14. 2 — 3a + a^ a^—x^ a? + ax + 01? g^ + l ^ a- + fa + !)'(«- ly _ «" + ! a^-1 a-(a2+l) +1 ' a?-l' 171. When addition and subtraction with multiplication and division occur in the same fractional algebraic expression the additions and subtractions are first performed. Illustration : Simplify f^-^Vf 3.- 6+ ^^ > a; — 2 a; + 2/ V a' + 2_/ lx + \ x-l\./.,^ g 12 \ r (3:+1)Cr»: + 2')-(a;-l)(a -2)-[ Vs-2a: + 2JV x + lj \_ x''-i J ■ L a; + 2 J 6x 3a;2 a;2 - 4 x + 2 6x x + 2 "x2-4 ' 3a;2 = ?^ • Result. a; (a; - 2 j DlVISmX OF FRACTIONS 143 Simplify : Exercise 48 (See also page 402) 1. 2. x + 1- X — S\ /'x — m _ o \ / 3m \ •' ! mj \m' — dm + 9/ V .(• / \ III- — _' vix — o x-j \u J 6. M>+i+~'V('--i+- 7. 8. 9. 1 c e^- oic- a + ; + l]f|±^_l _1> ((- — z^ { c + 2r-,r- 2 a 11 (c + 2)^-(c-2)^ . 8c c-2 ■ C--2 12. \. 2 (a; -2) ^- 3 2 X- + 2 « + 4 ■ 13. (m — 1 + m + t±f\ \/?)r wi. 1 \ . 2m^ + ! lj\6 4 12J ' r,m + 3 z m-> + » m 14. 15. r x" ~ |_(a3 _ ^jjy^ ;-l ; + l + S)- m m + 1 + 1 m — 1 + 1 W+1 y 16 rif^ L_^ «^--t ■ L2Va + 2 a-2j 2a-+ia m — 1 / V.m + 1 -(a + 2), 144 FRACTIONS 17. 18. 112/ ^ + -„ + — - i m' n' ^^+9(^+^1" c ahx^ -j- (g + b) x' + 1 ahc 19. . a + 20. ex + 1 cJKacx + a + fiacx + a + x) c*- 2c^- c + 2 _ (d - 1)^ ^ CT?- c -2d + 2 ^ d^ d FRACTIONS IK THE COMPLEX FORM 172. A complex fraction is a fraction whose numerator or denominator, or both, are fractions. Illustration : c' + d' — ■ c 1. Simplify r- + d^ „ _d 1 _ 1 d [■ C2 C^ + l d^ — cd d 1_ 1 d c Simplify 1 c'^ - d- . c' + d^ c-d cd . c'^ — cd + d- 'd c-^ - a^ C'' + d« X cd (c + d)(c-t?) ^^ c-d (c + d)(c2-cd + (Z^) -1 1+^ Exercise 49 (See also page 403. ) 5. a; + 3 + ■ a; r)-4 + - 4. X — — 1 X X — ? + l X a 2+- a; X a 2 + « a; a+2 (( a+1 a+4 10 a a;+l + a; + 2- a + 1 4 :i;+3 ~2~ ' x+S FRACTIONS IN THE COMPLEX FORM a-- 4 x^+4^^ 1 •••^-^ ,,■2 + 4 r + 2 SOM. HI,. AI.G. 10 1-X 140 FRACTIONS 27. 28. ■ + -1 1 + »» + 'iiv' 2 -\-l — m -^ ( I- + inf. 1 + '■ /I c 1 + ; 1 V 2 4 •2 r ci '^2 2 + 1 29. iiO. 31. '/ + 1 ■ + - a — 1 + 1 a-1 + 1 « + l -1 g-l a + 1 2c + a; — ■ 2 c + a; - 2 ex ■ _a_±_2._2 1 1 - 2 a ] + 2 (t :; c + a; a(2 - g.) a 1+2(0 + 2) 1 1-2 a 1+2 a -,, 2 a — 1 a + 1 a — 1 32. a + 1 a-1 a—1 a+1 a— ] a+l a + 1 « — 1 1 + 33. 34. a+1 a—1 c+l 9 3a a- Vr + 27 . a- + 9 a3_27 ■ a- -9 : + l 0! + ; ( a -3)^ + 3 a, ^(\__1\ _(a+3)2-3a ' 1^9 a 3 ~\ L (a; + 2)5 1-- m- + 1 + m (a; + 2)VV(^ + 2/ a; + 2 -1 CHAPTER XIV FRACTIONAL AND LITERAL LINEAR EQUATIONS PROBLEMS 173. To clear an equation of fractions is to change its form so that the fractions shall disappear. This change is ac- complished by the use of the L. C. D. of the given fractions. Given: ^±6^x+i0. 4 6 Multiplying both members by 12, 12(a: + 6) ^ 12(a;-H0) _ 4 6 Reducing, 3 (k + 6) = 2 (a; + 10). And the original equation is merely changed in form and is free from fractions. In general, to clear an equation of fractions : 174. Multiply both members of the equation by the L. C. D. of the fractions, remembering that the sign of each fraction becomes the sign of its numerator. Solve the resulting integral equation. Illustrations: 1. Solve2^±l_3^:^ = 6-^. 3 5 2 The L. C. D. = .SO. Multiplying both members by 30, 10(2ir + l)-6(3x-2)=15(6-x). 20 a; + 10- 18 a; + 12 = 90- 15a:. 20a;- 18a; + 15a; = -10-12 + 90. 17 a; = 68. y = 4. Result. 117 148 FRACTIONAL AND LITERAL LINEAR EQUATIONS 2. Solve ■'■-1 x + 2_., From which x = -- Result. Multiplying both members by the L. C. 1)., (x + l)(a; - 1), (x- ])2-(x+ l)(x + 2)=2(x2- l)-2x--'. x2 _ 2 X + 1 - x2 - 3 X - 2 = 2 X- - 2 - 2 x2. 1 5' Exercise 50 Solve : , x + 1 . x — 1 . ^ 2x — l i-x 2a; + l 3 2 3 5 10 4 3 3 5 3 ^L±i_^zd = li 7 ••!■-« +2 S.v + 1 2a; + l ^„ 3 5 15 ■ 4 3 2 4 2« + l_3* + 1^7 3.); + 2 x-G^l-2.«^ 4 3 6 ■ 5 2 3 9. i(a;-l)-|(«; + l) + 2=0. 10. |(2a;-l)-J^ = i(5 + 3a;). 11. (x+y}(x~-]^) = x^. 12. (X - 1) (..+ ^) = (,«-!) (a; -i). 13. l(.^._2)-Ka; + l)-i(a;-2) = 0. 14. 2i + ^J^ = 2a;-i^:^. = 2 5 15 C-^+l)' (a; + 2)^ ^ 14-a;^ 3 2 6 16 ^ — '^ — ^ — 6 . 19 i^Zll _ 3jB^ll _ 1 a; + 5 x+3 ' x + 1 x — 1 17. pJZ^^p^. 20. -i^ E 3 = 0. 3a; + 4 6a;-7 a; — 5 x — 3 18. ^:!ill|_|ir:zi' = o. 21. ^i^^ + 2 - 1^^ = 0. 6.i'-l 4.1- + 1 ,0+3 .c + 1 Sl'KCIAL FORMS OF I'RACTIONAL LINEAR EQUATIONS 149 22 ^^ _ 3a; x , 2x 5 23. a; + l ,1+1 2.i- + 2 3x + 3 6 .)- + 2.f +4 _ 2x ^ .x"-2a; + 4 x + 2 4-a;- x-2 24. ^ 2 6_^o. 2x,+ r, :\-2x 4a-2-9 25. ■" .r - 4 _ IS.i-^ + fl- ^ 3a! + 4 3 a' + 4 9.i'--16 4 -3 a;' 26 g_xl+.S.v + l g^._ 9.r--3a; + l ;!.i- + i ■ 1-3 a; 2 3 a. 27. + - <^ ^ 4 .>■- + ;! .!• + 2 a,- + 4.f + ;! .r- + 5a; + 6 28. ■^• + 1 + ^z:J =— ■^•ii 2 .1- - .)• - 1 2 X- - 3 ,<■ - 2 X- - 3 X + 2 SPECIAL FORMS OF FRACTIONAL LINEAR EQUATIONS (n) The Independent Monojiial Dknominator Illustration : 175. Solve 3a; + 4 a;-l _ 2a;-l _7 3 2a; + l 2 6' The L..C.D. of the monomial denominators, 3, 2, and 6, is 6. Multi- plying both membei-s of the equation by 6, we have, 2(.3x-|-4) _ 6(a:-l) _3(2a;_i) = 7. 2x + l From which o a- + 8 - ^ '^^ ~ ^) - 6 x + 3 = 7 • 2x+ 1 At this point note that all x-tenns outside of the fraction disappear. In any simple equation of this type the unknown term similarly disappears when the equation is cleared of monomial denominators. If the unknown term does not disappear except- ing from the fraction having the binomial denominator, error has been made in the work. 150 FRACTIONAL AND LITERAL LINEAR EQUATIONS 'J'ransposing and collecting, Dividing ty — 2, Clearing of fractions, 6 (a: - 1) - i 2x + 1 8(a;-l) =2(2x + l). 3a;-3 = 4x + 2. a; = — 5. Result. Solve : 3. Exercise 51 x+ 1 a?— 1 _ a;— 1 3 a,' + l~ 3 3a;-l a' + 1 ^ 2a; + l 3 a; + 2 2 ' 4a;+5 a; + l _ 2a; — 1 4 x-l~ 2 23; + ! a;— 1 Sa' + l ^A 2 .(:-f-10 5 ~ ' 2a?-l , x-2 x-1 ■ + = 0. 4 2 a; + 4 2 3a; — 4 9a; — 4 a; + ll a; 3 12 12a; -27 4 2a; + 7 a; _3a;-l_5a;4-6 = 0. 12 3a; — 3 10. 3a;4-5 4a;-12a; + 1 ^75 2a;-5 G 3 6* 8a;-5 2| 16x+3 ^ a;-l^ 8 2 16 a!-5' 2a; — 5 3a; + l _ 3a; + i , 3a; — 7 as — -J + 24 SPECIAL FOIIMS OM FRACTIONAL LINEAR EQUATIONS 151 (6) Forms having Four or More Dissimilar Denominators 176. Illustration : Solve ^^ - ?^ = ^^ - 5^ . x — 2 a;— 1 x — 6 x — 5 To avoid the use of a common denominator having four binomial factors, eacli member of the equation is firet simplified independently of the other. Combining separately, (x -l)(x-S) -(x- 2) (x-2) _ (z- 7)(x - 5) - (x - 6) (a; - 6) _ (x-l)(a;-2) (x-6)(x-5) Simplifying, ^^ = ^i . (x - 1) (x - 2) (x - 6) (x - 5) The fractions being equal and having the same numerators, it follows that the denominators must be equal. Therefore, (x - 1.) (x - 2) = (x - 6) (x - 5). x2 _ 3 X + 2 = x2 - 11 X + 30. 8 X = 28. x= h Result. Solve : 2. Ezercise 52 1 1 1 1 x—5 x — 4, x — 3 x-2 111 1 x+1 x+2 x+3 x + i 2 11 2 2a;-3 x-4: x-1 2a;-7 11 11 7 7 x + 3 x — 4: x+2 x-9 x + 2 x+3_x + 4: x + 5 x+ 5 x+ G x + 7 x + 8 x—S x—7 x—5 x — 4: •10 x-9 x-1 X-& 152 FRACTIONAL AND LITERAL LINEAR KQUATIONS- (c) Literal Fractional Equations 177. In literal equations either a part or all of the assumed known quantities are represented by letters, the first letters of the alphabet being ordinarily chosen for these known quanti- ties. Illustration : Solve ?^+i' + l = i?--^. X — a 2 X — a Multiplying both members of the equation by L. C. D. (_x — a)(2x — a), we have, (x + a)(2 X — a) + (j; — a) (2 .r — rt) = (-t ,. — a)(x — a). 2 x^ + ax — a^ + 2 x- — '■) ax + a'^ = 4 x^ — 5 ax + a^. 2 x2 + 2 x'^ — 4 x- + ax — 3 ax + -5 ax = a^ — a- + a^- S ax = aK X = - . Result. 3 Exercise 53 (See also page 404.) Solve : 1. 6a; + 7a = 5a; + 9a. 14, (c + d)x+ (c— d)x = 4:cl 2. 3 X — i a = o X + 3 a. 15. abnv' — bhn — umx -\- bx = 0. 3. 3a-icx = 7 a-6r.,: 16. (.k - cf = {x -2 cf - 3 c\ 4. r) 7nx -{- i u = 7 iiix — It. 17. (m — x){ik — x) = (711 + x)x. 5. 3x-ic' = x + 3c\ 18. (x + a){x + l) = (x + a + iy. 6. 4a; + 5a2=13o2 + 2a;. 19. (c^ + a;)^ = x^ + c2(c2 + 4). 7. o(a; + l) = a + r. 20. (.b+3 ?n.)2-(a;+»i) (x-m) =0. 8. 2ux-Au'' = x-a\ 21. (.v + a + 1)^ = (.0 - k. - 1)^. 9. ix + a)a = 3{x-a). 22. (a + 6)2x = «' + 6(ax- &''). 10. (a; 4- m) (a; + 1) = xl 23. (x + af + l =(x + a + lf. 11. cd(a; + l) = c2d^ + x. 24. (a + x){b + l) = {b + x){a + l). 12. a(x-a)=c(x-c). 25. (x — a)^ + (x + <()= = 2 (x -f c)^. 13. (n^ + x) (c^ + x) = x''. 26. (x-3 a - c)^ = (x-f a + 3 «)". LITERAL FUACTIONAL EQUATIONS 153 27. -5L + ^ = 5. 35. ^i±^^g + c + 2_ 2a 3a x—cx—c—2 28. 5^-^ = A. 36. ^-^•_g 3a;-p 3a4a6a2 m + a; a; + p 29. 1 + 1 = 1_1. 37. -1+^ , * + d_2 a X a X x—dx—c X C X 30. — + « = - + dm. 38 a; a; 31 i(l+-^)=Kl-> 39 2*-" 1 *-« a;— c a; + c 32. a; + m x + ?i a; + n a;4- m a; — e — 2 a; — c+1 33 x+2a X —4c .T+Sc x—3a ^^ c rf c-fZ •i; — c X -^ d. X 34. 2 a; 4- 5 c 5 .-; + 4 c 2a; — 5 c 6a;— 4 c 42.. °«^ ic + d)- -^^ c + a; ' " d + a; 13 /^a "*" V 1 ^a 1 ^^ ^ 1-a 43. ^a ^^ _ -^ j • 1 V ' a-lj ^ '*■ .. d . 1 _ d 1 46. .c — /;(. d-2 a; + l d + 2 a;-l . a;+ 2 m _ x — 2 // _ 4(wi, + w) X— '2 m X + 2 ^1 X x + 2m_ ■'trnx + 'dw? Qc'^'^^ H ex -\-d (ca; + cZ)(da; + c) dx + c a; —2c a- — (> c a' — 4c a; — 8 c x — 3m, X — im a;^ — 7 wia; + 12 m^ 164 rEACTlONAL AND LITERAL LINEAR EQUATIONS PROBLEMS LEADING TO FRACTIONAL LINEAR EQUATIONS Exercise 54 (See also page 405) 1. The difference between the fourth and the ninth parts of a certain number is 2 more than one twelfth of the number. Find the number. Let X = the required number. Then 5 _ ^ = the difference between the fourth and ninth parts. h 2 = one twelfth the number in- creased by 2. Hence, from the conditions, - -- = — + 2. 4 9 12 Clearing, 9 a; — 4 x = o a; -f 72. 2 :c = 72 X = 36, the required number. Verification :---=§?_ ^' = 9-4 = 5. 4 9 4 9 ^ + 2=— + 2=3 + 2 = 5. 12 12 2. Find that number the difference of whose fifth and sixth parts is 3 less than the difference of its third and fourth parts. 3. When the sum of the third, eighth, and twelfth parts of a number is divided by 2 the quotient is 1 more than one fourth the number. Find the number. 4. Find three consecutive numbers such that the first divided by 6, the second by 5, and the third by 2, give quo- tients whose sum is 4 less than the greatest number. 5. The first digit of a number of three figures is two thirds the second digit and 4 less than the third. If the sum of the digits is 18, what is the number? 6. The sum of two numbers is 38, and if the greater number is divided by the less increased by 2, the quotient is 3 and the ii'iuainder 4. Find the number. PROBLEMS W I'm FRACTIONAL LINEAR EQUATIONS 155 Let X =z the smaller number. 'I'lieii 38 — X = the greater number. Now DiYidend - Remainder ^ Q^otj^^t Divisor Hence, ^^'^Y :3. Or, 38-a;-4=3a; + 0. From which x = 7, the smaller number ; and 38 — a; = 31, the larger number. Verifying : ^38-^1-4 ^ 38-7-4 ^ 27 ^ , ° X + -2 7+2 9 7. The sum of two numbers is 57, and if the greater num- ber is divided by the less, the quotient is 3 and the remainder 9. Find the numbers. 8. The difference between two numbers is 23, and if the greater is divided by 4 less than twice the smaller, the quotient is 3. rind the numbers. 9. The sum of the third, fifth, and sixth parts of a number is divided by one half the number, and the qviotient is 1 and the remainder 6. Find the number. 10. If the sum of three consecutive numbers is divided by the smallest number increased by 7, the quotient is 2 and the remainder is 6. Find the three numbers. 11. In 3 years a certain man vrill be half as old as his brother, and the sum of their present ages is 69 years. What is the present age of each ? Let X = the man's age in years at the present time. Then 6Q — x = the brother's age in years at the present time. Hence, as + 3 = the man's age after o years, and 72 — a; = the brother's age after 3 years. Then a; + 3 = ll^z^. 2 and X = 22, the man's present age. 69 — a: = 69 — 22 = 47, the brother's present age. Verification: 2(22 + 3) = 47 + 3. 50 = 50. 150 FRACTIONAL AND LITKUAL LINKAH EQUATIONS 12. A is one third as old as B, but m 8 years he will be only one half as old. Find the present age of each. 13. A's age is one fourth that of B, but in 5 years A will be one third as old as B. Find the present age of each. 14. A child is li times as old as Tiis brother, but 2 years ago he was 1^ times as old. How old is each now ? 15. The sum of the ages of a father and son is 72 years, and if the son were one year older and the father one year younger, the son's age would be one third that of the father. Find the present age of each. 16. If the length and the width of a certain rectangular field were each increased by 10 feet, the area of the field would be increased by 800 square feet. If the length is now 10 feet more than the width, what are the dimensions of the field ? Let X r= the width of the field in feet. Then a; + 10 = the length of the field in feet Since the area of the field equals the product of the length by the width, x(x + 10) = a;''^ + 10 a; = the present area of the field in square feet. In like manner, (a;+10)(a: + 20) = a;2 4.30a; + 200 = the area of the field in square feet if length and width are increased. Hence, Qi? + 30 a; + 200) - (x^ + 10 a;) = 800. x* + 30 a; + 200 - a;2 - 10 a; = 800 20 a; = 600. a; = 30, the present width of the field ; , and a; + 10 = 40,thepresentlengthof theflekl Verification : (x + 10) (a; + 20) - a: (a; + 10) = (40) (50) - 30 (40) = 2000 - 1200 = 800. 17. The length of a certain rectangle is 10 feet greater, and the width 5 feet less, than a side of an equivalent square. Find the dimensions of the rectangle. PROBLEMS WITH FRACTIONAL LINEAR EQUATIONS 157 18. A square has the same area as a certain rectangle whose length is 20. feet more, and whose width is 12 feet less, than the side of this particular square. Find the dimensions of the rectangle. 19. The length of a certain rectangle is 12 feet greater than tlie width. If each dimension is increased by o feet, the area of the rectangle will be greater by 225 square feet. Find the dimensions of the rectangle at present. 20. A square is cut from a certain rectangular field, the length and width of the iield being respectively 50 feet and 30 feet greater than the side of the square cut out. If 9500 square feet remain in the field, what was the original length? the original width ? the original area ? 21. A can do a piece of work in 6 days, B the same work in G days, and C the same in 7 days. How many days will be re- quired for the work if all three work together ? Let X = the number of days all three together require. \ = the portion done by A in 1 day. \ = the portion done by B alone in 1 day. ^ = the portion done by C alone in 1 day. Hence, ^ + J + 7 = the amount all together can do in 1 day. Therefore, 7 + :^ + ;; = - • 5 6 7 X 42x4-35 a; + 30 a; = 210. 107 X = 210. X = l}Jf days. Result. 22. A can do a piece of work in 4 days, B in 5 days, and G in 8 days. How many days will be required if all three work together ? 23. A and B can together build a wall in 7 days, and C alone can build the wall in 15 days. How many days will be required if all three work together ? 1.58 iUAUTIONAL AND LlTJiliAL LINEAR EQUATIONS 24. A and B can paint a house in 8 days, A and C to- gether in 9 days, and A alone in 12 days. In how many days can B and C together do the work ? 25. Two pipes enter a tank, the first of which can fill it in 7 hours while the second pipe requires 9 hours to fill. How many hours will be required to fill the tank if each runs alone 1 hour, and then both run together until filled ? 26. At what time between 4 and 5 o'clock are the hands of a clock together ? At 4 o'clock the hour-hand is 20 minute-spaces ahead of the minute- hand. Hence, Let X = the number of spaces tlie minute-hand passes over, and a; — 20 = the number of spaces the lioui-hand passes over. Now the minute-liand moves 12 times as fast as the hour-hand. Therefore, 12 (a; — 20) = the number of spaces the minute-hand passes over. Hence, 12 (x — 20) = x. 12 X- 240 = X. \\x= 240. X =21jV That is, the hands of the clock will be together at 213^ minutes after 4 o'clock. 27. At what time between 3 and 4 o'clock will the hands of a clock be together ? 28. At what time between 7 and 8 o'clock will the hands of a clock be together ? 29. At what time between 10 and 11 o'clock will the hands of a clock be at right angles to each other ? (Hint : In this case the position of the minute-hand will be 15 minute-spaces behind the hour-hand.) 30. Find the time between 4 and 5 o'clock when the hands of a clock are at right angles to each other. (Hint: Two pos- sible positions may be found in this case, for the minute-hand may be 15 minute-spaces ahead or behind the hour-hand.) PROBLEMS WII'H FRACTIONAL LINEAR EQUATIONS 159 31. The denoininator of a certain fraction is greater by 2 than the numerator. If 1 is added to both the numerator and denominator, the fraction becomes f. Find the original fraction. 32. Out of a certain sum a man paid a bill of f 30, loaned \ of the remainder, and finally had left $56. How much had he at first ? 33. The largest of three consecutive odd numbers is divided into the sum of the other two, the quotient being 1 and the remainder 9. Find the numbers. 34. The sum of the ages of a father and son is 80 years, but if each were 2 years older the son's age would be |- the father's age. How old is each ? 35. A certain number is decreased by 12 and the remainder is divided by 4. If the resulting quotient is increased by 7, the sum is the same as if the original number had been in- creased by 7 and then divided by 3. Find the number. 36. A man gave f of a certain sum to relatives, y^ to each of two churches, ^ to a library, and the remainder, $ 6000, to a hospital. What was the total bequeathed ? 37. In a certain baseball game a total of 13 runs was made by both teams. If the winning team had made 2 more runs, and the losing team 3 less, the quotient obtained by dividing the winning runs by the losing runs would have been 5. How many runs did each team make ? 38. The distance around a rectangular field is 96 rods, and the length of the field is f the width. Find the length and the width of the rectangle, and the number of square rods it contains. 39. In eight games a certain fielder made 2 less runs than hits. If 5 times the number of hits he made is divided by the number of runs increased by 3, the quotient is 4. How many hits and how many runs did he make ? 160 FRACTIONAL LITERAL AND LINEAR EQUATIONS 40. Find three consecutive even numbers such that 3 less than one half the first, plus 2 less than one half the second, plus 1 less than one half the third equals 15. 41. In going a certain distance an automobile moving 20 miles an hour required 3 hours less time than a second auto- mobile making 16 miles an hour. What was the distance in miles ? 42. A number is composed of two digits whose difference is 4. If the digits are reversed, the resulting number is ^ the original number. Find the number. 43. A can run 10 yards in 1 second, B 8 yards in 1 second. If A gives B a start of 3 seconds, in how many seconds will A overtake B ? 44. The length of a rectangle is 9 rods more than its width. If the length is increased by 6 rods and the width decreased by 3 rods, the area is unchanged. Find the length and breadth of the rectangle. 45. An automobile going 26 miles an hour is 40 minutes ahead of one going 30 miles an hour. In what time will the second automobile overtake the first ? 46. At what time between 8 and 9 o'clock do the hands of a clock point in opposite directions ? 47. A freight train goes from A to B at 15 miles per hour. After it has been gone 4 hours an express train leaves A for B, going at a rate of 45 miles per hour, and the express reaches B \ hour ahead of the freight. How many miles is it from AtoB? 48. In traveling a certain distance a train going 45 miles an hour requires 5^ hours less time than an automobile going the same distance at 27 miles per hour. What is the distance between the two points ? CHAPTER XV APPLICATIONS OF GENERAL SYMBOLS. REVIEW STATEMENTS. PHYSICAL FORMULAS. DERIVED EXPRESSIONS THE GENERAL STATEMENT OP A PROBLEM 178. From the following illustrations it will be seen tliat when the given numbers of a problem are literal quantities, the statement and the solution result in a formula or general expression for that particular kind of problem. Illustrations : 1. If A can mow a field in m days, and B can mow the same field in p days, in how many days can both together mow the field ? Let X = the number of days both working together require. Then - = the portion of the work both together can do in 1 day. X Also — = the portion of the work that A alone can do in 1 day. m — = the portion of the work that B alone can do in 1 day, o Hence, -!- + - = the portion of the work both together can do in 1 day ; m p and — I- - = - is the required general equation for the condition. m p X ' Solving, mx + px = mp. (m +f>)x = mp. a;=-™2-. Result. m +p This expression is, therefore, a formula for finding the time in which two men whose individual ability is known, can, work- SOM. EL. ALG. 11 161 162 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS ing together, accomplish a given task. By substituting in this formula, any problem involving the same condition can be solved. For example : A requires 4 days to do a certain task, and B requires 5 days for the same work. In how many days can both working to- gether complete the work? Here we have A's time alone (or m) = 4 ; B's time alone (or p) = 5. In the formula x = ^^ = , ' ^ = -tt = 2| days, the time in which both together can do it. 2. Divide the number a into two parts such that m times the smaller part shall be contained q times in the larger part. Let X = - the smaller part. Then a — X - = the larger part. From the conditions, a — X ■mx = ?. id a — X = mqx. mqx + X = a. (mq + 1)0; : = a. a , mq + 1 , the smaller part required. Also, a-z = a a_^amg + «-a^ amq ^ the larger part mq + 1 mq+1 mq + 1 ^,^^^5^^^ To use this formula : Suppose we are required to divide 60 into two parts such that twice the smaller part shall be con- tained 7 times in the larger part. We have = 4, the smaller part required. = 56, the larger part required, = 60. 2. 60 m = "'' 2. ■7 + 1 3 = 7. a - ■'^-f. .2.7 7 + 1 And , verifying. 56- ^2(4): = 7. GENERAL STATEMENT OF A I'KOBLEM 168 Exercise 55 1. Divide the number c into two parts such that m times the larger part shall equal n times the smaller part. 2. Divide a into two parts such that the sum of -th of the n larger part and -th of the smaller part shall be q. T 3. The sum of two numbers is s, and if the greater number. g, is divided by the less number, the quotient is q and the remainder r. Find the numbers. 4. If A and B can together mow a field in t days and B alone can mow the same field in h days, find the number of days that A working alone will require to do the work. 5. Show that the difference of the squares of any two con- secutive numbers is 1 more than double the smaller number. 6. m times a certain number is as much above A; "as d is above c times the same number. Find the number. 7. A and B are m miles apart, and start to travel toward each other. If they start at the same time and A goes at a rate of k miles an hour while B goes at the rate of s miles, how far will each have gone when they meet ? 8. When a certain number is divided by a, the quotient is c and the remainder m. Find the number. 9. The front wheel of a wagon is m feet in circumference, and the rear wheel n feet in circumference. How far has the wagon gone when the rear wheel has made r revolutions less than the front wheel ? 10. A and B can together build a barn in r days, B and C the same barn in s days, and A and C the same in t days. In how many days can each alone build it ? 104 rUACHCAL APPLICATIONS OF GENERAL SYMBOLS THE USE OF THE LABORATORY FORMULAS OF PHYSICS 179. Density is defined as mass per unit of volume, and may M be calculated by the formula -D=^, in which D is the density of the body under examination, M its mass, and V its volume. Volume is another name for cubical contents. The mass of a body may be found by weighing it, and the formula for density W may be written D = t-^, in which W is the weight of the body to be considered. Exercise 56 1. A block of iron 3 feet 6 inches long, 2 feet wide, and 6 inches thick weighs 1706.25 pounds. Calculate its density in pounds per cubic foot. 2. The density of water is approximately 62.6 pounds per cubic foot. What is the volume of a ton of water? 3. Acubical block of wood 9 centimeters on an edge has a mass of 371.79 grams. Calculate its density in grams per cubic centimeter. 4. A block of iron having an irregular cavity weighs 3.265 kilograms. When the cavity is filled with mercury, the whole weighs 3997 grams. The density of mercury being 13.6 grams per cubic centimeter, calculate the volume of the cavity. 5. The density of a certain substance being a grams per cubic centimeter, calculate the mass of this substance necessary to fill a vessel of 1000 cubic centimeters' capacity. 180. To change a reading on a Centigrade thermometer to a corresponding reading on a Fahrenheit thermometer use is made of the formula in which F is the reading in degrees on the Fahrenheit scale, and C the reading in degrees on the Centigrade scale. To LABORATORY FORMULAS OF PHYSICS 165 change Fahrenheit readings to Centigrade readings we use the formula lixercise 57 1. Change the following Centigrade readings to correspond- ing Fahrenheit readings : 20°;40°;0°; -15°; -273°; -180°. 2. Cliange the following Fahrenheit readings to correspond- ing Centigrade readings : 212°; 32°; 70°; -10°. 3. At what temperature would the reading on a Centigrade thermometer be the same as the reading on a Fahrenheit thermometer ? 4. The Centigrade scale is marked 0° at the freezing point of water, and 100° at the boiling point of water. The Reaumur thermometer is marked 0° at the freezing point of water and 80° at the boiling point. Prepare (1) a formula for changing Reau- mur readings to Centigrade readings, and (2) a formula for changing Centigrade readings to Reaumur readings. 181. In the figure, we have a straight bar whose weight is „„, ^^ to be neglected. The bar is sup- j I (/- \z.'I^r.l. ! ported at the point O and is _L \-t/-'r-" — I i acted upon by the several forces, 4/" f> f) f"> etc., in the directions indicated by the arrows. These ' forces tend to cause rotation of the bar about the axis 0. The tendency of a force to produce rotation is called its moment, and the moment is calculated by multiplying the magnitude of the force by the distance of its point of application from the axis. Moments tending to rotate a body clockwise are given a positive sign ; those tending to produce I'otation in the opposite direction are given a negative sign. In order that a body under the influence of moments mar be in equilibrium, i.e. stationary, the algebraic sum of the IGO PRACTICAL AITLICATIONS OF GENERAL SYMBOLS moments acting on it must be 0. Thus in the figure, if the bar is in equilibrium, f'd"+f"'d">-fd-fd<-f»"d""=Q, or, better : (/"d" +/"'d"') - (/d +/'d' +/""d"") = 0. (/'i /")/'". etc., are read "/prime," "/second," "/third" etc.) Exercise 58 1. Two weights of 4 and 12 pounds respectively are bal- anced on a bar, the distance from the support to the further ■weight being 6 feet. What is the distance from the support to the nearer weight ? 2. A bar 11 feet long is in equilibrium when weights of 15 and 18 pounds are hung at its ends. Find the distance of the point of support from each end. 3. If weights are distrib- i uted upon a bar as in the | i i figure, where must a weight ^ ^' \ ,o: of 40 pounds be placed to keep O J!"' "^ the bar in equilibrium ? ' — I LiJ 4. A 10-pound weight hangs at one end of a 12-foot bar, and a 15-pound weight hangs at the same side of the support- ing point, but 2 feet nearer it. If a 40-pound weight at the other end keeps the bar in equilibrium, at what distances from the ends is the point of support located ? THE VALUE OF ANY ONE ELEMENT OF A FORMULA IN TERMS OF THE OTHER ELEMENTS 182. The statement of a mathematical law by means of a formula always gives an expression for the value of the par- ticular element to which the law refers. By transpositions and divisions we are able to derive from any formula another expression or formula for any one of the other elements. TRANSFORMATION OF FORMULAS 167 Illustrations : 1. Given the formula, li = ^^ . Derive a formula for s. g + s Clearing of fractions, {g + s) R = gs. Multiplying, Eg + Bs = gs. Transposing, Bs — gs = — Bg. Dividing by — 1, gs - Bs = Bg. Collecting coefficients, {g — B)s = Bg. Dividing hy (g — B), s = — ^, the required formula for s. g- B 2. Given the formula, l — a-[-(n — l) d. Find an expression for n. I = a + (n — 1) d. Multiplying, I = a + nd — d. Transposing, I — a + d= nd. Dividing by d, n= ~ '^ — , the required formula for n. THE TRANSFORMATION OF FORMULAS Ezercise 59 PHYSICAL FORMULAS 1. Given v = at, find the value of t in terms of a and v. 2. Given S = ^g(2 t — 1), find a formula for t in terms of s and g. 3. Given C= — , derive a formula for S in terms of E, b, P, C, and B 4. Giv( ]}, and p'. 4. Given -= -H — , find an expression for each element, /, f P P' 16S PKACTICAL APPLICATIONS OP GENERAL SYMBOLS 5. Given J' = | (7 + 32, derive a formula for C in terms of F. 6. Given F = — , v = gt, and w = Fs, find a value for w in terms of m, g, and s. 7. Given ^J-i = ^^-^, derive expressions for Ti and Tj- MISCELLANEOUS FORMULAS 8. Given C=2irB, obtain a formula for B. 9. Given l = a+(n — l) d, derive an expression for a in terms of /, n, and d. 10. Given S =-(a + I), find the value of a in terms of I, n, and S. 11. Given T= nR(B + L), find L in terms of the elements involved. 12. Given S = , find each element in terms of the r — 1 others. 13. Given d = ^y~f) , find a formula for S. n{n — l) 14. Given a =p +prt, find each element in terms of the others. GENERAL REVIEW Exercise 60 1. Simplify 2a- [3-2 Ja-4(a-a + l)J]. 2. Solve (a;-l)(a; + 3)-2(a;-l)(3a;+l) = (3-a;)(2+5a!). 3. Show that A +-i ^^\(J^ ^.iV 1=0 V a+1 a + 3Aa-3 ct-l ) GENERAL REVIEW 169 i-^ + il a' a -H '\ 1 a 2 la 2) 4. Simplify 5. Factor a^ — 8 {a? — a^) — aV. 6. What number added to the numerators of the fractions - and -, respectively, will make the results equal? Is there h d an impossible case ? 7. Factor Gay' + Gd-x — 37 ax-. 9. Solve- ■ + : 6 l_2aT l-4x l-3a; 10. Factor (a' - 9)(a + 2) - a - (4 + a) (3 + a) - 3. (3 1 x + m 3 x — m 11. Solve — = — - — . 3 x + m 3 12. By three different methods factor (a^—6y — «P. 13 Solve i^-?^-i ly 5 14. For what value of aisa;* — 3a^4-2a^ + 21a! — 3a divis- ible by cb^ + a; - 3 ? 15. The sum of the numerator and the denominator of a certain fraction is 39. If 3 is subtracted from both numerator and denominator, the result is ^. Find the original fraction 16. Factor 225 - 4 a^ (9 + x) (9 + a;). 17. Find the H. C. F. and the L. C. M. of a;* - aa;' - 2 a^.T-, 2 a!^ - 2 a% and 3 ar* + 12 aa^ + 3 a^*. 170 PRACTICAL APPLICATIONS OF GBNKRAL SYMBOLS 19. Solve and verify 2a;-l x-1 ^ 8a;-l _ •^ 7 2a; + 2 28 20. Prove that ~ = — = cd- IH — 1 cd cd cd cd 21. Simplify («-l)'-4^r_3 2_n (o-S)^. 3 'a; + 3' 23. A Doy^iiaS'a dollar, and his sister has 28 cents. He spends three times as much as she spends, but has left four times as much as she has left. How much did each spend ? 24. What is the value of m if — — - = 1 when 71 = —1. m — n 2 3 7 X X — - 0= a 1 3 4 3 1 25. Solve + - = -_-. -(.r-l) -(1+x) 1-i- "^ 4^ •^ 4^ •^ of 26. Prove that the sum of any five coiisecutive numbers squals 6 times the middle one. 27. Factor (a' + 5 a- 10)^ + 2{a^ + 5a- 10)a - 8 a'. 28. Find three consecutive numbers such that if the second and third are taken in order as the digits of a number, this number will be 7 more than four times the sum of the 1st and 3d given numbers. 29. If a = 2, 6 = 3, and c = — 4, find the value of {a- + 0^(3 a - c)V(7a + c)(4 6-a). 30. Simplify - GENERAL REVIEW 3 3a; + l 171 +■> 31. At what time between 8 and 9 o'clock are the hands of a watch 5 minutes apart ? 32. Find the H. C. F. and the L. C. M. of a^-4:,S-a^, -i+4.x~-3f, (2-a;)2, and (x-2)(l-x). X 33. Solve and verify ^^^^^ ^+ - = 34. Simplify /^i + 1 + - ) -5- \x or ' 1 a?+l . a;(a;-l)+l ' V-l' 35. What value of a; will make (4a; + 3)(3 a; — 1) equal to (0 .V + 5)(2 a; - 1) ? 36. Simplify (6-c)(c-a)(a-&) 6-c c-a a-6^ (6 + c)(c + a)(a+&) & + c c + a a+b 37. By what must a^ + 2ar' + 3a^ + 4a; + 5 be divided to give a quotient of a^ + 4 a; + 14 and a remainder of 44 a; + 47 ? 8a;=-l 38. Simplify _. 2-a;- 1- 39. Show that if a = - 1, 1 +a; i(l-^°)4 K-")^''- 40. Factor mV — mV — 27 mV + 27. 41. Solve 2^ + (iL±ii'-ca; = a+3ca;. w a 42. Factor 6 a'' — 6 rip — (4 n — 9 p)a. 172 PRACTICAL APPLICATIONS OF CxENERAL SYMBOLS 43. The head of a certain fish weighs h pounds, the tail weighs as much as the head and ^ the body, and the body as much as the head and tail. What is the weight of the fish in terms of h ? 44. Simplify 6a?-6 n-(«^ a + 1 4: a J ■ 48 a 45. In how many years will s dollars amount to a dollars at r per cent, simple interest ? 46. Solve form: (2 m + 2)(3 6 -c) +2 ac = 2 c(a-m). 47. If a certain number of wagons is sold at f 80 each, the same amount is received as when 10 less are sold at $ 100 each. How many are sold in each case ? 48. Find the value of /^1_1+1V r«I^-.«i^4l)-hl1^^ .,en « = ^. Va^ a/ |_ a" a' + l J a^—1 m 49. Factor 3^ + a' — 2(l — ax) — (x + a). 50. Two bills were paid with a lO-doUar bank note. One bill was 25% more than the other, and the change received was if the smaller bill. Find the amount of each bill. 51. Solve for s: (s - a) (s + 6) _ (s + a) (s - &) - 2 (a - 6) = 52. Factor a:«-a^ — 2(3 a^ + 4). 53. Solve for n, Z = (m — 1) (a — Z) + a. 54. An orderly is dispatched with an order, and 3 hours after he leaves a second orderly is sent after him with instruc- tions to overtake the first in 6 hours. To do this he must travel 4 miles an hour faster than the first traveled. How many miles an hour does each travel ? 55. Factor 50 a; + 2 ar' - 38 a;^. CHAPTER XVI SIMULTANEOUS LINEAR EQUATIONS. PROBLEMS 183. If X and y are two unknown quantities and their sum equals 7, we may write x + y = l. Clearly, an unlimited number of values of x and y will satisfy this equation. For example : Ha; = 2, 2/ = 5. Ifa; = l, 2/ = 6. If a; = 0, y = 7. Jtx= - 1, y = 8. \tx=- 2, 2/ = 9, etc. 184. Such an equation in two unknown quantities, satisfied by an unlimited number of values for the unknown quantities, is an indeterminate equation. 185. If, however, we have with this equation a second equa- tion stating a different relation between x and y, as a; - 2/ = 3, then the pair of equations, x + y = 1, x — y = Z, is such that. eac^ is satisfied only when x = 5 and y = 2. For a; + 2/ = 5 + 2 = 7, And x-y = 5-2 = S. 173 174 SIMULTANEOUS LINEAR EQUATIONS No other values of x and y -will satisfy this pair of equations. Hence, 186. Simultaneous equations are equations in which the same unknown quantity has the same value. 187. A group of two or more simultaneous equations is a system of equations. 188. Two possible cases of equations that the beginner may confuse with simultaneous equations must be carefully noted. (a) Inconsistent Equations. Given : x + y = 9, x + y = 2. It is manifestly impossible to find a set of values for x and y that shall satisfy both given equations. The equations are inconsistent. (b) Equivalent Equations. Given : z + y = i, 3x + 3s/ = 12. If the first equation is multiplied by 3, it becomes the same as the second equation, and every set of values that satisfies the second satis- fies the first as well. The equations are equivalent. 189. For a definite solution of a pair of simultaneous equa- tions we must have a different relation between the unknown quantities expressed by the given equations. 190. Equations that express different relations are inde- pendent equations. 191. Simultaneous equations are solved by obtaining from the given equations a single equation with but one unknown quantity. This process is elimination. Each of the three methods of elimination in common use should be thoroughly mastered. ELIMINATION BY SUBSTITUTION ELIMINATION BY SUBSTITUTION Illustration ; Solve the equations : 5x + 2y = ll, (1) 3a! + 4?/ = l. (2) From (2), x = — i Substituting in ( 1 ) , rJ L^Al\ + 2 j/ = 11 . From which 5-20y ^. 2 j, = 11. Clearing, 5 - 20 2/ + 6 j> = .33. - 14 2^ = 28. ■2/ = - 2. Substituting in(l), 5x + 2(-2) = ll. 51-4 = 11. = Ml = - 2. J x= 3,1 ^Eesult. y Check : Substituting in (1), 5 (3) 4- 2 (- 2) = 15 - 4 = 11. Substituting in (2), 3 (3) + 4 (- 2) = 9 - 8 = 1. From the illustration we have the general process for elimi- nation by substitution : 192. From one of the given equations obtain a value for one of the unknown, quantities in terms of the other unknown quan- tity. Substitute this value in the other equation and solve. The method of substitution is of decided advantage in the solution of those systems in which the coefficients of one equa- tion are small numbers. In later algebra a knowledge of this method is indispensable. In applying this process of elimination care should be taken that the expression for substitution is obtained from the equa- tion whose coefficients are smallest. The resulting derived equation will usually be free from large numbers. 176 KIMULTAXEOUS LINEAR EQUATIONS ELIMINATION BY COMPARISON Illustration : 5x + 2y = 9, (1) Solve the equations : , ^ [2x + 3y = 8. (2) From (1), x= ^~^^ ' From (2), x = ^— y. 5 2 By Ax. 5, 9-2y^8-5y_ •' ' 5 2 Clearing, 2 (9 - 2 y) = 5 (8 - 3 j/). 18 - 4 !/ = 40 - 15 !/. 11 ^ = 22. 2/ = 2. Substituting in (1), 5 x + 2 (2) = 9. 5x + 4 = 9. Hence, x = 1,1 jjg^^^,^ , !/ = 2. J a; = 1. Check : Substituting in (1), 6 (1) + 2 (2) = 5 + 4 = 9. Substituting in (2), 2 (1) + 3 (2) = 2 + 9 = 8. la general, to eliminate by comparison : 193. From each equation obtain the value of the same unknown quantity in terms of the other unknown quantity. Place these values equal to each other and solve. The method of comparison is particularly adapted to those systems of simultaneous equations in which the coefficients are literal quantities. ELIMINATION BY ADDITION OR SUBTRACTION Illustrations : 1. Solve the equations : |3a'-4=2/ = 5, (1) ^ \5x + 3y = 18. (2) Choosing the terms containing y for elimination, we seek to malie the coefficients of y in both equations equal ; for, if these coefficients were ELIMINATION BY ADDITION OR SUBTRACTION 177 equal, adding the equations would cause ij to disappear. The L. ('. iM. of tlie coefficients of y, 3 and 4, is ]■!. Dividing eacli coefficient into \'l, we obtain the multiijliers for the respective equations that will malis Ihe coefficients of y the same in both. Multiplying (1) by 3, x - 12 y = 15 (3) Multiplying (2) by 4, ■ 20 a: + 12 y = 72 (4) Adding (3) and (4), 29 a; =87 X =3. Substituting in (1), 3 (3) - 4 ?/ = 5. 9 -42, = 5. Hence, x = 3, 1 ^^^^^^ It is to be noted that the signs of y, the eliminated letter, being unlike, the process of addition causes the y-term to disappear. 2. Solve the equations : J ^ * - ^ 2/ = 1, (1) Multiplying (1) by 3, 6 x - 9 ;/ = 3. (3) Multiplying (2) by 2, 6 x + 14 j/ = 26. (4) Subtracting (4) from (3), - 23 !/ = - 23. j/ = l. Substituting in (1), 2 x -3 (1) = 1. 2x-3 = l. Hence,x = 2,^jjg^^,^ r = 2, :?:! In this example since the signs of the x-term are like, the equations are subtracted to eliminate the x-term having the same coefficients. In general to eliminate by addition or subtraction : 194. Multiply one or both given equations by the smallest num- bers that ivill make the coefficients of one unknown quantity equal. If the signs of the coefficients of the term to be eliminated are unlike, add the equations ; if like, subtract them. Ttie necpssitry multipliers for the elimination of a letter may be found by dividing each coefficient of that letter into the lowest common multiple of the given coefficients of that letter. The method of addition or subtraction is of advantage when it is desirable to avoid fractions, and is used in the solutions of systems involving more than two unknown quantities. SOM. EL. ALG. 12 178 SIMULTANKUUS LINEAR EQUATIONS Exercise 61 (See alsd page 40()) Solve : 1. .i- + jy = 3, 10. ;!.s + 7?/ = -8, ;;.(• + 2// = 7. -s- + .'/ = < I. 2. a; — y = 4, 11. r> x + o i/ = 10, 2x + 3y = 13. S.c-5// = 1d. Z. 4:X + 3y = 7, 12. 7'-;] = 4//, 3x-y = 2. .'■ + ,'/ - -,= 0. 4. 6x-5y = 7, 13. 5 .!• + 8 .»/ - 2 = 0, .r + 9 2/ = 11. 7 )/ + 71.' - ;; .V = 0. 5. 3.j; + 4z = 2, 14. 10 //, + 11 ((' = 32, 4.C — « = 9. 15 m + 31 = 23 40. 6. 3s — 5< = 13, 15. 3// — 4.v = 77, 2 s + 7 ? = - 12. 6 ?/ + .V = 1. 7. 3 »; + ,!/ = ^10, 16. 5 s + 3 ^ = 112, 2x-5y = -l. 4 « - 5 .s - 49 = 0. 8. 4« — 3 /'• = (), 17. n)i+13'V = 4, ""■"^ 2 y + 6 ('■ = 26. 7 /t + 19 (' = 4. 9. 5a; + 72/ = 24. 18. 12?}-7/-9 = 0, x-y = 0. ir>? -77< = -300. SIMULTANEOUS LINEAR EQUATIONS CONTAINING THREE OR MORE UNKNOWN QUANTITIES 195. A system of tliree independent equations involving thi-ee unknown numbers is solved by a repeated application of the process of elimination. The method of addition or sub- traction is most commonly used with systems having three or more unknown quantities. THREE OR MORE UNKNOWN QUANTITIES Illustration : 1. Solve the equations Multiply (1) by 3, Multiply (2) by 2, Subtract (5) from (4), Divide by 5, Multiply (2) by 5, Multiply (3) by 3, Adding (8) and (9), Multiply (7) by 19, Subtracting, ITS) 3x-2y+3z = ll, (1) IS : 2 X - .1 // + 1' « = 9, (-') 3 .(-• + 5 ?/ + 4 z = 6. (•"J 9 ,r - ;/ + 9 z = 33 (4) 4 J- — li ;/ + 4 z = 18 (-5) 5 x + 5 z = 15 (6) X + z = 3. (7) 10,r-15j/+10z = 45 (8) <) .,• + 18 (/ + U z = 18 ('•') 19 a; + 22 z = 03 (10) VJz 4 l'.iz=57 3 z = (i z=2. X + (2) = 3, X = 1. X = 1, 3(1) -2 2/ + 3 (2) = 11. y=-l, Result. y^d-ny_ a m Clearing, cm — bmy = ad— any. Transposing, any — hmij = ad — cm. Therefore, (an — bm)y = nd — cm, , ad — rm anil y = an — hm The labor of substituting a root found for one unknown and reducing the resulting expression for the value of the other unknown, is frequently as great as that of making a new solution for the value still undetermined. Therefore, in practice it will be well to make a separate elimination for oai'h unknown. lixercise 65 Solve: 1. x-\-y = a, 7. ax + by = '-■, 13. ax + by = m, x-y = c. mx + ny = c?. ex + cly = »r. 2. X + ;/ = m, 8. ax + by = -. 14. mx + ny = m', X — // = )l. rx — c1y = )n. iix + 7ny = «'. 3. x + !/ — 2a, 9. ex — ily = 0, 15. a(x + 2y) = 4, X — >/ = 2 c. dx — cy = «. e(x-2y)^X 4. cx-\- (h/ = 1, 10. KX +&// = !, 16. m{x + y)=r-,. ex - (hj = 2. ,(.',r + b'y = 1. n{x-y)=10. 5, mx + ny = l. 11. ax -\-by = be, 17. (C-(x + ey) = e-. nx — my= 1. bx — ay = erf. <^{x - ay) = a\ 6. mx + ny = 3, 12. ax + by = e. 18. d-x+c-y = cl-, sx + ty = .'i. mx — ny = ti'x -f- &y =: d'. 186 SIMULTANEOUS LINEAR EQUATIONS 19. x+y = a + l; x — y = a + l. 20. x + y = 2a + b; x — y = a + 2b. 21. ax + cy=a + '2c; cx+ay=a—2c. 22. (m + 2)x— (m — 2)y = 0; x — y = n. 23. mx + ny = (wi + ny ; mx — ny = m^ 24 (a + ni) a; — (a — m) .'/ = 4 am ; (a — 25. cx + dy = c^ — dP; dx-\-cy = i/h means " Required : One of the three equal factors of ?)." ■v'c means " Required : One of the four equal factors of c." Vx means " Required : One of the n equal factors of a;," etc. THE GENERAL PRINCIPLES OF EVOLUTION In the discussion of these principles both m and n are posi- tive and integral numbers. (a) The Root of a Power By Art. 215, (a™)" = a"". Therefore, by definition (Art. 60), a™ is the mth root of a"". For a™ is one of the n equal factors of (o™)". That is, a" = y/a^. The Fifth Index Law. The conclusion is a direct result of a division of the exponent of the given quantity by the index of the required root. Or, mn ■y/a'"" = a " = a". Hence : 226. Any required root of a power is obtained by dividing the exponent of the power by the index of the required root. (b) The Root of a Root By Art. 226, ("^o)*"" = a. Extracting the nth root, ("y/a)" = Va. Extracting the mth root, ("H/a) = '^y/a. GENERAL PRINCIPLES OF EVOLUTION 211 Hence : 227. The mnth root of an expression is equal to the mth root of the nth root of the expression. (c) The Root of a Product By Art. 226, {y/aby = ah. Therefore, (^a x v'S)" = ( v'b)" x i^)". Or, {y/aby = {Va X y/by. Hence, y/ab = v^a x Vb. Therefore : 228. Any required root of a pi-oduct of two or more factors is equal to the product of the like roots of the factors. w The Root of a Fkaction By Art. 217, \b) b"' Therefore, ^V b' Hence : 229. Any required root of afractioji is obtained hy finding the like roots of its numerator and denominator. THE SIGNS OF ROOTS (a) Positive Even Powers By Art. 2^8, (+ a){+ a) = + a^ and (- a)(- o) = + o= Therefore, V+a^ = + a or — a. Hence : 230. Every positive number has two square roots whose abso- lute value is the same, but whose signs are opposite in kind. The double sign, ±, is used to indicate two roots. Thus, y/a' = ±a. The sign ± is read, " plus or minus." •212 INVOLUTIUM AND EVOLUTION (b) Negative Even Powers Since — a^ = (+ a)(— a), we have a product of unequal factors. Hence ; 231. At this point we have no expression for the even root of a negative number. (c) Positive and Negative Odd Powers By Art.219, (+a)(+a)(+a) = + a3. Also, (- a)(- a)(-a) = -(i3, etc. Therefore, V+a^ = +a, V— a^ = — a, etc. Hence : 232. The odd roots of a positive quantity are positive, and the odd roots of a negative quant it;/ are negative ; or, briefly, the odd roots of a quantity bear the same sign as the given quantity. THE EVOLUTION OF MONOMIALS Illustrations : 1. Kequired the cube root of 8 a'''bV. The root is odd ; the given quantity, positive ; the sign of the result, + . Dividing each exponent by the index of the root. ia<^¥c^ = 2 rfibcK Result. 2. Required the fifth root of - 243 x'Y'' The index is odd ; the given quantity, negative ; the sign of the result. Dividing each exponent by the index of the root. V - 248 ajioj/i^ = V-Six^Oyii> = _ 3 xhf. Result. 3. Required the fourth root of 16 a*6Vl The root is even ; the given quantity, positive ; the result bears the double sign. Hence ; \/l6lfi¥c^ = \/¥a^b^ = ±2a''bc\ Result . SQUARE ROOT OF POLYNOMIALS 213 4. Eequired the square root of 1587600. A root of a large number may frequently be obtained from its prime factors. Hence, V1587600 = VPT34752772 = ± (2^ . 32 . 5 • 7) = ± (4 . 9 . 5 . 7) = i 1260. Result. In the consideration of numerical quantity we shall consider only the positive roots in our results. Find the value of : Oral Drill 1. V121 aV. 2. VeimVi)". 3. 4. -^27 a^f. 5. ■v'-343cW. 6. -^16 m«ft^. 7. •- ^ 16 «" 12. V-10247fti^»2o_ j9_ 3/27V 10 ' Ma u 18 ^64 d" 13. v64a^w . 14. V729«i2z™ 15. •vl28 d^d^*. 20. 21. 5 / 32 a:^" V243m'^" s/126^ \ 27 ^3 16. V256m«;r«. V 27 c- yMR. 22. A ^64m'' \: 17 243 m^ '64 m'' \1024a?» 23. V1296. 24. -^'3375. THE SQUARE ROOT OF POLYHOMIALS 233. If a binomial, (a + h), is squared, we obtain (a^ + 2 ah + Ir). We have in the following process a method for extract- ing the square root, (a + h), of the given square, (o^ -\-2ab + b-). a2 + 2 a6 + 62 \a + h square root. The first term of the root, a^ a, is the square root of the 2 06 + 6 +2(1(6-1- 62 first term of the given ex- 6 -I- 2 a6 -I- 62 pression, a-. Subtracting a^ from the given expression, the remainder, (2 ab + 62), results. Dividing the first term of this remainder, 2 ab, by twice the part of the root already found, 214 INVOLUTION AND EVOLUTION 2 a, 'the quotient is 6, the second term of the root. This second term, b, is added to the trial divisor, 2 a ; and the sum, 2 a + 6, is multiplied hy 6. The result, (2 a + 6) 6, = 2 ah + h', and completes the process. It will he seen that much of the work depends directly upon the trial divisor, 2 a. The reason for this prominence of the trial divisor will be seen from the following Graphical Representation of a Square Root 234. In the accompanying figure we have a graphical repre- sentation of a square constructed upon a given line, a + b. If the square whose area is a^ is subtracted from the whole area, there remain the areas ab + ab+b-=2ab + b\ (Note that this corresponds with the sub- traction above.) Now the length of the side of the square removed being a, we are to provide for a remaining area built upon two sides of that square, or a + a = 2a (the trial divisor). Now ff^ ab ab l^ ab ab Area Length = Width. Hence, — = 6. la That is, b is the width of the remaining area whose length is known. Now, because of its position in the original square, there still remains unprovided for the square b^, whose side is b in length. Hence the length of the total area necessary to complete the square is 2 a + 6. (This explains the addition of the second term of the root to the trial divisor above.) Multiplying our known length by the width ascer- tained by division, we have, as above, {2a + b)h = 2ab-\-b% which area completes the square required. SQUAKE HOOT OF POLYNOMIALS 215 235. By the principle of Art. 233, we obtain the square root of any polynomial, the trial divisor at any point being in every case twice the part of the root already found. Illustration : 1. Extract the square root of a;^ + 6 a;' + 19 x^ + 30 a; +25. A polynomial must be arranged in order if its root is to be found with- out difficulty. x* + 6 .-■» + 19 x2 + 30 x+25 [^+3x+5 X* Result. First Trial Divisor, 2(3?) = 2 x^ First Completion, (2a:'-'-|-3a;)(3x) = +6x3+ 19 r'^ +6x'+ 9x2 Second Trial Divisor, 2(.r^ + 3x) =2x'2+6x Second Completion, (2x'''+t) r + 5)(+5) = + 10r^+.30x + 25 + 10x2+.%x + 25 236. From the process we have the general method : Arrange the terms of the given expression according to the powej's of some letter. Extract the square root of the first term, write the result for the first term of the root, and subtract its square from the given expression. With twice the root already found as a trial divisor, divide the first term of the remainder. Write the quotient for the second term of the root, annex it to the trial divisor to form a complete divisor, and multiply the com- plete divisor by it. Subtract the product for a new remainder. Continue the jjrocess in, the same order of operation until all the terms of the root are found. In extracting the square root of fractional expressions care must be taken that the expression is properly arranged. The descending powers of a letter occurring in both numerator and denominator of a fraction are written thus : ^ , ,11,1,1 a a^ a^ a^ , 1 2 T\ CI '7* O "Y 2. Extract the square root of — + llH — ^" + '^H 216 INVOLUTION AND EVOLUTION In descending powers of a : 2f -1-5-2-1- 11 -L 5-2-1-^ x"^ X a a'' ^ + 3 + *. Result a;2 \xl X + 12-1-11 X + ^+9 X 9^ + z\=?^+6 X / X \ X al \ aj a a' — 4- — a a?' 4-2-I-— -f — Ezercise 74 Extract the square root of : 1. x* + 4a,'^+6a^+4a;+l. 2. X* - 4:01? + 10 9?- 12 X + 9. 3. a*-6a'+12a + 4 + 5al 4. a;« + 4a^-2iB^-16ar' + a^ + 12a; + 4. 6. 13m^-30m + 4m^+20m' + 9. '. 4 X* + 4 a!^2/ + 9 a^'^y^ + 4 a:^ + 4 ?/^ 7 . 9 a* - 12 a^6 + 34 a^S^ - 20 a&= + 25 h\ 8. 26c*-20c^d-4cd''+14c2d2^d*. 9. i-12x + nx'-32a?+3ix''~20af + 25a^. 10. 16 a/' - 16 0^2/ - 28 ar'2/' + 30 x?/^ - 20 a^t/^ +29 a^y* -f 25 j/". 11. ?! + 3a;+9. 4 12. aj* + 2»'+2a^ + a;+^. a^ a« . 13 a' 13. — — — + 4 2 4 ■3a +9. SQUARE ROOT OF ARITHMETICAL NUMBERS 217 14. 9l,^^ + 3 + ^-^ + t ,- ix* 4.0? ,1a? X ,1 9 9 9 3^4 16 9 a^' &a? IZa? a; 1 ■ 25 5 "*" 10 2"*" 16" 17 4a° 2 a. a^ 19 g 9a^ 3a; a^ 12 a' 18. i^_2^-lI^ + 9^ + i^V9- + I^. -s, 9 9 4^9 ^16^ 4 Find three terms of the square root of : 19. l+9a;. 21. a? + ax. 23. 9a;* — 1. 20. x^—%. 22. 4a;2-5a;. 24. 36 -12a;. THE SQUARE ROOT OF ARITHMETICAL NUMBERS 237. Since an arithmetical square integer is the result of the multiplication of some integer by itself, we are assisted in obtaining arithmetical square roots by noting a certain relation that exists between such numbers and their squares. 1^ = 11 A number of one place has not more than two 92 = 81 J places in its square. 10^ = 100 1 A number of two places has not more than four 992 = 9801 J places in its square. lOO'' = 10000 1 A number of three places has not more than six 9992 = 998001 J places in its square. Conversely, therefore : If an integral square number has two figures, its square root has one figure. If an integral square number has four figures, its square root has two figures. If an integi-al square number has six figures, its square root has three figures. 218 INVOLUTION AND EVOLUTION Hence : 238. Separate any integral square number into groups of two figures each, and the number of groups obtained is the same as the number of figures in its square root. Illustrations : 1. Find the square root of 1296. Beginning at the decimal point, separate into periods of two figures Faballbi. Alqebbaic Process each. 0= Cay = a + 2a6 + 6'-'|a + 6 2 302 = a + b 1296|.S0 + 6 Result. 900 2(a) =2(1 (2a + 6)(6) = + 2 a6 + 62 2 (30) = 60 + 2 a6 + 62 (60 + 6) (6) = 396 396 In the square root of 1296 : The greatest square in 1296 is 900. ITie square root of 900 is 30. The trial divisor is 2 (30) = 60. The second term of the root is (396 -=- 60 =6). "For the completion, (60 + 6) (6) = 396. The process is repeated in the same order if the given integer is of a higher order. 2. Mnd the square root of 541,696. The foUovvfing process is given in a form commonly used in practice. Separating into periods of two figures each : Explanation : The greatest square contained in the fiist period (54) is' 49. The square root of 49 is 7. 7 is, there- fore, the first figure of the root. Subtracting 49 from 54, and bringing down the two figures of the next period, we have 516, the remainder. Annex- ing a to the first figure of the root, 7, our trial divisor is 2 (70) = 140. Dividing 516 by 140, we obtain H, the second figure of the root. (140 -(- 3) 3 = 429, which product is subtracted from 516. With the remainder (87), we bring down the last two figures (96), and the new remainder is 8796. Annexing a to the figures of the root already found, our trial divisor is 2 (730) = 1460. Dividing 8796 by 1460, we obtain 6, 140 + 3 3 541696 [736 49 516 429 1460 + 6 6 8796 8796 SQrARi: ROOT OF AKITIIMETICAL NUMBERS IW the third flgm-e of the root. (14i!0 + 6)6= 8796, which product, sub- tracted from STSIiJ, gi\es a remaindei' of 0, mid the square is completed. Tlie addition of the to the tlgures of the root already obtained gives a trial divisor of the same order as the remamder, or of the next lower oitler. The process gives fewer figures and less likelihood of error. If a r/iren s(juare nmnber has tlecimal places, we point off by beginning at the decimal point, first separating the whole nmnber as before, finally sejiarating the decimal into periods from left to right. Ciphers may be annexed, if necessary, to complete any period. If a giren number is not a perfect square, its approximate square root can be found to any desired number of places. T7te square root of a common fraction is best found by chang- ing the fraction to a decimal and extracting the square root of the decimal to the required number of places. Illustrations : 1. Find the square root of 2. Find, to 19920.4996. places, a IL'^. three decimal 19920.4996 (141.14 1 Result. 20-1-4 99 4 |96 280-1-1 |:«0 1 1-281 2820 -I- 1 •2S2-20 + 4 4 3949 2821 112896 1 1289t) f = .37o. 12f = 12.375. The required tliree decimal places necessitate six decimal figures in the square. Hence, with three ciphers annexed, we have to obtain the square root of 12.376000 13.617+ - 9 Result The decimal point in the residt is located easily by noting between which periods of tlie given example tlie given decimal point lies. In the example above there are two periods to the right of the decimal point given ; therefoi-e, thei-e wU be two decimal places in the root obtained. 60-1-5 5 700 + 1 1 7020 + 7 337 325 1250 701 54900 49189 ■2'20 INVOLUTION AND EVOLUTION Exercise 75 Find the square root of : 1. 9216. 5. 186624. 9. .717409. 13 .00002209. 2. 67081. 6. 4202500. 10. 9617.7249. 14. .0001752976. 3. 32761. 7. 49.434961. 11. 44994.8944. 15. .009409. 4. 182329. 8. 9486.76. 12. .00119716. 16. .0000879844. Find, to three decimal places, the square root of : 17. 3. 19. 7. 21. .5. 18. 5. 20. 10. 22. .05. Find, to three decimal places, the value of : 27. 6 + 3 V'2; 29. 2V3-V5: 28. V7 + V10. 30. 3V7-7V3. -'■33. How many rods in the side of a square field whose area is 2,722^500 square feet? 34. Simplify and extract the square root of af{10!«' + 13)-2{2x* + 3 + 7a^)x+{x^ + l) {x* - a^ + 1). 35. Simplify and extract the square root of (a^ - 2a; - 3) (a;2 - a; - 6) (a^ + 3 a; + 2). / 36. Show that the required square root in the preceding example can be readily obtained by factoring and inspection. 37. If a, 6, and c are the sides of a right triangle, and a lies opposite the right angle, we may prove by geometry that a" = b^ + c^. Find the length of a in a right triangle in which b and c are 210 feet and 350 respectively. 38. Applying the principle given in the preceding example, find, to three decimal places, the distance from the " home " plate to the second base ; the four base lines of a regulation baseball diamond forming a square 90 feet on each side. 23. f 25. .037. 24. ue I- of: 31. 26. .0037. V6 + V5. 32. ^j^5 -V2. CHAPTER XIX THEORY OF EXPONENTS 239. Tlie Index Laws for Positive Integral Values ofm and n have been established : (1) For Multiplication : «"• x a" = a»>+'' (61) (2) For Division (in greater than «) : a"" -^ n" = a"-" (79) (3) For a Power of a Power : (a")" = a""* (216) (4) For a Power of a Product : (a6)" = a"6" (216) (5) For a Root of a Power : Vo^ = <^"' (226) 240. The extensi6ii of the practice of algebra requires that these laws be also extended to include values of m and n other than positive integral values only; and the purpose of this chapter is to so extend those laws. We shall, therefore : - 1. Assume that the first index law (a" X a" = «"+") is true for all values ofm, and n. 2. Define the meaning of the new forms that resuU under this assumption, m or n, or both, being negative or fractional. 3. Shoiv that the laws already established still hold true with our new and broader values for m and n. THE ZERO EXPONENT If m and n may have any values, let n = 0. (61) Then, a" X a"* = a"* Dividing by o", o» 0^ That is, a» = 1. Heuue, we define a^ as equal to 1. Or: 221 2-22 THEORY OF EXPONENTS 241. Any quantity with the exponent equals 1. Illustrations : 1. a;» = 1. 2. {mn)0 = 1. 3. 3 m" = 3 • 1 = 3. 4. 4o0 + (2 2/)+da\ 11. 2 a" + (a" - 1). 3. 3iB»?/2". 6. 4a°-3. 9. 4 a;" -5/. 12. 4°a; + (a + 3 a;)". THE NEGATIVE EXPONENT If m and n may have any values, let n be less than and equal to — m. (61) (241) Then, a" X = «o whence, a"* X a-» = 1. Dividing by a •», a" = Dividing by a", 1 a-™ = — • a'" Hence, we define a~" as 1 divided by a'". From this definition we obtain an important principle of con- stant use in practice : 242. Any factor of the numerator of a fraction may be trans- ferred to the denominator, or any factor of the denominator may be transferred to the numerator, if the sign of the exponent of the transferred factor is changed. Illustrations : 2 ^-2„j,-i=X Q 2-hnx->' ^ 3 mra ^ PKACTIOXAL FORJI UK THE EXPONENT 223 Oral Drill Transfer to denominators all factors having negative exponents : 1. ax-^ 4. 2a-'b-'-. 7. a-'-bc-l 10. 2 a- Va;-^ 2. cd-'a;-=. 5. 2-'a^-6-'' 8. x~hj-h^'. n. 5 x~^y-h*. 3. 2a6-l 6. 3a-^x'y-\ 9. 3^'tr-6^V-'. 12. 3~'a-'marhj. Read the following withont denominators : , „ a^ cd 3 a c^ 1 13. — • 15. — ■■ 17. -VT,- 19. -~ — • 21 — = . ¥ mn a?y- a--b~- 2-'^ary 14. ^^ 16. ^. 18. ^. 20. 4^, 22. ,? • Read the following with all exponents positive : 23. 5a-6c. 2r. #^. 31. ^~^«^ 2a;V ■'^- 'Z-'m'-rr^z 24. 12x-^y-h. 28. „ _, , _, ■ 32. r — ^j—.-. 2x ^z^ 3 c~^da;"' 25. -50./.^. 29. 3^=j^,. 33. ^3^. 26. 3a-6c-U 30. 1^^. 34. " ^ «"^" S-'x^y'z -3 m-'x-^2 y-'z- THE FRACTIONAL FORM OF THE EXPONENT If the expression a^ can be shown to conform to the first index law, we may find a definition for the fractional form of exponents. By the first index law, (a^)2 = ai \ a^ = a^"*"^ = a. Hence" the meaning of ci^ is established, and the exponent in this form still agrees with the fundamental index law. That is: 224 THEORY OF EXPONENTS (fi = Va is one of the two equal factors of a. Similarly, (a's)" = a^ x cfi x a^ = a'i'^s+J _ „-3 - a'. 2 That is, a'^ is one of the three equal factors of a'^. In like maimer, aJ = Va = one of the three equal factors of a. dfl = \/^ =.three of the four equal factors of a. Therefore : 243. In the fractional form of an exponent we may define the denominator as indicating a required root, and the numerator as indicating a required power. Illustration : v^ =8* =22 = 4. And 4 contains too o/ffte Jftree equal factors of 8. VsP = 81*=3'=27. And 27 contains three of the four equal factors of 81. In general : m m m mm m o" • o" • a" •■• to n factors = a" " " ■•• to n terms mn m That is, a" is an expression whose mth power is a"' m And a" Is the nth root of a""- Or, as above, a" is one of the equal factors of a". It is understood that while a'^ must equal either + Va or — Va, we 1 _ consider the positive value only ; and a^ is defined as + Va, the principal square root of a. In future operations we may apply tlie definition of Art. 243 to expressions given in radical forms, observing that 244. The index of a radical may be made the denominator of an exponent in the fractional form, the given exponent' of the power of the quantity becoming the numerator of the fractional form. FRACTIONAL FORM OF THE EXPONENT 225 Illustrations : 1. Vm6 = m^. 2. VxT = x'^. 3. \/b^ = b^. And in the converse operation of the principle of Art. 244 : 4. x^ = = Va" ■ Vb" = a^b^ 3. When n is negative and fractional. Let n = —^,p and q being positive and integral. -?- 1 1 _2 __. (abY = (ab) " = = ■ = a ib i (aft)' alb Let the student discuss this law (1) when m and n are both positive and fractional, and (2) when m and u are both negative and fractional. APPLICATIONS OF THE PRINCIPLES OF EXPONENTS (a) Simple Forjis involving Integral Exponents 246. In processes with exponents no particular order of method can be said to apply generally. Experience with dif- ferent types will familiarize the student with those steps that ordinarily prodiu'e the clearest and best solutions. As a rule, results are considered in their simplest form when written with positive exponents. 228 THEORY OF EXPONENTS 247. Illustrations: 1. Simplify x~* • ar* • a;"'. X-* ■ x^ ■ x-'' = a;-*+6-' = »-»= — • Eesuhi. afi 2, Simplify m^n'" • mr^n" m^n-^ • m-2ji-6 = m^-^n,-'^'^ = ra»-' = — • Result. 3. Simplify - (a-r x"'^ {mx~^) ~'-' O^TO-S ^ (a-2)3 _a2TO-3a;4ar8 x-i (j»a;-i)-2 x-^m-V = a2-6OT-5 +2a;4+i-« = oj-^m-ia;' = *^ . Result. Simplify : Exercise 76 1. a' X a'K ' 5. m^ • m"^ • m'. 9. a)-'-5-x-*. 2. C-' X c'. 6. a;* ■ a;" • a;-*. 10. m-'-T-m-'. 3. m"'' X m*. 7. a-'-x-^ ■ a-^x*. 11. a~%~^ H- a^TO"', 4. a;-* •a;'. 8. c-^d-c^d-*- 12. x-^y~^-T-x-^y-^. 13. ..?.^. . 15. apy-'m^n'^ m^n-^aPy-^ 14. a a' "«-''^. 16. a^m-hi-^x-'^ 17. (a-y. 22. (m-'w)-*. 27. {a'^ + x'^f)-*- 18. (c-^-». 19. (a)-*. 23. (a->c-^». 24. (a-'m-*/. 28. (m^ji)" X (3 ai°)». 20. (a-y. 21. (»-•&)'. 25. (aVa-^)--l 26. (cVZ . cOrf-i)- \ Of 30. r'^^-T'. APPLICATIONS OF PH1X0[I'LES OF EXPONENTS 229 (ft) Types involving the Fractional Form 248. In the following illustrations attention is called to each important feature of the process, and the order of the principles that is emphasized in each is such as will, under similar conditions, produce the best form of solution. Illustrations : 1. Simplify (aV^x-^)~\ (a26-Vi)-2 = o-*6^a;2 = ^- Result. Note (1) that the firet step is the application of the law (a™)" = a"", and (2) that the result is given with all exponents positive. 2. Simplify (c= Vc='^)^ (c2 VFi)3 = (c2 . c~^)3 = (Jf = el Result. Note that the law a'" x a^ = 0""+" is first applied so as to unite c-factors. 3. Simplify {>/(Vo^)-i|"^_ Result. Note that the reduction is accomplished outward. 4. Simplify '^•'^^' 25 Vm > I9j/m^\'^_ / t)»i~n ~S _ / 9 \~t _ / 25m^ \^_ 25? mj _ 125 m^ Result. Note that in the third step inverting the fraction changes the sign of the exponent of the fraction. In general, \b) b' a'~[a •1-M) THEORY OF EXPONENTS Exercise 77 (See also page 109.) Simplify : 1. {2a-^x^y. 2. (Sak-y. 3. (2a'V)-2. 4. (-Sahr'y. 5. 2(-32a-i°)*. 10. vSllr 6. a{-16(r"x)i- 11. 3V-27-V. 7. «-'(27a;-ia-^)i 12. 1^16 /«";-». 8. ■v'-8c«d-l 13. |-v^ 27 c'-^a;. 9. V - .'!2 :»-i»//^. 14. — J-v^— 32c'^ ■8«2 16. (2.r^)-N/-4.r 17. {2chi)-^^8^. 18. (,ir 'Vx^'xT)8. 19. (V' . c'. 15. (3a)-iV-27a^ 23. [mV m-X™"^)']" 24. ["\/a-'c*(c"*ai)^]-8. 25. [A/a.f-'\/mFi]-2. 26. (V27a?)-2^(V81a;')-'- 20. (v4«Wc-i)^'". 21. v^?(l(ir.r-^)l 27. '\/a= y\ay' ^)-\yHt. hj-^). 30. 22. ^{cex-Wcw-y. 3/ — 33. 28. (VlL'^aS/'M'-')-^. 29. 29. ,-7\G tV/'t V 36. f^27;^y. 31. ^^"y«. 34, ("V ai^a"" 32. \ = a"Vc-i' a~'c 39. J c'Vd . , 35. Vr-/-^ cVd 'S/cd' ««•" 37. fV"^'''l'- 38. 3V12, Vc-VrW ' "" V 5^/81 <( _ V»^ v'x^ v.?'' 40. 41. [ V m'a;"^ V mlT^^VMi'a:"^] 42. (8 a^a;^H-37a!--3by c-2 + 2c->-3. 7. 2a-a* + 4a^ + 4ai-3 by a^-ai+3. 8. 35+4a-"'-16a-='"+19o-'"'— Ga-*" by 7 + .Ta "'-3a-^"'. 9. x^ + 2xi-7x~^-8x''^ + 12x-'hy x-3x-' +2x~K 3). 34. (2a-' + a-' + 3)-. 38. (27 0"^ 125) by (3 0"^+ 5). Find the value of x in each of the following : 39. x-i = y, and y- = 9. 42. x^ = y"', and y = 3. 40. a; = y-i, and y- = 3. 43. a;^ = y^, and 2/» = 9. 41. x-^ = y, and y-' = 2. 44. a:"^ = y-% and ?/"* = 9, Simplify : 45. (a-i - x-y - (a-' + a;-^) (a-' - a;->). 46. (a-"' + 3a'»)^ — (a-" — 3 0"")^ 47. (a-' + l)a-5-(a-3-l)2 + (l_a-»). 48. 49. '+- O^'d^ m di- c^d^ ^+i d'5" i _1_ a;* + 13 a;^?/' _ a;* + 4 y^ _.r* + 2 a-^3/^ ~3y^ x^ + 3 y^. ■5xM-_2yi" . a;5 — 9 y^ CHAPTER XX RADICALS. IMAGINARY NUMBERS. REVIEW 252. A radical expression is an indicated root of a number or expression. Thus : y/2, y/l, VlO, and Vs + 1 are radical expressions. 253. Any expression in the form ^x is a radical expression, or radical. The number indicating the required root is the index of the radical, and the quantity under the radical is the radicand. In \/7, the index is 3, and the radicand, 7. 254. When an indicated root of a rational number cannot be exactly obtained, the expression is called a surd. 255. A radical is rational if its root can be exactly obtained, irrational if its root cannot be exactly obtained. Thus : V25 is a rational expression ; VlO is an irrational expression. 256. A mixed surd is an indicated product of a rational factor and a surd factor. Thus : SVS, 4v'7 X, aby/a + 6 are mixed surds. 257. In a mixed surd the rational factor is the coefficient of the surd. Thus : In 4 y/Sx, 4 is the coefficient of the surd. 258. A surd having no rational factor greater than 1 is an entire surd. Thus : VFoc is an entire surd. 236 TRANSFORMATION OF RADICALS 237 259. The order of a surd is denoted by the index Ojf the required root. Thus : V5 is a surd of the second order, or a quadratic surd. V7 is a surd of the third order, or a cubic surd. 260. The principal root. Since (+ ay = + a^ and (— a)^ = 4- a% we have V+^= ± a. That is, any positive perfect square has two roots, one + and the other — , but in elementary algebra only the + value, or principal root, is considered in even roots. THE TRANSFORMATION OF RADICALS TO REDUCE A RADICAL TO ITS SIMPLEST FORM 261. A surd is considered to be in its simplest form when the radicand is an integral expression having no factor whose power is the same as the given index. There are three com- mon cases of reduction of surds. («) WJien a given radicand is a power whose exponent has a factor in common with the given index. By Art. 244, v^ = a^ = a* = Va. Hence, to reduce a radical to a radical of simpler index : 262. Divide the exponents of the factors of the radicand by the index of the radical, and write the result with the radical sign. Illustration : WcM^ y/^ofiifi = 2ToM = ■S^2^. Result. 9. K/m. 10. V9^*. 11. ■v/144 m»n*. 4. ^/a?fz'. 8. „ 19. 20. iVlGa^c. 7. V64. V54aV. 21. ^ VI 02;) Vfi7.') //t"'/t'. •''.'/'■ 23. 24. 9to2 V54 mW. 22. V48 cM".v :". TRANSFORMATIDN OF RADICALS 28'J 25. V 16 V^Zc" 13 V 16 m'.r 28. 6 3/ 125 a^y% 15^ sfjsr a; \ 216 m« * ^°- 14 ^'125a;''■ -■ >3lf3- - ^'M a? '{a-xf ' ' '\{x + Vf (c) TFAeJi >'!«- 4^^ +s;/128- 2-2/1- + 6^1 -9\^. 15. 8V| + 14V|- V75-^V512 + 5V||-6V||. MULTIPLICATION OF RADICALS 269. Any two radicals may be multiplied. Illustrations : 1. Multiply 3V6 by 2Vi5. (3 V'B)(2\/T5) = (:; • 2)( V6 ■ Vl5) = GVQO = ISVlO. Result. 2. Multiply V28 x V42 X VlS. ICxpressing each radical in prime factors, V28)(Vi2)(Vl6) = \/(7.22)(7.2.3)(.5.3) = \/(72 • 22- 32)(6 .2) = (7 • li • 3) VIO = 42 VTO. Result. 3. Multiply 2-s/l by 3\/8. Changing to radicals of the same index, 2\/4 = 2 • 4^ = 2 • 4T^ = 2 v^F. 3\/8 = 3 • si = 3 • 8" = 3 v/88. (,2v'HK3'-0'85) = (2•:^)('^'4^T83)=6v/p^=G\/2I^=0\/p26■• = 12v^32. 244 KADICALS. IMAGINARY NUMBERS. REVIEW Hence, to multiply two or more monomial radicals : 270. Multiply the product of the given coefficients by the prod- uct of the given radicands, first changing the radicals to radicals having a common index. The principle is in no way changed in application to frac- tional radicands. It is usually best to multiply fractions in the form given, reducing the product to its simplest form. Exercise 87 (See also page 411) Multiply : 1. V3by V6. 5. |V6by^Vl5. 9. |V^ by |\/28|. 2. Vl2 by V8. 6. 75 by ^2. 10. V35^ by Vlia^. 3. 2V7by4.V2i. 7. V6 by ^/9. 11. ^SiS? by^SIi?^. 4. 3-\/i hy ,/216 '27 3 240 RADICALS. IMAGINARY NUMBERS. RKVIEW 3. Divide ^^- by '\63y \10l >/? X 32 102 Exercise 89 Divide : 1. VSbyVi2. 5. 6V5by2Vl5. 9. ^/36 by -J/54. 2. V6 by VT8. 6. 4^4 by 3^l2. 10. i/\ by VI. 3. V27by VT2. 7. -^ V72 by fVJ. 11. V|| by ^|.. 4. 4Vl6by2V5. 8. -x/tS by v 50. 12. V^ by v^. DIVISION BY RATIONALIZATION 273. If a given divisor involves quadratic surds only, a division is really accomplished by the process of rationalization ; or, by multiplying both dividend and divisor by the eaypression that will free the divisor from surds. (a) When the Divisor is a Monomial. Illustrations : 1. Divide 8 by 2 VS. _l_ = _l_xVl = 84 = 8^ = 4V3. Result. 2V3 2V3 V3 2.3 6 3 2. Divide 2 \/8 by /3- (\/2- 1) ^^ VS - (\/2-l) ■\/3+V2-l VS+CVl-]) V3-(V5-1) _ 6 -2V6 + 2v'8 -2V2 2V2 _ 3-V6+V3--\/2 3 _ V6 + V3 - V2 ^ Va _ 3 V2 - 2\/3 + V6 - 2 V5 V2 2 Bsercise 91 (See also page 411) Rationalize the denominators of: 1. — = 4. 2 + V2 2. 3 V2-2 V2+ V3 V^- V3 6- V2 2 + V2 3V2 + I 3 ^^ 6 ' V7-V2 ■ V2-1 Va— V. X 11. 2Vm+V^. jg 3Vm— Vn 2_Va_+V2^_ 17_ 12. 3Va— V2a , „ a VS + c Va; lo. ^- ig aVx — cVa; 14. V^+_2. ^g Va; + l-2 4_V3-V5 ^ 2V2-4 3V2+2 g 4V3-3V2 2V3 + 3V2 <,_ 5V6-2V3 vs 3V2 + 3 - 2 - V.r V5 — 2+ Vx v« + 1 + V2 a -1 Va + 1 - V2 a V3 + V2 - 1 -1 V3 3- -V2 + 1 V2+V3 3 + V2-V3 4 - V3 + V5 INVOLUTION AND EVOLUTION OF RADICALS 249 mVOLUTIOH AMD EVOLUTION OF RADICALS 276. By the aid of the principles governing exponents, we may obtain any power or any root of a radical expression. Illustrations : 1. Find the value of (2 v^(?)'. (2\/a5)2= (2 a}y = 2^0^ = ia^ = 4 v^. Result. 2. Find the cube root of VaV. V Vn%^ = V a^a;i = a^x^' = VoP = xVkc. Result. Exercise 92 Find the value of : 1- (V«)'. 5. v'VZ 9. -i/^(a + by*. 2. (VW)'- 6. '-"■ 26 That is, we have a quadratic surd equal to a rational expression, an impossible condition because of the definition of a surd. Therefore, Va cannot equal 6 + Vc. 250 RADICALS. IMAGINARY NUMBERS. REVIEW 278. Given, a + V6 = c + V^, V6 and s/d being surds. Then a = c and b = d. If o is not equal to c, we have, by transposition, a = c + y/d- Vft, which, by Art. 377, is impossible. Therefore a must equal c. Hence it follows that Vfe = y/d. 279. Given, VaTVf=Vi+Vy. Then, v a — V6 = V» — y/y, when a, b, x, and y are rational expressions. Va+ Vft = Vx + y/y- Squaring, a + VS = x + 2 Vxy + y. By Art. 278, a = x + y. (1) And V6 = 2 Vxy. (2) Subtracting, a — y/h = z — 2 Vxy +y. Extracting sq. rt., v a — V6 = Vx — y/y. THE SQUARE ROOT OF A BINOMIAL SURD The square root of an expression consisting of a rational number and a quadratic surd is obtained by application of the principles of Arts. 277, 278, and 279. (a) The General Method 280. Illustration: Pind the square root of 11 + V96. We may assume that, v 11 + \/96 = Vx + Vy. (1) Then (Art. 279), Vll - VOfi = Vx - Vy. (2) Multiplying (1) by (2), \/l21 - 96 = x - p. Or, X - 2/ = 5. (3) SQUARE HOOT t)F A BINOxMlAL SURD 251 (4) Squaring (1), 11 + \/96 = .>■ + 2 V^ + y. Hence (Art. 278), 11 =a; + j^. From (3) and (4), x-y = o. x + y = U. Whence, X = 8, ;/ = 3. Therefore, Vx+ y/T/=V8+ Vii= 2 V2+ Vs. That is, Vn + V^; = 2\'2+ vTi. Resvdt, (6) The Method, of Inspection 281. If a binomial quadratic surd can be written in the form a + 2 V6, its square root may be obtained by inspection. Illustration : 1. Find the square root of 2,S + 6 VlO. (j VIO may be written thus : 6 VIO = 2 (3 VTo) = 2 VOO. Then 23 + 6 VW = 23 + 2 VSO. (The required form of a + 2^6.) The factors of 90 whose sum is 23 are 18 and 5. Hence, the square root of (23 + 2 V90), or the square root of (18 + 2 VOO + 5) = Vis + v^ = 3 V2+ Vs. Result. Exercise 93 Find the square root of: 1. 7+4 VS. 9. 146-56V6. 2. 11-6V2. 10. 124-30Vii. 3 17 + 12V2. 11. 3a; + 2a;V2. 4. 28 + 6 V3. 12. c2 + m + 2cVm. 5. 33-8V2. 13. 2m + 2 Vm^-l. 6. 30-12V6. 14. 2m'-l + 2m^m^-l. 7. 67 + 16V3. 16. c2 + c + l-2cV^+I. 8. 138 + .S0\/2T. 16. a=+4rt+l + (2a+2)V2^ 252 RADICALS. IMAGINARY NUMBERS. REVIEW EQUATIONS INVOLVING IRRATIONAL EXPRESSIONS 282. Equations in which the unknown number is involved in radical expressions are called irrational equations. 283. An equation containing radicals is first rationalized by involution. If there are two or more radicals in the given equa- tion, it may be necessary to repeat the process of squaring before the equation is free from radicals. 284. Jloots of Irrational Equations. It can be shown that (1) The process of squaring may introduce a root that is not a root of the given equation. (2) An irrational equation may have no root whatever. Hence, we conclude : 285- -Any solution of an irrational equation must he tested, and a root that does not satisfy the original equation must be rejected. Illustration : Solve Va; + 3 — 5 = Va; — 2. Vx + 3 - 5 = Vx-2. Squaring, a; + 3 - 10 Vx + 3 + 2.j =x-2. Transposing, x — x— 10 Vx + 3 = — 3 — 25 - Collecting, - 10 Vx + 3 = - 30. Dividing by - 10, y/x + 3 = 3. Squaring, x + 3 = 9. Whence, x = 6. Substituting in the original equation, Ve + 3 - 5 = Ve^^. V9-5=V4. 3 - 5 = 2. Since the original equation is not satisfied by the solution x = 6, this solution must be rejected. It will be found by trial that the solution, x = 6, satisfies the equation 5 — Vx + 3 =Vx — 2. EQUATIONS INVOLVING IRRAtlONAL EXPRESSIONS 268 Exercise 94 (See also page 411.) Solve and test the solutions of : 1. ViTl = 2. 15. 2(V« + o)(V«-5)= -35. 2. Vx — 1 = 3. 16. -y/x + 12 = Va; -3+3. 3. ■\/'x—l = V5. 17. V36 + a; - Va; = 2. 4. V5 a; + 2 = 3. 18. 2 Va? - 1 = V4a;-ll. 5. Va? — l = x — l. 19. Va; — 16 + Va; = 8. 6. 2VS— 3 = Va;+2. 20. V:^: — 1 + Va; = V4 a; + 1. 7. V«' + 3 .x-2 = ,r + 1. 21. V.I- + 12- Va;-3 = 3. 8. Va;2 - 11 = .« - 1. 22. V.k + 3 + Va; -2 = 2 Va;. 9. Va; + l = o — 1. 23. V.r — 5 + Va; + 7 =6. 10. V4.T' + 7a;— 7 = 2.1;. 24. Va; + 5 — Va;= V4 a; + 1. 11. Vl2 +a; — Va; = 3. 25. V.i;+a4- Va;+2 a = V4 x— a. 12. V«^ - 9 + a; = 9. 26. v.f-o - V4 .t;-2 = — V;«+3. 13. V7Ta;=7-Va;. 27. V2 a; + 6 - V2 .i; - 1 - 1 = 0. 14. V| + a;=Va; + |. 28. l+3Va:= Vg .« + V36 a; + 13. 29. V4 x + c— V9 x — c= V25 a; — 3 c. 30. Va; + 1 + Va; + 2 = Vc - 1— Va; - 3. 31. Va; - 5 + V.e + 8 = Va; - 3 + Va; + 2. 32 — — Va;— 7= V-c. 35. Va; = — V.t— 9. Va; — 7 Va;-9 o, /— Tc , /- 15 ^-i" - 9 + Va: iJd. Va;+5+va;= — . 36. - , -=—1. Vas+i") Va; — 9— Va; 34. Va; + V.c-4=- 4 Va = . 37. J^ + JM = 0. -4 \.i;-5 >.i; + 5 40. 41. 254 RADICALS. IMAGINAIiY NUMBERS. REVIEW 38. V^ + 5^V5 + 9. 44. ^^^+^^-^=2^. Vx+l Vx + 3 -Vc + x 39 vV+28^V^+38. 45. 3Vl--4^ 7V^-26 _^^_ V^ + 4 ^^ + 6 Va;-1 7Va;+21 2V^-l _4Vx + l 46. V5 + a^- V5-a; ^l_ 3Vx + ^ eVi+s' V5+a;+ V5-a; 2- 2V^ + 3 4Va+l ^Q 47. \/4x + 1 + a/ J^^ 9^ ^Vo;-! 6V;J;-2 v'4 .1; + 1 - V4 a; 42 2Vx-14 ^ 2V;^-2 ^g 3Vx-7-6 ^ 2Va-7-4 V«— 6 Va; + 2 BV.i;— 7-1 4Va;-7-7 43 5x-9 ^^ V5^-3 49 « + l a-1 ^4a V5^ + 3 2 ■ ■ V^- + l V^-1 »^-l" 50. V2li - Vai = V3 *; + 2Va[6a;-(4 — a)]. IMAGINARY AND COMPLEX NUMBERS 286. The factors of + a^ are either (+a) and ( + «), or (— «) and (—a). The factors of — a^ are (+a) and (—a). Clearly, therefore, no even root of a negative number is pos- sible. Hence : 287. An imaginary number is an indicated even root of a negative number. V— 2, V— 5, V— 10, are imaginary numbers. 288. The symbol, V— 1, is called the unit of imaglnaries. 289. The rational and irrational numbers hitherto con- sidered are known as real numbers in contradistinction to this new idea of imaginary numbers. 290. Imaginary numbers occurring in the form of V— '', where b is the real number, are called pure imaglnaries. OPERATIONS WITH IMAGINARY NUMBERS -255 291. An imaginary number occurring in the form of (' + V— &, where a and b are real numbers, is known as a complex number. 292. The Meaning of a Pure Imaginary. By definition, ^ by V^^. Exercise 96 12 by V- 3. 4. V-3(J by -V-16. 5. -V^lSby -V^54. 6. V^^ by V^^ by V^^28. 7. V^^ by V^^ by V-16. 8. 9. 10. 11. 12. 13. 14. 15. 16. _ V- 9) (- V- -T.) (V- 100). V- a-) (- V=:T?) (- V- IC'b'). -V^a*) (- aV- a-) (- aV- a*). -V^l) (2V^ (-SV'^^). - 2 V'^^Tl) ( - 3 V^ (5 V^32). V^c"^') (- V - />'^) ( - V - c-x). aV - ((-) (rf-^V^^T:) ( - a-'V^^). — A- /;( — // ) (V» — m). V-.r-2,r-l) (VS^Ti) (vTiri-)- MULTIPLICATION OF COMPLEX NUMBERS 297. We assume that the principles underlying the multipli- cation of real numbers still hold true in the multiplication of complex numbers. 90M. Kr.. ALO. — 17 :i:j8 KADICALS. IMAGINARY NUMBERS. REVIEW Illustrations : 1. Multiply 3 + V^2 by 2 - V^^. 2-6V^2 - 15v'^l-10(-2) 6-ll\/^2 + 20 = 26- llV^. Result. 2. Expand (3 - V^^)^ In processes involving frequent repetitions of the imaginary unit, V— 1, it is convenient to substitute i in place of V— 1, replacing the imaginary unit after the simplification is com- plete. Let (3 - y/^Ty = (3 - i)*. Then : (3 - iy = 81 - 108 i + 54 i^ - 12 j« + i* = 81 - 108\/^1 + 54(-l) -12(--v^ri) + (1) (Art. 292) = 81- lOSV^Hl -54 + UV^ + l = 28 - 96\/^I. Result. Exercise 97 Simplify : 1. (2 + a/^T) (3 + V^l). 2. (3-V^l)(4-V^. 3. (2 + 2V'=n:) (3 + 2V^. 4. (V^ri+v^=^)(V^T:+V^. 5. (2V^T - 3V^3) (2V'=T: - 4V^. 6. (3V2 + 2V^5) (2V2 - 3V^^). 7. (V— a — x) (V— a + x). 8. (V^=i: + 2 V^^ + 3V^r3)2. 9. (2 + 3V^^)^ 11. (3V2-2V^T)l 10. (3 + 2^ 12. (aV- 1 - bV- If. OPERATIONS WITH IMAGINARY NUMBERS ■259 DIVISION OF IMAGINARY AND COMPLEX NUMBERS BY RATIONALIZATION OF THE DIVISOR 298. Illustrations: 1. Divide V- 28 by 8. 28) (V- 1) 128 [14 1 ^ ^ , 2. Divide V- 12 by VS, >/^iri2 _ (V- 12) X V^ _ V- 36 _ 6\/^n! ^ 2 V^ VS ^ V3x VS " 3 3 3. Divide 2 + 2 V^^ by 2 -2V^^. 2 + 2V^3 2 + 2 V^l _ (2 + 2V^^y ^4: + : 1 • Result. 3-12 2_2V-3 2 + 2-1/- 3 22 - (2 V- 3)2 4 + 12 8\/-3-8 _ V- 3- 1 16 2 Result. Exercise 98 Simplify : 1. ^ 4. 13, 14. V- -2 Vl8 V- rs V- -12 V' =^ V- -12 5. 2V^3 2V33_v: X + V^^ _ V-27 _ 2V^=^ 2V^^ -14 -4V20 -V-5 15. 16. 9. 10. 11. 12. 5_V32 2+V=^' 3+V33 3-V^^ V2 + V3-1 V2-V^^ V^-f 3V-2c_ Vc— V— c 2 m - 3 a;V^l 2m — .V V — 1 260 RADICALS. IMAGINARY NUMBERS. REVIEW 299. Conjugate Imaginaries. Two complex numbers differing only in the sign of the term containing the imaginary are con- jugate imaginary numbers. Thus, a +b^/—l and a — bV—l are conjugate imaginaries. By addition, (« + ftv/^H^) + (a - fiV^^) =2 a. By multiplication, (a + b V- 1) (a - 6 V- 1) = a^ + bK 300. I7ie SMWi and the product of two conjugate complex num- he)-s are real. MISCELLANEOUS PROCESSES WITH IMAGINARY NUMBERS Exercise 99 Simplify : 2. (-V^T)'-(- V^^/. 5. (- v'=^)=-(-V^^ 3, (_2V^2)2 + 3(V^3)l 6. (V^^-l)^-(l + -V^)''. 8. (1 + v^^/ + (1 - v^)' - (1 - v"=t:). 10. (m + n V — 1) (m — Ji V^^). 11. V-1+1 I V-1-1 V^-1 ■1+1 12. 13. (^ 3)''+(V-i)« 14. (2 + V-iy + (2-V-l)^ 3V^1 i_ a + a;V— 1 , a — a;V — 1 15. . - + - 2V-2+ V-1 (V3i)4_(V32y V-l-V-2 ■kV— 1 a + ajV— 1 16. ( iV=^-i)^-V^ . 17 (V-i)°-(V-i)°+(V-iV V^2 GENERAL REVIEW 261 GENBRAL REVIEW Exercise 100 1. Expand (xk - 2 Vxf^ 2. Find the square root of (a^ + 3 x + 2)(a;- — l)(ar' + a; — 2). 3. Simplify v'S + V^ + V| - V^ - V8. 5. Factor 27-'.r" - 8-'. 6. Find the approximate numerical value of 2V'2 + 1 . 3V2-1 _ V2 - 1 ' V2 + 1 ' V. Simplify (a'bc-r(a'b-^oT . L (a-'6c-)'"(a6-^c)" a''<»-">&-''c'» 8. Solve irnx-ny^m'-n^ my — nx = m^ — n'. 9. Find the square root of a* + a' — 4 «« -f- 4 a + 2 a^— 4 al 10. Simplify fV^^ + Vi^T \ ^ 2 11. Simplify (- V^^)' - (- V^T)* - (- V=T)». 12. How may the square root of (a;^— a;— 2)(a;-— 4)(;r +3a;+2) be found without multiplying ? ^(x + y)-z = -3, 13. Solve • 2x + (z — y) = 9, z + ^(x + y + z)=2. 14. Factor a; (a; +1)+ a (2 a; + 1) + a''. 15. Simplify (—r)---(^-:^x + 2r^_ ,^.,_l)-i_^.,_2)-i-(x--3)-i 262 RADICALS. IMAGINARY NUMBERS. REVIEW 16. Simplify " y + ^P c-2 (c3 + (f)-i(c'-d») 17. Simplify (V^-V^-V^)(V^T+V=^-V^). 18. Solve 9;-2?/ = 3,32/ + « = ll, 2z-3x = l. 19. 20. Draw the graphs of 2 a; + 3?/ = 13 and 3aj — y = 3, and check by solving the system. 21. Solve 5x + 2y — z = 3x + 3y + z=^7x + iy — z = 3. 22. Factor x-* — 4 a;-^ — 21 . 23 . What is the square root of x(x^ — 4) + l — 2a^(2x-3)? 24. Find the cube of - 2 V^^ + ( V^^)"'. ixl6"x2-1--^ 2" 8'" 25. Simplify L ^77^ • 26. Simplify x+lx-\-x(x + a;~')~^]~^ 27. Simplify (- 2 V^H. - V^^) (3 V^l + V=2). 28. Expand (2a=-3a;)=. „ „ , J, 1 m — n + l 2 ^ 3m+?i— 1 ^ 29. Solve for m and n : ■ — - = -~ and ' r- =6. 2m + ?(— 1 13 m — n + 2 30. Simplify (^^""jiis ^ C^yr, . \m m' 31. Factor c{c^-\-x + 2c) + c{3? + 1) + {x + l)x. 32. Solve = 1, - + - =3, - + ~ =1. X y X z z y 33. Explain the meaning of a"' = — • GENERAL REVIEW 263 34. Simplify— -3 a;" +27'^- 1**. 35. Simplify ieo- + 8c--2c-1.9a'-l. A 1\ 36. Solve for s : V4 s — 9 — Vs + 1 = Vs — 4. 37. Find two numbers whose sum is m and whose differ- ence is n. ~1 — c~^ (c — c" 38. Simplify I —- 1 + c-i (c 39. Find to three decimal places the value of _ • 3V3-2V2 40. Solve 2x = 5ij,y — x~2z = l,12y — 3 + z = 4:X. 41. Simplify f-^-4-- "^ 2 + V-l 5 2-V-l 42. Find the square root of 1 - a^'X ^ - a^) Ux |_^g_ ar- (1 — a^)~^ 43. Is the expression a*' +2a^ + «-' + 2 a" + 2 + a'"^ a per- fect square ? 44. Simplify -^("^-^) -F ^ (1 - m?f (1 - m'^y ,, -c , 1 , 2(3 + a;) , 3 a; + 4 46. Solve 1 = m, = n- ex dy dx cy 47. If a~^ = c"', and c* = f, iind the numerical value of a. 48. Simplify ( L^ + L^ Y L^ L^^ Vs + v-i 3-V-i/V3-fV-i 3-V-iy 49. Collect 3Vl|-4V3- 6 V5| + 9V8l. 264 RADICALS. IMAGINARY NUMBERS. REVIEW 50. Under what condition is a solution of mx + ny = a and kx + ly = b impossible ? 51. What must be the values of m and n in order that the division of x* + af^ + ma? + nx—'6 by a^ — a; + 1 may be exact ? 52. Extract the square root of 4 a?" - 24 a-» + 9 a-^" - 16 a" + 28. 53. Simplify V(a;+l)(a^-l) - Vse^-o^- Va;-!. 54. A discount of 10 a' + a-' 61. Solve(a; + l)(a;-l)-i-__J__ = 8a;-'. (a; + 3)(a! — 3)^ 62. Introduce the binomial coefficient under the radical and simplify: (V3 - 2) Vj -t-4V3. GENERAL REVIEW 265 63. Simplify 1^ - 3 «» + 3(3 xf - 1» + (1 + a'>){l - a«). 65. Find the value of (a— V6)(a+ V6)(a— V— 6)(a+V' — 6) when a = 2 and 6=3. 66. Solve and verify [^ x + y = m'-mn + n% 67. Simplify a ' a* f ' V V«-'c-iy 68. Show that (m^ + n-) ?i-'— m ■nv'—ir 11 - -mn -m 69. Solve and test the solution of 2 « V.'i X — (1 — v.' 5 3:4-—: — - — =0. V6 70. Draw the graphs of 2x + !iy = 7 and 4 a; + 6?/ = 14, and discuss the result. 71. Simplify |V1 — x + a+(x- a)(l - a; + a)~K 72. Kationalize the denominator of — —=1 — 3 V m + w — V m — n 73. Solve V2 a; + 5 + V8 a; — 3 = V'2 a; + 1, and test the solution. 74. Simplify [a^^l^^^'] ' [-\/a*a;-'^m^'] ^ [iK/a;-^Vm]- 75. Simplify V2+ V^l • V2- V^^ • V3 + V^^ • V3- V^2. CHAPTER XXI QUADRATIC EQUATIONS 301. A quadratic equation is an eqxiation that, in its simplest form, contains the second power, but no higher power, of the unknown quantity. A quadratic equation is an equation of the second degree (Art. 64). PURE QUADRATIC EQUATIONS 302. A pure quadratic equation is an equation containing only the square of the unknown quantity. Thus : aa^ = a, ar^ = 16, j/' = 20 al A pure quadratic equation is called also an incomplete quadratic equation. 303. Every pure quadratic equation may be reduced to the form x' = c. In this typical form we note : (1) The coefficient of x^ is unity. (2) The constant term, c, may represent any number, posi- tive or negative, integral or fractional. Extracting the square root of both numbers of the equation a? = c, we obtain x= ± ■y/c', and both values, + Vc^and — Vc, satisfy the given equation. 266 PURE QUADRATIC EQUATIONS 267 The omission of the double sign before the square root of the left member has no effect on the result, no root being lost by the omission. For: (1) +x= + Vc, (3) +x = — Vc, and (2) -x=-Vc, (4) -a;= + ^/^. From (1) and (2), a; = Vc, and from (3) and (4), x= — Vc. Clearly, nothing is gained or lost by the omission of the double sign before the square root of the left member, and it is not customary to write it. From the foregoing we have the method for solving pure quadratic equations : 304. Reduce the given equation to the form o? = c. Extract the square root of both members, observing both posi- tive and negative roots of the right member. Illustrations : 1. Solve 9 a!^ - 7 = 2 x-= + 21. 9x2-7=2x2 + 21. Transposing, 9a;2_2a;- = 7 + 21. Uniting, 7a;2rr28. Dividing by 7, x^ _ 4. Extracting square root, x = ± 2. Result. 'Verifying, 9 (±2)^-7 = 2 (±2)2 + 21. 9.4-7=2.4 + 21. 29 = 29. a-\-x X —2a 2. Solve a — x x + 2a Clearing of fractions, 2 a^ + .3 ax + x^ = - x^ + 3 ax - 2 a*. Transposing, x^ + x^ = - 2 a" - 2 a^. Uniting, 2x2= -4a2. Dividing by 2, x^ = - 2 a^. Extracting square root, x = ± V — 2o2. Or, X = ± a V — 2. Result. 2U8 QUADRATIC EQUATIONS Exercise 101 Solve: 1. a!= = 49. 4. 4a.-2 = 9. 7. 3a^ + l = 0. 2. 3a,-2 = 75. 5. 6a" -18=0. 8. 4. ax- - 16 a? = 0. 3. 5a;2-80 = 0. 6. 2a.-2 = 5. 9. 72/^ = 35a<. 10. 5af + 3 = 2x^ + 27. 11. 2»2 + 7 = 5a,-^ - 20. 12. 3*2 4- 18 - a; (a; + 1) + a; = 0. 13. a;(3a; + 2)-3= (x + l)2. 14. 4a;= - 3.v + 2 = (2a; - l)(a; - 1). 15. (x + ly -{x- ly = 26. 16. (3 + a;) (a;2 - 2) = 3 + x (a^ - 2). 17. (a; - 2)= - 6 (a; - 2)^ = a; (a; - 4) (a; - 9). x'-S 1 1 ■ ar' + a;-6 a; + 3 x-2' 19. cc + 1 x-2^Q x-3 x+i 20. x' + X—1 x^ ~x — l_Q x+2 x-2 ~ ' 21. x^ + x-2_x^ + x-l a? — X -\-2 a;- — a: + 1 AFFECTED QUADRATIC EQUATIONS 305. An affected quadratic equation is an equation con- taining both the square and the first power of the unknown quantity, but no higher power. a:2+12a;=13. 3x2-2x-7 = 0. aa;^ + 6x + c = 0. An affected quadratic equation is called also a complete quadratic equation. AFFECTED QUADRATIC EQUATIONS 269 Completing the Square 306. The solution of an affected quadratic equation is based upon the formatiou of a perfect trinomial square. In the per- fect trinomial square, we note (1) The coefficient of uf is unity. (2) The third term, a^ is the square of one half the coefficient ofx. That is, a^=(^^. Similarly in ai^ + 12 ax + 36, 36 = /^— Y. Ina^-8aa; + 16a^ l&a^=(-—^. 307. Given, therefore, an expression containing the first power of X, and the second power of x with the coefficient unity, we may form a perfect trinomial square by adding to the ex- pression the square of one half the coefficient ofx. Thus, given, x^ + 6x, x^ + 6x+(3y, orx^ — 6a; + 9, a perfect square. a;2_i8a;, a;^- 18a;-t-(-9)^ or a;2_ ig a; + 81, a perfect square. Sx'^ + ix, x'' -i- (.i) X + (_^y, ora;2_|a! + |, a perfect square. Oral Drill With each expression in the proper form, give the term necessary to make a perfect trinomial square. 1. a^ + lOa;. 5. y--32y. 9. 3a? + 4x. 2. a? + 14:X. 6. x' + lx. 10. Sa.-^ — 8 a;. 3. aP-lGx. 7. a?--5x. 11. 2ar-3a!. 4. a? + 24:X. 8. y'+lly. 12. 6«2_]i7a._ The principle of completing the square is applied to the solu- tion of affected quadratic equations in the following : 270 QUADRATIC EQUATIONS Illustrations: 1. Solve a^ + 12 a; = 13. Adding to both members the square of one half the coefficient of x, a;2 + l2a;+ (6)2 = 13 + 36 = 49. Extracting square root, x+ 6 = ± 7. Whence, a; = ± 7 - 6. That is, a; = + 7 - 6 or a; = - 7 - 6. From which, x=l or x = — 13. Result. Both values of x are roots of and satisfy the given equation. Verification : Ifa; = l: 12+12(1) = 13, 1+12 = 13. Ifa;=-13: (-13)2 + 12 (-13) =13, 169-166 = 13. 2. Solve2a;2_3a._20 = 0. 2a;2-3a;-20=0. Transposing, 2 a;'^ — 3 a; = 20. Dividing by 2, x^- lx = 10. Completing the square, a;^ — f a; + (|)'^ = 10 + ^5 = W- Extracting square root, a; — | = ± '/-. Whence, a; = '^, or - J^. Or, a; = 4, or — |. Result. Verification : If X = 4 : 2 (4)2 - .^ (4) - 20 = 0. 32 - 12 - 20 = 0. Ux = -i: 2(-e2-3(-e-20=0. ^ + -'^-20=0. _, a? — x+l x' + X — 1 ^ 3. Solve ^ , „ =1- X— 2 x + 2 Clearing of fractions, (x2-a;+ l)(a;+2) - (a;^ + cc - l)(x - 2) = (a; + 2)(a!-2). a;3 + x2 - a; + 2 - x3 + a;2 + 3 a; _ 2 = x2 - 4. Whence, x^ + 2 x = — 4. Completing the square, x2 + 2x+l = — 4 + 1 = -3. Extracting square root, a;+l = ±V— 3. From which, x = - 1 ± V- 3. Result. AFFECTED QUADRATIC EQUATIONS 271 Irrational roots result when, after completins the square, the rii;ht member of an fijuatiouis irrational. The orijtinal equation will be satisfied b}' either (— 1 + V— 3) or (-1- V-:3). From these illustrations we may state the general process for completing the square in the solution of affected quadratic equations : 308. Simplify the given eqxtation and reduce to the form X- + hx = c. Add to each member the square of one half the coefficient ofx. Extract the square root of each member, and solve the tivo resulting simple equations. Exercise 102 (See also page 412) Solve and verify : 1. .r=-4.i-=12. 2. .i- + 2.r = ?,5. 3. .i-" - f) .»• = 27. 4. a- + 3. 1' = 10. 5. .r + ■'■ = 50. 6. 2.v--3a: = 2. 7. oar- 7. (• = 6. a o.i- + 17.r + 20 = 0. 9. 4.V- — o.r-6 = 0. 10. 6.i-'-.i- = 2. 21. 2.f= + 3.r= -10. 22. (.•!.»• + 1)^ - (4 .»' + l)(2.r-l) = 11. 23. (.r* - 1) - (ar^ + 2) (.i-' - ;',) - (x + 5) = 0. 24. (2 a; + 3) (a; - 2) - (3 a: - 1)^ = ,,• {x - 3) + 1. 25. (2a; + l)(3a;-2)-(a; + l)(2.r-l) = (.f + l)(3a;-l). 11. 15a;2 + 14a:=-3. 12. 14.1"- 5.^ = 24. 13. 20.r^' + .r = l. 14. 12.r- + 23.r + 10 = 0. 15. X- + 4 .r = 3. 16. .,■■-' -Qx = 6. 17. .1" + ,5 .,■ - 3 = 0. 18. 2 .r - S .!■ = 1. 19. .r- - 4 ,1' = - 9. 20. 3.c--a' + l=0. 72 QUADRATIC EQUATIONS 26. a;' - 2 (.t; ~ 1 ) + .B (.r - 1 ) - (x - 1 ) (.B - 2) = 0. 27. X {x- — X -3) + {x + 3)(x + 11) = (x — l) (x^ + 1). 28. 29. 30. 31. 32. 33. ZX — 1 3; ar' + 2a;+4 2-a; a^-8 The Quadratic Formula 309. Every affected quadratic equation may be reduced to the form ax' + 6a; + c = 0, in which form the coefficients, a, b, and o, represent numbers, positive or negative, integral or fractional. 310. Solving this affected quadratic equation by completing the square, we have, ax^ + bx = — c. a a 2 + 3 _,_ a; + 1 a; — 1 34. x' + x-l_x^ + x + 2 a--' + a; + 2 ar= + a;-3 a; + 2 1_1 a; + 3 a; 3 35. a; + 3 ^_x-3 x-3 x+3 -+1+ ; =1. X x + 1 36. ^ - ' +1 2 _. a; — 4 a; + 4 a---16 x + 1 x-1 .. x-1 x+2 ■ 37. X X a,-^ + 6ft'+ 8 a; + 2 a; + 5 .v- + 7 a; + 10 x-1 x-2 3 x-2 x-3 4 38. a;+3 a;-2_2(a;-5) of x' + l X^ + x^ 2a;4-l 3a; + 2_ 2a!-l 3a;-2 = 0. 39. 2a; + l 3a!-l_5a;-2 a; + l a; — 1 a; +2 40. -^ + 4 2 x' + x-l a \2a/ a ' 6^ — 4 ac 4a2 AFFECTED QUAIMIATIC EQUATIONS 27o a; + b 2a X = ± 2a b y/b^ -4ac 2a 2a b±Vb2-4ac 2a This value of x from the general equation, aa^ + &» + c = 0, serves as a formula for the solution of affected quadratic equa- tions. The formula is expressed in terms of the coefficients of the given general equation, and by substitution of particular values for a, b, and c from a given equation we obtain the roots of that equation. The formula is the most practical of the many methods of solution and it should be memorized. To obtain the solution of an affected quadratic equation by means of the general formula : 311. Transpose all terms of the given equation to the left memher, and reduce to the form aa? + hx-\- c = 0. In the formula substitute the coefficient of the given a? for a, the coefficient of the given x for b, and the given constant for c. Simplify the resulting expression. Illustrations : 1. Solve by the formula, 2a^ + 5x = 12. Transposing, 2x2 + 5a;- Then in - 12 = 0. For the formula : a = 2, - 6 ± V62 -4ac 2a 6 = 5, c =-12. -(6)±Vf5)2_4(2)C-12) 2(2) - 5 ± V25 + 96 4 _ -5±n 4 = i, or - 4. Result. SOM. EL. AT,G.- -18 274 QUADRATIC EQUATIONS 2. Solve by the formula, a;^— -y-a; = |. Transposing and clearing of fractions, 3 ri;2 - 10 a; - 8 = 0. For the formula, a = 3, 6 = — 10, c = — 8. Substituting, ,^ -(-10)±V(-W-4(3)(-8) . 10±14. ''- 6 X = 4, or - |. Result. Exercise 103 Solve by the formula : 1. 0)2-11 a; + 24 = 0. 2. a!'-9a; = 22. 3. a.-2 + 9a; + 14 = 0. 4. a^-lla;=-28. 5. a^ + 9a; = 52. 6. a;2-x-72 = 0. 7. ar'-21a; = 46. 8. 6a?-x = 2. 9. 2ay'+7x = 15. 10. 8a^ + 2a; = 3. ^ 9 20. a)2_^_£^0. 11. 15a^+44a; = 20. 30 3 The Solution by Factoring 312. Many affected quadratic equations may be solved by an application of factoring. This method is based upon the principle that : 313. The product of two or more factors is zero when one of the factors is equal to zero. 12. 21:B^ + 29a;-10 = 0. 13. 14a^ + 53a; + 14 = 0. 14. 30 a^- 11 ."• = 30. 15. 42 a^ - 59 a; = - 20. 16. x'-i^x + 2 = 0. 17. ^ + ^^=i- 18. a^-- = 2. 6 19. x'4.^-^. AFFECTED QUAUKATIC EQUATIONS 275 To solve an affected quadratic equation by factoring: 314. Reduce the given equation to the (jvnerul quadratic form, u.tr + hx + (• = 0, and factor the rc^ultiiK/ trinviuial. A^Kiiine that each factor in turn cijnalK zero, and solve the other factor for the value of the unknown quantity. Illustrations: 1. Solve by factoring, a^ + 6 a; = 7. Transposing, x'^ -{- Gx — V = 0. Factoring, (a; + 7) (a; — 1) =0. From which, a; + 7 = 0, or x — 1 = 0. Solving, x = — ~i and a; = 1. Result. 2. Solve by factoring, 2 .xp- + W- .c = t. Simplifying, 10a;2 + 11 a;- 6 =0. Factoring, (2 .r + 3) (5 a: - 2 ) = 0. Whence, 2 a; + 3 = and 5 .r -2 = 0. And .r = — 5, or a; = f . Result. Exercise 104 Solve by factoring: 1. .i--5;r = 14. 2. .«2-S.r + 15 = 0. 3. .xf-3.i- = 4. 4. .ir-13.r + 12 = 0. 5. .1-- 12.1- -13 = 0. 6. .1;= - 19 .r = 42. 7. 2.i-^-9.r + 4 = 0. 8. 4.r=-r).v=6. 9. 6. r-- 17 .(- = -12. 10. 8.-e2-38.c = -35. 11. 15 x" - 77 X + 10 = 0, 12. 12a-"-23.r + 10 = 0. 13. •'^-¥«' = f 14. «'+^''-a; + l=0. 15. ar"-? = I. 4 2 16. ■|a; = a,-^-f|. 276 QUADRATIC EQUATIONS LITERAL AFFECTED QUADRATIC EQUATIONS 315. Any one of the three methods given applies readily to literal affected quadratic equations. Exercise 105 Solve: 1. x'-2ax = 3a'. 2. x^ = 5 ax— 6,a\ 3. x'^ + ax=2a\ 4. 3 aV + acx = 2 c''. 5. 35 cV — c'x= 6. 6. nV + ninx = 2m. 7. 6 x' + acx = 12 fi'd'. 8. 2 aV — 5 acx = 3 cl 9. cV — c- 4- 4 c = 4. 10. 12 a^x^ + 25 acx = 7, :\ 11. 9mV-(3m + iy = 0. 12. 10cV-21c2a; + 9 = 0. 13. x^ + {a + l)x + a = 0. 14. x'+{a + b)x + ab = 0. 15. x' — 2ax + a^ = l. 16. cV + c(m — 7i)x = mn. 17. mx'-{m + l)x + l = 0. 18. acx' + aHK = cmx + mn. 19. cV-c(a + l)a; + a = 0. 20. (a; + a+m)2=(a-m)2. 21. c2(a;2-l)=a;(a; + 2c). 22. (a2-l)a;2 + 2aa; + l = 0. 23. a;2 + 4:aa; — 4a-l = 0. 24. (o2-4)a;2-(a-2)a!=12. 25. (2a; + a)(a; — c) = (a; — o)(a; + c). 26. a;2-2aa;-26a; = c2-a2-6(2a + 6). 27. aV + 2 aca? + &x^ + aa; + ca: = 12. 28. 6a;''+ (5a + 7)a;4-a' + 3a + 2 = 0. 29. {x + a){x + c) + {x — a){x-^c)=c\ 30. (a + a;+iy — (a + a;)(a + l)=l. 31. {a — x){a — 2 x) — {2 a — x){a — 3 x)=2 a'. 32. (a=' + a-2)a;''-(2a + l)a; + l = 0. 33. (x + m + 1) (m - 1) +(35 - m - 1) (m + 1) = 2(05*- 3 m^- 1). 34. x' — 2{a + c)x + (a + c + l){a + c—l)=0. 35. x2-(2m-3)a; + m''-3m-4 = 0. 36. {2x + a + iy-2(2x + a){a+l)=4:x-2a'x. DISCUSSION OF THE AFFECTED QUADRATIC EQUATION 277 37 g_+« I « ^5 44 1 1 _ 1 1 a a; + a 2 »t — j; //t c — a; c „„ a; . 2 a— 1 ;! aa; ^- c^ a^ n oo. -\ = . 45. = 0. a+1 X 9j(a+l) (c + xf (c—xy 39, - ^_ + ' = ;; 46. ^ *^:^"^ +. ^±A"i^ = 2. 2 r -\-x 2 c — x 2x + ac x + 8 ac -1 g— 1 __ a;2 — 2 .V a; + a _ a; + 3 a _ 1 40. 1 = . -±1. = -. 2«4-a; 2a— a; 4a-— .r- a; + 2 a a; + 4a 3 41 ■^' — « I a; + g _ o g.v .r-2a a;+2 g _ 2a!^+3 a'' a;+g a; — g .c- — a^ a;+oa a;— 3 a a;^— 9a^ 42. 1 c + cJ _ 1 ^g a' + m j^ _ a; + 3 m c -\- X cri rf + ■'• a; + 2 m a; + 4 m . a! + a + l a; + a— 1 _„ c + rf 4c + 5 a;__3c — 2a; a; + g— 1 x + a + 1 c — d c + x c — x discussion of the affected quadratic equation Character of the Roots 316. If the two voots of the affected quadratic equation, aa;^+ 6a; + c = 0, be denoted by i\ and r^, we have (Art. 310), - & + V^^ - 4 gc _ Z, _ V&^ - 4 , r, = , r., = \ia J, a The character of the result in each root depends directly upon the value of the radical expression Vft- — 4 ac, for the radicand, 6^ — 4 ac, may be positive, zero, or negative. (I) ^YIlen 6-— 4 ac ix •poaWu^e. (g) The roots are real, for the sqxiare root of a positive quantity may be obtained exactly or approximately. (6) The roots are unequal, for V&^ — 4 ac is + in rj and — in rj. (c) The roots are rational if 6^ — 4ac is a perfect square, irrational if not. For each root reduces to — —. 2 a 278 QUADKATIC EQUATIONS Illustrations : (1) In 2x2+ lla;+ 12=0, a=-', 6 = 11, c = 12. 62-4ac = 121-iH;=25. Hence, the roots of 2x2+ 11 x+ 12=0 are real, unequal, and rational. (2) In5x2 + llx + 3 = 0, « = 5, 6 = 11, c = 3. h'^-i ac = \-2l~G0=e\. Hence, the roots of6x2 + llx + 3 = fl are real, unequal, and irrational. (II) When b" — iac equals 0. (a) The roots are real, and ' (6) The roots are equal. Illustration : In4x2-20x+2.5=0, a=4, 6 = -20, c=25. 62-4ac=400-400=0. Hence, the roots of 4 x- — 20 x + 25 = are real and equal. (III) W]ien 6^ — 4 ac is negative. (a) The roots are imaginary, for the square root of a negar tive number is impossible. Illustration : In3a;2 + 2x + 2 = 0, = 3, 6 =2, c = 2. b"- -4 ac = i -2i=-20. Hence, the roots of 3 a;- + 2 x + 2 = are imaginary. 317. We may, therefore, by inspection of the discriminant, 6^ — 4ac, summarize the foregoing conclusions as follows: I. li b^ — i ac > 0, the roots are real and unequal. II. If 6- — 4 ac = 0, the roots are real and equal. III. If 6^ — 4 ac < 0, the roots are imaginary. Illustrations : 1. Determine the character of the roots of a;- — 5 a; = 6. x2-5x-6 = 0. = 1, 6 = -5, c = -6. 62 - 4 ac = [(- 5)2 - 4 (1) (- 6)] = (25 + 24) = 49. Therefore, the roots are real, unequal, and rational. (316, I.) 2. Show that the roots of a;' — 4 a; -|- 5= are imaginary, 62-4ac=[(-4)2-4(l)(5)] = (lfi-20) = -4. Hence, the roots are imaginary. DISCUSSION OF AFFECTICD QUADKATIC EQUATION 279 3. Determine the value of m for which the roots of 4 a;- + 10 a; + m = are equal. By Art. 316, II, tbe discriminant must equal 0. Hence, (10)2 _ 4 (-4) („i) = q, 100 - 16 m = 0, m = s^. That is, the roots of 1 x^ + 10 a; + ^ = are equal. 4. For what value of m will the roots of (m + 1) .ir' + (6 9)1 + 2) a; + 7 m + 4 = be equal ? The discriminant must equal 0. a = (?n+ 1), h = (() m + 2), c=(7ni + 4). Then (6 m + 2)2 - 4 (m + 1) (7 m + 4) = 8 m"- - 20 m - 12. Solving, 8?»2 -20?» - 12 =0, TO = 3 and - 1. Kesult. By silbstitutinj:; these values, 3 and — J-, in the given equation, two dif- ferent equations result, both of which have equal roots. Exercise 106 By inspection of the discriminant, determine the character of the roots of : 1. a;2 + 7.'c=8. 7. 2a!=-|-7a;-|-3 = 0. 2. X- -6.c = 40. 8. 3a^-5x=-12. 3. :r + 5x-84: = 0. 9. a^ + 10a;-|-l = 0. 4. ,r'_3a; + 54=,o. 10. 3a;2-12«=-17. 5. .1;= + 12.1; = -36. 11. 5a;2-|-40a; = l. 6. .T a;2 + 8 a; + 4 = 0. 12. 7 ar^ + 11 a; + 12 = 0, Determine the values of m for which the two roots of each )f the following equations are equal : 13. ix^ + 20x + m = 0. 16. a!^ + (m-|-5)a; + 5m+l=0. 14. 9a;= + ma; + 2o = 0. 17. (m+l)a,'2-(m-2).'»; + l =0. 15. 2 ma^— 30. V — 15 = 0. 18. 2 ma;- + 7 ma; =a;^ -I- 6 a; — 5 m. 19. (m-f-l)a;^ + ma;= — 9(a;-|-l). 20. m—7 + mxP — mx = — 2 a; — a;^. 280 QUADRATIC EQUATION'S Relation or the Roots and CoEFificiENTS 318. If the roots of av? + 6a; + c = 0ora!^+-a;+- = a a are obtained by Art. 310, and are denoted by r^ and r.2 respec- tively, we have -h+ y/b'^-iac „„j - ft - Vfe" - 4 ac ri = ■ — j; and r^ = — 2o 2a By addition, ri + r^ ft + V62 _ 4 ac - 6 - V62 - 4 ac 2a Or, r,+r^ = ~^-. By multiplication, nr. = {- h +^W^4nrc)(- V -^iF^iHi) 4 a^ _ ft' - ft^ + 4 ac 4a2 Or, ri)-2=-- a Hence, we may state : 319. In any affected quadratic equation of the form a a (a) The s%im of the roots equals the coefficient of x with its sign changed. (6) The product of the roots is equal to the constant term. Illustrations : 1. Find by inspection the sum and the product of the roots of 6 a;^ + 5 a; = 6. Changing to the required form, x'^ + ^x — 1 =0. From Art. 319 : Sum of the roots = — | . Product of the roots =—1. Result. 2. One root of 2 a;^ + 5 a; = 12 is — f . Find the other root. Transposing and dividing, x^ + ^x — 6 =0. Dividing the product of the roots, — 6, by tlie known root, — |, we have, (- 6) H- (- f) =+ (6 X f) = 4. Result. DISCUSSION OF AFKECTKD QUADHATIC EQUATION 281 Exercise 107 Find by inspection the sum and the product of the roots of : 1. a^ + 9a- + 14 = 0. 5. 3 a;^- 10 a; + 3 = 0. 2. a;--21.« = 46. 6. 2af + 3x = 2. 3. 2.'c2+7a; = 15. 7. 7 ar' + 9 a;-10 = 0. 4. 6af-x = 12. 8. 15a^ + 14a; + 3 = 0. 9. One root of 5a;^ — 26a;+ 6 = 0is5. Find the other root. 10. One root of 8 a^ = 15 a; + 2 is — |. Find the other root. 11. AN'ith the values of r^ and r^ from the equation ax' + 6a; + c = 0, find the value of ,.2 „2 11 Also the value of — H '•i + '-2 n rj Formation of an Affected Quadratic Equation with Given Roots b c Consider again the form af + -x + - = 0. (1) a a If, as before, i\ and j-j denote the roots of this equation, we have (Art. 319) : b »'i + ''2= a Whence, - = - r, - r^ (2). Also (319), -= r^r^ (3) a a Substituting in (1) the values found in (2) and (3), x'+( — ri — r2)x + rir^ = 0. x^ — r-iX — r,^ + r-iT^ = 0. (a? — r^) — (ViX — r^r,) = 0. a!(a; — r.j) — /•, (x — ?\,) = 0. (:x-r,)(:x~r,)=0. 282 QUADRATIC EQUATIONS Therefore, to form a quadratic equation that shall have any given roots : 320. Subtract each root from x, and equate the product of the resulting expressions to 0. Illustrations : 1. Form the equation whose roots shall be 3 and 7. By Art. 320, {x - 3) (cc - 7) = 0. Or, a;2 - 10 a; + 21 = 0. Result. 2. Form the equation whose roots shall be f and — |. By Art. 320, (x-\){x-{-l)-) = 0; (a;-f)(^ + f) = 0; 15 x2 + a; - 6 = 0. Result. Exercise 108 Form the equations whose roots shall be : 2 8. a— 1, a+1. 1. 2, 5. 5. -2, 2. 3, - 8. 3. -4, - 7. 6. 5 2' . ^ -1. 7. 2 3' 3 9. rr — 1, 2 a. 7 10. 2±V-1. 11- V2, -1. 8 3 j2 Va + 1 V^t-1 4' ■ 2 ' 2 ■ GRAPH OF A QUADRATIC EQUATION IN ONE VARIABLE 321. The graph of a quadratic equation is obtained by application of the principles governing the graphs of linear equations. The given equation is written in the typical form, ax^ + bx + = 0, and the left member is equated to y. By as- suming successive values for x, the corresponding values of y are obtained as before. GRAPHS OF QUADRATIC EQUATIONS 283 GRAPHS OF QUADRATIC EQUATIONS HAVING UNEQUAL ROOTS 322. Plot the graph of x- + 2 a; — S = y. I ^ ^ i Since y = x- + 2 x — 8, in tlie figure : If K = 3, y = 7. P. x= 2, 2/ = 0. Pi. X = 1, 2/ = — 5. P-,. x= 0, ;/ = - 8. Ps. x=-l,2^=-n. 7-'4. X = - 2, y = - 8. Pj. X = - n, 2/ = - 5. Pb. ar = - 4, )/ = 0. P^. x=-5,y= 7. Pa. etc. The curve representing the equation, a" + 2 x — S = y, might be indefinitely extended by choosing further values of .r. In- termediate points on the curve may be obtained by assuming fractional values of x and obtaining corresponding values of y. 323. TTie loicest point of the grajyh of a quadratic equation in one variable may, in general, be obtained by completing the square. Tlius, a;2 _,_ 2 X + 1 = 8 + 1 ; (x + 1)^ = 9 ; (x + 1)2 - 9 = 0. Now (x + 1 )2 — lias its greatest negative value when x = — 1 . Hence, the coordinates of the lowest point of the curve are (— 1, — 9). Plot the graph oi2aP + 7x — 4: = y. y = 2x^ + 1 x — 4. Then in the first figure on p. 284 : (The student will note that, in the figure, the scale of the graph is so chosen tliat one unit of division on the axis corresponds to two units from the solutions. ) 284 QUADRATIC EQUATIONS X = 2, 2^ = 18. P. x=-2, 1/ =-10. Pi. x= 1,2/ = 5. 1\. x = -?,, y =—7. Fs. x= 0,y=-i. P2. x=-4,y = 0. Pe. x=-l,y=-9. Pa. X =— 5, y — 11. Pi. ::::::s3:::_: ::::-::: 'i i::: :h;:::: ^:::|[::::? :::iZpciEEE: 5 On completing the square the lowest point in the curve is (— J, — -»ji). Exercise 109 Plot the graphs of: 1. ay'— ()x + 5 = y. 2. r' + 6x + 8=y. 3. x' + x — 12 =y. 4. iaf — 5x = y. 5. 2x'-5x=-3 + y. 6. 8x^ + 2x-3 = y. GRAPHS OF QUADRATIC EQUATIONS HAVING EQUAL ROOTS 324. Plot the graph of a^— 6x + 9 = y. Since y = x^ — 6x + 9, in the figure : If X = 0, y = 9. P. x=l,y = 4. Pi. X = 2, 2/ = 1. P;. X = 3, 2/ = 0. Pa. X = 4, ?y = 1. P4. X = 5, 2/ = 4. Ps. X = 6, 2/ = 9. Pe. i ^ a 45 Now, x2 — 6 X + 9 = 0, may he written (x — 3)^ = ; and X in this equation can never he 0. Hence, the curve cannot cut the XX axis, but is tangent to it at (3, 0), the lowest point of the graph. GRAPHS OF QUADRATIC EQUATIONS 285 ORAPHS OF QUADRATIC EQUATIONS HAVING IMAGINARY ROOTS 325. Plot the graph of a^ - a; + 2 = y. Since y = x^ - x + 2, in the figure : lfx=0,y = 2. P. x = 3, y = 8. Ps. x = l, y = 2. Pi. x = - 1, y = i. Pi. x = 2, y = i. Pj. x = -2, y = 8. Pi. X = -3, 2/ ^U. Pb. Completing the square in .1- — a; + 2 = : And this expression not being or negative for any value of x, it follows that y cannot be or nega- tive ; that is, the graph cannot touch the XX' axis. Important Conclusion. The graph of any quadratic equation in one variable and in the form aaf +bx + G = 0, intersects the XX' axis at two points whose abcissas are roots of the equation, provided that the given equation has reed and unequal roots ; has one point in the XX' axis if the roots are equal; and does not cut the XX' axis if the roots are imaginary. :z:i::i;"i zzlz ::: ::"::|i::ii:::i::: 3 ^ IL IE \ T *\ 5 _, C-K ^ : ___: _^. Ezercise 110 Plot the graphs of : 1. ar' — 4a;+4=2/. 2. 4:x'' + ix+l=y. 3. a? + >3x + %=y. 5. a;^ 4- a; + 6 = _?/. 6. 23?-&x+n = 'i CHAPTER XXII THE QUADRATIC FORM. HIGHER EQUATIONS. IRRATIONAL EQUATIONS EQUATIONS IN THE QUADRATIC FORM 326. An equation in the quadratic form is an equation having three terms, two of which contain the unknown num- ber; the exponent of the unknown number in one term being twice the exponent of the unknown number in the other term. Thus: a;*-13a;2 + 30 = 0; x^+7x^ = S;- -J/x'^ - 4 v'k = 12. HIGHER EQUATIONS SOLVED BY QUADRATIC METHODS 327. It will be seen at once that many etjuations in the quadratic form must be of a higher degree than the second. The method of factoring permits the solution of many such equations, and is generally employed in elementary algebra. 328. If quadratic factors result from the application of factoring to higher forms of equations, they may ordinarily be solved by completing the square or by the quadratic formula. Such factors most frequently occur in connection with binomial equations. Illustrations : 1. Solve X* - 13 X- + 36 = 0. Factoring, (x^ - 4) (x- - 9) = 0. And, (x + 2) (a; - 2) (a; + 3) (x - 3) = 0. Whence, x = - 2, 2, - 3, and 3. liesult. 286 EQUATIONS IN THE QUADRATIC FORM 287 2. Solve .r" + 7 ar' - 8 = 0. Factoring, (x" + 8) (k^ - 1) = 0. And, {x + 2) {x^ - 2x + i) (x - \) (a:- + 2 + 1) = 0. By inspection, x = — 2 anrl 1. Solving, ■ 'x^ -2x + 4=0. x = l± V^. Solving, .X- + x+1 =0. ^ ^ - 1 ± V- 3 2 Therefore, a- = - ^1, 1 ± V^l, 1, and^= 2 ~ ' ^*^""- 3. Solve \/^ - 4\/x = 12. Expressed with fractional exponents and in the transposed fonn, we have, xi _ 4 x^ - 12 = 0. Factoring, {x^ - 6) (xi + 2) = 0. Whence, x' = d, or x' = — 2. From which, x = 216, or — 8. Result. When tested both roots prove to be solutions. 4. Solve2\/ocF-5x-3 = 10 + r,x~x'. Transposing, k^ — 5.x + 2 Vx'-* — 5 .i; — 2 =r 10. Adding -2, x2-ox-2+ 2Vx^-5x-2 = 8. Whence, (x^ - 5 .r - 2) + 2(i2 - 5 x - 2V2 - 8 = 0. Factoring, [(x^ - 5 x - 2)^ - 2][(x2 - 5x - 2)^ + 4] = 0. From which (x^ - r, / - 2)^ = 2 and (x^ - 5 a: - 2) ^ = - 4. Squaring, x- — 5 x — 2 = 4, and x^ — 5 a: — 2 = 16. Solving, x = 6, -1, and 5±_V97 Results. 329. Eor convenience of solution a single letter may be substituted for a compound expression in equations in the quadratic form. Care should be taken that no solution is lost in the final substitutions. 288 THE QUADRATIC FORM. HIGHER EQUATIONS Exercise 111 (See also page 413.) Solve : 1. a;^-5a;2 = 36. 2. a!< - 24 a;2 = 25. _§, a;-2 - 2 a;-> = 8. 4. x^ + 9a^ = -8. 5. £»-« — 5 a;-2 = - 4. 6. x-* — x-'' = 2. 7. a;-"= + 7 a;-^ = 8. 8. x^ + xi = 2. 9. a;^ — a;J = 6. 10. x^ + 7x^ = 8. 11. a;-^+x"* = 6. 12. a;-' + 2 k"^ = 8. 13. a;-3 - 26 a!~^ = 27. 14. Vx + ■\/x = 6. 15. a;5 _ 5-v^a^ = — 4. 16. »^' + 7^/aP = 8. 17. \/a^ — 7 a;^ = 8. 18. 3Va^+26a;^=-16. 19. 17V»-4a;=-15. 20. a;-i + 3 = 4VK^ 21. 2a;-^-3+V^=0. 22. a:^» + 5a;" — 4 = 0. 2 ^ 23. asn — 6a;n = — 5. 24. (a; + l)2_(a; + l)=6. 25. (a; + 2)2 -(a; + 2) = 20. 26. (a;-3)2 + 5(a;-3)=6. 27. 2(a; + l)^-(a; + l)=6. 28. 3(a;-5)2-7(a;-5)=-2. 29. (2a; + l)=i-9(2a;+l)+18=0. 30. V2a; + l--^2a!+i = 6. 31. ■y/x-5 — -\fx^ = A2. 32. Vx + 10- + 24 = 0. 56. (2 a;2 + 3 a;)^ - 3 (2 a;'' + 3 a;) = 4. 57. (aj' + 2 a; - 1) + 5 V.a;'' + 2 a; - 1 = 6. 58. a;2 + 2 a; - 6 - Vx^ + 2 a; - 6 = 6. 59. x^ - 10 a; + 18 - 2 Var" - 10 a; + 18 = 3. 2a;2 + 3a; + 9- 5 V2 a;^ + 3 a; + 9 = 6. 2a;2-3a;+7 + 2\/2 a;'' - 3 a; + 7 = 24. 3 a;2 - 16 a; + 21 + 3V3 a;"- 16a; + 21 = 28. ._. 7a;2-5a; + 7V'7a;2-5a!+l = -7. 64. a!'^ + 8 a; - 2v'a;2 + 8 a; + 40 = - 5. 65. a;'' - 2 a; + 6 Va;" - 2 a; + 5 = 11. 66. a;'' + 5 a; — 2 Va;^ — 4 a; + 7 = 9 a; — 7. x-lj x-l x^-2 ylx^-2 69. 71. 60. 61. 62. 63. 67. ■'+?T-f«+!V*^- »■ ?±l+<5-^J*=«- xj \ xj x—1 \x'' + 2 J 70 /'a; + ?Y + 2('a; + |)=35. 75. 2('x + 5y+3g + a;^ = 20. -^J-<-!)— «■ (l-!)"-) + ■'■ — 5 = x. Transposing, — 2 \/'Z x^ — 3 x — 36 = — 2 a; — 2. Dividing by - 2, \/2x'-3x-3& = x + l. Squaring, 2x'^ -3x - 35 = x'^ + 2x + 1. From which, x" — 5 a; - 36 = 0. And, X = 9, or — 4. Result. The solution, x = 9, satisfies the given equation. Note. If the solution, x =— 4, is tested and, in extracting the square root of the right member the negative value of the root is taken, we have : V2{ -4) +7 - V( - 4) - 6 = -/^^. v^n; - 3V^ = - 2>/^n;. But this process is at variance with the accepted condition that positive square roots are to be taken through elementary algebra. IRRATIONAL EQUATIONS 293 Exercise 113 (See also page 413.) Solve and test the solutions of : 1. v'a; — 3 = .v— 5. 13. Va; + 1 — »; + 1 = 0. 2. Vlt)-.s^ = 4^x. 14. Va; + 1 + « - 19 = 0. 3. a, + V25-..= =7. ^^_ x + V2=\/a. + 2xV2 + 4. 4. A .!■+ l + a; = ll. 5. "V i"> — .V + .^' = 5. 16. x'' + x-3 = Vx*+2x'+10. ^^ Va-^-a; + 2 _1 6. ■V'2x^-x + 55 = x+5. ''■ V5.v^_--^r^ 7. V2 a;'' + 15 a; + 28 = 4 + x. 8. 2 + V2a; + 8=2V7TT.. ^^- '/S-. , v^^^r^T 3 VS .r - V2 g - 1 _ 1 V3"a^ + V2a;-1 9. Vi'" + A- + 1 = V3 a- + 4. 19. V.r+1-Vx^l^3^^g_ 10. a'2 + a; + l=Va;^ + 2a'^ + 9. ' V5+T+v'a;-l 11. V21x^-4-x + 2 = 0. V 2a^-a; + 3 20. \ =Va; + 2-l. 12. V4a;-3-l = Va; + 1. Va; + 2+l „, 3 a;— VSa-^— 4a; z-? — s 21. =^ — = x — vaJ^ — A .)'+Va;2-2 22 -t-V:i _V4^- + 2 2^ Sx-l ^ o_ V5^ + 5 Va; 4+Va; ' V5a,- + 1 ^ 4A/2a;-6 ^ 6- V2a; ^^ 5a!4-Va;'-5 ^^^p 3 + V2a; V2a;-3' ' 5x-^/j.VZ^ 13 26. "-^ + ^ =2. c + Vc^ — x^ c — Vc^ — x' CHAPTER XXIII SIMULTANEOUS QUADRATIC EQUATIONS. PROBLEMS 332. In the solution of simultaneous quadratic equations we have particular methods for dealing with three common types ; but no general method for all possible cases can be given. SOLUTION BY SUBSTITUTION 333 Wlien one equation is of the first degree and the other of the second degree. 1. Solve ]!-'-3^r2/^ = l' (;) (2cif-3xy- [5x+y=3. From (2), y = 3-5x. (3) Substituting in (1), ix'^-SxCS-bx) - (3 - 5 xY = 1. 2 a;- - 9 X + 15 a;2 _ 9 + 30 X _ 2u X- = 1. 8x--21a; + 10 = 0. (4) Factoring, (8 a; — 5)(i — 2) = 0. From which, z = J, or 2. Substituting in (3), If 1 = 2, )/ = 3-5(2) = -7. Hence, the corresponding values of x and y are * = ^' and * = ^' 1 Result. y = ~i ^.1 334. Corresponding values must be clearly understood as to meaning. From (4) in the above solution tico values of x result. Each value of x is substituted in (3), and two values for y result. Therefore, we associate a value of x with thai value of y result- ing from its use in substitution. 29-1 SOLUTION BY COMPARISON AND FACTORING 295 Exercise 114 Solve ; 1. x + 2y = 5, 6. .v' + xy + y- = 19, 3? + ^xy=l. ■" + ?/ = 5. 2. x-3y=2, 7. .if-3xy-y^ + 3 = 0, xy + y^ = i\. 3x + 2y-5 = 0. 3. 2 a; +3 2/ = 12, 8. 3 a; + 2 7/ = 5, xy-& = 0. 2-a;!/-2/2 = 0. 4. 3x + 2y = 9, 9. j;2 + 2/2 + a,+2/ = 6, xy — x—2. .T — 42/ = 6. 5. a; + 2?/ + l = 0, 10. 2x + 3y + 7 = 0, 3xy-f- = -l. 3x^+2 rf-x + Ay=:12. SOLUTION BY COMPARISON AND FACTORING 335. When both equations are of the second degree and both are homogeneous. Solve P^-^^ = ^«' (1) [ 3.x- -2/2 = 11. (2) A factorable expres.sion in x and y re.sults if the constant terms are eliminated from (1) and (2) by comparison. Multiplying (1) by 11, Multiplying (2) by 10, Equating left members. From which, Factoring; (5), Therefore, And, 22 x^ - llxy = 110. (3) 30X2_ 10 1/2=: 110. (4) 30 a;2 _ 10 y^ = 22 x^ - - 11 zy. 8x^ + 11 xy - 10 y' =0. (5) (ix + 2y){8x-5y) =0. •=-|- (6) .='t- (7) 296 SIMULTANEOUS QUADRATIC EQUATIONS By substitution in (1): Again in (1) : From (6) : If2/=-| From C7): M2/-Y' 2x2- x(-|) = IO. 2x2-x(«^)=i;. 2x2 + ^ = 10. 2x2 ^^^=10. 5 8 x2 = 20. 2 x2 = 50. x2 = 4. x2 = 25. Hence, in (6), x=±2. x=± 5. Hence, in (7), "-1- -^(±-2)=Tl. y Tlierefore, X = i 2, X = ±5;1 Result. 2/=:F1, 2/ =±8. J Note. The sign ^ is the rtsult of a subtraction of a quantity having the sign ±. Thus, — (± a) = — (+ a) or — (— a) =— aor + o, =T a. Solve : 1 . a? + xy =Q, »2/ -2/^ = 1- 2. 3? — xy = 4, 2 3/2 — a;;/ = 6. 3. ic2 _ 2 a;;/ = — 8, 2/2 — 3 a^ = — 9. 4. a52 + 2«2/ = 9, 3 «!/ — 2/2 = - 4. 5. :r? -\-hxy = — 6, .^2 + 2/2 = 5. Exercise 115 6. v'-xy->ry^ = \, ^-\-xy.-\-f=\- 7. a;2_3a;2/ + 2/-' = -5, 2 a;2 + a;)/ - 2 z/- = - 4. 8. a;2 ^ a;,^ _ 4 y^ — ig^ a;2 — an/ — 3 2/^ = — 9. 9. 10af + 7xy-y' = -n, 12x^ + 9 xy-y- = -lb. 10. 2x^ + 4:xy + Ty^ — ^~> = 0, .5ar' + 3a;2/-32/'-15 = 0. SOLUTION OF SYMMETRICAL TYPES 336. When the given equations are symmetrical witli respect to X and y ; that is, ivhen x and y may he interchanged without changing the equations. SOLUTION OF SYMMETRICAL TYPES 297 1. Solve . xy — 10. Squaring (1), Multiplying (2) by 4, By subtraction, Extracting square root, Adding (1), a;2 + 2 a;!/ + y2 = 49. 4 xy = 40. a;2 _ 2 xj; + ?/2 = 9. Subtracting (1) from (3), 2x=+3 + 7, or-3 + 7 2 X = 10, or 4. X = 5, or 2 ; 1 2? = 2, or I 2;1 5. I Result. 2. Solve ■xy = 12, .0^ + 2/^ = 25. Multiply (1) by 2, 2 xj^ = 24. Add (2) and (3), x^ + 2xy -^ y^ = 49. Subtract (3) from (2), x^ - 2 xy + y^ = 1. Extract square root of (4), x + y =±7. Extract square root of (5), ^ — 2/ = ± 1. Four pairs of equations result in (6) and (7), viz. : x + y=7 x + y= 7 x + y = -7 X — y = 1 x — y =—1 x — y= 1 x = 4 x=3 x=— 3 y=3 y= i y=-4 Hence, the corresponding values for x and y are a;=±4, 8/=±3,i a; = ±3; 2/=±4.J (1) (2) <8) Result. (1) (2) (3) (4) (5) (6) (7) x + y = x — y = x = y = -7 -1 -4 Solve: 1. x + y = S, xy = 6. 2. .v + .v = 5, .1-^-4 = 0. Exercise 116 3. ar' + y^=.8, x + y = 3. 4. xy—3=0, x^+rf = 10. 298 SIMULTANEOUS QUADRATIC EQUATIONS 5. 3? + xy + y'^ = 1, T. a^ + y^ = 5, x + y = 3. (x+yy = 9. 6. x + y — 5 = 0, 8. x^ + 3xy + y^ = 59, x^ —xy + y'^T. af + xy+f = 39. SOLUTIONS OF MISCELLANEOUS TYPES 337. Systems of simultaneous quadratic equations not con- forming to the three types already considered are readily recognized, and the student will gradually gain the experience necessary to properly solve such systems. No general method for these types can be given. 338. It is frequently possible to obtain solutions of those systems in -which a given equation is of a degree higher than the second, derived equations of the second degree resulting from divisions and multiplications. Illustrations: 1. Solve l^ + f + ^- + ^y = ^^> (1) [xy-6 = 0. (2) Transposing and multiplying (2) by 2, 2xy = 12. (3) Adding (1) and (3), x^ + 2 xy +'y'^ + 3 x + 3 j/ = 40. Whence, (x + y)^ + 3 (x + y) - 40 = 0. Factoring, (x + ;/ + 8) (x + J/ - 5) = 0. Hence, x + y=—%, OTx + y = b. Combining each of these results with (2), we have two systems : (a) X + 2/ = - 8, (6) X + !/ = 5, xj/ - 6 = 0. xj/ - 6 = 0. The two systems (a) and (J)) may be readily solved by the principle of Art. 340. 2. Solve .■-V + 5-2/ = H (1) -.7 = 1. (2) From (1), xV + 5x?/-14=0, {xy + 7) {xy - 2) = 0. SolA'TlONS OF MISCELLANEOUS TYPES 299 Whence, xy = 2, ot xy = — 7. Coinbiuing each of these derived equations with (2), we have, (a) xy = 2, (6) xy=-l, X — y = 1. c — y = 1. Solving these two systems, we find from (a), -j- = 2, or — 1, 1, or - 2 ; from (6), x=^±^ -l±3V-3 ^ 2 ■ Result. Solve 1^-^ + 2^ = 82, I a; + « = 4. Let and Then in (1), Or, Whence, x = u + v, y = u — V. (u + vy + (tt - vy = 82. u* + 4u'v + 6 uH^ + iuv^ + »* + u* — ^u^v+ 6u2b2 — 4?(iv' + V* 2 II* + 12 «V^ + 2 D* = 82 u* + 6mV -f-«4 = 41. Substituting in (2), (1) (2) u + v-\-u — v = 4. u = 2. (5) Substituting (5) in (4), (2)« + G (2)2c2 + «< = 41. Or, V* + 24 v2 - 25 = 0. Whence, v=±'[, or ±5\/— 1. Therefore, if « = 2 and w =it I, And, if u = 2 and t) =± 5v'- 1, K = !( + u = 2±l=3orl, x = u + v = 2± SV— 1, !/ = « — M = 2Tl = lor3. y = u—v = 2^ 5V— 1. Hence, in brief form, x=S, 1, or 2^5^— 1 ; 2/=l,3, or2T5v''^. Result. 300 SIMULTANEOUS QT^ADRATIC EQUATIONS Ezercise 117 (See also page 414.) Solve : 1. x + y = 7; xy = 12. 9. x + ij = 10; x' +y^ = 58. 2. x + y = 13; xy = 36. 10. x — y^i; x' + y'' = iO. 3. x-y = i; xy = 21. 11. a;-2+?/-'=-^f ; a; '+2/-i=|, 4. a;-i+?/-' = 7; a!-'2/-i = 12. 12. a; + ?/ = 7; af" + .V* = 91. 5. x'^ + y^ = 4:0; xy = 12. 13. a^ — 2/^ = 19; x — y = l. 6. •a;-2+2/-2=13 ; a;-V' = 36. 14. .x' + 2/" = 17 ; a;?/ = 2. 7. xi+yi = 5; a;V = 6. 15. a;* + 2/'' = 82; 3- + 7/ = 2. 8. x~^+y~^ = 7; x~'-y-i = 10. 16. a:^ + «/* = 211; a; + 2/ = 1. 17. x'+y'' = 13; 2x-3y = 0. 18. a;2 + 3a;?/4-2.v^ = 3; 2a;2 + 2/2 = 6. 19. a;2-a;2/ + ?/2=7; 3 a;^ + 13 a;,?/ + 8 2/' = 162. 20. 0)2 + a;;/ = 56; xy + 2y^ = 60. 21. a;22/ + a;2/=' = 30; .ry + a;y = 468. 22. x^' + xz + z^'=30■, a'= - .ra + z^ = 18. 23. a;^ + 2/" = # a;.?/ ; a; — ?/ = i xy. 24. a;2 + a;2/+2/2=91; x + y/xy + y = 13. 25. a;2 + 2/2 + a: + 2/ = 2, a:2/-2 = 0. 26. (a; — 2/^ — a;y =5; K — 2/ + a;2/ = l. 27. a;^ + a;y + y^ = 19; x''~xy + y^ = 7. 28. a; + 2/ = 4; a;^— 2 a;2/ + 3 !/ — a; = 3. 29. .'B2 + 2/2 + a; + 2/ = 18; a;^ - 2/^ + a> - 2/ = 6. 30. x + y = 7; (x -If + (y ~2y = 28. 31. a;2 — a;?/ + 2/'' = 3 ; x — xy + y = l. SOLUTIONS OK MISCELLANEOUS TYPES 301 32. x^+y^ = xy + 7; x+y = xy — l. 33. a? + f = 35; x''y + xy' = 30. 34. x'' + y''-x—>/ = 18; xy + x + y = 19. 35. x^ + y^=20; xy — x — y='2. 36. X — y/xy + y = 7; x+ ^xy + )/ = 19. 37. x-\-y + y/x + y = 12; xy = 20. 38. x^ + y^ = 7 + xy; ii? + f = 6xy — l. 39. x"^ — xy + y = — G; y — xy — x = 12. 40. a;^-2/2_(a; + 2/) = 8; {x-yy(x+ y) = 32. 41. a;'' — a; + 2/2-2/-22 = 0; a;+?/ + a;?/ + l=0. 42. 6a;2 + 2/- = 5a;(/ + 12; 2 xy + 3x^ = 3 i/ - 3. 43. .c2 + 2a-(/ + 2/2=120-2a;-22/; xy-y^ = 8. 44. 3 a;2 + 2 a;^/ - 2 2/2 = 14(a; - y) ; 2 a;^ + aiz/ - 3 y^ = 7(a; - i/). 45. x'^ + y'' + xy = 39; x — y = 3. 46. .); + 2/ = *2/'~lj 5 x — y — xy = 1. 47. a;-^+2 2/-* = ^; 2a;-^ -2/-^- = |. 48. (a;-2/) + (a;-.v)i = 12; (x + y)-{x + y)i = 2. 49. p2 + g2 +-^ + g + 14 = 0; />g = 7. 50. a;y + a;-2/ = 5; a;]/(a; — 2/) = 6. 51. y{x-y)=2; x{x+y)=15. 52. ?/ = a;2-3a; + 2; 3y-2a;=-2. 53. a? + y'^ - x — y = 78; xy-\-x + y = 39. 54. 4a;2-25?/'' = 0; 2 a;^- 10 2/2_32/ = 4. 55. 2a;'' + 3.r]/ + 5)/2 + 3 2/ = 8x + 2; 5a; + 2?/-l = 0. 302 SIMULTANEOUS QUADRATIC EQUATIONS GRAPHS OF QUADRATIC EQUATIONS IN TWO VARIABLES TYPE FORMS OF EQUATIONS AHD CORRESPONDING GRAPHS (1) Type Form r ^ ::::: ::-:f|^ :::::: :::::-z!::::!l3::::: 2 % ^ ^ )j:___: X L_ _j A ' : :: _s-::: iciz ::: : -'•Se^^ ?" jr- +/- = c. Illustration : Plot the graph of a^ + y" = 25. Prom the equation we have If ' = ± V2o - a;=0, y = ± V'25, or 5. P. a;=l, ;/ = ±-\/24, or ±4.89+. Pj. a;=2, ;/=± V^r, or ±4.06+. ^2. a;=3, ?/= ± v'lO. or ±4. P3. a;=4, ?/=±\ ii, or ±3. etc. Plotting these points, we ob- tain a c!)T?p as the graph of the equation, :^-yy- = 25. 339. It will be seen that the coordinates of any point (a;, y) are legs of a right triangle whose hypotenuse is the dis- tance from the origin to the point (a;, y). That is, for any point on the curve we have (see figure), or, ^-\-y' = 2h. q In general, therefore : 340. Tim graph of any equation in two variables in the form ay' + y'—c is a circle. GRAPHS OF QUADRATIC EQUATIONS 303 (II) Type Form ■■■ y'= ax -\-c. Illustration: Plot the graph of If j;=0, y = ± V8, or ±2.8+. Pa. x= l,2/=±vT2, or ±8.4+. P3. =i:=-l, (/=±V4, or ±2. Pj. a;=-2, !/=0; etc. P. (Note the enlarged scale.) Plotting the.se points, we ob- tain a parabola as the graph of the equation, x/' = ix + %. Y — g ::::-:-::±:^ =^^f? c / r SI E P B L. ^^ ^ '''^^ ^•H Y" It will be seen that if x is less than — 2, y is imaginary ; hence, no point in the curve lies to the left of P. From this type form and graph we have, in general : 341. ,,2 77ie graph of any equation in tiro variables in the form ')f = ax-{-c is a jjarabola. 1 1 1 1 1 ^ Ti n — I III g--jB^ , (III) Type Form ••• ax^ + hy^ = c. Illustration : Plot the graph of 4a^-l-9t/2 = 36. If!/ = 0, x2 = 9, a; = ±3. A. x = 0,yi = 4,y=±2. B. Hence, s = + 3 and — 3, the points where the graph cuts XX ; and, ?/ = + 2 and — 2, the points where the grapli cuts YY'. 304 SIMULTANEOUS QUADRATIC EQUATIONS For any other points let x = ± 1 . Then, 9 ifl = 32, ?/ = ± * ^21 or ?/ = ± 1 .9+. Hence, for P, Pi, Pa, and Pa, we have (1, 1.9), (1, - 1.9), (- 1, 1.9), (— 1, — 1.9) respectively. Plotting these points, we obtain an ellipse as the graph of the equation 4 a;3 + 9 !/2 = 36. 342. Assuming values sufficiently greater or less than those by which the points above are obtained, we can show that no points in the graph can lie to the right or left, above or below, the intersections with the axes of reference. From this type form and graph we have, in general : 343. The graph of any equation in two variables in the form ax' -\-hy^ = c is an ellipse. (IV) Tjrpe Forms • • • ax'^ — b/' = c and xy = c. Illustrations : 1. Plot the graph of From the given equation ,,■2 a;2 - 1 " 4 If a; = ±l, 2/2 = 0, y=0. a; = ±2, !/2=J, 2/=±|V3, or ±.86+. For the values of x we plot A, (1,0), andP, (-1, 0). AlsoP,(2, .86);Pi,(-2,.86); Pa, (-2, -.86);P3,(2,-.86). With these points we obtain an hyperbola as the graph of the equation, x'^ — iy'^ = l. 344. It is found by trial that any value of x between + 1 and — 1 gives an imaginary value for y, hence no part of the curve can lie between the points A and B. s ^^s ^^^ *»je p^'' .- iir -^^ s^: 5B__ftt X / 5 S, x-"^ ^V ^^:^ n"''^^ X ^s^ r- GRAPHS OF QUADRATIC EQUATIONS 305 2. Plot the graph of If 2. = -4, ... ". + 4, etc., y = - 1, ... ±co ... + 1, etc. Plotting the points, ( — 4, — 1), etc., we obtain an hyperbola whose branches lie in the angles rOXand T'OX'. From the two cases : 345. TJie graph of any equation in two variables in Y S •v ^ ^ - X ' X ^ ^ s \ ^ \ 1 r Y' the form ax' — by^ = c, or in the form xy = c, is an hyperbola. Plot the graph of : 1. af + y^ = 16. 2. ar^ + / = 49. Exercise 118 3. 2/^ = 4 a; + 4. 4. x'' + 6y' = 9. 6. a^-3f.= 2. 6. xy = 10. S SOLUTION OF SIMULTANEOUS QUADRATIC EQUATIONS BY GRAPHS Illustrations : m 7 ¥ -^^. ^ 1. Given a;2 + 2/2 = 25, 3a;-22/ = 6. From the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. In the figure the graph of a;^ + 2/2 = 25 is a circle, and the graph of 3x — 2y = 6 is a straight line. SOM. EL. ALG. 20 306 SIMULTANEOUS QUADRATIC EQUATIONS m By measurement of the graplis we find the coordinates of /\ and Po to be (4, 3) and (— y|, — jf )• Solving the equations, we obtain x = 4, y = 3; or a; = — J[|, ?/ = — ^|. The accuracy of this and of subsequent cases may be increased by plotting on a larger scale. 2. Given x- + y^+9x + U = 0, y'=ix + 16. From the intersections of the graphs of the given equations obtain the roots, and cheek J I I I I \ I I I / I I |~ | I -\-\~\ the results by solving the J^ t7 equations. Z_5sl — ^3^v-------- ^^ *^® figure the graph ofar' + 2/^ + 9a; + 14 = is a circle and the graph of y' = ix + lQisa, parabola. By measurement of the graphs we find the coordinatee of Pj and P^ to be (— 3, 2) and (—3, —2) respectively, and these coordinates correspond to the real roots obtained from the solution. For the solution of the system gives x = — 3, ?/ = ± 2, or a; = — 10, 2/ = ± V— 24. We cannot find a point corresponding to the imaginary root, and we make the important conclusion that : 346. Since there can be no point having one or both coordinates imaginary, the graphs of two equations can have no intersections corresponding to imaginary roots. 3. Given ^2 + 4 2/2 = 17, xy = 2. From the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. GRAPHS OF QUAUKATIC EQUATIONS 307 In the figure the graph of 05^ + 4^^ = 17 is an ellipse, and the graph of xy ■= 2 is an hyperbola. By measurement of the graphs we find the co- ordinates of Pi, P2, Pjj and 1\ to be (4, ^), (1, 2), (-4, -h), (-1, -2) respectively. The solu- tion of the equations gives a;=±4, 2/=±|, ora5=±l, 2/= ±2. It will be noted that this last case is the first in which both equations are homogeneous and of the second degree, and that the graphs serve to emphasize once more the importance that attaches to the association of corresponding values of x and y in the solution of a system of simultaneous quadratic equations. 4. Given 5a; + 22/ = 10. From the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. In the figure the graph of 2x'-\-2y^ = 125 is an ellipse, and the graph of 6 a; + 2 ?/ = 10 is a straight line. !¥ ^ r :"::::;5b;:::::::: :::^2--:::^:::-s-::: .E^E:E:::-5--:5:. :E^,:EEE:-::^-^^?E: EEEE--EE-E:E|eEEE ?^ 308 SIMULTANEOUS QUADRATIC EQUATIONS Y "'s, S^ x^ " " sP Z :^,, 7 ^s ^t ^ 5s S± Q_ 5-___x ± ^v t J ^ 3 -.t _ _S^ ___^Z ?^ By measurement of the graphs we find the coordinates of Pi and Pj to be (0, 5) and (3.7+, 4.4-) respectively. The solution of the equations gives X = 0, y = 5, or .c = 3.75+, 2/ =4.38+. 5. Given, x' + y' = 100, 3a; + 42/ = 60. Trom the intersections of the graphs of the given equations obtain the roots, and check the results by solving the equations. In the figure the graph of x^ + y^ = 100 is a circle, and the graph of Sx + iy = 60 is a straight line. Solving the system, we find that the equation resulting from substitution gives equal roots for y. Hence, for x we find but one value, the solution of the system being x = 6, y = 8. It will be seen from the graph that the circle and straight line have one point only in common ; that is, the line is tangent to the circle. The coordinates of the point of tangency are found to be (6, 8) or P. In general : 347. If, in the solution of a system of equations, a derived equation has equal roots, the graphs of the equations are tangent to each other. 348. It will be helpful to the student in the exercise follow- ing if the type forms of equations and the corresponding graphs are remembered. 1 . ax + by = c, The Straight Line. 2. jr2+/ = c, 3. y = ax + c, The Circle. The Parabola. GRAPHS OF QUADRATIC EQUATIONS 309 4. ax'^ + 6/ = c, The Ellipse. ■^ ' t ve Hyperbola. o. xy = c Exercise 119 Plot the graphs of the following, and check the roots deter- mined by a solution of each system : 1. a?-{.y'- = 25, 5. a;2 + ?/H- « + 2/ = 18, a; — ?/=l. xy -\- X -\- y ^ 11. 2. a;2+t/2=74, 6. ?/- = 4.e + 8, !CT/ = 35. a; + 2/ — 6 = 0. 3. 2/^=8 a;, 7. 9 a;^ _ 16 7/^ = 144, X — y=&. a? -\- y- = 36. 4. a; + 2/ = 12, 8. a^ + 4 ?/ = 4, a;?/ = 32. a:^ + y^ = 17. 9. ar^ +2/2 = 32, a;2-a;2/ + 2/2 = 28. 10. Solve the system, x+ y = and xy = 4, and determine if the solution alone will show that the graphs intersect, or do not intersect. 11. How do the graphs of xy = i and x^ — y^=l& differ in their positions relative to the axes of reference ? 12. Can you describe the position of the graph of the equa- tion, 3? -\- y^ — 2x = Q, without plotting it ? 13. How do the graphs of the equations, a;^ 4- 2?/^ = 32 and 4 x^ + 2/2 = 16, differ in their relative positions ? 14. Plot the graphs of a;^ _ a;^^ -^ ^^2 = 28 and x — y = Q, and show that their intersections check the roots found by solution. 15. In how many points may the graphs of a!^ + t/^ = 25 and x^ 4- 2 2/2 = 64 intersect? Prove your answer by plotting. 310 SIMULTANEOUS QUADRATIC EQUATIONS 16. Show the positions of the graphs of x' + y^ = 100 and .!•/■ -|- 4 y^ = 100, and determine the number of real roots. 17. Show by solution that j/^ — 4 a; — 12 and a;^ + y^ + 3 a; = can have but one point in common, and prove your answer by a graph of the system. PROBLEMS PRODUCING SIMULTANEOUS QUADRATIC EQUATIONS Exercise 120 (See also page 415.) 1. Divide 20 into two parts whose product is 96. 2. The sum of two numbers is 16, and the difference of their squares is 32. Find them. 3. Find two consecutive odd numbers, the sum of whose squares shall be 74. 4. Divide 9 into two such parts that the product of their squares shall be 196. 5. Find the two numbers whose squares, added, equal 130, but whose squares differ by 32. 6. The prodvict of two numbers is 20, and their sum multi- plied by- their difference is 9. What are the numbers ? 7. Divide 18 into two parts such that their product plus the sum of their squares shall be 244. 8. There are two numbers whose difference is 2, and the difference of whose cubes is 56. Find them. 9. The hypotenuse of a right triangle is 25 feet, and the area is 160 square feet. Find the lengths of the sides. 10. Find the two numbers whose difference multiplied by the smaller produces 54, and whose difference multiplied by their sum produces 144. 11. The sum of the squares of two numbers is 2a^ + 2, and the sum of the numbers is 2 a. Find them. PROBLEMS PRODUCING QUADRATIC EQUATIONS 311 12. Two cubical coal bins together hold 280 cubic feet of coal, and the sum of their lengths is 10 feet. Find the length of each bin. 13. There are two numbers whose difference is 4, and the difference of their cubes exceeds the difference of their squares by 276. Find them. 14. A number of two digits is equal to 4 times the sum of its digits, and the sum of the squares of the digits is 45. What is the number ? 15. 168 feet of fence inclose a rectangular plot of land, and the area of the plot is 1440 square feet. Find the dimensions. 16. The sum of the squares of two numbers is 25, and the sum of their fourth powers is 337. What are the numbers ? 17. The product of two numbers is 16 more than 5 times the larger number, and 6 less than 16 times the smaller num- ber. Find the numbers. 18. In a number of two digits the units' digit is 3 times the tens' digit, and if the number is multiplied by the sum of the digits, the product is 52. Find the number. 19. If a number of two digits is multiplied by the tens' digit, the product is 96; but if the number is multiplied by the units' digit, the product is 64. Find the number. 20. The simple interest on $600 for a certain number of years at a certain rate is $120. If the time were 2 years shorter, and the rate 2 oj^ more, the interest would be f 108. Find the time and the rate. 21. Two men working together can paint a house in 6 days, but one of them working alone would require 5 more days than the other if he worked alone. How many days would each working alone require for the task ? 312 SIMULTANEOUS QUADRATIC EQUATIONS 22. If the dimensions of a rectangle were each increased 2 feet, its ai'ea would be 165 square feet. If each were de- creased 2 feet, the area would be 77 square feet. Find the dimensions of the rectangle. 23. If the sum of two numbers is added to their product, the result is 47. The sum of the squares of the numbers is 62 more than the sum of the two. What are the numbers ? 24. The area of a rectangular field is 15 acres, and the perimeter is 200 rods. What are the dimensions of the field, and what is the length in feet of the diagonal ? ' 25. The combined capacity of two cubical tanks is 637 cubic feet, and the sum of an edge of one and an edge of the other is 13 feet. Find the length of a diagonal of any face of each cube. 26. The sum of the areas of two square floors is 1300 square yards, and the combined length of the two perimeters is 600 feet. Find the area of each floor in square yards. 27. A web of cloth costs 50 cents more per yard than an- other, and each web costs $ 63. One web is 8 yards longer than the other. How many yards of cloth are there in each web ? 28. The sum of the squares of the two digits of a number is 13. If the square of the units' digit is subtracted from the square of the tens' digit, and the remainder is divided by the sum of the digits, the quotient is 1. Find the number. 29. The sum of the squares of two numbers is increased by the sum of the numbers, and the result is 18. The difference of the squares of the numbers is increased by the difference of the numbers, and the result is 6. Find the numbefs. 30. A certain floor having an area of 50 square feet can be covered with 360 rectangular tiles of a certain size ; but if the masons use a tile 1 inch longer and 1 inch wider, the floor can be covered with 240 tiles. Find the sizes of the different tiles. PROBLEMS PRODUCING QUADRATIC EQUATIONS 313 31. The sum of the volumes of two cubical blocks of stone is 2240 cubic inches, and the sum of an edge of one and an edge of the other is 20 inches. Find the volume and the total suiface of each cube. 32. A field is 30 rods long and 20 rods wide, but its length is decreased and its width increased so that its area is 44 square rods greater. The change increases the perimeter by 2 rods. What amount is added to the length and subtracted from the width ? 33. A bicyclist starts on a 12-mile trip, intending to arrive at a certain tima After going 3 miles, he is delayed 15 min- utes, and he must complete the journey at a rate 3 miles an hour faster in order to arrive on time. Find his original rate of speed. 34. If the difference of the squares of two numbers is di- vided by the smaller number the quotient and the remainder are each 4. If the difference of the squares of the numbers is divided by the greater number, the quotient and the remainder are each 3. What are the numbers ? 35. A boatman rows 18 miles downstream in 6 hours less time than it takes him to return. If he were to double his ordinary rate, his rate downstream would be 10 miles aii hour. Find his rate of rowing in still water, and the rate of the stream. 36. Find the sides of a rectangle whose area is unchanged if its length is increased by 4 feet and its breadth decreased by 3 feet ; but which loses one third of its area if its length is increased by 16 feet and its breadth decreased by 10 feet. 37. Two bodies move toward each other from A and B and meet after 35 seconds. If it takes one 24 seconds longer than the other to move from A to B, how long does it take each to traverse the distance ? 314: SIMULTANEOUS QUADRATIC EQUATIONS 38. A and B start simultaneously from two towns to meet one another. A travels 1 mile an hour faster than B, and they meet in 4 hours. If B had increased his rate by 1 mile an hour and A had traveled at f of his former pace, they would have met in 4 hours and 16 minutes. How far apart are the towns? 39. If the length of a certain rectangle is increased by 2 feet, and the width is decreased by 1 foot, the area of the rectangle will be unchanged. The area of the rectangle is the same as the area of a square whose side is 3 feet greater than the width of the rectangle. What are the dimensions of the rectangle ? 40. A certain principal at a certain rate amounts to $ 1560 in one year at simple interest. If the principal were greater by $ 100 and the rate 1^ times as great, the amount at the end of two years would be $ 1792. What is the principal and what is the rate ? 41. A rectangular piece of tin is made into an open box by cutting a 5-inch square from each corner and turning up the sides and ends. If a 3-inch square were cut from each corner, the box, made in the same way, would hold the same amount. The tin is 4 inches longer than it is wide. How much does the box hold ? 42. A certain farm is rectangular in shape, and its length is four times its width. The farm cost as many dollars per acre as it is rods in length, and the number of dollars paid for it was four times the number of rods in its perimeter. Find the length and the width of the farm in rods, and the price paid for it. 43. Twenty persons contributed to the purchase price of a gift, one half of the whole amount being furnished in equal portions by the women of the partj"^, and the other half by the men. Each man gave one dollar more than each woman, and a total of $48 was paid for the gift. Find how many men and how many women contributed to the cost, and the amount that each gave. CHAPTER XXIV RATIO. PROPORTION. VARIATION. RATIO 349. If a and b are the measures of two magnitudes of the same kind, then the quotient of a divided by b is the ratio of a to 6. Ratios are expressed in the fractional form, -, or with the b colon, a: b. Each form is read "a is to b." 350. In the ratio, a : b, the first term, o, is the antecedent, and the second term, b, is the consequent. THE PROPERTIES OF RATIOS 351. The properties of ratios are the pivperties of fractions; for the ratios, -, m:n, (x + y): (x — y), etc., are fractions. b (a) The Multiplication and the Division of the Terms of Ratios 352. The value of a ratio is unchanged if both its terms are multiplied or divided by the same number. 353. A ratio is multiplied if its antecedent is multiplied, or if its consequent is divided, by a given number. 354. A ratio is divided if its antecedent is divided, or if its consequent is multiplied, by a given number. S15 816 RATIO. PROPORTION. VARIATION (6) Increasing or Decreasing the Terms of a Ratio 355. If a, b, and x are positive, and a is less than b, the ratio a:b is increased when x is added to both a and b. -p, a + x a _ x(b — a) °' b + x b^blb + x)' And, since a 6. 357. An inverse ratio is a ratio obtained by interchanging the antecedent and the consequent. Thus, the inverse ratio of m : n is the ratio n-.m. 358. A compound ratio is a ratio obtained by taking the product of the corresponding terms of two or more ratios. Thus, mx ; ny is a ratio compounded from the ratios, m ; n and x : y. 359. A duplicate ratio is a ratio formed by compounding a given ratio with itself. Thus : a^ : 62 jg the duplicate ratio of a : 6. In like manner, a^ : b'^ is the triplicate ratio oi a:b. Exercise 121 1. Write the inverse ratio of a : a; ; of m : n ; of 7 : 12 ; of 3a;:5a;; of (2a + l) : (2a- 1); oi (x^ + xy + y'): (x^-xy +f). 2. Arrange in order of magnitude the ratios 2:5, 3:7, 4 : 9, 5 : 8, 10 : 17, 12 : 19, 21 : 27, 32 : 39, and 40 : 51. 3. Compound the ratios 3 : 7 and 10 : 17. 4. Find the ratio compounded of 3 : 8, 4 : 9, and 6 : 11. PKOPORTION 317 5. Compound the ratios (a^ — 9) : (a;" + 8) and (x + 2):{x—3). 6. What is the ratio compounded from the duplicate of 2 : '3 aud the triplicate of 3 : 2 ? 7. Find the value of the ratio (a; + 6) : (x' + 7 a; + 6). 8. Two numbers are in the ratio of 4:7, but if 3 is added to each number, the sums will be in the ratio of 5 : 8. Find the numbers. 9. The ratio of a father's age to his son's is 16 : 3, and the father is 39 years older than the son. Find the age of each. 10. In a certain factory 5 men and 4 boys receive the same amount for a day's work as would be paid if 3 men and 12 boys were engaged for the same time. What is the ratio of the wages paid the men and the boys individually ? PROPORTION 360. A proportion is an equation whose members are equal ratios. Thus, the four numbers, a, 6, c, and d, are in proportion if — = - • d 361. A proportion may be written in three ways : (1) - = -. (2) a:b = c:d. (3) a:b : :c:d. b d Each form is read " a i.s to & as c is to d!." We understand the mean- ing of a proportion to be that the quotient of a -f- 6 is the same in value as the quotient of c -=- [■ •••)• {h + d + f+—)=a:b. Proof: Since ? = ^ = ^ = ..., oaf then, ^ = r, - = r,^ = r, etc. b d f Whence, a = br, c = dr, e =fr, etc. Adding, a + c + e + ■■■ = br + dr -\- fr + —. Whence, a + c + e + ■■• =(6 + d +/+ •••)»•. And, ° + '= + ^ + - = r=g. Or, (a + c+e+ •••): (6 + (^ +/+•••) =0:6- That is : /n a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. 382. Given a:b = b:c. Then a : c=a^ :b\ Proof : Since It follows that. Whence, Or, a _b b c' O V, b. _a a _ - X b c ~6 b' a a2 c 62' a : c = a2 :62. APPLICATIONS OF THE PKOPKRTIES OF PROPORTION 323 That is : If three numbers are in continued proportion, the first is to the third as the square of the first is to the square of the second. 383. Given a:b = b:c = c:d. Then a.:d = a^:h^. Proof : Since o _ '< _ o bed it follows that, «x-x^"='?x5x-- h c d b b b Wlience, Or, a : d = (i3 : 63. That is : If four numbers are in continued proportion, the first is to the fourth as the cube of the first is to the cube of the second. 384. Giren a:b = c: d, e:f=g:h, k:l = m:n. Then aek: hfl = cgni : dhn. Proof: Since 2=^, 5 = 2, ^ = B. b d f h I n Multiplying, ?t = f^- bfl dhn Or, aek : bfl = cgm : dhn. That is : The product.^ of the corresponding terms of ttvo or more pro- portions are in proportion. APPLICATIONS OF THE PROPERTIES OF PROPORTION 385. Of the important properties of proportion in common nse in the solution of problems involving certain relations, we ma,y note briefly the following : (1) Since, in any proportion, the product of the means equals the product of the extremes (Art. oTl), we may readily find any one term of a proportion when three terms aie known. 324: RATIO. PROPORTION. VARIATION (2) Since — = - = r, we have a = br, and c= dr ; and the sutstitution 6 d of these values for a and c will be of frequent service in reductions. (3) Composition and division (Art. 379) are of frequent use in mini- mizing the work necessary for the solution of certain types of equations. (4) An assumed identity involving any four numbers, a, b, c, and d, is shown to be true if, by transformations, we obtain a proportion, a : 6 = c : d. Illustrations : 1. Find the ratio of a; to « when — — — ^ = - . ^ ix-Sy 6 In the form of a proportion we have 3x + 2y:4:X-3y = 5:6. By Art. 371, e(Sx + 2y) = 6{ix-3y). Whence, 18x + 12j/=:20x- 15?/. 2x = 2~y. Therefore (Art. 374), x:y = 21:2. ResiUt. 2. li a:b = c:d, show that o + 3c: & + 3^ = 2a4-c : 26+d Since a : 6 = c : d, we nave - = - = r. Whence, a = br, and c = dr. b d Reducing each ratio separately, a + Sc ^ {br) + 3(dr) ^ r(b + 3d) _^ b+3d b+Zd 6+3d 2 g + c ^ 2 (6)-) + (dr) _ r(2 6 4 d) _ ^ 26 + d 26 + d 26 + d a + 3c _2a+c Therefore, 3 Solve the equation 6 + 3d 26 + d 2a; — 1 _ x + A + 2a! — 1 ar + a; + 4' By composition and division (Art. 379), (2 3;-l) + (a:-^ + 2a:-l) _ (x + 4) + (a:2 + a: + 4) (2x-l)-(a:2 + 2K-l) (x + i)-{x:^ + x + i) ci.„„iif„:„„ x'^ + ix-2 x2 + 2a; + 8 Smiplifying, —^ — =^ — -i— . — x^ —x^ Whence, x^ + i:X — 2=x'^ + 2x-\-%. 2x = 10. x = 5. Result. APPLICATIONS OF THE PROPERTIES OF PROPORTION 325 4. Find two numbers whose sum is to their product as 3 is to 20; and whose sum increased by 1 is to their difference increased by 1 as 7 is to 1. Let a; and y = the two required numbers. Then, x + y:xy = Z : 20. (P.y the first condition.) (1) And, x-\y-\-\ -.x — y -\-l='i -.1. (By the second condition. ) (2) From (1), 20z + 20^ = 3a;2/. {?,) From (2), Zx-i:>j = -'A. (4) 42^-8 From (4), 3 Hence, in (3), 20/' ^^ ''^ \-\-Wy = Zy ( ^^ ~^ \. Simplifying, 12 y'^ - 149 ?/ + 60 = 0. (5) Solving (5), y = 12, or j^. X = 1-5, or — ^. It will be found that all four values satisfy the given conditions. Therefore, 12 and 15 are the required numbers. Exercise 122 Show that the following are true proportions : 1. 4 2. 2 3. 3 8 = 5:10. 4. 3:9.125 = 2:6^2-. 7 = 2.6:8.75. 5. 4x : 3 i/ = 8 a;;/ : e?/^. 5^ = 6:100,. 6- 2^ = ^=2c^.:^. Determine whether the following are true proportions : 7. 34:53 = 19:39. 9. 12.1 : 4.4 = 2.2 : .8. 8. 4J:4 = 8f:8i. 10. a^ - 1 : « + 1 = a;- 1 : a;. Find the value of the unknown in each of the following : 11. 8: 10 = 12: a;. 13. 32: 12 = a;: 6. 12. 4:a; = 16:15. 14. ?/:8 = 12:4. 32G RATIO. PROPORTION. VARIATION 16. 5: 2 = 10.5:. 3. 18. ic : a = 3x: 6ii. lo. 2a;:y = 6:3. ,, „ 1« o o 20. -^■.a + l = x:a'-l. 17. 3 a; ; 16 = .9: .2. a — 1 Find the ratio of a; to y in : 21. 5x = 7y. 2x + y _ m 3 a; — 2/ h 22. 3x + 2y=5y-3x. 25. a;- — 9 ?/"' = 8 a;?/. 23. 2x + y: 3x — y=^. 26. x — y:3y = y — x:x-\-y Find the value of a; and y in : 27. a; - 1 : 2/ + 1 = 2 : 3, 30. a; + 1 : 2/ + 1 = 3 : 4, a; + y = 5. a; : 2 = 2/ : 3. 28. a; + 2:y + 2 = 2:3, ^^ x + y+1 ^2 x+o ^^ 3 a; - ?/ + 1 = 0. • a; + 2/-I ■'!' y+2 29. X + y -.x — y = 3:2, x + 2y _ 5 a; + l _ 44 a; + l:2/-l = 3:2. 'a!-22/~C' 2/ + l~2l' Find a mean proportional between : 33. 20 and 5. 34. 3 and 27. 35. 4 a^^^ and 16 a^a;*. 36. ^ + 3- + ^ and ^•^'-^-- x2-3a- + 2 ar+2a;-3 37. ^-^ + ^ and ^ ■ x^-1 38. ^^^-l_and ^^ 5+4V2 4+3V2 Find a third proportional to : 39. 12 and 16. 42. (c + xf and c^ - r'. 43. ar' - 1 and x^ + a; + 1. 40. 3 a^ and 2 a?. 44. ^"^ + "rand("^'-"')- 41. 2: 8 and 3:6. ji^ 71^ 51. 2 : .-) = 3 — .1- : X. 52. .-1 : 3 = 4 - .r : x. 53. 6 : 7 = 12 ~y:y. 54. ia:3a = 10-x 55. a^: a^c = c — x : x APPLICATIONS OF THE PROPERTIES OF PROPORTION 327 Find a fourth proportional to : 45. 4, 7, and 9. 48. a;* — 1, a; + 1, and a; — 1. 46. 3 X, 2 (/, and 6 z. 49. c^ — d», c^ — cP, and c' + cd + d*. 47. I, f, and 4. 50. 1 +V2, 2 + V2, and 2 -VST Change the form of each of the following proportions so that the unknown quantity shall occur in but one term : 56. m : n =p — z : z. 57. 3 : 2 = a; + 1 : «. 58. 15 : 7 = a; + 1 : a;. 59. 3 a : 5 c = .« + 1 : a; — 1. 60. a''+l:a-— l = a;+l:«— 1. 61. c + d:c — d = c + y:c — y. 62. »)- + m + 1 : 7Ji- — m — 1 = .« — 2 : a; — 8. 63. Find two numbers in the ratio of 2:3, such that if each is decreased by 1 their ratio becomes as 3 : 5. 64. Find two numbers in the ratio of 4 : 7, such that if each number is increased by 2 the ratio becomes as 5 : 8. 65. If ad = bc, write all the possible proportions whose terms are «, b, c, and d. 66. Find two numbers whose sum is to their product as 8 : 15, and whose sum is to their difference as 4:1. 67. What number must be added to each of the numbers, 4, 14, 10, and 30, that the resulting sums may be in proportion ? 68. Separate 10 into two parts such that their product shall be to the difference of their squares as 6 : 5. 69. Two rectangles have equal areas, and their bases are to each other as 5 : 16. What is the ratio of their altitudes ? 70. The lengths of the sides of three squares are in the ratio of 2, 3, and 4. Find a side of each if the sum of the three areas is 725 square units of area. 328 RATIO. PROPORTION. VARIATION 71. If a : 6 = 2 : 3, and b:c = 3:5, find o : c. 72. If a; : 2/ = 3 : 4, and y: z = 5 : 6, find x : z. 73. If c:d = 3.2:4.8, and d: A: = 3.2 : 8, find c : A;. 74. If a : 6 = & : c = 2 a; : 3 y, find a : c in terms of x and y. 75. If m:m = 5:2, find (2m + «):(2m-n). 76. li a + h:h = x + y.y, find {x -\-y):(x — y) in x and j/. If a : 6 = c : d, 77. Prove that ab:cd = ¥: d''. Also prove that ab:cd = a^: c^. 78. Prove that cM -.h = hc?: d. Also prove that a + 6 : a = c + d : c. 79. Prove that a + c:h + d = c:d. Also prove that ■\/a?- — ¥ : Vc^ — d^ = b:d. 80. Prove that 3 a'' - 2b^:2b'' = Sac - 2bd: bd. Also prove that a^ + 2 a& : a& = c'' + 2 cd : cd. 81. Prove that a : & = c : d = VoM^ : V6' + dl 82. Prove that a'^ + ft^ : c^ + d^ = a^ - 6" ; g^ _ dl 83. Prove that {ma + nb): {ma — nb) = {mc + nd) : (mc — nd) 84. Prove that a:a + c = a + 6:a+6 + c + d. If a : 6 = 6 : c, 85. Prove that a — b:b — c = b:c. Also prove that a:a + b = a — b:a — c. 86. Prove that ab — b^:bc — d' = b^: c\ Also prove that a^ + ¥:b^ + d' = a: o. 87. Prove that a%a + &) : c(& + c) = a''(a — 6) : c{b — c). 88. Prove that a + b:b + c= Va^ - b'^ : V6^ — c\ 89. Prove that (a^ + &2)(62 + c2) = (a6 + fic)^. 90. Prove that a^ — W -.a^ + ¥ = a — c:a -\- c. APPLICATIONS OF THE PKOPERTIES OK PROPORTION 329 Solve the equations : 91. 2x4-3 :2ai + 5 = 3a; + 2 :3a; + 4. 92. 2a^ + 2x + l:23f-2x-l=^x + l:x-l. 93. X + m : X — m = m + II : m — n. 94. or — 1 : .« + 1 = .r + 1 : a- — 1. 95. a;^ + 2 a; + 4 : a;- — a; — 1 = u; + 2 x — 2. si? + x—'[ X- — x + 2 96. 97. 98. 99. 100. 3?—X+'6 a; + w + 1 _ x + n — 1 .g= + 2.-c + 3 a? +'6 x + 2 3a;^ + a; + l 4ar' + a; + l VSTl X 3 - + X-2 X — )l + 1 3 X — n — 1 .!•- + 3 X - 4 X- + 4 a; - 3 .),•= -X- 1 4ar'-.<;- -1 6 - V'.c + 1 4 + Va; + -' 3 + Va; + 1 101. Separate 100 into three parts which shall be in the ratio of 2 : 3 : 5. 102. Three angles of a certain triangle are in the ratio of 1:2:3. If the sum of the angles of a triangle is 180°, find the number of degrees in each of the angles. 103. The sum of three sides -of a triangle is 240 feet, and the ratio of the sides is as 3:4:6. Find the length of each side. 104. For what value of a will the quantity a — 1 be a mean proportional between the quantities a + 3 and a -|- 2 ? 105. If 2 is subtracted from the smaller of two numbers and 4 is added to the larger, the ratio is as 3 : 6, but. if 1 is subtracted from the greater while 1 is added to the smaller, the ratio is as 10 : 9. Find the numbers. 330 RATIO. PROPORTION. VARIATION 106. Two trains approach each other between two points 100 miles apart, and their rates of traveling are as 4 : 6. How many miles will each have traveled when they meet ? 107. Find two numbers whose sum is 10, such that the ratio of the sum of their squares to the square of their sum is 13 : 25. 108. Find the length of the two parts into which a line I inches long is divided if the parts are in the ratio of a : 6. 109. What must be the value of x in order that 2 a; — 1, 2 a; + 1, 2 a; + 5, and 2 (2 a; + 1) may be in proportion. 110. Find each side of a triangle whose perimeter is n inches, the ratio of the sides being as a : & : c. VARIATION 386. A variable is a quantity that, under the conditions of a problem, may have many different values. 387. A constant is a quantity that, in the same problem, has a single fixed value. 388. If the ratio of any two values of a given variable equals the ratio of any two corresponding values of a second variable, then the first quantity is said to vary as the second quantity. Illustration : Suppose an automobile is running at a speed of 10 miles per hour. The total distance traveled at the end of any hour depends upon the total number of hours that have elapsed since the start was made. If it runs 6 hours, the distance covered is 60 miles ; If it runs 9 hours, the distance covered is 90 miles. Clearly, therefore, the ratio of the two periods of time is the same as the ratio of the two distances covered. That is : 6 : 9 = 60 : 90. We may say, therefore, that the distance (d) varies as the time («), or d a: t. The symbol oc is read "varies as." VARIATIONS UNOEK DIFFERING CONDITIONS 331 389. If I'^-y, then x equals y multiplied by some constant quantity. I'roof : Let a and 6 denote any one pair of coiTesponding values of X and y. Then by definition, - =-. y b From which, x=-y- b Denoting tlie constant ratio. 5- bv c, x = cy. b In general : We may change any variation to an equation by the introduc- tion of the constant factor or ratio. 390. If one pair of coiTesponding values for the variables in a given rariation is known, the constant ratio is reaildy obtained. Illustration : If a; varies as y, and a; = 12 when y = S, find the value of x when y = 10. We have x = cy. Substituting, 12 = c x 3. Whence, c = 4. Hence, when «/ = 10, x = i y = 4 xlO = 4(1. Re.sult. VARIATIONS UNDER DIFFERING CONDITIONS (n) Direct Variation 391. The simple form, xxy, is a direct variation. Illustration: The distance covered by a train moving at a uniform rate of speed varies directly as the time elapsed, That is : dxt OT d = ct, where c is the constant ratio. 332 RATIO. PEOPORTION. VARIATION (6) Inverse Variation 392. A quantity is said to vary inversely as another quantity when the first varies as the reciprocal of the second. Illustration : The time («) required by an automobile going from New York to Phila- delphia varies inverselSr as the speed (s) ; for if the speed is doubled, the time required will be but one half the former time. That is : 1 c tx-OT t = -, where c is the constant ratio. (c) Joint Variation 393. If a quantity varies as the product of two quantities, the first quantity is said to vary jointly as the other two quantities. Illustration: The number of dollars (iV) paid to a motorman for a certain number of trips {t) varies jointly as the number of doUare paid to him for one trip (re) and the number of trips made. That is : Nxnt OT N= cnt, where c is the constant ratio. (rf) Direct and Inverse Variation 394. One quantity may vary directly as a second quantity, and, also, inversely as a third quantity. In such a case the quantities are said to be in direct and inverse variation. Illustration : In mowing a field, the time required for the work (J) varies directly as the number of acres in the field (A), but inversely as the number of mea engaged at the task (re) . That is : A cA to: — or « = ii=, where c is the constant ratio, n n VARIATIONS UNDER DIFFERING CONDITIONS 333 («) Compound Variation 395. The result obtained by taking the sum or the differ- ence of two or more variations is a compound variation. Illustration : If y equals the sum of two quantities, a and 6, and if a varies directly as a;' while b varies inversely as x^, then r' a = C3? and b = —■ Adding (since y^a + h), y = cx^ + —. It is to be noted that in such cases two different factors are necessary (/) Important Principle 396. If X depends only upon y and z, and if x varies as y when z is constant, and x varies as z when y is constant, then x varies as yz tvhen both y and z vary. Let X, y, z (1), Xx, y\, z (2), and xi, yi, zx (3) be three sets of corre- sponding values. Then, Since « has the same value in (1) and (2), £- = ?^. (I) a;i yi Since y, has the same value in (2) and (3), ^ = — . (II) X^ Zx Multiplying (I) by (II), 2. = Jl . Xi yizx Or, x:OCi = yz: yxZ\. That is : The ratio of any two values of x equals the ratio of the corre- sponding values ofyz, and, by definition, x varies as yz. Illustration : The area {A) of a rectangle varies as the base (S) when the height (S) is constant, and the area varies as the height when the base is constant. Therefore, when both the base and height vary, the area will vary as the base and height jointly. 33-4 KATIO. PROPORTION. VARIATION II III IV 8 8 s 5 10 IS 10 12 A = BM A = Bn A = BH A = BH = 10.5 =12.5 =10-8 =12-8 = 60 = 60 = 80 = 96. In II. B changes, ff constant, ^ changes (vlxjB, fi^ constant). In III. H changes, B constant, A changes (AxH, B constant). In IV. B changes, S constant, A changes {A cc BH, B and H vary). APPLICATIONS OF VARIATIONS ■397. Illustrations: 1. If X varies as 2/^ a^nd x = 2 when y = ^, find x when 3=16. Since we have And, if a; = 2 and 2/ = 4, Whence, Then, when y = 16, xxy', X = my^. 2 = m X 42. TO = f X = J (16)2. x = ^^. X = 32. Result. 2. If X varies inversely as y^, and x = 2 when ^ = 4, find X when y =2. Since *"^' we have Whence, and 2-i-8' TO = 128. Then, when y = 2, x = l??. 23 X = 16. Result APPLICATIONS OF VARIATIONS 335 3. If s is the Huin of two quantities, one of wliich varies directly as .r- and the other inversely as .-•, and if .s = 6 when X = 2, and s = 'i when a; == — 2, find s when a; = — 1. Let u and v represent the two quantities. Tlien s=u + »; From which, soca;2+ — X S = OTX-^ + ^. X 0) Substituting : Ifa; = 2, (2) If X = - 2, 2 = m(-2)2 + -^ = 4m 2 (3) From (2) and (3), m = 1, 11 = i. If a; = - 1, in (1), S=l(-l)2+-i- = 1-4 = - 3- Result. 4. The volume of a sphere varies as the cube of its diame- ter. If three metal spheres whose diameters are 6, 8, and 10 inches, respectively, are melted and recast into a single sphere, what is its diameter ? Let V denote the volume of the required sphere, and D its diameter. Then Vxl)\ Whence, V = mD\ (1) Denote the volumes of the three given spheres by Fi, Fa, and Vs We have, therefore, Fi = m(6)3 = 216 m, F2 = m(8)3 = 512j», F3 = m(10)3 = 1000 m, Whence, by addition, Fi + Fi + Fs = 1728 nt. But, by the conditions, Fi + V^+Vs^ V- in 7)s. (From 1.) 330 KATIO. PROPORTION. VARJA'l'ION Hence, mD'=n28m, D» = 1728, D =12. Result. That is, the diameter of the sphere obtained from the given spheres is 12 inches. Exercise 123 1. If a; varies as y, and a; = 10 when y = 2, find, x when y = 5. 2. If X varies as y, and x = 3.2 when y = 0.8, find x when y = 6.6. 3. If 0! + 1 varies as y — 1, and a; =6 when 2/ = 4, find x when y = 7. 4. If 2 05 — 3 varies as 3 y + 2, and y = 2 when x = 0.2, find y when x = 1.5. 5. If a;'' varies as y", and a; = 3 when y = 2, find y when a; = 4. 6. If X varies inversely as y^ and a; = 2 when y = 4, find x when 2/ = 3. 7. If a; varies inversely as y", and a; = 2 when y = ^, find ?/ when a? = 1-^. 8. If X varies jointly as y and z, and a; = 3 when 2/ = 4 and « = 2, find a; when y = 5 and 2 = 4. 9. If X varies inversely as y^ — 1, and a; = 4 when y=5, find X when y = 15. 10. If X varies as -, and y= —2 when x = 7, find the equa- y tion joining x and ?/. 11. If the square of a; varies as the cube of y, and if a!= 6 when 2/ = 4, find the value of y when x = 30. 12. If s is the sum of two quantities, one of which varies as X while the other varies inversely as x ; and if s = 2 when x = ^ and s = 2 when a; = — 1, find the equation between s and x. APPLICATIONS OF VARIATIONS 337 13. If w varies as the sum of x, y, and z, and w = 4 when •>:='', y = — -!, and z = 5, lind x if lo = — S, y= 2, and «= -6. 14. Given that s = the sum of three quantities that vary as X, a^, and a;'', respectively. If u; = 1, s = 3 ; if a; = 2, s = 6 ; and if a; = 4, s = 16. Express the value of s in terms of x. 15. The area of a circle varies as the square of its diameter. Find the diameter of a circle whose area shall be equivalent to the sum of the areas of two circles whose diameters are 6 and 8 inches respectively. 16. The intensity of light varies inversely as the square of the distance from the source. How far from a lamp is a certain point that receives just half as much light as a point 25 feet distant from the lamp ? 17. The volume of a sphere varies as the cube of its diameter. If three spheres whose diameters are 3, 4, and 6 inches, respec- tively, are melted and recast into a single sphere, what is the diameter of the new sphere ? 18. The vplume of a rectangular solid varies jointly as the length, width, and height. If a cube of steel 8 inches on an edge is rolled into a bar whose width is 6 inches and depth 2 inches, what will be the length of the bar in feet ? 19. If the amount earned while erecting a certain wall varies jointly as the number of men engaged and the number of days they work, how many days will it take 4 men to earn f 100 when 6 men working 9 days earn $ 135 ? 20. The pressure of the wind on a plane surface varies jointly as the area of the surface and the square of the wind's velocity. The pressure on a square foot is one pound when the wind is blowing at a rate of 15 miles an hour. What will be the velocity of a wind whose pressure on a square yard is 81 pounds ? son. Kl,. AI.G. 22 CHAPTER XXV PROGRESSIONS ARITHMETICAL PROGRESSION 398. A series is a succession of terms formed in accordance with a fixed law. 399. An arithmetical progression is a series in wliich each term, after the first, is greater or less than the preceding term by a constant quantity. This constant quantity is the common difference. 400. We may regard each term of an arithmetical progres- sion as being obtained by the addition of the common difference to the preceding term ; hence, An increasing arithmetical progression results from a positive common difference, and a decreasing arithmetical progression re- sults from a negative common difference. Thus : 1, 5, 9, 13, . . . etc., is an increasing series in .which the common differ- ence is 4. 7,4, 1, —2, . . . etc.. Is a decreasing series in which the common differ- ence Is —3. 401. In general, if a is the first term, and d the common difference, a, a-\-d, a + 2 d, a + S d, a + i d, . . . etc., is the general form of an arithmetical progression. 338 ARITHMETICAL PROGRESSION 339 402. The nth Term of an Arithmetical Progression. In the general form, a, a + d, a + 2cl, a + 3 d, a + id, . , . etc., it will be seen that the coefficient of d in any term is less by 1 than the number of the term. Hence, the coefficient of d in the nth term is n — 1. Therefore, If a = the first terra of an arithmetical progression, d = the common difference, n = the number of terms in the series, I = the last term (that is, any required term), then, Z = a + (n - l)d. (I; 403. The Sum of the Terms of an Arithmetical Progression. If s denotes the sum of the terms of an arithmetical progression, we may write the following identities, the second being the first written in reverse order : s = a + (a + d) + (a + id) + + (l--2d) + (? - f?) + I. S-1+ {l-d) + (l-2d) + + (a + 2d) + (a + d) + a. '■i .s- = (o + z) + (a + i) + (a + Z) + ... + (a + i) + (a + + (a + I). Whence, 2s=H(a + Z)* Or, s = 5(a + I). (II) 2 404. Substituting for I in (II) the expression for J. in (I), we itain s = Simplifying, we obtain obtain „ s=5[a + a + (n-l)d] s = 5[2a + (n-l)a]. (Ill) 405. The first term (a), tlie common difference (d), the number of terms (u), the last term (I), and the sum of the terms (s), are the elements of an arithmetical progression. APPLICATION OF THE FORMULAS FOR ARITHMETICAL PROGRESSION 406. Illustrations : 1. Find the 10th term and the sum of 10 terms of the arithmetical progression, 1, 4, 7, 10, . . . 340 PROGRESSIONS We have given, a = 1, d = 3, n = 10. In the fonnula, I = a + (n — l^d. Substituting, i = 1 + (10 - 1) (3) = 28, the required 10th term. In the formula, s = - (a + Z)- Substituting, s = — (1 + 28) = 5 x29 = 145, the required sum of 10 terms. 2. The first term of an arithmetical progression is 3, the last term, 38, and the sum of the terms, 164. Find the series. We have given, a = 3, I = SS, s = 164. Then, s=~{a + l) or 164 = - (3 + 38) , whence n=8. Also, l = a+{n-l)d, or 38 = 3 + (8 - 1)(J, whence d = 5. Therefore, 3, 8, 13, 18, 23, 28, 33, 38, is the required series. 3. How many terms of the series, 2, 5, 8, •••, will be re- quired in -order to give a sum of 126 ? We have a = 2, d = 3, s = 126. Then i = a + (« - l)fZ = 2 + (» - 1)3 = 2 + 3 n - 3 = 3 « - 1. This value for I in terms of n is substituted in the formula s=|(« + 0- Whence, 126 = " (2 + 3 ra - 1), or 3 n^ + n = 252. Solving this quadratic equation in n, we have (using the quadratic formula), ^^^_1±^^T3024^ „^-l±55^ n = 9,or-9J. 6 6 Therefore, the required number of terms for the given sum is 9. Formula III (§ 404) may be used here without finding I. ARITHMETICAL PROGK KSSIUN 341 Ezercise 124 1. Find the 16th term of 7, 10, 13, ■••. 2. Find the 10th term of 2, - 1, - 4. .... 3. Find the 12th term of 4, 3.2, 2.4, ••.. 4. Find the 10th term of i, |, |, .... 5. Find the 20th term of |, |, ^, .... 6. Find the 10th term of - 12, - 9.5, - 7, - 4.5, .... 7. Find the 14th term of a; + 1, a; + 3, a; + 6, •••. 8. Find the 11th term of a; — 5 a, a; — 4 a, x— 3 a, •■•. 9. Find the 10th term of 4 + V2, 3 H- 2V2, 2 + 3 V2, .... 10. Find the 358th term of .0075, .01, .0125, .015, ■••. Find the sum of : 11. 3, 8, 13,18, ••• to 24 terms. 12. -2, -9, -16, •■■to 12 terms. 13. .25, .3, .35, •■■ to 40 terms. 14. I, ff,i|, ••• to 16 terms. 15. 20V2-10V3, 18V2-9V3, 16V2-8V3, ... to 21 terms. Find the first term and the common difference when : 16. s = 297, ?(=9, Z=16. 19. Z = 0, s = 100, 9i = 25. 17. s = 294, 91 = 12, Z = 41. 20. n = 6, s = 20.4, Z = 4.9. 18. M = 13, s=260, Z = -f. 21. s = 0, Z = 34.5, m=24. Find the common difference when: 22. a = 4, Z = 40, n = 13. 24. Z= 2, m = 7, s= 19.25. 23. s=-27.5, a=4.5,ii = n. 25. Z = .97, a = .8, s = 35.4. •26. a = .08, s = — 25, n = 25. 27. « = 30, a=-13V2, Z = 16V2. 342 PROGRESSIONS Find the fii'st term when: 28. n = 12, 1 = 10, s = 60. 29. d = — l,n = 10,s = ~ 100. 30. d = -.d,n = li,l = -8.3. Find the numbei' of terms when: 31. « = 7, d = — 3, Z = — 23. 32. Z = f, rt = |, s = 17. 33. a = --l,l = -J>-^,d = -^^. How many consecutive terms of 34. 2,-3, - 8, ••• will give a sum of - 205 ? 35. 1, 1^, If, ••• will give a sum of 83^ ? 36. — 3, -3^, —4, ••• will give a sum of - 97^? 37. 0.36, 0.32, 0.28, ••• will give a sum of -.4? 38. 12V3, 10V3, 8V3, 'Will give a sum of 0? THE DERIVATION OF GENERAL FORMULAS FROM THE FUNDAMEHTAl FORMULAS 407. From the fundamental formulas (I) and (II) in Arts 402 and 403 we may derive a formula for any desired element in terms of any other three elements. Illustrations : 1. Given a, n, and s; derive a formula for d. From the fundamental formulas, I = a+ (n — l)cl and s = " (a + I), we must eliminate the element I ; and by substituting the value of I from the first formula for I in the second formula, we have $ ='^la Jf a ■{■ (n-l)^]. ARITHMETICAL PROGRESSION 343 Erom which, 2 s = )i[2 a+ (n — l)d], 2 s = 2 a« + n{n — l)d, 2 s — 2 an = n(n — l)d, d = -1- i, the required formula for d. n(n — 1) 2. Given d, I, and s, find a formula for n. From (I), Art. 402, I = a + {n - l)d. Hence, a =1 — (n — l)d. Substituting this value for I in (II), Art. 403, we have, n s = -^\_2 a+ (n — l)d] = I [2 ? - 2(re - l)d + (n - !)(*]. Simplifying, 2 s = 2 nl ~ n{n — \)d. Whence, dn^ - (21 + d)H+ 2s = 0. Solving for n, n = — ^ ^ ' ^ 2d Exercise 125 1. Given d, n, and I ; derive a formula for a. 2. Given n, I, and s ; derive a formula for a. 3. Given a, n, and s ; derive a formula for I. 4. Given a, I, and n ; derive a formula for d. 5. Given a, n, and s; derive a formula for d. 6. Given a, I, and s; derive a formula for d. 7. Given rf, ?, and n ; derive a formula for s. 8. Given ?t, ?, and s ; derive a formula for d. 9. Given a, I, and s; derive a formula for n. 10. Given d, m, and s; derive a formula for a. 11. Given a, d, and s; derive a formula for I. 12. Given a, d, and s; derive a formula for m. 3rt4 PROGRESSIONS ARITHMETICAL MEANS 408. If we know a and 6, the first and last terms respec- tively in an arithmetical progression, we may form au arith- metical progression of m + 2 terms by inserting m arithmetical means between a and b. Illustration : Insert 7 arithmetic means between 3 and 43. We seek an arithmetical progression of (m -|- 2) = (7 -|- 2) = 9 terms. (For two terms, the first and the last, are given.) Therefore, a = 3, I = 43, n = 9. It remains to find d. In the formula, 1= a + (n — l)d. Substituting, 4,3 = 3+ (9 - l)d. Whence, d = 5. Therefore, 3, 8, 13, 18, 23, 28, 33, 38, 43 is the required series. 409. To insert a single arithmetic mean, m, between two numbers a and b, we have as a result the series a, m, I. Therefore, m — a = I — m. And, 2m = a-\- I. a + I Or, the arithmetic mean between two numbers equals one half the sum of the mimbers. Exercise 126 1. Insert 6 arithmetical means between 13 and 37. • 2. Insert 9 arithmetical means between 6 and 11. 3. Insert 7 arithmetical means between i and f. 4. Insert 15 arithmetical means between — f and |. 5. Insert the arithmetical mean between 12.4 and 13.2. 6. Insert the arithmetical mean between a + 2 and a — 2. 7. Insert the arithmetical mean between and • a+ c a — c ARITHMETICAL PROGUESSION 345 PROBLEMS INVOLVING ARITHMETICAL PROGRESSIONS 410. Illustrations : 1. The 4tli term of au arithmetical progression is 11, and the 9th term 26. Find the first 3 terms of the progression. The fourth term is o + .3 d, and the ninth term a + 8d. Therefore, a + 8 d = 26. And, a + 3d=n. Whence, 5 (J = 15. d = 3. a = 2. Hence, the required terms of the progression are 2, 5, 8. Result. 2. Find 5 numbers in arithmetical progression, such that the product of the 1st and 5th shall be 28, and the sum of the 2d, 3d, and 4th shall be 24. Let X —2y, x — y, x, x+ y, and x + '2y represent the numbers. Then, (x-2y) (x + 2y) = 28. (1) And, x-y + x + x + y = 24. (2) From (2), 3 a; = 24. x = 8. (3) From (1) , 0:2 - 4 !/2 = 28. (4) Substituting from (3) , 8^ - 4 y^ = 28. 4 y2 = 36. y=±3. Hence, the required numbers are 2, -5. 8, 11, 14 ; or 14, 11, 8, 5, 2. The symmetrical forms, x — 2y, x—y, x, x + y, x + 2y, are chosen merely for convenience. Exercise 127 1. How many numbers between 50 and 500 are exactly divisible by 6 ? 2. Find the sum of all the numbers of two figures that are exactly divisible by 7. 346 PROGRESSIONS 3. Find the sum of the first 20 odd numbers. 4. Find X so that 2a; — 1, 2a; + 2, 3a; — 2, and 3 a; + 1 shall be in arithmetical progression. 5 Are the 3 numbers, 6a; — 3?/, x-\-2y, and 7?/ — 3a;, in arithmetical progression ? 6. What will be the value of x if the numbers 2 a; — 3, a;, and 10 — a; are in arithmetical progression ? 7. Find the sum of 1 + 2 + 3 + 4 + 6 H ton terms. 8. Find the sum of 1+3 + 5 + 7+ •••torn terms. 9. Find the sum of 21 terms of an arithmetical progression whose middle term is 23 and whose common difference is 2. 10. Find the sum of 2 + 4 + 6 + 8+ •■•to n terms, and compare the result with that of example 8. 11. Find the sum of the first 25 numbers that are divisible by 7. 12. The sum of 3 numbers in arithmetical progression is 24, and the product of the 2d and 3d is 88. What are the numbers ? 13. The 5th term of an arithmetical progression of 49 terms is 3, and the 16th term, 63. Find the 34th term. 14. The 6th term of an arithmetical progression is —19, and the sum of the first 18 terms, 36. Find the common dif- ference and write the first 6 terms of the series. 15. Prove that the sum of n consecutive odd numbers begin- ning with 1 is w^. 16. A clerk's salary is increased $ 60 every 6 months for a period of 8 years. At the end of the 3d year he was receiving $ 1000. At what salary did he begin and what will he receive during the last half of his 8th year ? 17. Prove that the sum of the terms of an arithmetical pro- gression in which a, n, and d are all equal, is equal to — ^^ — - — ^■ ARITHMETICAL PROGRESSION 347 18. Find 4 numbers in arithmetical progression such that the sum of the 1st and 3d shall be 44, and the product of the 2d and 4th, 572. 19. How many strokes are sounded in 24 hours by a clock striking hours only ? 20. A boy saves 25 cents the first week of a new year, and increases his savings 5 cents each week through the entire year. How much will he have saved by December 31st ? 21. The sum of three numbers in arithmetical progression is 27, and their product, 693. Find the three numbers. 22. Show that if every alternate term of an arithmetical progression is removed, the remaining terms will be in arith- metical progression. 23. A laborer agreed to fill 40 tanks with water, the tanks being placed in a straight line and at a uniform distance of 10 feet from each other. Each tank holds 18 gallons and he carries at each trip 2 pails holding 3 gallons each. If the source of supply is 10 feet from the first tank in the row, how far does he travel before the 40 are filled ? 24. A man travels 210 miles. The first day he goes 12 miles, and he increases each succeeding day's distance by 2 miles. How many days will he require to complete the journey, and how far will he be obliged to go on the last day ? 25. There are 2 arithmetical progressions, 13, 15, 17, ■.• and 37, 35, 33, ••■, in one of which d = 2, and in the other d= — 2. What must be the number of terms for both series in order that the sums for both series shall be equal ? 26. A contractor failing to complete a bridge in a certain specified time is compelled to forfeit % 100 a day for the first 10 days of extra time required, and for each additional day beginning with the 11th the forfeit is daily increased by $ 10. He loses in all $2550. By how many days did he overrun the stipulated time ? 348 PROGRESSIONS GEOMETRICAL PROGRESSION 411. A geometrical progression is a series in which each terra, after the first, is obtained by multiplying the preceding term by a constant quantity. This constant number is the common ratio. Thus, 3, 6, 12, 24, 48,.-. is a geometrical progression in which the common ratio is 2. 2, i, h h- is a geometrical progression in which the common ratio is J. 412. In general, if a is the first term of a geometrical pro- gression, and r the common ratio, a, ar, ar^, ar^, ar*, ar^, •■• etc., is the general form of a geometrical progression. 413. The nth Term of a Geometrical Progression. In the general form, a, ar, ar'^, ar'^, «»■*, ar^, ■■■ etc., it will be seen that the exponent of r in any term is less by 1 than the number of the term. Hence, the exponent of r in the nth term is n - 1. Therefore, if a = the first term of a geometrical progression, r = the common ratio, n = the number of terms in the series, I = the last term (that is, any required term). Then, / = ar"-i. (I) 414. The Sum of the Terms of a Geometrical Progression. If s denotes the sum of the terms of a geometrical progression, we may write s = a + ar+ ar^-\ — ar"-" + ar'^-- + ar"-^. (1) Multiplying (1) by r, rs = ar + ar^ + ar^ + ■■■ ar"-^ + ar"-' + ar". (2) Subtracting (1) from (2), rs — s = ar" — a. From which, s = ^'^"~^ . (II) r - 1 GEOMETRICAL PROGRESSION 349 415. Multiplying (I) by r, we obtain r/ = ar". Substituting rl for ar in (II), we obtain „ r/-a r-1 (III) 416. The first term (a), the common ratio (r), the number of terms (n), the last term (J), and the sum of the terms (s), are the elements of a geometrical progression. APPLICATIONS OF THE FORMULAS FOR GEOMETRICAL PROGRESSION Illustrations; 1. Find the 8th term and the sum of 8 terms of the pro- gression, 2, 6, 18, 54, ... . We have given, a = 2, r = 3, n = 8. In the formula, I = aj'"-i, we substitute I = (2)(3)^ Whence, = 2 . 2187, Or, = 4374, the required 8th term. rl — a In the formula, r-1 we substitute ^ ^ (■3)(4374) - 2_ 3-1 Whence, s = 6560, the required sum of 8 terms. 2. Find the 12th term and the sum of 12 terms of 6, — 3, f, ... . We have given, a = Q, r = — \, n = li. In the formula, I = ar"~\ I = 6(- i)" = 6(- jjVi) = - i^2i- In the formula, s = tLll«. ^ r-l -^-1 ~ -I ~1024~^"' Hence, the required 12th term is — xj'm and the sum of 12 terms, 3 1023 350 PROGRESSIONS Exercise 128 1. Find the 7th term of 2, 6, 18, ... . 2. Find the 8th term of 3, - 6, 12, ... . 3. Find the 6tli term of — 3, — 6, — 12, ... . 4. Find the 9th term of \, i 1, ... . 5. Find tlie 9th term of — |, |, — |, ... . 6. Find the 8th term of 3, 3V2^ 6, ... . 7. Find the 6th term of 2.4, 0.24, 0.024, ... . 8. Find the 10th term of 0.0001, 0.001, 0.01, , 9. Find the 15th term of a% ai'x-, aV, ... . Find the sum of: 10. 2, 4, 8, ... to 8 terms. 11. 40, 20, 10, ... to 8 terms. 12. 120, 12, 1.2, ... to 6 terms. • 13. I, |-, |, ... to 5 terms. 14. — 27, 9, — 3, ... to 6 terms. 15. 21 1^, y'^, ... to 10 terms. 16. 1 + m + m^ + m^ + ... to (( terms. 17. 64 — 32+16— 8 + 4+ ... to 71 terms. Find the common ratio when : 18. a = 4, ^=1024, n = 9. 19. Z = -729, a = 3, »=6. 20. a = - 4, Z = - 512, s = - 1020. 21. a = 4, ? = — Y^i! s = f 2^. 22. a = 4, Z = gL 11 = 9. 23. a==3,s = -\%Sl=2h- 24. 7i = 10,l = ^t^,a = 2. GKOMETRICAL PROGRESSION 351 Find the number of terms when: 26. ft = - 3, r = - 2, s = 102;!. 27. r=-2, s=-22, a=-2. 28. a=-2, l=-128, r = 2. 29. Z = - 160, a = 5,r=-2. 30. s = -jfY^, r = J, Z = ^Y- 31. a = .2, s = 204.6, I = 102.4. How many terms ofythe series : 32. f, \, I'j, ••• will make a sum of f|-|-? 33. 18, — 6, 2, ••• will make a sum of 4^* ? 34. V2, 2, 2V2, ••• will make a sum of 62 +31V2? GENERAL FORMULAS DERIVED PROM THE FUNDAMENTAL FORMULAS 417. From the fundamental formulas (I) and (II) in Arts. 413 and 414 we may derive a formula for any desired element in terms of any other three elements. Illustrations : 1. Given a, I, and s ; derive a formula for r. From (III) (Art. 416), s =''~^, we obtain rs — s = rl — a. Whence, rs — rl = s — a, r(_s — I) = s — a, s -I 2. Given a, n, and I ; derive a formula for s. rromEx. 1, y = lZL^. (1) s — I From (1) (Art. 413), I - ar«-\. r"-i = - , r = yj-- (2) 352 PROUHESSIONS From (1) and (2), s —a s-l s — a s- I a" -V-a. -"-^ = s'-i'l- "-i/T", «(" -^a -"-%/?) = "-^'^ - ■'-^», s = "-■i^"- "-■J/T" »-^a- "-^i Note. The general formulas for n involve logarithms, and are ordi- narily reserved for advanced students. 1. Given r, 2. Given r, 3. Given a, 4. Given a, 5. Given r, 6. Given r, 7. Given »-, 8. Given a, 9. Given w, 10. Given w, Xizerciae 129 n, and Z; derive a n, and s M, and I r, and s n, and s Z, and s n, and Z n, and s Z, and s Z, and s derive a derive a derive a derive a derive a derive a derive a derive a derive a formula for a. formula for Z. formula for r. formula for Z. formula for a. formula for a. formula for s. formula for r. formula for a. formula for r. THB INFINITE DECREASING GEOMETRIC SERIES 418. If the absolute value of r in a geometrical progression, a, ar, ar', ••-, or"~\ is less than 1, the successive terms become numerically less and less ; hence, by taking a sufiS.ciently great number of terms, that is, by making n sufficiently great, the ?ith term becomes as small as we may choose, although never equal to zero. GEOMETKICAL i-UOGRESSlUN 35iJ Thus, if a = ] and r = i, the series, J-, J, i, ^'j, ..., may be continued until the Jith term is so small as to have no assignable value. Hence, we may say that is the limiting value of the isth term. ^ It follows, therefore, that if I approaches 0, rl approaches 0. Hence, s = , or approaches — ^ as a limiting value. r — 1 1 — ;■ 1 — /■ Hence, s = — 5— 1 — r is the formula for the sum of the terms of an infinite decreasing geometric Illustration : Find the siim of the infinite series, l+i+i + ^+ •••• 1 13 We have a = 1, r = I. Hence, s = = - = - . Result. ^ 1-i I 2 THE RECURRING DECIMAL 419. A recurring decimal is the sum of an infinite decreasing geometrical progression in which the ratio is 0.1 or a power of 0.1. Thus, .272727 ■•• = .27 + .0027 + .000027 + ••• is a geometrical pro- gression in which a = .27 and )• = .01. Illustration : Find the value of .292929 ••• . We have .292929 ... = .2— , jf ,., = 2 . ,-«-i, ^h = r^, \ = r. Hence, the required progression is 2, ?, |, ^, /j-, ^fj. 421. To insert a single geometrical mean, m, between the two numbers, a and I, we require the value of m in a : m = m : I. From which, m = Val. Tlie geometrical mean between two numbers equals the square root of their product. PROBLEMS INVOLVING GEOMETRICAL PROGRESSION 422. Illustrations : 1. The 3d term of a geometrical progression is 2, and the 7th term 162. Find the 6th term. We have ai'^ = 2, (1) ar» = 162. (2) Dividing (2) by (1), r* = 81, r = 3. Therefore, ar^ = 2, a(Sy = 2, a = f Hence, the 6th term, or*, = (|) (.3)^ = (2) (S^) = 18. Eesult. 2. Find 3 numbers in geometrical progression, such that the sum of the 1st and 3d decreased by the 2d shall be 7, and the sum of the squares of the 3 shall be 91. GEOMETRICAL PROGRESSION 355 Let a, ar, and ar- represent the numbers. Then, a — ar + ar'^ = 7, (1) ai + ai)-2 + dh-^ = 91. (2) Dividing (2) by (1), a + ar + ar'-' = 13. (3) Subtracting (1) from (3), 2 ar = 6. Whence, ar = 3, 3 >• = -. a Substituting in (1), a - a(A + «f jj j = 7, a-3 + '-=7. a a = 1 or 9. Hence, the required numbers are 1, 3, and 9. Result. lizercise 130 1. Find the first 3 terms of a geometrical progression whose 3d term is 9 and whose 6th term is 243. 2. Find the first 2 terras of a geometrical progression whose 5th term is ^ and whose 10th term is 16. 3. Insert 4 geometrical means between 1 and 243. 4. Determine the nature, whether arithmetical or geometri- cal, of the series, \, i, ^, ... . 5. Find the first 2 terms of a geometrical progression in which the 5th term is \ and tlie 12th ^erm 16. 6. Which term of the geometrical progression, 3, 6, 12, ..., is 3072 ? 7. Find to n terms the sum of the series, 1, 3, 9, 27, ... . 8. Insert 4 geometrical means between j\j and 32. 9. Find, to infinity, the sum of 2, 1^, |, ... . 10. Find the value of the recurring decimal, 0.1515 ... . 11. Find the value of x such that .r — 1, x + 3, x + 11 may be in geometrical progression. 356 PROGRESSIONS 12. Insert a single geometrical mean between 6^ and 5^. 13. Find the value of the xecurring decimal, 2.214214 ... . 14. Find 4 numbers in geometrical progression such that the sum of the 1st and 3d shall be 15, and the sum of the 2d and 4th, 30. 15. Find to infinity the sum of 4, — f, |, ... . 16. Find to infinity the sum of 1.2161616 :.. . 17. Insert a single geometrical mean between 3 V2 — 2 and 3 V'2 + 2. 18. The 1st term of a geometrical progression is 10, and the sum of the terms to infinity is 20. Find the common ratio. 2 11 19. Find to infinity the sum of — -::> — =) — -=> ••• • V2 V2 2V2 20. Insert 3 geometrical means between a~* and a*, and find the sum of the resulting series. 21. Insert 5 geometrical means between -j and —^ and find the sum of the series. 22. What must be added to each of the numbers, 5, 11, 23, that the resulting numbers may be in geometrical progression ? 23. The sum of the first 3 terms of a decreasing geomet- rical progression is to the sum to infinity as 7 : 8. Find the common ratio. 24. If the numbers, x — 2, 2 a; — 1, and 5 a; + 2, are in geo- metrical progression, what is the common ratio of the series ? 25. The sum of the first 8 terms of a geometrical pro- gression is equal to 17 times the sum of the first 4 terms. Find the common ratio. 26. The population of a certain city is 312,500, and it has increased uniformly by 25 % every 3 years for a period of 12 years. What was the population 12 years ago ? GEOMETRICAL PROGRESSION 357 27. A ball on falling to the pavement rebounds | of the height from which it was dropped, and it continues to succes- sively rebound | of each preceding distance until it is at rest. If the height from which it originally fell was 60 feet, through how great a distance does it pass in falling and rebounding ? 28. What is the condition necessary that a + 1, a + 3, a+7, and a + 15 shall be in geometrical progression? For what value of a is this condition true ? 29. Show that if 4 numbers, m, n, x, and y, are. in geo- metrical progression, then m + n, n + x, and x + y are also in geometrical progression. 30. A sum of money invested at 6 % compound interest will double itself in 12 years. What will be the amount of f 10 invested at compound interest at the end of 60 years ? 31. If 4 numbers are in geometrical progression, the sum of the 2d and 4th divided by the sum of the 1st and 3d is equal to the common ratio. 32. Find 3 numbers in geometrical progression whose sum is 21, and the sum of whose squares is 189. 33. Find an arithmetical progression whose first term is 2, and whose 1st, 3d, and 7th terms are in geometrical pro- gression. 34. If the alternate terms of a geometrical progression are removed, the remaining terms are in geometrical progression. 35. Find to infinity the sum of the series x+i^yx+ij ^\x+ij ^ 36. Find the 4th term of an infinite decreasing geometric series the sum of whose terms is ^, and whose first term is .25. CHAPTER XXVI THE BINOMIAL THEOREM. POSITIVE INTEGRAL EXPONENT 423. A finite series is a series having a limited number of terms. 424. The binomial theorem is a formula by means of which any power of a binomial may be expanded into a series. 425. By actual multiplication we may obtain: (a + &)2 = a2 + 2 db + h\ ia + 6)8 = a' + 3 a% + 3 aV^ + 6'. (a + &)* = a< + 4 a% + 6 a^fts + 4 afts + j4 . etc. In the products we observe : 1. TJie number of terms exceeds by 1 the exponent of the binomial. 2. The exponent of a in the first term is the same as the exponent of the binomial, and decreases by 1 in each succeed- ing term, 3. The exponent of b in the second term is 1, and increases by 1 in each succeeding term until it is the same as the expo- nent of the binomial. 4. The coefficient of the first term is 1, the coefficient of the second term is the same as the exponent of the binomial. 5. If the coefficient of any term is multiplied by the exponent of a in that term, and the product is divided by the exponent of b in that term increased by 1, the result is the coefficient of the next following term. 368 THE BINOMIAL THKOKEM 369 426. ]>y observing these laws we may write the expansion of (a + by thus : (a + by = a* + i a»b + ^a^b^ + ^^a¥ + H4^ ^*- 1-2 r-2-S 1.2-.5-4 In like manner, we may write the expansion of (a + 6)" in the form : (« + by = «„ + „a— 6 + »^^i^a.-262 + »(«- l)(n - 2) ^„_3. '+■ This expression is the binomial formula, and we will new prove that it is a general expression for any power of (a + b), for positive integral values of n. PROOF OF THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS 427. We have shown by actual multiplication that the laws governing the successive expansions of (a+b) are true up to and including the fourth power. If, now, we assume that the laws of Art. 425 are true for any power, as the ?ith power, and if, furthermore, we show the laws to hold for the (»i + l)th power, then the truth of Art. 425 for all positive integral values of n is established. This method of proof is known as mathematical induction. Both members of the formula (1) belo\v are multiplied by (a + 6). (a + b)« = a" + na>-i6+ »(^I^ a^-^^ „,(n - 1)(» -2) ^„_3^3 ... ^^^ ^ ^ 1-2 1-2-3 (a + b) (a + b) (a + 6)«+i = a-*! + 7ia''b + "('^ - ^) gn-ift'^ + n(n-l)(n-2) ^,_^^, ^ ... 1 • 2 \ • 2i • o + a"6+ wa"-i62+ ZLlz^ a^-v + - (a + 6)«+i = a"+i +(n + l)a"6 + f^^I^rd} + mla-'-ife^ r n(«-l)(w-2) <«- 1)13.-268 , ... L 1-2-3 1-2 J 360 THE BINOMIAL THEOREM 1 • ^ ± • Z • o It will be observed that (2) is identical with (1) excepting that every n of (1) is replaced by ji + 1 in (2). That is, we assumed the laws of 426 to be true for n, and have shown tliein to be true for n + 1. Similarly, we might show the laws to be true for n + 2, and so on, indefinitely. Henae, the laws of Art. 425 being true for the 4th power may be shown to hold true for the 5th power ; and holding for the 6th power, may be shown to hold for the 6th power. Therefore, for any positive integral values of n : Ca + b)» = a" + na''-ib + °C"--^) a"-2b^ + 1(°-^)(°-^) a"-8bs + •• ■ 1-2 lS-3 428. The Factorial Denominator. In practice it is convenient to write [3_ for 1 • 2 • 3, [6^ for 1 • 2 . 3 • 4 • 6 • 6, etc. We read [3 as "factorial 3," \6_ as "factorial 6," etc. In general, \n_ means the product of the natural numbers 1-2-3-4-6---71 inclusive. 429. An expansion of a binomial is a finite series when n is a positive integer. For, in the coefficients, we finally reach a factor, n — n, or 0. And the term containing this factor disappears. Moreover, every following term contains this factor, hence each term following disappears. 430. The Signs of the Terms in an Expansion. If both a and b are positive in (a + 6)", the signs of all the terms in the expansion are positive. If b is negative, that is, given (a— b), all terms involving even powers of — 6 are positive, while all terms involving odd powers of — 6 are negative. Therefore, the signs of the terms of the expansion of («—&)" are alternately + and — . APrLICAl'lONS OF THE BINOMIAL FORMULA 361 APPLICATIONS OF THE BINOMIAL FORMULA 431. Illustrations : 1, Expand (2 a + 3 &)». (2 a + 3 6)6 = (2 a-)H5(2 ay (3 6) + f^i (2 ay {3 by+ ^lil| (2 ay (3 bf 1 • J ■ 1 ■ 2 ■ 3 = 32 a5 + 240 a*b + 720 a»b^ + 1080 a^fts + 810 ab* + 243 bK Result. 2. Expand /"--.^^y. Changing to a form best suited to tlie binomial formula, (2 a ■i-at)6 = (2 a-i)6-6(2 a-i)S(at) + 15(2 a-i)*(at)2-20(2 a-i)3(at)3 + 15 (2 a-i)2(al)4 - 6 (2 a-i)(o^)6 + (ai)6 =64 a-6 - 192 a-^ai + 240 a-*ai - 160 a-^a^ + 60 a-^at - 12 a-ioV+ai = 64_192^240_160^g0„|_^2aUa'. Result. 432. To find Any Required Term of (a + b)". In finding the rth or general term in an expansion, we ob- serve the general formation of any term of (a + 6)". If r be the number of the term required ; 1. The exponent of h is less by 1 than the number of the term. 2. The exponent of a is m minus the exponent of 6. 3. The last factor in the numerator of the coefficient is greater by 1 than the exponent of a. 4. The last factor in the denominator of the coefficient is the same as the exponent of h. 362 THE BINOMIAL THEOREM Therefore, In the rth tei'm : The exponent of b = r — 1.. The exponent oi a = 7i — (r — 1) = n — r + 1. The last factor of the numerator in the coefficient = ?i — r f 2. The last factor of the denominator in the coefficient = r — 1. Or, the rth term = n(n_ - l)(n - 2)- (n - r +2) ^„_,+,b-i \JLzl Illustration : 1. Find the 7th term of (2 x'-x-y. We have r = 7, n = 10. Then, (2 x^ - x-i)i» = [(2 x^) + (- x"!)]", . whence, for the formula of Art. 360, 01 = (2x2) and b = (- x-i). The exponent of6 = r— 1 = 7 — 1=6. The exponent of a = !j — »• + 1 = 10 — 7 4- 1 = 4. The last factor of the numerator of the coefficient = re - )■ + 2 = 10 - 7 + 2 = 5. The last factor of the denominator of the coefficient = )--l=7-l=6. Then the 7th term = W -9 -S -7 ■()■■> ^^ ^2y(^_ ^-i)* 1 • 2 . 3 ■ 4 . 5 • 6 = 210 . 16 x8 • x-6 = .3810x2. Kesult. The same principle enables us to find the number of a term containing a required power of a letter when that letter occurs in both terms of tlie given binomial. 2. Find the term containing a;" in ( 2 .^'' — — - ] . Let )• = the number of the required term. Disregarding all but the literal factors of the term, we may write : X'l = (x3)12-r+l(a;-2)'-l = (a;8S-3,-+ii) (-j;-2r+2) _ a41-6r_ Therefore, r = 6, the number of the required term. That is, on writing the 6th term, we shall find the exponent of x to be 11. APPLICATIONS OF THK BINOMIAL F(JliMULA MCo Exercise 131 Expand : 1. (d -->)'. 5. (5a;' + 3a;)l 9. (4V2 -SVS)". 2. (1 - li xf- 6. (2 d'x - 3 ?/)'. 10. (2 o v'^b"- v':i^i*. 3. (.'!.« -2)''. 7. (K + Vay. 11. {ax-' — ^ax-y. 4. (4o-cY/j\ 8. (.rVj^ + 'v^a;)''. 12. {2x~^-Zx^y. Find the 4th term of : The oth term of : The 6th term of : 16. {u + 2xy. 20. (a -8 a;)'. 24. {a^ ■Jr^^ay. IT. {l + 2xf. 21. ( V^- - ^/'a?')!^. 25. (x--' - 3 a;^)!^. 18. (a-2Va?)'° 22. (VO - VSy. 26. (V2-3a;^y^ 19. A' + ^Y- 2^. a-2xi/ar- 27. ^'^^"^ ^^'^ Va; a / The 7th term of : The 8th term of : The 9th term of : 28. {a^'-x-y. 31. (l-wVa;)". 34. {x-2y^y\ 29. (f(-2-2a;)«. 32. (a;-i + 2 a^y. 35. {(?-x-^cy\ 30. (a^-.'l-C/o)" 33. (a2-aV2)". 36. (Vm + 27n"i)". Find the term containing : 37. a;* in (x^ + a;-!)'". 40. a;= in (a; — 3 »^')". 38. a;" in (a; + 3V5y. 41. x* m {x^ -y/oF^y. 39. a;' in (1 — a-Va;)'^. 42. a;~^ in (aa;^' — Vaa;~')l Find the term not containing x in : 43. {a?-2x-''f'- 44. rt^a.'"' -2 a"=a;i)«. 45. (a^a;-2-^^)'2. CHAPTER XXVir LOGARITHMS 433. By means of the exponents, 2, 3, 4, etc., we may express certain numbers as exact powers of 10. Thus, 100 = 102, 1000 = 108, 10000 = lOS etc. 434. Suppose, now, that 10 is given any real exponent, as x. Then some positive number, iV, results as the ath power of 10. That is, JV"=10'. 435. If, therefore, we knew the necessary approximate values for X, we might express any number as an approximate power of 10. 436. By a method of advanced algebra, these approximate values for x have been obtained. For example, it has been found that 180 = 102-2553^ 4600 = 1038582, 19600 = 10«-292s^ etc. 437. These exponents are called the logarithms of the num- bers they produce. 438. The exponent that must be given 10 in order to produce a required number, N, is called the logarithm of H to the base 10. The expression 10»" = N is usually written x = logio iV, and is read, " a; is the logarithm of N to the base 10.' 364 Tllli PARTS OF A LOGARITHM 365 The object and use of logarithms is to simplify numerical work in the processes of multiplication, division, involution, and evolution. 439. Logarithms to the base 10 are known as common log- arithms, and are in universal use for numerical operations. Unless otherwise stated, the discussions of this and subsequent chap- ters refer to common logarithms. Any positive number, except unity, may be taken as the base of a system of logarithms. A negative number Is not considered as having a logarithm. THE PARTS OF A LOGARITHM 440. Consider the results in the following : 103 = 1000, log 1000 = 3, 10'^ = 100, log 100 = 2, 101 =10, log 10 =1, 10° =1, logl =0, 10-1 = .1, log.i =_i, 10-2 = .01, log .01 = - 2, 10-3 = .001, log .001 = - 3 ; etc. From these results it is evident that (1) The common logarithm of a number greater than 1 is positive. (2) The common logarithm of a number between and 1 is negative. (3) The common logarithm of an integral or a mixed number between 1 and 10 is + a decimal, between 10 and 100 is 1 + a decimal, between 100 arid 1000 is 2 + a decimal, etc. (4) The common logarithm of a decimal number between 1 and 0.1 is — 1 -f- a decimal, between 0.1 and 0.01 is — 2 + a decimal, between 0.01 and 0.001 is — 3 + a decimal, etc. 866 LOGARITHMS 441. The integral part of a logarithm is the characteristic The decimal part is the mantissa. In log 362 = 2.6465, 2 is the characteristic and .5466 is the mantissa. I. To obtain the characteristic of a logarithm. (a) Wlien the given number is integral or mixed. 442. By Art. 440 (3), the characteristic of the logarithm of a number having one digit to tlie left of the decimal point is 0, of a number having two digits to the left of the decimal point is 1, and of a number having three digits to the left of the decimal point is 2. In general, The characteristic of the logarithm of a number greater than unity is 1 less than the number of digits to the left of the decimal point. (b) When the given number is a decimal. 443. By Art. 440 (4), the characteristic of the logarithm of a decimal having no cipher between its decimal point and its first significant figure is —1, of a decimal having one cipher between its decimal point and its first significant figure is — 2, and of a decimal having two ciphers between its decimal point and its first significant figure is — 3. In order to avoid writing these negative characteristics — 1, — 2, — 3, etc., it is customary to consider that - 1 = 9- 10, -2 = 8-10, - .3 = 7 - 10, etc. With the negative results written in this form, we have, in general, The characteristic of the logarithm of a number less than unity is obtained by subtracting from 9 the number of ciphers bettoeen its decimal point and the fir.^t significant figure, nunexing —10 al'ier the mantissa. USK OF P'OITR-PLACK TABLES 36T II. To obtain the mantissa of a logarithm. (a) Important jirinciple governing the finding of all mantissas. 444. It has been computed that the logarithm of 36,200 is 4.5465, and from our discussion we have seen that the logarithm of 100 is 2. We may write, therefore, 104.6466 _ 36200 (1) and 102 = 100. (2) These logarithms being, by definition, exponents, we may treat them as such in the following operation. Dividing (1) by (2), we have : 104.6466 ^ 36200 102 100 ' From which, io2-6465 = 352. That is, lofe 852 = 2.5465. Clearly, therefore, the mantissas of the logarithms of 362 and 35,200 are equal. In like manner, we may show that log 35.2 = 1.5465, log 3.52 = 0.5465, log .352 = 9.5465 - 10, etc. In general : If two numbers differ only in the position of their decimal points, their logarithms have the same mantissas. THE USE OF THE FOUR-PLACE TABLE 445. The table of logarithms is used for two distinct and opposite operations. (1) Given a number, to find the corresponding logarithm. (2) Given a logarithm, to find the corresponding number. 368 LOGARITHMS I. To find the logarithm of a given number. 446. (a) Numbers having three figures. Illustrations : 1. What is the logarithm of 247 ? On page 370, in the column headed " N," we find " 24," the first two figures of the given number. In the same horizontal line with 24, under the heading corresponding to the last figure of the given number, " 7," we find the mantissa, 3927. Since the given number has three figures to the left of the decimal point, the required characteristic is 2 (Art. 442). Therefore, log 247 = 2.3927. Eesult 2. What is the logarithm of .0562 ? Opposite "56" of the "N" column (p. 371), and under "2," we find the mantissa 7497. Since the given number is a decimal 'and has one cipher between its decimal point and its first significant figure, we subtract 1 from 9, and annex — 10 to the mantissa (Art. 443). Therefore, log .0562 = 8.7497 - 10. Result. (6) JSuiribers having two figures. Illustrations : 1. What is the logarithm of 76 ? Opposite "76" of the "N" column, and under "0," we find the mantissa 8808. The characteristic is 1 (Art. 442). Therefore, log 76 = 1.8808. Result. 2. What is the logarithm of .0027 ? Opposite " 27 " of the " N " column, and under "0," we find the man- tissa 4314. The characteristic is (9 — 2), or 7, vnth —10 annexed (Art. 443) . Tlierefore, log .0027 = 7.4314 - 10. Result. INTERPOLATION 3tj9 (c) Numbers having one figure. Illustrations : 1. What is the logarithm of 7 ? Opposite "70" of the"N" column, and under "0," we find the mantissa 8451. The characteristic is (Art. 442). Therefore, log 7 = .8451. Result. 2. What is the logarithm of .00008 ? Opposite "80" of the "N" column, and under "0," we find the mantissa 9031. The characteristic is (9 — 4), or 5, with — 10 annexed (Art. 443). Therefore, log .00008 = 5.9031 - 10. Result. 447. It is evident that the logarithms of numbers having one or two figures have mantissas from the " " column of the table. Exercise 132 Find the logarithms of the following numbers : 1. 124. 7. 84.2. 13. 6.73. 19. .0642. 2. 283. 8. 39.6. 14. .829. 20. .0006. 3. 689. 9. 2.85. 15. .342. 21. .00016. 4. 676. 10. 6.76. 16. .676. 22. .0676. 5. 643. 11. 3.70. 17. .037. 23. .00809. 6. 540. 12. 6.89. 18. .0681. 24. .00000734. INTERPOLATION 448. Interpolation is based upon the assumption that the dif- ferences of logarithms are proportional to the differences of their corresponding numbers. While the assumption is not absolutely correct, the results obtained are exceedingly close approximations. The process is necessary in obtaining the logarithms of numbers having four places by means of the four-place table. SOM. EL, ALG. — 24 .'!70 LOGARITHMS N 1 2 3 4 5 6 7 8 9 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0792 0828 0864 0899 0934 0969 1004 1038 1072 1106 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 14 15 1461 1492 1523 1553 1584 1614 1644 1673 1703 17; a 1761 1790 1818 1847 1875 1903 1931 1959 1987 2U14 10 2041 2068 2095 2122 2148 2175 2201 2227 2253 2279 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 25-9 18 2553 2577 2601 2625 2648 2072 2695 2718 2742 2705 19 20 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 3222 3243 3263 3284 3304 3324 3345 3365 3386 3404 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 25 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4200 4216 4232 4249 4265 4281 4298 27 4314 4330 4:M6 4362 4378 4393 4409 4425 4440 4456 28 4472 4487 4502 4518 4533 4548 4564 4579 4694 4609 29 30 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5092 5105 5119 5132 5145 5169 5172 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 6302 34 35 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 5441 5453 5465 5478 5490 5502 5514 5527 5539 6551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 6775 5786 38 5798 5809 5821 5832 5843 5855 5866 5877 5888 5899 39 40 5911 5922 5933 5944 5955 5966 5977 5988 6999 6010 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6222 42 6232 6243 6253 6263 6274 6284 6294 6304 6314 6325 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 45 6435 6444 6454 6464 6474 6484 6493 6603 6513 6522 6532 6542 6551 6561 6571 6580 651)0 6599 6. Whence, log jjip = px. But, from (1), x = log m, (Art. 438) Hence, log m" =p(log m). 456. The logarithm of a root of a numher is equal to the logarithm of the number dioided by the index of the root. Let 10" = m. (1) Raising both members to the power indicated by -, Whence, log m» = -. (Art. 438 ) But, from (1), x = log m, (Art. 4381 Therefore, 10' = : m'. 1 log m» = q X = : log m, log 1 m' = log m 2 376 LOGARITHMS 457. These demonstrations may be briefly summarized as follows : (1) To multiply numbers, add their logarithms. (2) To divide numbers, subtract the logarithm of the divisor from the logarithm of the dividend. (3) To raise a number to a power, multiply the logarithm of the number by the exponent of the required power. (4) To extract a root of a number, divide the logarithm of the number by the index of the required root. The antilogarithm of a result obtained by one of these processes is the required number. THE COLOGARITHM 458. The cologarithm of a number is the logarithm of the reciprocal of the number. By its use an appreciable saving of labor is made in computations with logarithms. By definition, cologiV=logi- -Zv = log 1 - log iV (Art. 454) = 0-logiV (Art. 451) = -logJV. To avoid this negative form we may write, colog iV= 10 - log JV— 10 459. In general, to obtain the cologarithm of a number : Subtract the logarithm from 10 — 10. In practice the subtraction is usually accomplished by be- ginning at the characteristic and subtracting each figure from 9, excepting the last significant figure, which is subtracted from 10. LOGARITHMS IN COMPUTATIONS 377 Illustration : Find the cologarithm of 16. The log of 1 is 0, which we may write as 10 — 10. Then, colog 16 = log 1 — log 16. log 1 = 10 - 10 log 16= 1.2041 Subtracting, colog 16 = 8.7959 - 10. Result. 460. If the characteristic of a logarithm is greater than 10 but less than 20, we use in like manner, colog iV = 20 — log iV- 20. Similarly, 30 — 30, 40 — 40, etc., may he used if necessary. 461. Negative factors in computations with logarithms. The logarithms of negative factors in any group are found without regard to the negative signs, and the result is positive or nega- tive according as the number of negative factors is even or odd. USE OF LOGARITHMS IN COMPUTATIOHS 462. Illustrations : 1. Multiply 6.85 by 37.8. log 6.85 =0.8357 log 37.8 = 1.6775 (Art. 453), 2.4132 = log of product, antilog 2.4132 = 258.9. Result. 2. Divide 30,400 by 1280. log 30400 = 4.4829 log 1280 = 3.1072 (Art. 454), 1.3757 = log of quotient, antilog 1.3757 = 23.75. Result. 3. Divide 2640 by 36,900. log 2640 = 3.4216 (Art. 458), colog 36900 = 5.4330 - 10 8.8546 — 10 = log of quotient, antilog 8.8546 - 10 = 0.07155. Result. 378 LOGARITHMS 4. What is the value of ^/846 ? log 846 = 2.9274. (Art. 456) 3)2.9274 .9758 = log of cube root, antilog .9758 = 9.458. Result. 5. Find the value of V.00276. log .00276 = 7.4409 - 10. When a negative logarithm occurs in obtaining a root, the characteris- tic must be written in such a form that the number subtracted from the logarithm shall be 10 times the index of the root. The divisor in this case being 6, we change the form of the logarithm by adding 60 — 50, and the subsequent division gives the required negative characteristic, log .00276 = 7.4409 - 10. Adding, 50 — 50 57.4409-60 (Art. 456) 6 ).57.4409-60 9.5735 - 10 = log of sixth root, antilog 9.5735 - 10 = .3745. Result. 4322x27.1'' 6. Find the value of 35.1^ (Art. 455) 2 log 432 = 2 (2.6355) z= 5.2710 (Art. 455) 3 log 27.1 = 3 (1.4330) = 4.2990 (Arts. 455, 458) 4 colog 35.1 = 4 (8.4547-10) = 3.8188 - 10 3.3888 = log of result, antilog 3.3888 = 2448. Result. 7. Simplify V/ 2720^ X V288 x 432^ .068' X V27.8' X 624^ I log 2720 = i (3.4.346) = 1.7173 I log 288 = ^ (2.4.'-)i)4) =1.2297 flog 4.32 =-+(2.6.355) =3.5140 7 colog .068 = 7(1.1675). =8.1725 1 oolog 27.8 = I (8.5560 - 10) = 7.8340 - 10 2 colog 624 = 2 (7.2048 -10 ) = 4.4096 - 10 6.8771 (Art. 456) 4)6.8771 1.7103, log of result, antilog 1.7193 = 52.40. Result. MISCELLANEOUS APPLICATIONS OF LOGARITHMS 379 Ezerciae 135 Find by logarithms the vakie of : 1. 4600 X. 85. 2. 72x380. 3. .28 X. 00012. 4. .017 X .0062. 5. 4.96 X 58.4. 6. .00621 X .000621 7. 73400 X .00811. 8. .0293 X .000602. 9. 691 X .0000131. 624 X 372 X 891 19 20. 21. 457 X 196 X 583 43.2 X 3.28 X .246 .537 X 3.41 X 56.8' 630 X 2100 X .007 10. 1280 -.0064. 11. 68.5 -=-6.12. 12. 2.741 -- .00822. 13. .00431 H- .0931. 14. .07241 -f- .3623. 15. (2741 X 3.623) -^ 242. 16. (4.625 X .5821) -- 2.067. 17. 34.74 -- (2.8.51 X 4.309). 18. 6.904 -=- (3.676 x .00275) ,0372 X .584 X .00027 22 23. 24. 3.25 X 472 X 6500 25. 4.522. 28. -s/Sft. 26. 2.74'. 29. -v/28:06. 27. .0276*. 30. A/.00724. .273 X .00042 X .0121 .007 X .07 X .7 X 7 35.7 X 7.14 X .1428" .643 X. 0468x2760 , 346 X .0072 X .01 ' 31. 3.207f. 32. 5.6O2I. 31 .0027" X .' ' 'V3.207*x6 5.82^X762" 6.42" X 26.7* 36. 4 33. .000754t. ' VJ809x-C^.003241 V;0962 X v'.OOOOTi MISCELLANEOUS APPLICATIONS OF LOGARITHMS 463. I. Changing the base of a system of logarithms. Let a and b represent tlie bases of two systems of logarithms, and m the number under consideration. We are to show that log^TO = — sS--> log,6 380 LOGARITHMS Let o' = 'm, (1) and 6" = m. (2) Then, x = logajji and y = logjw. (Art. 438) From (1) and (2), a' = V. Extracting the yth root, a^ = 6. Therefore, loga6=- or y = ^ • 2/ loga6 That is, iog,m=l2S2»J. log.6 Illustration: I. Find the logarithm of 12 to the base 5. By Art. 463, log6l2 = }£Sioll logio6 ^ 1.0792 .6990 = 1.5439. Result. II. Equations involving logarithms. 464. An equation in which an unknown number appears as an exponent is called an exponential equation. Thus : a" = b is a general form for such equations. Illustrations : 1. Solve the equation, 4* = 64. If, in an equation in the form of a^ — b, b is an exact power of a, the solution is readily obtained by inspection. From 4»' = 64, we have 4"^ = 4^. Therefore, x = 3. Result. 2. Find the value of a; in 5"^ = 13. Using logarithms, x log 5 = log 13. 3-^ log 13 log 5 _ 1.1139 .6990 = 1.5935. Result. MISCELLANEOUS APPLICATIONS OF LOGARITHMS 381 3. Solve the equation 2 V» = ■v^3. 2 v^ = ■v'^, \/x = — - , « = — (changing f onn and squaring") Then, log a; = | log 3 + colog 4 |log3=|(.4771)= .3181 colog 4= 9.3979 - 10 Therefore, 9.7160 - 10 = log of x a; = .52. Result. 465. Certain forms of equations involving logarithms may be so transformed as to give results without the aid of loga- rithm tables. 1. Find a; if log^ 32 = 5. 2. Find a; if log3a; = 4. By logarithms, 32 = a^. By logarithms, x = 3*. Whence, a; = 2. Result. Whence, x = 81. Result. (That is, the base of that system (That is, the number, whose log in which the log of 32 is 5, is 2.) to the base 3 is 4, is 81.) III. Use of formulas for compound interest and annuities. 466. If P is a given principal, n the number of years during which the interest is compounded annually, and r the given rate per cent, the amount, A, can be obtained from the formula A = P(l + r)\ Illustration : 1. Find the amount of $ 1500 for 8 years at 5 %, compounded annually. In the formula A = P(l + r)". By substitution, A = 1500 (1.05)8. By logarithms, log A = log 1500 + 8 (log 1.05) log 1500 = 3.1761 8 (log 1.06) = 8 (.0212) = .1696 )og^= 3.3457 .A = $2216.50. Result. 382 LOGARITHMS 467. An annuity is a fixed sum of money payable yearly, or at other fixed intervals. If the amount of an annuity is represented by A, the number of yearly payments by n, and the rate of money at the present time by r, the present value of the annuity, F, can be obtained from the formula P=4i L-l. »•[_ (l+»-)»J Illustration : 2. Pind the present value of annuity of $ 900 for 20 years at 4%. Af-, 1 -|^900rj_ I -[ lu the formula, By logarithms, Hence, Then, By logarithms, '=f[' (1.04)2' (1 + r)20j '.04 L 20 (log 1.04) = 20 (.0170) = ..S400. (1.04)2' ^ 2.1879. p_900n_ 1 n_900x 1.1879 .04 L 2.18' .879.J 04 X 2.1879 log 900 =2.9542 log 1.1879= .0747 colog .04 = 1.3979 colog2.1879 = 9.6600 -10 4.0868 = log P P= $12,211, approximately. Besult. Calculate : 1. log, 12. 2. logs 28. 3. log; 42. 4. log,2 64. Solve : 13. a; = 12Vl4. 14. a;V5 = 70. Exercise 136 5. Iogij60. 6. logs 256. 7. logs 9.6. 8. log,6.7. 15. x-\/3=-s/7. 16. 12V^=19. 9. log,., .48. 10. logig,5.416. 11. log.5.875. 12. log.9.007. 17. 2V3x=5VM 18. 10a;"i = 54 19. ^4a;2=VlO. 20. (3a;)'=. 20. — 9 erf - cd + 11 cd + cd. 10. 10 — 11 — 7 + 9 — 6. 21. — S a.vy + 16 axy + axy. 11. 3 a + 4 a — 5 (( + 7 ((. 22. cxz + 7 cxz — 3 cxz + 7 cxz. 23. 5 (i — 8 rt + 3 a — 4 a + 6 a — 2 a + 11 a. 24. 4a + 7a — 5a — 9o— 10a + 4a— 9a. 25. 5c — 8c + 3c — 4c + 8c — 5c — 3c + 4c. 26. —3xy — 4:Xy + 8xy — llxy — 5xy — 3 xy — 7 xy. 27. 5 ahc — 3 ahc + 7 ahc — 8 a6c — 10 ahc + 9 ahc. 28. 3a+5— 4a+3a — 7 + 5a-9 + 7o + ll — 3a. SOM. EL. ALG. — 25 385 386 SUPPLEMENTARY EXERCISES Written Addition Find the sum of : 29. 5a + 36 + 4c, 3a — 46 + c, and 2a + 5b — 6c. 30. 6 a; — 7 2/ + 3, 8 03 — 4 2/ + 6, and Sy — Sx — i. 31. 2c + 3a; — 4, oc + 6 — 3 a;, and — 7 — 4 c + «. 32. 8 m — 4 w + 6p — 2, 3 m + 7, 5 w — 2p, and 3 j? — 5 — 11 m. 33. 4«6+3ctc+5 6c, 7&c— 2a&+6ac, and —2ac+3ab—9bc. 34. 7 aa; — 3 6a3 + 9, — 2 aa; + 7 — 6 6a;, and 3 6a; — 4 ax — 12. 35. 2 ac — 5 6cd + 1 — a;^/, bed — 2 + 4 a;?/, and 7 + 2 6cri + 3 xy. 36. -4 a6c + 3 bed — 10, 2 6cd — 6, — 5 bed + abc, and 2 a^+16. 37. 5a — 4c + 3a;— 4?/ + 7c — 5a; — 32/4-lla + 7a! — 6c + 32/ — 4a; + 5c — 3a + 43/ — 6a + 5a; — 7 c. 38. 5 xyz — 9 + 2/2 + 3a;y— 4as2; — xyz + 16 — 6 «/« — 4 xyz + 4 a;2 — 2 a?i/ — 3 x?/ + 17 — 4 2/2; + 11 a;2 — 10. 39. |a + Sa, + 32, + ^^ 1 a- 1 a; + -i-2/-i, f a + |a;-i2/ + ^, and|a + |x-|2/-|- 40. J-a6 — |ac + f ad + 3, — f a6 + |ac — iad-2, f a6 + 1 ac — ^ ad + 5, and — ^ a6 — ^ ac + ad — 5. 41. 0.3 a;+0.4 y-0.5 z, 0.12 x-0.7 y+0.1 », 0.35 x+.ll 2/+.3 «, 1.2 a; + 0.2 2/ - 3.1 2, and -1.1 x +1.3 2/ + 2. 42. 0.15 m - 0.25 n + 0.15, - 0.4 m + 0.17 n - 0.7, 3.1 m - 0.14 w - 0.2, and 3.1 m - 10.5 n + 7.1. 43. 2(a + 6)-3(a + c)+4(a+d), 5(a+6)+7(a+c)-3(o+d), -4(a+6)-2(a+d)— 3(a+c), and (a+6)-(a+c)-(a+d) 44. 4(x+l)+7(x+2)+3(x+3), -3(x+l)-2(x+2)-6(x+3) -4(x+l)-7(x+3)-4(x+2), and 2(x+2)-(x+l)+4(x+]) 45. c((x+2/)+6(a!+«)+c(?/ + z), 6(x + 2/) +a (x + 2) +a(2/ + »); and c(x +2/) +c (x + «) + 6(2/ + z). 46. 4 a+5(m+n— x)— 5(m— M+x), 2(m+M — x) — (m— w+a;); + 3 (m + n — x) — 3 a + 2 (m — »i + x), and — 3(m — n + k) — 7(m + m — x)'+ 2 a. SUBTRACTION 387 Oral Subtraction 47. 48. 49. 50. 51. 52. From 15 -15 15 -15 17 -17 Take 7 -7 -7 7 -10 10 53. 54. 55. 56. 57. 58. From ' 8a 7 a 14 a -14 a 17 a -17 a Take 7 a 8a -3a -3a 20 a -17 a 59. 60. 61. 62. 63. 64. From — 7 ax -7cd 15 ac -17ac Take 3 ax — icd 5 ax 66. 67. — 5aa; -13ac 69. 15 ac 65. 68. 70. From -9abc — 5 xyz 8 abni amx — amx Take 11 abc — ijxyz — 11 ahm 15 a6c — 9ama; 9ama; Written Subtraction 71. From 4 a — 5 a; + 17 subtract 3 a — 9 a; — 4. 72. From 8 a6 + 7 ac — 3 ad subtract 4 a& + 7 ac — 10 ad. 73. From axy —19 ay + 17 ax subtract 4 axy — 12 aa; + 15 ay. 74. From 2 cdn — 4 adn + 11 subtract 3 adn — 5 acn — 19. 75. Subtract 2 ax + 17 ay — l&az from 9 aa;+ 10 a?/ + 4 a2. 76. Subtract 3ccZ+7aa;-3cy+9 from 2crf+3cy4-7aa;— 16. 77. Subtract 2 am — 3 an + 11 from 4 — 5 aw + 3 am. 78. Subtract 5 aa; — 7 a^ + 3 a« — 17 from 2ay-\-2. 79. From 5 mn — 19 + 6 mx take 3 mn — mx — 19. 80. From |a + -|a; — |?/ subtract \a — ^x — ^y. 81. Subtract | mn — f mx + f from f mn — ^mx — ^. 388 SUPl'LEMENTARY EXERCISES 82. rrom 0Ax + 0.32y—lAz subtract 0.15a;-0.17 2/+2.7 2. 83. Subtract the sum of 4 a + 3 a; — 15 and 7 a — 5 a; — 9 from the sum of 2 a + 7 a; — 12 and 14 — 12 a — 3 ic. 84. Subtract the sum of 3 c + 7 d — 19 and 9 c — d — 5 from the sum of 4 c - 8 r? + 16 and 8 c + 1-1 d - 40. 85. Subtract the sum of 1.2 x + 3Ay— 1.5z and 0.8 a; — 0.4 2/ + 0.6 a; from the sum of 2.7 a; + 7.2 2 and —0.7y + 0.3z. 86. What expression subtracted from 12a + 76 — 3c + 10 will give the expression 5 a +3b — 7 c + 7 ? 87. To what expression must you add 5 ni + 3 ji — 7 a; + 17 to obtain the expression 3m + 9»i — 8a; — 12? 88. By how much does the expression fa; — 1-3/ + |2 — | exceed the expression ^x — ^y -\-^z — 1-? 89. What is the excess of the expression 1.4 a + 3 b — 5.1 c + .08 over the expression .7 o + l..j & — 2.66 c + .04 ? 90. What is the remainder when (a + 2) a; + (a + 3).y + (a + 4) 2 is subtracted from (a + 3) a; + (a + 4) y + (a + 5) 2 ? Oral Multiplication Find the product of: 91. 4(-5). 97. (-5)(-2). 103. -(-9) (-6) 92. (-5) (4). 98. (7)(0). 104. _(-6)(14). 93. (-6)(-4). 99. -(-2) (-5). 105. -(0)(-7). 94. (3) (-10). 100. -(-3) (4). 106. -(-10)(0). 95. (-3) (-10). 101. -(4) (-15). 107. (2) (3) (7). 96. (-7) (5). 102. (-9) (-6). 108. (-2) (3) (7). 109. (-2)(-3)(-7). 112. ( -5)(-7)(0). 110. -(3)(-2)(-6). 113. - _(3)(-9)(-10). Ul. _(-6)(-3)(-4). 114. - -2 (3) (-5) (-3). MULTIPLICATION 389 115. 116. 117. 118. 119. 120. ()« — 7x -3c — 5 mn 8xy — 7a6c -7 -4 5 -7 -3 -5 121. 122. 123. 124. 125. 126. 2 ax -4ac2 3ax^ — om^a; - 7 a'a; 16 a*xy -Sax 2ac — 'kv'x — 4 ma;* - 4 a V - 3 ax'^y 127. 6 a&c by 5 ac. 134. — aVo; by 3 ac''x\ 128. Soma; by —2 ma;". 135. 2ahlxy by —SccPxyK 129. 4cda;^ by 3cda;'. 136. — 4aa;y by 3a*xh/. 130. 5ama; by — 2 o^ma;^. 137. — 11 cd^??!' by — ScW?/. 131. 4 ca;2i/ by 3 cxy''. 138. 14 af'y^z by 3 ma;y. 132. — 5 arfrti by 2 a^dm'. 139. — 8 aVi/ by — 4 a^xhjK 133. — Sac^a; by - Sa^cy. 140. 3 a^xy^z by — 5 a^a^z\ Written Multiplication Multiply : 141. ifl^' + io + i by ia + i. 142. |a^-|a-i by |a + ^. 143. f a=-|a='4-ia-f by 3a2 + 2a-f 144. I a;' — ^ a;^ + ^ xy^ — i 2/^ by ^ x^ — xy + ^ y\ 145. 0.3 a;2 - 0.2 a; + 0.1 by 0.5 a; + 3. 146. 0.4 a' + 0.3 a^ - 0.2 « + 0.5 by 0.3 a + 0.4. 147. 1.5x''-3.2xy + 2y^ by 0.2 a; + 0.4?/. 148. 4 a;' - 0.3 35^ + 0.04 a;?/"- 0.1 2/' by x'' + 0.2xy + 0.3y^. 149. a;^+6a; + c by a!+a. 151. a;'' — ca; — 1 by x + a. 150. a;^ — ca; + d by x + 2. 152. ax^ — bx + c by a; + n. 153. mx'' + nx + l by a;— p. 154. a;^ + (a + '*)a; — an. by »: + ', 155. ax^ + {b + c + l)x + d hy x+'. -m. - m. 390 SUPPLEMENTARY KXERCISES Written Division Divide : 156. ^a'-^a + ^ by f r, - f 157. J^a!»-^a;2 + J^.T- J^ by ^x-\. 158. |^3-J^a + | by fa- 2. ,=083 . , 9mf 27 , . 2x 3 27 ^^^"^ 8 ~6i'' ^ T^i^' 160. c*_|3c^ + iic'-|c^-JiV-c + | by c^-ic + 5. 161. 5.4 a;< - 3.3 a^ - 10.3 a^ + 3.2 a; + 0.8 by 1.5 a; + 0.25. 162. .04 a;* - .04 0^2/ - .09 a;y + .01 a;/ + .02 y by 0.2 x^ + 0.3 xy + 0.1 yl 163. (x + cy — 2(x + c)d + d^ by a; + c — d. 164. (a + iy + 3(a + l)2a; + 3(o + l)a;2 + a;' by a + l + a;. 165. (m + xf + i{m + xf + 5(m + x) + 2 by m + x + 2. 166. ma? + (w — mp) a;^ + (1 — np) x—p by a; —p. 167. x' —(171 + 11 +p) a? +{mn + np + mp)x-- mnp by a;— m. 168. c* + (5 + a!)c' — (4 — 5 a; + d)c2 - (4 a; + 5 d)c + 4 d by c2 + 5c-4. Linear Equations Solve : 169. 2 a; + 15 = 19. 179. 4a;— 7 = 3a! + 2. 170. 3a; + 7=28. 180. 5 a; + 14 = 2 a; + 20. 171. 4a;— 9 = 15. 181. 4a; + 8 — s + 17. 172. 3 a; + 11 = 32. 182. 6a; - 5 = 2 .a;- 25. 173. 8 a;- 7 = 49. 183. 10 + 4 a; = 19 + 3 a;. 174. 7 a; = 15 + 2 a;. 184. — 7 + 8a! = 7-6a;. 175. 9 a; = 4 a; + 10. 185. 10 — 5 a; = 6 a; + 10. 176. 3 a; + 13 = 4. 186. — 8 .1; + 7 — 2 a; = - 3. 177. 7x + 15 = 8. 187. 4a; + 13-5a! + 8 = 0. 178. -3a; + 7 = 19. 188. 6a;-14-2a;-7 = 0. LINEAR EQUATIONS- 391 189. 4a; + 3-(3x-4) = 17. 190. 10 X - (3 a; + 1) - (5 a; + 7) = 6. 191. 15-(a; + 14)-(2a; + l)=3a! + l. 192. (3 a; + 7) - (2 a; + 1) - (5 a; + 3) = 14 a; - 2. 193. 82/ + l-[42/-l-(32/-l)]=0. 194. 5(a; + 2)4-2(a;-3) = 4(a;-5)-2(a; + l). 195. (2/ + 1)^ -{y + 2)2 =(y + 3)^ - (y + 5)^. 196. (a; + 3)(a; + 4)-(a; + 2)(ar + l)-30 = 0. 197. 2(2/ + l)(2/ + 2)-2(2/-l)(2/-2) = 36. 198. 3 a;(a; + 1) - 2 .K(a; + 1) - a; (a; + 6) = 21. 199. 32/(2/-l)-3(y + l)(2/-l)=4(2/-3). 200. 5 a;(a; — 4) — 7 a;(a;— 5) + 2 a;(a; + 6) = 15. 201. (Ay-iy- (2y -iy-{2 y + iy -Sf-2 = 0. 202. {3z + lf-4:z(z + 3)-5z{z-3)=0. 203. (2 a; + 1) (3 a; + 1) - (4 ;»; + 1) (2 a; - 1) = (4 + a;) (1 - 2 a;). 204. 3x2(a; + l)-3a;2(a;-l)-6a;(a; + l)-l = 0. 205. 5(2/ + l)-5[2/-(2/ + l)]-5[2 2/-3(2/ + l)] = 0. 206. (a; + 1)' - (,e - 1)' - 6(a; + 1) (a; - 1) = a;. 207. (4a; + 3)(2a;+l)(a;+l)-4(2a;+l)(a;+l)(a;-l) = 14a;='. Problems leading to Linear Equations State and solve : 208. Find five consecutive numbers whose sum is 116. 209. Find five consecutive even numbers whose sum is 210. 210. Fiiid seven consecutive odd numbers whose sum is 245. 211. Find two numbers whose sum is 70, and whose differ- ence is 6. 212. Divide 125 into two parts, one of which is 43 more than the other. 392 SUPPLEMENTARY EXERCISES 213. Divide $ 1400 between A and B, so that A shall re- ceive $ 230 more than B. 214. Divide 120 into four parts, each of which shall be 10 more than the next smaller part. 215. A and B together have $390, and A has $90 more than B. How many dollars has each ? 216. The difference between the squares of two consecutive even numbers is 52. What are the numbers ? 217. The sum of two numbers is 16, and the difEerence of their squares is 32. What are the numbers ? 218. Divide 70 into two parts, such that ten times the smaller part shall equal eight times the larger part. 219. Divide 31 into two parts such that 1 less than 8 times the smaller shall equal 6 times the larger part. 220. Ten times a certain number is as much above 77 as 43 is above five times the number. What is the number ? 221. How can you pay a bill of $5.95 with the same num- ber of coins of each kind, using only dimes and quarters ? 222. A man divided $1500 among four sons, each receiv- ing $50 more than the next youngest. What amount did each receive ? 223. A, B, and C together have $213. C has twice as much as A, and $57 more than B. How many dollars has each? 224. A certain number exceeds another number by 3, and the difEerence between their squares is 51. What are the num- bers ? 225. A boy has $2.25 in dimes and nickels, and has the same number of coins of each kind. How many coins has he in all ? 226. Divide one dollar between two boys so that one of them shall receive 10 cents more than twice the amount the other receives. PKOBLEMS LEADING TO LINEAR EQUATIONS 393 227. A certain man is three times as old as his son, and in 5 years the sum of their ages is 66 years. What is the age of each now ? 228. Divide 74 into three parts such that the second shall be 6 more than the first, and the third 10 less than the second part. 229. The combined ages of two brothers is 50 years, and in 5 years the older will be twice as old as the younger. How old is each now ? 230. The sum of the ages of a father and son is 80 years, and twice the son's age exceeds the father's age by 1 year. Find the age of each. 231. The distance around a rectangular hall is 720 feet, and the length of the hall is double its width. Find the dimensions of the room. 232. A baseball team played 140 games in a season, winning 42 more games than it lost. What was the number of games won? 233. Two workmen receive together $45 for a certain task, and one receives twice as much as the other. What amount does each receive ? 234. One number is three times another, and if the smaller is subtracted from 34, and the larger subtracted from 70, the remainders are equal. Find the numbers. 235. The annual salaries of A and B together amount to $6700, and A receives $100 more than twice the amount B receives. Find the annual salary of each. 236. The sum of the ages of a father and son is 96 years, but if the son's age is trebled, it will be 8 years greater than the father's age. What is the age of each ? 237. There are three numbers, the third being 21 more than the second, while the first is 5 times the second and also 4 times the third. What are the numbers ? 39-1: SUPPLEMENTARY EXERCISES 238. The oldest of three boys weighs twice as much as the youngest, and 15 pounds more than the second boy. If all three together weigh 335 pounds, what is the weight of each ? 239. A boy has two dollars in quarters, dimes, and nickels. He has 3 times as many dimes as quarters, and 3 times as many nickels as dimes. How many coins has he of each kind ? 240. A bookcase, a table, and a chair together cost $ 77. The bookcase cost 7 times as much as the chair, and $ 13 more than the table. What was the cost of each ? 241. In a certain high school class of 42 students the num- ber of girls is 2 less than three times the number of boys. How many boys and how many girls are there in the class ? 242. One number is three times another number, and the remainder when the smaller is subtracted from 19 is the same as the remainder when the larger is subtracted from 43. Find the numbers. 243. The total number in attendance at a church fair was 285, and 3 times the number of adults present equaled twice the number of children. How many adults and how many children attended ? 244. After giving 14 cents to each of a group of boys I had 5 cents left, and I found that if I had given each boy 16 cents I would have lacked 5 cents of the necessary amount. How many boys were there ? 245. In a certain family the father is 7 years older than the mother, and the sum of their ages is 9 times the age of their son. The combined ages of all three is 70 years. How old is each ? 246. A walks over a certain road at the rate of 3 miles an hour. Two hours after he leaves, B walks after him at a rate of 4 miles an hour. How many miles will A have gone when B overtakes him ? 247. Three men, A, B, and C, are paid all together $7200, annually, and the difference between A's and B's salaries, and PROBLKMS LEADING T(1 LINEAR EQUATIONS 395 also between B's and C's is $800. Find the amount that each receives annually. 248. Three people observe that the sura of their ages is S."! years, that the oldest is three times as old as the youngest, and that he is 20 years older than the second. How old is each ? 249. A field of 30 acres is planted to corn and to wheat. Four times the number of acres planted to wheat equals six times the number planted to corn. How many acres are planted to each grain ? 250. A and B are 60 miles apart, and start to travel toward each other. They leave at the same time, A going 4 miles an hour, and B going 5 miles an hour. In how many hours will they meet ? 251. Two farms together cost $9600. For one of them a price of $ 70 an acre was paid, and for the other the price was S 80 an acre. Both farms together contain 130 acres. How many acres are there in each ? 252. Two automobiles start at the same time from towns 70 miles apart, and go toward each other. One goes at the rate of 20 miles, the other at the rate of 15 miles per hour. In how many hours will they meet ? 253. S 75.60 was taken at the gate at a ball game. Adults paid 50 cents each, and children 20 cents each, for admission. If 5 times as many adults as children attended the game, what number of each attended ? 254. Two automobiles start at the same time from the same town and travel in opposite directions. One of them goes 20 miles an hour and the other 15 miles an hour. After how many hours will they be 70 miles apart ? 255. A man has 3 hours at his disposal and walks out into the country at a rate of 4 miles an hour. How many miles may he walk so that, returning on a trolley car running 12 miles an ho>u', he will return in just 3 hours ? 396 SUPPLEMENTARY EXERCISES 256. A man walks 5 miles on a journey, rides a certain dis- tance on a trolley car, then takes an antomobile for a distance 4 times as far as he has already traveled. His total journey being 75 miles, how far does he ride in the automobile ? Oral Multiplication Give the product of : 257. {a + xf. 267. (2 a + 3)1 277. {c-Sxy. 258. (c + my. 268. (4 a + 7)^. 278. (c - 5 xy. 259. (m + yf. 269. (5( i + sy. 279. {a^-5y. 260. {a + 2y. 270. (3 a + 2 xy. 280. {x^ - 9)2. 261. {c+5y. 271. (4 a + 5 a;)''. 281. {v? - iiy. 262. {a + 2xy. 272. («- -xy. 282. (3 a - 2y. 263. (m + 5 yy. 273. (c- -my. 283. (5 a - 9 xy. 264. (c + 6 xy. 274. (m -yf- 284. {la- 10)-. 265. (a^4-3)l 275. (X- -sy. 285. (3a2-6a;)= 266. {x' + iy.' 276. (x- -ly 286. (7a^-3a;2) 287. (a + x){a- -X). 294. (2 c I + 3) (2 a - 3). 288. (c + w) (c - -m). 295. [3o t + 4)(3a-4). 289. {m + y){m -y)- 296. (4 a + 5) (4 a -5). 290. (c + 3)(c- -3). 297. (So (2 + 5) (3 a^ - :>). 291. (a + 4) (a - -4). 298. [4fl 1^-9) (4 a'' + 9). 292. (m — 5) (m + 5). 299. [5fl t^ + 7)(5a'-7). 293. (a; + ll)(a; -11). 300. (4 a* - 11) (4 a' + 11). 301. (a + 4) (a + 5). 306. [X- -7) (a: -3). 302. (a; + 3) (a; + 7). 307. [y- -9) (2/ -10). 303. (c + 5)(c + 9). 308. [ah -8)(a6-9). 304. (a + 7) (a + 8). 309. [^ -5){xy-l\). 305. (m + 6) (m + 7). 310. (m»i - 10) (mn- 12). MULTIPLICATION AND DIVISION 397 311. (a -3) (a -4). 316 . (a; -9) (a; + 8). 312. [x - 4) (a; - 6). 317 . (d-8)(d + 7). 313. [m + 7)(m — 5) 318 (ac - 10) (ac + 5). 314. :c-6)(c + 9). 319 . (aa;-13)(aa; + 10) 315. [n + Q){n-7). 320 {xy + ll)(xy-12). 321. (2a + 3)(3a + 2). 331. [a + b + cy. 322. [3 a + 4) (4 a + 3). 332. [x + y + zy. 323. ^4 a; + 5) (2 a; + 3). 333. [a + 2b + cy. 324. [6 n - 5) (2 n - 3). 334. [x + y + 3zy. 325. ;7a;-4)(3a;-4). 335. [m + 2n + 3y. 326. [3 a + 7) (5 a- 2). 336. [m-n- ly. 327. ;5a; + 3)(4a;-5). 337. 'x-2y-iy. 328. (3c + 7)(5c-2). 338. ( 2a-Sh + 2y. 329. :4 2/-9)(9 2/ + 2). 339. ;3a + 6-2c + dy. 330.' ^7 a; -8) (4 a; + 9). 340. ^4a-36 + 2c-3)l Oral Division Give the quotient of : 341. {a?' — m})^{a + 'm). 342. {a^ - x") ^ {a - x). 343. (m^ — M^) ^ {m — n). 344. {6? — im}) ^ {d + m). 345. (x'^ — c')^(x-c). 346. (ri.2 — 2/2) -f. (n + 2/). 347. (c^ - 9) -- (c + 3). 348. (a;2 - 4) ^ (a; - 2). 349. (9 a;2 - 1) H- (3 a; + 1). 350. (25 a;2 - 4) -=- (5 a; - 2). 351. (36 a;^ - 25) -=- (6 a; + 5). 352. (49 a;^ - 100 y") -=- (7 a; - 10 y). 353. (a^ - a^) -- (a - a;). 357. (a^ - 27) ^- (a - 3). 354. (ffi - //r') -t- (c - m). 358. (8 a' - 1) -^ (2 a- 1). 355. (d'^y>)-i-(d-y). 359. (27 - S .f^) - (3 - 2 «). 356. (a' - 8) - (a - 2). 360. (27af'-125)-f-(3a!-5). 398 SUPPLEMENTAKY EXERCISES 361. (m' + ?i') - T- (m + ?i). 365. ( 362. (c' + a^)^ (c + a;). 366. (! 363. (a' + f)-^ - (a + y)- 367. ( 364. (a' + 8)^ (a + 2). 368. (. 369. (a* - a;^) -; -(a -a;). 377. (( 370. {a' - x') - -(a + x). 378. (( 371. (a'-x^)-i - (a - x). 379. (( 372. (a" - x^) - - (a + a;). 380. (( 373. {a' - 16) - ^(a-2). 381. (' 374. (a' - 32) - ^(a-2). 382. (< 375. (a'-a;0^ -ia-x). 383. (( 376. (a' - 128) ^(a-2). 384. (. (64 + a?) H- (4 + a). (8 a' + 1) -- (2 a + 1). (125 + 27 a') ^(5 + 3 a). (216a'+343)^(6a+7). (a^ + «*)-=- (a + a;). {al + x') -i-(a + x). (a^ + l)-=-(a + l). {a' + 32) -=- (a + 2). (1 + aV) -7- (1 + ax). (32a^ + l)-=-(2a + l), {a' + 128) H- (a + 2). (32 a' + a^) -- (2 a + x). Give for each of the following a divisor and its correspond- ing quotient : 399. a^ — 16 X*. 400. a* — x^. 401. a' + xK 402. a^-32. 403. 32 + a;^ 404. a' — x'. 405. a^ + x\ 406. w' — af. 407. a' + 128. 408. a'— a;'. 385. a' — x\ 386. a? — x^. 387. a^ — 4. 388. a' — 8. 389. 8 a' -27. 390. 64a2— 1. 391. 64 a' — 1. 392. 64a'— 1. 393. 27a^ + l. 394. 8a^ + 27. 395. 125 a;' -27. 409. a*—x\ 396. 125a;3^216. 410. a^ + x\ 397. a* — x*. 411. 4a''— 25. 398. a* — 10. 412. 8 a' + .c". 413. m'' + 27. 414. c'' + 64. 415. 8a^-a?. 416. c'd' + a;i2. 417. a^a? — y^. 418. Qi:a? + 21yK 419. 125c« + 8. 420. 216-27a;i2- 421. aV-8l2/^. 422. cV + 125. 423. 343 a'- 27. 424. 512 + 125a:«. 425. 625o»-l. 426. 81a'^-16. FACTORING 399 Factoring Find the factors of : 427. 8(^-27. 437. 81x*-'Sx. 447. 3 m' - 3 m^ 428. 16 m* -81. 438. 250 a;^ - 2 a;. 448. 32 a' + a^. 429. 16 m'' — 81. 439. 125 a^ — 5 a^. 449. 216 a; — a;'. 430. m* — 27 m. 440. 8 a' — 8 al 450. 96 a;Y — 3 xij. 431. 27a^ + a;. 441. 169 a;' - 144 ,r. 451. 243 a* - 48 a. 432. 54 a;' -16. 442. 2a;^-256. 452. 169 3;^ + 169. 433. 81 a^- 25 2/2. 443. 3 c*^ - 729 c. 453. 64:xY + 2y\ 434. 50 a' -18 a. 444. 125aV + l. 454. 162 a;'' - 486 a;. 435. a^-64ar'. 445. 3 aj^ + 3 x*. 455. 375 ai^^ - 243 a;*. 436. a;«-64. 446. 5 a' - 5 a;'. 456. 200 a;*- 2 a;^ 457. 2a2 + 20a+50. 474. 2aa; + 6 a+4a! + 12. 458. 2a2 + 24a + 70. 475. c* + 7c^ + 16. 459. a? + 2a;2 + 3a;4-6. 476. 12y^--20y + 3. 460. 3 a;''' + 5 a; + 2. 477. 50 a;^ - 140 a;* + 98 a;. 461. m* + mW + a!^ 478. 81 a;^ — (a + 1)^. 462. 12 a;2 - 36 a; + 27. 479. cW - aj*- 4 a;-4. 463. a'-x^ — 2x — l. 480. 3 a^ + 6 0==— 24 a. 464. 6 m^- 11m + 3. -481. 10 0^ + 23 + 12. 465. 5 a' + 25 tt2 + 30 a. 482. 16 m^ — 44 m^ + 25. ■X 466. 4(a + a;)2 — 9. 483. 45 - 4 a; 467. 16*2 -56 a; + 49. 484. 16 a;^ -4 a*- 9 + 12 a. 468. 4 a!* — 4a!'' — 4 m'' + 1. 485. a;' — a;^ + a; — 1: 469. 15 02-380 + 11. 486. 36 (.u + 1)" - 1. 470. (a + l)2 + 5(a+l) + 6. 487. 200 a!^ + 18 + 120 a;. 471. 4 a;* + 36 a^ + 56 a;2. 488. 12 a!^ + 28 ayy + 15 ?/2. 472. 16 (a + 1)2 -9. 489. (a; + 1)2-10 (a: +1) + 25. 473. m2 — n2— l + 2n. 490. a* + 15 a' - 34 a2. 400 SUPPLEMENTARY EXERCISES 491. 100 x'' - 9 (a + 1)^ 511. 25 7rv'-9x'^+12xy-4:y\ 492. a? + 2x' — x — 2. 512. cV— c' - ix^ + 4:. 493. ISO tt' + 125 x^ - 300 ax. 513. 169 a;^ + 100 / - 260 scy. 494. 30^ + 82z'' + 56z. 514. (m+xy-hl7(m+x)-3S. 495. 24:a-2x'' + 2a' + 72. 515. 4 (a + a;)^ - 9 (a; + y)^. 496. SOc'-STc'i-TTc. 516. 99 a? + 39 x'^y - 18 xy\ 497. 100 cW + 144 cW + 81. 517. 4: d' -9 c? - 4: c^ + 9. 498. 25 (a + «)''- 9x2. 5ig 64 a;^ + 87 xy + 49 2/1 499. a* + a^ + a + l. 519. 56 m^ + 8 mwa; — 10 nV. 500. 5a!5-5a;2/''+30a;2/-45a;. 520. ~40xy-a^b^ + aP+i00y\ 501. 4a2 + 4a-288. 521. 2 rn'riV - 19 mux - 33. 502. 56 n^ + 45 - 107 «. 522. 25(a; + 1)^- 36(a; + 2)1 503. 4a* + 8aV + 9x\ 523. a' + 8-aW-8a?. 504. {a + xy + 6{a + x)+9. 524. 7 a;^ + 21 af - 490 asl 505. 15mV — 23 mn + 4. 525. 135 a;^ + 320 a;. 506. (c + a;)2 - 7 (c + a;) - 30. 526. a;^ + 27 a;^ - a^ - 27. 507. 3a^ + 24a-99. 527. (a + a;)'' - (a + x) - 12. 508. mV — 4 a;'' — m2 + 4. 528. a^ + ax + bx + ab. 509. 512x5 — 162 a;. 539. m?+ {m + n)x-\- mn. 510. 4 aV - (x^ + a'' - 1)^ 530. 4 m''- 9 x^- 1 - 6 a;. 531. a'' — 2 aa; — 2 C2/ + x^ - 2/2 - c\ 532. ax + ay + a% — mx — my — mz. 533. a2 + 2ax + x2 + 4a+4x + 4. 534. 4 m" — x^ ^- 2/2 + 2 mx — w^ _ 4 my. 535. o2 _ d2 + x2 - 2/2- 2 ax - 2 d2/. 536. m2 — x2 — a2 + 2/2 — 2(m2/ — ax). 537. x2 — (c + d! + l)x + c +d. 538. (a + l)2 + 5(a + l)(a-l) + 6(a-l)2. 539. 6(a-4)2-5(a2-16) + (« + 4)2. 540. 6 a2 — 12 ax + x2 — 7 a + 7 X + 1. ADDITION OF FRACTIONS 401 Addition of Fractions Simplify : 541. ^±3_^:z_2. 548. 2i«±3 2 + 3X-5 4 6 4 .3 ■ 542. 2x-3_2x-5 g^g 5^:rl_2-3^±i 6 9 6 5 ■ 543. 3x_f2_5^ gg^ 5-3^_g^2a,-3 544. 2^_^. 551. ^^ + 2- 4 a 3 a as + 1 545. 3a_+1^2a-3_ ^^^ «-«^_i + 6 a^ 4 ^3 a + a; 546. i^& + £:^. 553. l^_2a- ab ac a + 2 547. ^L^_?!:=^. 554. «^ + 2 ' 5 x + 1 x-1 a + x a — 5" a2-2 a'^a; aa? a^ — 1 a^-fl 555. ^^ + 1_^^±A_?!^M^. x + 1 x — 1 x^ — 1 ,„„ o + 4a; a — 3x , a^—3ax~2x'^ 50D. . a + 2x a-2x a?—4.x^ , __ a — m,a + m . 2 m^ 557. 1 I — « + ?ft a — wi a^ — m^ 558 tt — 2 a + 2 5a' + 3a + 4 ■2a + 2 3«-3 6a2-6 c-n 3 as , 4 a; 7 a; 55y. h ■ a^ + aa; a^ — ax a' — x^ 1 ' 1 560. h - a2 + 3a + 2 a^ + 4:a + 3 a2 + 5a + 6 561 a^' + Jg + 1 . a;-! a;^(2a;-l) ■a;^-l-2a; + 4a;-2 3^-8 SOM. EL. ALG. — 2G 402 SUPPLEMENTARY EXERCISES Multiplication and Division of Fractions Simplify : 562. 563. 564. 565. 566. 572. 573. 574. 575. 576. 577. 578. 579. 580. 581. a' -x-- a^-^asc ax (a + xy m' + 4 m + 3 m" — 3 m m^ — 9 ■ni' + m ' (? + ex 567. 568. irn^ + m,— 6 . m-t-^4 to'' — TO — 2 ' TO + 1 ' c' - x" . c2 + 2ca; + a;2 CO! - 2 a;2 c'' + ca; — 6 a;'' cx-\-Sx'^ a x\ ax + x^ 569. c' — af c^ + ca; + a;2 g^-a;^ . a' + a;' a^ + aa; + a;'' a' — af e- ay a'' + 2aa;+a; c 570. «_2 + ^ a; a + 1+- (^ — a^ 571. TO- -6+- TO frn_5\ \5 mj a -12 a' + 4. a- 5 g' + 5 g + 4 g2 + 8g + 15 ' a^-l ' 2 g' + 7 g + 3 _ 3g' + 7a + 4 3o2 + 5a + 2 ' g2 + 8g+ 15 ' 2a'i-a-l' g" 4- gy + gx + xy a^ + am + an + ran _^ a'' + 2 ay ^- ?/ a'-lQ a2 + 6g+.5 rt^ + OTO + ax + mx a^ + an + aa; + nx a^+ 2ax+ a;"' (g + a;)' - 1 _ (a - xf -\ ^ o' - a;' + 2 .t - 1 a^-(x + Vf-' (a + a;+l)2 ' a2 + 2a + l-a!2' g- 1 + g + a; , g g'+l a + 1 X — X & + cP a + x^ c — d 1 + ^^) g+iy a+x _ a — x a — x g + a; c I \d c 4 a' rt(a + 2)+l .2-1 a; g g X c' — cx + x' c^ — ex a;_g_2 g a; g ga; c + x c' + a;' + 2(? (? + 3? 11- X— g + l X + a — 1 2 ca; 0^^ + 03; + .x' a2 + ^V('«'+--2a a) \ a gV + 1)4-1 .«'-!' _g' + 2g2-g-2"g + 2 COMPLEX FRACTIONS 403 Complex Fractions Simplify : g + l a x + 1 .)• — 2 a a +2 x-2 7+1 582. —i ^- 586. -. —■ a + 11 a; — 1x4-2 a + 2a a; — 2a; + l X ^ + 1 a^ + l (a-lY x-1 a^-la'^ + l 583. :z 587. — ^ 1 + ■a' + l x—1 a* + a^ + l c + d d_ fc^+icr ^\ caf 584. \ ' + '. 588. V^r , ^^ + -^. 1,1 icx(c + x) c d c^ — cx + a? 1 2__ oj^ ^ ax -\- ay -\- xy x — 2 x + 2 a? + am ■]- an + mn 5o5. — qI 1 — ■ 589. ■ • o . 4 ac + ga; + cy + jcy a; — 2 a; + 2 ac + aa;4- cm + mx 590. 591. 592. 593. -2 a;-— ^-4 I a; — 4 -2 a;- a; — 2 a; — 4 (\ _ l-« , 1 + 2 g^ N ^ 2a + l \ 1 + g 1 — ct-y" g + l l-[a(a-2) + l] c^ — 6 ca; + 9 .x'^ c='- 4 ca; + 4 a;^ c^ -9a;^ c^ + ca; — 6a^ c^ -4a;^ c^ — ca; — 6a^ 1 1 1 g + l 1 1 ''+^ a-1 1 ' g ' u g+l g— 1 g+l 404 SUPPLEMENTARY EXERCISES Fractional and Literal Equations Find value of a; in : 594. ^-"^ x-6 3 a;-2 x~l x+2 3 4' 2 7 5 595. 3^ + 1 4^-1 S 597. 2^ ^^-6 3 i«+9 598. 2a;-l 3a;-5 2a;-6 ^ 4 5 3' 599. 16a;-8 10a;+l . 7a; + 3 7 4 8 ■ 600. a;^ 03 + 1 6 _Q x^ — 1 x — 1 X + 1 601. x+1 1 _x—l x + 5 a;2 — 25~"a; — 5' 602. 2 a; 5a;2 _ 3 a; 3 a; + 6 6 :«2 _ 24 ~ 2 a; - 4' 603. 4(a;-l) 5(x--2) _ 9(a;-3) a;2 + 3a; + 2 a;2+5a; + 6 x'' + 4a; + 3 604. mn + ni^x = mhi — mx. 607. cx + 2 cd = dx + c^ + d?. 605. (c + d).« = c2 + cd. 608. 2ax— 9 m'=3m.');-4a2. 606. ax-\-bx = a?- — V. 609. 4aa; + 4 m = 2 ca; + 2m. 610. aa; + 6a; + ca; = (a + hf — cl 611. (m + >t)a; = m^ + »i't — TO — «• 612. 2 ma; + 4 m = (m + 1)^ + 2 a;. 613. (m^ 4- xy = x^ + i m^ + m\ 614. (a;+l)2 + 3a=(a; + a)2 + 3. 615. (a; — l)(a;— a) = (a; — a — ly. 616 ?-^ — - — E!~i2 = i+_^ 618 ^ + ™ I X 4x— m _Q a a; aa; a x — m 2 a; — m „.„ a; 5 a 2 a; „,„ a;— a .i- 4- (» 2 aa; 617. -■=^— -. 619 a + 1 «2-l a-1 a + 1 a^-l PROBLEMS LEADING TO FRACTIONAL EQUATIONS 405 Problems leading to Fractional Equations State and solve : 620. Find that number, the sum of whose third, fourth, and fifth parts is 47. 621. Three times a certain number exceeds one fourth the number by 77. What is the number ? 622. Divide $100 between A and B, so that two thirds of A's share shall equal B's share. 623. Find the number whose double is 34 more than the sum of its third and fourth parts. 624. Divide 80 into two parts such that one seventh the smaller part shall be equal to one ninth the larger part. 625. A is 15 years older than B. 6 years ago B's age was I of A's age. How old is each at the present time ? 626. Find the two parts of 150 such that the greater de- creased by 60 shall equal one half the smaller part. 627. A is now half as old as B, but 10 years ago he was one third as old. What is the age of each ? 628. What number added to both numerator and denomi- nator of the fraction ^ will produce a fraction equal to ^| ? 629. A man is three times as old as his son, and f of his age plus f of the son's age is 70 years. Find the age of each. 630. One half of the number of days in a certain school year is 31 days more than one third the number. How many days are there in the school year ? 631. Find three consecutive even numbers such that one half the sum of the first and second shall exceed by 6 one third the sum of the second and third. 632. Three boys together earn a sum of money. A was paid $5 less than one half the total, B received $5 more than one fourth of it, and C received the remainder, $15. How much did each receive ? 406 SUPPLKMKN'l'AliV EXRUCISES 633. The distance from Philadelphia to Jersey City is 20 miles more than the distance from New York to New Haven, and the first distance is ^ of the second. Find each of the distances. 634. The paving brick necessary for a square courtyard would exactly cover a rectangular yard whose' length is 6 yards more and whose width is 4 yards less than the side of the square. How long is the square ? 635. A company of men is arranged in a hollow square with 7 rows of men on each side. If there are 840 men in all, how- many men are there on each outside row of the square ? 636. The denominator of a certain fraction exceeds the numerator by 12. If 5 is added to both numerator and de- nominator, the fraction becomes |. What is the fraction ? 637. In one day a boy reads 3 pages more than ^ of a book, and on the second day he read 2 pages less than i of it. The number of pages still remaining was 42. How many pages were there in the book ? 638. Four men pay $ 6200 for a lot. A pays f as much as B. C pays half as much as A and B together, and D pays $800 less than A and B together. What amount did each pay? 639. The distance around a rectangular field is 152 rods, and one third the length plus one half the width equals 29 rods. Find the length, the width, and the area of the field. Simultaneous Linear Equations Solve : 640. 4 a; - 3 2/ = 30, 642. 2 a; - 2/ = 9, 644. 2 a; - 9 2 = 11, 3x — 5y=17. 5 X— 3 y = 14:. 3 a; — 4 2 = 7. 641. 2 a; + 4 2/ = 54, 643. 6 s - 13 f = 2, 645. 8s + 3a; = 3, 3x-3y = 45. 5s-12< = 4. 12s + 9a; = 3. SIMULTANEOUS LINEAlt EQUATIONS 407 646. Sx = 5y, 649. 2x — z = 0, 652. 5x — 7y = 0, ix-y = n. 3z + l=x. 3a;-l = 4)/. 647.^ + 1=7, 650.^ + ^ = 11, 653.^ + 1 = 10, 54-^ = 5 ^-U^S ^4-^=-1 3 4 ■ 4 2' 43 648.^ + 1 = 10, 651.^-1 = 8, 654. ^ + f = -6, 4 o 3 2 3 4 6 4' 24' 32' 655. ^i±^+-i±l = 5, 656. 2^i±i + 3y^ = 5, 3 4' 54 ' x-1 y + 1 ^ 25 4 .r - 13 _ ,v+2 ^ , 4 3 6" 3 5 ' 657. (x + 5) (y -^ 2) - 23 ?/ = x>/. 2X + 1/-1 _ 3 .-e - 2 y - 2 ^ _ 9 4 2 ~' 2' 4a;-;'^+y , 2a; + ,y + l 2 ^'^- 3 + 5 = 5- 3x + 7y-l 4,r + 3?/ + 2 ^ a 2 3 659. x + y +z = 6, x — y + z=:4:, x — y—z = 2. 660. a; + 22/ + 2; = 8, 2a;-2/ + 2 = 3, a; — ?/ + 22 = 5. 661. x + 2y + 3z = -4:, 664. 2 a! + 3 y + 5 z =!39, a; + 32/-2« = 0, 4a;-5y + 32 = 53, 2x-y + 3z = 0. 2y-5a; + 4« = 5. 662. 2x + 3y+-2z = li, 665. 8 a; + 4^-3 z = 6, 3.'B-22/-4« = -7, 7a;-52/ + 6z = 23, 4a; + 32/-22 = 4. 4a; — 5y + 42 = 8. 663. 3x + iy-2z = -li, 666. 15 a; + lOy — 92 = 16, 2 a; + 5 .V - 3 2 = - 20, 19 z-3x + iy = 20, 3a; + 2y + 42 = 18. 2\ y + 5z~17 x = 9. 408 SUPPLEMENTARY EXERCISES Problems leading to Simultaneous Linear Equations State and solve : 667. Find two numbers whose sum is 87 and whose differ- ence is 31. 668. One half the difference of two numbers is 9, and ^ their sum is 26. Find the numbers. 669. The sum of two numbers is 35, and their difference is 2 more than one fifth the smaller number. What are the numbers ? 670. 3 boys and 4 mea working together are paid $ 15. At another time 4 boys and 3 men receive at the same rate $13. How much is paid to each man and to each boy ? 671. Find two numbers, such that the greater is 6 times the difference of the numbers, and the sum of the two is 2 more than twice the smaller number. 672. The numerator of a fraction is 3 less than the denomi- nator, and 5 times the numerator is 1 less than 3 times the denominator. What is the fraction ? 673. Seven years ago a man was 4 times as old as his son, and in 7 more years the man will be twice as old. Find the present age of each. 674. If the greater of two numbers is divided by the smaller, the quotient is 1 and the remaind,er 5. Three times the greater number contains the smaller 4 times. What are the numbers? 675. A and B buy a horse for $ 180. A can pay for it if B will loan him half what he has. B can pay for it if A will loan I of his money. How much money has each ? 676. The age of a bride is f the age of her husband, but after 12 years pass her age will be | of his age. What is the present age of each ? 677. A field is 16 rods longer than it is wide and the total distance around it is 192 rods. Its shape being rectangular, find its dimensions. EXPONENTS 409 678. The sum of the three angles A, B, and C, of a triangle is 180°. The angle at A plus the angle at B equals 110°, and the angle at A plus the angle at G equals 135°. Find each angle in degrees. 679. Divide 144 into three parts, such that the third divided by the sum of the other two gives a quotient of 2 and a remain- der of 6, while the first contains the second 3 times with a remainder of 2. 680. Three pounds of sugar, four pounds of tea, and two pounds of coffee sell for S 3.28 ; five pounds of sugar, two pounds of tea, and three pounds of coffee for $ 2.55 ; and four pounds of sugar, three pounds of tea, and four pounds of coffee for $ 3.44. Eind the price of each per pound. 681. A man so willed his property that his eldest son received one half the amount given to the other three sons ; the next oldest received one third the sum of the other three; the third received one half the sum of the other three ; and the fourth received S 4000. What was the amount of the estate ? Exponents Simplify ; 682. (a"V- 690. (3 mhiar'^)-'. .698. (-7a->)-^. 683. (a?x-y. 691. ■(2a--3a-i)2. 699. {-2a')-\ 684. {a-^x-^y. 692. {?,x-r-Ax^y. 700. (2 a~^xy. 685. («»i-2)-2. 693. (3 a?y-hy. 701. (3-1 a^y. 686. (a^2/'z-')i 694. (-3 ay. 702. (-8"¥ 687. (27ca;-2)"i 695. {-axy. 703. (-27-). 688. (2a-itT)^ 696. {-Sw}xy. 704. (-125)-*. 689. {2xhj-hY. 697. (- ':, mx^y. 705. (^_64)^ 410 SUPPLEMENTARY EXERCISES 706. (ix-^y-^yi. 707. (9a-Vy^)-i. 708. (a;Va;-l)^. 709. {c-^-^e=')l 718, 710. (K-'VicVa;)-'. 711. (c-VF^)". 712. (3 a;~iA/»*)-2. 713. {cx-^Vc^y\ 714. (m"^VmV16)^ 715. |(v'a=3S)'Si 716. (m-''^m^Vm*y. 719. 720. 721. 722. xVa c~^'\/cx d-Wc^' X' /x~H z~'-y/x~''z 3 m~'Vf'~ •,n ■\^. Ttf Simplify: 733. V12. 734. V20. 735. ^54. 736. -v'96. 737. 075. 738. VlOSa^ 723. 724. 725. 726. 727. 717. [_{-\/27 d?yj. 2 a-VFi 5 a m'n ^y/n 2 aVI^'N -' ,3a:»V8¥^ 2a^^2Ta 3 Va;Va^\ Radicals 740. 2121 x" vf. Vf. 3 m 3/64 «■- 4 a V 9 m- 12m^S-V. 762. 9V2 + 6V3+V50-Vl08-2V98-f-3V75. 763. 2VT_V| + fV| + -V''-V|-iV24 + |V20. 764. V24 a' + VrA a' + V6^ - 2 Vl50^ + v'294^, 765. ^ + 3vT8¥'-?V501? + ^a/1 + ?J^. V2 a a 20 Va 3 V 2 766. «V| + iV|-|V^i-iV2i + |V40. 767. 2Vi8x3Vl2. 772. *^Xx|-^|. 768. 3Vl5x4V35. 773. VSrfi x ^9^. 769. iVi8x|V24. 774. ^18 x V3. 770. iVUxfVf. 775. \/l2xV2r. 771. l^fxeVf. 776. V^Xa/|. Rationalize the denominators of : ,,, 3+V2 ^gQ 2V5+4V3 _ ^33 2V2-V3 3V5-V3 ' " 2V2 + Va" ^gj 3V^-|-4V2 ^g^_ 3V2-V6 " 2Va-3V«' ' 4V2+V6" 782. ^^ + 3^1 785. 4Vi5-VIO . 3 V3 - 4 V6 2 V3 - V2 786. Va'^ + 5 = a; + 1. 787. V.r + U = 11 — Vcc. 3-V2 77ft 4 + V2 4- V2 77Q V3 + 5 V3-2 Solve : 412 SUPPLEMENTARY EXERCISES 788. V.i; — 5+Vx = 5. 791. V'-!.'-- 1 — T= \/2 3;+6. 789. y/x'-~-~8x + 6 = X. 792. VS^ + l- V^ .(;-6=l. 790. Vx^^ + y/x + 5 = 7. 793. Vx+l»t- Vx-6=2V2. 794. V^Zr2 _ V4 a- - 3 + Va- + 1 = 0. Miscellaneous Quadratic Equations Solve : 795. 2a? + 5x = 12. 801. 4 ar^ + 12 ma; = 7 m^. 796. Sx"— ■a;=2. 802. 3aV + 8aa; = 3. 797. 5m^ + 18x=8. 803. 2 aV — 11 aca; = - 15 c^. 798. 8 a;'- -10 a; = -3. 804. x^ + {a-h)x = ab. 799. 2 a-^ + .(' + 7 = 0. 805. a;" — ma; + nx = mn. 800. 4a;2 + 5;r + 3 = 0. 806. aca? — adx = bcx — bd. 807. (4a; + l)2- (3a;+2)2=(2a;-3y + 4. 808. (3 a; + 2) (2 a; - 1) - (2 a; + 1) (3 a; - 2) = (a; - 3)1 809. (5 a; - 1)2 - (4 .t; - 1)= - (3 a; - 1)^ = :^ (x- + 7) 810. 2(2a;+l)(a;-3)-3(3x+2)(a;-2)+3(a>-l)(a;+7) = 0. 811. ^±3 + 1 = ^+1. 816. -1 ^ = 1-1. m — X n — X m n 812. ^^^--^ = -^^. 817. ^^+|±i^ = 2. 2 + .1' 3 — a; 81.3. ?^ + ^±i-10 = o. 818. 2aa;4c_^3aa; + -.^Q ' ■* - '' " 2 ax— c 3 ax — 2 c x + i ^-a; + 2 x + 1 8 a; + 3 x + 2 3 x + A a; — 4 a; + 4"^ a; + 4 10 a; -4 3 = 0. a;— 7 a; + 4 (2x-l)_ a; — 7 :1. a;-l , - + 1_ 25 a; 814. ■■'i^S_{ --^--> ^l, 819. ^_ + -ili_ = l. a; + 4 a; — 7 a; — ma; — '/ 815 a;-l , a; + l ^ 25 a; ^^o a;-2c x+2c 10 ' 2a;+2 ' 3a;-3 63^-6 ' x + 2c x-2c 3 821. a; — 2 = Va;— 2. 822. V5 a;— 4= Va;+2. PROBLEMS PRODUCING QUAURATIC EQUATIONS 413 823. Va; + 1 = 11 - x. 827. V? a; + 1 - \/3 :i' + 1 = 2. 824. ViV — 5+a; = 5. 828. Va; ^3 = V4 a; + 1 - 2. 825. .1- - 7 = 3 Va; - 7. 829. Va; + 5 — Va; — 3 = V^. 826. 2 Va; - 6 + 5 = a;. 830. Va; + 8 + Va; + 3 = 5 V«. 831. VaT- 12 - Va; + 9 = VaT^^. 832. Va;-2+V5a;- 1 = V4 a; + 1. 833. x*~5.-r + i = 0. 839. 8 a;^ - 3 a;-^ = — 10. 834. ,i-'-8a- = 9. 840. a;-' - 3 a;"^ - 4 = 0. 835. .i;" - 7 a.-3 = 8. 841. 6 a;"^ + 13 a;~J + 6 = 0. 836. X-'' - 3 a;-^ = 10. 842. v^ + 3 Va; - 10 = 0. 837. x-' + 7 x-' = lU. 843. V*"' - 6 -^a? = 24. 838. .i;-' + a;"^ = 20. 844. 3 Vk^ + ^^ = 2. 845. (a; + 3)--5(a; + 3) = 14. 846. (a,-^ + 2)^-6(a,-" + 2) = 66. 847. (a;^ + a;)- - 8 (a;^ + a;) + 12 = 0. 848. (a;^ + 3 a; + 6)2 - 2 (a;= + 3 a; + 6) = 8. Problems producing Quadratic Equations State and solve : 849. Find the two factors of 600 whose sum is 49. 850. Find three consecutive numbers, the sum of whose squares is 194. 851. Find the number that, added to seven times its recipro- cal, gives a total of 8. 852. Find two factors of 357, one of which exceeds the other by 4. 853. Find two numbers whose sum is 30, and whose squares when added give a total of 458. 414 SUPPLEMENTARY EXERCISES 854. Divide 15 into two parts whose product shall equal ten times their difference. 855. One number is 1 more than another, and their sum plus their product equals 19. Find them. 856. A square floor would contain 31 square feet more if each side were 1 foot longer. Find its area. 857. By what number must you divide 266 so that the divisor and quotient are equal and the remainder is 10 ? 858. One leg of a right triangle exceeds the other by 2 feet, and the hypotenuse is 10 feet. Find the length of the legs of the triangle. (See p. 220. Ex. 37.) 859. A garden 40 feet long and 28 feet wide has around it a path of uniform width. The area of the path is 960 square feet, ^hat is the width of the path ? 860. A company of men hired a hall for $ 96, but 4 of the party failed to pay their share, and each of the others paid $2 more than they had agreed to pay. How many men paid the bill ? 861. Two men started at the same time to go from Philar delphia to Jersey City, 90 miles. A reached Jersey City 1 hour before B, for he traveled 1 mile an hour faster. At what rate did each travel ? Simultaneous Quadratic Equations Solve : 862. x + i/ = iO, xy = 300. 864. x + y = 12, x^ + y'' = 80. 863. x — y = 6, xy = — 8. 865. x — y = 9, 05^4-^^ = 45. 866. 2x + y = 5, 2x' + y' = 9. 867. 3 a; + 4 2/ = 24, x' + y^ = 25. 868. x' + xy + y'^ = 13, ar'-t/' = 26. 869. 3? — xy = 15, 2xy — y^= 16. 870. 3x^ — 2xy = 5, x — y = 2. 871. x'^ + 3xy = 28, xy + iy'^^S. .SIMULTANKOUS QUADRATIC EQUATIONS 416 872. a;^ + .?/' = 28, a; + ?y = 4. 873. a!-i + r' = fV. a;-^ + r' = ^• 874. a«/ + 3a;2 = 5, 'd xy + A y"" = 22. 875. x-y-2 = 0, a^ _ yf _ 886 = 0. 876. 3 a;2 - 4 ay + 2 2/2 = 17, a!^ - y^ + 16 = 0. 877. 2a;2_3a.T/ + 2?/2 = 8, .v-?/ = l. 878. a!'' + ^ = 97, x + y = -l. 879. a!^ - 2 a!!/ = 3 2/, 2a;2-9/ = 9?/. 880. x + y = 2, 12-xY = 4:xy. 881. a^ + a!V + ?/* = 133, a;^ + a;?/ + ly'' = 19. 882. x-'-y-' = l, a;-'-2/-=i-. 883. a;2 + 3/2 + 2.i; + 22/ = 23, x + y = 5. 884. a;2 + 2/2 4- a; + 2/ = 14, a;^ — y^ ^ a; — .*/ = 10 885. a;2 - a;?/ - 2 2/2 = 0, 3 2/^ - IO2/ = - 8. 886. a!2 + 2/2 = a;)/ + 10, a; — 2/ = a!2/ + 2. 887. x^ + 3xy = 27, 3 y'^ + 2 xy = 24:. 888. xY + xy==72, x + y—6 = 0. 889. a;2 + 2/2 = 52, a; + 2/ + a^ = 34. 890. 4 ar" + 25 2/2 = 2, 2a! + 52/-2 = 0. 891. 2 a!2 - 2 9:2/ — 2/2 = 3, a;2 + 3a;2/ + 2/2 = ll. 892. i 35 + 2 2/ = 18, ^a;j/ = 8. 893. a?* 4- a;2/ + 2 2/2 = 44, 2 a^ — xy + rf = 16. 894. af' + 2/* = 97, a;2/— 6 = 0. 895. y + Va;2 — 1 = 2, Va; + 1 — Vaj - 1 = Vy. 896. a;2 + .'j;2/+4 2/2 = 6, 3 a;2 + 8 2/2 = 14. 897. x^ + xy + y^ = 39, a;2 — a!2/ + 2/2 = 19. 898. x + y = 3xy — l, x'^ + y^ = 8 xY — 3 xy — 3. 899. a;2 + 2/2 + a; + 2/ = 32, xy + 16 = 0. 416 SUPPLEMENTARY EXERCISES Problems producing Simultaneous Quadratic Equations. 900. The sum of the squares of two numbers is 130, and their product is 63. Find the numbers. 901. The sum of the cubes of two numbers is 126, and the sum of the numbers is 6. Find them. 902. One of two numbers exceeds 30 by as much as the other is less than 30, and their product is 875. Find them. 903. The diagonal of a rectangle is 100 feet, and the longer side exceeds the shorter by 20 feet. Find the area of the rectangle. 904. The sum of the squares of two numbers is 13 times the smaller number, and the sum of the numbers is 10. Find them. 905. The sum of the squares of two numbers is 2 a^ + 2, and the sum of the numbers is 2 a. What are the numbers, and what is the diilerence of their squares ? 906. How many yards of picture molding will be needed for a room whose ceiling area is 1200 square feet, the diagonal of the ceiling being 50 feet ? 907. If twice the product of the ages of two children is added to the sum of their ages, the result is 13 years. One child is 3 years older than the other. Find the age of each. 908. If the sum of two numbers is multiplied by the lesser the product is 5. If their difference is multiplied by the greater, the product is 12. Find the numbers. 909. The difference between the numerator and denominator of a certain improper fraction is 2, and if both terms of the fraction are increased by 3, the value of the fraction will be decreased by ^\. Find the fraction. 910. A certain floor having an area of 50 square feet can be. covered with 360 rectangular tiles of a certain size ; but if the masons use a tile 1 inch longer and 1 inch wider, the floor can be covered with 240 tiles. Find the sizes of the different tiles. CHAPTER XXIX SUPPLEMENTARY TOPICS THE REMAINDER THEOREM 468. If any polynomial of the form, Cia;"+ 023;""'+ G^x''"'' + ■■• 6'„, be divided by x — a, the remaiiider will be CiU" + Cjfii'' '+ Cja""^ 4- ■ • • C„ ; which expression differs from the given expression in that a takes the place of x. Proof : Let Q denote the quotient and B the remainder when Cix" + dx"-^ + C3X"-2 + ... C„ is divided by s - a. Continue the division until the remainder does not contain x. Then, Qix-a)-\- B= Cix« + C^x^-^ + C'sX"-^ + ... a„. Since this identity is true for all values of x, let x = a. Then, Q{a-a)+ B = da" + O^a"-' + Ozo?'-' + •■■ C„. And, B = da" + da"-'^ + Gsa"-^ + •■■ 0„. Application : 1. Without division obtain the remainder when 7oi? + ^v? — 13 a; + 8 is divided by x — 2. We have given, x — a = x — 2, hence a = 2. Hence, /; = 7 ■ 2^ + 3 • 2^ - l.S • 2 + 8 = 50. Result. THE FACTOR THEOREM 469. If any rational and integral expression containing x be- comes equal to when a is substituted for x, the expression is eooactly divisible by x — a. Proof : Let £ be the given expression, and let E be divided by x - a until the remainder no longer contains x. Let Q denote the quotient ob- tained and B the remainder. SOM. EL. ALG. 27 417 418 SUPPLEMENTARY TOPICS Then, E= Q{:c-a) + M. This' identity being true for all values of x, we may assume that x = a. By the hypothesis the substitution of a for x makes E equal to 0. Therefore, = £(a-o) + JJ, = + B, B = 0. Therefore, the remainder being 0, the given expression is exactly divisible by x — a. Or, X — a is a factor of the cjivpn expression, E. Illustrations : 1. Factor ar''- 12 a; + 16. By trial we find that the exp session, a;' — 12 a;+ 16, equals when a; = 2. Therefore, if a; = 2, a; — a — x — 2, and a; — 2 is a factor. Then (a^ - 12a; + 16) -=-(x - 2) = x^ + 2x - 8. The factors of x^ -|- 2 x - 8 are found to be {x + 4) and (x - 2). Therefore, x^ - 12 x + 1 6 = (x - 2)(i + 4)(x - 2). Result. 2. Factora^+9a;2 4-23.r + 15. (In an expression whose signs are all plus, it is evident that no positive number can be found the substitution of which will make the expression equal to 0.) By trial we find that if x = — 1, the expression becomes 0. Hence, ifx = — 1, x — a= [x — (— l)] = x + l, a factor required. Then, (x^ + 9x2 + 23x+ 15)-s-(x + 1) = x-^ + 8x+ 15. Furthermore, x^ + 8 x + 15 = (x + 3) (x + 5). Therefore, x' + 9 a;^ + 23 x + 15 = (x + 1) (a; + 3) (a; + 5) . Result. Ezercise 138 Without divisioa show that 1. a^ + 3 a^ + 3 a + 2 is divisible by a + 2. 2. a;< - 8 k' + 24 K^ - 32 a; + 16 is divisible by a; - 2. 3. c^-c^-Sc^ + Qc — 9is divisible by c + 3. 4. 27 a'' + 9 «2 - 3 a - 10 is divisible by 3 a - 2. 5. jft" — 5 m^ + 9 TO^ — 6 m^ — m + 2 is divisible by m — 2. DIVISOKS Oh- BINOMIALS 419 Factor : 6. (^ + 0-2. 12. c'+c'-oc + S. 7. 3^-7x + 6. 13. a? + 33^-6x-S. 8. ?>i'— 8m + 3. 14. m^ + 5m- — 13m + 7. 9. a» + 3a2 + 7a4-5. 15. 2 f + of -37/-2. 10. x^ + AccP + 5x + 2. 16. 3a'-10(r + 4a + 8. 11. TO5-2m=-7m-4. 17. a;*-ar'-3a^ + 5a;-2. THE THEORY OF DIVISORS OF. BINOMIALS 470. The following proofs depend directly upon the prin- ciples established in Arts. 468 and 469. If n is a positive integer, we may establish as follows the divisibility of the binomials, af — y" and xf + y", by the bino- mials, x — y and x + y. I. of — y" is a! ways divisible by x — y. For, it ij is substituted for x in a;» — ij", we have x'^ — y" = y" — y" = 0. Therefore, x — y is always a divisor of x' — y". II. of — y" is dici.sible by x + y if n is even. For, if — !/ is substituted for a; in a;" — a5 + 102a*-171a5 -36a5+ 54a<-27a3 3(2a2-3a)2=12a<-3HflrH27a--' 3(2a2-3a)(4)= 2ia^-?,e>a 42= +]( 48 a<-144 0^+204 fl!2_l44a+64 48a^-144aH204a2-144a + 64 475. The distribution of the volume of a cubic solid whose edge is a 4- h, and the relation of the separate portions to the separate terms of the polynomial representing its cube, can be readily seen from the following illustrations. The planes cutting the cube pass at right angles to each other and parallel to the faces of the cube ; each at a distance of 6 units from the faces of the cube. 1 U JB. JS In the figure let the edge of the cube in (I) be a and the edge of the complete cube in (IV) be a + &. I. a\ II. a? + ?,a?h. III. a? + ^a^h + 3ah\ IV. a^-\-^a% + ?jab'^ + W. ) 428 SUPPLEMENTARY TOPICS THE CDBE ROOT OF ARITJIMETICAL NUMBERS 476. It will be observed that 1^ = 1 A number of one figure has not more than three 93 = 729 figures in its cube. 10^ = 1000 A number of two figures has not more than six 998 = 970299, etc. figures in its cube. Conversely, therefore: If an integral cube has three figures, its cube root has one figure. If an integral cube has six figures, its cube root has two figures, etc. Hence : 477. Separate any integral number into groups of three figures each, beginning ai the right, and the number of groups obtained is the same as the number of integral figures in its cube root. It is to be noted that the process of cube root of numbers as given in arithmetical practice is based directly upon the al- gebraic principles learned in Art. 474. Illustration : 1. Find the cube root of 421875. Separating the number into groups of three figures each, we haye 421875. 421876 1 70 + 5 = 75. Result. gs = 70° = 343000 a+b 8(a)2= 3(70)2 = 14700 8(a)(6) =3(70) (5)= 1050 (6)2 = (5)2 = 26 (15775) (5) = 78875 78875 The analogy between the formation of the cube of (a + 6) and the cube of 75, or (70 + 6) , may be seen in the following : (a + 6)8 = (js + 3 0,26 _,_ 3 362 + fts (70 + 5)3 = 70S + 3(70)2(5) + 3(70) (5)2 + 52 = 343000 + 73600 + 5250 + 125 = 421876. CUBK ROOT OF AKITIIMKTICAL NUMBERS 429 2. Find the cube root of 12812904. In practice the process is usually abbreviated as follows: 12812904 123 4. Result. 2-' = 8 3(20)-'' = 3(20) (3) = 32 = 1200 180 9 3(1389) =^ 4812 4167 3(230)2 = 3(230) (4) = 42 = 158700 2760 16 4(161476). = 645904 645904. Exercise 142 Find the cube root of : 1. cv' + ea' + 12 a + 8. 2. 27 c^- 54c- + 36c -8. 3. 8a'-36a^ + o4a-27. 4. 27 x^ - 135 X* + 225 x' - 125. 5. a^ + 6a' + 15a* + 20a^ + 15a' + 6a + l. 6. a^-6x^ + 21x*-Ux» + 63x'-5ix+27. 7. a' + 9a' + 21a*-9a?-A2a' + 36d,-8. 8. 102.T;* + 204 x' - 171af' - 144a; + 64 - 36a:' + 8a;^ 9. 27 wi.'" - :^r, m^ + 12m2 - 8 + 30m^ - 45m» + 27m'l 10. ,■" - 3c' + 6 c^ - 4c« + 6c^ - 2c= + 3c + 1. 11. 74088. 16. 3112.136. 12. 389017. 17. 14706125. 13. 658503. 18. 48.228544. 14. 912673. 19. .559476224. 15. 1953125. 20. 000138991832. 430 GENliliAL KEVIKW GENERAL REVIEW Exercise 143 1. Find the numerical value of 81"^, I62, (|)-*, 5", 25~t Detine briefly tlie law that governs each reduction. 2. Solve X + Vn^5^ + 0.06 = 0. 3. Eorm the equation whose roots shall be a — V — 1 and a + V— 1. 4. Solve for x : 2(4 - 3a;)^ - 12 = x. 5. Factor 125 a!"'- Va?. 6. Solve X + 10 a;"' + 7 = 0, and test the roots obtained. 7. By inspection determine the nature of the roots of 3 a;- — 7 a; = — 5. 8. Solve 4 a;^ — 4 aa; + cv' — c'^ = 0. 9. Solve and verify both solutions of (a; + l)^(a; — 2) = a;(a; - 3) (a; - 1) - 4. 10. Solve - — %— = a;(a.- + 2) + 1. (a; - 1)-^ 11. Find the square root of (aj^ + l)-' x-^ (x^ + l)-! 12. Find the value of m for which the equation, 3a;- — 6a; + m = 0, has equal roots. 13. Factor 2[2a2 _ (a; _ a) - 2 a(2a! - a)]. 14. Write the 8th term of the quotient of a;''— y" divided by x - y. 15. Solve a;^ + 2a;3 — 3 = 0. 16. Solve ^-2m _^ 3m ^ x + c . c + cZ a; + 2 m a; + 2 m 17. Find the five roots of aj^^ - 4 af - .-c^ + 4 = 0. GENERAL UEVIEW 431 18. Factor IT) .1,1'' — 15. 19. Solve x{x — y) = -', (a; + y)° = 9. 20. Form that quadratic equatiou whose roots shall be - i and - f 21. Obtain the factors of x' — 5x + 2 by solving the equa- tion, x? — 5 x + 2 = 0. 22. Divide 16 into .'J parts in geometrical progression, so that the sum of the 1st and 2d shall be to the difference of the 2d and 3d parts as 3 : 2. 23. Find log of 7, given log 5 = .6990 and log 14 = 1.1461. 24. Solve and test the solutions of V2m + a; — V2m — x — y/2x = 0. 25. Factor 4 + --^^-! — - -\ — • X p~^ 26. Form the quadratic equation whose roots shall be 1+ V^^ and 1 - V^^. 27. Solve a^ + xy + y^ = 7x, x^ — xy + y^ = 3x. 28. What is the sum of the first n numbers divisible by 5 ? 30. Solve 2^' X 2^ = 40. 31. Givenl+V«: l+3Vi = 2:5; finda;. 32. Factor x-^-2+x*. 33. If n is an odd integer, which of the following indicated divisions are possible ? a;" + jv" x'+y" x^ — y" af — f x+y ' x—y ' x+y ' x—y 34. Form the quadratic equation whose roots are 1 -f- a V« and 1 — n V oc. 432 GENERAL REVIEW 35. Under what condition will the roots of (u^-\-hx + c = be iinagiuary ? JProve your answer. 36. If the length and breadth of a certain rectangle are each increased by 2 rods, the area will become 48 sq. rd. ; but if each dimension were decreased by 2 rods, the area would become 8 sq. rd. Find the dimensions of the rectangle. 37. Factor a^ + 2a.'2-5a;-6. 38. Write in logarithmic form 27' = 81, and find x. 39. Solveforw: ^^-2c = -^'* n n — 1 40. What is the price of candles per dozen when 3 less for 36 cents raises the price 12 cents per dozen ? 41. Show that the product of the roots of »'' — 5a;— 2 = is -2. 42. Prove that — =, 3, ^ , are in geometrical progression, and find the sum of 10 terms. 43. It m : x = n : y, show that mx — ny : mx + ny = m^ — n^ : m? + f^- 44. Find an expression for the nth odd number, and illus- trate your answer by a numerical substitution. 45. Find n in the formula I = ar'""^. 46. Solve for a and m : 2 a^ — am = 12, 2 am — m^ = 8. 47. Factor (x* + 3)-^ (l + 3a;)-i 48. Eationalize the denominator of "^ — _ ~ — ^ • 2 - V6 + 2 V2 49. Insert 6 geometrical means between J^ and 12|-. 50. Find X m — - — = -^^ 12 Vx GENKUAL REVIEW 43:5 51. Without solving, prove that the roots of 6 .r- + 5 x= ^1 are real and rational. 52. Solve m + V;! x + .x-' : m + 1 =m — V.'i x + xr : m — 1. 53. The difference between the 5th and the 7th terms of an arithmetical progression is 6, and the sum of the first 14 terms is — 105. Find the first term and the common difference. 54. Without solving, determine the nature of the roots of 16 .X- + 1 = 8 X. 55. Find the 5th term of («2-a;-2)'^ 56. Show that the sum of the squares of the roots of a^-3a;+l = is 7. 57. If m^ — rv' varies as a^, and if a; = 2 when m = 5 and n = 3, find the equation between m, n, and x. 58. Find x and y if 2^+" = 16 and 3^-" = 9. 59. Find a 4th proportional to x* — 1, a;^ + 1, and a;^ — 1. 60. Construct the quadratic equation, the product of whose roots shall be twice the sum of the roots of a^ — 7a; + 12 = 0; and the sum of whose roots shall be 3 times the product of the roots of a^ + 2 a; = 3. 61. Factor 3 a^ - 2 4- .T^ + a;^ — 3 a;. 62. Find by logarithms the value of ■\/2 x (i)* X .01 x Si 63. The sum of two numbers is 20, and their geometrical mean increased by 2 equals their arithmetical mean. What are the numbers ? 1 _ 64. What is the interpretation of a;^ = Va;? 65. Write the 5th term of (a + 6)"". 66. Solve c^ (af + 1) = m' + 2 c'x. 67. The sum of the last 3 terms of an arithmetical progres- sion of 7 terms equals 3 times the sum of the first 3 terms. The sum of the 3d and 5th terms is 32. Find the 1st term and the common ratio. SOM. EL. ALG. 28 434 GENERAL HKVIEVV 68. How many digits in 36*^ ? 69. Expand and simplify {ar'^a — a^^ir'^)^. 70. What must be the equation between m and n if the roots of m:i?+nx-\-p are real ? if equal ? 71. Find the (?• + l)th term of (1 -x)^. 72. Find two numbers in the ratio of 3:2, such that their sum has to the difference of their squares the ratio of 1 : 5. 73. What is the sum and product of the roots of 5 x~- a;"' 74. Solvea;2 + 2/^ = 26, 5 a; + 2/ = 24. Plot the graphs of the system and verify the solutions. 75. Show that either root of a? — c = is a mean propor- tional between the roots of a;^ + 6a; + c = 0. 76. Solve a;^-2a- + 3= Var' — 2a; + 5. Are all the solu- tions roots of the given equation ? X I when a; = 2? 78. The intensity of light varies inversely as the square of the distance from its source. How far must an object that is 8 feet from a lamp be moved so that it may receive but \ as much light? 79. Find an expression for x in a-"^^= 3 c. 80. Solve a; — 1.3 = .3 x-\ 81. lia:h — c:d, show that (ma — nb)(ma +n6)~' _ (mc — nd)(mc + wd)~' mn mn 10 82. Solve V7 .r - V3 a; + 4 = V3a;+4 GENERAL REVIEW 435 83. If r, and r., -represent the roots of a^ + bx + c = 0, find ''i" + '''2' and j-jVa-. 84. Solve x~^ +x~i + 1 = 0. 85. Find the geometrical progression whose sum to infinity is -^^, and whose 2d term is .002. 86. Expand (V2 - V^^f. 87. Solvefors: -2 -2 ^+-^7 + 2 s + i s" -2s = 2 s^+ 5. 88. Find the value of k in order that the equation (k + &)x'-2k{a?-l)-2kx-3 = may have equal roots. 89. The floor area of a certain room is 320 sq. ft., each eud wall 128 sq. ft., and each side wall 160 sq. ft. What are the dimensions of the room ? 90. For what values of m are the roots of the equation ( III + 2)3? + 2 «ix + 1 = equal ? 91. Given a:b = c:d; prove that 3a + 2c:3a~2c =zl2b + 8d:12b -8d. 92. Plot the graph of 3 x^ + 10 a; = 12, and check the result by solving. 93. Find the ratio between the 5th term of ( 1 + - ) and xV- the 4th term of ( 1 + - (-0' 94. Calculate by logarithms the fourth proportional to 3.84, 2.76 and 4.62, and, also, the mean proportional between -^12 and Vi2. 95. Solve for s and t: ^ + f = 91,s = 7 — t. ar'^ — a 96. Find n in s = r-i 436 GENEKAL REVIEW 97. The velocity of a body falling from rest varies directly as the time of falling. If the velocity of a ball is 160 feet after 6 seconds of fall, what will it be at the end of the 10th second ? 98. In an arithmetic progression, a= — V— 1, d=l-|-V— 1, n = 20. Find I and s. 99. Write the (r + l)th term of (a + &)". 100. Find the middle term of ( "^ - ^ ~^ Vv-1 ^ 101. Plot the graphs of 4 ar^ + 9 y^ = 36, x + 2 ?/ = 3. Check by solution. 102. Insert 4 geometrical means between V— 1 and — 32. 103. Prove the formula for I in each of the progressions. 104. Form the quadratic equation which has for one root the positive value of "^'7 + 4V3, and for the other root the arithmetic mean between 4 — 2 V3 and zero. 105. If m : w = ;i : s = s : ?, show that n + s is a mean pro- portional between m + n and t-\- s. 106. Solve a;?/ = 2 m^ + 5 m + 2, a;^ + j/^ = 5 m^ + 8 m + 5. 107. Solve and test the solution : Va;^ — mx + n — x = m. 2\^^ — I that does not contain x. X'J 109. Solve for s and t: s^ + s« + 2 f- = 46, 2s2-si + <==29. 110. Plot the graphs oi xf + y"- + x -\- y =M, x + y — 1 = <). 111. What is the value of 1.027027 • ■ • ? 112. Find x if 3^"-^ = {^-y-^. 113. If a + & = 61, and a^ — fti = 1, find the values of a and 6. GENERAL REVIEW 4o1 GENERAL REVIEW OF COLLEGE AND TECHNICAL ENTRANCE EXAMINATION QUESTIONS Exercise 144 This -exercise includes the preliminary requirements, quad- ratic equations, ratio and proportion, the progressions, and the binomial theorem. 1. Simplify I l i) + Vi^' - q _,. I p-Vp'-g p+Vq {Princeton.) 2. Find the simplest forms of: («) 2 x»V9 x^ + 81 + 27 V4a?' + 30, I (Coll. Eld. Exam. Board.) 3. Rationalize the denominator of — "-^t V3-V2 + 1 Also find its value correct to two decimals. (3/. I. T.) Solve for t : t 15 (Pennsylvania.) ^2_2f_1.5 2(<2+2;-35j ?2 + 10? + 21 5. Two men, A and B, start at the same time from a cer- tain point and walk east and south respectively. At the end of 5 hours A has walked 5 miles farther than B, and they are 25 miles apart. Find the rate of each. (Coll. E'ltt. Exam. Board.) 6. Solve V(a; - ay +2 ax+ V(a; + a)^- ax=2 a. (M. I. T.) 7. If a and b are roots of the equation px^ -\-qx-\-r = 0, give the value of the following expressions : a + h, ab, (a — b)% and jxi'^ + qa -|- r. {Princeton.) 438 GENERAL REVIEW 8. Simplify ( V6 - 2)( -^ ^ = + — ^ vV2 + V3-V5 V2+V3 + V5y (M. I. T.) 9. (a) Solve by factoring acx^ + 6m + adx = bd. (b) Solve a;^ + 7 a;' = 8. {Coll. Ent. Exam. Board.) 10. The radius of a circle is 3 feet. By how much must this radius be increased in order that the area of the circle may be increased by 100 square feet ? (Jf. I. T.) 11. Solve V2a;— 3a + V3a; — 2a = 3Va. {Pennsylvania.) 12. If a : & = c : d, show that ma-\- lib _mc + 7id ,-. ma — lib inc — nd 13. Solve x''-3xy + 2if = 3, 2 .r" + ?/" = 6. {Princeton.) 14. Solve V3 a; + 10 = VIO a; + 16 + V-r + 2. Which value of X satisfies the equation ? {Coll. Ent. E.vam. Board.) 15. Solve x'i + 2 y~i = i, 2 a:"* - ?/"^ = |. ( Yale.) 16. A motor ear traveled 20 miles at a certain uniform rate and then returned over the same route at a different uniform rate, the running time for the round trip being 2 hours 15 min- utes. One third of the outward trip and one half of the return trip together occupied 55 minutes. Find the two rates of traveling. {Coll. Ent. Exam. Board.) 17. Solve 3 a;"' +20 a;"* = 32. {Yale.) 18. Solve: {3x + 2y- 13)(a; -2y-2) = 0, ,(2a!-72/ + 8)(4a;-32/-23) = 0. {Haverford.) 19. Beef dealers in a certain town used to sell 6000 pounds of beef per week. After raising the price 5 % they find they lost, by so doing, 16 % of their receipts. How many pounds per week do they now sell ? {Harvard). GENERAL REVIEW 439 20. Solve a;2 - 8 a; + 6 Va;^ - 8 a; + 24 = 3. (M. I. T.) 21 Solve 12^ + ^*2' + ^2'' + ^2/ = 8a;+2, |5a; + 2y-l = 0. {Cornell.) 22. Solve 1^ + 2^=1?^ [x + y+Vx + y = 2. (Yale.) 23. If m + n : m — n =x + y : X — y, show that 0?" + ?>i^ : a;^ — im? = y- + n^ -.y'^ — n^. ( YixZe.) 24. The journey between two towns is made by stage, and the return trip by a train traveling 16 miles more per hour than the stage. The whole round trip could have been made in the same time at an average rate of 12 miles an hour. Find the rate of the stage per hour. {Coll. Ent. Exam. Board.) 25. Solve (1 - lf)x' - 2(a - bc)x + {a^ - c'') = 0. {M. I. T.) 26. One ship is 60 miles east of another. The easterly ship sails due west at the rate- of 12 miles an hour, and the other sails due south at the rate of 16 miles an hour. After how many hours will they be 50 miles apart ? Verify your answer. (Yale.) 27. Solve aa;^ + 6a; + c = 0. When will it have equal roots ? When real roots ? Determine fc so that a;^ + 2 fca; 4- 9 = will have equal roots. (M. I. T.) 28. Solve ^ + ^ = 1, 1 + 1 = 1. {Pennsylvania.) 29. On the same day A, B, and C started to solve a certain number of problems. A solved 6 a day and finished 4 days after B. C solved 3 more a day than B, and finished 2 days before he did. Find the number of problems and the number of days each worked. {Coll. Ent. Exam. Board.) 30. The corners of a square, the length of whose side is 2, are cut ofE in such a way that a regular octagon is formed. What is the length of a side of this octagon? {Cornell.) 4-10 GENERAL REVIEW 31. Find the value of a such that one root of 100 x^ + 60 » + a=0 shall be double the other. (M. I. T.) 32. Solvex + y=7; {x-lf -\-{y -2y = 28. (Princeton.) 3.3. Solve 2a; - 3 2/ = 12, 4 a;'' + 9 / + 21 2/ = 30. Associate each value of x with the proper value of y. (Coll. Ent. Exam. Board.) 34. Form the equation whose roots are the arithmetic and geometric means between the roots of a;'' — 10 a; + 9 = 0. {Cornell^ ' a , b 35. Solve (1) va; + V2/ = 7. - + ~ = c, X y i + ^ = w. {M.I. T.) X y 36. Construct with respect to the same axes of reference the graphs of 2/+2a;''— 3 a;— 9=0, and y-\-x—S = (). Estimate from the figure the values of x and y which satisfy both equations. {Coll. Ent. Exam. Board.) f a;' J- «2 := 4 37. Solve •! „ „' and illustrate the solution by means [y^=x~2, of graphs. (M. I. T.) 38. A man worked for a number of days and earned $75. If he had received 50 cents more per day, he would have earned the same amount in 5 days less. How many days did he work ? (Coll. Ent. Exam. Board.) 39. Find the sum of n terms of «_1 ^ n_2 ^ n-3 _^ . . . _ ^p^i^,,ton.) 40. Write the first three and the last three terms of 1 \' 2 j:' H ) , and simplify the result. ( Yale.) VxJ GENEKAl. KKVIEW 441 41. Solve a!^ + 2/2 = 13, y^ = i{x-2) and plot the graph of each. (Cornell.) 42. Solve /3a; j x + y _^ xy~{x+y)= 54. (M. I. T.) 43. Solve 2x'^— 1 =2/^ + 3 xy, 5 — x- = y^. Write results so that with each value of x the proper value of y is associated. (Coll. Ent. Exam. Board.) 44. Find the sum of all positive integers of three digits that are multiples of 9. {M. I. T.) .. a , !'a;y + 28a;(/-480 = 0, ^'- «°^^^|2. + , = 11. {Yale.) 46. Show that if c^ = a^ + W, a + b + c:b-\-c — a = a + c — b:a + b — c. (Coll. Ent. Exam. Board.) 47. The successive swings of a pendulum in a resisting medium form a geometrical progression. If the iirst swing is 2 centimeters, and the second 1.9 centimeters, what will be the length of the sixth swing ? (Harvard.) 48. Solve the following and illustrate by graphs : ly' = Sx, ff = 8x, ^""^ {y=2x'-l. ^^^ [y = 2x'+3. (M. I. T.) 49. The arithmetic mean between two numbers is 42^, and their geometric mean is 42. Find the numbers. (Coll. Ent. Exam. Board.) 50. Find the geometrical progression whose second term is — 2, and whose sum to infinity is 4|-. (Princeton.) 51. The numbers a, b, c, d, e, f, g are in arithmetical pro- gression. If a = — 2, and g = 12, iind the value of c. (Ha7-vard.) 442 GENERAL REVIEW 52. The sum of three numbers in geometrical progression is 70. If the first be multiplied by 4, the second by 5, and the third by 4, the resulting numbers will be in arithmetical pro- gression. Find the three numbers. {Coll. Ent. Exam. Board.) 53. Solve 4a;"^— 5 a;"^-- 9 = 0, and ^'x+-\/x — VI — x= 1. Check the solutions. (M. I. T.) 54. Find the sum to infinity of the series 1, — \, \, etc. Also find the sum of the positive terms. ( Yale.) 55. Derive the formula for the sum of n terms of a geo- metrical progression. The sum to infinity of a geometrical progression is ten times the first term. Find r. {Princeton.) 56. Two trolley cars of equal speed leave A and B, which are 20 miles apart, at different times. Just as the cars pass each other an accident reduces the power, and their speed is decreased 10 miles per hour. One car goes from A to B in 56 minutes and the other from B to A in 72 minutes. What is their common speed ? ( Yah.) 57. Insert five geometrical means between 1 and ^f^. {Pennsylvania.) 58. What must be the value of a in the equations a;^ = 4 ay, x + ij = l, in order that the two sets of solutions shall be the same. Draw the graphs of the equations using the value of a found. {M. I. T.) 59. Define an arithmetical progression, and state with proof the formula for the sum of the first n terms. Can the arith- metic mean of two numbers be equal to their sum ? If so, when ? {Harvard.) 60. (a) Construct with respect to the same axes of refer- ence the graphs oi x — y = 2 and y = x'—4:. (6) Solve these equations and show the relations between the roots and the graphs. {Coll. Ent. Exam. Board.) 61. Expand (1 + 2* + i82)^ {M. I. T.) r.RNRRAL URVIKW 443 62. Divide 38 into three parts which are in geometrical progression, and such that if 1, 2, and 1 are added to these parts respectively, the results shall be in arithmetical pro- gression. {Cornell.) 63. Show that the coefficient of the middle term of (1 + x)'^ is equal to the sum of tlie coefficients of the 8th and 9th terms of (1 + xy\ {Coll. Ent. E.ram. Board.) 64. Plot the graphs of y- — ^ .v = and y + 2x — 4: = 0, and find from the graphs the common solution. (Haverford.) 65. (a) Obtain the formula for the sum of the terms of a geo- metrical progression. (6) In an arithmetical progression the number of terms is 20 times the common difference, the difference of the first and last terms is 12, and the last term is the square of the first term. Write the first three and the last three terms of the progressions that satisfy these conditions. (Fate.) 66. In the expansion of {3x write down the term V 2a;y that does not contain x. (Cornell.) 67. (a) The sum of the first eight terms of an arithmetic progression is 64, and the sum of the first eighteen terms is 324. Mnd the series. (6) Sum to infinity the series 1.35 -|- 0.045 + 0.0015 + etc. (Coll. Ent. Exam. Board.) 68. Write the term of (5a^ — ) which contains cc". X J {Princeton.) 69. What distance will an elastic ball travel before coming to rest, if it is dropped from a height of 15 feet, and after each fall it rebounds two thirds of the height from which it last fell? {M. I. T.) 444 GENEKAL REVIEW 70. The digits of a number of three figures are in arithmeti- cal progression and their sum is 9. If 198 be added to the number, the digits will be reversed. Find the number. (Pennsylvania.) 71. A man starts from home to catch a train, walking at the rate of 1 yard in 1 second, and arrives 2 minutes late. If he had walked at the rate of 4 yards in 3 seconds, he would have arrived 2^ minutes early. Find the distance from his home to the station. {Coll. Ent. Exam. Board.) 72. If the middle terra of (3 a; ) is equal to the fourth V 2^xJ term of f 2 V« + o"- ) ' ^^*^ *^® value of x. (M. I. T.) 73. Find the sum of an infinite number of terms of 3, 2, f . . . . Find an arithmetical progression such that the sum of the first three terms is one half the sum of the next four terms, the first term being 1. {Haverford.) 74. Solve x'-\-5x + d = Q. What is the least integer which, when substituted for d in the equation, makes the roots of the equation imaginary? {Coll. Ent. Exam. Board.) 75. Compute the larger of the two roots of the equation x^ -\- 1.3 X + 0.2 = 0, correct to three significant figures. {Harvard.) 76. Write tlie first and the last of the n arithmetic means between a and h, and find the sum of the n means. {M. T. T.) 77. The illumination of an object varies inversely as the square of its distance from a light, and directly as the candle power of the light. If the illumination on the page of a book is just right for reading at a distance of 6 feet from a 16 candle-power light, at what distance from a 25 candle-power light should a book be held? {Cornell.) INDEX (Numbers refer to pages.) Abscissa, 197. Addends, 22. Addition, 22. Affected quadratic equation, 2118. Aggregation, signs of, 29. Algebraic expression, 20. Algebraic fraction, 122. Algebraic number, 16. Alternation, 320. Annuity, 382. Antecedent, 315, 318. Antilogaritbm, 373. Arithmetical means, :U4. Arithmetical progression, '■'•'■iS. Ascending order, 44. Associative law, 22, :<9. Axes of reference, 197. Axiom, 13, 65. Base of logarithm, 3(14. Binomial, 21. Binomial formula, Jtt'l. Binomial theorem, 358. Brace, 29. Bracket, 29. Characteristic of logaritlim, 366. Checking results, 25, 46, 59. Clearing of fractions, 147. Coefficient, 12. compound, 35. Cologarithm, 376. Collecting terms, 25. Common difference, 338. Common factor, 119. Common multiple, 127. Common ratio, 348. Commutative law, 22, 39. Complex fraction, 144. Complex number, 255. Composition, 321. Compound ratio, 316. Compound variation, 333, Conditional equation, 64. Conjugate imaginary, 260. Consequent, 315, 318. Constant, 199. Continued proportion, 318. Coordinates, rectilinear, 197. Cube root, 99. Definite number symbols, 9. Degree, of an expression, 44. of a term, 44. Denominator, 122. factorial, 360. lowest common, 129. Density, 164. Descending order, 44. Difference, 32. common, 338. Direct and inverse variation , 332. Direct variation, 331. Discriminant, 278. Distributive law, 39, 52. Dividend, 52. Division, 52, 321. Divisor, 52. Divisors, theory of, 419. Duplicate ratio, 316. Elements, of an arithmetical progres- sion, 339. of a geometrical progression, 349. Elimination, 174. Equality, 12. Equation, 64. affected quadratic, 268. complete quadratic, 268. conditional, 64. equivalent, 174. exponential, 380. identical, 64. in the quadratic form, 286. incomplete quadratic, 266. inconsistent, 174. 445 446 INDEX Equation, independent, 174. irrational, 252. linear, 65. pure quadratic, 2()6. quadratic, 266. simple, 65. simultaneous, 174. solution of an, 64. system of, 174. Equilibrium, 165. Equivalent equation, 174. Exponent, 41. in the fractional form, 223. negative, 222. zero, 221. Exponential equation, 380. Expression, algebraic, 20. homogeneous, 45. . integral, 99. mixed, 125. prime, 100. rational, 99. Extremes, of a proportion, 317. Factor, 12, 9