Sttfum, New ^arh .XQ.h..n H.eKir^:..Ta.*n.n.er 3 1924 031 273 588 olin.anx Cornell University Library The original of tliis book is in tine Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31924031273588 MATHEMATICAL TEXTS FOE SCHOOLS Edited by Percey T. Smith, Ph.D. Professor of Mathematics in the Sheffield ScientiHc School of Yale University First Course in Algebra Second Course in Algebra Complete School Algebra By H. E. Hawkes, Ph.D., W. A. Luby, A.B., and F. C. Touton, Ph.B. Plane Geometry By William Betz, M.A., and H. E. Webb, A.B. Advanced Algebra By H. E. Hawkes, Ph.D. Plane and Spherical Trigonometry and Four-Place Tables of Logarithms Plane and Spherical Trigonometry Plane Trigonometry and Four-Place Tables of Logarithms Four-Place Tables of Logarithms By W. A. Granville, Ph.D. COMPLETE SCHOOL ALGEBRA BY HERBERT E. HAWKES, Ph.D. PROFESSOR OF MATHEMATICS IN COLUMBIA UNIVERSITY AND WILLIAM A. LUBY, A.B. HEAD OF THE DEPARTMENT OF MATHEMATICS, CENTRAL HIGH SCHOOL KANSAS CITY, MISSOURI ERANK C. TOUTON, Ph.B. PRINCIPAL OF CENTRAL HIGH SCHOOL, ST. JOSEPH, MISSOURI GINN AND COMPANY BOSTON • NEW TOEK • CHICAGO - LONDON COPYKIGHT, 1909, 1910, 1911, 1912, BV HERBERT E. HAWKES, WILLIAM A. LUBY, AND FRANK C. TOUTON ALL EIGHTS RESERVED 912.5 t£f)t lattenaiuth 8reg« GINN AND COMPANY • PRO- PRIETORS ■ BOSTON ■ U.S.A. PREFACE This book, designed for at least one and one-half year's work in algebra, is intended for the use of those schools which find a one-book course best adapted to their needs. Experience thoroughly justifies a careful review at the beginning of the third half year's work. As the first twenty-three chapters contain the greater portion of the usual first year's work in algebra, it seemed best to follow them with the review ma- terial. Here each topic is given a broader and more advanced treatment than is permissible in first-year work. New matter is used throughout, and many new applications are given in order to make a fresh and inviting appeal to the student. In the remaining chapters the aim has been to include those topics considered necessary by the best secondary schools, and to treat each in a clear, practical, and attractive manner. A painstaking attempt has been made in this book to select and arrange the material so that no student will find the work insipid, while the student who might find the subject unduly hard will be prepared for approaching difficulties, and encour- aged into interested and successful efEort. To accomplish this, every group of problems or exercises is graded with exactness ; lists of exercises which afford extended practice in the mere translation of English into algebraic language are given at proper intervals ; the explanations are many and unusually full ; and the exceptionally numerous hints arid illustrative examples are placed precisely where needed. The authors do not believe it is desirable to attempt to cor- relate algebra with other subjects than arithmetic, geometry, and physics. They have no hobbies in the way of " practical " vi COMPLETE SCHOOL ALGEBRA problems, no aversion to those that are informational. They have, however, a strong preference for a problem that is " thinkable." Experience shows that long lists of problems manufactured from tables of statistics are not necessarily interesting. It will be found that there is a close connection with geometry through the problems on factoring, proportion, variation, quadratic equations, and the exercises and prob- lems on radicals. A correlation with physics equally close is impossible and undesirable. Yet at many places there is defi- nite preparation for the point of view needed by the student of physics. The exercises and problems have been tested by class use. While their number is ample it is not exces"sive. On account of the grading, omissions should preferably be made from the ends of the various lists. Particular attention is called to certain other important features, namely : the prominence everywhere given to the equation ; the continued emphasis on checking ; the numerous problems and their varied character; the frequent changes from exercises to problems ; the short reviews in factoring ; the extent of the work with graphs, — ■ its detailed explana- tion, its position before the process it illustrates, its use in connection with logarithms ; the introduction of the idea of a function and its graphical treatment ; the accuracy of the defi- nitions ; the historical notes ; the portraits of eminent mathe- maticians ; and the biographical notes which accompany them. Of late years the teaching of algebra has been widely and thoughtfully discussed. Every development of this period has been carefully studied by the authors. It has been their object to discern all the sound, progressive ideas in the recent evolu- tion and to utilize them in the preparation of a text which will develop both the student's reasoning power and the need- ful facility on the technical side of algebra. Other constant aims have been to make the book teachable, and to reduce the explanation on the teacher's part to a minimum. Scrupulous PREFACE vii care was besto-wed on the English, in order to give the defini- tions accurately, to state the problems clearly, and to formu- late the rules with simplicity and precision. Along with the endeavor to accomplish these various ends, a continuous effort has been made to produce a text which is distinctly modern, lucid, interesting, and mathematically correct. The present volume is in many parts identical with either our " First Course in Algebra " or our " Second Course in Algebra," in which a detailed acknowledgment is made of our obligation to those who have given us the benefit of their criti- cism and suggestion. That acknowledgment will not be repeated here, but to all of the many teachers who have so materially aided our efforts, we wish to express our hearty thanks. CONTENTS CHAPTER PAGE I. Introductiok to Algebra (Sects. 1-10) .... 1 II. Positive and Negative Numbers (Sects. 11-16) . 13 III. Addition (Sects. 17-18) 27 IV. Simple Equations (Sect. 19) 33 V. Subtraction (Sects. 20-21) 39 VI. Identities and Equations of Condition (Sects. 22-23) 43 Vll. Parentheses (Sects. 24-25) 55 VIII. Multiplication (Sects. 26-34) 59 IX. Parentheses in Equations (Sects. 35-36) ... 66 X. Division (Sects. 37-39) 74 XI. Equations and Problems (Sects. 40-41) . . .82 XII. Important Special Products (Sects. 42-46) . 93 XIII. Factoring (Sects. 47-56) 101 XIV. Solution of Equations by Factoring (Sect. 57) 122 XV. Highest Common Factor and Lowest Common Multiple (Sects. 58-59) 131 XVI. Fractions (Sects. 60-68) 135 XVII. Equations containing Fractions (Sects. 69-74) . 156 XVIII. Ratio and Proportion (Sects. 75-76) 176 XIX. Graphical Representation (Sects. 77-82) . . . 187 XX. Linear Systems (Septs. 83-88) 203 XXI. Square Root and Radicals (Sects. 89-96) ... 228 XXII. Graphical Solution of Equations in One Un- known (Sects. 97-99) 259 XXIIL Quadratic Equations (Sects. 100-102) .... 267 XXIV. Fundamental Operations (In Part Review) (Sects. 103-108) 280 XXV. Factoring and Fractions (In Part Review) (Sects. 109-121) 290 XXVI. Linear Equations in One Unknown (In Part. Review) (Sects. 122-123) 308 ix X COMPLETE SCHOOL ALGEBRA CHAPTER PAGE XXVIL Determinants and Review of Linear Sys- tems (Sects. 124-131) 318 XXVIII. Roots and Radicals (In Part Review) (Sects. 132-142) 337 XXIX. Exponents (Sect. 143) 352 XXX. Graphical Solution of Equations (In Part Review) (Sects. 144-149) 361 XXXI. Quadratic Equations (Review) (Sects. 150- 151) 369 XXXII. Irrational Equations (Sect. 152) . ... 376 XXXIII. Graphs of Quadratic Equations in Two Variables (Sects. 153-154) 383 XXXIV. Systems Solvable by Quadratics (Sects. 155- 161) 391 XXXV. Progressions (Sects. 162-171) 407 XXXVL Limits and Infinity (Sects. 172-177) .... 424 XXXVII. Logarithms (Sects. 178-192) 431 XXXVIII. Ratio, Proportion, and Variation (Sects. 193- 197) 455 XXXIX. Imaginaries (Sects. 198-205) 467 XL. Theory of Quadratic Equations (Sects. 206- 210) 478 XLI. The Binomial Theorem (Sects. 211-216) . . 486 XLII. Supplementary Topics (Sects. 217-220) ... 493 INDEX 503 LIST OF PORTRAITS PAGE Sir William Rowan Hamilton 60 Sir Isaac Newton ... . . 84 Ren^ Descartes 200 Francois Vieta . . 258 Gottfried Wilhelm Leibnitz . 324 John Wallis 356 Felix Klein 366 Sir John Napier ... 438 Karl Friedrich Gauss 468 COMPLETE SCHOOL ALGEBRA CHAPTER I INTRODUCTION TO ALGEBRA 1. Algebra deals with many topics which are new to the student, yet he will find the subject a continuation of his previous work in arithmetic. 2. SjTnbols. Symbols are employed far more extensively in algebra than in arithmetic, and many new ideas arise in con- nection with their meaning and use. Some symbols represent numbers, others indicate operations upon them, and others represent relations between numbers. Arabic numerals, and letters are used to represent numbers. The following symbols of operation, +, — , X , and h-, have the same meaning as in arithmetic. The sign + is read plus ; — , minus ; x , multi- plied hy ; and -=-, divided by. The sign of multiplication is usually replaced by a dot or omitted. Por example, 3 x a is written 3 -a, or 3a, and 2 x « X ^ is written 2 ah. Also a -f- 5 is often written y- The sign = is read equals, or is equal to, EXERCISES 1. Express Ah + Zm in seconds, if h and m stand for the number of seconds in an hour and in a minute respectively. 2. Express 6 ?/ + 4/ in inches, if y and / stand for the num- ber of inches in a yard and in a foot respectively. 3. If q and d represent the number of cents in a quarter and in a dime respectively, find the value in cents oi 4:q + Qd. 1 2 COMPLETE SCHOOL ALGEBRA 4. If t and h represent the number of pounds in one ton and in one hundredweight respectively, express 4 ^ + 6 A in pounds. 5. 3 a; + 5 a; = how many x ? 6. 4a; + 5a;=(?) x. 7. 2a; + 3a; + 6a: = ? 8. 2a; + 2 + 3a; + 4 = (?)a;.+ ? 9. x + x + 2 + x + x + 2 = ? 11. 5a + 18-3a-7 = ? 10. n + n + l + n + 2 = ? 12. 8a; - 3 + 18 - 5a; = ? 13. 4m;-8 + 3w + 20 = ? 14. If m represents one month and y = 12 m, express y + 5m in terms of m. 15. If y (one yard) equals 3/ (/= one foot), express 2 y+ If in terms of/. 16. Express 4 g' + 3 w in terms of n, if q (one quarter) equals 5 m (re = one nickel). 17. Express 2d+15 h in terms of h, if cZ = 24 h. 18. Express 15 A + 60 »i in terms of m, if h — 60 m. 19. The side of a square is 5 inches. What is its area? its perimeter ? 20. The side of a square is s inches. What represents its perimeter? its area? 21. The base of a rectangle is 12 feet and its altitude is 4 feet. What is its area ? its perimeter ? 22. If b represents the number of feet in the base of a rect- angle and a the number of feet in its altitude, what is its area ? its perimeter ? 23. A rectangle is twice as long as it is wide. Let x repre- sent the number of inches in its width. Then express (a) the length in terms of x; (V) the perimeter; (c) the area. 24. A man is three times as old as his son. If y is the number of years in the son's age, what will represent the father's age ? INTRODUCTION TO ALGEBRA 3 25. A father is 30 years older than his- son. If y represents the son's age in years, what will represent the father's age ? 26. A rectangle is 12 feet longer than it is wide. Let w represent the width in feet. Then represent the length and the perimeter in terms of w and some numbers. 27. A rectangle is 18 feet narrower than it is long. If w represents the width in feet, what will conveniently represent the length ? the perimeter ? 28. A rectangle is 4 feet longer than twice its width. Ex- press the width, the length, and the perimeter in terms of a letter, or a letter and numbers. Origin of symbols. Many of the symbols that are in common use in algebra at the present time have histories .which not only are interesting in themselves, but which also serve to indicate the slow and uncertain development of the subject. It is often found that symbols which seem without meaning represent some abbreviation or suggestion long since forgotten, and that operations and methods which we find hard to master have sometimes required hundreds of years to perfect. In the early centuries there were practically no algebraic symbols in common use ; one wrote out in full the words plus, minus, equals, and the like. But in the sixteenth century several Italian mathema- . ticians used the initial letters p and m for + and — . Some think that our modem symbol — came into use through writing the initial m so rapidly that the curves of the letter gradually flattened out, leaving finally a straight line. The symbol + may have originated similarly in the rapid writing of the letter p. But in the opinion of others these symbols were first used in the German warehouses of the fifteenth century to mark the weights of boxes of goods. If a lot of boxes, each supposed to weigh 100 pounds, came to the ware- house, the weight would be checked, and if a certain box exceeded- the standard weight by 5 pounds, it was marked 100 + 5 ; if it lacked 5 pounds, it was marked 100 — 5. Though the first book to use these symbols was published in 1489, it was not until about 1630 that they could be said to be in common use. Both of the symbols for multiplication given in the text were first used about 1630. The cross was used by two Englishmen, Oughtred and Harriot, and the dot is first found in the writings 4 COMPLETE SCHOOL ALGEBRA of the Frenchman, Descartes. It is interesting to note that Harriot was sent to America in 1585 by Sir Walter Raleigh and returned to England with a report of observations. He made the first survey of Virginia and North Carolina and constructed maps of those regions. It is strange that the line was used to denote division long before any of the other symbols here mentioned were in use. This is, in fact, one of the oldest signs of operation that we have. The Arabs, as early as 1000 a.d., used both - and a/h to denote the quotient of a by 6. The symbol -^ did not occur until about 1630. Equality has been denoted in a variety of ways. The word equals was usually written out in full imtil about the year 1600, though the two sides of an equation were written one over the other by the Hindus as early as the twelfth century. The modern sign = was probably introduced by the Englishman, Recorde, in 1557, because, he says, "Noe. 2. thynges can be moare equalle " than two parallel lines. This symbol was not generally accepted at first, and in its place the symbols ||, x, and oc are frequently met during the next fifty years. 3. The usefulness of symbols. Symbols enable one to abbre- viate ordinary language in the solution of problems. For example: Three times a certain number is equal to 20 diminished by 5. What is the number ? If n represents the number, the preceding statement and question can be written in symbols, thus : 3w = 20-5. n = ? The symbolic statement, 3 w = 20 — 5, is called an equation and n the unknown number. If 3 ?i = 20 - 5, 3n = 15, and n = 5. The preceding example illustrates the algebraic method of stating and solving the problem. The method is brief and direct and its advantages will become more apparent as the student progresses. INTRODUCTION TO ALGEBRA 5 EXAMPLE 1. The sum of two numbers is 112. The greater is three times the less. What are the numbers ? Solution: By the conditions of the problem, greater number + less number =112. If -we represent the less by I, then 3 1 must represent the greater, and the above statement becomes 31 + 1 = 112. Collecting, 4 1 = 112. Whence Z = 1 1 2 = 28, and 3 Z = 3 X 28 = 84. Therefore the greater number is 84 and the less 28. PROBLEMS 1. The sum of two numbers is 160. The greater is four times the less. I'ind each. 2. A certain number plus five times itself equals 216. Find the number. 3. One number is seven times another. Their sum is 72. Find each. 4. The first of three numbers is twice the third, and the second is four times the third. The sum of the three numbers is 105. Find the numbers. 5. The sum of three numbers is 117. The second is twice the first, and the third three times the secon(J. Find each. 6. There are three numbers whose sum is 192. The first is twice the second, and the third equals the sum of the other two. Find the numbers. 7. The sum of three numbers is 324. The second is five times the first, and the third is six times the second. Find the numbers. 8. The sum of three numbers is 104. The second is three times the first, and the third is the^ sum of the other two. Find the nunibers. 6 COMPLETE SCHOOL ALGEBRA 9. What sum of money placed at interest for 1 year at 5% amounts to |378? Solution : From arithmetic, principal + interest = amount. By the conditions of the problem, principal + .05 principal = $378. If p represents the principal, this last statement becomes ■p + .05j9 = 378. Collecting, 1.03^ = 378. 378 Whence p = = 360. ^ 1.05 Therefore the required sum is |360. 10. What sum of money placed at interest for 1 year at 6% amounts to $265? 11. What sum of money placed at simple interest for 3 years at 4% will amount to $700? 12. In how many years will $225, at 6% simple interest, gain $27 ? 13. In how many years will $520, at 6^-% simple interest, gain $169 ? 14. At what per cent simple interest will $375 gain $75 in 2 years ? Solution : Let x = the rate of interest. Then |375 x = the interest for one year, and $750 x = the interest for two years. But 1^5 = the interest for two years. Therefore 750 x = 75, and X = ^L.. Hence the money is lent at 10%. 15. At what per cent simple interest will $825 gain $165 in 4 years ? 16. At what per cent simple interest will $250 amount to $317.50 in 6 years ? 17. In how many years will $200 double itself at 4% simple interest ? INTRODUCTION TO ALGEBRA 7 18. In how many years -will $150 treble itself at 5% simple interest ? 19. The perimeter of a certain square is 160 feet. Find the length of each side. 20. The perimeter of a certain rectangle is 256 feet. It is three times as long as it is wide. Tind its dimensions. 21. TJie perimeter of the rectangle formed by placing two equal squares side by side is 198 inches. Find the dimensions and the perimeter of each square. 22. Two equal squares are placed side by side, forming a rectangle. If the perimeter of each square is 120 inches, find the perimeter of the rectangle. 4. Representation of numbers. In algebra numbers are rep- resented by one or more numerals or letters or by both com- bined. Thus 3, 25, a, 2 5, 4a;y, and 2x+Z are algebraic symbols for numbers. Precisely what numbers 4 xy and 2 a; + 3 represent is not known until the numbers which x and y stand for are known. In one problem these symbcjs may have values quite different from those they have in another. To devise methods of deter- mining these values in the various problems which arise is the principal aim of algebra. 5. Factors. A factor of a product is any one of the numbers which multiplied together form the product. Thus 3 ab means 8 times a times h. Here 3, a, and h are each fac- tors of 3 db. Similarly the expression 4 (a -I- 5) means 4 times the sum of a and J. Here 4 and a -H 6 are factors of 4 (a -I- 6). 6. Exponents. An exponent is a number written at the right of and above another number to show how many times the latter is to be taken as a factor. (Later this definition will be modified so as to include frac- tions and other algebraic numbers as exponents.) 8 COMPLETE SCHOOL ALGEBRA Thus 3= = 3 ■ 3 ; 5' = 5 • 5 • 5. Also a^-a-a-a-a, and 4 xy' = 4 -x -y ■ y -y. In a"', 6 is the exponent of a. If a is 4 and b is 3, a'> = i^ = 4. ■ i ■ 4. The exponent 1 is not usually written. 7. Coefficients. If a number is the product of two factors, either of these factors is called the coefficient of the other in that product. Thus in 4 x^y, 4 is the coefficient of x^y, y is the coefficient of 4 x^, and iyis the coefficient of x^. The numerical coefficient 1 is usually omitted. If a numerical coefficient other than 1 occurs, it is usually ■written first. For instance, we write 5 x, not x 5. The following examples illustrate the difference in meaning between a coeflB.cient and an exponent : 3x=:x + x + x. x' = X ■ X ■ X. If, in each case, a; = 5, 3 a; stands for the number 15, while a;* stands for 125. If a; is 10 in each case, 3 a; = 30, while x" = 1000. 8. Use of parentheses. If two or more numbers connected by signs of operation are inclosed in a parenthesis, the entire expression is treated as a symbol for a single number. Thus 3(6 + 4) means 3-10, or 30; (17 - 2) -=- (8 - 3) means 15 -^ 5, or 3 ; (5 + 7)^ means 12^, or 144 ; and 6(x + y) means six times the sum of x and y. As in arithmetic, the symbol for square root is "v , or V^, and the symbol for cube root is -s/. The name radical sign is applied to all symbdls like the fol- lowing : V , "V^, ■v^, "V , etc. The small figure in a radical sign, like the 3 in "v , is often called the index. Note. There has been a considerable variety in the symbols for the roots of numbers. The symbol V was introduced in 1544 by the German, Stifel, and is a corruption of the initial letter of the Latin word radix, which means "root." Before his time square root was denoted by the symbol B , used nowadays by physicians on prescrip- tions as an abbreviation for the word recipe. Thus -v/5 would have been denoted by IJ^S. Some early writers used a dot to indicate square root, and expressed -n/2 by • 2. The Arabs denoted the root of a number by an arable letter placed directly over the number. INTRODUCTION TO ALGEBRA 9 EXERCISES Write in symbols : 1. The sum of three times a and five times 5. 2. Three times a subtracted from five times h. 3. The square of a subtracted from the square of b. 4. Two times a squared subtracted from three times a squared. 5. The quotient of a divided by h. 6. The product of four times a squared and b. 1. The sum of a and b divided by their difference. 8. The product of a and 2b — c. 9. The product of a by the sum of b and c. 10. The result of subtracting a — b from 7 x. 11. The product of the sum of a and b by their difference. 12. The sum of the square root of 5 a and the cube root of 7 J. 13. The product of x — y and the square root of 7 a. 14. The square of the sum of a and b. 15. The square of the difference of a and b. 16. The quotient of three times a times the square of b by four times c times the cube of a. 17. The sum of the quotients of a by 3 a; and 42/ by c. 18. Eead Exercises 1-14, page 11. 9. Order of fundamental arithmetical operations. If we read the expression 6 + 4-9— 12 ^3 from left to right, and perform each indicated operation as we come to its symbol, we obtain 10, 90, 78, and a final result of 26. If we perform the multipli- cation and division first, the expression becomes 6 + 36 — 4, which equals 38. These results show that the value of the expression is determined largely by the order in which the operations are performed. When there is no statement to the contrary, it is understood that : 10 COMPLETE SCHOOL ALGEBRA I. A series of operations involving addition and subtraction alone shall he performed in the order in which they occur. Thus 8 + 12 - 10 + 6 = 20 - 10 + 6 = 10 + 6 = 16. It is incorrect to say 8 + 12 - 10 + 6 = 20 - 16 = 4. II. A series of operations involving multiplication and division alone shall be performed in the order in which they occur. Thus 8 • 12 -^ 6 ■ 4 = 96 H- 6 • 4 = 16 • 4 = 64. It is incorrect to say 8 • 12 -=- 6 • 4 = 96 -h 24 = 4. III. In a series of operations involving addition, subtraction, multiplication, and division the multiplications and divisions shall be performed in order before any addition or subtraction. Then the additions and subtractions shall be performed in ac- cordance with I. Therefore 6 + 4-9- 12 ^3 = 6 + 36 -4 = 38. It is incorrect to say 6 + 4- 9 - 12 -^ 3 = 10 - 9 - 4 = 86. In a series of operations an expression inclosed in a paren- thesis is regarded as a single number. Obviously, within any parenthesis I, II, and III apply. Therefore (3 + 2) 6 = 5 • 6 = 30, and 8 + (7- 3)(9 - 6 -^ 2) - 4 = 8 + 4 • 6 - 4 = 8 + 24 - 4 = 28. EXERCISES Simplify the following : 1.20-5 + 6-10. 6. 18 -=-(2-3). 2. 16 - (8 - 2). 7. (6 - 3)(17 - 2 ■ 5). 3. 14 - (16 - 8)+(12 - 4). 8. 23 - 2 • 6 - 4 -f- 2 + 16. 4. 6^-3-2. 9. 18-r-(9-3). 5. 8 • 6 H- 3 - 10. 10. (10 - 3)(16 -3-2 + 8^4). 11. 14-3-(16-2-5)-r-6 + 8.2. 12. (18 - 2) - 6 H- (4 + 2 • 8 - 18 -=- 9). 13. (16 - 6) (18 - 8) -- 100 • 5 - 5. 14. (5 + 3)(5-3)--5-3. INTRODUCTION TO ALGEBRA 11 10. Evaluation of algebraic expressions. Finding the numer- ical value of an expression for certain values of the letters therein is frequently necessary. This process "will be used later to test the accuracy of the results of algebraic operations. EXERCISES In Exercises 1-23 let a = 3, & = 1, c = 5, t^ = 7, and e = 2. Substitute for each letter its numerical value and then sim- plify the results according to the rules of § 9 : 1. 4:a + 3d. g i + i + ^ 2. a" + 2o. 3. ah + cd. 4. c^ — 5 ab. 5. ahcd — 4 6\ 6. d + c e 7. cde ace 5e 2a 8. 10 5 c 6 h 15. 3 5= 16. a\ 17. c". 18. d' + e". 19. b" + c". 20. d'-c'' + P. 10. 11. 12. 13. cde 6^ 4tcd ac ae cd + ae + ce a — h + 6 g'' -f- 5^ + c' + t?" ' a -\- b + c + d a' — ce^ + 3 cd d — a + c 14. 55' -f 45^-25-5. ■ 14 5* + 11 5« -t- 11 5^ + 13 5 - 20. 21. e^-c". d + e" 22. 23. 2c + a a" + 35 a - d — e — b Find the numerical value of the following expressions -when = i,b = 0,c = 5,d = 7, and e = 8 : 24. ■ -K 26. abo + acd — be. c + e + 2 5 97 «*j.Mj.^. OK . ii- 1 1 ; ^^- a + c + d "' "d 12 COMPLETE SCHOOL ALGEBRA 28. 3d-2e + o 29. Va + V26. 30. 2Va + ^. 31. b Va + cd + Ve ■ Va. 32. -V2ae + d -y/^ae. 33. -y/b" + c^ + d + a. 34. -^Sd + ec + a — l. 35. de + ac -y/ao^ + 2 e + c. 36. (a + c) c. 37. (a + e)(c + (-2). 14. -18-12. 22. (- -5)0- ■12) 30. — 12-2. 15. 15-14. 23. - 3 (-8). 31. — 39 -=- (- 3). 16. +7-0. 24. - 5-4. 32. 45 ;-H(-i5). 17. -0-3. 25. 6 0. 33. 0- ■^(-6). 18. (-3) (6). 26. 0. (-9). 34. -3. 19. (-5)6. 27. 4- 8. 35. — 27^9. 20. (7) (-5). 28. - 3-6. 36. 3 -5 + 6. 37. - 4 + 6 -2 + 1. 38. -4 + 6 + 2-1. 39. 2 -3 + 4- 5-6. Add: 40. 7 41. 6 42. -8 43. 4 -2 -2 6 -9 3 -3 2 -3 -5 4 -6 6 Simplify : 44. 3-6 H- 3. 49. 3''. 45. -4(7)-^(- -2). 50. (- 3)^. 46. 3(-6)--2. 51. 2«. 47. 4-6(-8)-r- (-16). 52. (- 2y. 48. 18-H(-3)- 6h-4. 53. (-Ay+(iy. 54. (- ly +.(- 1)« + (- 2y + (- 2)". 55. 6 + 3-2 + 18 -^(-3). 56. 52-4-=-(-2)+6(-3). 57. 3-6-^9-2-6^4 + (-3)^. 58. 6 + 6-3^-5''.2-v-10. If a; = 3 and y = — 2, find the value of : 59. y\ 61. /. 63. 2y\ 65. SjcV. 60. y'. 62. y^ 64. 2/. 66. 4a;y. POSITIVE AND NEGATIVE NUMBERS 25 67. x'-y'. 69. x" -\- 2 xy -\- y\ 68. «" — y'. 70. x^ — 2xy-\-y\ 71. {x-\-y-){x-y). 72. x' + Sx'^y + Sxy^ + f. 73. /-3a;y2 + 3a;V-a:«. 74. Does Ax—2==2x + 8,iix = 5? 75. Does 3a; — 6 = 2a; + 8, ifa;=— 9? 76. Doesa;=-a;-12 = 0,ifa! = 4? ifa; = -8? if a; = -4? 77. Does 3a3= + 19a;=14, if a; = f ? if a; = 2? if a; =-7? 78. At 7.00 A.M. on a certain day the thermometer registered 15 degrees above zero. The mercury then fell at the rate of 3 degrees per hour. What was the temperature at 9.00 a.m. ? at noon? at 3.00 p.m.? 79. With reference to a certain assumed level, a surveyor found the heights of 5 points to be + 30 feet, — 7 feet, + 18 feet, — 10 feet, and + 16 feet respectively. What was the average height of the 5 points ? What meaning has the result ? 80. A sixth point whose height was — 38 feet was later included in the preceding survey. Find the mean height of the 6 points. What meaning has the result ? 81. Later a seventh point was added to the preceding survey. The average height of the 7 points was then zero. Find the height of this last point. 82. Euclid lived about 300 b.c. Sir Isaac Kewton died in 1727 A.D. If dates before Christ are considered negative and those after Christ be considered positive, how might these dates be written? 83. What is the meaning of the date —450? of +1910? What is the difference in time between these two dates ? 84. In still water a gasoline launch can travel 8 miles per hour. Using positive and negative numbers, represent its rate both up and down a river which runs IJ miles per hour. . 26 COMPLETE SCHOOL ALGEBRA 85. A boat is traveling 12 miles per hour. A man on its deck is walking 3 miles per hour. Using positive and nega- tive numbers, represent the rate at which he approaches his destination when he walks toward the bow and when he walks toward the stern. 86. A balloon capable of supporting 500 pounds is held down by 10 men whose average weight is 150 pounds. Using posi- tive and negative numbers, represent the weight of the balloon, the men, and the balloon and men together. 87. A man swims in still water at the rate of 1-^ miles per hour. If he swims in a river which flows 2 miles per hour, represent his rate (a) when he swims downstream; (b) when he swims upstream. What is the practical meaning of the last answer ? CHAPTER III ADDITION 17. Addition of monomials. A number symbol consisting of a numeral, or a letter, or a product of letters alone, or the product of a numeral and one or more letters is called a monomial or term. Thus 5, —a, b\ a^x, —icy^, faV?/, x", and a;" + 2 are terms. Frequently, where no confusion would arise, expressions like (a + b), Z(x — y), 5 "y/ifi, and ■y/a — x are also called terms, for often they may be replaced by a single letter. The literal part of a term is the portion composed of letters. Similar terms are integers and fractions, like numerical roots, or such terms as have like literal parts. Thus 3, — 7, and 9 are similar terms as well as Vs and 3 V2. Also a, 4 a, and — 10 a are similar terms as are a^x, — 3 a^x, and 7a^x. Dissimilar terms are unlike numerical roots or such terms as have unlike literal parts. Thus 4, "vS, VS, and VS are dissimilar terms as well as 3a, 45, and 6 c^. Also 7 a^x, 3 ax^, and 5 ax are dissimilar terms. We know that 6 acres and 3 acres = 9 acres. Similarly 6a + 3a = 9a, and 6as + (— 3a) = 3a., and 5xy+ &xy=:1Xxy. In like manner the sum of 8 y, — 3 y, 2 y, and — y is 6y. Such terms as — y, x, ay, and — c'^x are equivalent to —ly, + lx, + lay, and —Ic^x, the coefScient 1 being always under- stood if no numerical coefficient is written. Thus for adding similar terms we have the Rule. Find the algebraic sum of the numerical coefficients and prefix this result to the com,mon literal part. 27 28 COMPLETE SCHOOL ALGEBRA EXERCISES Pind the algebraic sum of : 1.18-5 + 6. i. 12a^-7a^ + 3a'' + 4:a\ 2. 18a- 5a + 6a. 5. 6 - 4 - 17 + 20. 3. 12-7+3 + 4. 6. 6ab-4:ab — 17ab + 20ab. 7. 15-17 + 8-12-25. 8. 8 abo — llabo — iabc + 15 abc — 12 abc. 9. 11 + 5-9-3+16-25. 10. 16 ao — 9 ac + 5 ao — 2 ac — Sao + Hoc. 11. 7a; + 4 a; — 15 a; — 8 a: + 3 x. 12. 12y-172/ + 102/+20y-25y. 13. 4 CBy — 8 a;?/ — 12 a;y + 13 a;y — a;y. 14. 14 a;2 - 13 a;2 + a;2 - 5 a;2 + 4 x^. 15. 5if-2y^-ll2/ + if-Ty^. 16. Ta'^b -5 a'b + 8 a^S - a^b + 9 a%. Obviously 2 + 3 = 3 + 2, and 2 — 3 + 5 = 2+5 — 3 = — 3 + 5 + 2, etc. This illustrates the law that in addition the terms may be arranged in any order. Hence 6 a! + 7 c = 1 c -\- 6d, and the sum of 3 and x is either a; + 3 or 3 + a; ; also 0. + 6 = 5 + a, and a + 5 + (! = 5 + c + a = c + a + 5, etc. Algebraic expressions for numbers with unlike literal paris, such as 6 cZ and 7 c, may be added by writing them one after the other with a plus sign between them; thus 6d -\-l c. The addition ol 6d and — 7c is indicated by writing 6d -\- ( — 7c), which is the same as 6 c? — 7 c. Similarly the sum of 3 a;, — 2y, and — 7s may be written 3 x + ( — 2 y) + (— 7«), or, more simply, 3a;-2y-7«. Thus for adding dissimilar terms we have the Rule. Write the terms one after another in any order, giving to each its proper sign. If similar and dissimilar terms are to be added, the two preceding rules must be observed. ADDITION 29 EXERCISES Pind the algebraic sum of : 1. a, 3 b, and — c. 2. 4 cc, — 2 6, 3 y, and 10. 3. 3 ab"^, 2 bx, — cy, and 4 a^j. 4. 5 x^y, — 5 xy'^, c^y, and — 2 c^'. 5. 4a;, — 3 ct, 2 6, — 5a;, and By. 6. 5 a, — 4 J, + 3 c, 6 5, and — 2 cl 7. 3 a^ + 2 S^, - 6 c2, - 4 5^, and 5 a\ 8. 4 a%, - 4 a6», - 3 a%, 3 a*^ 4 a'5, and 2 aJ». 9. - 4 a^6 + 6 a%^ - 15 a^i" + 3 a%^ + a^6^. 10. -SS-23S= + 17i^ + J»-0S8 + 135l 11. 12 a'S + &a%- a'b + 16 a=6 - 13 a% - 25 a^b. 12. llVa-14Va + 2lVa- Va. 13. 3 -y/x — y — Va; — 2/ + 9 Va; — y — 7 Va; — y. 14. 3(a + 5)-2(a + J)+8(a + 5). 15. -7(a-25), (a-2J), 12(a-2J). 16. 8(a + 5)+3(a-S)-4(a + 5)-2(a — 5). 17. 4(a; - y) - 3(a; + 3) - 6(a; - y) + 5(a; - 3). 18. (2 a; - y)2 - 3(2 a; - y)2 + 4 (2 a; - yy - 7(2 x - y)». 18. Addition of polynomials. A polynomial is an algebraic expression consisting of two or more terms. It is not usual to call an expression a polynomial if any of its terms contain a letter under a radical sign. Thus we shall not call expres- sions like Vx — 3 + 4 polynomials. A binomial is a polynomial of two terms. A trinomial is a polynomial of three terms. 30 COMPLETE SCHOOL ALGEBRA EXAMPLE Add the following polynomials : 4 a — 6 6 — a^c ; 3 6 + 4 0% ; -3a-7a^G + 10; Ba + 3b-6. Solution : 4 a — 6 J — aV - 3 a - 7 a^e + 10 5a+ 35 - 6 Sum, 6 a — 4 a'c + 4 For the addition of polynomials we have the E.ULB. Write similar terms in the same column. Find the algebraic sum of the term,s in each column and write the results in succession with their proper signs. A check on an operation is another operation which tests the correctness of the first. For example, in arithmetic the result of division is checked by multiplication ; thus the check for 132 h- 6 = 22, is 22 • 6 = 132. In the following example the letters a, b, and c represent any numbers whatever. Therefore in order to check the result we may give them any numerical values we please. Let a = 1; & = 1, and c = 1. EXAMPLE Add the polynomials 5a — 3b — 3c, 2a — 56 + 6 c, and — 3a + 2J — 4 c. Solution : Check : 5a-36-3c = 5-8-3=-l 2a-5b + 6c = 2-5 + 6= 3 • -3a+2J-4c = -3+2-4=-. 5 Sum, ia — 6b — c — 3 But 4a-6J- c = 4-6-l=-3 We conclude that the addition is correctly performed, since the numerical value of the sum is — 3 and the sum of the numerical values of the three polynomials to be added is — 3 also. ADDITION" 31 A numerical check will usually detect errors, though not always. Two errors may be made, one of which offsets the other ; these errors would, not be detected by a numerical check like the preceding one. Thus if in the preeediiig exercise the incorrect sum 4a — 56 — 2c had been obtained, the substitution of 1 for a, h, and c in this result would have given — 3, an apparent check. The number of times that errors will thus balance one another, however, is small compared to the total number of errors made ; hence the check illustrated is practically very useful. If a more reliable check is desired in similar exercises, it can be obtained by the substi- tution of a different number for every letter. EXERCISES Add the following polynomials and check the results : 1. X + y + z, X — 2y + 3s, and 3 as + 4 y — 7s. 2. 2x + Sy — s, 3x — 8y + 6s, and x — y — z. 3. 3 a; + 6?/, 4a: — 7y + 6s, and 3x — 3y — 3z. 4. 7x — y + 3e!, 5a; — 4 s, and 2 a; + 6 y — 5 s. 5. 4a! — 3y— 5s, 6y— 2s, and 7 x — 6y — As. 6. X + 2z + 3y, y — 3s + x, and s — 2x — iy. 7. 5x — y + 3s, 2y — lis + x, and 9s — 7y. 8. Sa — 7b — 6c, 5c — 4a. — 3 6, and 36 + 7c. 9. 9 ac — be, Sab — 4: ac, and — 12 a6 — ao. 10. a" -4a + 10, 6 a - 6 a^ + 4, and 3a -16 + 2a". 11. 9 — a" + a, —7+ Ga^ — 4a, and 5a"— a. 12. ^a;- Jy + |s, x — y + 2s, and ix — iy + -^s. 13. J a; - ^^ s, ix + ^y, and f s - | y. 14. a-3(x-2/)+s, 5-10a + 4(a;-2/), and-2(a!-2/) + 6. 15. a + )]. 58 COMPLETE SCHOOL ALGEBRA 25. Inclosing terms in parenthesis. Obviously 16 + 9 - 5 = 16 + (9 - 5) = 16 + 4 = 20. Similarly a -{- b — o = a + (b — c). From this we have the Peinciple. Two or more terms m,ay be inclosed in a paren- thesis preceded by a plus sign, without changing the sign of any of the terms. The expression 17 + 8 - 3 = IT - (- 8 + 3) = IT - (- 5) = ir + 5 = 22. Similarly a -\- b — c = a — (— b + c). From this we have the Peinciple. Two or more terms m,ay be inclosed in a paren- thesis preceded by a minus sign, provided the sign of each term thus inclosed be changed. EXERCISES In the following inclose in a parenthesis preceded by a plus sign all the terms containing the letters x or y, and inclose in a parenthesis preceded by a Ainus sign all the terms contain- ing the letters a ov b: 1. x''~a''~2ab-b\ 3. y' - ib^ -{- iab - a\ 2. 12 ab + x^-Ob^-i a\ 4. 10 ab + x^— a''- 25 b\ 5. a;^ — J^ — 4 0." + 4 y" — 4 a& — 4 a;y. 6. Jtab + x'' — Ab^ + y^ — a^ — 2 xy. 7. l&x''-a''-l&xy-b'^-\-2ah-\-4.y\ 8. x^-b^- 10 xy + 12 ab- 36 a= + 25 y\ 9. (a'-x^-if-b^ 11. (da-Sx + y)-(4.b-7y). 10. {x-a)-{b-y). 12. {b - 2x)-(a - 2y - Qx). 13. (x" - 4.ab -b')-{a' + ^xy + if). 14. (2x - 2,y -7 a)-{x - 2y ~7 b). 15. -(Zx^ + 2 a"" - b^)-(a' - b^ - if + x-^. CHAPTEK VIII MULTIPLICATION 26. Product of terms containing unlike letters. We assume that the factors of a product may be written in any order. This principle is called the Commutative Law of Multiplication. That is, 2-4 = 4-2. Similarly a xh = h x a. As 3 X & is written Sb, a x b is written ab, x^ X y^ is written xhf', and a xb x ohs, written ahc. Further 2 a^ x 3 = 2 x 3 x a^ = 6 a^ and 2ax3J = 2x3xaxJ =6a6. Similarly C a;" . 5 yS _ g . 5 . ^.2 . ^,3 = 30 j;y. Also 4 aJ- 3 £2 = 4. 3. a6. ;j2 -12 aJz2. We have also assumed that the various operations of multi- plication in any product may be. performed in any order. This principle is called the Associative Law of Multiplication. That is, (3 ■ 2) 4 = 3 (2 . 4) . Similarly a {b ■ c) = (a ■ b)c. This merely tells us that a multiplied by the product of b and e is the same as the product of a and b multiplied by c. Biographical note. Sik William Rowan Hamilton. It is strange that of all the topics treated in this book, the last to be thoroughly under- stood by mathematicians are those appearing in the first chapters. But in all the sciences it is often most difBcult to answer the questions that at first sight seem quite obvious. Any child can ask what electricity is, but the wisest scientist cannot tell. He can only explain what electricity does. It is easy to ask how the earth came to be revolving around the sun with the moon revolving around it, but even the deepest students of astronomy differ in their theories of how it came to be. And so in mathe- matics, long after many of the more complicated processes of algebra were completely understood, the simple laws of operation of numbers were surrounded with haze. One of the men who did most to clarify the 59 60 COMPLETE SCHOOL ALGEBRA nature of these laws was Sir William Rowan Hamilton (1805-1865). He ■was born in Dublin, Ireland, where he lived most of his life. He was a precocious boy, and at the age of twelve was familiar with thirteen lan- guages. He devised kinds of numbers that do not follow the same laws as those that we use in algebra, and so threw a flood of light on the nature and properties of these common numbers. He was the first to recognize the importance of the Associative Law, and called it by that name. Most of his works are very advanced in character and are difficult to read. 27. Product of terms containing like letters. By the definition of an exponent (§ 6), a^ = a -a, and a^ = a- a- a. Therefore a^xa^ = a-axo-a-a = a° = a^ + ^. Similarly I xh" y.h'^ = h xh-l-hy-h -h-h ■h-h = h^ = l)'^ + ^ + K In like manner 32x3^x3= = 3-3x3-3-3-3x3-3-3-3-3 = 3" Also ay^x y^ = ay^ = ay^ + ^j and 2abxZa^ = 6a% = 6a'^ + %, and 4 x'yz x 5 xy" = 20 x^y*z = 20 a:^ + i^i + s^. Therefore we have the Principle. The exponent of any letter in the -product is equal to the sum, of the exponents of that letter in the factors. This is expressed in general terms, thus : The law of signs for the multiplication of positive and nega- tive numbers, given in § 15, applies to literal terms as well. Thus +2a2 X (+ 3a=) = + 6a'. + 2a2 X (-3a=)= - 6 o'. - 2 a^ X (+ 3 a6) = - 6 a'. - 2 a2 X (- 3 a^) = + 6 a'. For the multiplication of two monomials we have the EuLE. Keeping in mind the rule of signs for multiplication, write the product of the numerical coefficients followed by all the letters that occur in the multiplier and the multiplicand, each letter having as its exponent the sum of the exponents of that letter in the multiplier and the multiplicand. SIR WTLI.IAM KOWAN HAMILTON MULTIPLICATION 61 . ORAL EXERCISES Perform the following indicated multiplications : 1. (3) (-8). 9. (-9 a) (-10). 17. (5a*)(ra«). 2. (-2) (5). 10. (-3axy. 18. (-4^=)=. 3. (-7) (-3). IL (4«)(-2a). 19. (a^) (- 20 a). 4. (-4^) (3). 12. (6 a6c)2. 20. (_4a«)(-6a''). 5. (-4.xy.. 13. (-llx)(2cc). 21. i+er/y. 6. (7) (-5 a). 14. (7x)(-3x). 22. (ix)(5y). 7. (3 a) (-6). 15. (-2 ay. 23. (-Sw'xy. 8. (-2yr. 16. (-2a)(-3a^). 24. (3 «=')(- y)- 25. (5x^y)(- 2x^. 28. ■(2aa!2)8. 26. (-6xY) 2 29. (5 a ')(- 4a2)(-3a). 27. (-A)(- -xY) 30. (3aa:)(- 4 a^a;) (- 2 csa;'). 28. Multiplication of a polynomial by a monomial. Clearly 2 (5 + 3) is equivalent to 2 • 6 + 2 • 3, each being equal to 16. Similarly a(b + c)= ab + ac. This principle is called the Distributive Law of Multi-plication. Therefore, for the multiplication of a polynomial by a monomial, we have the Rule. Multiply each term of the polynomial by the monomial and write in succession the resulting terms with their proper Example : Sx^ — 2xy + iy — 5a — G 2x1 Product, Qxiy -^ x^y^ + & xy^ - Id axy - 12 xy Note. It should be kept in mind that the laws of operation that have been mentioned in this chapter, though evident from arithmetic only when the letters represent positive integers, are also valid when the letters stand for negative numbers, fractions, algebraic expres- sions, or other kinds of numbers that we shall introduce later. The principle which states that the operations on all numbers follow the rules expressed by the commutative, associative, and distributive laws is often called the Law of Permanence. 62 COMPLETE SCHOOL ALGEBRA EXERCISES Multiply : 1. x + 3hj2x. 5. -4a;= + 5a!-6by 605*. 2. 7x'-5\ij 3x^ 6. a:' - 3 a;'' + 4 by - 6 x\ 3. 5x^-2xhj -ix\ 7. x'^-2xy + y''\ij — 3xy. 4. 7 a;y - « by 3 xy. 8. a* - a^S^ + 5^ by - a^J". 9. _ aV + 2 ax — rS" by — 4 aSa;. 10. Tx'^ -Sx''-12x + &hj -%x\ 11. — 9 a^ — 12 aa; + 42 a;^ by | ax^ Perform the multiplication indicated : n. 4(2x-3). 16. -3x(2x-7). 13. 2x(x-y). 17. -3(a!^-2x-6). 14. -8(3x-7). 18. 5xy(a;^-6x + 9). 15. _9(_4a + S), 19. — 3 x (ax — 5x + 3 ex''). 20. — 7 a6 (ax^ + bx + c). 29. Multiplication of polynomials. Clearly" (5 + 3) (7 — 4) = 8.3 = 24. The multiplication may also be performed as fol- lows : (6 + 3) (7 - 4) = 5 (7 - 4) + 3 (7 - 4) = 35 - 20 + 21 - 12 = 24. Similarly (2x + 3)(4x - 5) = 2x(4x - 5) + 3(4x - 5) = 8x2-10x + 12x-15, or 8x2 + 2x-15. In general terms (a -\- V) (c + d) = a (c -\- d) + h (c + d) = ac + ad + be + hd. This gives for the multiplication of polynomials the EtTLE. Multiply the multiplicand by each term of the multi- plier in turn, and add the partial products. Example : 3x —2y 2x +3y Multiplying by 2 a;, 6x^ — ixy first partial product. Multiplying by 3 y, + dxy — Qy^ second partial product. Complete product, . 6x'^ + 5xy — Qy^ sum of partial products. MULTIPLICATIOlf 63 30. Powers. A power of a number is the product obtained by using the number as a factor one or more times. For example, 8, or 2^, is the third power of 2 ; 81, or 3*, is the fourth power of 3, and 32 x^, or (2 x)^, is the fifth power of 2 x. 31. Arrangement. A polynomial is said to be arranged accord- ing to the descendinff powers of a certain letter when the expo- nents of that letter in suecessiYe terms decrease from left to right. Thus 2 a;* — 5 a;'' — 6 a; + 8 is arranged according to the descending powers of x. Again, 4 — 2 y -|- y° and a;' — 3 xhj -f- 3 xy^ — if^ are arranged according to the ascendinrj powers of y. Whenever it is possible to arrange the multiplier and the multiplicand in a similar order it should be done, as the addi- tion of the partial products is then much more easily performed. 32. Check of multiplication. The work of multiplication can be checked by giving a convenient numerical value to each letter involved and finding the corresponding numerical values of the multiplier, the multiplicand; and the product. The prod- uct of the numerical values of the multiplier and the multipli- cand should equal the numerical value of the product. The number 1 is more convenient than any other number to use in checking, but it will not check exponents, since x^ = x^ = a;", etc., if a; = 1. It checks merely the coefficients. If a check on both coefficients and exponents is wanted, the number 2 is the most convenient. EXAMPLES 1. Multiply 3 x^ — 5 + x^ — 2 X hj 6 + x" — 5x. Solution : Arranging both multiplier and multiplicand in descend- ing powers of x and multiplying, we obtaiu : Check: x = 1. Sx^ + x^-^x-S , =-3 x'^-5x + 6 = + 2 3x^+ X*- 2x^- 5x^ -6 -15 a;*- ox^+10x'' + 25x -Fl8a:g+ 6a''-12a;-30 Product, 3 z= - 14 z^ 4- 11 i'^ + 11 a,-^ -M3 a; -'30 = - 6 64 COMPLETE SCHOOL ALGEBRA 2. Multiply 10 x?y2 _ 2 y= + 5 a;*y - 4 xY by 3 x^ - 7 y' - 6 a; V Solution : Arranging terms and multiplying, Check: x = y = \. 5a;*y+ 10a:y-4a;y-2y= = 9 3 x^ - 6 g'y - 7 y^ = -10 15 xiy + 30 ay - 12 x^y^ - 6 x^y^ - 90 - 30 kV - 60 xSf + 24 x'^y^ + 12 ar^j/^ - 70 a:«y° - 35 xY + 28 a^V + 14 y^ 15 a;'y - 72 x'^y^ - 76 a;Sj/= - 11 x^^ + 40 xV + 14 ,,8 _ _ go 33. Degree. The degree of a term with respect to a certain letter is determined by the exponent of that letter in the term. ' Thus X, 3 xy, and 4 a^xz are of the first degree in x, and 3 xy' is of the second degree in y. The degree of a term with respect to two or more letters is determined by the sum of the exponents of those letters in that term. Thus 5 x^y is of the fourth degree in x and y ; 4 aV)c^ is of the sixth degree in a, b, and c. 34. Homogeneous expressions. Terms are homogeneous if they are of the same degree with respect to the same letter or letters. Thus 3 a%^, 4 ab^, and a^b are homogeneous terms. A polynomial is homogeneous if its terms are homogeneous. For example, x^y — 8 x^y^ and 3 a* + a%^ + b* are homogeneous polynomials. An important property of homogeneous expressions is : The sum, the difference, the product, or the quotient of any two hom,ogeneous expressions is a homogeneous expression. This propertjr is useful in checking exponents in multipli- cation. Thus, if it be required to multiply x^ — ixy + y^ hj 3? — Zx^y + 3 xy'^ — y^, we know beforehand that every term of the product will be of the fifth degree. MULTIPLICATION 65 EXERCISES Multiply and check results : 1. a; + 4 by a; + 3. 9. 3x - ^ hj 2x - i. 2. 2a; + 3 by a; + 3. 10. - Sa; + 11 a by 5a; - a. 3. 4 a; + 7 by 3 a; + 2. 11. ax — bx by ex + dx. 4. 3 a: — 6 by 3 a; + 8. 12. — ca; + t^ by 5a; — cx\ 5. 3a; -2 by 2a; + 3. 13. 4a! - J by 6a; + f. 6. 6 — 4 a by 5 a — 7. 14. a;^ — 6 a; + 6 by a; — 3. 7. 2a! + ?/by a; + 3y. 15. 3a;2 - 3a; - 7 by 2a; + 4. 8. 2 a; — 3 y by 3 a; — 2 y. 16. a!^ — a;y + y*" by a; + y. 17. a'x^ — 2 a^x + 4 a' by ax -\-2a. 18. 3a;»-a;^-6a;by 2a;«-5a;l, 19. 2a;= - 7a; + 12 by a;^ - 3a; - 5. 20. a"" - \ a + ^hj a^ - a + \. 21. a;'' — a;y + y" by x^ + xy + y\ 22. 3x'^ + 5x''-x + 2'bj x''-2x + l. Expand : 23. (a;3-a;-5)(2a!^-3a;-4). 24. (3x-x' + x^-6)(5-x''-3x). 25. (4a - 5a2 + 7 + a')(6 4- «' - a + a^. 26. (6a; -4 + 8a;«)(8-5a;= + 2a;'- 9a;). 27. (a;^y — y^a;) (Axy — 5 x^y) (3 x'y — 7 xy^). 28. (x" + y^ + s' — xy — xs — yz)(x + y + z). 29. (a + h+ cy. 36. (a; + 2 y)^ - (a; - 2 yf. 30. {c + d-\y. 37. (4a;-3y)^-(3a;+4y)». 31. (a-25 + 3c-4e^)'. 38. (a; - 3)» - (2 a; - 1)^. 32. {x + y + zf- 39. (a;" - 3)(a;<' + 4).- 33. (a; + y)' + (a; - y)». * 40. {a?" + 5)\ 34. (2a -h+ 3y. 41. (2 a;" - 3)». 35. (2a;-3ay)». 42. {2x^''-3xy. CHAPTER IX PARENTHESES IN EQUATIONS 35. Simple equations involving parentheses. The removal of parentheses is really an easy matter -which is governed by simple rules. In handling parentheses, however, it' is very easy to acquire careless habits, -which are difficult to over- come. Accuracy in such -work can be attained only by espe- cial care in removing each parenthesis that is preceded by a minus sign. EXAMPLES 1. Solve the equation 6 (2 a; - 1) - 3 (4 a; - 6) = 7.' Solution: Multiplying by the coefficients 5 and 3, this becomes (10 3;-5)-(12x-18) = 7. Kemoving parentheses, 10 a; - 5 - 12 a; -I- 18 = 7. Combining, — 2 a; -(- 13 = 7. Transposing, — 2a; = 7 — 13 = — 6. Dividing by — 2, x=Z. Check: 5(2 ■ 3 - 1) - 3(4- 3 - 6) = 7. Simplifying, 25 - 18 = 7, or 7 = 7. Sometimes the square of the unknown number appears and then vanishes, as in the following. 2. Solve the equation 4-l-(re — 3)(re — 5)=15— (7— «)(2+m). Solution: Expanding, 4 + («2_8n-(- 15) = 15-(14 4-5n-n2). Removing parentheses, 4 -1- ri2 - 8 n -f 15 = 15 - 14 - 5 n -I- ««. Subtracting n^ from each member and combining, 19- 8n= l-5n. 66 PARENTHESES IN EQUATIONS 67 Transposing and combining, - 3ji=-18. Dividing by — 3, n = 6. Check: 4 + (6 - 3) (6 - 5) = 15 - (7 - 6) (2 + 6). Simplifying, 4 + 3 = 15 - 8, or 7 - 7. EXERCISES Solve and check : 1. 5{x- 1) = 30. 2. 3 + 2(a;-3)=l. 3. 7 (3 a; - 2) + 11 = 60. 4. 4 (2 a; -5) + 15 = 3 (a; + 10). 5. 12 y - 2 (4 y - 7) - 16 = 0. 6..9y-3(2 2/-4)=2(5-4y) + 2, 7. 4-2(4y-3)=3(y-5). 8. 7(2/-3)-2(4 + 2/) = 9. 9. 6(w-7)+24 + 4ra=0. 10. 6w- 9(2ra + 4) = 2(re-9). 11. 7n-12-2(n-5) = n-19. 12. 4(2w-7)-3(4«-8) + 4 = 2ra-3. 13. 3 7i - 2(4 A + 8) = 3 A - 24. 14. 5(3A + l)-77i = 3(7i-7) + 4. 15. (A - 2) (7j - 6) = (k + 3) (A + 2). 16. (A + 4) (A + 3) -(A + 2) (A + 1) - 42 = 0. 17. (x + 4) (a; + 6) = (x + 18) (a; + 13). 18. (k - 7) (6 + A) - (7c - 5) (A + 7) + 5 = 0. 19. (2a;-6)(4a;-7) = 8a;= + 52. 20. (3y + 5)(42/+7)-(2y + 3)(62/ + ll)-2 = 0. 21. (w + 3)(6» + 5)-(2w + 4)(3»i-8) = 38. 22. (a: + 3)= -(a; + 6)2=- 40. 23. (x + 2)2 - (a: - 4)^ + 48 = 0. 68 COMPLETE SCHOOL ALGEBRA EXERCISES 1. The length of a rectangle is a and its breadth is h. What is its area ? its perimeter ? 2. The length of a rectangle is a; — 4 and its width is 3; What is its area ? its perimeter ? 3. What is the area of a rectangle whose length is 2x — i and whose breadth is a; + 2 ? the perimeter ? 4. -Each of four horses cost flOO. What was the cost of aU? 5. Each of n horses cost $80. What represents the cost of all? 6. Each of a books cost b cents. What represents the cost of all ? 7. What is the total cost of x hats at a cents each, and y hats at b cents each ? 8. What is the cost of x horses at 5 + 10 dollars each ? 9. Represent the total cost of x chairs at 6 + 2 dollars each, and y chairs at a cost of c — 3 dollars each. 10. What is 5% of 16 ? of x ? 11. What is 3% oix + 120 ? of 12a; - 300 a ? 12. A is M. years old. What will three times his age 4 years from now be ? 13. If two sums of money are x dollars and 1000 — x dollars respectively, express the following as equations : (a) 4% of the first sum equals |180. (6) 3% of the first sum equals 5(/o oi the second. (c) 5«;^ of the first sum is $20 less than 4% of the second. 14. A picture is 10 inches wide and 12 inches long and has a frame 2 inches wide. What are the outside dimensions of the frame ? 15. If the frame in the preceding were x inches wide, what would represent the outside dimensions of the frame ? the PARENTHESES IN" EQUATIONS 69 area of the picture and frame ? the area of the picture ? the area of the frame ? 36. Problems involving parentheses. The following problems involve two or more unknowns and the use of parentheses. One of the unknowns can always be represented hy a single letter and the others by binomials involving this letter and one or more numbers. It may be necessary in some of the problems to inclose each of these binomials in a parenthesis and to think of them and use them as if they represented a single number. When the student can use a binomial in this way as readily as he uses a single letter, like x, he has made considerable progress in the algebraic way of thinking. PROBLEMS 1. The sum of two numbers is 88. Three times the less equals twice the greater, plus 29. Find the numbers. Solution: Here are two unknowns, the greater number and the less. Each can be represented in terms of a single letter as follows : Let n represent the less number. Then 88 — n must represent the greater. By the conditions of the problem : Three times the less = twice the greater + 29. Hence 3 n = 2 (88 - n) + 29. Simplifying, 3 n = 176 - 2 n + 29. Combining, 3 ra = 205 — 2 ». Transposing, 3 n + 2 ra = 205. Whence n = 41, the less number, and 88 — ra = 47, the greater number. Check : 41 + 47 = 88 ; 3 • 41 = 2 • 47 + 29, or 123 = 123. 2. The sum of two numbers is 49. Twice the greater, minus 13, equals five times the less. Find the numbers. 3. The sum of two numbers is 143. Ten times the less added to five times the greater equals 950. Find the numbers. 4. Separate 45 into two parts such that five times the greater plus four times the less may equal 207. 70 COMPLETE SCHOOL ALGEBRA 5. The sum of two numbers is 88. Three times the greater equals five times the less, plus 29. Find the numbers. 6. Separate 93 into two parts so that seven times the less, minus 7, equals six times the greater. 7. Separate 48 into two parts so that twice the greater, minus 7, equals three times the less, minus 5. 8. The sum of two numbers is 12|-. Seven times one num- ber miniis ten times the other equals 45. Find the numbers. 9. Separate 121 into two parts so that four times the one, increased by 8, equals three times the other. 10. Twice a certain number minus five times another number equals 240. The sum of the numbers is 15. Find the numbers. 11. The sum of two numbers is 14. Nine times the one minus eleven times the other equals zero. Find the numbers. 12. The square of a number plus the square of the next consecutive number is 17 greater than twice the square of the smaller number. Find the numbers. 13. The difference of the squares of two consecutive numbers is 75. Find the numbers. 14. The difference of the squares of two consecutive numbers is 23. Find the numbers. 15. The difference of the squares of two consecutive odd numbers is 104. Find the numbers. 16. The difference of the squares of two consecutive odd numbers is 40. Find the numbers. 17. The product of two consecutive eveh numbers is 56 less than the square of the greater number. Find the numbers. 18. The product of two consecutive odd numbers equals the square of the smaller increased by 46. Find the numbers. 19. A square has the same area as a rectangle whose length is 8 inches greater and whose breadth is 4 inches less than the side of the square. Find the area of each. PARENTHESES IS EQUATIONS 71 -Solution : There are three unknowns in this problem, — the side of the square, the length of the rectangle, and the breadth of the rectangle. The three can be represented in terms of the same letter as follows : Let s = the side of the square in inches. Then s + 8 = the length of the rectangle in inches, and s — 4 = the breadth of the rectangle in inches. Now the area of the square is ss, or s^ square inches. Similarly the area of the rectangle is (s + 8) (s — 4), which, ex- panded, equals s' + i s — S2. By the conditions of the problem the area of the square equals the area of the rectangle. Therefore s^ = s^ + 4 s - 32. Subtracting s^ from each member, = 4«-32. Whence s = 8, the side of the square, and s + 8 = 16, the length of the rectangle, and « — 4 = 4, the breadth of the rectangle. Therefore the area of the square is 8 ■ 8, or 64, square inches, and the area of the rectangle is 16 • 4, or 64, square inches. The check is obvious. 20. A square field has the same area as a rectangular field whose length is 30 rods greater, and whose breadth is 20 rods less, than the side of the square. How many acres are there in each field? 21. A tennis court, for two players, is 24 feet longer than twice its breadth. The distance around the court is 210 feet. Find the length and the breadth of the court. 22. A tennis court, for 4 players, is 6 feet longer than twice its breadth. The perimeter of the court is 228 feet. Find the dimensions of the court. 23. The breadth of a basket-ball court is 20 feet less than its length. The perimeter of the court is 80 yards. Find the dimensions. 24. The perimeter of a football field is 780 feet. Its length is 50 yards less than three times its breadth. Find the length and the breadth. 72 COMPLETE SCHOOL ALGEBRA 25. The value of 15 pieces of money, consisting of nickels and dimes, is 90 cents. Find the number of each. Solution : There are two unknowns in this problem, the number of nickels and the number of dimes. Since their sum is 15, the two can be represented in terms of one letter, thus : Let d = the number of dimes. Then lo — d = the number of nickels, and 10 • rf = the value of the dimes in cents. Also (15 — c?) 5 = the value of the nickels in cents. The value of the nickels and dimes together is represented by 10 d + (15 - d) 5. By the conditions of the problem the value of the nickels and dimes together is 90 cents. Therefore 10 d + (15 -d)5 = 90. Solving, d — Z, the number of dimes, and 15 — rf = 12, the number of nickels. Check: 3 • 10 + 12 • 5 = 30 + 60 = 90. 26. The value of 38 coins, consisting of dimes and quarters, is |5.30. Find the number of each. 27. A collection of nickels, dimes, and quarters amounts to $6.05. There are 5 more nickels than dimes, and the number of quarters is equal to the number of nickels and dimes together. Find the number of each. 28. The value of 40 coins, consisting of nickels and dimes, is |2.90. Find the number of each. 29. A is 20 years older than B. In 10 years A will be twice as old as B. Find the age of each now. 30. A is four times as old as B. In 20 years A will be twice as old as B. Find the present age of each. 31. A's age is 8 years more than twice B's age. Sixteen years ago A was four times as old as B. Find the age of each now. 32. A part of $800 is invested at 3% and the remainder at 4i^. The yearly income from the two investments is |30. Find each investment. PARENTHESES IN EQUATIONS 73 Solution : Let x = the number of dollars invested at 3%. Then 800 — x = the number of dollars invested at 4%. Hence .03 x = the yearly income from the 3% investment, and .04 (800 — x) = the yearly income from the i% investment. Therefore, by the conditions of the problem, .03 X + .04 (800 -x) = 30. (1) Multiplying each member of (1) by 100, in order to free the equation of decimals, we obtain 3 X + 4 (800 -x) = 3000. ' (2) Solving (2), X = 200, and 800 - a; = 600. Hence the 8% investment is $200 and the 4% is |600. Check: 200 600 ■03 .04 6.00 24.00 |6 + $24 = |30. 33. A part of |1400 is invested at 5% and the remainder at 6%. The total annual income from the tvro investments is $76. Find the amount of each investment. 34. A sum of money at 6% interest and a second sum. at 8% yield a total annual income of $53. The first sum exceeds the second by $125. Find each. 35. A 5% investment yields annually $15 less than a 6^^ investment. If the sum of the two investments is $1240, find each. CHAPTEE X DIVISION 37. Division of monomials. Division of numerical terms was explained under Positive and Negative Numbers. On page 22 will be found the rule for this division. Just as 2 -=- 3 is written §, so a -j- 5 is written as a fraction, - ) and this result can be simplified no farther. Similarly 2 2 « a-" H- a;-' = — > x^ and 2.-.3. = |f. But 12 c^ : 4.P ^^f In like manner, — 12a -i- 6b = — ■ Here the quotient is a fraction, and the minus sign indicates that the fraction is negative. Similarly 9a;-H(— 3y) = > and -24aV-f-(-6a»)= + i^. By the definition of an exponent, a^ = a- a- a-a-a and Thai ''' a^^a' = ^■^■a-a-a ^ ^^ ^, _ , Similarly 2" ^ 2" = ^X^X^x2x2x2 ^ ^b = 2-', and ax^ -^ x"^ = "^ "1^ — "-^j o^ ax^-\ In like manner, 6 6?/^ -;- 2 y' = 3 hy^, or 3b-y^-^. U DIVISION Y5 These examples illustrate the Peinciple. The exponent of any letter in the quotient is equal to its exponent in the dividend minus its exponent in the divisor. The foregoing principle expressed in general terms is : 7V-T-n'> = n"-*. What this equation means when 6 = a and when b is greater than a will be explained later. The law of signs in division may be indicated as f oUows : + ab ^(+a) = + b. + ab -i- (— a) = — b. — ab -i- (+ a) = — S. — ab -i- (— a) = + 6. Prom what precedes we see that ax^ -r- x^ = — ^— ^ = a. Hence a letter which has the same exponent in divisor and dividend should not appear in the quotient. Therefore for the division of monomials we have the KuLE. Divide the numerical coefficient of the dividend by the numerical coefficient of the divisor, keeping in mind the rule of signs for division. Write after this quotient all the letters of the dividend except those having the same exponent in divisor and dividend, giving to each letter an exponent equal to its exponent in the dividend minus its exponent in the divisor. If there are any letters in the divisor unlike those in the divi- dend, write them under the preceding result as a denominator. ORAL EXERCISES Perform the indicated division : 1. _io--2. 3. -16-h(-4). 5. -4a«^2a=. 2. 12 -=-(-3). 4. 8a»-=-a2. g. 6a;2-^(-3x). 7. — 18 a;' -=- (— 6 a;*). 8. — 25 ew;' h- 5 ax. 76 COMPLETE SCHOOL ALGEBRA 9. 12ax^---(-3bx'). 13. - 36 a;y -5- (- 6 a;y). 10. - 28 a/ -=- (- 7 y). 14. 63c»(i= -j- (- 9 6crf^. 11. 70a;y -f-(-10a!y). 15. 6i a^b'' -i- (- 16 a¥). 12. 48 ax'' -f- (- 16 6a;*). 16. - 28 a^i" -=- (_ 7 a*6'). 17. 15 xYz^ 75 xy^s ■ 21. - 17 aWc'' 25. x"*" 51 aVc ^ 18. 42 x^^j/^" -6a;y 22. 39a;iy°»'' -13a;iy'«" 26. X' X 19. a;" x^' 23. - 11 a^J^c" 66 aci^ 27. SaV a^x" _^Sa 24. - 121 a"62V» - 11 «"S"cii 28. 6a.2a+8 x" -2a;* 38. Division of a polynomial by a monomial. The division of the binoniial (18 — 12) by 3 can be performed in two ways : Thus (18 - 12) -H 3 = 6 H- 3 = 2, or (18 - 12) -r- 3 = Jj4 - J/ = 6 - 4 = 2. Similarly (ax + bx) -i- x = 1 = a + b. Therefore, for the division of a polynomial by a monomial we have the EuiiB. Divide each term of the polynomial by the m,onomial and write the partial quotients in succession. EXERCISES Perform the indicated division : ^ 6 g^ - 4 X g 4a;y-12x^ 25a;V+30a!y° 2a; ■ — 4a; ' — 5 xy „ 9a;-18x» , 9 aa;= - 12 a;= 16bx^-36x^ 2. 4 6 • -3 a; -3ax^ 4:bx^ 14xy-28a;y 4 a^y - 8a;y + 12a;y 7a;y ■ ■ 4a;V DIVISION 77 ^^ 15 g'6° + 9 a*b^ - 30 a°&'' 12. - 3 ai'i^ 16 g^&V - 24 a°& V - 48 a%-'o' 8 a'^^c jg 85 xyz - 51 X V + 102 a: V° - 170 x^i^z — 17 icys 14. Hx-B)+a{x-Z) X — 3 3a;(3a: + 4)-4y(3a; + 4) 3x + 4 5a(2x^-y)-35(2x''-y) 2x^-y (a + hy-3(a + by 21(x~yy-35(x-yY (a + by ■ '"■ -7{x-yy 16(3 a: - 4)^ - 24(3 a; - 4)° - 48 (3 a; - 4)^ -8(3x-4)'' - 5(ac° -2d)^ + x(ao^ -2d) 21. 22. 23. 5(ao'-2cr) 4:x*-8x^''-6x^'-^ 2x^ 3 a;" — 2 a:°+^ — x''+^ + a;" a;2 6a;2»-3 _l2a;*"+^ — 18a;»''+5 — 3 a;' ,2n 39. Division of one polynomial by another. Division is the reverse of multiplication, and the process of dividing one polynomial by another ivill be best understood by finding the product of two polynomials and then dividing it by one of them ; the other, of course, will be the quotient. A close inspection of the steps in the multiplication (A) which 78 COMPLETE SCHOOL ALGEBRA follows will make clear the necessity for each step in the division (B). 4 x2 _ 5 x + 6 2x-Z SaH"- 10a:='+ 12a; -12x^+15 a; -18 (A) 8a;8- 22a;2 + 27a:-18 Now let 8 x' - 22 a;2 + 27 a; - 18 be the dividend and 4 x" - 5 a; + 6 the divisor. Then the quotient must be 2 a; — 3. Dividend, 8 x^ - 22 x^ + 27 x - l&H x'^ - b x + Q, Divisor (4x2_5j; + 6)2a;, 8 x^ - 10 a:" + 12 x |2x-3 , Quotient -12x2 + 15x-18 (B) (4x2-5x+6)(-3), -12x'' + 15x- 18 The term having the highest power of x in the dividend, 8x^, was obtained by multiplying the term having the highest power of X in the multiplicand by 2 x. If the multiplication were not before us, we could obtain the 2 x by dividing 8x^ by 4x^;- that is, by dividing the term of highest degree in the dividend by the term of highest degree in the divisor. Multiplying the entire divisor by 2 x, we get the first partial product of the multiplication (A). Subtract- ing, we get — 12x^ + 15x — 18. If the multiplication (A) did not tell us that the second term of the quotient was — 3, we could ob- tain it by dividing — 12 x^ by 4 x^; that is, by dividing the term of highest degree in the remainder by the term of highest degree in the divisor. Multiplying the entire divisor by — 8 and writing the product under the remainder, we get (4x'' — 5 x -1- 6)(— 3), or — ISx'^ -f 15x — 18, for the second partial product of the multiplication (A). As there is no final remainder the division is said to be exact. The process of dividing one polynomial by another is ex- pressed in the EuiiE. Arrange the dividend and the divisor according to the descending (or ascending) powers of some common letter, called the letter of arrangement. Divide the first term of the dividend by the first term of the divisor and write the result for the first term, of the quotient. Multiply the entire divisor hy the first term of the quotient, write the result under the dividend, and subtract, being careful DIVISION Y9 to write the terms of the remainder in the same order as those of the divisor. Divide the first term of the remainder by the first term of the divisor for the second term of the quotient and proceed as before until there is no remainder, or until the remainder is of lower degree in the letter of arrangement than the divisor. If there is no remainder, the result of division may be expressed as follows : Dividend ^ T,. . = Quotient. Divisor If there is a remainder, the result is expressed as follows : Dividend t. ^. , ^ , . , , Remainder -,. ■ = Partial Quotient -I =-^^ Divisor Divisor This last corresponds to what is done in arithmetic in dividing 17 by 5, which is written ^^ = 3f . This means that J/- = 3 + f , the plus being understood. Note. "We saw on page 1 that it is customary to represent the product of two letters by placing one after the other with no sign between them. Thus ab means a times b. But addition, not multi- plication, is implied by placing the fraction | after the number 3. This practice comes down to us from the Arabs, who denoted all additions by placing the number symbols in succession without any sign of operation. The later Greeks also had the same notation. EXAMPLES 1. Divide 38a; +- 2a;* -Tx^ - 24 - 7a;» by 6 + a;= -5x. Solution : Arranging terms and dividing, Dividend, 2 X* - 7 x^ - 7x2 + 38x-24 2x<- 10x3+12x2 3x3- 3x»- -19x2 + 38x -15x2 + 18x - - 4x2 + 20x- - 4x2 + 20x- -24 -24 x^ — 5x + 6, Divisor 2x2 + 3x-4, Quotient Check: Let x = l. Then the dividend = 2, the divisor = 2, and the quotient = 1 ; and 2-^2=1. 80 COMPLETE SCHOOL ALGEBRA The student must avoid checking by any number which makes the divisor zero. 2. Divide 8 xy^ + 8 a;' — 7 y' — 12 x'y by 4 a;^ + y'' — 4 xij. Solution: Arranging terms and dividing, Dividend, 8 a;^ - 12 a;"?/ + 8 x'f 4:x'^ ~ ixy + y'^, Divisor 2x — y. Partial Quotient — 4 x^y + 6 xy^ — 7 y^ — 4 x^y + 4 xy^ — y^ 2 xy^ — 6 y^, Remainder 2 XV 6 V The total quotient is 2 x — w + - — „ . — - ■ ^ " ix^-ixy + y^ Check: Let x = y = 1. Then the dividend .= — 3, the divisor = 1, and the quotient = — 3 ; and — 3 -^ 1 = — 3. General Check foe Division, (a) When the division is exact. Multiply the divisor by the quotient. The product should be the dividend. (5) When there is a remainder. Multiply the divisor by partial quotient and add in the remainder. The result should be the dividend. Divide : EXERCISES 1. x^ + 7 a; + 12 by a; + 3. 2. a;='-2a;-15 by x-5. 3. a;'' + 6a; + 6 by a; +'3. 4. — 7 X + 6 + a;2 by X — 1. 5. 6x=-13x + 6by 2x-3. 6. 26x^ + 30x2 -7 by 7 + 6x2. 7. 12 a= - 21 + 19 a by 4 a - 3. 8. - 8 + x' + 4x - 2x2 by X - 2. 9. a' + 3 a% + 3 aS^ + 5=' by a + b. 10. x» - 16x= + 65x - 63 by X - 7. 11. 5x= + 5x-25x»-l by 6x^-1. 12. 2x»-14x2 + 14x+12by 2x-4. DIVISION 81 13. 3 a= + 28 a^ + 89 a - i40 by 3 a - 5. 14. 37 X + Gx" - 24, - 23x^ hj 2x - S. 15. 53 a + 8 - 63 a^ + 12 a* by 4 «= - 7 a - 1. 16. 15 a' - 56 a'' + 99 a - 70 by 3 a^ - 7 a + 10. 17. 23 a^ + a* - 55 a + 11 a^ - 140 by a^ - 5. 18. 4 a' + 1 + a* + 4 a + 6 a" by 1 + a= + 2 a. 19. a* - 8 a» + 24 a^ - 32 a + 16 by a^ - 4 a + 4. 20. 40 a; — 31 x^ + 21 + cc* + 4 a;' by a;^ — 3 — 7 a;. 21. 11 a; - 42 a;^ + 10x^-27 a;' -36 by 9 + 2a;2 - 5a;. 22. a;* — 3 a^x^ + a* by a;^ — aa; + a^. 23. a* + 4 6* + 3 a'b^ hj 2P + a^~ ab. 24. 16 a;* - 60a!y + 25 y*' by 4 a;'' - 10 xy -5y^ 25. 9 a* + 49 6* + 26 a%^ by 7 &^ + 3 a^ + 4 aS. 26. 4 a« - 44 a*&* + 100 b" by 2 a* - 10 5* + 2 a%\ 27. 25a;^-10a;= + 40ar-18by 5a!-6. 28. a;' — y' by a; — y. 34. a;* + y* by a; + y. 29. a' — 125 5' by a — 5 6. 35. x* — y* by a; + y- 30. a^ + 343 5= by a^ + 7h. 36. a;* — / by a; — y. 31. x^ - 16 by a; + 2. 37. a:^ + 2/^ by a; + y. 32. /- 52/''-3000 by y - 5. 38. a;« + y° by a; - y. 33. a;* + y* by a; — y. 39. a;^ — y^ by a; + y. 40. a;^ — 2/^ by a; — y. 41. ai^" — 5 a;" +6 by a;" — 3. 42. a!«" — 7a;"' + 12 by a;*" — 4. 43. a;2» + 2a;''+i + 3a;'' + 3a! by a;»4-2a;. CHAPTER XI EQUATIONS AND PROBLEMS 40. Equations involving literal coefficients. The most general form of a simple equation in one unknown is one in which the unknown occurs with literal coefficients. The solution of such an equation frequently involves division of polynomials. EXAMPLE Find the value of x in aa; + 4 a. = a^ + 2 a; + 4 and check the result. Solution : ax + 4 a = a^ + 2 a; + 4. Transposing, aa; — 2 a: = a^ — 4 a + 4. Writing the coefficients of a; as a binomial, (a-2)a; = a2-4a + 4. Dividing both, members by the coefficient of x, a^ - 4 a + 4 x = = a-2. a — 2 Check : Substituting a — 2 for x in the original equation, it becomes a (a - 2) + 4 a = a^ + 2 (a - 2) + 4. Simplifying, a2-2a + 4a = a2 + 2a-4 + 4. Combining, a^ + 2 a = a^ + 2 a. EXERCISES Solve for X and check : 1. x + 2a = 6a. 4. ca; + c= = 6 c'. 2. x + a = b. 5. 5(b-x)=10b. 3.bx + b = Ab. 6. bx-(b + c)=5b-c. 7. Sax — ab = 2ax — ac. 8. Abx — 7a^b = 6 ab^ + 3 bx. 82 EQUATIONS AND PROBLEMS 83 9. ax + bx ^= ac + bo. 10. a^x + l — a* — x = 0. 11. ax+2ab = 2a^ + bx. 12. ax — a' — 4: = 3a -- X. 13. 4 bV + (a + bx)G=(a,- bx) c. 14. ax — ao + be = 2 ac — 5 be + 2 bx. 15. (x + a)(x + b) = x^ + 2 a^ + Sab. 16. 15(a;-a)-6(a; + a) = 3(5a-3a;). 17. 4:X — cx — 8+2a + 6c = 6a — Sao + 2ex. 18. 9 ab +(x - 3 a)(x - 3b) = (x + 3 a) (x - 3 a) - 9 aK 19. (5a ~ ib)x - 5{P + Aa^ + 6ab) = 10b'-3(2a^ + 3 bx) - a(2x - b). 20. a'^x + 3 aa; + 10 a = a= + a; + 3. 41. Uniform motion. If a train travels for 8 hours at an average rate of 40 miles an hour, the total distance traversed is 8 X 40, or 320 miles. This illustrates uniform motion involving : 1. Time measured in seconds, minutes, hours, etc. 2. Rate (velocity), or the distance traveled in a unit of time, one second, one hour, or one day. 3. Distance (total) measured in standard units of length as feet, or inches, or meters, or kilometers, etc. Time (t), rate (r), and distance (d) are connected by the relation d=rxt. On this simple equation a large number of problems in algebra and in physics are based. Biographical note. Siu Isaac Newton. Sir Isaac Newton (1642-1727) was probably the keenest mathematical thinker who ever lived. He was the son of a farmer of slender means, and as a boy was rather lazy. It is said, however, that his complete victory over a larger boy in a fight at school led him to feel that perhaps he could be equally successful in his studies if he really tried. His ambition and interest being once roused, he never ceased to apply himself during the rest of his long life. 84 COMPLETE SCHOOL ALGEBRA His most important scientific achievement was tiie discovery and verifi- cation of tiie laws of motion. In liis great worlc called the " Principia " * he showed by mathematical reasoning that all bodies, great and small, — the planet revolving around the sun, as well as the apple falling from the tree, — follow the same . laws. His greatest discovery in pure mathematics was that of a method called the calculus, which is the basis of most of the advances in mathematics and in theoretical physics made since his time. But important as was Newton's mathematical work, his most signifi- cant contribution to mankind was an idea, — ■ the idea that the world in which we live is not independent of the rest of the universe, but that every smallest particle of matter is connected with the most remote planet and star ; that we cannot think of ourselves as the center of all things, but that we merely occupy our place in a system of universal law. EXAMPLE A pedestrian traveling 4 miles per hour is overtaken 14 hours after leaving a certain point by a horseman who left the same starting point 8 hours after the pedestrian. Tied the rate of the horseman. Solution : This is a problem in uniform motion, involving the distance, the rate, and the time of a pedestrian and a horseman respectively. By a careful reading of the problem one discovers that the time for each was a different number of hours, that each went at a different rate, but that each traveled the same distance. Hence the equation will be formed by expressing d in terms of r and t for both the pedestrian and the horseman and then equating the two expressions for d. By the conditions : i, or time in hours /', or rate in miles per hour Distance, rf-rx< Pedestrian 14 4 56 = 4x14 Horseman 6 X Q-x Hence and Check: 6 a; = 56, 4-14: 56; 91-6 = 56. * A copy of this book, presented to the College by Newton himself, may be seen in the library of Yale University. SIE ISAAC NEWTON EQUATIONS AND PROBLEMS 85 PROBLEMS A and B start from the same place at the same time and travel in opposite directions : 1. A goes 8 miles per hour and B 10 miles per hour. In how many hours will they be 120 miles apart ? 180 miles apart ? 2. A travels twice as fast as B. In 6 hours they are 135 miles apart. Find the rate of each. 3. A travels 2 miles per hour more than B. After 8 hours they are 96 miles apart. Find the rate of each, 4. A goes 4 miles per hour more than B. After 6 hours the distance between them is 168 miles. Find the rate of each. 5. B goes 3 miles per. hour less than A, and travels | as fast as A. Find the rate of each. After how many hours will the distance between them be 168 miles ? 6. A travels 3 hours and stops, and B travels 5 hours. Then they are 77 miles apart. A's rate is twice B's. Find their rates and the distance each has traveled. 7. B travels 9 hours at a rate of 4 miles per hour less than A. A travels an equal distance in 3 hours less time, and then stops. How far are they apart at the end of 9 hours ? A and B start at the same time from two points 144 miles apart and travel toward each other until they meet. Find the rate of each : 8. If they travel at the same rate and meet in 8 hours. 9. If A travels 2 miles per hour less than B and they meet in 9 hours. 10. If B travels three times as fast as A and they meet in 12 hours. 11. If they meet in 6 hours and B travels 24 miles more than A. 12. If they meet in 8 hours and B goes 2 miles per hour more than A. 86 COMPLETE SCHOOL ALGEBRA 13. If they meet in 9 hours and A travels 4 miles per hour more than B. Find the number of hours from the start until the time of meeting : 14. If B goes 6 miles per hour more than A and travels twice as far as A. 15. If A travels 6 miles per hour and B 9 miles per hour, but B is delayed 4 hours on the way. 16. If A is delayed 3 hours and B is delayed 5 hours, and their rates are 7 miles and 9 miles per hour respectively. 17. A and B start from the same place at the same time and travel in opposite directions. A travels 3 miles per hour and B 4 miles per hour. In how many hours will they be 42 miles apart ? 18. A and B start at the same time from two points 72 miles apart and travel toward each other. A travels 8 miles per hour and B 10 miles per hour. In how many hours will they meet ? ■ 19. The distance from Kansas City to St. Louis is 285 miles. A train running 37 miles per hour leaves Kansas City for St. Louis at the same time a train running 38 miles per hour leaves St. Louis for Kansas City. In how many hours will they meet ? 20. A starts from a certain place and travels 8 miles per hour. Four hours later B starts from the same place and travels in the same direction at the rate of 10 miles per hour. How many hours does B travel before overtaking A ? 21. Two bicyclists 108 miles apart start at the same time and travel toward each other. One travels 10 miles per hour, the other 12 miles per hour. The latter is delayed 2 hours on the way. In how many hours will they meet, and how far has each traveled ? 22. A passenger train starts 2 hours later than a freight, train, from the same station but in an opposite direction. The rate of the passenger train is 42 miles per hour and EQUATIONS AND PROBLEMS 87 the rate of the freight train is 24 miles per hour. In how many hours after the passenger train starts will the two trains be 246 miles apart ? 23. A messenger going at the rate of 8 miles per hour has journeyed 2 hours when it is found necessary to change the message. At what rate must a second messenger then travel to overtake the first in 6 hours ? 24. A passenger train and a freight train start together from the same station and move in the same direction on parallel tracks at the rate of 45 miles and 18 miles per hour respectively. How much time will have elapsed before the passenger train will be 144 miles ahead of the freight train? (Problems 25-32 may be solved without using equations.) The distance from P to Q is 108 miles. A and B leave P at the same time and travel at different rates toward Q. The one who reaches Q first at once returns. Find the distance each has traveled when they meet : 25. If A's rate is 9 miles per hour and B's is 15. 26. If B travels five times as fast as A. 27. If A travels 66 miles more than B. 28. If they meet 24 miles from Q and B travels faster than A. Find the rate of each if A travels faster than B : 29. If they meet in 6 hourshalfway between P and Q. 30. If they meet in 6 hours f of the way from P to Q. 31. If they meet in 6 hours 12 miles from Q. 32. If they meet in 6 hours 96 miles from P. 33. If A travels 4 miles per hour more than B and meets B in 12 hours. Find the distance each travels : 34. If they meet in 6 hours and A travels 2 miles per hour more than B. 35. If A travels 4 miles per hour more than B and they meet in 9 hours. 88 COMPLETE SCHOOL ALGEBRA The Telocity of a bullet continually decreases from the in- stant it leaves the gun. This is due to the resistance of the air. In the following problems consider the velocity of sound as llO'O feet per second. 36. Two and one-half seconds after a marksman fires his rifle he hears the bullet strike the target which is 550 yards distant. Pind the average velocity of the bullet. 37. One and three-fourths seconds after a marksman fires his revolver he hears the bullet strike the target 50 rods distant. !Find the average velocity of the bullet. 38. A marksman fires at a target 1000 yards distant. The bullet passes over a boy, who hears the sound of it striking the target and the report of the gun at the same instant. The velocity (average) of the bullet is 1650 feet per second. Find the distance of the boy from the target. REVIEW EXERCISES AND PROBLEMS 1. Simplify 5a — 7x+3b — 10c—lid + 12y — 8x + 12a-llx + Qb. 2. Simplify 2 cd — a% + 7 cd^ — 12 ab + 17 ad — cd + 4: cH. 3. Add 10 a; — 9 -1- 4 ax — 2 c(Z, — 6 aa; — 16 a; -f 12 c(Z, 10 - ax + 5 cd, and — 6 cii + 11 a; — 19 — 7 ax. 4. AMa^-3 + 4:a^-a, - a + Sa'' ~ 5 a', - 3 a^ + 6 a' - 2a -f 8, - a - 5 a^ - 2 a^ -h 5, and 2 a^ - 3 a^ + 4 a - 6. 5. Add Ax^ + Bx^ + Cx and Ax^ + Bx+ C. 6. Add 7(a-h)-lQ{o-d)+d,{x~y), -&{o-d)- 4 (a - Z-) - 7 (a; - y), and 12 (a; - y) - 16(c - cZ) -f 7(a - b). 7. Add -2{x + y) + 3{x-y)-4.{a + b), 8(x-y)- 9(x + y) + 4:{a — b), and — (a + b) + (x — y) --10(x + y). 8. Add J aa; + I ax^ — | aV + a, ^ ax^ — ax — 9a + 2 a^x% and a'x^ + ^ ax^ — ^ a. 9. What must be added to2x^ — 3xy + y'' — Ito give x^ + 10 xy - 9 - 5 y2 ? EQUATIONS AND PROBLEMS 89 10. What must be added to 3 a^y — 7 ay'^ + « — 5 a?/ to give - ah/ + 12 ay"^ -ay -15? 11. What must be added to — 17 a^bccP — 4 ahVd'^ + 3 cAcd + aa; to give 3 aJVc?'- 12? 12. What must be added to 27xy + 13 « — 12 + a& to give ? Remove the parentheses and collect terms : 13. {a + h)-{a — h) + 4.a —(a + dh) + c. 14. 15x-2(12a;-16?/)+a;y + 3(3a;-4a;2/)-a;. 15. ah -(J X + 4.y - 2 z)+ A2 + o{5 ab - X - ij). 16. [x-(-4a; + 8y + 2x:)+5-4(3a;-«)]-5 5. 17. {ah — {—5ac+l{d — a) + ica + ld) — &ba'\ + 13 a. 18. {5 x^ - 2 [- 4 - (2 a;2 - 3 x)] + 5} - (a:' + 4 a; - 2). Perform the indicated multiplication : 19. (a;* — 4 a;^^ + 4 y*) (a;* + 4 a; V'' + 4 /)• 20. (4a;*-a;2 + 6)(3x=-a; + 7). 21. {a + b + c) (a^ + b^ + c^ - ab - ac - be). 22. (x + y) (a;^ — x^y + a:^y^ — x^y^ + ai^i/* — a;j/^ + y'). 23. (5x-9a;= + a:^-4a;*-3x= + 10)(-a;= + 3a;-a;2+2x^). 24. (3a;^-42/)(9a;* + 16y^)(3ai^ + 4y). 25. (a - 3 5 + 2 c)^. 27. (x" - if) (a;» + /). 26. (x-y + s)'- 28. (ai« + 2y'' + 3«'')(a!»-3y"'). Divide : 29. 6 a^ - 13 a* + 4 a' + 3 a^ by 2 a'' - 3 a^ + a. 30. - 30 a^ - 11 a« + 82 a^ + 12 a - 48 by 3 «== + 2 a - 4. 31. 10a;2 + 20x^-lla;= + 10a;8-3a;* + 2by -3a;2-2+,v. 2a; + 5a;^ ^ 32. a« - 3 «« + a^ - 7 «== + 3 by (a" - 1)=. 33. a^-5 a*b + 10 a%^ - 10 a^b^ + 7ab^- h^ by {a - by. 34. (6 a^ + 5 a - 6) (2 a^ - 13 a + 20) by (3 a - 2) (2 a - 5). 35. a;» — 272/^ + 64 + 36a:y by a;-3y + 4. 90 COMPLETE SCHOOL ALGEBRA Mnd the value of y in the following equations : 36. 62/-20 + 62/-18 = 36y-4-402/+9. 37. 42/-3(6-2/)+2 = 6(2/-2)-13y + 24. 38. 62/-4(l-22/) + 10 = 4(22/-5)-4y + l. 39. (2/ + 7)(y-ll) = (y + 8)(2/-5)-2. 40. (y-3)^ = (2/^-6y+9)(5 + y)-82/^ 41. 2/2 + a^ -(y + 1)-^ - (« + 1)=^ = 0. 42. 3 ay + 6 ac + 4 6^ = 3 aS + 4 Sy + 8 Jc. _ J -I- V^^ — 4 ac Find the value of a;, if a; = ^^ : ' 2a 43. When a = 1, J = + 10, and c = — 11. 44. "When a = 6, 5 = — 6, and c = — 8. Find the value of x, \1 x= t, : ' 2a 45. When a = 3, 5 = 5, and c = — 8. 46. When a = 2, J = 7, and c = - 22. 47. li f= 12, m = 18, and v = 20, find the value of s in the equation /s = ^ mv^. 48. If a = 32.2, m = 7, if = 40, find the value of e in the equation e = J in{atY. 49. If s = ;r > find the value of Vs (s — a) (s — 6) (s — o) Li ■when a = 3, 5 = 4, and c = 5. 50. If the square of a certain number is increased by 38, the result is equal to the product of the next two consecutive numbers. Find the numbers. 51. The square of a certain odd number is 98 less than the product of the next two consecutive odd numbers. Find the ■ numbers. 52. If the square of a certain odd number is increased by 47, the result is equal to the product of the next two consecutive even numbers. Find the numbers. EQUATIONS AND PROBLEMS 91 53. The product of two consecutive numbers is 42 less than the product of the next two consecutive numbers. Find the numbers. 54. The ages of two persons are respectively 42 years and 15 years. In how many years will the elder be twice as old as the younger ? 55. The ages of two persons are respectively 36 years and 16 years. How many years ago was the older person three times as old as the younger ? 56. The ages of two men are respectively 62 years and 21 years. How many years hence will the older man be twice as old as the younger ? 57. The ages of two persons are 87 years and 42 years respectively. How many years ago was the elder four times as old as the younger ? 58. A collection of quarters, dimes, and nickels, containing 32 coins, is worth |3.60. There being twice as many nickels as quarters, find the number of each. 59. A bullet is fired from a rifle at a speed which would average 1280 feet per second. Six seconds later the marksman hears it strike the target. The velocity of sound is 1120 feet per second. Find the distance to the target. 60. The leader in some games proposed to tell the age of the others thus : each was to add 12 years to his age, to multiply the sum by 3, to subtract 36 from the product, and then to add his age. Each in turn announced his final result and the leader at once gave the correct age. What did he do to each result to obtain the proper age ? 61. The per cent of the population of the United States under 20 years exceeds by 5% the population between 20 years and 60 years. The per cent between 20 years and 60 years is nine times the per cent above 60 years. Find the per cent of the population under 20 years, between 20 years and 60 years, and over 60 years. 92 COMPLETE SCHOOL ALGEBRA 62. The length of the St. Gothard tunnel exceeds that of Mount Cenis (Italy) by 9000 feet. The length of the St. Goth-' ard tunnel is 1320 feet less than twice the length of the tunnel at Hoosac, Massachusetts. If the sum of the lengths of these tunnels is 113,760 feet, find the length of each. 63. The height of the first cascade of the Yosemite water- fall exceeds that of Staubbach Falls by 520 feet and is 60 feet more than four times the height of the falls of the Zambezi. The height of the latter is 32 feet more than twice the height of Niagara Falls. The sum of the four heights is 3004 feet. Fiad each. 64. The combined weight of a cubic foot of mercury, a cubic foot of water, and a cubic foot of alcohol is 962.2 pounds. Alcohol weighs 11.5 pounds per cubic foot less than water, while a cubic foot of mercury weighs 32.7 pounds more than sixteen cubic feet of alcohol. Find the weight of each per cubic foot. 65. If the average annual rainfall at Boston were 3 inches less, it would be one third the rainfall of Neahbay, Wash- ington. The annual rainfall of Boston is 1 inch greater than that of St. Louis and 1 inch less than five times that of San Diego, California. If these places together have a total annual rainfall of 219 inches, find the rainfall at each place. 66. From tables which have been compiled it is found that a person 10 years old may expect to live 6.39 years longer than one 21 years old and 15.31 years longer than one 45 years old ; and a person 45 years old may expect to live .94 of a year less than twice as long as one 65 years old. If four people (one at each of these ages) may expect to live a total of 109.42 years, how many years may each expect to live ? CHAPTEE XII IMPORTANT SPECIAL PRODUCTS 42. The square of a binomial. The multiplication a +b a +b a? + ah ab + V a^ + 2ab + b'' gives the formula (a + bf = a' + 2ab-\- b\ This may be expressed, in -words as follows: I. The square of the sum of two terms is the sum of the squares of the terms plus twice their product. Similarly ia- bf = a^ -2ab + V^, which may be expressed in words as follows : II. The square of the difference of two terms is the sum of their squares minus twice their product. EXERCISES Square the following either by I or II : 1. 16. Solution : 16^ = (10 + 6)" = 100 + 36 + 2 ■ 10 • 6 = 256. 2. 49. Solution : 49= = (40 + 9)^ or (50 - l)^. Using the latter, (50 - 1)^ = 2500 + 1 - 2 • 50 ■ 1 = 2401. 3. 15. 6. 19. 9. x + 1. 4. 17. 1. x-\- y. 10. a + 2. 5. 18. 8. x-o. 11. b'' + 3. 93 94 12. c-4. 13. 2 c^ + ^. 14. Zx-y. 15. x + 2y. 16. a-4.b. 17. f-2. 18. 2a + 1. 19. 3 a" + 2. COMPLETE SCHOOL ALGEBRA 20. ia + 3. 28. 4a;y— 2aj''. 21. 2a2_35. 29. 3 a;* — 5 a;«. 22. 3c'-4a'^. 30. 2a;^-6a:V 23. 2a;=-6. 31. 41. 24. Sxy+y'. 32. 92. 25. 7a!»-3/. 33. lOL 26. 8y^ + 2bx. 34. 202. 27. 6xy-2a\ 35. 1001. Perform mentally the indicated division : a^ + 2ab + b^ .. ^" + 402/ + 400 ^^- ^+6 20 + 2/ a=-2ac + c^ ,, 9 a;" - 12 ay + 4 y° 37. 41. ^ 3 a — e 2y — 6x a^-Aa + i ^„ 4 a^- 20 a + 25 38. K^ — 42. ^—-^ a — 2 — 5 + 2a 9a:^-6x + l Ao^j--12xy + 9j^ ^^- 3a; -1 3y-2a; rind a binomial divisor for each of the following trinomials : 44. x^-2xy+4/. 48. 16a;2-8x + L 45. o^ + 2a+l. 49. 16a;= -8a;y + /. 46. a^ + 4a + 4. 50. 9a;^- 6x^ + 1. 47. 4a:' + 4a;+l. 51. 16 y« - 40 y' + 25. State the two binomials whose product is : 52.c^ + 2cd + d\ ' 55. ia'-ia + l. 53. a^-2a + L 56. 4x^ + 12a! + 9. 54. x" -10a! + 25. 57. 9 a^b" - 6 ab + 1. 58. 16 a;y — 24 xV + 9 <*'■ 59. 25 a'b^ - 60 a^b'c^ + 36 c\ It is often convenient to use the word term in a broader sense than that in which it has been used in the work thus far. IMPOKTANT SPECIAL PRODUCTS 95 For example, in the expression (a + i) — 3 (a + 2 5) + 4 (a + 3 6) the binomials (a + 6), — 3 (a + 2 6), and 4 (a + 3 6) are often spoken of as terms, and the entire expression is then called a trinomial. If in (a + b')(a + b) = a^ + 2ab + b^ we substitute x + y ioi a, we get l(,x + y) + b] [(x + y) + b-] = (^x + yy + 2 (x + y)b + b\ Expand- ing (x + yy by the formula, and expanding 2b(x + y) also, we obtain x^+2xy + y^ + 2bx + 2by + b'^. This means that if we regard (x + y) as a term and apply the formula for squaring the sum. of two terms, we can square the tri- nomial x + y + b mentally. Similarly (x + y -hy = [(s -f- y) - &J [(x + y)-b'\ = {x + yy-2b{x + y) + b^ = x^ + 2xy + y^-2bx~2by + h\ EXERCISES Using one of the formulae on page 93, square the following trinomials : \. a + h + e. 5. a — h + c. 9. 2 a -|- S — 4 c. 2. a + h — G. 6. a — b — 0. 10. 3a — 2b + 5 e. 3. a + b-1. 7. a + 2b-3. 11. 2a-3&-4c. i. a + b+1. 8. a — 2b + o. 12. ic — 1 +2 ab. 43. The product of the sum and the difference of two terms. The multiplication a +b a—b a^ + ab -ab-b^ gives the formula This may be expressed in words as follows : III. The product of the sum, and the differenoe of two terms equals the difference of their squares taken in the same order as the difference of the terms. 96 COMPLETE SCHOOL ALGEBRA EXERCISES Expand by the preceding formula : 1. (6 + 3) (6 -3). 4. (10 - 4) (10 + 4). 2. (8 + 4) (8 - 4). 5. (7 + 2) (7 - 2). 3. (9 -2) (9 + 2). 6. (12 -3) (12 + 3). 7. 22-18. Solution : 22 • 18 = (20 + 2) (20 - 2) = 400 - 4 = 396. 8. 35-25. 11. 36-45. 14. 72-68. 9. 33 - 27. 12. 52 - 48. 15. 75 • 85. 10. 36-44. 13. 65-75. 16. 97 103. 17. (x + 3) (x - 3). 26. (4 + y^) (4 - y^). 18. (a + 5) (a -5). 27. (4 - cc) (a; + 4). 19. (2 a; + 4) (2 a; -4). 28. (2o + a) (2 c - a). 20. (3n + 5)(3n- 5). 29. (3a + b){b-3 a). 21. (x + y){x- y). 30. (4 6 -f 2 c) (4 5 - 2 c). 22. {x -a)(x + a). 31. (3 xy — 2) (3 xy + 2). 23. (a;-l)(a;+l). 32. (4 a6 - 3) (4 a5 + 3). 24. (a + 2)(«-2). 33. (a* - 6^) (a* + S=). 25. (a^ + 3)(a^-3). 34. («« + «»)(-«» + «''). 35. (x' - 2 2/) (a:» + 2 y). 36. (4 a6 — a^) (4 a6 + a^). 37. (3 cc^^ + 2 c?) (- 2 c? + 3 ctZ'). 38. {Gcd + S)(-3 + Qcd). 39. (4a;y + 2 2/)(2y-4x2/). 40. (3 abc - 2 6c) (2 Jc + 3 abc). Perform the indicated division : 41. (a^-&^) -=-(« + 6). 44. (36 - a^) -f- (6 - a). 42. {o''-d')-^(c-d). 45. (9a;=-16)-^(4 + 3a!). 43. (9 - J2) ^ (3 + 5). 46. (a;^ _ l) ^ (a;2 _ l). IMPORTANT SPECIAL PRODUCTS 97 Eind a binomial divisor for each of the following binomials : 47. x^ - y\ 49. 4 a;^ - 9. 51. 16 - x\ 48. x" - 1. 50. 25 - 16 x". 52. / - 4. State the two binomials whose product is : 53. o-'-d-'. 55. w^-ie. 57. 36^^-1. 54. vP' - 4. 56. 9 - 4 a? 58. a? - 9. 59. 25-4 ji^. 60. 100-9 a;2. If in (a + 6) (a — J) = a^ — J^ -^g replace a by a; + 2/, we get [(a; + y) + i] [(a; + ?/) - 5] = (a; + y)^ - 6^ ; which, when (x + jr)'' is expanded, becomes a;^ + 2 xy + y^ — 6^. Similarly, replacing 6 by (z + y), we get [a + (a: + 2/)][a - (a; + y)] = a^ - (a: + j,)^. Expanding, = a^ — (a;^ + 2 a;?^ + y^). Removing the parenthesis, = n^ — a;^ — 2 a;;/ — ^2. Perform the indicated multiplication : 61. [(x + y) + l][(x+y)-l]. 62. (x + a + 3) (a; + a — 3). 63. [(a; - a) + 3] [(a; - a) - 3]. 64. (x + 4 + c) (a; + 4 — c). 65. (2 a - J + c) (2 a - 5 - c). 66. [x + (5 + c)][a;-(5 + c)]. 67. [x + (5 - c)] [a; - (S - c)]. 68. [3M-(a;-y)][3-(a;-2/)]. 69. [4a; + (2y-a;)][4a;-(2 2/-a;)]. 70. [10 -(a- 5)] [10 + (a - 5)]. State the two binomials whose product is : 71. 49a;2-l. 76. (3 y - s) 72. .64x^-25. 77. h''-{x + y) 73. (a + J)^-l. 78. V-ix-y) 74. (a; - y)^ - 4. 79. c^ - (a; + c)''. 75. (^x-Vf-a\ 80. 4 -(a; -2)1 ,2. 98 COMPLETE SCHOOL ALGEBRA 44. The product of two binomials having a common term. The multiplication X + a X +b x^ + ax + bx + ab x^ + (a + b)x + ab gives the formula (x +a){x+b) = x' + {a+b)x+ ab. This may be expressed in words as follows r IV. The product of two binomials having a common term equals the square of the common term, plus the algebraic sum of the unlike terms m,ultiplied by the common term, plus the algebraic product of the unlike terms. EXERCISES Expand by the preceding formula : 1. (x + l)(x + 2). 6. (n + l){n-2). IL (a-l)(a-2). 2. (x + 2)(x + 3). 7. (re-2)(w+'3). 12. (a -2) (a -3). 3. (a; + 3) (a; + 4). 8. (n-3)(n + A). 13. (a -3) (a -4). 4. (x+4:)(x + 5). 9. (n-4:)(n + 5). 14. (a - i) (a - 5). 5. (x + 5)(x + 6). 10. (n-5)(n + G). 15. (a- 4) (a -6). 16. (2tj + 3)(2y + i). 23. {3a-S)(3a + l). 17. (2ij + 2)(2ij + 3). 24. (4«,+'3)(4a-5). 18. (3 a + 1) (3 a + 4). 25. (4 a - 5) (4 a - 6). 19. (2n + 3){2n + 5). 26. (4 aJ + 1) (4 aj - 6). 20. (2n + 3)(2n-5). 27. (3 a; - 2) (3 a; + 5). 2L (2 Ji + 3) (2 M - 4). 28. (4a - 36) (4a + 55). 22. (2 a + 1) (2 a -5). 29. (5 a -6b)(5 a -7 b). Perform mentally the indicated division : 30 ^!_±3^J-2_ g^ x' + 5x + 6 x' + ix + 3 x + 1 ' x + 2 " a; + 3 IMPORTANT SPECIAL PRODUCTS 99 aj^ — 5 a; + 4 x^ + 6a; + 5 35. x + 1 x" + 7 a; + 12 36. a; + 3 .^-7. + 6 ,,X±I^±^. 3e.^'-'\+^. 38. 40. x-1 a;' — 6 a; + 5 3 ""' X — 5 7a; + 12 a; — 4 Pind an exact binomial divisor for each of the following trinomials : 41. a;2 + 3a; + 2. 45.. a;^ + 7 a; + 10. 49. a;^ - 6 a; + 8. 42. 3!^ + 5 a; + 6. 46. a;^ + 8 a; + 15. . 50. a;^ - 8 a; + 15. 43. a;= + 7a;+12. 47. a;^ - 3a; + 2. 51. a;^ - 8 x + 12. 44. a;2 + 6a; + 8. 48. a;2-5a; + 6. 52. a;''-9a!+14. State the two binomials whose product is : 53. a;'' + 8a; + 7. 56. a;^-lla; + 10. 59. a;'=4-9a;4-20. 54. a;2 + 9a; + 8. 57. a;2-10a;+16. 60. a;='-9a; + 18. 55. a;2-10a; + 9. 58. a;2 + 10a; + 21. 61. a;2 + 9a;+14. 62. a;2-12a; + 32. 63. a;^+lla; + 10. 45. The square of any polynomial. The multiplication a +b — e a + b — a^ + ah — ao ab +h^— be — ac — he + (^ a^ + 2 ab - 2 ae + b^ - 2be + c"' gives the formula (a+b — cf = a' + b^ + c^ + 2ab—2 ac— 2 be. This may be expressed in words as follows : V. The square of any polynomial is equal to the sum of the squares of the terms, plus twice the algebraic product of each term by each term that follows it in the polynomial. 100 COMPLETE SCHOOL ALGEBRA EXERCISES Expand by the preceding formula : 1. (a + b + ef. 2. (a + b+lf. 3. {a + b + 2f. 4. {2a + h-cY. 5. (a-h- cY- 6. {2x + y-iy. 7. (4a; + 2/-2f- 8. {a — ho + dy. 9. (3a-3S + l)=. 10. {a — h + c—d)\ 11. (x + 2/-a + l)2. 12. {a-ij+b-By. 46. The cube of a binomial. The multiplication a +b a +b a^ + ab ab + b'' a^ + 2ab + b'^ a +b a" + 2 a% + ah'' a% + 2ab^ + b' a'^ + Sa^ + S a¥ + h^ gives the formula (a + 6)' = a* + 3 a'^ft + 3 a&" + &'. Similarly (a - 6)' = a' - 3 a"* + 3 aV - 6». EXERCISES Expand by the preceding formula: 1- {x + yf- 6. (a!-2)^ n. (2a:-iy. 2. {x-yf. 7. (a: -3)1 12. (3£c + 2)=. 3. {x^Vf. 8. {x + Zf. 13. (3 a; -2 2/)=. 4. (r«-l)^ 9. (2a; + yy. 14. (4a; + 3y)^ 5. (x-\-2)\ 10. {x-2y)\ 15. {5x-2yy. 16. Express in Tfords the formula for the cube of ' the binomial a + S. 17. Express in words the formula for the cube of the binomial a — h. CHAPTER XIII FACTORING 47. Definition. Factoring is the process of finding the two or more expressions "whose product is equal to a given ex- pression. The subject of factoring is very extensiTC. In this chapter we shall consider only the more common forms of factorable expressions and only such factors as do not contain irrational numbers and fractional terms (see § 91). Thus fractional expressions like a^, 4, etc., will be 9 a^ regarded as prime,* though the student can readily prove that (i + ")(-3-")=h°^' ^--^ *^^* {l^'){l-']=t^-'--^°^- sibly he can see that (V3 + a)(V3 — a) = 3 — a^, and perhaps that {x + ^y){x — -\/2y) = x^-2 yK But here also 3 — a^ and i^ _ 2 j,2 are considered prime because their factors contain the irrational numbers Vs and "V^. 48. Roots of monomials. In factoring it is often necessary to find the square root, the cube root, and other roots of vari- ous monomials. The square root of a monomial is one of the two equal factors whose product is the monomial. Since + 2 • -1- 2 = 4 and — 2 ■ — 2 = 4, the square root of 4 is ± 2, which means plus 2 or minus 2. Similarly the square root of 9 is ± 3 and the square root of a^ is ± a. That is, Every positive number or algebraic expression, has two square roots which have the same absolute value but opposite signs. It is customary to speak of the positive square root of a number as the principal square root, and if no sign precedes » An integral expression wiU be considered prime when no two rational integral expressions (see § 57) can be found (except the expression itself and 1) whose product is the given expression. 101 102 COMPLETE SCHOOL ALGEBRA the radical the principal root is understood. "When both the positive and the negative square roots are considered, both signs must precede the radical. Thus Vi = 2, not - 2 ; - Vi =- 2, not + 2. Since a'-a^ = (- a') (- a^) = a", ± Va" = ± a\ Similarly a%^ • a%^ = a^%'^, and ± Va^ = ±a*6'. That is, The exponent of any letter in the square root of a monomial is one half the exponent of that letter in the monomial. Hence for extracting the square root of a monomial we have the EuLE. Write the square root of the numerical coefficient preceded by the double sign ± and followed by all the letters of the Tnonomial, giving to each letter an exponent equal to one half its exponent in the monomial. A rule much like the preceding one holds for fourth root, sixth root, and other even roots. Thus ± -^Sla* = ± 3 a, and ± -^/^ = ± x\ In the chapters on Factoring and Fractions where square roots arise only the positive square root will be considered. According to the definition of square root the two factors of a term, either of which is its square root, must be equal. Consequently they must have the same sign. Since the product of two terms hav- ing like signs cannot be negative, we cannot extract the square root of a negative term. Hence we do not consider the square root of such terms as — 4, —9 a^, aud — 16 x^y* in this chapter. The cube root of a monomial is one of the three equal factors whose product is the monomial. In this chapter only a single cube root of a number is considered ; that is, the principal cube root. Since 3 • 3 ■ 3 = 27, ^^27 = 3. And as - 3 ■ - 3 • - 3 = - 27, "v'- 27 = - 3. That is. The cube root of a monomial has the same sign as the m,onomial. FACTORING 103 Since a*-a*-a* = a}\ "v^ = ^4. Similarly a^fts • a%^ ■ a%^ = a%o, and -v'^^ = a^b^ That is, The exponent of any letter in the cube root of a , term is one third of the exponent of that letter in the term. Hence for extracting the cube root of a monomial we have the Rule. Write the cube root of the numerical coefficient pre- ceded by the sign of the monomial and followed by all the letters of the monomial, giving to each letter an exponent equal to one .third of its exponent' in the m.onomial. A rule much like the preceding one holds for fifth root, seventh root, and other odd roots. Thus V- 8 = - 2, VxW = x^, and Vl28 EXERCISES Find the value of the following : a;" = 2a;a. 1. V4a2. 15. V3246^ 29. VaW. 2. V9a*. 16. V625 5y«»». 17. Va;2". 18. Va;*". 19. Va;«». 30. •v'343 2/». 3. V25 a'a;*. 31. _ J6n 7. 4a2-9. 18. 16 X* - y'. 29. 4a;2» — /». 8. 25 - a;''. 19. «" - y"- 30. 9 a;"" — 4 /». 9. A* -81, 20. a;16_4yl6_ 31. (« + 6)=' - 4. 10. 16a;^-25. 21. 144a^J*-9. 32. (a: — y)^ — 4 c". 11. 36c«-49d '. 22. 196 - xy. 33. (a _ 25)2 -9*". 34. 4 (a; +i/r- -1. 35. 25(3 S -0) i'' - 64. 36. 4 - (a + 5)' I Solution : 4 — (a + ft)' = E2 + (« + 6)][2-(a + (2 + a + 6) (2 - a - 6; .6)] 37. 81 -(a; - -yy- 38. 4a2- -(2 x + yf. Some polynonjials of four or six terms may be arranged as the difference of two squares and factored as in the preceding exercises. 116 COMPLETE SCHOOL ALGEBRA EXAMPLES 1. Factor: a^ + 2 ab + b^ — c^ Solution : a^ + 2ab + b^ - c^ = (a + b)''- c^ = la + b + c)(a + b — c). 2. Factor: a^ + 2 ab + h'^ — c'^ — 2od — d\ Solution : a^ + 2ab + b'^ - c^- 2cd - d'^ = a2 + 2 aft + 62 - (fiS + 2 erf + d^) = (a + by - (c + rf)2 = la + b + c + d){a + b — c — d). EXERCISES Factor : 1. (a-by~9x^ 2. 16f-(h + 2 ky. 3. 25a*-(2b-3cy. 6. (5x + 2yy-(3x -7 yy. Solution : (o x + 2 yy— (Z x - 7 yy = l(5x + 2y) + (3x-7y)2[(5x + 2y)-(3x-7y) = (5x + 2y + 3x-7y)(6x + 2y-3x + 7y) = (8x-5y){2x + 9y). 7. (a-by-(a + by. 12. (ib- ay-(7 a - &by. 8. (^a-by-(x + yy. 13. 4:(a - 2by-(2x - yy. 9. (a + 2by-(a-3by. 14. 9(a + J)2-(2a-56)^. 10. (2a-3by-(3a -2by. 15. (« - 2&)2- 4(a + 6)^. 11. (5h + 37cy-(2o-9d)^ 16. y= - c^ + a;= - 2 aiy. 4. 49; x«-(7c- - 2 tZ)". 5. 1- - (5 A - S ■ ky. 6. (5. c + 2y)2. -(3x- Solution : y^ - c'' + x^ - 2 xy = x^-2xy + y'-c'^ = {x- yy - c2 = (:c - !/ + c) (a: - 2/ - c). 17. x^+2ax + a''- y\ 22. F-^*-4tt + 4A2 18. a;2 + 2a;+l-4«2. 23. 6 ay — c"" + 9 a;2 + ?/l 19. c'-2cd-{-d^-V6a\ 24. 12aS- 4^2 + 452 + 80. 4 a:=^ - 12 ca; + 9 c^ - 25y^ 25. 1 - 4 ax + 4 a=a;^ - x^. 21. x^ - y^ - 2 ax + a''. 26. 40^ - 20 ccZ + 25 ci^- 9 «;*. FACTORING 117 27. 9d^-25a''-6cd + c^ 28. 12ab + x^ - 4.0" -9b\ Solution : 12 ai +. a;^ — 4 a^ _ 9 J2 = a:2 - 4 a^ + 12 a6 - 9 62 = a:2 _ (4 a2 _ 12 afi + 9 6^) = (2 + 2 a - 3 i) (a; - 2 a + 3 J). 29. a=_j2_2Sc-c^ 33. Qx^ - 42/^ - a^ _ 4^^^ 30. a;2 - a^ - 2 ac - c^. 34. 6 a; + 9 y^ - 9 - xl 31. y2_i2 + 4 6c-4c2. 35. 45c + 1 - 4c^ - §2. 32. 2 5c — c^ — 52 + a*. 36. ibo — Ab^ + ix<' — o\ 37. a;2 - a^ + 2/2 _ 4 - 2 a;y + 4 a. Solution : x^— a^-^r'f — ^ — 2xy-\-A^a = x^ - 2 xy + y^ - a^ + 4:a - i -(x^ -2xy + y'')-(a^-4:a + 4:) = (^ - yy - (a - 2)2 = (x — 2/ + a — 2) (a; — 2/ — a + 2). 38. a'' + 2ab + b^-c''-2cd- d\ 39. 9 ^2 _ 6 AA; + A2 _ 4 (.2 _ 4 cc? - t^" 40. a;2 — l + 2/2_a2 + 2a;y + 2a. 41. 1 + 25c + 2 a - c2 - &2 ^ ^2 REVIEW EXERCISES Factor : 1. a?-x. 3. a;«-2a;* + l. 5. a^ - x. 2. a;*-2a;= + l. 4, a;^ - 8 a;' + 16 a;. 6. a;" - x^. 7. a;^-10a;2 + 9. 13. 4 w« + 48 w^ _ 28 ?i*. 8. a;^— 13a;2 + 36. 14. 16 a;* + 8 a;^ - 3. 9. 3 a^x* — 12 a -^ ! and -y/ia^ are rational, while -\/2x, Va — b, and V^ are irrational. An algebraic expression is rational if its terms are rational. An irrational term or an irrational expression may be- rational with respect to a certain letter or letters. Thus ix^ — -y/a a; + 2 a is rational with respect to x. The equation ix^ — Va a; + 2 a = is an equation rational in x. A term is integral if it has no literal denominator and the exponents of the letters are positive integers. 3 x*^ Thus — — and 4 ca-^ are integral terms. (It will be seen later what o x-^ and x' mean, and why they are not integral terms.) An expression is integral if its terms are integral. A nonintegral term or expression may be integral with respect to a certain letter, if that letter does not occur in any denominator. S X b Thus the left member of a-' = is integral with respect a a^ to X and the equation itself is integral in x. The degree of a rational integral equation in one unknown is the same as the highest power of the unknown in it. An equation of the second degree is called a quadratic equation. For example, x^— Qx = 18 and ax'^ + bx ■{■ c = Q are quadratic equations. 122 SOLUTION OF EQUATIONS BY FACTORING 123 An equation of the third degree is called a cubic. For example, x^ = 1, x^ - 5 x^ + 6 x + 2 = 0, and ax^ + bx^ + ex + d = are cubic equations. An equation of the fourth degree is called a biquadratic. Thus a:* = 16, x* + Sx'^ = 4, and ax* + bx' + cx^ + dx + e = are biquadratic equations. One important application of factoring is determining the roots of equations of the second or a higher degree. In the solution of equations by factoring use is made of the following Peinciple. If the product of two or more factors is zero, one of the factors must he zero. Two or more, or even all of the factors may be zero, hut the vanishing of one is sufficient to make a product zero. EXAMPLES 1. Solve the quadratic equation x^ + 5x = &. Solution : Transposing, a:^ + 5 a; — 6 = 0. Factoring, (x - 1) (a; + 6) = 0. Suppose the first factor, a; — 1, has the value zero. Then its product with the second factor is zero, no matter what value a; + 6 may have. Hence the value of x which makes x— X equal to zero is a root of the quadratic. Setting a; — 1 = 0, we obtain x = 1. Then (x - 1) (a; + 6) = (1 - 1) (1 + 6) = (0)-(7) = 0. Similarly if the second factor, x+ 6, has the value zero, its product with the first factor is zero, ho matter what value a; — 1 may have. Setting a; + 6 = 0, we obtain x =— 6. Then (a; - 1) (a; + 6) = (- 6 - 1) (- 6 + 6) = (-7).(0) = 0. Check : Substituting 1 for a: in a;^ + 5 a; = 6, 1 + 5 = 6. Substituting — 6 idi x in x^ + 5x = 6, 36 - 30 = 6. 124 COMPLETE SCHOOL ALGEBRA 2. Solve the cubic equation a;' + x^ = 4 a; + 4. Solution : Transposing, x^ + x^ — ix — 4: = 0. Grouping, (r* + a:^) + (- 4 a; - 4) = 0. x^(x + l)-4(x + l) = 0. (x + 1) (a;2 - 4) = 0. (x + 1) (a; + 2) (s - 2) = 0. Setting each factor equal to zero, a; + 2 = 0, -whence a; =—2; 1—2 = 0, -whence x = +2; a; + 1 = 0, whence a: =— 1. Therefore — 2, +2, and — 1 are the roots of the equation a;= + a;2 = 4a; + 4. rWhenx=-2, _8 + 4=-8 + 4. Check : i When a; = 2, 8 + 4 = 8 + 4. [when a; =-1, _l + l=_44.4. For solving an equation in one unkno-wn by factoring -we have the Rule. Transpose the terms so that the right member is zero. Then factor the expression on the left, set each factor which coiv- tains an unknown equal to zero, and solve the resulting equations. It must be kept in mind that a root of an equation is a number "which satisfies the equation. Thus the equation a;^+3a; + 2 = has the t-wo roots — 2 and — 1, since each of these numbers, if put for x, reduces the equation to an identity. The t-wo preceding examples indicate that in solving equations by factoring, t-wo factors will arise for a quadratic, three for a cubic, four for a biquadratic, etc. Further, if the factors of a given equation are unlike, each will yield a different root. Some of the factors, however, . may be alike. Thus a:^ — 6 x + 9 = can be written (a; — 3) (a; — 3) = 0. Here each factor gives the same root, 3, and though this equation is of the second degree, it has only one number, 3, for a root. Therefore, an equation usually has the same numher of distinct roots as the number rep- resenting its degree, hut never more than that number. One should never divide each member of an equation by an expression containing the unknown, for in this manner roots may be lost. SOLUTION OF EQUATIONS BY FACTORING 125 EXERCISES Find the roots of the following quadratic equations : 1. a;2-9 = 0. 5. a;'' -7 a; = -12. 9.8-9x=-x\ 2. a;2 = 16. 6. a;= - a; = 20. 10. 12a;-28 = -a;2. 3. a;''-3a; = 0. 7. 3 «'' - 15 a; = 0. ' 11; 4 a;^ = 16 a;. 4. x^ = 7x. 8. 5a;2+35a; = 0. 12. x'^ - a'' = 0. 13. x^ = 9b\ 20. 9 a;2 = 3 a; + 2. 14. a;2 - 2 aa; + a^ = 0. 21. 16 a;^ - 12 a; = 10. 15. a;^ + 4 &^ = 4 5a;. 22. 60 a; + 24 = 25 al 16. a;2 + aa; + 3a;+3a = 0. 23. x^ + bx = 0. 17. x^ + hx = 4:X + 4:b. 24. a;^ — aa; — 6a; = 0. 18. x^ + ax + bx + ab = 0. 25. a;" + 3 a; = ax. 19. 4a;2 + 8a; + 3 = 0. 26. 4x = a:2 + 4. Solve the following cubics : 27. a;' — 9a; = 0. 29. a;' — Sa;^ + 6 a; = 0. 28. a;= + a;2 = 4x + 4. 30. 2 a;» - a;^ = 32 a; - 16. 31. 5a;'= + a!' = 45 + 9x. Find the roots of the following biquadratics : 32. a;*-6a;2 + 4 = 0. 34. a;*-36a;2 = 0. 33. 9 + «* = 10x^ 35. a;* + 15a;2 = 8a;». 36. a;^-2a;'' + l = 0. 37. Point out the error in the following : Let a; = 1. Then x^ = x. Subtracting 1, a;^ — 1 = a; — 1. Factoring, (a; — 1) (a; + 1) = a; — 1. Dividing by a; — 1, a; + l = l. Therefore 1 + 1 = 1, or 2 = 1. 126 COMPLETE SCHOOL ALGEBRA PROBLEMS 1. The square of a certain number plus the number itself is 90. Find the number. 2. If from the square of a certain number twice the number be taken, the remainder will be 35. Pind the number. 3. If to the square of a certain number the sum of twice the number and 5 be added, the result will be 148. Knd the number. 4. Four times the square of a certain number is equal to seven times the number. What is the number ? 5. A certain number is added to 20, and the same number is also added to 21 ; the product of the two sums is 930. What is the number ? 6. A certain number is subtracted from 17, and the same number is also subtracted from 23 ; the product of the re- mainders is 216. Find the number. 7. From 27 a certain number is subtracted, and the same number is added to 21 ; the product of the results thus obtained is 640. Find the number. 8. If a certain number be added to 15, and the same number be subtracted from 22, the product of the sum and difference thus obtained will be 70 more than 23 times the number. Find the number. 9. The difference of two numbers is 6, and the difference of their squares is 120. Find the numbers. 10. If from the square of three times a certain number, five times the number be taken, the result will be eight times the square of the number. Find the number. 11. The depth of a certain lot whose area is 2500 square feet is four times its frontage. Find its dimensions. 12. The area of the floor of a certain room is 24 square yards. The length is 6 feet more than the breadth. What are the dimensions of the floor ? SOLUTION OF EQUATIONS BY FACTORING 127 13. The area of a rectangular field is 216 square rods. The field is 6 rods longer than it is wide. Eind its dimensions. 14. The sum of the squares of two consecutive numbers is 145. Find the numbers. 15. The sum of the squares of two consecutive odd numbers is 290. Find the numbers. 16. The sum of the squares of three consecutive odd numbers is 251. Find the numbers. 17. An uncovered square box 8 inches deep has 185 square inches of inside surface. Find the other inside dimensions. 18. Remembering that the faces of a cube are squares, find the edge of a cubical box whose entire outer surface is 294 square inches. 19. A rectangular box is four times as long and three times as wide as it is deep. There are 608 square feet in its entire outer surface. Find its dimensions. 20. A box is 3 inches longer and 1 inch wider than it is deep. There are 62 square inches in its entire outer surface. Find its dimensions. The altitude of a triangle is the length of a perpendicular from any vertex to the side opposite. This side is called the base. B S In the adjacent figures BD is the altitude and AG is the base of each triangle. If a is the altitude of a triangle and b its base, the area of , • a& the triangle is — • 128 COMPLETE SCHOOL ALGEBRA 21. The area of a triangle is 30 square feet ; its altitude is 6 feet. Find the base. 22. The altitude of a triangle is three times the base and the area is 64 square feet. Find the base and the altitude. 23. The base of a triangle is five times the altitude and the area is 40 square feet. Find the base and the altitude. 24. The area of a triangle is 75 square meters ; the base is six times the altitude. Find the altitude and the base. 25. The area of a triangle is 24 square feet ; the altitude is 2 feet longer than the base. Find the altitude and the base. Hint. Let x = the base in feet. Then x -\- 2 = the altitude in feet, J x(x + 2') x^ + 2x ^, and ^ , or = tne area.. Therefore x^ + 2x ^ ^^^ 2 Multiplying each member by 2, this equation becomes ^2 + 2 a; = 48. 26. The altitude of a triangle is 3 feet longer than the base and the area is 6 square yards. Find the base and the altitude. 27. One leg of a right triangle is 2 feet longer than the other and the area is 24 square feet. Find the legs. 28. The area of a right triangle is 30 square feet and one leg is 7 feet longer than the other. Find the legs. 29. The area of a triangle is 2| square feet and the base is 6 inches longer than twice the altitude. Find the base and altitude. 30. The area of a triangle is 4 square yards and the altitude is 6 feet more than three times the base. Find the base and the altitude. 31. The area of a triangle is .015 square meters. The altitude is 5 centimeters shorter than the base. Find the dimensions. SOLUTION OF EQUATIONS BY rACTOKING 129 A trapesoid is a four-sided figure, two of whose sides are unequal and parallel. The bases of the adjacent trap- c ezoid are the two parallel sides h and c. The altitude, a, of the trapezoid is the perpendicular distance be- tween the bases. The area of a trapezoid is '■ . • 32. Find the area of a trapezoid whose bases are 10 and 18 and whose altitude is 12. 33. The altitude of a trapezoid is 8 inches, its area is 96 square inches, and one base is 4 inches longer than the other. Find each base. Hint. Let x = the length of one base in inches. Then a; + 4 = the length of the other base in inches, and — ^ ;- , or 8 a; -t- 16 = area of the trapezoid. Therefore 8 a; + 16 = 96. 34. One base of a trapezoid is 12 feet, the other base is twice the altitude, and the area is 112 square feet. Find the altitude. 35. The altitude of a trapezoid is ^ the shorter base and the latter is | of the other base. The area is 360 square feet; Find the bases and the altitude. 36. One base of a trapezoid is 4 feet longer than the altitude, the other base is 6 feet longer than the altitude, and the area is 66 square feet. Find the bases and the altitude. 37. The bases of a trapezoid are respectively 8 feet and 12 feet longer than the altitude, and the area is 16 square yards. Find the bases and the altitude. 38. One base of a trapezoid is 4 feet longer than the other, the altitude is \ the sum of the bases, and the area is 4 square yards. Find the bases and the altitude. 130 COMPLETE SCHOOL ALGEBRA 39. The area of a trapezoid is 10 square yards, the altitude equals one base, and the other base exceeds the altitude by 2 feet. Find the bases and the altitude. 40. One base of a trapezoid exceeds the other by 10 feet, the altitude is 2 feet longer than five times the shorter base, and the area is 22 square yards. Find the altitude and the two bases. 41. The area of a trapezoid is .09 square meters. One base' is twice the other and the altitude 10 centimeters less than the longer base. Find the bases and the altitude. CHAPTER XV HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE 58. Highest common factor. The degree of a rational, integral monomial is determined by the sum of the exponents of the letters in it. Thus ax^ is of the fourth degree, and 4 aVy^ is of the eighth degree. The degree of a rational, integral polynomial is the same as that of its term of highest degree. Thus G axy^ + 4 a^x* — 3 ax^yi is of the seventh degree. The highest common factor (H.C.r.) of two or more rational, integral expressions is the rational, integral expression of high- est degree with the greatest numerical coefficient which is an exact divisor of each. Thus the II.C.F. of 36 a^ft^ and 48 a% is 12 a^. The H.C.F. of x^ — ix and x^ — 5 x^ + Q x is x^ — 2 x. The problem of finding the H.C.P. of polynomials which can- not be factored by inspection will not be considered here, as it is not necessary to find the H.C.F. of such expressions in elementary work. EXAMPLES 1. Find the H.C.F. of 24 xys,. 48 ajy.^', and 12x^y^s\ Solution: Factoring, we have 24 x^yH = 2= • 8 arVz, 48 xVz^ = 2« • 8 xhj^z^, 72 x^fz^ = 2^ ■ S^x^yh\ Here the highest power of 2 common to each expression is the third, of 3 the first, of x the third, of y the second, and of z the first. Therefore the H.C.F. of the three expressions is 2^ • 3 • x^y^z, which equals 2i,x^^z. 131 132 COMPLETE SCHOOL ALGEBRA 2. Find the H.C.F. of 2 a;* - 12 a;' + 18 x^ and ix^ - 36 x\ Solution : Factoring, -we have 2 k' - 12 a;S + 18 a;2 = 2 2:2 (a; _ 3)2^ 4 x^ - 36 a;s = 2vlx - 3) (a: + 3). Therefore the H.C.F. is 2 a;^ (a; - 3), -which equals 2 3:= - 6 a;^. The method of the preceding solutions for finding the H.C.F. of two or more rational, integral expressions is stated in the Rule. Separate each expression into its prime factors. Then find the product of such factors as occur in each expression, using each the least nuwher of times it occurs in any one expression. EXERCISES Find the H.C.F. of the following : 1. 12, 18. 3. 96, 144. 5. 125, 225. 2. 24, 56. 4. 84, 196. 6. 64, 96, 266. 7. 90, 108, 324. 11. 32 a%c' and 48 a%H. 8. 12 x^ and 18 a^ 12. 125 m'n^p and 100 nf. 9. 16a:y and24a:y. 13. 18 A^A;^, 36A*/fc,and24A=A;=. 10. 21 cH and 21 cc^^ 14. ^xr/, 54 a;^ and 15a;y. 15. 27a*6V, 54a»ra, and ?,la%\^. 16. a;2 - 9 and a;^ - 6 a; + 6. 17. a;2 + 3a;-10 anda;2 + 6a; + 5. 18. a;' — 4 a; and a^ — 8 a;2 + 12 x. 19. 2c' + 12c2 + 18c and c^-2c2- 15c. 20. 8 + y' and y2 + 4 y + 4. 21. X* - 2x2 + 1 a^jjj 3.2 _ 2x + 1. 22. a5 + 3 J + ac + 3 c and 2 ai + 6 & — 2 ac — 6 c. 23. c' + Zcd^^d-', c''-\-hcd + %d\ and c2 + c(^ - 2 c?2. 59. Lowest common multiple. The lowest common multiple (L.C.M.) of two or more rational, integral expressions is the rational, integral expression of lowest degree which will exactly contain each. LOWEST COMMON MULTIPLE 133 EXAMPLES 1. Find the L.C.M. of 24cBy, 36a;*y, and 64a;ys. Solution : 24 x^y^ = 2' • 3 • xhf, 36a:''y=22.32-a;*^, 54 x^y^z = 21 • 32 • x^yH. Since the L.C.M. must contain each of the expressions, it must have 2^ as a factor. If the L.C.M. has 2^ as a factor, it will con- tain 2^ and 2^ which occur in the second and third monomials respec- tively. Similarly, the L.C.M. must contain as factors 3^, a;^, y'^, and z. Therefore the L.C.M. is 2' • 3^ • x^y'^z, which equals 216 T^yH. 2. Find the L.C.M. of 9 ahj + 18 ahy + 9 Ihj, 15 ax^ - 15 Ix", and 2 a^ - 4 ai + 2 V^. Solution : 9 a^j/ + 18 aS^^ 4- Ihj = ^hj (a + hf, 15 ax^ - 15 bx^ = 3 • 5 a;2(a _ j)^ 2 a^ _ 4 a6 + 2 J2 = 2 (a - b)^. In order that the L.C.M. may exactly contain each of the three expressions, it must have 2^, 3^, 5^, x^, y, (a 4- b)^, and (a — i)^ as factors. Hence the L.C.M. is 2 ■ S^- 5^ xhj(a + bf{a - by, which equals 90 a*x^y - 180 a^bVy + 90 b^x'y. The method of finding the L.C.M. of two or more rational, integral expressions is stated in the following Rule. Separate each expression into its prime factors. Then find the product of all the different prime factors, vsing each fac- tor the greatest number of times it occurs in any one- expression. EXERCISES Find the L.C.M. of : 1. 12, 18. 3. 20, 28. 5. 64, 120, 216. 2. 32, 48. , 4. 96, 144. 6. 128, 160, 200. 7. xhj, xif, xf. 12. 36 aV', 42 abc, 63 h\. 8. 6 c(i», 4 cHe, lOe'^dV. 13. 4 a, a^ - ab. 9. 8 abc, 3 b% 12 c^ 14. 12 ax, 3 a^x^ - 3 aa;^. 10. 18 m^ 15 mn^p, 20 m'p^. 15. ccc + cy, dy + t^a;. 11. 20, 18 xy*, 27'ax'y. 16. 3 x -f- 3 s, 6 a^a; + 6 a'^z. 134 COMPLETE SCHOOL ALGEBRA 17. x^ — xy, ax + ay. 18. a;2 - 9, a;^ - 5 a; + 6. 19. c^ - 4, c'^ - 8 c - 20. 20. 4 ax, 4 a;2 — 1, and 4 a;'' + 4 a: + 1. 21. x^ + 1, cc* - 1, and x* - 2x^<\- 1. 22. 4 - c^ c= + 8, and c^ + 6 c + 8. 23. ae — Ihd ^-2ad — he, a^ -\- ab — 2 h\ 24. a;^ — y, a;' — ?/', and x^ + 2xy + y\ 25. aa;^ + foy, 2 ax + 2 by, and a'^a;^ — 6y. 26. 2a;» - 2x, 3a;^ + 15a;' - 18 a!^, and x^ - 36. 27. 8 - 2/^ 2/= - 4, and 42/^ + 2^' + y*. 28. 2a;' - 6'a;» + 4 a;^ and a;" - 4 y'' - 6 a; + 9. CHAPTER XVI FRACTIOHS 60. Algebraic fractions. The expression - > in whieli a and b represent numbers or polynomials, is an algebraic fraction. It is read " a divided by b," or " a over b." A fraction is an indi- cated quotient in -which the dividend is called the numerator and the divisor the denominator. The numerator and denomi- nator are often called the terms of a fraction. As division by zero has no meaning, the denominator of a fraction can never be zero. The reduction of a fraction to lower or to higher terms, the addition of fractions, iand the subtraction of fractions in both arithmetic and algebra depend on the Principle. The numerator and the denominator of a fraction may be multiplied by the same expression or divided by the same expression without changing the value of the fraction. „, 8 3-4 12 , 18 18^6 3 Thus - = = — , and — = = - ■ 4 4-4 16 30 30-1-6 5 „. ., , a a-n an , a a -i- n a/n Similarly - = ■ = -— , and - = = — — h ■ n on b -i- n b/n 4 6 n Since -r ' ~ ' ^^^ ~ ^^^ each equal to 1, each of the four preced- 4 6 n ing illustrations is really a multiplication or a division of a fraction by 1. This produces no change in the numerical value of any frac- tion, though it may change its form. 61. Reduction of fractions to lowest terms. A rational fraction is in its lowest terms when no rational factor except 1 is com- mon to both numerator and denominator. Cancellation is the process of dividing the numerator and the denominator of a fraction by a factor common to both. 135 136 COMPLETE SCHOOL ALGEBRA EXAMPLES Reduce to lowest terms : 72 a^xhf 108 aVv/-^' 2 a ' «„wif.n- 72 aVy' _ ^X-^-^jf _ 2 ay' 3 ai' 2 c^ - 32 c 2. 4 c" + 16 c* - 128 c' . 2cS-32c ^ j:> as the resulting fraction "•"^ x + a + Ae'^ would be |. Similarly, in the fraction ^— ^, no can- cellation is possible. We have seen that we may multiply or divide both numerator and denominator of a fraction by the same number without affecting the value of the fraction. But we should never forget that adding the same number to or subtracting the same number from both numerator and denominator changes the value of the fraction. Also squaring both numerator and denominator leads to a different value. Compare this statement with the operations that may be performed on each member of an equation as given on page 33. FRACTIONS 137 EXERCISES Reduce to lowest terms : 1. 2. 46 cei^e^ 20 c^ci' 3a;y + 3y= '• 3y^ 12a;y 18 xy-' 36 a'^b^ „ 21c^2 + 14c *■ 14c ■ 54 a^Sc' 32 a*h^o 48 a^b^o" ^' 2a 1 ^°-^ 2a + 2 a'^-2a + l 4.x' + ix+l Ax^-1 18. a;2 - y^ (x-yf 2a? -2m 4«^-8a6 + 4J2' 19. a'-b^ {a - by 18 aa;3 + 9 ay 12te» + 6Jy 20. c^-d^ cO - d" 13- a;^ _ a; - 30 21. y' — s' c^ - 16 d'' 22. x'-8 x^-4 2*2 -.-2 a- -180 2 a;2 - 162 1 23. x^-l iC^-1 _ 21 + 10a; + a;^ ^^- ^^ - 9 ■ 24. 8c*-10c» + 12 c + 5 25. 64a;= + l ! f - 15 1 + 8 a; + 16 x^ 62. Changes of sign in a fraction. The siffn of a fraction is the plus or minus sign placed before the line separating the numerator from the denominator. Hence there are in a frac- tion three signs to consider, — the sign of the fraction, the sign of the numerator, and the sign of the denominator. Now in division the quotient of two expressions having like signs is positive, and the quotient of two expressions having unlike signs is negative. 138 COMPLETE SCHOOL ALGEBRA Therefore +i|= + 4; +ZL^ = +4; Or, m general terms, + — - = + — r = 7 — T • ^ + b —h +J -b These examples illustrate the Peinciplb. In a. fraction the signs of both numerator and denominator, or the sign of the numerator and the sign before the fraction, or the sign of the denominator and the sign before the fraction may be changed without altering the value of the fraction. Hence any fraction may be -written in at least four ways, if proper changes of sign are made. rr, ,2a ,-2a -2n +2a Thus H = + ~ — 37—3 — X + 'i X —'3 — a; + 3 Similarly 2 a; -5 _ -2a; + 5 -2a: + 5_ 2x-5 ■22/ + 4 — X + 2y- ^ x-2ij + i -x + 2y — 'i EXERCISES Write in three other ways each of the following : — X _ « _ c^ 1. ■d y ' a — b ' — 2o + d „a; .a; — 3 _ x — y — y 2a; +4 a! + y — 3 g° — 5 a; + 6 _ — a; + a;' — 2 a;^ - 7 a; + 12 " " " a;= - a;^y - 1 ' 63. Equivalent fractions. Two fractions are equivalent when one can be obtained from the other by multiplying or by divid- ing both of its terms by the same expression. For example, - and — are equivalent fractions ; also — and - ■ The lowest common denominator (L.C.D.) of two or more fractions is the L.C.M. of their denominatoi's. FRACTIONS 139 EXAMPLES Reduce to equivalent fractions having the lowest common denominator: , 5a ,35 1. -^^j- and -; — ; • Solution: The L. CM. of the denominators is 12aJV. Multiplying both numerator and denominator of the first fraction by the factor 2ac, -which is found in the L.C.M. but not in the denominator of 10 a'c the fraction, gives ■ Multiplying both numerator and denomi- nator of the second fraction by the factor 3 b^, ■which is found in the 9 b' L.C.M. but not in the denominator of this fraction, gives • 10 a'e 9 b' ' Hence the required fractions are r-- and — - — — - • ^ 12 abV 12 a6V 3a;-l 5 Solution : Factoring the denominators and rewriting gives 3r-l , 5 ■ and ■ 2a;(z + 3)(a;-3) (a; -3) (a; -2) By inspection the L.C.D. is seen to be 2x(x + Z^^x— 3)(x — 2). Multiplying both terms of the first fraction by the factor x—2, ■which is found' in the L.C.D. but not in the denominator of the frac- (3 a; - 1) (a; - 2) 3x^-1x^2 *'°'^' ^^'^ 2:.(z.f3)(^-3)(;-2) ' °' 2:.^-4:.''-18:c^ + 36:.- Multiplying both terms of the second fraction by 2 x (x + 3) , found in the L.C.D. but not in the denominator of the fraction, gives ^ — : — L , or (r - 3) (K - 2) 2 a; (2 + 3) 2 a;* - 4 a;^ - 18 s^ + 36 a; Therefore, to change two or more fractions (in their lowest terms) to equivalent fractions having the L.C.D., we have the Rule. Rewrite the fractions with their denominators in fac- tored form- Find the L. CM. of the denominators of the fractions. Multiply the numerator and the denominator of each fraction hy those factors of this L.C.M. which are not found in the denominator of the fraction. 140 COMPLETE SCHOOL ALGEBRA EXERCISES Change the following fractions to equivalent fractions having the lowest common denominator : 111 x + y x-2y 1. 7;' :;'■;;■ 10- 2 3 6 Zxi/ Ix'y 5 7 , 2 + m= 4 + m' ^' 24' 32' ^^' 2m ' 4™ „ 4 2c ,„ 2 4 3. :— :; 1^- 12. 3rf 6t^ ■ a; + 2 3a; + 6 ^ 3a + 5 3a ,0 * 6 12 16 a; + 1 a; - 1 5a; + y 4a: + 3y 3c 4 21 ' 35 ■ c^-c^^' c + c^' 4 3 _ 3a; 5 6. —^^ -—-■ 15. 6 a% alio x + 3 a;^+5x + 6 ^c Vld , 2x By 16. '^ 3 TO?i° 11 mV ' x" — xif x^ — 2xy -\- y'^ 3a 2h ^ ,„ 3c+2 4-c and - — -• 17. 2¥o Ba'o Q,ab c" ~ d'' c^-lcd + Qd'' 3b 9a , Tab 2a; + 5 2a;-4 9. TT—r.' -. — T' and ^ , „ • 18. 2od^ icde bode" ' x^-1 a;^ - 3 a;" + 2 a; Note. The problem of operating with fractions presented great difficulties to all the early races. The Egyptians and the Greeks, even down to the sixth century of our era, always reduced their fractions to the sum of several fractions, each of which had 1 for a numerator. For instance, I would be expressed as ^ + -|. The Romans usually expressed all the fractions of a sum in terms of fractions with the common denominator 12. The Babylonians resorted to a similar device, but used 60 for the denwninator. In some way they all attempted to evade the difficulty of considering changes in both numerator and denominator. The Hindus seem to have been the first to reduce fractions to a common denominator, though Euclid (300 B.C.) was familiar with the method of finding the least common multiple of two or more numbers. FRACTIONS 141 64. Addition and subtraction of fractions. If two or more frac- tions have tte same denominator, their sum is the fraction obtained by adding their numerators and writing the result over their common denominator. ■p ,1,2,47 .a 2a 8a 6a For example, - + - + -=-, and - + — + — = ^9999 b b b b If two fractions have the same denominator, their difference is the fraction obtained by subtracting the numerator of the subtrahend from the numerator of the minuend and writing the result over their common denominator. ■r, ,523 ., a b a — b i! or example, - — - = -, and = . Ill c c c If it is required to add or to subtract two fractions having unlike denominators, the fractions must be changed to equiva- lent fractions having a common denominator ; then theii sum or their difference is obtained as above. For example, to find the sum of | + | + f , we reduce the fractions to equivalent fractions having a common denominator by multiplying both terms of ^ by 2, of | by 3, and of | by 4. The fractions become A' T?k' ^^^ TJ respectively, and their sum is ^|. o c In adding unlike algebraic fractions, as - and - , we treat them in b d a similar way. Multiply both terms of - by d, and both terms of - d by b. The fractions become — - and — respectively, whose sum is bd bd ad + be „. ., , a c ad be , . , , ad — be • Similai-ly = , wnicri equals . bd ^ b d bd bd ^ bd EXAMPLES 1. Simplify -g-^— + ^^^ • Solution: The L.C.D. is lix^y. The work of reducing the two fractions to equivalent fractions whose denominator is 12 x^y, and of adding the resulting fractions, follows. 6a; + 1^3a;-5iy_(ea; + l)4.y (8 x - 5 y) 3 a; 3 a;2 ^xy ix'^-^y ^xy-Zx 142 COMPLETE SCHOOL ALGEBRA 6x+l 3x— 6y _ 24:xy + iy 9x^ — 15xy 3 1-2 "•" 4.xy 12x^y 12x^y _ 24: xy + iy + 9x^ — 16 xy _ 9xy + 4:y + &x^ ~ Wi^ 12x^y Check : Setting the original expression equal to the final result and substituting 1 for both x and y we obtain : 6a: + l Sx-5y _ dxy + 4:y + Qx^ Sx^ 4:xy ~ 12x^y 6+13-5 9+4+9 3 4 12 7 -^_22 22_22 3 4 "12' ""^ 12" 12' At this point the student should read the rule on page 143 and then solve Exercises 1-20, page 144. „ T.. J 0.1. 1 u • • „2a!-3 3a! -4 , -2 2. Find the algebraic sum of -^-^ - ^,_g^^g + ^qjg • Solution : Rewriting the fractions with their denominators in the C ^ Af ^■ 2X-S &X-4: ,-2_,, factored form, we get The ^ (:i:+3)(a;-3) (a;- 3) (a; - 2) a; + 3 L.C.D. is (x + 3) (a; — 3) (x — 2). The work of reducing these frac- tions to equivalent fractions and finding their algebraic sum follows. 2x-3 3a;-4 -2 (a; + 3) (a; - 3) (a; - 3) (x - 2) a,- + 3 (2 a; - 3) (x - 2) _ (3 a: - 4) (a: + 3) - 2 (a; - 3) (a: - 2) (a; + 3)(a;-3)(a:-2) (x- 3) (a;- 2) (a: + 3) (z+ 3)(a;-3)(a;-2) 2 x' - 7 X + 6 - (3 a:' + 5 a: - 12) + (- 2 x' + 10 a: - 12) (x + 3) (x - 3) (x - 2) 2x''-7x+6-3x^-5x + 12-2x^ + 10x-12 (x + 3) (x - 3) (X - 2) -3x2-2x+6 -3x2-2x + 6 (x + 3) (x - 3) (x - 2) x» - 2 x^ - 9 X + 18 Check: Proceed as in Example 1, substituting 1 for x. 2x-3 ■3x-4 -2_ -3x2-2x + 6 -9 x2-5x + 6 x + 3 xs-2x2-9x + 18 2-3 "3-4 -2 _ -3-2+6 1-9 1-5 + 61 + 3^1-2-9+ 18" FRACTIONS 143 In checking work in fractions, stick values must be chosen for the letters as will make no denominator zero. This pre- vents the substitution of 2, 3, or — 3 for x in checking the foregoing example. 3. Simplify 3a! -4 --~^* Solution : This may be ■written 2 3a;_4 a;2_5a; 1 x + 2 The L.C.M. of the denominators is a; + 2. Multiplying both terms of the first fraction by a; + 2 and leaving the second unchanged, we get ■ jii- 5a: _ (3a:-4)(x + 2) x^-^x X : + 2 l(x + 2) _ 3 a:2 + 2 a; - 8 - (a;' a: + 2 3a;2+2a;-8-a;2 x + 2 2a;2 + 7a;-8 a: + 2 a; + 2 2 -ox) + 5x Check : Let X = 2. 3a;- 4- .r2_5a; 2x2+7x- a; + 2 a: + 2 8_ 6- -4- 4-10 8 + 14-8 2 + 2 2 + 2 2 -6-14. 4 4 ¥ = ¥- Therefore, to find the algebraic sum of two or more fractions (in their lowest terms), we have the KuLE. Reduce the fractions to equivalent fractions having the lowest common denominator. Write in succession over the lowest common denominator the numerators of the equivalent fractions, inclosing each numerator in a parenthesis preceded hy the sign of the corresponding fraction. Rewrite the fraction just obtained, removing the parentheses in the numerator. Then combine like terms in the numerator and, if necessary, reduce the resulting fraction to its lowest terms. 144 COMPLETE SCHOOL ALGEBRA EXERCISES Find the algebraic sum of : , 2'x 3x „3c,c 1 _ x — 3 , 2x- + -F-- 3. — + — -TJ- 5.—; h- 3 ' 5 7 21 3 4 ' 6 2a 5a a 9a 12?» 5m 5a + 7 9a + x 1 6" i' 16~~~8 24" ~14 21 _ 5 c 3o — x,x — So 8. 3 5 10 4m — 3 7— 9 m 3a4-5m — 4 6 9 27 9.? + ^. 12. f + ^-^. a 4 ffi-' boa ,« 3 , 4 8 ,,7 5,6 10. -^ + --;5-5- 13.— ■ + -• jj ^__5__l_«. j4 _i__^ §_ ' 2/ 2 y' 4 ' a;y 2 a;^ 5 xy' 3a;-l 7-23;" , 5a;=-8 15. 16. &x^ ix' 9x* 4c' -9 6-0 c^-4 2ed 5c2 c(^2 n.3+£,3 19.^-1. a; — o 4 a;"" + a;y a; 18.^ 2_. 20. i^^ + -^. a; — 5a; + 6 a!^ — V a — b In the solution of exercises similar to 21-30 the student should follow the method of Example 2, page 142. 10 1 61. ^^- a^ - 16 ' a' - 23.'r/+ . a;^ — 1 x^- ■ 25 -m^ 5 -6a + 8 4 -3a; + 2 m^ + 16m + 65 c-5 2c-3 ""■ c^-6c c^-8c + 12 „ x^-Sxi/ + y'' bx-3y •*"■ 9_6a;+a;2 9-3x FRACTIONS 145 26.44^ + i+ '-^ 27. x'^ + X X x^ + 2x -\- 1 a + h a — 2b 2a • a'-ah a?-2ab-\-W a"- ¥ m — 2n m^ — Sn' , 3 m — n 88. 1 m^ + mn + n^ m' — n' m — n c^ + cd + cP c-d 2c'^ + 5cd 29. cd + d^ 2c + 2d o^ + d^ 30. -r^ - 7 fs- 33. =^-- + n + m. ^ — 2/ (^ — 3/) m — n 31. iJ^ + ^- 34. 3a + &-^"^^' Hint. See Example 3, page 143. 32. X - 3 - ^^- 35. x-' + y^- ^^°~^° - xy. 5 m* — 2 m'' , ^ 36. »?i^ ^-; — .- + 1 - m. m'' + wi + 1 38. a" ^— — ■ , ,„ ab + b\ a^ + ab + ¥ 39. x^ + xSj - ^ •' + xif + 2/'. x-y 2c-' + d^ c 41.6+ , 6;'-/ ^ -^^ + 5. r" — 9 rs + 14 s^ r — 7s ''■{^-'-^^-{'^-'-^} 3 4 Hint. Kemoving parentheses, we get x—S — 2 a; + 7 + . X -p O X "V ^ 3 4 Combining like terms, gives — a; + 4 — ^ + > etc. 146 43. 44. 45. 46. 47. 48. COMPLETE SCHOOL ALGEBRA 6 m\ / m — 3n\ "2m \ / m — 3n\ 3 dej \ 3 dej \ a + b) — 5a a — b 2a-3-b + 4^2 3a 2r + s 2-x x^-i 2a + h 2r- a + b^ - ( 3 a + 2 6 5a \ ~ 2a-b)' 2 s - 3 )■ + 2s + 3 ')■ Hint. These fractions may be ■written 1 a,2-3 2-x (x + 2) (x - 2) Apparently the L.C.M. of the denominators is (2 — x)(x + 2) (z — 2), but if both terms of the first fraction be multiplied by — 1, we 1 — X The L.C.M. of the denominators of the fractions obtain x-2 1- X , a;2 - 3 . -2 49. 50. ■ and - x^-i + ■ is (x + 2) (a; - 2). X — 3 3 — X 6 3_ a;2-25 53. 5 — X X 51. 52. 3a 2a-l a= - 4 2 - a 3c 4c-2 a; 2a:-l l + 2a! l-4a!2 54. 55. : + 4a; — 1 2a; + 3 a;2-13a: + 42 ' 1 -x 6-x 3a;-l 4a;-l x + 2 a;2 + 7a; — 8 1-a; "^S+a;' Multiplying one factor of an indicated product by — 1 changes the sign of every term of the expanded product. Thus (x - 2) (x — S) = x^ - o X + G. Multiplying the terms of the factor a; — 2 by — 1, we have (2 — x) (x — S), or — x^ + 5 x — Q. FRACTIONS 147 Multiplying the terms in two factors of an indicated product by — 1 does not change the sign of the expanded product. Thus (z - 2) (r - 3) = a;2 _ 5 :r + 6. But (x - 2) (- 1) (I - 3) (- 1) = (2 - X) (3 - X) = z^ _ 5 a; + 6. In general, changing the sign of an odd number of factors in an indicated product changes the sign of every term of the expanded product ; but if an even number of factors are thus treated, the expanded product is unchanged. 1.1 1 56. (a — b)(a — c) (b — c) (b — a) (o — a) (o — b) Hint. Apparently (a — 5) (a — c) (6 — c) (5 — a) (c — a)(c — h) is the L.C.D. ; multiplying both terms of the second fraction once by — 1 and those of the third twice by — 1 the three fractions become 1 , -1 1 (a -h)(a- c) (6 - c) (a - h) (a - c) (6 - c) ' of which the L.C.D. is (a - 6) (a - c) (6 — c). 2 1 57. 58. 59. (05 -y){x- e) (x - y) {z -x) 3a 2a (a -3) (a- 4) (3 -a) (A -a)' 2 3 4 (in — n)(m + n) (n — m)(m — 7) (n — m)(l — m) c 1,3 60. -„ — -r — -777 - t; -o ^ + ■ c2_10c + 24 6c -c^- 8 (6-c)(2-c) 65. Reduction of a fraction to a mixed expression. A mixed expression is an expression consisting of a rational, integral part and a rational, fractional part. , In arithmetic 4| means 4 + f , while in algebra a~ means a times -> or — • Hence in algebraic mixed expressions, the c c integral and fractional portions must be connected by a plus or a minus sign. 148 COMPLETE SCHOOL ALGEBRA If the numerator of a fraction (in its lowest terms) is of the same degree as the denominator, or of a higher degree, the fraction may often be reduced to a mixed expression. Obviously such fractions as — , - , j and — cannot ■' a" + b^ x^ + y^ xy be reduced to mixed expressions. 4 3;4 _ g a;3 _ 3;2 _ X Example : Reduce to a fraction. x^ ^ ^4 — Q ™3 /j;" — \ — ^" — J_ Solution : Dividing, = = 4 x — 6 + = 4a;- 6 -— • To reduce a fraction (in its lowest terms) to a mixed expres- sion we have the Rule. Perform the indicated division, thus obtaining a partial quotient, until the remainder is of lower degree than the divisor. Write the remainder over the divisor and connect the resulting fraction hy a plus sign to the partial quotient, thus forming the complete quotient. The reduction of a mixed expression to a fraction is per- formed as in Exercises 31-41, page 145. EXERCISES Reduce to mixed expressions : lBx^-Wx + 2 24 »° - 6 g^ - 14 5x ' Qa - a" + aW + ¥ 5 a° -f 3 g" - 6 ^- ■ -■ 7. „ , , '- ■ "• Sa^i + Sa-f 2' x^ + y^ 0+1 d^ + 1 d-1 27 x' - y« 3x+y y* + y^ + l a^ + ab- ¥■ 32/«-ll 2/ + 3 (a + Vf a' + P 16 a* + ¥ 12. 9. '" '. "' ■ 13. x^ — y^ x + 1 (a" + by y^ — y —1 2a — 1 a^ + FRACTIONS 149 66. Multiplication of fractions. In algebra as in arithmetic the product of two or more fractions is the product of their numerators divided by the product of their denominators. Thus f.6 = 15. Similarly a c _ ac b"d~bd' and 5.- 4 = S • I = iA. Inlikemanner n-^ = ^.^ = !^. bib b EXAMPLES 1. Multiply -^-T- by ^„ ". ., - ^ -^ 5 2/». -^16 aV Solution : Canceling common factors, n ( X If 4aV &5f^ _ #p:y ^/ _ 1 xy^ ^f ' 16 aV " ^/ ' >6^ ~ T7 ■ A 15 16 /2aY^ (zlI^Y ( 3c' Y \a: / ■ \ 5a; / \10aa!/ (-)-(^)'©-(^)' 2a; + In. 19 „„ 2a -6 4a'-25a2 / «» - 8 a' 2a2 + 2 ■ 4-4a^ \ a-\ ,, 8m''-125w'' /2m 5»i\ f. /„ _^ . ^25«\ 21. ^- :- r^ ;r— H- Smw 2m + 6ra + -s — ) \ a; a;'' / \ a; ^ )\ 2x-\-by ) / 9c^-4rf' \ / , 15c^\ 15 c^<£ + 6 cd'' ^^■\ 6c^ j-^^r/ + 4c+ 4^ j-9,.^ + 24ccZ^-20ci»' ..(e._.._I).(.,a,|).(_A_). 68. Complex fractions. A complex fraction is a fraction con- taining a fractional expression either in its numerator or in its denominator or in both. 154 COMPLETE SCHOOL ALGEBRA EXAMPLE Simplify ^^g^^ - X x^ Solution : Reducing the numerator and the denominator to simple fractions j -.n X — 3 X x^- -Zx- 10 X x^- -2x- ■15 2_15 X x^ X' Performing the indicated division, z^-Sa-lO J, X (x^— -5T(x + 2) -^. x^-2x-15 ~ ^. '-(a— ^(x + 3) x^ ~ a; + 3 " Check : Let x = 1. TV,. 1-3--Y- 1 + 2 -12 ^ 3 3 -Inen l = or = -, or — = -. l-|-i_6 1 + 3 -16 4 4 4 To simplify a complex fraction we have the Rule. Reduce both the numerator and the denominator to simple fractions, then perform the indicated division. EXERCISES Simplify: 1. ^-i 3 ^-(*)^ 5 (*)=-2, 2 + ^ 4-J '• 3-1 2. * + l 4 t + l . 6 ^-t + f. 1-2 2-1-:^ 3-t + l 7 i^ + 1 Q 2-1 '■ f-1 a)^- -a)=-i2a/ 8.2i-3i + ^^ 10 a/5 ^■^■ - 4| cjd FRACTIOKS 165 1+x 9 , b 20 a a IH - 11. -JL^. 15. 2_. 19. 1 L. 1-- l_l._iH i_li_2 a;" a^ a" a* a b , c m', ^ 16 ^ 12. -• 16. 20. . , 1 ,n^ 24 , 10 , 1 1 + -J in-\ n 7 + -? + T 9-f. 2 + ^4-. x + ^'^y ia' ,_ a — 2 y X 13. 17. 21. 1 — 7— a 4 r-H — -^ 1 ba a + 2 y — — — - 94-2 14. — • 18. -• 22. a iab • ( 2^' V 2x^Y • ?a;-^-^ 23. ' : • 27. 8 2« 8 a;' i^ . 9 ,W {x-yY 4 "^4^''" 3a; a — b a + b x a b^ 2 ax + Bbx + Q> ab + x' a — b a + i x + 2a x + 3b 25. 1--^ — ^^ -■ 29. .^ ^- 1 ^"^ «„ (3a + 2bY a 1 ^-" -c ^ + " a a 1-a ^ x^-1 30 5x^-6a; + l. ^ , 3a;+2 1 + a a 5a5 — 1 CHAPTEE XVII EQUATIONS CONTAINING FRACTIONS 69. Monomial denominators. Equations containing fractions with monomial denominators are easily solved. Yet unless each fraction preceded by a minus sign is handled with care, errors will be frequent. EXAMPLE Solve the equation ;^ I 9 + -^ 1 ^ — z ^ + -tt = 2x. Solution: Performing the indicated multiplication, 18 , 10a; 6x-3 , 5a; „ T + ^ ^ + T = ^"- Multiplying each member by the L.C.M. of the denominators, 105, and canceling, 1^ ..ier+ i°^.-ier- ^^:^ ^65-+ — •J«r= 2 x ■ 105. 270 + 50 X - (6 a; - 3) 21 + 175 a; = 210 x. 270 + 50 a; - 126 a; + 63 + 175 a; = 210 X. Combining like terms, 99 x + 333 = 210 x. Then -lllx=-338. Whence x = 3. 4 - 3 + 5 = 6, or 6 = 6. For solving equations containing fractions with monomial denominators, we have the Rule. Free the equation of any parentheses it may contain. Find the L.C.M. of the denominators of the fractions and •multiply each fraction and each integral term, of the equation by it, losing cancellation wherever possible. Transpose and solve as usual. 156 n^. , 2/. , 15\ 3(6-1) , 15 . Check: -(9 + - -^-^+- = 6. EQUATIONS CONTAINING FRACTIONS 157 EXERCISES Find the roots of the following and verify results : , 1 + 1 = 10. 4.^«-i(.-3)... 2.-x + -x = 5i. S. ^^-g(a; + 5)=0. „ x + 5 2a; + 4 ^ „ 3, , ^, 5a;-7 7 .3. -^ gt- = l. 6. -(a; + l) ^ = -■ 10 2 + -(a; + 4) = -3. 8. 6a; -12 4 1 6 11 (2^ ^>=3- 9. . 12a; -7 1 2a; 1 g g-0. 10. 10a; -7 , 5/2 \ 15x- 6 ' 2^5 7~ 3 11 11. 5a; 1 3/ 5\ 7 6 2 Sl^"' 3;"*'32~"- 12. 5 4_9 a; 3 x' 13. 9x 3 3a;-7 17 4 4 3 ' 24 14. eel 3x 5x 15 15. 1 13 8 2a; 24 3a; 2a-3a;,5a-2a;,41 . 6 a 5 a 30 17. -(a-3a;)+-=— = -6. ,„ ex n,^ „ . /2c n\ 18.---(3x-5on)=cn[j--^j- 158 COMPLETE SCHOOL ALGEBRA 19. 2x-b--(Sx-^h)+2a = ^^^^^^=^- a^ ' a 20. (x + 5) (x — 6) = x(x — f). 21. (x + ^)(3 + !)=!(.- 6)+ 8J. 22. (x _ ^)(a; + 3) = (a; _ 1) (x + 2)+ 1|J. 23. (X -i)(x + j)-(x - iy - H = 0. 2i. (x + iy-{x-i)(x + i)+U=0. PROBLEMS 1. One fourth of a certain nurqber plus -^ of that number equals 16. Find the number. 2. The difference between J of a certain number and -^j of it is 70. Find the number. 3. The sum of two numbers is 38. One tenth of the greater number equals J of the less. Find the numbers. 4. The width of a rectangle is f of its length. The perim- eter is 216 centimeters. Find the area of the rectangle. 5. What number must be added to the numerator of the frac- tion I so that the resulting fraction will be J of the number ? 6. Three fourths of a certain number is J the sum of the next two consecutive numbers. Find the numbers. 7. A certain odd number divided by 11 is equal to ^^ of the sum of the next two consecutive odd numbers. Find the numbers. 8. What number added to both terms of the fraction -j-f gives a fraction whose value is f ? 9. Separate 42 into two parts such that ^ of their differ- ence is J. 10. One fourth the difference of three times a certain number and 4 equals ^ the difference of live times the number and 4. Find the number. EQUATIONS CONTAINING FRACTIONS 159 11. Separate 112 into two parts such that their quotient is |. 12. There are two numhers whose sum is 24. If their differ- ence be divided by their sum, the quotient will be 3| less than the difference of the two numbers. Find the numbers. 13. The quotient of 27 plus seven times a certain number, divided by twice the number equals the quotient of 90 plus five times the number, divided by three times the number. Find the number. 14. A's age is f B's age. In 10 years A's age will be twice B's age. Find their ages now. 15. The age of A is § that of B. Fourteen years ago A's age was ^ B's age. Find their ages now. 16. A is 16 years older than B. Eight years ago B was f as old as A. Find their ages now. 17. Jupiter has 4 more moons than Uranus, and Saturn 2 more than twice as many as Uranus ; Mars has 6 fewer than Jupiter, and Neptune haK as many as Mars. These planets have together 25 moons. How many has each ? 18. A triangle has the same area as a trapezoid. The alti- tude of the triangle is 30 meters and its base is 8 meters. The altitude of the trapezoid is ^ that of the triangle, and one base equals the base of the triangle. Find the other base of the trapezoid. 19. A marksman hears the bullet strike the target 3 seconds after the report of his rifle. If the average velocity of the bullet is 1925 feet per second and the velocity of sound is 1100 feet per second, find the distance to the target and the length of time the bullet was in the air. 20. A gunner using one of the best modem rifl.es would hear the projectile strike the target 2640 yards distant in 9f seconds after the report of the gun, provided the projectile maintained throughout its flight the same velocity it had on leaving the gun. Find this velocity if sound travels 1100 feet per second. 160 COMPLETE SCHOOL ALGEBRA 70. Equations containing fractions with polynomial denominators . The method of solving equations of this type is illustrated in the examples which follow. EXAMPLES L Solve the equation ^-^ = ^-j-2|_^ - ^-1-^ . Solution : Factoring the denominators and rewriting, x^ 2x 1 2(x-l)(x^ + x + l) 3 (x2 + a; + 1) 6(x- 1) Multiplying both members of the equation by the L.C.M. of the denominators, 6(2: — 1) (x^ + a; + 1), and canceling, 2 ^^ /^x-iyi^£^HH^-^-^~--X.i^,-^(x^ + x + l). Then Zx^ = 4^x''- i.x - x'^ - x - I. Transposing, Sx'^ — ix^ + ix + x'^ + x^— 1. Combining like terms, 5a;=— 1, or x =— \. Check : ^V _ -1 1 -A -2 5^-1 + 3 - 5 - 10 - 5 252 63 36 -1-6 Multiplying by 252, - 5 =- 40 + 35 or - 5 =- 5. In solving equations containing fractions with polynomial denominators, the student should write the denominators and their L.C.M. in factored form, as in the preceding solution. With this exception, the rule on page 156 applies to all equa- tions containing fractions. Whenever both members of an equation are multiplied by an expression containing the unlfnown, roots may be introduced by the process. In fact, an apparent root may thus be obtained for a statement which no number whatever can satisfy. Such statements are frequently called " impossible equations," although, strictly speaking, they are not equations at all. EQUATIONS CONTAINING FRACTIONS 161 „ „ , 3a;-2 4a;-4 , . 2. Solve -^-2- = -^32-4-1. (A) Solution: i£^.x^,_2r= ^^,^^^^+ 1 . (^ _ 2). Then 3x - 2 = ix - 4: + x - 2. Transposing, 3a;— 4a; — «=— 4 — 2 + 2. Combining like terms, — 2a; =— 4. Whence a; = 2. On attempting to check, the fraction — becomes - • Since division by zero has no meaning, 2 is not a root of (A), nor can any number be found which is. The preceding example illustrates the need of checking; for an equation has a root, and a false statement in the form of an equation has none. Moreover the example emphasizes the point that any result ■we obtain from the solution of an, equation is a root, not because ■we obtain it by correctly performing certain operations, as clearing of fractions, transposing, etc., but because it satisfies the original equation. 3x-2 The reason that no root can be obtained for the statement — = f- 1 is because an impossible number relation is implied therein. a;— 2 This can be sho'wn by solving the equation as follows : Transposing, 1 = 0. X— 2 a;— 2 Keducing the left-hand member to a fraction, 3r-2'-4a; + 4-a; + 2 x-2 -2a; + 4 0, = 0. a; -2 Factoring the numerator and reducing to lo^west terms, or -2 = 0. That is, the impossible condition that —2 = ■was implied in stating (A). The reason that the impossible condition appears by this method of solution before ■we check, is to be found in the fact that we did not multiply both members of the equation by any expression con- taining X, as we did in the first solution. 162 COMPLETE SCHOOL ALGEBRA EXERCISES Solve and check : , 32 ^ ^ 4 12 + a: 4 1. — = 5. 5. = 5 ■ x X ox o 2. ^- = 5. 6. h -5 — = 0. 3x 6 4a; 3a; „^. . x-2 , ^ „a: + 5 3(a; + l) „, 3. 6a; H r— = 4a;+7.. .7. -r^^ '' =3^. 4 6a; a; 4. 3a;-^^-^ = 6. 8. 6a; - 4x^3 - -W 7 = 3. 4 5 ~ V «; ' 4 10. 8x-7 _i-a; 4a; ' ^^- X ■• 17. X — 3 X — 9 X + 4 X + 6 11. 2x-3 3 + 2x 18. J^ + STi + ^ = <'- 12. x-2 16 x-3 16 19. 3x 2 3x-2 4 x-2 4 13. 7 X + 6 8 ~ ^" 20. 4 3x + 4 X x-3 ' 6 ~2 14. 3a;2-7x-4 3 4x2-10x-8 4 21. 2x + 3 1 ^ x + 8 X — 6 X — 6 15. 1 3 x-2 x-3 22. X 6 2x + f 4 4x-12 3 16. 1 X — 6 X x-3 x+3x+3 • 23. x-4 7 3 X + 6 ' 5 X + 5 24 ^ + ^ 15x-5 X 2 6~ X + 2/- 10 25 * 1 5 _L /iQ — n "^- 3x + 6 ' 7X + 14 ' ^" "■ 7 OR 1 47 -3 4x-12 ' 220 6x-15 EQUATIONS CONTAINING FRACTIONS 163 2^ 4a; 6 _ 10a; + 11 x + 3 2a; + 6 3x + 9 3 x-2 5 3a; a:^-25 ' ic^ - 25 a;-3"^ 2 -3 a; + 3 a;2 - 9 x + 2 10 - a;2 10 x-'2 4 - x^ a;2 - 4 a;-4 a: -15 2x2 - 10a; -1 ^-5 ' cc + 4 a;^ - a; - 20 £c + 2 x^3 4a; + 9 a; + 3 ' a; + 2 a;2 + 6a; + 6 28. 29. 30. 31. 32. 71. Equations containing decimals. The method of solving an equation containing decimals is illustrated in the foUowing examples. EXAMPLES 1. Solve the equation .4 a; + .7 = 9.7 — .05 a;. Solution : Multiplying by 100, 40 a; + 70 = 970 - 5 x. Transposing and collecting, 45 a: = 900. Dividing by 45, z = 20. Check : .4 x 20 + .7 = 9.7 - .05 x 20, 8.7 = 8.7. In equations containing fractions, if decimals occur in any denominator, multiply both terms of such fractions by such a power of ten as will reduce the decimals in the denominators to integers. Then clear the equation of fractions and proceed as in the foregoing example. 4.3. 3g 15a; 2. Solve the equation — ^- + -^ + 10 a; = 9.08. .0 .t>o Multiplying both terms of the first fraction by 10, and both terms of the second fraction by 100, 40.-38 150. 5 38 The equation can now be cleared of fractions and then solved as usual. 164 COMPLETE SCHOOL ALGEBRA EXERCISES Solve and check : 1. .3a; + 4 = .25. 2. .15x-Ax = 235a; - 2352.5. 3. L3x + 8.24 = -5.26 -3.2 a;. 4. 3a; — 1.245a; + .6a; = 1.5 + .355a;. 5. 3.5 X + .0564 - .1 a; = 4.9128 - .02 x. 6. .12 (2 a; + .06) - .15(1.5 a; - 2) = 0.246. „ .01 a; + .003 , .02 a; + .0008 ..,_ 7. —^ + ^ ■ = .0017. 8. 9. .3 (a; + 5) 4 (.25 x - .35) _ 14.325 8 7 56 ■ 0.5(6 -.2a;) .3(.4a;-3) ^ .80 .16 _ .32x , .045a; ^ „ ^„ ,, 33 , 3.75 "•:or+-l2r = ^2-^'- ^^•^T^ + .5(.-.8.5) =^- Note. The introduction into Europe of the Arabic notation for numbers was one of the important events of the Middle Ages. This notation originated among the Hindus at least as early as 700 a.d. It was adopted by the Arabs, and was introduced by the Moors into Spain during the twelfth and thirteenth centuries. Any one who has tried to multiply two numbers in the Roman notation, like MDCC VII by MCXVIII, will realize the difficulties that surrounded arithmetical operations before the Arabic system was taught. Before the introduc- tion of this system, one of the principal uses for arithmetic was the determination of the day of the month on which Easter came.^ Roger Bacon in the thirteenth century urged the theologians " to abound in the power of numbering," so that they might carry out these com- putations. Business accounts were kept on the abacus, a contrivance of wires and sliding balls on which arithmetical operations can be performed with great rapidity. Though computation in the decimal system was common in Europe from the thirteenth century, the final step in perfecting the notation was not taken until about 1600, when Sir John Napier made use of the decimal point in the modern sense. It was not until the beginning of the eighteenth century that it came into general use. EQUATIONS CONTAINIKG FRACTIONS 165 72. Literal equations. At this point the student should review the solution on page 82. EXERCISES Solve and check : 1. 5 ca; — 8 c^ = 4 c'' — ex. 4. ax -{- hx = a? -\- ah. 2. 2(a; + l)-4A; = 2. 5. ex + b^ = bx + be. 3. 3 (2a: — a) = 2 (a; — 2 a). 6. mx + n^ = m^ — nx. 7. 6 ac + ca; + 4 a^ = 2 aa; + 3 c^ -t 2 c«. 8. bax-5a^ + &b'^ = +7 ah + Sbx. 9. 7^ = b. 13. - + - = a + b. jia a b 10. = a. 14. c = — + d. X XX. a 3a 5 ax 2bx 11. - + ^r- = - • 15. — — 4 6^ = a^ x 2x 4 2b a — -4- — — -4-— IB ^ , " — ■^^ Q __4. '3a;a;26a; c 3 17. - H 1- ac = Jc + a5 + T ■ a c 56 46 ~20' ,„ aa; 36/ 2a6\ ,/a b\ ^^■Y-t("--3-) = '^^2-5)- a b\2 J a b X — m? _n oo 5 ^ — ^ -^'^'^ _ Q 21» "^ — * 22. — , o. a; — w' ?» c c a b{b — x) a(b — X) lab c. a _ c" — ae + 2 a^ ' a(x + c) c(x — a) 2 ac(x — a) 166 COMPLETE SCHOOL ALGEBRA ,5.i- + l-^ + J- = 0. 26.^ + ^ = ^ + 1- ab X abx bx ax a c^ ip S,c-Zd o-d 27. -; 28. dx ex X od a -{- X a — X 2 b + X h — X x^ — h^ X — a a — b ! + "" ax ^„ 2"''3 , 7a 30. h 31. a X ia — 6 X a^ + ao 2cx(a + c) _ a" + 2 «c + c^ a; + 3c x^ + 5cx + 6o^ x + 2e 73. Meaning of primes and subscripts. Different but related values are often represented by the same letter, with smaller figures or letters written at the right and above or below the letter used ; as, y', y", x„, 4 x^, t,l t„. These ate read y prime, y second, x sub zero, 4 x sub three, the square of t sub m, and t sub w respectively. Primes and subscripts must not be treated as exponents, and the student should carefully note that x^, and Xg are as different numerically as a and b. The notation just explained is very convenient in physics, where L^ and L^ may denote different but related lengths ; W^ and W2 may represent two different weights ; and t,^, t^, and t^ may mean three unequal but related intervals of time. Primes are cumbersome and easily confused with exponents; hence subscripts are preferable. The following equations are taken from algebra, geometry, and physics, where it is often necessary to express one of the quantities (weight, time, distance, etc.) in terms of the others. EQUATIONS CONTAINING FRACTIONS 167 EXERCISES 1. Solve for R, K = 2 ttRH. 2. Solve for a, A = —■ 3. Solve for R, C = 2 -irR. 4. Solve for r and t, d = rt. 5. Solve for a and A,— = -—r- A 360 6. Solve for C, 3^ = |. 7. Solve for r, C R + r E 8. Solve for r and n, C = R + Mr n-e 9. Solve for r and re, C = ^, , R + nr 10. Solve for F, C = |(F - 32). 11. Solve for Trj,J]^ = ^- 12. Solve for r and i, ^ = P(l + rt). 13. SolveforPj, 5 = §- „ , „ , , n(a -{- V) 14. Solve for n and i, s = _ — - ■ 2 7-Z 15. Solve for a, I, and r, s = _ _ 16. Solve for 6, zr^ = — ■ 180 TT 17. Solve for h, V^ = Fo(l + .00366 «i). (bi + 6«) a 18. Solve for b^, A = ^ ^ ' 19. Solve for x, - b c — X 168 COMPLETE SCHOOL ALGEBRA 20. Solve for F, D^, and Z)o, - = — + 77 ■ J" X*! Da 21. C (i„ - — , ) and are ratios. 2 2x a-b ^ The dividend, or numerator, in a ratio is called the anteced- ent, and the divisor, or denominator, is called the consequent. We may speak of the ratio of two concrete numbers if they have a common unit of measure. The ratio of 6 feet to 3 feet is I, the common unit of measure being 1 foot. Obviously no ratio exists between 5 years and 3 feet. Measurem.ent is the process of finding the numerical rela- tion (ratio) of whatever is measured to a standard unit of measure. Thus, when we say a distance is 100 yards, we mean that it is 100 times the length of the standard yard. For the United States the standard yard is the distance between two scratches on a certain gun-metal bar. This bar, along with the standard pound, the standard gallon, etc., is kept at the Bureau of Weights and Measures in Washington, D.C. If we say a piece of paper contains 54 square inches, we are expressing by the number 54 the ratio of the surface of the paper to the surface of a square whose side is one inch. Every measurement, then, is the determination of a ratio, either exact or approximate. 176 RATIO AND PROPORTION 17T Note. Until comparatively recent times there was, no unity among the various nations in regard to the standards of measurement. Just as we now have English, French, and American money, and are obliged to change when we go from one country to another, so until recently the different countries had their own standards of measurement. The yard and the foot are now in common use in English-speaking coun- tries, but in France and Germany the meter is the standard. In earlier times there was even greater confusion. Among the Hebrews the unit of length was the cubit, which, tradition tells us, was the distance from the end of the king's longest finger to the point of his elbow. Our word foot is a reminder of the time when the length of the king's foot was the standard. But with the advance of civilization and the increase of trade between different nations more or less uniformity in standards of measurement has been secured. EXERCISES Simplify the following ratios by writing them as fractions and reducing the fractions to their lowest terms : 1. 5 : 10. 4. 3tV : 3^. 7. 160 lb. : 1 ton. 2. 10:5. 5. 8§:5f 8. {x^-y^:{x-\-y). 3. 16 a^ : 8 a. 6. 3 days : 9 hours. 9. (a'-f- ¥) : (a + b). 12. 1 £c-3 K^ — 6a; + 6 13. (a;2 -xy + y"): (x^ + y^). ».(«.^|): X ". fi?-VfS+S+^ X 16. Separate 40 into two parts which are in the ratio of 2 : 3. Hint. Let 2 a; = one part, and 3 a; = the other. Then2a;+ 3x = 40, etc. 17. Separate 16 into two parts which are in the ratio of 5 : 3. 178 COMPLETE SCHOOL ALGEBRA " 18. Separate 84 into two parts which are in the ratio of 3 : 11. 19. Separate 36 into three parts which are to each other as 2:3:4. 20. Separate 135 into three parts which are to each other as 4 : 6 : 6. 21. What number added to both terms of the ratio | gives as the result the ratio || ? 22. What number subtracted from both terms of the ratio H gives as the result the ratio | ? 23. If a is a positive number, which is the greater ratio, 4 + 2a 4+3a„ or ! 4 + 3a 4 + 4a Hint. Reduce the fractions to equivalent fractions having a com- mon denominator, and then compare the numerators of the resulting fractions. 24. If a and i are positive numbers, which is the greater a + ib a + 6b . ratio, zT or ^rr ■ ' a + 5b a + 7 b 25. If a positive number is added to both/terms of a proper fraction, what change is produced in the numerical value of the fraction ? \ 76. Proportion. Four numbers, a, b, c, and d, are in proportion if the ratio of the first pair equals the ratio of the second pair.- a c This proportion is written a : b ^ c : d, oi - = —. Da The first and fourth terms (a, d) are called the extremes, and the second and third terms (h, c) are called the means. Since a proportion is an equation, any operation which may properly be performed on an equation may be performed on a proportion. Then in the proportion r = ^ both members may be multi- plied by bd, giving ad = be. Therefore, In any proportion the product of the means equals the product of the extremes. RATIO AND PEOPORTION 179 EXERCISES Find the value of x in the proportions : 3 6 4 1 1 1. 4~iB" 6. Z~^'-x' 9. -:2 = 3:4i. 2. 4 16 7~ x' 3 X 7. 1 2 X 3 4' 10. a c , 1 a: = c : - • 3. 2~ 6' 3:a; = 7; 11. 4. ;9. 8. 2 3J 1 4 ' X a 1 , 5. a;:4 = 3; :6. a; 12. x' X 13. 4: 3^ = 3: a; -3. 14. 5 : a; - 3 = 7 : 2a; + 6. A mean proportional between two numbers, a and b, is the number m, if — = — • This means that m^ = ab, ot tn = ±. -^ab. m b 2 ±4 Since = > + 4 is a mean proportional between 2 and 8, as ±4 8 is also — 4. A third proportional to two numbers, a and b, is the number . -t a, b *''^b=T In I — Yjj 12 is a third proportional to 3 and 6. A fourth proportional to three numbers, a, b, and c, is the number /, if - = — ■ Since -f-^ = ||, 24 is a foui-th proportional to 5, 12, and 10. EXERCISES Find the mean proportionals between : 1. 1 and 4. 4. 3 and 12. 7. ^ and ^j. 2. 4 and 9. 5. (a — by and 4. 4 9 8. -: and — 5 • 3. 16 and 4. 6. \ and ^. a* aa;^ 9. Find a third proportional to the numbers in Exercises 1-7 which precede. 180 COMPLETE SCHOOL ALGEBRA Find a fourth proportional to : 10. 1, 2, and 3. 15. a, a^, and a". 11. 4, 5, and 6. 16. a^ a^, and a*. 12. 7, 14, and 5. 17. a + b, a — b, and a^ — b^. 13. 6, 12, and a. ,„ «— ^ 1 i o 70 ' ' 18. —pr—> r J and a^ — b^ 14. 7, 21a;, and 6 a;. 2 a. + 6 If />s = qr is divided by qs, we obtain 4 = ^, or £ = r. (1) Also ps = 2'' divided by rs gives £ = 2. (2) (3) And g'r = ps divided by jjr gives p r Therefore, If the product of any two numbers (ps) equals the product of two other numbers (qr), one pair may be made the means and the other pair the extremes of a proportion. If - = - ; then from (1) and (2), - = ^- Here - = - is said b d ^ ^ ^ ' c d e d to be obtained from - = -\>y alternation. b d b d If - = - ) then from (1) and C3), - = - • Here - = - is said b d ^ ^ ^ ^' a c a G to be obtained from v = -; by inversion. b d •' EXERCISES Write as a proportion in three ways : 1. 3-4 = 2-6. 3. 3-6 = 2-a!. 2. 5 ■ 6 = 3 • 10. i. ad = b-o. 5. (a + &)(a-S)=2-3. 6. (a + b)(a - b) = (a + 2)(a + 3). RATIO AND PROPORTION 181 Write as a proportion : 7. a^ - S^ = 2 • 3. 9. a'' - 5 a + 6 = 4 -2. 8. a^-2ab + b^ = 3-6. 10. a^ - 7 a + 12 =(a + by. 11. o^ - 6 a + 9 = cf2 - 10 a + 16. 12. x^ - ixi/ + iy^ = a^ + 10 ab + 25b\ 13. xy = 4. 15. ab = 1. 17. m?ip = a;y». 1^. xy = 3. 16. abc = rfe. 18. ab = a + b. Write as a proportion so that x is the fourth term : 19. 3 • 4 = 5 • x. 22. px = «;■. a6 ■^ ^ 25 X = 20. 4a: = 9-7. 23. acx = bd. '. c 21. a6 = ca;. 24. 1 = az. 26. xy = y + l. Write by alternation : 27. 1 = |. 29. 4 : 6 = 6 : a;. 28. 15 = — 3 y 30. ^=f- Write by inversion : 31. 4:8 = 3:6. „« 3 a ^^■2 = Y 33. P ■.Pi=Wt,:W. 35. 36. X a 1.1_1_ 37. If four numbers, a, b, c, and d, are in proportion, they are in proportion by addition, subtraction, and addition and subtraction. Addition. Let 1, = T ^^^ Adding 1 to both members, ^ + 1=^+1, (2) ^=£±i. (3) Here (3) is said to be obtained from (1) hy- addition. 182 COMPLETE SCHOOL ALGEBRA (I c Subtraction. Let T = "; " 0-1 d Subtracting 1 from both members, or — — = — — ■ (3) b d Here (3) is said to be obtained from (1) by subtraction. Addition and Subtraction. Let - = - . (1) b d Then ^^-±-5 = ^^t_l' (addition), (2) b d and ■ = (subtraction). (3) Dividing (2) by (3), "-±1 = £±i. (4) a — o c — d Equation (4) is said to be obtained from (1) by addition and subtraction. Addition, subtraction, and addition and subtraction are often called composition, division, and composition and division respectively. EXERCISES Write by addition : 1. § = |. 3. a : a3 = 1 : 2. ^ _ £f 2. 4 : 12 = 8 : 24. 4. 4 : 3 = / : .-r. ®' I^ ~ 5| ' 6. Write Exercises 1-4, preceding, by subtraction. 7. Write Exercises 1-4, preceding, by addition and sub- traction. „,„« c a -\-h c 4- d 8. it - = - , prove = a a . c ^ T^„ a, c a — h c — d 9. Ii-r = -? prove = a a c ,nif"' " 2a + b 2c + d 10. It - = - ) prove ; = ; b d '^ b d ,, Tc a c a + 3b c + 3d 11. it-=-, prove ; — ; u d ^ b d RATIO AND PROPORTION 183 Write Exercises 12 and 13 by addition and subtraction and solve the resulting equations for x. 12 3x4-4 ^ 3 + 2 6a: + 3 ^ 2a, + 5 ■3a;-4 3-2" 6a;-3 2a-5' A series of equal ratios. If ? = £ = — , ri'i h d f ^' then ° + "+'^=g = £ = l. C2) Proof: Let H = £ = l = ^. C3-, I d f . ^> Then a = 6r, (4) c = dr, (5) e = /r. (6) Adding (4), (5), and (6), a -^ c ■\- e = lr ^ dr +fr. (7) Factoring in (7), a + c + e = (b + d + fy. (S) Therefore a + c + e^ ^ ^. ^^^ Hence, by (3). ,^J|f;=M=f C^^) This result may be expressed verbally : In a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent. EXERCISES Test the truth of the preceding result in Exercises 1^4 : 1. ^ = 3 = ^5^. 3. 3 : 4 = 6 : 8 = 12 : 16. „12'36 . I a b 2. - = TT- = TT-T ■ 4. 2a 3a6 x — y ax — ay hx — hy 5. Taken in the same order, the sides of two triangles are 3, 4, 5, and 9, 12, 16 respectively. What is the ratio of the sides of the first triangle to the corresponding sides of the second ? Compare this ratio with the ratio of the perimeter of the first triangle to the perimeter of the second. _„ a G e 2a + 3c + 4e a '• ^6=5 = 7' P^°^^ 26 + 3ci + 4/=6- 184 COMPLETE SCHOOL ALGEBRA MISCELLANEOUS EXERCISES IN PROPORTION 1. li a:b = 4::x, and a : 5 = 4 : - j find x. 2. If 7 = - ) and 7 = 7; find y. b y b 1 y 3. li p : r = 7 : &, and y : r = 3 : 5, find the ratio p : q. 4. li p : q = ^•. a, and q : r = a:l, find the ratio p : r. 5. The sides of a triangle are 8, 10, and 12. The side 12 is divided into the ratio of the other two sides. Pind the two parts. 6. The perimeter of a triangle is 63. Two sides are 18 and 24 and the other side is divided in the ratio of these two. Find the two parts of the third side. 7. A flagstaff casts a shadow 12 yards long ; at the same time a man 6 feet 10 inches tall casts a shadow 35 inches long. How high is the pole ? Fact from, Geometry. If one triangle is similar to another, the sides of the first taken in any order are proportional to the sides of the second taken in the same order. 8. The sides of a triangle are 10, 15, and 20 respectively. In a similar triangle the side corresponding to 10 is 12. Find the other sides. Compare the ratio of the two corresponding sides with the ratio of the perimeters. 9. The sides of a triangle are 9, 10, and 17. The perimeter of a similar triangle is 108. Find the sides of the second triangle. Fact from Geometry. A line parallel to one side of a tri- angle divides the other two sides into four proportional parts. Thus in triangle ABC which follows, line DE is parallel to ., „^ ., AD AE side BC, and -— - = — — • ' DB EC Also a line parallel to one side of a triangle forms with the other two sides a second triangle similar to the first. RATIO AND PROPORTION 185 In the following figure triangle ADE is similar to triangle ABC. Therefore = = AB AC BC 10. In triangle ABC : (a) If AD = 6, DB = 4, and AE = 10, find EC. (b) If DB = 4, AD = 8, and DE = 6, find BC. (c) If DB = 4, AD = 8, and ^C = 10, find AE. (d) li AB = AC = 16, and AD = 10, find AE. 11. Draw a triangle and letter the vertices F, G, and IT respectively. Draw B.K parallel to FG, R being on side HF and K on side HG. Then if: {a) FG = 15, BK = 10, HR = 8, find HF. (b) FG = 20, iJ/i: = 16, i/F = 10, find RF. (c) FG = 20, iJif = 16, 7JF = 6, find HR. (d) FG = 18, RK = 15, A'G = 4, find ZTA'. Fact from Geometry/. The line joining the middle points of two sides of a triangle is parallel to the third side. 12. Two sides of a triangle are 30 centimeters and 24 centi- meters respectively. The line joining their middle points is 12 centimeters long. Find the third side of the triangle. 13. The sides of a triangle are 10, 12, and 16 centimeters respectively. Find the lengths of the lines connecting the middle points of its sides. 14. In the right-angled triangle ABC, line BD is perpendicular to AC. Then BD is a mean propor- tional between AD and DC. (a) li AD = 9 and BD = 6, find DC. (b) If AD = A and AC = 20, find BD. 186 COMPLETE SCHOOL ALGEBRA -ir^ 15. In the semicircle ABC, line CD is perpendicular to AB. Then CD is a mean proportional between AD and DB. (a) If AD =2 and DB = 18, find Ci>. (b) If Ci) = 8 and ^£ = 34, '•^ find .4Z>. 16. The distance AB between two points on opposite banks of a river was wanted. Stakes were set at E, B, D, and C, so that BE was parallel to CD, and so G that ABC and AED were straight lines. The measured values of DC, CB, and BE were 480 feet, IGO feet, and 420 feet respect- ively. "What was the computed value of BA ? r 17. The perimeters of two sim- ilar triangles are 45 and 135 re- spectively. One side of the first is 11 and a second side is 19. Find the sides of the second triangle. 18. Two men start at the same time and travel in opposite directions. The ratio of their rates is 2 : 3. In 5 hours they are 100 miles apart. Find the rate of each. 19. A, B, and C ate equally of a stock of provisions which A and B furnished. The values of what A and B contributed were in the ratio of 7 to 8 respectively. C paid |30 for what he ate. How should A and B have divided the money ? 20. A clock provided with hands to indicate the minute, the hour, and the day of the month showed correct time at 4 p.m. on February 21, 1900. The clock gained 10 minutes daily. What was the correct time when the clock indicated 4 p.m. on the 28th of the next month ? CHAPTER XIX GRAPHICAL REPRESENTATION 77 . Temperature curve. The curve A B CDEF is called a graph. It was made by a recording thermometer. Such instruments are provided with an arm carrying a pen, which moves up as the temperature rises, and down as it falls. A clock movement runs a strip of cross-ruled paper under the pen and thus a con- tinuous line }S traced on the paper. The following record ex- tends from 2 P.M. of Wednesday, February 19, 1908, to 10.30 a.m. WEDNESDAY FRIDAY -m. M ■m of the Friday following. The numbers 60, 60, 70, 80, and 90 denote degrees Fahrenheit. There are 5 spaces from 50° to 60°. Hence one space corresponds to 2 degrees. The numbers 2, 4, 6, 8, and 10 indicate the time of day. Whether this is a.m. or P.M. can be determined by the position of these numbers with respect to the heavy curved lines marked noon. The point A 187 188 COMPLETE SCHOOL ALGEBRA on the graph informs us that at 2 p.m. Wednesday the temper- ature was 80 degrees. The point B between 6 p.m. and 7 p.m. Wednesday marks the highest temperature recorded. The point C tells us that the temperature was about 66^ degrees at 6 a.m. Thursday. The preceding record was made indoors, and the sudden fall from D to E was caused by the opening of a door leading into a cold hallway. The portion of the graph from D to E shows that the temperature of the room fell approximately 18 degrees in about 30 minutes. EXERCISES By reference to the graph (page 187) answer the following : 1. With what temperature does the record begin ? end? 2. What is the highest temperature recorded ? the lowest? 3. About what time was the highest temperature recorded ? the lowest ? 4. How often did the instrument record a temperature of 80 degrees ? 72 degrees ? 78 degrees ? 62 degrees ? 5. At what times did it record a temperature of 80 degrees ? 72 degrees ? 78 degrees ? 62 degrees ? 6. To what practical use can a graph such as the one here explained be put ? 78. Falling body curve. The curved line OABDC in the adjar cent figure is another graph. It represents closely the rela- tion between the distance a sphere of lead, if allowed to drop through the air, will fall in any number of seconds from one to eight. Time measured in seconds from the instant the sphere begins to fall is represented on the line OX. One inch on OX corresponds to 4 seconds of time, ^ inch to 2 seconds, ■^j; inch to f of a second, etc. The distance measured in feet through which the sphere falls is represented on the line OY. One inch on OF corresponds to 320 feet, -^^j of an inch to 32 feet, etc. The point A on the curve, just above 2, corresponds GRAPHICAL REPRESENTATION 189 to a time of 2 seconds. A is opposite the number 64 on OY. This means that in 2 seconds the lead sphere falls 64 feet. Similarly B corresponds to a time of 4 seconds and a dis- tance of 266 feet. That is, the lead sphere falls 256 feet in 4 seconds. > f r i / ^ 1 / ^ to 0- ^ - ; m r / 1 C ( lj 1 ^-K' 1 4E 0- ? ■V / c « ' / ' / ° / B / ,/ r /' / / A, / /^ ^ Tir ne in se CO ds 1 ^ 3 ^ 5 i 7 i ca le % n. = 1 E ec ond The question, " How far does a body fall in 6 seconds ? " can be answered by reference to the graph, thus : The point on the curve corresponding to 6 seconds is C, just above 6 on OX. The point on OY opposite C corresponds to 576 feet. Therefore in 6 seconds a body falls 576 feet. 190 COMPLETE SCHOOL ALGEBRA EXERCISES By reference to the graph (page 189) answer the following : 1. How far does a body fall in 8 seconds ? 3 seconds ? 7 seconds ? 2^ seconds ? 6.2 seconds ? The question, "Plow long will it take a body to fall 480 feet? " can be answered by reference to the curve, thus : Opposite 480 on Oy is the point D on the curve. D is directly over a point mid- way between 5 and 6 on OX. Therefore, to fall 480 feet a body requires 5^ seconds. By reference to the graph, answer the following : 2. How many seconds does a body require to fall 400 feet? 196 feet? 100 feet? 26 feet? 120 feet ? 750 feet? The two preceding graphs are pictorial representations of the relation between two variables. In the first graph the variables were time and temperature, both of which, in the period under consideration, were constantly changing. In the second graph the variables were time and distance. It must be borne in mind that the correctness of any graph is limited by the fact that we cannot measure any physical quantity with perfect accuracy, and that we cannot draw the graph itself with absolute pre- cision. This makes results obtained graphically only approxi- mately correct, but close enough, nevertheless, to be extremely useful for many purposes. 79. Graph of an equation. A relation between two variable numbers not connected with physical quantities, such as tem- perature and time, can also be represented b/ a graph. The question, "What two numbers added give five?" may be ex- pressed by the equation x + ij = 5. Here x and y are ani/ two numbers whose sum is 5. It can be seen by inspection that if x is 1, y is 4, and if a; is 2, y is 3. Or we may proceed as follows : Give x any value, say 3 ; then the equation becomes 3 + y = 6. Trans- posing and solving, y=2. Similarly give x the value 3J ; then ^2 + 2/ = S> whence y = 1^. Proceeding in this way, we may GRAPHICAL REPRESENTATION 191 obtain a few of the many related pairs of values of x and y, which may be tabulated as follows : X A D c D JS 7'' G •/ // / AT Ji 1 H 2 8 H 4 5 C 7 -1 -2 y 4 ^ 3 2 If 1 5 - 1 -2 6 7 Now in the figure we lay off equal spaces on OX from 0, and on O F from O each j\ of an inch, and agree to have the values of X correspond to distances measured from OF parallel to OX, and ~ ~ \ F ^ ( J / V B f F -x- V' ~" "y -', -I i C a_ —\ ~ f H ~ " "■ ~ ~ 1 ~ "" ■ — — — ~ — — r ~ " "" _ _ _ _ the values of y to distances measured from OX parallel to OY. Then the point A corresponds to the first pair of numbers, a; = 1, y = 4. In like manner, B corresponds to the second pair of num- bers, X = 11,1/ = 31- Similarly C, D, E, and F correspond respec- tively to the third, fourth, fifth, and sixth pairs of numbers. 192 COMPLETE SCHOOL ALGEBRA Apparently A, B, C, etc., are points on a straight line. Even points B and E, corresponding to fractional values of x and y, are in line with the others. The inference seems warranted, then, that a straight line drawn from ^1 to i^ would be a j>07'- tlon of the graph of the equation a; + y = 5. If the line AF is continued, it meets OX at G. The distance of G to the right of F is 6 and its distance above OX is zero. Evidently point G corresponds to the seventh pair of numbers, x = S, y = 0. Similarly FA extended cuts OF at point / whose distance from OX is 5 and whose distance to the right of OF is zero. Therefore this point corresponds to the eighth pair of numbers, a; = 0, y = 5. If AF is extended below OX, it passes through points H and /. The point II is just under the sixth space mark on dx and 1 space below OA', and the point / is just under the seventh space mark on OX and 2 spaces below OA'. The point H must correspond to the ninth pair of numbers, x = 6, y = — 1, and / to the tenth pair, k = 7, y = — 2. This leads us to extend the line YO downward and divide it into spaces equal to those above O, and to number the consecutive points of division with the negative numbers, — 1, — 2, — 3, etc. Since the point K is opposite the sixth space mark on F and 1 space to the left of OY, it corresponds to the eleventh pair of numbers, a; = — 1, i/ = 6. Similarly li corresponds to the twelfth pair, a; = — 2, y = 7. This leads us to extend XO to the left and, dividing it into equal spaces, to number the consecutive points of division with negative numbers, —1,-2, — 3, etc. Then the line III extended indefinitely in both directions would be the complete graph of the equation a; + y = 5. More- over, every point on this line would correspond to a pair of numerical values of x and y which satisfy this equation. These numerical values would include all the possible integers and fractions both positive and negative. The truth of this will become clearer as we proceed. GRAPHICAL REPRESENTATION 193 80. Definitions and assumptions. The preceding explanations and questions should tend to make clear that in constructing the graph of an equation in two variables a number of assump- tions must be made. These assumptions and some necessary- definitions are now stated. It is agreed : I. To have two lines at right angles to each other, as X'OX, called the jr-axis, and Y'OY, called the i/-axis, as in the follow- ing figure. II. To have a line of definite length as a unit of distance. Then the number 2 will correspond to a distance of twice the unit, the number 4^ to a distance of 4^ times the unit, etc. ; '■ Quadrant II Quadrant- 1 {+' +) p X,- y<+ x,+ P, y,+ y- y,- x,+ r (-'-) ^^a:,- (■!-'-) Quadrant III Quadrant IV . " III. That the distance (measured jiarallel to the x-axis) from the ?/-axis to any point in the surface of the paper be the A'-distance (or abscissa) of the point, and the distance (measured parallel to the y-axis) from the a;-axis to the point be the y-distance (or ordinate) of the point. IV. That the a;-distance of a point to the right of the {/-axis be represented by a positive number, and the ir-distance of a point to the left by a negative number ; also the y-distance of 194 COMPLETE SCHOOL ALGEBRA a point above the a;-axis be represented by a positive number, and the y-distance of a point below the a;-axis by a negative number. Briefly, distances vieasured from the axes to the right and upward are positive, to the left and downward, negative. V. That every point in the surface of the paper corresponds to a pair of numbers, one or both of which may be positive, . negative, integral, or fractional. VI. That of a given pair of numbers the first be the measure of the a;-distance and th-e second the measure of the y-distance. Thus the point (2, 3) is the point whose a;-distance is 2 and whose 2/-distance is 3. VII. That the point of intersection of the axes be called the origin. The values of the x- and the ^/-distances of a point are often called the coordinates of the point. Though not an absolute necessity, cross-ruled paper is a great convenience in all graphical work. Excellent results, however, can be obtained with ordinary paper and a rule marked in inches and fractions of an inch for measuring distances. Hence the graphical work which follows should not be omitted because it is found incon- venient to obtain cross-ruled paper for class use, EXERCISES Draw two axes and locate the following points, using \ inch or 1 centimeter as the imit distance. 1. (3,6); (-3,5); (-3,-5); (+3,-5). 2. (4,-2); (-6,4); (-1,-2); (+2,-4), 3. (0,4); (0,-4); (4,0); (-4,0). 4. (2,2); (0,2); (-2,6); (2,0). 5. (0,-5); (-5,0); (0,0). 6. If one coordinate of a point is zero, where is it located ? Where, if both are zero ? Locating points as in the piieceding exercise is called plotting the points. GRAPHICAL REPRESENTATION 195 EXERCISES 1. Find and tabulate six pairs of values of x and y which satisfy the equation a; + 2 y = 8. Draw two axes and, using \ inch as the unit distance, plot each of the points. Are the six points in a straight line? Where do all the points lie whose X- and y-distances satisfy the equation a; + 2 y = 8 ? "What, then, is the graph of the equation a; + 2 ?/ = 8 ? Does a; = 4, ?/ = 4, satisfy this equation ? Plot the point (4, 4). Is it on the graph of the equation ? If the x- and y-distances of a point satisfy the equation a; + 2 y = 8, where is the point located ? If the x- and y-distances do not satisfy the equation a; + 2 y = 8, where is the point located ? Find and tabulate six pairs of values for x and y which sat- isfy each of the following equations. Use numbers not greater than 10. Have at least one negative value for x and one nega- tive value for y. Then plot the six corresponding points. 2. 3 a; -f 2 y = 6. 4. a; + y = 0. 6. a; = 3 y. 3. 3a;-4y = 12. 5. a; - y = 0. 7. y = 2a:. The preceding work should be enough to convince the student that the graph of an equation of the first degree in x and y is a straight line. It can be proved that the graph of any equa- tion of the first degree (linear) in two variables is a straight line, but the student would not understand the proof were it given now. Therefore it will be assumed that the graph of every linear equation in two variables is a straight line. And as a straight line is determined by any two of its points, it will be sufficient in graphing a linear equation in two variables to plot any two points whose x- and y-distances satisfy the equar tion, and then to draw through these two points a straight line. The two points most convenient to plot are usually the two in which the line cuts the axes. Occasionally these points come very close together, and consequently they will not determine accurately the position of the line. In such cases one should decide on two values of x rather far apart (such as and 5, or 196 COMPLETE SCHOOL ALGEBRA and — 5) and compute the corresponding values of y. Two such points will fix the position of the line more accurately. If a line goes through the origin (as in Exercise 6 preceding), x = Q, y=Q, will do for one point, but a point outside the axes must be taken for the second one. Example : Graph the equation 2 a; + 6 y = 10. In this equHr tion if a; = 0, y = 2 ; and if y = 0; a; = 5. Here the point (0, 2) is on the y-axis in the adjacent figure, 2 units above the origin, and the point (5, 0) is on the a;-axis, 5 units to the right of the origin. The straight line through these two points is the graph of 2x + 5y = 10. > r ^ ■^ ' ■^ ^ * ^ <^ »• ■ ^ ^- K, ■«. ,.' »» *, / >. •.. . ,/ Check : If an error has been made in obtaining the value of X on y from the equation, or in plotting the values found, it can be quickly detected by plotting a third point, the values of whose X- and y-distahces satisfy the equation. If this third point lies on the line .determined by the first two points, the line has been correctly located; if it does not, a mistake has been made. GRAPHICAL REPRESENTATION 197 EXERCISES . Graph the following linear equations : 1. a; + y = 6. 4. 3a; + 4y = 12. 'i.x-2y = (i. 2. a; — y = 5. 5. 4 a; — 3 ?/ = 12. 8. 3 a; — y = 0. 3. a; + 2y = 8. 6. 2a; + 42/ = 9. 9. x = 4. Hint. The equation z = 4 is equivalent to the equation a; + y = 4. This last is satisfied by a = 4 and any value of y. Thus the pairs of values (4, 3); (4, 6); (4, 0); (4, - 2), etc., satisfy the equation a; + y = 4. Plotting these points, it is evident that the required graph is a line parallel to the y-a-aa and 4 units to the right of it. 10. a; =-6. 12. 2/ =-2. 14. y = Q. 11. 2/ = 6. 13. a; = 0. 15. x=±3. 16. If a point is on a line, do the values of its x- and its y-distances satisfy the equation of the line ? 17. If the values of the x- and the y-distances of a point satisfy the equation of a line, is the point located on the graph of the equation ? 18. Is the point (3, 4) on the line vrhose equation is 3a;-4y = 12? Is (0, 4)? Is (4, 0)? 19. Can you determine without reference to the graph itself if the point (2, 6) is on any of the graphs of the equations in Exercises 1-9 above ? If so, on which ones ? 20. Which of the graphs of the equations in Exercises 1-16 pass through the origin? In a linear equation containing one or more variables the constant term is the term which does not contain a variable. Thus in the equation Sx + 4:y= 12, 12 is called the constant term. Also in ax + by = c, the constant term is c. 21. What is the value of the constant term in the equations whose graphs pass through the origin? What can be said of its value in those equations whose graphs do not pass through the origin? 198 COMPLETE SCHOOL ALGEBRA 22. Can you tell, then, by looking at a linear equation whether its graph goes through the origin or not? Explain. 23. Can you tell from the equation when a line is parallel to the a;-axis ? the y-axis ? Explain. It should now be clear that: The equation of a line is satisfied by the values of the oi>-distance and the y-distanoe of any point on that line. Any point the values of whose x-distanoe and whose y-distance satisfy the equation is on the graph of the equation. 81. Graphical solution of linear equations in two variables. If we construct the graphs of the two equations x + 2y = 8 and , i^ ^ ^^f . ^^ f •j ^ 1 ^xb' '(4.2 1 ^ r ^S / "N N / ^ / ^^ , , / ■' 1^3 2 + 6 y = 24," = 0. Here the second equation divided by 3 gives the first. There- fore any set of roots of the first is a set of the second. If we choose to regard the two equations as really different, which is not at all necessary, we say that they havfc an infinite (unlimited) number of sets of roots. EXERCISES Solve the following systems of equations and check results : ^ x + 2y = 7, 8a; + 2/ = 7, ' 5x-2y = ll. ' llx + 2y = 2S. ^ 2x + y = i, 5l + 2p = 0, ' Zx~y = 21. '• 5l+p = S. ^ 7m-n = 2, IQv +2u = 22, ' n — 2m = — 3. ' u + 5 v = 11. 10h-k=-3, 12x + Sy = U, ■ 12 A + 12 A = 102. 3a; - lOy = 8. g 7r--8.=-30, 2x-y=-l, ■ r + 11 s = 20. 15x-Qy = 20. LINEAR SYSTEMS 207 jj 3s -^ = 12, 27 A + 32 A = 6, ■2!t-6s = 10. 16/c-9A = 8. j2 5a; -3m; = 2, 2r + 25ri = 15, ■ 15a; + 12m; =—6. ■3r = 10ri— 44 j3 a:i-6a;2 = 7, 12w-2w = 18, ■ 12x2 - a;i = 0. ' 3»t = 18w + 10. 85. Solution by substitution. The method of solving a system of two linear equations by substitution is illustrated in the EXAMPLE a 1 4-1. ^ (3x-13y = 41, (1) Solve the system H „ . ^ ^ '^ •^ 1 8 a; + 11 y = 18. (2) Solution : From (1), 3 a: = 13 y + 41. (3) ■ Solving (8) for x in terms oi y, x = — ^-^ (4) Substituting ^^y + ^^ for x in (2), Sa^l±m.^ny=lS. (5) (5) -3, 8(13 2/ + 41) + 33y = 54. (6) Simplifying, 104 y + S28 + S3y= 54. (7) Collecting, 137jf=-274. (8) (8) -^137, y=-2. (9) Substituting — 2 for y in (4), x = — = 5. Check : Substituting 5 for x and — 2 for y in (1) and (2) gives the obvious identities 15 + 26 = 41 and 40 - 22 = 18. The method of the preceding solution is stated in the BuLE. Solve either equation for one variable in terms of the other. Substitute this value in the equation from which it was not obtained and solve the resulting equation. Substitute the definite value just found, in the simplest of the preceding equations which contains both variables, and solve, thus obtaining a definite value for the other variable. Check. As on page 206. 208 COMPLETE SCHOOL ALGEBRA EXERCISES Solve by the method of substitution a;-2y^=8, 3a; + 2y = 7. x-2y=-12, 4:X — y = 1. 14m — 2 re = 1, w. — 6 m = 0. 2. 4. 5. 6. 7. 6 A + 10 ;b = 19, 2 A; = 3 7i. 3s + 12 = 3 + *, lS + 2p== q, p + q = -9. 3r + 15s = 7, 12 + 5s=-r. 8. 9. 10. 11. 12. 20 y - 3 « = 1, « — 6 y = 0. .75^ + 1.5^ = 3, q=p 16. 3a;-20 _ 2x + 5y . 10 = a; — y. 1 _^ 7 + 2 mj ?«,; 9Mj = TWj + 3. 5Ri + 2^ -7 iJi iJa = 2(iJ2 + 2), = 0. 86. Simultaneous equations containing fractions. The method of solving a system of two linear equations containing frac- tions is illustrated in the EXAMPLE rSa; 69_3y Solve the system - 3 6 - 2 ' 3a; „ 9 4-22/2- Solution: (1) ■ 6, 16x-5Q = 9y. Transposing in (3), (2) -4, Transposing in (.5), (4) -3, (6) ■ 16, 16a;-9y = 59. 3x=-8y- 3a;+ 8y =- 18. A8x- 27 y= 177. 48a; + 128y =-288. (7) -(8), (9)^-1.55, -155jr= 465. y=-3. 18. Substituting - 3 for ^ in (4), 16 a; + 27 = 59. Whence x = 2. (1) (2) (3) (4) (5) (6) (7) (8) (9) LINEAR SYSTEMS 209 Check : Substituting 2 for x and — 3 for j^ in (1), 16 59 _ - 9 3 6 ~ 2 ' or -9^-9 2 2 Substituting 2 for x and — 3 for y in (2), 6 — fi _ 9 or 8 — 8 - As in the foregoing solution, it is usually best to clear the equations of fractions and write them in the form of (4) and (6) before attempting to eliminate one of the variables. Equa- tions (4) and (6) are in what is called the general form of a linear equation in two variables. This form is represented for all such equations by ax + hy = c. Here a, h, and c denote numbers, or known literal expressions, EXERCISES Solve the following systems of equations and check results : 2a; , , 26 2x y -z+^y=-3' g T-2 — ^' '■3.-^=-4. ' ^ = '-f-^. .4a; + .9 y = 5.7, 2x-y = l. 12m = i^ + 17, ■ 7. ^ 2m 49 _ 5 y 3 ~i2~12' 9. = 3 + ^, 12 A * li 9 _ -Bs 37c--g- = 18, 3 4- 4 • *■ llA ITA^ .04m + .75w = 10, 10 "^ 2 ^' ^- .8m-1.25n = 5. 2. 3 re 18 '^ 5-5' ^^ + 7m- 16. 3. 6 ' 4 ^' 2-Ri i?2 Q 3 8-^- 210 COMPLETE SCHOOL ALGEBRA 28a; -162/ = 56, - + 12 = 17 10. 2x + l . .. X — 15 2/ = 0. '• + 3-^ = X 11. 7 5' 4j- + ^_2 5)- + 2 = 7;. ,„ 6r + ^ 5 19 ^ -2 = 5. 12. a; + 3 , y + 4 _3 ' 2r ~6~+ 10 ~ 2' 3 7x + l lly-4 _^^ ^ 2^40 2 '^ 20. t^ ti~ hh 20m + 9 _ »i , . 2 ^2 — 3 ii = 0. 13. 7 ~ 7 + ' 5m + 3w. — 1 ,1 ™ + ¥- = »*• 3 ^ ** "*" 3 ' 21. „^ 25 m 115 1_1^1 a; 2/ 6 Hint. Solve Exercise 14 with- 22 = 0- Jc + 5 ^ 1 + 2 k + l~ 1-2' out clearing of fractions. ~^ = 4 l + k + 1 mi ma 15 2 a?! — 3 X2 _ „ 11 1 „, a:2-4a:i m.1 9»2 " " I " /? Y o x^ -f- Xi Xi a '^ + ^=^2' 3a;-2y = 7, 7 6^ „ 24. 2 5 |-5i = 0. ^= T- X y ^ X — 5 y — 4 ,, ^^^ = ''' ,. 6T^-4^ = «' IV- 00 25. ^ ^ - + - = 14. 1 + 1 + 2 = 0. X y 4 2 2 1 26. — j^ = -; ) wi + 2 n, = 4v W1 + 2 + M2 59i2 + 3%-5 ^ '^ LINEAR SYSTEMS 211 In the following problems the student should state two equations in two unknowns. Instead of using x and y, the first letter of the word denoting an^ unknown should be used to represent that unknown. Thus in Problem 6 below, n would represent the number of nickels and q the number of quarters. The course here suggested is desirable for many reasons, and it should be followed in all problems containing two or more unknowns, unless the words denoting two of the un- knowns begin with the same letter. PROBLEMS 1. The difference of two numbers is 25 and their sum is 46. Find the numbers. 2. The quotient of two numbers is 6 and their sum is 49. rind the numbers. 3. Find two numbers whose difference is 36 and whose quotient is 3. 4. The value of a certain fraction is |. If 3 be added to the numerator and 1 to the denominator, the value of the result- ing fraction is |. Find the fraction. 5. The greater of two numbers divided by the less gives a partial quotient of 3 and a remainder of 6. The less divided by the greater gives a fraction which is .7 less than 1, Find the numbers. 6. A collection of nickels and quarters, containing 77 coins, amounted to $9.85. How many coins of each kind were there ? 7. If J be subtracted f;-om the numerator and \ added to the denominator of a certain fraction, the value of the result- ing fraction is f . The sum of the numerator and the denomi- nator of the original fraction is 19. Find the fraction. 8. The difference between the numerator and the denomi- nator of a certain proper fraction is 24. If | be added to the 212 COMPLETE SCHOOL ALGEBRA numerator and ^ be taken from the denominator, the value of the resulting fraction is ^\. Find the fraction. 9. Two weights balance when one is 12 inches and the other 8 inches from the fulcrum. If the first weight increased by 2 pounds is placed 10 inches from the fulcrum, the balance is maintained. Find the two weights. 10. Two weights balance when one is 12 inches and the other 18 inches from the fulcrum. If the first weight is de- creased 12 pounds, the other weight must be moved 3 inches nearer the fulcrum to balance. Find the weights. 11. Two weights balance when one is 15 inches and the other 10 inches from the fulcrum. The smaller weight is moved one inch nearer the fulcrum and decreased 6 pounds. Then the larger weight is decreased 12 pounds and a bal- ance results. Find the two weights.. 12. A's age is now twice B's. Seven years ago B was ^ as old as A. Find the age of each now. 13. In 5 years A will be twice as old as B. Five years ago A was three times as old as B. Find the age of each now. 14. The perimeter of a rectangle is 232 feet and the length is 8 feet more than twice the width. Find the dimensions of the rectangle. 15. A past of $1000 is invested at 6% and the remainder at 5%. The yearly income from both is |54. Find the num- ber of dollars in each investment. 16. A part of |2000 is invested at 44% and the remainder at 3-^%. The yearly income from the 3^% investment exceeds the other yearly income by $10. Find the number of dollars in each investment. 17. A part of $6000 is invested at 4% and the remainder at 6%. The i^fo investment yields $126 more in 5 years than the one at 6 »^ does in 3 years. Find the number of dollars in each investment. LINEAR SYSTEMS 213 18. Ten rubles are worth 10 cents less than 20 marks, and 12 marks are worth 4 rubles and a dollar. Find the value of a ruble and a mark in centa 19. Five francs are worth 19 cents more than 2 florins, and the sum of 3 francs and one florin is worth 5 cents less than one dollar. Find the value of a florin and a franc in cents. 20. The sum of the two digits of a 2-digit number is 9. If 45 be subtracted from the number, the result will be expressed by the digits in reverse order. Find the number. Solution: Let < = the digit in tens' place, and « = the digit in units' place. Then i + u = d. (1) But t standing in tens' place has its numerical value multiplied by 10. Therefore the number is represented by the binomial lOt + u, and the number formed by the digits in reverse order is represented by the binomial 10 m + ^ Hence lOt + u- i5 = lOu + 1. (2) Simplifying (2), t - u = 5. (3) Solving (1) and (3), t = 7, and « = 2. Hence the number is 72. Check: 7 + 2 = 9. 72 - 45 = 27. 21. The sum of the digits of a 2-digit number is 7. If 27 be added to the number, the result is expressed by the digits in reverse order. Find the number. 22. The tens' digit of a 2-digit number is twice the units' digit. If 36 be subtracted from the number, the result is expressed by the digits in reverse order. Find the number. 23. If a 2-digit number be divided by the sum of its digits, the quotient is 4. If 36 be added to the number, the result is expressed by the digits in reverse order. Find the number. 24. If a 2-digit number be increased by 3 and then the result be divided by the sum of its digits, the quotient is 9. If the number be divided by three times the units' digit, the quotient is 17. Find the number. 214 COMPLETE SCHOOL ALGEBRA 25. If a 2-digit number be divided by the sum of its digits, the quotient is 7. If the number formed by the digits in re- verse order be divided by the sum of the digits and 3, the quotient is 3. Eind the number. 26. The sum of the reciprocals of two numbers is ^j, and the difference of their reciprocals is ^^. Find the numbers. 27. The difference of the reciprocals of two numbers is 1^. The quotient of the greater number divided by the less is Ij. Find the numbers. 28. If 15 grams be taken from one pan of a balance and placed in the other, the sum of the weights in the first will be ^ the sum of those in the second. But if 85 grams be taken from the second and placed in the first, the sums of the weights in each pan will then be the same. Find the weight in each pan at first. 29. A gives B $20 ; then B has twice as much money as A. B then gives A $75 and has left J as much as A. How many dollars had each at first ? 30. The circumference of the fore wheel of a carriage is 2 feet less than that of the rear wheel. The fore wheel makes as many revolutions in going 155 feet as the rear wheel in going 186 feet. Find the circumference of each wheel. 31. If the length and the width of a rectangle be each in- creased one foot, the area will be increased 18 square feet. But if the length and the width be each decreased one foot, the area will be decreased by 16 square feet. Find the length and the breadth. 32. A and B working together can do a piece of work in 2f days. A works 50% more rapidly than B. How many days would each require alone ? 33. A and B together can do a piece of work in 1\ days. They work together for 5 days, and A finishes the job by him- self in 3§ days. How many days would each require alone ? LINEAR SYSTEMS 215 34. If the length of a rectangle be increased by 4 feet and the width decreased by 2 feet, the area i§ increased 8 square feet. But if the length be decreased by 1 foot and the width increased by 3 feet, the area is increased 33 square feet. Find the dimensions of the rectangle in feet and its area in square yards. 35. A rectangle has the same area as one 10 feet longer and 6 feet narrower. It also has the same area as one 4 feet longer and 3 feet narrower. Find the dimensions of the rec- tangle. 36. The products of three pairs of numbers are equal. One number in the second pair is 2 greater, and one in the third pair 3 greater, than the first number in the first pair. The other numbers in the second and third pairs- are respectively 15 less and 18 less than the second number of the first pair. Find each pair of numbers. 37. If the number of men who together purchased a piece of land had been 3 more, each would have had to pay $200 less than he did ( but if the number of men had been 4 less, each would have had to pay |600 more than he did. Find the num- ber of men and the price of the land. 38. A man rows 10 miles downstream in 2 hours and returns in 2 hours and 30 minutes. Find the rate of the river and his rate in still water. Hint. Let x = the man's rate in still water in miles an hour, and y = the rate of the river in miles an hour. Then his rate down- stream is a; + y miles an hour, and upstream x — y miles an hour. 39. A boat goes downstream 36 miles in 3 hours and up- stream 24 miles in 3 hours. Find its rate in still water and the rate of the current. 40. The rate of a boat in still water is 8^ miles an hour. It goes down the river from A to B in 14 hours. It returns one half the distance from B to A in 10 hours. Find the rate of the river and the distance from B to A. 216 COMPLETE SCHOOL ALGEBRA 41. A boat which runs 12 miles an hour in still water goes downstream from A to C in 7 hours. It returns upstream to B, 36 miles below A, in 6 hours. Find the distance from A to C and the rate of the stream. 42. A train leaves A one hour late and runs from A to B at 25 ofo more than its usual rate, arriving on time. If it had run from A to B at 24 miles an hour, it would have been 10 minutes late. Pind the distance from A to B and the usual rate of the train. 43. A train leaves A 40 minutes late. It then runs to B at a rate 20% greater than usual, and arrives 16 minutes late. Had it run 15 miles of the distance from A 'to B at the usual rate and the rest of the trip at the increased rate, it would have been 22 minutes late. Find the usual rate and the dis- tance from A to B. 44. The rate of a passenger train is 66 feet a second and the rate of a freight train 44 feet a second. When they run on parallel tracks in opposite directions they pass each other in 15 seconds. The length of the freight train is twice the length of the passenger train. Find the length of each. 45. The rate of a passenger train is 45 miles an hour and that of a freight train is 30 miles an hour. The freight train is 350 feet longer than the passenger train. When the trains run on parallel tracks in the same direction they pass each other in 1 minute and 15 seconds. Find the length of each. 46. The length of a freight train is 1540 feet and the length of a passenger train 660 feet. When they run on parallel tracks in opposite directions -they pass each other in 20 sec- onds, and when they run in the sam'e direction they pass each other in 1 minute and 40 seconds. Find the rates of the trains. 47. Two bicyclists travel in opposite directions around a quarter-mile track and meet every 22 seconds. When they travel in the same direction, the faster passes the slower once every 3 minutes and 40 seconds. Find the rate of each rider. LINEAR SYSTEMS 217 87. Literal equations in two variables. Linear systems in which, the variables have literal coefficients are solved by the method of § 84. EXERCISES In Exercises 1-16 consider a, b, c, d, and these letters with subscripts, as known numbers ; solve for the other letters in- volved and check. Solve Exercises 17-20 for x and y. 3 a; + 7 y = 17 a, 10a; -42/ = 2a. Sx-y = lQb, 4.x + 9y = 3b. 5m — 4?i = 10a — 4, m — 2na = 0. 11 A + 67c = 33 c, 4. 7i k _ . 12Ri — llRz = a + 12b, ■ R^ + R2 = 2a + b. 8^ + 9 y = 4 B + 9 »!, 6. » „ a — 12 «! 2-3? = 4 Tex 5y_ x+^^=llG+3. 4 7.6 X + 3 y = 6 a, .25x + .5y = 0. 2a a ' a 4b 4 9. 18- ^."^^T"' !»• d(x — d) — y — 0. 10. £;r + 3s = l — c?, 7£?r + 36s = 7-12d A - A; = 0, 11. •A + A; A — A; 5c ~ 2o 12. (a + 6) re = 1 — em, (^a + b)m — 1 = — en. m n 13. b,-b,^b, + b, ' m + w = 2 6i. 14. ^ + i = 25, a; y 2" 3 „ , =5c — 6. X y 1 1 Bi + ^2 15. « + *i ' 2/ 2 B1B2 Bl "2 _ Q a; + Bi y 16. m + 2b _ m — B |w — B 2re 3w — 2b 5m 17. kx — ry = 0, a: + y — A = 0. ax - -by = G, 2^ Ba;4-&2/ = c, y = 6. ■ dx + ey=f. a; + 218 COMPLETE SCHOOL ALGEBRA GENERAL PROBLEMS 1. If one book costs a dollars, what will c books cost ? 2. If a. books cost b dollars, what will one book cost? c books ? 3. (a) Find the perimeter and the area of a rectangle whose length is a and whose width is b. (b) Then find the perimeter and the area of a second rectangle whose dimensions are three times the first, (c) The perimeter of the second is how many- times the perimeter of the first ? (d) The area of the second is how many times the area of the first ? 4. The base of a triangle is 8. The altitude is 10. Find the area. 5. The base of a triangle is b. The altitude is 8. Find the area. 6. The base of a triangle is b. The altitude is a. Find the area. 7. The base of a triangle is a + b. The altitude is a ~ b. Find the area. 8. The base of a triangle is a; — 2 y. The altitude is a; + 2 y. Find the area. 9. The area of a triangle is Jc. The base is b. Find the altitude. 10. The altitude of a triangle is a inches and the base is 10 inches. If 2 inches be taken from the altitude, how much must- the base be increased so that the area will be the same as before ? 11. The altitude of a triangle is a feet, the base is b feet. The altitude is increased h feet and the base decreased so that the area is the same as before. How many feet are taken from the base ? 12. The sum of two numbers is s and their difference is d. Find the numbers. LINEAR SYSTEMS 219 13. The first of two mimbers is a times the second, and the first minus the second is h. Find the mimbers. , 14. The sum of two numbers is h, and the quotient of the first divided by the second is a. Find the numbers. 15. If a be added to the numerator of a certain fraction, the value of the resulting fraction is 2. If h be added to the denominator, the value of the resulting fraction is 1. Find the fraction. 16. If the numerator of a certain fraction be increased by 1, the value of the resulting fraction is x. If the denominator of the fraction be decreased by 2, the value of tie resulting fraction is y. Find the numerator and the denominator. 17. The value of a certain fraction is h. If 2 be added to the numerator, the value of the resulting fraction is o. Find the numerator and the denominator. 18. A boy who weighs a pounds and one who weighs h pounds balance at the opposite ends of a teeter board whose length is I feet. How far is the fulcrum from each end of the board ? 19. A certain number of books at 80 cents each and another number at fl.lO each cost together A dollars. If the price of the books had been interchanged, the total cost would have been It dollars. Find the number of each kind. 20. Two books cost c dollars. The first cost d cents more than the second. Find the cost of each. 21. A and B have together h dollars. A gives h dollars to B and then they have equal sums. How many dollars had each at first ? 22. If A gives h dollars to B, they will have equal sums. If . B gives k dollars to A, A will have twice as much as B. How many dollars has each? 23. If A gives $10 'to B, B will have A dollars more than A. But if B gives Ic dollars to A, A will have three times as much as B. How many dollars has each ? 220 COMPLETE SCHOOL ALGEBKA 24. A and B have together $40. A gives h dollars to B, after ■which B gives k dollars to A. Then they have equal sums. How many dollars had each at first ? 25. A gives r dollars to B and then has -J- as touch money as B. Then B gives |8 to A and has left f as much money as A. How many dollars had each at first ? 26. A part of flOOO is invested at a% and the remainder at ft%. The yearly income from both investments is c dollars. How many dollars are there in each investment ? 27. A portion of x dollars is invested at 5 % and the remainder at4i^. The yearly income is ?/ dollars. How many dollars are there in each investment ? 28. A works three times as fast as B. Together they can do a piece of work in o days. How many days would each require alone ? 29. A works h times as fast as B. Together they can do a piece of work in 4 days. How many days would each require alone ? 30. A and B together can do a piece of work in h days. A can do f of the work in 6 days. How many days does each require alone ? 31. A and B together can do a piece of work in 5 days. A can do | of -it in k days. How matiy days does each require alone ? 32. B requires twice as much time as A to do a piece of work which they can do together in mdays. How many days does each require alone ? 33. A and B together can do a piece of work in p days. A works q times as fast as B. How many days does each require alone ? 34. A man travels n miles and then returns to his starting point. Going, his rate is 3 miles an hour ; returning, it is 4 miles an hour. How many hours did the entire journey take? LINEAR SYSTEMS 221 35. A and B start at the same time from two towns k miles apart and travel toward each other until they meet. A travels 3 miles an hour and B travels 5 mijgs an hour. In how many hours do. they meet? How far does each travel? 36. In Problem 34, what would the required time have been, if the rate going had been p miles an hour and the rate return- *ing q miles an hour ? 37. In Problem 35, what would have been the respective dis- tances if A had rested h hours on the way before he met B ? 38. A and B start at the same time from two points c miles apart and travel toward each other until they meet. A travels p miles an hour and B travels q miles an hour. In how many hours do they meet ? 39. In Problem 38, how many miles does each travel ? 40. A man rides in a carriage d miles and returns on foot at the rate of 3 miles an hour. The time of riding is h hours less than the time of walking. Find the rate of the carriage. 41. In the preceding problem, if the rate of walking had been c miles an hour, what would have been the rate of the carriage ? 42. A man rides a disJtance of p miles and walks back at the rate of q miles an hour. The entire trip took t hours. Find his rate of riding. 43. A and B start from the same point at the same time and travel in opposite directions for n hours. They are then 50 miles apart. A trayels 2 miles an hour more than B. Find the rate of each. 44. In Problem 43, what would have been the respective rates if A had traveled k miles an hour more than B, and at the end of n hours they were h miles apart ? 45. A man has just t hours at his disposal. How far can he ride in a carriage which travels p miles an hour, and yet have time to walk back at the rate of q miles an hour ? 222 COMPLETE SCHOOL ALGEBRA 88. Determinate systems in three and four variables. Consider the equations : m + n + p = 6. - (1) 2m + 3m + 4^ = 16. (2) Sm + An + 5p = 22. (3) m + 2n + 3p:=10. (4) ' 6m + 9m + 12^ = 48. (5), Equation (3) is (1) plus (2); (4) is (2) minus (1); (5) is (2) multiplied by 3. Hence we speak of (3), (4), and (5), with respect to (1) and (2), as derived equations. Equations (1) and (2) are spoken of as independent with, respect to each other, because neither can be derived from the other as (3), (4), and (5) were derived from (1) and (2). A system of three independent equations of the first degree in three variables, no two equations being incompatible, has one set of roots and only one. The method of obtaining the set of roots of a determinate system is illustrated in the following EXAMPLE fm + 6n — 5p = 23, (1) Solve the system J 3 to — 8 m + 4p = — 1, (2) [Tm-lOn + lOp^O. (3) Solution : Eliminate one variable, say/), between (1) and (2) thus : (1) ■ 4, 4 m + 24 n - 20.p = 92. (4) (2) ■ 5, 15m-40n + 20p=- 5 . (5) (4) + (5), 19 m- 16 n = 87. (6) Now eliminate p between (2) and (3) as follows : (2) -.5, 15TO-40n + 20j9=- 5. (7) (3) • 2, 14to-20w + 20jo= . (8) (7) -(8), m-20n =-5. (9) The equations (6) and (9) contain the same two variables, m and re. (6)-l, 19m- 16n= 87. (10) (9) • 19, 19m-380ra=- 9.5 . (il) (10) -(11), 864 n= 182. (12) (12) H- 364, 71 = 1. (13) LINEAR SYSTEMS , 223 Substituting ^ for nin(9), m - 10 =— 5. (14) Solving (14), m = 5. (15) Substituting ^ for n and 5 for m in (1), 5 + 3 - 5;j = 23. (16) Solving (16), p=-3. Check: Substituting 5 for m, \ for n, and — 3 for j3 in (1), (2), ^°d(3), 5 + 3 + 15 = 23, or. 23 = 23. 15-4- 12 =-1, or-l=-l. 35 - 5 - 30 = 0, .or = 0. For the solution of a simultaneous system of linear equa- tions in three variables we have the EuLE. Decide from an inspection of the coefficients which variable is most easily eliminated. Using any two equations, eliminate that variable. With one of the equations just used, and the third equation, again eliminate the same variable. The last two operations give two equations in the same two variables. Solve these two equations by the rule, pages 205-206. . Substitute the two values found in the simplest of the origi- nal equations and solve for the third variable. Check. Substitute the values found in each of the original equations and simplify results. Four or more independent equations in three variables have no common set of roots. In general a system of n + 1 independent linear equations in n variables has no set of root") ; a system of n independent linear equations in n variables, no two of which are incompatible, has one set of roots ; and a system of ra — 1 independent linear equations in n variables, no two of which are incompatible, has an infinite number of sets of roots. A system of four independent equations in four variables may be solved as follows : Use the first and second equation, then the first and third, and lastly the first and fourth, and eliminate the same variable each time. This gives a system of three equations in the same three variables, which can be solved by the rule given above. 224 COMPLETE SCHOOL ALGEBRA EXERCISES Solve the following systems : m + n-2:p = 13, lii_l = i 1. m — 3n—p=—3, m n p ' m — n + 4:p=—n. iilj.l_?. x + y + 3s = ^, m n p 3 2. x-2y + 4« = 7, i_l + i = o/ 2a; -112/ -24s = 5. 'm n p 3. 3x — y-5s = 13, Hint. Solve Exercise 11 with- out clearing of fractions. 13. 5x + 3ij + 2z = l. 1_3 10__o 2A + 3 7c-4Z=-26, m~ n p ~~ ' i. 3h-k + 27l = S7^, 12 i + --t-- = 15 7i + 5k + 33l = 74^. ' m'^ p'^ n 2m + 3n-ip=-3, l_i + :5=_l. 5. m + n + 3p=-9, m n p 2 m + 2n — 1 p = 6. x + 82/ + 6» = l, 6. 3x + lQz + 42/ = — 5, a; + 4g = 0. 2h-3l + 4.k-2 = 0, 7. 3^-3^-15 = 0, 7 /i - 4 /c - 31 = 0. 4 r — 10 s = 5, 8. Q,r — t = 3, 5s + 2t = -%. 2 «! — 3 0.3 = 4, 9. 3 % + oss = 5, ^2 — 2 a^ = 2. 3ri + 5r2 = 74, 10. ri-2r8 = -16, 7 rs — 4 ra = 44. 14. 15. 1 -i = 2, 1 A -1 = 3, 1 -h = 4. r + S +!; + M = = 2.8, r — S + i — M = = 7.2, r + 2s H-Si!- 6m = = 7, r + s- -Sif + M = -■ 1.7. 2 X +5= y =•26, 4 y _10 z = 3, 1 a; 25 " 2 ■ LINEAR SYSTEMS 225 Note. Perhaps tte student wonders if a linear equation in three variables has a graphic representation. It may partially satisfy his curiosity to say that by means of three axes at right angles to each other such a representation, though beyond the scope of this book, is possible. Further, the points whose x, y, and z values satisfy the equation lie in a flat surface called a plane. Two such surfaces may intersect in a straight line, and the system of two equations which the surfaces represent is satisfied by the x, y, and z values of any point on this line. Three such surfaces may intersect in a single point, and the system which the surfaces represent is satisfied by the X, y, and z values of this point. In the systems of equations in three variables on page 224 the student is really finding the coordi- nates of the point of intersection of three planes. Those who desire more information on this subject are referred to books on analytic geometry. Since space has but three dimensions, this method of representa- tion of linear equations in two or three variables cannot be extended to equations containing four or more. PROBLEMS 1. Find three numbers of which the sum of the first and second is 76, the second and third 54, and the first and third 68. 2. The sum of three numbers is 58. The sum and the quotient of two of them are 24 and 2 respectively. Find the numbers. 3. The perimeter of a triangle is 64 feet. Two of its sides are equal, and the third side is 10 feet longer than either of the first two. Find the length of each side. 4. The sum of two sides of a triangle is 52 feet and the difference is 12 feet. The perimeter of the triangle is 93 feet. Find the length of each side. 5. The sum of the two sides of a triangle which meet at one vertex is 41 feet, at another vertex 48 feet, and at the third vertex 43 feet. Find the length of each side. 6. The sum of three numbers is 26. The quotient of two of them is 9, and the sum of these two divided by the third is 3J. Find the numbers. 226 COMPLETE SCHOOL ALGEBRA Fact from Geometry. The sum of the three angles of any tri- angle (each angle being measured in degrees) is 180 degrees. 7. Two of the angles of a triangle are equal, and their sum is equal to the third. Eind the number of degrees in each angle. 8. Two angles of a triangle are equal, and their sum is ^ the third. How many degrees are there in each ? 9. Angle J. of a triangle is 17 degrees larger than angle B, and angle B is 20 degrees larger than angle C. How many degrees are there in each ? 10. The sum of two angles of a triangle is 36 degrees more than the third, and the third is six times the difference of the first two. How many degrees are there in each ? 11. A and B together can do a piece of work in 2 days, A and C in 3 days, and B and C in 4 days. Find the time re- quired by each alone and by all together. 12. Two pumps together can fill a tank in 4 hours. The first of these and a third together can fill the tank in 6 hours. All three together can fill the tank in 3^j hours. Find the num- ber of hours required by each alone. 13. The sum of two fractions having the same denominator is 6. If 1 be added to the numerator of the first, and 1 be subtracted from the numerator of the second, the resulting fractions ■will be equal. If 22 be added to the denominator of ' each fraction, the sum of the resulting fractions is \. Find the fractions. 14. The sum of the digits of a 3-digit number is 15. The units' digit exceeds the tens' digit by 5. If 396 be added to the number, the result is expressed by the digits in reverse order. Find the number. 15. If the tens' and units' digits of a 3-digit number be in- terchanged, the resulting number is 64 greater than the num- ber. If the tens' and hundreds' digits be interchanged, the LINEAR SYSTEMS 227 resulting number is 360 less than the number. The sum of the digits is 10. Find the number. 16. The sum of the 4 digits of a 4-digit number is 9. The units' digit is twice the thousands' digit, and the tens' digit equals the hundreds' digit. If 2997 is added to the number, the result is expressed by the digits in reverse order. Find the number. Fact from Geometry. The sum of the angles of any quadri- lateral (a closed figure bounded by four straight lines) is 360 degrees. 17. Find the number of degrees in each angle of a quad- rilateral in which the sum of the first and second angles is 200 degrees, the sum of the second and third 180 degrees, and the sum of the second, third, and fourth 256 degrees. 18. The sum of two opposite angles of a quadrilateral is 180 degrees and their difference is 30 degrees. The difference of the other two angles is 36 degrees. Find each angle. 19. The sum of two opposite sides of a quadrilateral is 30, the sum of the other two sides is 35, and two adjacent sides are equal. The sum of the equal sides is less by 17 than the sum of the other two. Find each side. 20. A, B, and C had together $300. A gave to B and C as many dollars as each of them had, after which B gave to A and C as many dollars as each of them then had. They then had equal amounts. How many dollars had each at first ? 21. A, B, and C had together |192. A gave to B and C as many dollars as each of them had, after which B gave to A and C as many dollars as each of them then had; and, lastly, C gave to A and B as many dollars as each of them then had. They then had equal amounts. How many dollars had each at first? CHAPTEE XXI SQUARE ROOT AND RADICALS 89. Square root of algebraic expressions. Since then the square root of f + 2tu + u^ = ±{t+u). A study of this last form will enable us to extract the square root of any polynomial which is a perfect square. Obviously the square root of t^ (the first term of the trinomial) is t, the first term of the root. If if^ is subtracted from the trinomial, the remainder is 2 tu-\- u^. The next term of the root (u) can be found by dividing the first term of the remainder (2 tu) by 2t, (twice the term of the root already found). The work may be arranged thus : i^ + 2tu + u- \t + u Trial divisor, 2 1 Complete divisor, 2t + u 2tu + u^ 2 fa + w^ = (2 i + u) u . Therefore the required roots are ±{t+ u). The foregoing process is easily extended to extracting the square root of the polynomial 4 a;* — 20 x^ + 37 a;^ — 30 a; + 9, whose square root contains three terms, as follows : 4a;*-20a;8+37a;2_30a;+9[2K2--5x + 3 (2a;2)2= 4x4 First trial divisor, 2 ■ 2a;2=4a;2 First complete divisor, 4x^ — 5x Second trial divisor, 2(2x^-5x)=iz^-10x Second complete divisor, 4a;2_i0a;+3 ' 228 -20x8 + 37 a;2 -20x^+25x^={ix'^-5x){-5x) 12a;2_30a; + 9 12xg-30a:+9=(4a:g-10a: + 3)3 SQUARE ROOT AND RADICALS 229 Therefore the required roots are ±{2x^ — 6x + S). The term 2a;2 -jyas obtained by taking the square root of 4 a;*; the second term, - 5x, by dividing - 20a;3 by the first trial divisor, 4a;2; and the third term, 3, by dividing l^afl by 4x2, the first term of the second trial divisor. The method just illustrated may be stated in the EuLE. Arrange the terms of the polynomial aocordinff to de- scending powers of some letter in it. Extract the square root of the fir^ term. Write the result {with plus sign only) as the first term of the root, and subtract its square from the given polynomial. Double the root already found for the first trial divisor, divide the first term of the remainder by it, and write the quotient as the second term of the root. Annex the quotient just found to the trial divisor, making the complete divisor ; multiply the complete divisor by the sec- ond term of the root, and subtract the product from, the last remainder. If terms of the polynomial still remain, double the root already found for a trial divisor, divide the first term, of the trial divi- sor into the first term of the remainder, write the quotient as the next term of the root, form the complete divisor, and proceed as before until the process ends, or until the required number of terms of the root have been found. Inclose the root thus found in a parenthesis preceded by the sign ±. rrote. The process of extracting the square root of numbers was familiar to mathematicians long before they knew how to find the square root of polynomials. This is consistent with the fact that the development of the methods of performing operations on literal number symbols generally followed and grew out of the similar operations on numerals. The application of the rules for extract- ing the square root of numbers to that of polynomials is generally ascribed to Recorde (1510-1558), who was the author of the earliest English work on algebra that we know. This book, which bears the title "The Whetstone of Wit," gives an accurate idea of the alge- braical knowledge of the time, and had a very wide influence. 230 COMPLETE SCHOOL ALGEBRA EXERCISES Extract the square roots of : 1. a^ + 3a^ + 2a^ + 2a + l. ' 2. 24:x^ - 32x + 16 + x^ - 8x^ 3. 21 c2 + c^ + 20 c -10 c' + 4 4. n^ + 9n^ + 10n' + 25 - 6n* - 30n. 5. 19 a^ - 11 a* + Aa<'- 30 a + 4:a^ + 14a» + 25. 6. c* - 4 o'd + 6 c^t^^ - 4 crf« ,+ e^*. 7. 30 a;?/' + 25 y* - 11 xY - 12 a;'y + 4 x*. , 8. - 36 a*x + 36 aV + 9 a« - 24 aV + 16 a;* + 48 ax^. 9. 9 c^ - 2 a^b^c" + 4 a'd'c + a*b^ - 12 abo^. 10. 2 a^Kc' — 4 Kc' - 4 a^a;^ + 4 a;^ + c« + a^x"". IL 4- a^ 4 4c^ c^ 5 "'"25' 13. x^ — 4.x^ + 5x^ — 2x + \. ^z"" 4 a;-' y a; Solution : Arranging terms in descending powers of x and applying the rule just stated, we obtain the following : 4£2 12a y"^ y .yT jt 4x \y / y 2f + 3. J' 2a; y ix + 3 y 12 a: + 7 _1 \y I y 2a: a; 4a:^ X 4a:2 \ y 2 a;/ V 2 a;/ Therefore the square roots are zt (?-f+3-X). SQUARE ROOT AND RADICALS 231 15. x* + 6x'> + ^ + 2x + i-- o 9 3 9 3 ^ 9 127 m^ , 25m* „ , 10m» 18. 9e*-12c8 + 4c2-- + i + 6. e c* m\ ?i2 17 5m , 5n n^ m^ i n m ■ 26c^'^ c» "^5c2 a^c"^ a*' 21 -^-^q4.1^_^4.-2- 12c • 4c* "^"^ 25a* c^ "^5ac 5a'' 22. Extract the fourth root of the expressions in Exercises 2 and 6 on the preceding page. Hint. The fourth root of a number equals the square root of its square root. 90. Square root of arithmetical numbers. Since 1 = I'', and 81 = 9^, a 1-digit or a 2-digit square has only one digit in its square root. And as 100 = 10^, and 9801 = (99)=, a 3-digit or a 4-digit square has iwo digits in its square root. Also 10,000 = 100^, and 998,001 = (999)^ ; hence a 6-digit or a 6-digit square has three digits in its square root. The preceding examples illustrate the relation between the number of digits in a number and the number of digits in its square root. They also suggest a method of obtaining the first digit in the square root of any number. Eor example, take the four numbers 78'43'66, 7'84'35, .98'01, and .03'27'40, and beginning at the decimal point in each number, point off periods of two digits each, as indicated. Any incomplete period on the right, as in ,03'27'4, should be 232 COMPLETE SCHOOL ALGEBRA completed by annexing one zero ; thus, .03'27'40. Now the first digit in the square root is the greatest integer whose square is less than or equal to the left-hand period. This is true whether the latter contains two digits or one. Therefore the first digit in the square root of 78'43'56 is 8, in the square root of 7'84'35 is 2, in the square root of .98'01 is 9, and in the square root of .03'27'40 is 1. Moreover the number of digits in the square root of a per- fect square is equal to the number of periods, provided any single digit remaining on the left is counted as a period. Just how t and u are involved in the square of (t + m), or t'^ + 2 tu + u^, is obvious on inspection, because the parts t^, 2 tu, and u"^ cannot be united into one term. In the square of an arithmetical number, however, the parts are united. Thus (53)2 ^ (50 + 3f = 2600 + 300 + 9 = 2809. Now it is clear how 50 and 3 are involved in 2500 + 300 + 9, but it is not plain from 2809 alone. Pointing off, however, enables us to discover at once the first digit, 5, which is equivalent to 5 tens, or 50. With the exception of pointing off, the method of ex- tracting the square root of an arithmetical number does not differ greatly from the method of extracting the square root of an algebraic expression. In fact, the formula, the square root of t'^+ 2tu -\- u^ = ±(t -\- u), can be used to explain the two processes. If t denotes the tens and u the units, t'^+2tu + lu^ is closely related to 2500 + 300 -\- 9, f being 2600, or (50)^ ; u" being 9, or 3^ ; and 2 tu being 2 • 50 • 3. Therefore the process of extract- ing the square root of 2809 may be based on these relations and the work arranged as follows : 2809|50j-3 f = 2500 2!; = 2-60 = 100 2!!-|-M = 100-f3 309 309 = (lQ0-f3)3=(2i;-[-w)w = 2to+t Therefore ± 53 are the two square roots of 2809. SQUARE ROOT AND RADICALS 233 If the number has three digits in its square root, the work and explanations may be arranged thus : l'74'24 |100 + 30 + 2 t^ = 10,000 10000 = 10 tens squared First trial divisor, 2 ^ = 2 • 100 = 200 First complete divisor, 2 i! + M = 200 + 30 = 230 Second trial divisor, 2 if = 2 • 130 = 260 Second complete divisor, ' 2 if + M = 260 + 2 = 262 7424 6900 = (2 • 10 tens + 30 units) 30 524 524 = (2 • 13 tens + 2 units) 2 Therefore ± 132 are the square roots of 17,424. When the method and reasons for the process have become familiar, the work may be shortened by omitting the explana- tions and unnecessary zeros as follows : 28'09[53 1'74'24[132 25 1 103 309 309 23 74 69 262 524 524 The method just illustrated for extracting the positive square root of a number is the one commonly used. For it we have the Rule. Begin at the decimal point and point off as many periods of two digits each as possible : to the left if the numiber is an integer, to the right if it is a decimal ; to both the left and the right if the number is part integral and part decim,al. Find the greatest integer whose square is equal to or less than the left-hand period, and- write this integer for the first digit of the root. Square the first digit of the root, subtract its square from the first period, and annex the second period to the remainder. 234 COMPLETE SCHOOL ALGEBRA Double the part of the root already found for a trial divisor, divide it into the remainder (omitting from the latter the right- hand digit), and write the integral part of the quotient as the next digit of the root. Annex the root digit just found to the trial divisor to make the complete divisor, multiply the complete divisor by this root digit, subtract the result from the dividend, and annex to the remainder the next period for a new dividend. Double the part of the root already found for a new trial divisor and proceed as before until the desired member of digits of the root have been found. After extracting the square root of a number involving deci- mals, point off one decimal place in the root for every decimal period in the number. Check. If the root is exact, square it. The result should be the original number. If the root is inexact, square it and add to this result the remainder. The sum should be the original number. Sometimes in using a trial divisor we obtain too great a quotient for the next digit of the root. This happens in obtaining the second digit of the square root of 32,301, where 2 into 22 gives 11. Obvi- ously 10 and 11 are both impossible. If 9 is tried, we get 9 • 29, or 261, which is greater than 223. Simi- 3'23'01 [1 larly 8 is too great. But 7 • 27 = 189, which is less 1 than 223. Therefore 7 is the second digit of the root. 2 1 223 With practice, in cases like the one Just explained, the student will be able to look ahead and decide mentally on the proper digit of the root. Occasionally the trial divisor gives a quotient less than 1. This indicates that the required root digit is 0, which should be written in the root. The next period should then be brought down. An instance of this kind occurs in finding the second digit in the square root of 9'42.49. The quotient of 4 -^ 6 9'42.49(3 is |, which is not an integer. Therefore the second 9 digit of the root is less than 1. Then the next period, 6 I 42^ 49, should be brought down. The new trial divisor will be 60, which will give 7 as the third digit of the root. The work can easily be completed, giving 30.7 as the square root. SQUARE ROOT AND RADICALS 235 An attempt to extract the square root of 2 by annexing decimal periods of zeros and applying tlie rule becomes a never-ending process : The number 2 has no exact square root, and no matter how far the work be carried, there is no final digit. As the work stands, we know that the square root of 2 lies between 1.414 and 1.415. It is correct to say that 1.414 is ap- proximately the square root of 2, or that it is the square root of 2 to three decimal places. If a closer approximation is desired, it can be obtained by extracting the square root to four or more decimal places. 24 2.00'00'00[L414 1 100 96 281 400 281 2824 11900 11296 A common fraction, or the fractional part of a mixed num- ber, should be reduced to a decimal before extracting the square root, unless the root is seen to be exact. EXERCISES Extract the square root, correct to three decimal places, of : 1. 6241. 5. 5. 9. .0035 2. 16129. 6. 7.135. 10. JgL. 3. 223,729. 7. .6279. 11. If 4. 2. 8. .0451. 12. ^. Fact from, Geometry. In the ad- jacent right triangle a? + V^ = c^ \ a and h are 'called the legs ; and c, the side opposite the right angle, a is called the hypotenuse. If leg a is 8 and leg h is 15, then substituting in a^ -\-W- = c^ gives 64 + 225 = a". Whence 289 = c^ and c = ± 17. Since — 17 is not a practical answer, it is rejected. rind the hypotenuse and the area of a right triangle whose legs are : 13. 84 and 13, 14. 133 and 156. 15. 646 and 812. 236 COMPLETE SCHOOL ALGEBRA rind the other leg and the area of a right triangle in which the hypotenuse and one leg are respectively : 16. 65 and 66. 17. 397 and 325. In rectangle ABCD, line DB is called a diagonal. 18. Find the diagonal of a rec- tangle whose adjacent sides are 24 feet and 143 feet. 19. One diagonal of a rectangle is 401 and one side is 399. Find iJ LI the other side and the area. 20. One diagonal of a rectangle is 677 and one side is 52. Find the perimeter of the rectangle. 21. A rectangle is 7 yards longer than it is wide. Its perim- eter is 102 feet. Find one diagonal. 22. One diagonal of a square is 74 meters. Find the side. 23. The side of a square is 52 inches. Find one diagonal. 24. A rectangle is 2.4 times as long as it is wide. One diagonal is 52. Find the length and the width. 25. The width of a rectangle is 25% less than the length. The diagonal is 100. Find the area. 26. The length of a rectangle is 10. The diagonal is twice the shorter side. Find the width. Fact from Geometry. A line drawn from one vertex of an equi- lateral triangle to the middle point of B the opposite side is perpendicular to it. Then in the equilateral triangle ABC, if D is the middle point of A C, BD is the altitude ; and b3'=ab'-ad' = ab'-(^\. \ 2 / J' S 'C 27. If BC in the adjacent triangle is 6, find BD and the area of the triangle. SQUARE BOOT AND RADICALS 237 28. If ^ C is 10, find BD and the area of the triangle. 29. If BD is 10, find AB and the area of the triangle. 30. The perimeter of an equilateral triangle is 36. Find the altitude. 31. The altitude of an equilateral triangle is 25 centimeters. Find one side. Note. A method of extracting the square root of numbers not unlike that in use to-day was employed by the Greek, Theon, about 350 A.D. In the Middle Ages square roots were extracted with a fair degree of accuracy by using the formulas of approximation : (1) Va' + a; = a + ^. (2) Va2 + X = a + ; X 2a ^ ■' 2a+l The true value of the square root of the number was proved to be between the results obtained by these expressions. Thus if V65 was desired, it was noticed that 65 = 64 + 1, and from (1) V65 = Vein = VS^+l = 8 + -A- = 8 1, while, from (2) , V^ = Vein = Vs^ + i = 8 + — ^—r = 8 1. ^■0 + 1 Thus the true value of VoS is between these two numbers. This method was known to the Ara,bs. It should be kept in mind that the use of decimal fractions and of the decimal point was not common until the eighteenth century. Consequently the complete application of the method of extracting the square root given in the text is comparatively recent. 91. Radicals. All the numbers of algebra are in one or the other of two classes — real numbers and imaginary numbers. Thus 3, — 5, V2, |, and 1.763 are real numbers. Eeal numbers are of two kinds — rational numbers and irrational numbers. A rational number is a positive or a negative integer or a num- ber which may be expressed as the quotient of two such integers. Thus 7, |, 4.237 are rational numbers. A rational number can be obtained from the number 1 by carry- ing out the operations of addition, subtraction, multiplication,- and division, which are therefore called rational operations. 238 COMPLETE SCHOOL ALGEBRA Any real mimber -wliieli is not a rational number is an irra- tional number. Thus V2> Vi, V7, are all irrational numbers, and cannot be expressed as the quotient of two integers.* The 'v2 to six places is 1.414213. And it can be proved that the digits in the decimal portion never repeat themselves in groups of digits which have a definite order, however far the process of extracting the root be carried. Hence the deci- inal portion of the root is said to be non-repeating. For exam- ple, .121212 ■ ■ • is a repeating decimal. As a never-ending decimal ■which does not repeat cannot be expressed as the quo- tient of two integers, the V^ is an irrational number. The V2, V4, etc., are also irrational numbers. It is beyond the scope of this book, however, to show how their approximate values are obtained. Symbols like V— 1, V— 4, "%— 16, were mentioned on' page 102. Such symbols arise when we express an even root of a nega- tive number. These indicated roots are called imaginary num- bers and will be treated later. A radical is an indicated root of any algebraic expression. Thus -yi, Vo, '\o,, and ■\^--Ex + Q are radicals. A. surd is an irrational root of a rational number. Surds are always irrational numbers. Thus V5, -\/7, etc., are surds. Though no irrational numbers can be expressed exactly in decimals, we can represent a few surds by the lengths of lines. Thus in the right triangle ABC, if AB = AC = 1 inch, BC = V2 inches. If AB were 2 inches and AC were 1 inch, BC would be VS inches. * See Hawkes's "Advanced Algebra," page 52. SQUARE ROOT AND RADICALS 239 An irrational number is not necessarily a surd. The length of the. circumference of a circle divided by the length of its diameter gives a number which is not rational. The symbol for this number is the Greek letter tt (pronounced pi). The approximate value of tt is ?y2- ; more closely it is 3.1416. The number which ir represents is a never-ending, non-repeating decimal whose value correct to ten decimals is 3.1415926535. Strictly speaking, the -s/tt is not a surd, nor iS an expression like V V3 -f- 2 a surd. The vVs is a surd, for it can be written -^3, as we shall see later. The index of a radical is tlie numerical, or literal, part of the radical sign. The index determines the order of the radical and indicates the root to. be extracted. In 5-^, 3 is the index, the radical is of the third order, and the coeflBcient is 5. The radicand is the number-, or expression, tinder the radical sign. In -y/d and -^mx, 9 and ax are the radicands. Ppr a given index the principal root of a number is its one real root, if it has but one ; or its real positive root, if it has two real roots. The principal root of "v'- 27 is - 3. That of -vTs is + 2, not - 2. Radical expressions may be written in two ways, with radi- cal signs or with fractional exponents. The relation between the two will now be explained. To do this it is necessary to extend the meaning of the term exponent, which, as defined on page 7, applied to integral exponents only. We shall assume that the laws which govern integral exponents hold for fractional exponents also. The fact that x^-x^ = x°, illustrates the more general law, x''-a^ = 83"+'', Vhere a and b represent either integers or fractions. Accordingly x^ ■ x^ = x^"^^ = x^ oy x. Since a;' multiplied by itself gives x, x^ must be another way of writing the square root of X. 240 COMPLETE SCHOOL ALGEBRA Hence VJ may be written a;^. Then 4^ = Vi = 2, and (25 a")^ = V25^ = 5 a. Further, a;* ■ a;^ • a;^ = a;'' = x. And since x^ is one of the three equal numbers whose product is X, x^ is another way of writing the cube root of x. Therefore Vx may be written x^. This means that 8^ = "v^ = 2, and 64^ = 4. Similarly ■\x = a;*, and ■yfx = a;«, etc. In general terms, -vGc = X". !Now xi = xi.x^.xi=.{xiy^i-^y, and xi=x^-i=(x^)i = V^^ Hence (A^y = V^, and 4i=(4^)»=(V4)',orV4S for both (a/4)' and Vi^ equal 8. In like manner, j = a^^.xh=(JY=(-^y. But also -^= (x^)i = {x-x)^ = x^-x^ = xi. Therefore i-^y=yc"-v/& = a^/b. Note. Although the Arabs ■were by no means able to state all the rules explained in this chapter, it is interesting to note that they did recognize the truth of a few of them. For instance, a writer about 830 A.D. gives, in his own notation, of course, the facts contained in the formulas a Vs = Va^, and Va ■\b = Voft. A radical is in its simplest form when the radicand : (a) Is integral. (6) Contains no rational factor raised to a power which is equal to, or greater than, the order of the radical. (c) Is not raised to a poifjer, unless the exponent of the power and the index of the root are prime to each other. For the meaning of (a), (b), and (c) study carefully the EXATHPLES Of (a) : 1. Vl = Vi = vT^ = Vi Ve = i V6. 2. 6Vj = 6Vt = 6v^ = 6-4 V3 = 2V3. 3- VA=VH=7^r i2=v^s /r2 = i^. A- IHE- \5x'~'\25x^~\ /4^ = V4a;^-a; = V(2a;2) ^-.15x-i- 25x' 5x Oi(b): 1. V4a;^ = V4a;^-a; = V(2a;2)2.a: = 2a;2V^. 2. 5 -v'24a;^= B-^Sx '- 3a:^ = 5 ^(2 xf ■ 3 x^= 10a; -V s^^ 3- Vie - 8 V2 = V4(4-2V2) = 2 V4 - 2 V2. Of (c) : 1. -Vi = -V2^ = 2* = 2^ = V2. 2. 79 = ^^=3* = 3^ = V3. 3. Va¥ = aM = a^6 = 6 V^. SQUAEE ROOT AND RADICALS 243 EXERCISES Express in s simplest form: 1. Vi2. 12. Vf- 2. V32. 13. 3V|. 3. V75. 14. 5Vf. 4. 2V60. 15. 7i. 5. -yiO. 16. 4^i. . 6.^^54. 17. 8>y|. 7. V20. 18. vi-ar- 8. -VSOOO. 19. vi+a)^- 9. 2-v^. 20. V2 +(§)'■ 10. VI- 21. |F^' ^ V5- iJ. Square : 47. ^. 48. 2^3. 49. 2-v'l2. 50. V2 - Vi Cube: 59. 2-v'3. 60.. 3 72. 61. (V2)l Simplify : 68. R'' +(j VJ^ 51. 3^0! + Vs. 52. (2^3)^ 53. 2a-^8x. 54. V2 + V3. 62. (V3)'. 63. VS-Vg! 64. 2V2+V3. 55. \/4 + 4V3. .V2 e,.^..I(fVB-fJ. 71. 72. R 56. ^\/9-9V2. 57. V2 + -v^. 58. 2 ^3 - V2. 65. V3-V2. 66. V2+V2. 67. ■v'2-V3. MAI R^ Ry/2- V3 70. 73. >J^^^^^. SQUARE ROOT AND RADICALS 249 -^ ^^\V/s_|V2" 75. (2 ^2 + a/2 2 / ' V 2 7?V2-V2 2 76. Pind the value oi x^ + 4:X + lit x =— 2 + VS. 77. Find the value of «=- 2 a; - 3 if a; = 2 + VS. 78. Find the values of a;^ - 4 a; - 1 if a; = 2 ± VS. 79. Find the values of3a;2 + 3x-6ifa; = ^ ± ^ Vi3. 80. Do the values x = 2 ± Vs satisfy the equation x^ — 4a; + l = 0? 81. Do the values a; = — 4 ± VS satisfy the equation x^ + 8a; + ll = 0? 82. What inference seems warranted as a result of Exercise 80 ? of Exercise 81 ? 95. Division of radicals. It is frequently necessary to find the approximate value of an expression which involves division by a radical expression. Thus 2 -j- Vs, (4 — V3)^(2 — Vs), —pz ! and — ;= := are types which often occur. , To find the approximate value of 2 -=- VS, we may extract the square root of 3 to several decimal places and then divide 2 by the approximate root obtained. Both of these processes are long and one of them is unnecessary. For, writing 2 -h VS in the form and multiplying both terms of the fraction by Vs gives — The process of finding the 2 a/S approximate value of — r — involves but one long operation, o Similarly the process of finding the approximate value of VT -r- ( Vt — V2) involves three rather lengthy operations, — the extracting of two square roots, and one long division. The labor of two of these operations can be avoided. 250 COMPLETE SCHOOL ALGEBRA — R Evidently V? -^ ( V7 — V2) = — ^ = • Multiplying both terms V7-V2 of this fraction by V7 + Vs gives •v7(y7+y2) ^ ^^ (V7-V2)(V7+V2) 7 + V14 7 + V14 „. ,. ,, , ,7 + Vl4. , , — - — - — ) or • i inding the value of — — involves only i — £t ■ o o one long operation, extracting the square root of 14. As in tlie two preceding illustrations, division of radicals is usually an indirect process performed by means of a rational- izing factor of the divisor. One radical expression is a rationalizing factor for another if the product of the two is rational. A rationalizing factor for Vs is Vs, for Vs • ■sfz = 3. For Vs a rationalizing factor is -vi, since V2 ■ V4 = Vs = 2. Similarly •v7 — V2 is a rationalizing factor for "vf + V2, as their product ( Vt - V2) ( V7 + Vi) = 7-2 = 5. In like manner (3 Vs - 2 V3) (3 V5 + 2 Vs) = 45 - 12 = 33. Therefore 3 Vo— 2 V3 is a rationalizing for 3 Vs + 2 Vs. The binomial radicals of the last two illustrations are of the general types Vo. + Vo and Va — Vj. Such binomials are called conjugate radicals and either is a rationalizing factor for the other. If a and h are rational, the product ( Va + Vo) ( Va — Vi), or a — J, is a rational number. There are many other types of radical expressions which have rationalizing factors. They seldom arise, however, and are too difficult for treatment in elementary algebra.* An irrational expression may have more than one rationalizing factor. Thus VI8, Vs, and V2 are rationalizing factors for Vl8. For VI8 • VI8 = 18 ; and vTs • Vs = Vl44 = 12 ; and Vl8-V2 = Vse = 6. Similarly, since ^4 • "v^ =^8 = 2, and -v/I • 4^l6 = "v'ei = 4, both -v2 and VlB are rationalizing factors for -ifi. In practice it is best to choose for monomials the rationalizing factor which has the least radicand. * See Hawkes's "Advanced Algebra," page 62. SQUAKE BOOT AND RADICALS 251 EXERCISES Determine a rationalizing factor for each of the following expressions and find the product of the expression and the factor : 1. V5. 4. Vs. 7. 2 2. Divide 6 V5 by 3 Vs. Solution : By direct division, coefficient by coefficient and radicand by radicand, 6 V5 -i- 3 V3 = ■= = 2 -y -, which becomes - Vl5. 3V3 >3 3 3. Divide 8 V12 by -2^6. Solution : Reducing the surds of the same order, and then proceed- ing as in direct division. ,_ ^,_ 8(12)^ 4(12)i . N: 3 V12 ^ 2 Ve = -^^ = -W- = 4 V- 12 ■ 12 • 12 2(6)* (6)i " 6-6 = 4V2-2-12 = 4V48. 252 COMPLETE SCHOOL ALGEBRA If the monomial divisor is a surd, it is always possible and often far more convenient to divide by means of a rational- izing factor of the divisor. 4. Divide 4 by Vs. 4 4V3 4V3 Solution : 4 h- VS ' V3 VsVs" 5. Divide 6 by -y/S. solution: 6^V3=A= ^^9 ^eVg^ej^^ ^ V3.V3-V9 V27 3 6. Divide Vs by "v^. solution: V3 -^ ^2 = ^ = ^ " ^^ = gJlJ! = gJliJ V2 V2-VI 2 2 = »V3^Tp= 1^432. When the divisor is a binomial (or polynomial) radical, the practical method of division is an indirect method by means of a rationalizing factor. 7. Divide 8 by 3 + Vt. Solution : 8 -^ (3 + Vt) = ? = B{S-^) 3 + V7 (3 + V7) (3 - V7) = ?i^=12-4V7. 8. Divide Vs + Vs by 2 Vs - Vs. Solution : ( Vs + Vs) -^ (2 Vs - Vs) ^ V5 + Vs ^ (V5 + V3)(2V5 + V3) 2V5-V3 (2V5-V3)(2V5+ V3) IO + 3VT5 + 3 13 , 3 /T- = 20^3 = i7 + 17^- For division of radicals we may use the Rule. Write the dividend over the divisor in the form of a fraction. SQUARE ROOT AND RADICALS 253 Then multiply the numerator and denominator of the frac- tion by the rationalizing factor of the denominator and sim- plify the resulting fraction. Every irrational algebraic expression containing nothing more com- plicated than rational numbers and radicals has a rationalizing factor. To find this factor for any given irrational expression is a problem which requires considerable algebraic training. At the present time it is wholly beyond the student to find the rationalizing factor of even so simple an expression as the denominator of the fraction — The approximate value of such a fraction can be obtained, however, by dividing the sum of the approximate values of the roots in the numerator by the sum of the approximate values of the roots in the denominator. EXERCISES Perform the indicated division. : 1. Vio--V2. 3V12 4ViO 2. (18)i-T-(3)K ■ V6 ■ 8V5. Ve 6. 8 Vl5 -=- 4 V5. 9. 8 -=-4 Vs. 3 V18 _io_ 3V2 6-2V2- ■ 2V5' ^°' 15V8" 11. (V6+ Vl8)^3V2. ^^ a'^Vo 12. (V12- V24)--2V3. " a-\/bo ^3 6V10 + 4V15-V20 21. •v'le^-v^i. 14. (12+V3+V5)-^V6. 2^ 15. -^^-^+^^ ^'- ^- 2V2 „i , , 16. (V6-|-2)-=- Vl25. 17. ^/^^Vi. 25.^^^^. 18. {xyf-^x^. 26. -v/4-f-72. 19. a-y/bc^dVo. 27. V2-^V4. 254 COMPLETE SCHOOL ALGEBRA 28. 4^ -^6^. 34 ^ _. 29. 6i^4i '2-^3 30. -^-.V27. 35. 4^(V3-V2). 31.^^^/!. 36. V5_-.(3V2 V5). 32. ' "■ ' * 37. (V5-V3)^(V5 + V3). ^72« • N81 38 ^^^ + ^^- ™o ,; / /^ , o^ 4V3-2V2 33. 5h-(V5 + 2). Hint. Study Examples 7 39. (2V5-_3V3)_ and 8, page 252. 5 V5 — 5 Find to -tliree decimals tKe approximate values of the fol- lowing : 40. 3 + V2. 7±V6 45 2V5+I . 41. 14-6 V7. 3 ■ " 3 V5-V 3 42. 6 ±2 Vs. 44. 3V6-H2V5. 46. V2 - VS. *''• "^i 8^7 v^ ' given Vi = 1.5874. V3 + V4 + V5 Change the following fractions to equivalent fractions hav- ing rational denominators : ( Va 4- Vi) ,, m-\/n + a'^b 48. )— = ;=:(- 51. (V^_ V6) ,„ 2VK-Va 49. -^ 7=- 52. Va; + 3 Va 50. ii:^±^. 53. VS- Vr Perform the indicated division : 54- (x — Va + i) H- (a; + Va+T). 55. ( Va + Vs) -H ( Vc + V5). 56. Is there any real distinction between the direction which precedes Exercise 48 and that which precedes Exercise 54 ? m V»i — a VS Va - 2 - 2 Va - 2 -F 2 2 -^2 + V2 SQUARE ROOT AND RADICALS 255 PROBLEMS (Obtain answers in simplest radical form.) 1. One leg of a right triangle is 10 and the other is 5. Find the hypotenuse. 2. The hypotenuse of a right triangle is 10 and one leg is 5. Find the other leg and the area. 73 3. The hypotenuse of a right triangle is R and one leg is -^ ■ Find the other leg and the area. 4. Find the diagonal of a square whose side is 10. 5. Find the sides and the area of a square whose diagonal is 10. 6. Find the sides and the area of a square whose diagonal is2R. 7. The side of an equilateral triangle is 12. Find the alti- tude and the area. 8. The side of an equilateral triangle is S. Find the alti- tude and the area. 9. The altitude of an equilateral triangle is 10. Find the side and the area. 10. The legs of a right triangle are equal. Its hypotenuse is 20. Find the legs and the area of the triangle. 11. The legs of a right triangle are equal and its area is 32. Find the hypotenuse. 12. The legs of a right triangle are -^ and -^ ■ Find the hypotenuse. 13. One leg of a right triangle is — ■ The hypotenuse is R. Find the other leg. 14. The legs of a right triangle are R and -^ ( Vs — 1 ). Find the hypotenuse. 7? ?? 15. The legs of a right triangle are -^ and R —-k "v3. Find the hypotenuse. 256 COMPLETE SCHOOL ALGEBRA 2 R I 16. The base of a certain rectangle is -r- V 4 — Vs and V4 + Vs the altitude is 9 22 — ITind the area of a second rectan- gle five times as long and three times as wide as the first. 96. Factors involving radicals. In the chapter on factoring it was definitely stated that factors involving radicals would not then be considered. This limitation on the character of a factor is no longer necessary. Consequently many expressions . which previously have been regarded as prime may now be thought of as factorable. Thus 3 a;^ - 1 = ( Vs x + l)(V3 a; - l) and 4 K^ - 5 = (2 a; + V5)(2 a; - Vs). It is not usual to allow the variable in an expression to occur under a radical sign in the factors. Hence, if a; is a variable, the trinomial a;^ + a; + 1 is not regarded as factorable into \x + Va; -\-\)\x — Va; +l), though the student can easily show that {x + Vx +l)(a; — Va; +l) = a;''+ x +1. Therefore in this extension of our notion of a factor it must be clearly understood that the use of radicals is limited to the coefficients in the terms of the factors. Such a concep- tion of a factor is a necessity for certain work in advanced algebra and geometry, and is very desirable in solving equar tions by factorings To restrict the use of radicals in the way just indicated is necessary for the sake of definiteness. Otherwise it would be im- possible to obey a direction to factor even so simple an expres- sion as x'^— y^; for if the variable is allowed under a radical sign in a factor, x"^ — y'^ has an infinite number of factors. Thus x^ — y^ = (x + y)(x — y) = (a; -H y){-\fx + V^)(V^ — s/y) = (a; + y){-y/x + ■\/y){-\fx + ■y/y){\^x —s/y) = etc. The extension of factoring explained above can be applied to the solution of equations as follows : SQUARE ROOT AND RADICALS 257 EXAMPLE Solve a;^ — 7 = by factoring. Solution : a^ - 7 = 0. (1) Then {x + -y/r) {x - Vt) = 0. (2) Therefore x + V7 = 0, or x =— a/T, and X — VT = 0, or a; = VT. It is apparent at once that these values check in a;^ — 7 = 0. In (2) it is obvious that if a root be substituted for the variable, one of the factors must become zero. EXERCISES Factor : 1. x^ — 6. 5. 4a;* -1. 9. a;' + 4. 2. 3a;^-4. 6. x" - 2. 10. 2a;8 + 8. 3. Sa!** — 1. 7. a;^ + 6. 11. 2a;=-8. 4. a;* -4. 8. Sx'-l. 12. 6x» + 24. Solve by factoring and cheek : 13. a;*" - 2 = 0. 18. 3a:* + 8 = Ux^. 14. a;2 - 6 = 0. 19. 5 a;* - 16 a;" + 3 = 0. 15. 2a;2-l = 0. 20. a;* + 8 a = 4 a;= + 2 aa;". 16. a;* + 6 = 5a;l 21. ax* - a;^ + 3a = 3 aV. 17. 4a;* + 5 = 12 a;'^. 22. ix* + a = x^ + Aax^ Biographical Note. Fkanqois Vieta. The reason that algebra is a uni- versal language which does not depend entirely on the nationality of the writer lies in the fact that the symbols used to indicate the various operations and relations are widely understood and adopted. This has not always been the case, and for a long time during the early history of the subject there was no accepted notation in algebra, but each man used any symbol that suited him. One of the men who did most to establish a fixed notation was Frangois Vieta (1540-1603), a French lawyer who studied and wrote on mathematics as a pastime. He was in public life during his whole career, and was well known for his ability to decipher the hidden meaning of dispatches captured from the enemy. It was he who established the use of the signs + and — for addition and subtraction, which, to be sure, had been used before his time, but were not generally accepted. He also denoted the known numbers in 258 COMPLETE SCHOOL ALGEBRA an equation by the consonants, B, C, Z), etc., and the unknowns by the vowels A, E, I, etc. He also recognized the existence of negative roots of equations, but rejected them as absurd. To denote the second and third powers of the unknown, he used the letters Q (quadratus) and C (cubus) respectively. Instead of using the sign =, he wrote aeq. {aequalis or aequatur). Thus Vieta would have written the equation k^ — 8 a;^ + 16 = 40 in the form 10-8Q + 16iVaeq.40. Before the time of Vieta this equation would have been written in a much more primitive notation. For instance, with writers only a little earlier it would appear as Cubus m SCensus p 16 rebus aequatur 40. It is easily seen that operations on equations in this form would be very hard to perform. Vieta is further distinguished as being the first man to obtain an exact numerical expression for the number tt, which occurs in geometry. His form of expression calls for an infinite number of operations which, of course, could never be performed, but the further one proceeded, the closer would be the approximation obtained. In a certain sense the familiar sign V implies an infinite number of operations, for one can never go through the process of extracting the square root of 2, for instance, and come out even. Vieta's method of denoting ir was, however, more involved than this, and made use of complicated irrational fractions. FRANCOIS VIETA CHAPTER XXII GRAPHICAL SOLUTION OF EQUATIONS IN ONE UNKNOWN 97. Graph of a linear function. An algebraic expression in- volving one or more letters is a function of the letter or letters involved. Thus 2x + S and a;^ + 5 a; — 6 are functions of one letter, x ; x^ — 2xy+ y^ and a;' + y^ are functions of two letters, x and y. . The letters of a function are usually referred to as variables. A function is called linear, quadratic, dr cubic according as its degree with respect to the variable (or variables) is first, second, or third respectively. After a function of any variable, say x, has once been given, it is convenient and usual to refer to it later in the same dis- cussion by the symbol f(x), which is read the function of x, or, more briefly, / of x. The numerical value of a linear function of x changes with every change in the variable. Thus if a; = 1, the linear function of a:, 2 a; + 3, equals 2 + 3, or 5; if a: = 2, 2x + 3 equals 4 + 3, or 7. The following table illus- trates this change further. When X = -3 -2 -1 1 2 3 f(x),2x + 3 = -3 -1 1 3 5 7 9 The relation between x and the function 2x + 3 may be represented graphically if OX continue to be the a;-axis and OF (or the function axis) replace the jr-axis of our previous graphical work. Beginning with (— 3, —■ 3) and plotting the points corresponding to the numbers in the table, we locate 259 260 COMPLETE SCHOOL ALGEBRA points A, B, C, D, E, G, and H respectively. The graph of the function /(cc) = 2 a: + 3 is evidently the straight line AH. G,L £.L &3 / 2 C,^1 ~:'' ~'- /- O I 2 3 ^ ac . --f fli. a/ g< 7 ^ Graphof/(a;)=2x + 3 This process of plotting the relation between a function and the variable contained in the function is called graphing the function. Care should always be taken in graphing to join the plotted points by a smooth curve or by a straight line, as the case may be. If the graph is not regular and graceful, it is almost certain that an error has been made in plotting the points. No equations in this book have graphs that are part straight line and part curve, or that pre- sent erratic changes in curvature. Although such curves have equa- tions, they are usually very complicated. GRAPHICAL SOLUTION OF EQUATIONS 261 EXERCISES (Exercises 1-8 refer to the preceding graph.) 1. Read from the graph the value of x when f(x) is zero. 2. Set 2x + 3 equal to zero, and solve. 3. Compare the results of Exercises 1 and 2. 4. Eead from the graph the value of x when f(x) is 4. 5. Set 2 a; + 3 equal to 4, and solve. 6. Compare the results of Exercises 4 and 5. 7. Can the value of x in 2 a; + 3 = 6 be read from the graph ? If so, read it. 8. Eead from the graph the root of 2 a; + 3 = — 2. 9. Graph the function f{x) = 6 — 2 a;. 10. Can the root of 5 — 2 x = be read from the graph just obtained ? If so, read it. 11. From the graph of Exercise 9 read the roots of : (a) 5 -2a; = 9; (c) 6^ 2a; =-3; (J) 6 -2a; = 5; (rf)5-2a;=-7. 12. Check by substitution the roots obtained in Exercise 11. 13. What, kind of a line do you expect the graph of any linear function of x to give ? 98. Graph of a quadratic function. The quadratic function /(a;) = 4 a;^ — 4 a; — 15 may be graphed as follows-: -Whena;=l,4a;'^-4a;-16 = 4-4-15 = -15,or/(a;)=-15. In like manner, the other numbers in the following table can be obtained. When X = -3 _ 2 -1 1 2 3 4 /(a;),4a:2_4a;_15 = 33 9 -7 -15 -15 -7 9 33 To represent the numbers in the preceding table conven- iently, it is necessary to use different scales 'ioi x and f(x). 262 COMPLETE SCHOOL ALGEBRA The difference can be seen from the numbers along the axes in the following figure. If we begin with (— 3, 33) and plot the points correspond- ing to the numbers in the table, we get the points A, B, C, D, E, G, H, and I respectively. Drawing a smooth curve through these points gives the graph of the following figure. This curve is called a parabola. f A L / \ / \ / \ / e 1 H \ / t \ / - \ / ■ c\ / V 7 \ s ^ / .' Graph oif(x)=ix^ — ix — 15 The student should note that it is often best to represent values of /(j:) on a different scale from values of x. He should always inspect his table of values and decide what scale to use before plot- ting a single point. The scale should be as large as possible and yet show enough of the curve to indicate its shape clearly. It will often be found convenient not to put the intersection of the axes in the center of the page. GRAPHICAL SOLUTION OF EQUATIONS 263 EXERCISES (Exercises 1,4, 7, and 9 refer to the preceding graph.) 1. Read from the graph, the values of x for which 4 x^ — 4 a; — 15 equals zero. 2. Set 4 a;^ — 4 a; — 15 equal to zero, and solve by factoring. 3. Compare the answers to Exercises 1 and 2. 4. Read from the graph the value of x for which f{x) equals 20. 5. What, then, are the roots of 4 a;^ _ 4 a; _ 15 = 20 ? 6. Solve 4 a;^ — 4 a: — 15 = 20 by factoring, and check your answers to Exercise 5. 7. Eead from the graph the roots of : (a) 4a;2-4a;-15 = 9. (&) 4a-^ - 4a; - 15 = - 7. 8. Check your answers to Exercise 7 by solving the equa- tions (a) and (i) by factoring. 9. Can you read from the graph the value of x which makes f{x) equal - 25? equal - 20? Explain. 10. Graph the function x^ — 2 a; — 4. First fill out the table : When X = -4 -3 -2 -1 1 2 3 4 5 6 f{x),x'-'ix-i = -1 -4 — 5 Then plot the eleven points and draw through them as smooth a curve as possible. 11. Is the curve obtained in Exercise 10 similar in shape to that of the preceding figure ? 12. Eead from the graph of Exercise 10 the approximate values of x which make the function a;^ — 2 a; — 4 equal zero. 13. What are the roots of the equation a;^ — 2a; — 4 = 0? 14. Can you solve the equation a;" — 2a; — 4 = by factor- ing? graphically? 264 COMPLETE SCHOOL ALGEBRA 15. If the terms of a quadratic equation be transposed so that the second member is zero, and then the function in the first member be graphed, can the roots of the original equation be read from the curve thus obtained? Explain. Solve Exercises 16 and 17 graphically and check each by sub- stituting in the original equation ; or solve the equation by factoring, and compare results with those obtained graphically. 16. a;2 — 3 a:.= 4. 17. a:^ — 4 a; + 4 = 0. 18. What peculiarity has the curve obtained in Exercise 17 ? How many roots has a:^ — 4a;+4 = 0? What values of x make a;2 — 4a; + 4 equal zero? equal to + 4 ? to — 1? to — 10? Exercises 19 and 20 cannot be solved by the factoring previously explained. They have irrational roots, but the approximate values of the roots can be obtained graphically. Solve Exercises 19 and 20 graphically. 19. x^-2x = 2. 20. x^ + 2x = 5. 21. Graph the linear function 3 a; + 4, using the same scale for X and/(x) ; then graph the function, using different scales. Compare the two lines obtained and the values of x and f(x) where each graph crosses the a;-axis and the J'-axis respectively. 22. Proceed as in Exercise 21 with the quadratic function a;" + 3 a; + 1. 99. Graphical illustration of imaginary roots. The solution of a;'' + 1 = gives x = ± V— 1. These roots are imaginary num- bers. Just now it is desirable to know that many quadratic equations have imaginary roots. It will be instructive also to see the result of an attempt at a graphical solution of a quad- ratic equation whose roots are imaginary. Tor this purpose we shall consider the three equations : I. x^-4:X-5 = 0. II. a;^ - 4 a; + 4 = 0. III. a;2-4a;+13=0. GRAPHICAL SOLUTION OF EQUATIONS 265 The graphs of the functions in the left members of I, II, and III are given in the adjacent figure. The three functions differ only in their constant terms ; for 9 added to the constant term of I gives the constant term of II, and 9 added to the constant term of II gives the constant term of III. Apparently, as the ■ constant term is increased, the graph rises. It does not change its shape, nor does it move to the left nor to the right. V \ 1 \ 1 1 \ \ 11 \ \ \ 1 \\ \ / 1 / \ \ » ^ i 1 / \ \ \ \ / 1 / \ i \ \ / t i \ V ^ \ / / J \ \ ^ J f / 1 V \ N i '^ f 1 / \ \ ^s ,J_ r'' •'' / / > V N / y \ •^V. ,1 [.- ,•'' / '' - : - \, 1 / i V y ■sj / From the graph the roots of a;^ — 4 a: — 5 = are seen to be 5 and — 1. These results are easily obtained by factoring : r'— ix— 5 = 0, or (a; — 5) (a; + 1) = 0. Therefore a; = 5 or — 1. If we imagine curve I to move upward, the two roots change in value and become the single root of curve II, which touches the a;-axis at a point where x equals 2. Solving x^— 4 a; + 4 = by factoring gives (a; - 2) (a; - 2) = 0. Therefore a; = 2. If we now imagine curve I to move still farther upward from its position II, it will no longer cut the a;-axis. Further, we do not expect that the graph when it reaches the position of III will show the roots of the equation a;^ — 4 a; + 13 = 0, as in fact it does not. The graph does show, however, that the value of a;^ — 4 a; + 13 at 266 COMPLETE SCHOOL ALGEBRA the lowest point of the curve is 9. This means that for every real value of X, positive or negative, x^ — 4 a; + 13 is never less than 9. The graph of III, then, makes clear that no real number, if substituted for X, will make x"^ — 4 x + 13 equal to zero. In the next chapter an algebraic method of solving any quadratic is given. That method wiU. later be used to show the roots of x^ — 4 X + 13 = to be the imaginary expressions 2+3 v — 1 and 2-3 V^. The preceding explanations show that the real roots of a quadratic equation can be obtained by the following steps. 1. Transpose the terms so that the second member is zero. 2. Graph the function in the first member. 3. Note the a;-distance of each point at which the curve crosses (or touches) the r-axis. The values of these x-distances are the roots of the quadratic. If the graph obtained in 2 does not cross the x-axis, the student knows that the roots are not real. He may still regard such equa- tions as having roots and refer to them as imaginary. Later he will learn precisely what an imaginary number is, and how to solve any quadratic whose roots are imaginary. EXERCISES If possible, solve graphically : 1. a;2-6a; + 7 = 0. i.2x^ + x + l = Q. 2. ic' - 6 a; + 11 = 0. 4. 2 a;^ + a; — 1 = 0. CHAPTER rXIII QUADRATIC EQUATIONS 100. Solution by completing the square. Before taking iip the work which follows, the student should review the exercises in forming trinomial squares, page 108. EXAMPLES 1. Solve a;2 + G a; — 16 = 0. (1) Solution : Transposing, x^ + G x = 16. (2) Adding 9 to each member of (2), x" + G X + d = 25. (3) Then (x + Sy = 5\ (4) Extracting the square root of each member of (4), x + 3=±5. (5) Whence a; =— 3 + 5, or 2, and a;=— 3 — 5, or — 8. Check : Substituting 2 for x in (1), 4 + 12 - 16 = 0, or = 0. Substituting — 8 for a; in (1), 64-48-10 = 0, or = 0. The 9 added to each member of (2) is the square of half the coefficient of x; that is, 9 = (§)'', or 3^. If the coefficient of a;^ is 1, the trinomial square can always be completed by adding the square of half the coefficient of x. If the coefficient of a;^ is — 1 or any num- ber other than + 1, the equation is solved as in the next example. The sign ± properly belongs to each member of (5). Thus ± (a; + 3) = i 5. This, however, gives precisely the same roots as a; + 3 = ± 5, a fact which the student can easily verify. For this reason the sign ± is put before one member only in equations obtained as was equation (5). . 267 X ■ ¥ (3) 268 COMPLETE SCHOOL ALGEBRA 2. Solve .3 a;^ - 7 a; - 20 = 0. (1) Solution: Transposing, Z x^ - 1 x = 20. (2) Dividing (2) by the coefficient of x'^, 2_7a;_20 ^ 3-3' Adding (J)^ to each, member of (3), ^'-|^ + (i)' = ¥- + il=-W- (4) Then (^x-\f = Q^-y. (5) Extracting the square root of each member of (5), r — 7 — _L 1 Y Whence a; = J ± i,'- = 4 or — |. Check: Substituting 4 for x in (1), 3 . 42 _ 7 . 4 _ 20 = 0. 48 - 28 - 20 = 0, or = 0. Substituting — | for 3; in (1), 3(-|)''-7(-|)-20 = 0. 2j5 + 85 _ 20 = 0. ly) - 20 = 0, or = 0. 3. Solve 4x^-4 a; -79 = 0. (1) Solution : Transposing, 4 x^ — 4 a; = 79. (2) Dividing each member of (2) by 4, x'^-x = -'/. (3) Adding Q)^ to each member of (3), x^ - X + ( 1 )2 = 7^» + 1 = 8^« . (4) Then (x-i)2=20. (5) Extracting the square root of each member of (5), i=±V20. (6) Whence x = 1 ± 2 Vs. (7) Now Vo = 2.2360+. Then 1 + 2 Vs = .5 + 4.472 + = 4.972+. (8) Also i-2V5 = ..5-4.472 + = -3.972+. (9) Check: Since (8) and (9) are not the exact values of x, they will, if substituted for x in (1), make its first member nearly but not quite zero. An exact check can be obtained by substituting the radical forms of the roots from equation (7) in equation (1). The check may be shortened by substituting both roots at the same time as follows : QUADRATIC EQUATIONS 269 Substituting ^ ± 2 Vs f or a; in 4 ar^ _ 4 a; _ 79 = 0. 4 (^ ± 2 V5)'' - 4 (1 ± 2 V5) - 79 = 0, 4 (I ± 2 V5_+ 20) - 2 T 8 VB - 79 = 0, 1±8V5 + 80-2t8V5-79 = 0. The radical terms vanish because the two upper signs before them must first be taken together and then the two lower signs. Therefore 1 + 80 - 2 - 79 = 0, or 81 - 81 = 0. In quadratic equations like the preceding the radical forms of the roots are often sufficient ; S.t other times values to two or three decimal places are necessary. Unless otherwise directed, obtain only the radical forms of irrational roots. The student should note that the equations like 4 a;^— 4a; — 79 = can be solved either graphically or by completing the square, but that their solution by factoring, though not impossible, is beyond him. The method of solving a quadratic equation in x illustrated in the three examples preceding may be stated in the EuLE. Transpose so that the terms containing x are in the first member and those which do not contain x are in the second. Divide both members of the equation by the coefficient of x^ (unless the coefficient of x^ is + !)■ Then add to both members the square of one half the coeffi- cient of x (in the equation just obtained), thus making the first member a perfect trinomial square. Rewrite the equation, expressing the first m,ember as the square of a binomial and the second member in its simplest form. Extract the square root of both members of the equation and write the sign ± before the square root of the second mender, thus obtaining two linear equations. Solve for x the equation in which the second m,ember is taken with the sign + and then solve the equation in which the second member is taken with the sign — . The results are the roots of the quadratic. Check. Substitute each root separately for x in the original equation. If the resulting equations are not obvious identities, simplify each until it becomes one. 270 COMPLETE SCHOOL ALGEBRA EXERCISES (Obtain the values of the radical answers in Exercises 10, IS, and 21 correct to three decimal places.) Solve by completing the square, and check results : 1. K^- 8a: -48 = 0. 3. x(x + 2)- 4. l-y = y\ 20. 25 a;2- 20 a; - 12 = 0. 5. 2 2/2_9y-|_4==o. 6. 2y2+5y = 0. 7. f-2t-15 = 0. 8. Sf-lt=&. 9. 'dv = 5v^~2. 10. a;2-2K-4 = 0. 11. ■y2-/jt;-l = 0. 12. s2-2s-3i = 0. 13. 71^ + 10/1 + 13 = 0. 14. 12!;2-25!5+12 = 0. 15. 42 + 2s2=_19«. 16. 15x2+ 4 a; = 4. 17. l-&v''=2v. 18. 20*2+ s = l. 19. 9a; + 4.= 9a;^ 30. (a; - 4)^- 3 (a; - 9) = 15. 31. (a; - 2) (a; + S) = a; (5 X - 9) - 2. 32. x--l- = 0. 35. ^ + '-^ = ^. x+2 s-2 s 2 33.-^ + 1 = 0. 36. ^--^ = 12. ?>— 62 2y — 31 — y 34 -^ + - = 37 3 + x x-5 1 38. If y = 2, solve for x the equation x^ — xy — 3 y" = — 12. 39. If X = — 3, solve x=— 4xy + x'+ i/+ 5 = for y. 2. a;2_5a;_14 = 0. 5(x + 2)=0. 20. 25x2-20x-12 = { • 21. 4x2=l-4x. 22. a;2_i_4V5x = 25. 23. x2-3 V2x + 4 = 0. 24. 1 ^^^ ^ _C 2R ' 2iJ2 25. 2x 2 7 9- 3~18x 26. 3 9 X 27. 2 +i 11 =0. 5^== 4 IQv 28. ^ + i--^ = o. 2^2 3a 29. 'l" ^« 40 = 0. 2 c QUADRATIC EQUATIONS 271 40. a;*- 5x^4 = 0. This is not a quadratic equation, but many equations of this form can be solved by completing the square. Solution : a,* — 5 a;^ + 4 = 0. X*— 5x^+ 2^5 = _ 4 + 2^5 - 9 . j-2 _ 5 _ I 3 a;2 = 4 or 1. Whence a; = ± 2 or ±1. Check as usual. 41. x^— 13 a;'' +36 =0. 44. 4 /c'' = 9 /c^ _ 2. 42. 4a;^- 5a;Hl = 0. 45. 9^4 + 12 = 31s=. 43. 9x^-37x^ + 4 = 0. 45. 4w* + 5 = 21i;l 47. Point out the error in the following : 9 - 30 = 49 - 70. 9 - 30 + 25 = 49 - 70 + 25. (3 - 5/ = (7 - 5)^. 3-6 = 7-5. 3 = 7. 101. Quadratic equations with literal coefficients. Such equa- tions are solved as in Exercises 1 and 15 which follow. EXERaSES Solve for x by completing the square, and check : 1. 2 a'^x^ — ax — 1 = 0. Solution: Transposing, 2aV — ax=l. X 1 Dividing by 2 a^ ^^ ~ 2~ ~ 2~2 ' X / 1 \^ 1 1 Completing the square, a:^ _ _ + (^- j = — + ^^ . Then /:,_iy= 9 . \^ 4 a/ 16 cr 1 3 Extracting the square root, x — -. — = ± - — ia ia 272 COMPLETE SCHOOL ALGEBRA Whence a; = - or — -— • a 2a Check : Substituting - for x in the original equation, a 2a2/iV-a/-) - 1 = 0, or 2 - 1 - 1 = 0. '"■©'-"© Substituting — — for x in the original equation, 2 a^ - -lY- a( -■^)-l = 0,orl + l-l==0. \ 2 a/ V 2 a/ '22 2. x''-ax-2a^ = 0. 9. a;^ + 4 Vaa; - 5 a = 0. 3. x^ + 2 aa; + a^ _ 4 = 0. 10. 2x'' + 9xV7i = 5 h. i. x^ + l=a + 2x. 11. aV + 2 aJ = a^ + b^ 5. 3x'' — ax = 10 a\ 12. 6 a;^ + aa; = a;. 6. 3mai + 2w2 = 2a:2. 13. x(x - b)= a(a + b). 7. a^a;" - 7 ax + 10 = 0. i _ f . A L 8. 4x=' + 4ax-3a2 = 0. ' ^ x^~ 4.x^ 15. x^ + 2 X = (XX + 2 a. Hint. x^ + (2 -a)x = 2a. .^ + (2-a).+(^J=2« + i^ 4-40 + 0" 4 2 - aV 4 + 4 a + a" / ^2-aV ■ 4 etc. 16. x= - (a + l)x + cf = 0. 1 1 ^ .5x ^ 20. ^- = 1 17. x" + te + ex + 6e = 0. « 5 a; a 18. x'^-ax + 4x-4a = 0. 21. x" + 5x + c = 0. 19. x^ + 2 a^J? = a^a; + 2 S^a;. 22. ax" + Sx + c = 0. 102. Solution by formula. The equation ax^ + bx+c=0 is the general' quadratic equation in standard form. The stu- dent solved this equation in the preceding exercise and found 2a ^ ' QUADRATIC EQUATIONS 273 The value (F) is a general result and may be used as a formula to solve any quadratic equation. The solution of a quadratic by formula requires less labor than any other method, except for such equations as can be solved by factoring at sight. Those -with considerable experiencie in algebra seldom solve a quadratic by any other method than by formula. EXAMPLES Solve by formula and check : 1. 3a;^-5a; = 8. Solution : Writing in standard form, 3 s^— 5a; — 8 = 0. Then 3 corresponds to a, — 5 to ft, and — 8 to c in the general quadratic ai? + fix + c = 0. Substituting these values in (F), where - 6 ± Vft" - 4 ac X = I 2a - r- 5) ± V25 - 4 ■ 3 (- 8) gives X = -^ i 273 ^ " 5 ± V25 + 96 5 ±11 8 , = = — - — = - or — 1. 6 6 8 Check as usual. 2. 2 k^x^=kx + l. Solution : Writing in standard form, 2 kV — ia — 1 = 0. Then a = 2k^, ft =- k, and c =- 1. Substituting these values in the formula (F), _ -(-K)± ■V(-k)'' - 4 • 2 fc'(- 1) ^ ~ 2-2k^ _ k± -y/k^ + 8fc'' _ ;i:±8^ _l _ J_ ^~ 4/fc2 ~ 4:k' ~k°^ 2 k' Check : Substituting - for x in the original equation, k ''fki=ii)^''°^'='^'- Substituting — — ■ for x in the original equation, 274 COMPLETE SCHOOL ALGEBRA EXERCISES Solve for x by formula and check : 1. 2a;^+6a; + 2 = 0. 10. 12 .t = 1 - 72 a;=. 2. 3 x'' + 5 a; = 2. 11. x^" + 2 hx - 3 A^ = 0. 3. a;2 - 3 a: - 10 = 0. 12. 2m^ = 9mx + 5 x\ 4. 2a! + 2 = x'. 13. 2x^+^2; -3;<;^=0. 5. £t;''-a; = l. 14. a;^ + 2a; Va — 3 a = 0. „ „ „ 11a; 1.5 „ 15! ma; = — m^ + 6 a;^. 6. 2.^-— --=0. 7. 2.^-3. = 1. 16.a;^ + -^V2-i.^ = 0. 8. 4a; + 6 = xl 17. ra'a;^— 3 Awx — 10 ^2= 0. 9. a;'' + a; VS = 10. 18. 6 mV + 19 mnx = 7 re''. 19. a:^ + 2 a; = hx + 2 A. Hint. x'^+ (2-h)x-21i = 0. Then a= 1, J = 2- A, and c = - 2A. Substituting these values in (F), (2 - /,) ± V(0 - ;i)2 - 4 ■ 1 ■ (- 2 A) ^ — 5 > etc. 20. a;^ + ra; — sx — rs = 0. 22. mnx^ + raa; = 3 wia; + 3. 21. 2 a;^ + rs = ra; + 2 sx. 23. TOAa;^ + 4 Aa; = 3 irex + 12. PROBLEMS (Reject all answers which do not satisfy the conditions of the problems.) 1. The sum of the square of a certain number and twice the number itself is 15. Find the number. 2. Find two numbers whose difference is 11 and whose product is 42. 3. If from twice the square of a certain number the number itself be taken away, the remainder is 45. Find the number. 4. Find two consecutive numbers whose product is 462. 5. Find two consecutive odd numbers whose product is 255. 6. Find three consecutive even numbers whose sum is J of the product of the first two. QUADRATIC EQUATIONS 21B 7. A rectangular field is 16 rods longer than it is wide. Its area is 32 acres (1 acre = 160 square rods). Find the dimen- sions of the field. 8. The sum of a certain number and its reciprocal is |^. Find the number. 9. The area of a triangular field is 5f acres. The base is 51 rods longer than the altitude. Find the base and the altitude. 10. Two square fields together contain 62.5 acres. A side of one is 20 rods longer than a side of the other. Find the side of each. >^ll. The area of a rectangle is 18 square inches less than twice the area of a square. The rectangle is 7 inches longer than the square, and a side of the latter equals the breadth of the rectangle. Find the side of the square. 12. The hypotenuse of a right triangle is 41 feet. One leg is 31 feet shorter than the other. Find the legs. 13. The legs of a right triangle are in the ratio of 3 : 4. The hypotenuse is 20. Find the legs. Fact from Geometry. If one angle of a right triangle is 30 degrees, the hypotenuse is twice the shorter leg. Conversely, if the hypotenuse of a right triangle is double one leg, one angle of the triangle is 30 degrees. 14. One angle of a right triangle is 30 degrees and its longer leg is 9. Find, correct to two decimal places, the other two sides. 15. The hypotenuse of a right triangle is 10 and one leg is 5 Vs. Show that one angle of the triangle is 30 degrees and find the number of degrees in each angle of the triangle. 16. The area of a square in square feet and its perimeter in linear feet are expressed by the same number. Find the side. 17. The area of a square in square feet and its perimeter in inches are expressed by the same number. Find the side. 276 COMPLETE SCHOOL ALGEBRA 18. The area of a square in square inches and its perimeter in feet are expressed by the same number. Find the side of the square. 19. The dimensions of a certain rectangle and the longest straight line which can be drawn on its surface are repre- sented in inches by three consecutive even numbers. Find its dimensions. 20. The dimensions of a rectangular box are in the ratio of 1:2:3. Find the edges, if the entire outer surface is 792 square inches. 21. The edges of two cubical bins differ by one yard. Their volumes differ by 61 cubic yards. Find the edge of each bin. 22. The rates of two trains differ by 5 miles per hour. The faster requires 1 hour less time to run 280 miles. Find the rate of each train. ^ 23. An automobile made a round trip of 160 miles in 9 liours. Eeturning, the rate was increased 4 miles per hour. Find the rate each way. 24. A page of a certain book is 2 inches longer than it is wide. The printed portion covers half the area of the page and the margin is 1 inch wide. Find the length and width of the page. 25. A man paid $16,000 for a farm. Later he sold all but 40 acres of it for the same sum, thereby gaining $20 on each acre sold. Find the number of acres in the farm. 26. The price of oranges being raised 10 cents per dozen, one gets 6 fewer oranges for 60 cents. Find the original price. 27. Two pumps together can fill a standpipe in 45 minutes. One pump alone requires 2 hours less time than the other. Find the time each requires alone. 28. Each of two trains ran 200 miles. One ran 7 miles per hour faster than the other and required 1 hour and 45 minutes less time. Find the rate of each train. QUADRATIC EQUATIONS 277 29. A and B leave point P at the same time, A going north and B east. Five hours later A has traveled 17 miles more than B and the distance between them is 53 miles. Find the rate of each. 30. A and B leave point P at the same time, A going north- west and B southwest. Five hourai later A has traveled 9 miles less and B has traveled 8 miles less than the distance between them. Find the rate of each. 31. A stone, dropped from a balloon which was passing over a river, struck the water 15 seconds later. How high was the balloon at the time the stone was dropped? Hint. The distance, S, through which a body falls from rest in t nfi seconds is given by the equation S = ^ , (/ being 32 feet. 32. A man drops a stone over a cliff and hears it strike the ground below 6^ seconds later. If sound travels 1152 feet per second, find the height of the cUff. GEOMETRICAL PROBLEHS (The circumference of a circle is 2 ttR, B being the radius. In the fol- lowing problems use '^ for ir.) 1. The circumference of a carriage wheel is 11 feet. How many revolutions will it make while the carriage goes 55 yards ? 2. The radius of a carriage wheel is 2 feet. How many revolutions does the wheel make while the carriage goes 132 yards ? 3. The circumference of a fore wheel of a carriage is 2 feet less than that of a rear wheel. In going 140 yards the smaller wheel makes 5 revolutions more than the larger. Find the cir- cumference of each wheel. 4. In going 100 yards a fore wheel of a carriage makes 5 revolutions more than one of the rear wheels. The circum- ference of one wheel is 2 feet less than that of the other. Find the circumference and the radius of each wheel 278 COMPLETE SCHOOL ALGEBRA 5. The circumferences of two wheels of a wagon differ by 2 feet. Together the two wheels make 11 revolutions while the wagon goes a distance of 20 yards. Find the diameter of each wheel. 6. The radii of two circles differ by 7 inches, and their areas differ by 770 squard* inches. Find their radii. If any two chords of a circle, A C and DE, cross at B, then ABxBC = DBx BE. 1. In the adjacent figure AC = 1S, DB = 4, and BE = 20. Find AB. 8. In the adjacent figure AC = 5 feet, DB = 18 inches, and BE = 48 inches. Find AB and BC. If AB is a line drawn from a point on the circumference perpendicular to the diameter, CD, of the circle, then AB-AB=CB- BD. 9. In the adjacent figure AB = 9 and CD = 30. Find CB and BD. 10. In the adjacent figure AB = 20 and CB = 2 BD. Find BD. 11. A line AB is 12 inches long. A point P is located on AB so that AB : AP = AP : PB. Find the length of AP. (Can a meaning be given to both answers ?) -», History of the quadratic equation. Though the development of the method of solving quadratic equations is closely connected with the general growth of algebra, yet it is possible to indicate the most important steps in the process rather briefly. The first writer on formal algebra waS: Diophantos, who lived at Alexandria, in Egypt, about 300 a.d. Most of his work that is preserved is devoted to the solution of problems that lead to equa- tions. So far as we know he was the first to indicate the unknown number by a single letter, in this respect being far in advance of . QUADRATIC EQUATIONS 279 many mathematicians who lived much later. It is a little remarkable, in fact, that as able and original a man as Diophantos should have exerted so little influence on his successors. He solved his quadratic equations by a method not unlike that of completing the square, but his imperfect knowledge of the nature of numbers made it impossible for him to understand the entire significance of the process. Though he made every effort not to consider equations whose roots were not positive integers, sometimes they would ci-eep in, and under such cir- cumstances, when his method led him to a negative or irrational root, he rejected the whole equation as absurd or impossible. Even when both of the roots were positive he took only the one afforded by the positive sign in the formula for solving a quadratic. The difficulties of Diophantos are typical of those encountered by mathematicians for the next fifteen hundred years. The difficulty lay, not in finding a formal method of solving the equation, but in understanding the result after it was obtained. The meanings of negative and of imaginary numbers have been two of the most diffi- cult of all mathematical ideas for men to grasp. Five or six hundred years later the Hindus devised a general solu- tion of the quadratic, but their chief advance over Diophantos lay in the fact that they did not regard an equation whose roots were negative ag necessarily absurd, but merely rejected the negative result of solving such an equation with the remark, " It is inadequate ; people do not approve of negative roots." The Hindus, however, did realize that a quadratic equation sometimes has two roots, a fact that Diophantos never comprehended. They even went so far as to illustrate the difference between positive and negative nimibers by assets .and debts. No material gain in the understanding of the solutions of the quadratic can be found until the seventeenth century. The keenest mathematicians of the sixteenth century, like Cardan and Vieta, rejected negative solutions regularly, though by this time irrational solutions were admitted. In fact, in 1544 Stifel, a German, pub- lished an algebra in which irrational numbers are included among the numbers proper. But he affirms that except in the case where a quadratic equation has two positive roots, no equation has more than one root. It was not until the work of Descartes and Gauss became widely known that the nature of the roots of all kinds of quadratic equations was completely understood. CHAPTEE XXIV FUNDAMENTAL OPERATIONS (In Part Review) 103. Order of fundamental operations. The numerical value of an expression which involves the signs of the four funda- mental operations depends on the order in which the indicated operations are performed. The assumptions on page 10 may- be stated as follows : In a series of operations involving addition, subtraction, mul- tiplication, and division, first the multiplications and divisions shall be performed in the order in which they occur. Then the additions and subtractions shall be performed in the order in which they occur or in any other order. Within any parenthesis the preceding rule applies. EXERCISES Simplify : 1. 3 - 5 + 6 - 8. 3. 24 H- 8 • 4 - 4 + 6. 2. 6^2 + 1-4. 4. (7-6)(18-2■4)-i-(20-^-4). 5. 42 -2(18 -2- 3)^ 4 + 3- 5. 6. 16 + 4--8-10 + 61^16-4-6-30.2+18-8 -- 48 - 2 • 18 -H 12. 7. (16 ^ 32 X 48 -=- 8 - 4 - 8 + 3) X [12 -=- 4 -- 3 - 1] + (42 -- 6 ■ 7 - 42 - 6) • 6. 8. Does a* = 4 a when a = 3 ? when a =2? when a = ? 9. What name is given to each 4 in o.^ = 4 a ? Define each. 10. Define power. Distinguish between exponent and power. 280 FUNDAMENTAL OPERATIONS 281 Find the numerical value of : 11. x^ — 5 a; + 6 when x = 5. 12. x' - 3a!^ + 3a; - 1 when x = 3. 13. x^ — 3 x^y + 3 xy'^ — y^ when a; = 4 and y = 2. , , a;* + xV + y^ x^ + y" ^ 14. —5 " , — when a; = 3 and y = 2. x^ — xy + if x + y 15. What is the absolute value of a number ? Illustrate. 104. Addition and subtraction. The order in which a series of positive and negative numbers is written does not affect the algebraic sum. This principle is called the Commutative Law of Addition. EXERCISES 1. Review the definition of term, of similar and dissimilar terms, and the rules on pages 17 and 30. 2. Review the rule for subtraction on page 40, and the rules and examples on pages 56-56. Add: 3. 16, - 3, + 2, - a, - 7, and 4. 4. 4 a, — 6 a, — 10 a, + 2 a, and 18 a. 5. 7x — 4:y — z,3x + z — ^y, and 18 «/ — 17 a; — 14 z. 6. 4 a^ — 3 ah — 4 ac^, 3 a^c — 8 ac^ — 8 a^, and 3 a^ — 6 ah. 7. lix = l, y = 2, and « = 3, find the numerical value of each of the three expressions and of the result obtained in Exercise 5. Compare the suan of the three numerical values with the numerical value of the result. 8. State a rule for cheeking work in addition of algebraic expressions. Write with polynomial coefficients : 9. ay + hy-\- cy. 12. Z{a + l)— c{a + h). 10. 3ax-4.bx + &x. 13. 46(3a;-2)-8c(3a;-2). 11. 4a: — ate — X. 14. 4m(5a — 3c) — 6?i(— 3c+ 5a) 282 COMPLETE SCHOOL ALGEBRA Subtract the first expression from the second in : 15. 4o, 6«. 17. 4a; + 3, 8a; + 6. 16. 8 a\ 6 a^ 18. 7 x" - 10, 6 a;^ + 20. 19. K — 3 2/^ + s — 4 ac + 7 ax, Ax — y^ + S — Sax + 9ac. 20. a' — c + 3 X — a^m — Sac, 4 a* + m — 8 a; — 10 ac + 4 a^m. Find the expression which added to the first will give the second in: 21. a;^-6a; + 6, 3«^-5a; + 2. 22. 4a;'^-3ca; + c^ 80^ + 7c£c - lOa;'^ + 8. Find the expression which subtracted from the first will give the second in : 23. Aa^ -2ab + b\ 7 a^ -lOab + 6b\ 24. c^ - lOca; + 8a;^ 9a;2 - lOca; + 4 + d 25. State a rule for checking work in subtraction suggested by the directions preceding Exercises 21 and 23. 26. From the sum of aa; — ac — 3 c^ and 4 c^ — 3 ac take the sum of 4 c^ — 8 ax + a^ and i ac -{- 3 ax — 5 c'. Remove parentheses and combine like terms : 27. 4x-3-(a-2a;) + (3a;-«). 28. 6x+(3c-8x + 2)-(c-a;-2). 29. 6x-[-(a-c) + (3c-4a)]. 30. 7c -[(3c -4)- 6 -(4a;- 3a -c)]. 31. 6 X - 4 (3 - 6 x) - 4[2 (x - 4) + 3 (2 X - 1) - (a: - 7)]. 32. 3x - 2[1 - 3(2x - 3 - a)- 5 {a -(3x - 2a)- 4}]. 33. State the rule for the removal of a parenthesis : (a) when it is preceded by the sign plus ; (b) when it is preceded by the sign minus. Inclose in a parenthesis preceded by the sign plus those terms which contain x and y, and inclose all other terms in a parenthesis preceded by the sign minus. FUNDAMENTAL OPERATIONS 283 34. x^-\-2xy-\-if- a\ 35. x' + liab- A9 a" - b\ 36. 2/^ + 6xy + 9x' - m^ - 10m - 26. 37. x' + 10 xY -c' + 12 c*d -36d^ + 25 y". 38. State the rules for inclosing terms in a parenthesis pre- ceded by (a) the sign plus ; (b) the sign minus. 105. Multiplication and division. Meaning of a zero exponent. The laws for exponents stated in the formulas of § 27, page 60, and § 37, page 76, are assumed to hold for all values of a and b. Then . x^ -i- x* = x*-^ = x". But x*^x^ = - = l. X* Therefore x" = 1. More generally, a;" -:- a;" = a;""" = x", and of -i- of = — = 1. a;" As before, x" = 1. That is, any number (except zero) whose exponent is zero is equal to 1. Hence, if x is not zero, 4° = (f )» = (- 6)» = (Bx)" = (x^ ~ 2x + ly, for each equals 1. 106. Meaning of a negative exponent. If in the formula of § 37, page 76, b is greater than a, we obtain a negative expo- nent for n. The meaning of such an exponent is illustrated as follows: „B ^ ^5 _ ^3-5^^-2. Obviously o' -^ a^ = -g = -^ • Therefore a~ " is another way of writing — • Tlien 4"' = 7i = ^- 4' 64 In general terms, x~° = — Consequently — i^ — — = x". 3C JL 284 COMPLETE SCHOOL ALGEBRA Similarly, we obtain the more general results bx- " = — and = bx". Xa x-" Hereafter it -will be assumed that. all the preceding expo- nential laws hold for positive, zero, and fractional exponents. EXERCISES 1. Eeview the definitions on pages 59-64, the principle on page 60, and the rules on pages 60, 61, and 62. Perform the indicated operation : 2. (4a;2-3a;)(2x). 3. (2 a; + 3) (Sec - 6). 4. Substitute 2 for x in each of the factors of Exercise 2, and in the product. Compare the numerical value of the product with the product of the numerical values of the factors. Then state a method of checking numerically work in multiplication : 5. {Zx'-lf. 8. (e- + 2e-^)l 6. (7 a;2« _ 8 cc" + 3)^- 9. (e^ - e""^)'.- 7. {x^ + x^f. 10. (e'^ - 3 e-^)*. 11. (x* + a;i + 1) (afl - x^ +1). 12. (4a;== - &x' + 3)(7k^'= - cc^" + 4). 13. {x^ - 2 X2/^ + y') {x^ + 2xf + y'). 14. (x-i-3a;-2a;-^)^. 15. (x-^ + 2x^ -dx^f. ,„ /2a? a 2\/2a? , a" 2a\ 17. (6a;2» — Scc-^"- 6a;-" + 3x'')2. 18. x^ - » - 90 = ? if a: = - 9. 19. x= - 3xV + 2ixy'^-f=? if a; = 2 and y = - 3. 20. x^ + dxhj + Zxy'^ + y^ ="? if x = - 4 and y = - 2. 21. State the Associative Law of Multiplication. Illustrate. 22. State the Distributive Law of Multiplication. Illustrate. FUNDAMENTAL OPERATIONS 285 23. Review the principle on page 75 and the rules on pages 75-76 and 78-79. 24. (Sx'-6x'-4:x)-i-(-2x). 25. (x^-7a; + 12)-=-(x-3). 26. (e^ - e-y =? if e = 2 ; if e = - 3. 27. e^^ -26° + e-^" = ? if e = 2 and a; = 2. 28. (x' - 64) -=- (x - 4:)(x^-4:x+ 16). 29. (x* - 8x^ + 33 X - 30)-=-(a:^ + 3a; - 5). 30. State a method of checking work in division similar to the check of multiplication. Find the remainder in : 31. (8x^-x''-5)^(2x-3). 32. (4:x*-x^-3)^(2x^-x^t). Divide : 33. x' + 8 / + 126 - 30 xy by cc + 2t/ + 5. 34. x^ + y^ + z' — 3 xyz by x + y + «. 35. 3x-" + x''-4x-«by 2x-2 + x= + 3a;-«. 36. x^ — y^ by [(x^ — y^) -=- (x^ + y^)]. 37. 9 TO + 4 m-i — 13 by 3 m* — 5 + 2 m~i 38. x^" + 4x-2« — 29 by x" — 2x-" — 5. a 39. Qx^" + 25x-*'' — IQx-" hj Bx-"" + Sx" ~7 X ". An I n a , /^ a , 36 Kx^ , 35 a^x\ /3 a , 2x\ 40. ( 6 a"* + 6 x" H -; h /9a= 2 3 / ■ \2 ' 3 ^, ,.a= , 243 a^ ,„ , 59-a 443«'' 43 a* /3 a' , 3 6o^ -^(-^\«-2-^j- 107. Detached coefficients. An algebraic expression is rational and integral if its terms are rational and integral. Such an expression is usually called a polynomial. 286 COMPLETE SCHOOL ALGEBRA An integral expression may not be rational. Nor is every rational expression integral. Thus, — + Vx + 8 is integral x^ 1 . . but not rational, while — H 1- 8 is rational but not integral. A rational integral expression is homogeneous if its terms are all of the same degree. Thus, a;^ — x^y — Ti^ and 3 x* + x^ — y* are homogeneous expressions of the third and the fourth degrees respectively. In the multiplication or division of polynomials which in- volve but one letter or which are homogeneous in two letters much labor can be saved by using the coefficients only. EXAMPLES 1. Multiply 3a;= - 4x + 6 by 2a;2 - 5a; + 3. Solution : Since x^ is missing in the first expression, its coefficient is zero. Inserting Oa;^ and detaching coefficients, the multiplication is as follows :- q+O — 4+6 2- 5 + 3 6 + 0-8 + 12 -15 + + 20-30 + 9 + 0-12+18 6-15 + 1 + 32-42+18 Supplying the powers of x, we obtain as the product 6 a;° — 15 a;* + a;3 + 32 a;2 - 42 a; + 18. 2. Divide 6a;* - llxhj + 2xhf ■\-21xf - 18 y* by 2a;^ - 6xy + 6/. Solution : 6- 6- -11+ 2 + 27- -15+18 -18 4-16 + 27 4-10 + 12 - 6 + 15 - - 6 + 15 - -18 -18 2 - 5 + i 3 + 2-3 Therefore the quotient is 3 .r^ + 2 a;!/ — 3 ■tf: In both multiplication and division by detached coefficients zero must be supplied for the coefficient of any missing term. FUNDAMENTAL OPERATIONS 287 EXERCISES Use detached coefficients and perform the indicated operation .- 1. (a;2-8a: + 16)(2x-3). 2. (x'-ix + 4)(a;2 + ix + 4). 3. (a^ -ab+ P) (a? + ab + V). 4. (2x' + nx + 2)^{2x+l). 5. (x' + 4a;-16)--(x-2). 6. (Zxy-&if-2x'^){?,x'^-&y''~Zxy). 7. (9a;*-4a; + 13x2 + 4-6x=)-=-(3a;--a; + 2). 8. (x^ + 4.y')^(x^ -2xy + 2if). 9. (81 a* - 171 aV + 25 h") -- (9 a^ - 5 i^ + 9 a5). 10. (4 a« - 2 a^ - 3 a-2 - 6 a-^ + 2 a) -V- (2 a^ - 2 - a-^). 11. (8 a; - 12a: V + 6 icV' - 2/"') "5" (2 a:* - y"')- 12. Which expressions in the preceding exercises are (a) not integral ? (S) not rational ? Note. It is interesting to observe that our ordinary decimal nota/- tion really involves the use of detached coefficients. The number 649, for instance, is an abbreviated way of writing 6 • 10^ + 4-10 + 9. In fact, the various digits in any number in the decimal form are the detached coefficients of some power of the number 10. 108. Important special products. Certain products are of fre- quent occurrence. These should be memorized so that one can write or state the result without the labor of actual multiplication. ORAL EXERCISES 1. Eeview the formulas, rules, and examples of pages 93-100. Perform the indicated operation : 2. (a; + 3)^ 5. (4 a; -3)1 8. (3 «' - 4 xcf . 3. (x-hf. 6. (x'-xf. 9. (3?-x-y. 4. (2x + 4)2. 7. (a;-3c^)^. 10. (x^-^x-y. 11. (16 a;^ + 8 x^c + a;V)^ (4 x + a;c) = ? Why? 12. (2 ft' — a-'') (2 a' — a-'') (2 a" — a-^) h- (2 ff' — fr="). 288 COMPLETE SCHOOL ALGEBRA 13. (x — e)(x + c). 20. (a; + 3) (a; + 4). 14. (x + 6)(a;-6). 21. (5 -3) (6 -4). 15. (a - 3 c) (a + 3 c). 22. (c-l)(c + 2). 16. (m — x)(x + m). 23. (a - 3) (a: + 5). 17. (4a!-3c)(3c + 4a:). 24. (a^ - 4 a) (a^ + 6 a). 18. {x^-cx){p? + cx). 25. (a^ - 2 or") (a' + 5 a-^). 19. (4c=-«')(a^ + 4c'). 26. (cx-4c')(ca; + 8c'^. 27. (a + 5 + c)l 29. (« - c + a;)l 31. (a - c + 2)1 28. {a + o + xf. 30. (« - c - xf. 32. (a; - c - 3 a)^. 33. o . FACTORIJSTG AND FRACTIONS 293 8. a^' — a-^''. 9. a^ - a-*. 10. (a + cy-1. 11. (a-x)*-4. 12. 9-(2 + a;)l 17. a^ + 2 aa + x^ - 9. 13. 16 - (« - a)"". Hint, a^ + 2 ox + x'' - 9 14. 5^« - (m - l)^"". " '^'^ "*■ ""^^ " ®' 15. (a + c)'' - (m - w)l 18. 25 - 10 a; + a;^ - 16m^ 16. (a-xy-(c~5y. 19. 9-12a + 4a2-&2» 20. m^-a''-6a-9. Hint, rrv' - a^ - 6 a- Q = m' - (a^ + 6 a + Q) = m^ - (a + Sy. 21. x'-4:tf + 20 tj- 25. 22. - 28cW - 49c^ + 1 - 4&y''. 44. 2/* - 18 1/^ + 81 - 16 a;* - 24a;y -^y\ 45. AV<;8 - 1024 k\ 47. x" - f - x - ij. 46. a;' - 83 cc^ + 289 a;l 48. 289 - 100 a^-J^- 20 ab. 49. 625 a« - 169 d' + 78 c(Z' - 9 cl 50. «"» - 125 1/"". 54. a!^-729. 51. 4 a;* -37x^+9. 55. a*^+8 + 64. 52. 256 - 16 k" + 8 A^/c' - A*. 56. a" + 226 a-' - 39. 53. x* + 4. 57. a;' - 6x2 + 12 a; .- §_ 58. 0,2 - 9 c^2 - 8 a6 + 6 C£Z - c^ + 16 h\ 59. 4 /i-" - 20/i-Vc + 257c2 - 6a6-^ - 9 a^ _ j-i. 60. a;2» - 2 a;" - 15. 64. 128 - x"'. 61. a'^ + a + b^ + b.. 65. 3 x^ + 10 a; - 8. 62. x"" - 12 a;" + 36. 66. «= - a? - a^ + 1. 63. 25a;^° + 50a;'^ - 39. 67. a;^ + 3a;2 + 9a; + 27. 68. a;^-6a;2 + 12a;-7. 69. 4x* - 25 2/« + 10 2/^ - 12 x" + 8. 70. 6^^ — 2 + e-2^. 75. x^-lxy'^ + Qif. 76. e^^ + e^o: + 6=^ + 1. 77. „4x-2_-Lo + 25a2-4-. 71. e^^ — 5 + 6e-2»^. 72. 3 a;2" + 6 a;" - 28. 73. e'^ — 2 e^ — 24 e-^ 78. e=^ — e""^ + 3 6-=" — 3 (f. 74. 6 6^^ — 5 6-2^^ — 13. 79. e=^+8 + e^^+a _ gS-^ — e^-^. 80. a;?/2 + a;«2 + x^y + a;^^ + y»^ + y^« + 2 xyz. 81. aS' — a^5 + ac^ — o?o + h(? — b^c. 119. Solution of equations by factoring. Study pages V2i2>-V2,'^. It is of particular importance that the principles on these pages be thoroughly mastered, and the significance of each step in the solution of the following equations completely grasped. FACTORING AND FRACTIONS 301 EXERCISES AND PROBLEMS Solve by factoring and check : L a;^-25 = 0. 6. a;* + 4 = 5a;l 2. x' + 10 = 7a;. 1. f = 13 t" - ^&t\ 4. r^ - ra = 30al 9. x^~2x^ = x-2. 5. 4»',-36s = 0. 10. a;2^-2x»' + l = 0. 11. x'^ - 8 a;^ + 16 = 0. 12. a;' + 5a;^- 18a; -72 = 0. Hint. Apply the Factor Theorem. 13. a;* + 3x»-8a2 + 16 = 12x. 14. a:' + 6 a= = 2 ax^ + 5 a'x. 15. a;*^ + 9 aa;^ = 9 a;% + x^. 16. x^ + 5x% - 16xc^ - 80c' = 0. 17. Separate 272 into two parts such that the greater equals the square of the less. 18. Twice the square of the least of three consecutive even numbers is 104 greater than the product of the others. Find each. 19. Find a number which when added to four times its reciprocal gives 20.2. 20. A and B together can do a piece of work in Ij days. If A requires 2 days more than B, how many days does each require alone ? 21. The hypotenuse of a right triangle is 40 feet. One leg is 8 feet longer than the other. Find the three sides. 22. The rates of two trains differ by 10 miles per hour. The slower requires 2 hours more than the faster to run 240 miles. Find the rate of each train. 23. The bases of a trapezoid are respectively 2 feet and 8 feet longer than the altitude, and the area is 14 square yards. Find the bases and the altitude. 302 COMPLETE SCHOOL ALGEBRA EXERCISES rind the H. C. F. of the following : 1. 28, 56, 84, 35. 2. 225, 120, 210, 135. 3. 198, 496, 693, 1155. 4. 816, 1224, 1360, 4080. 5. 91 x\f, 133 ay, 343 xhf. 6. a^ - 9 a + 14, a^ _ 4^ 5 ^2 _ -^q „. 7. x' + 27, 2x2 4- 3a; - 9, 5a;= + 15a;=. 8. 2 x^ + 8 X, 3 ax' + 6 ax + 6 ax^, 3 ax' + 12 ax'. 9. ■ [(x + y) (x - yy\\ x' - 2 xhf + y\ (x« - ff. 10. x' + a", x^ - a^'^ x^ - 3 xa' - 4 0"% x» + a'^" 11. a^' + 4 a<^&', a2« + i&2 + 2 a6^ - 2 a' + ^S^ a''^ - 16*^^ Find the L. C. M. of : 12. 24, 30, 54. 13. 105, 140, 245. 14. 174, 485, 4611, 5141. 15. 30 ax^ 225 a^xif, 75 aVy. 16. 12x2 + 6x, 12x'-3x, 16x^ + 2x. 17. a= - 8 b\ llf- a", a% + 4 ai' + 2 a%\ 18. x= - 2x2 - 6x + 6, 4 - x^ a - axl 19. a* + 4 a^ + 16, a^ - 4, a' + 8, a' - 8. 20. x= - 2 a^x + ax% 2 a^ + S a^x + ax'', 4 aV - a'x^ 21. m* - 3 m V + 9 ?^^ m' + 3 mre^ + 3 m\ 3n^+nm^-3 mn'. 22. e"' + e-"^ - 2, e^^ - 6"'% 6^^-3 + 2 e-^^ 23. x' - 2x2 - 2x - 3, x' - 27, x^ + x - 12. 120. Addition and subtraction of fractions. The student should make whatever review is necessary of the definitions, princi- ples, examples, and rules on pages 135-143. FACTORING AND FRACTIONS 303 EXERCISES Reduce to lowest terms : 18 aV 7 a^ — 14 b"^ -^a .26 24 aV ■ a^ + a2^,2 _ g j* "■ (a;" + c")^ 3x-6 x^-8 (x^-c^)- ■ a;^-4 *• (x-2y "• x* - c* ' y 2x^^-128 3a;^ + 2x-21 a;2'-4 ■ • 27 a;* -147^2 ' 8 ^°-l ,„ x^-x' + x-l ''■ x' + x^ + l •^"- a;* _ 1 By the use of § 62 -write in three other ways : 11 "—. 12 ^ ■ 13 2a:-3.v -2a ' a-2c ' Zx' - ^if - xy' Change to respectively equivalent fractions, writing the let- ters in the denominators in alphabetical order and making the first term in each factor positive : 14. , ^1 ,. 15. ^-^1 (c — «)(* — «) ' (y - a;) (s — x){z — y) Change to respectively equivalent fractions having the lowest common denominator : 2 3 ^°' 6a! - 12' a;^- 2a;' ^ a;^ - 5a; + 6' x^ - 9 22. Does t: z equal 7;— ? Explain. Define cancellation. 3a + 5 3a 23. If the same term occurs in each member of an equation, ' may it be canceled ? Explain. 19. ^^ + ^ a-2b 4:ab x-1 x 304 COMPLETE SCHOOL ALGEBRA Find the i ilgebraic sum of : 24 ^'^ a — x 3x _„5 — a: 5 24. g 9 4 X - 4 7 25 "'-^ 3a;2_2 a;'' -8 „„ 4m^ 14ai« 3a;^ rr2-9m2 a; - 3 3 a;'' - 9 a; + 14 a;^ - 4 „3 a; 1 2 + x 3m — a; X -\- 3 m ""'£■— 2 X X x^ — ix + 4: 30. 2 ^ + j^ ■ 31. a;^ + a; + 1 — 1 1 — v x" x-1 32. x^ f-l^:+.+i. a;^ — a; + 1 33. 2a:^-3a^ 2a; + a 27a;='-64a^ ^x^-lQa" 34. 2 3 (a — c) (a; — c) (c — a) (c — a) 35. 2x-4 3a; + l a; 9-a:^ a; -3 9-6x + x^ 36. 2a; + 12c 3a;-7c 6a;'-13fa;-5c-' Ax^ + Acx - 35c^ 37 (a — ^) (f — g) r?^ (a — m) (c - m) s (s — m) m.s m (m — s) ^ X — X X — O , O X — o QQ . ■ a;'' — a; + 1 a;* + a^ + 1 a;^ + 1 Show that ; 39. = 5i, when e^"^ = 4. 40. c (c — y') -i- d — (-1/ + <^ d -i- c = y, when y ^ c — d. 1 4; c = a; — 2, and d = x^ — 2x + i. "1 f\ "• ^-^ = a=* + 4x^ + 16'^^"^'" = ^ + 2'^ = «=' + 2a. + 4, FACTORING AND FRACTIONS 305 121. Multiplication and division of fractions. The student should make any necessary review of pages 147-154. The reciprocal of a number is 1 divided by the number. Thus the reciprocal of 2 is ^ ; of f is | ; and of 3f is ^^. Therefore the quotient of one fraction by another is the prod- uct of the first and the reciprocal of the second. Also the quo- tient of a fraction by an integral expression is the product of the fraction and the reciprocal of the integral expression. EXERCISES Perform the indicated operations : . 4aiV 21a' 9ia? 10x^0 , /2 aV /3 cM' . I- aV (- 2 xV- I- cV /ScV.. /9cVy. / -2aV Y (3a^* ^" \2a/ ■ \4aV ■ V ^^^ J O-Ox^ 4:x' — 2i/ 4 x^' — f' 2x' + ' ■ (2x''-y'y' (2x^ + 1/)^' 3 x^ + 1 ^ + x + l •4 9. 10. x^-1 \ x^ + 1/ x^ a-9Q . a'-j-Oa" . ia + G -100 ' a^ + lOa ' 2w'x + 3ax' 13 a \ a= - a - 42 2 a' - 12 a a + 6j ' a'' — 36 ax + 6x 7 _\ /„ , 5. ,11' a^ a ix\x m) ■ \ mV / \ m^l \ cx,~ax ) 13. — - + mx " 306 COMPLETE SCHOOL ALGEBRA x^ — 8 £c^ — 4 (x+2Y 14. Multiply ^,^^^,^^g-^^-2^,by the reciprocalof^,^^ ,^ 2a;^ + 5x + 3 3x^ + 3ax-.T-a 15. 6x^ + x-l 2ic^-2e« + 3a;-3c c + 2a + (2ffi + l + 2c)g \ x^ + X + ax -\- a I Simplify each term in : 16./^Y-5P^^^^^«^ I? I \ ^^ M find the difference between 2 + etc. four times this fraction and the value of tt, 3.1416. Note. William Brouncker, one of the brilliant mathematicians of his time, was an intimate friend of John Wallis (see page 356). These two scientists were among the pioneers in the study of expres- sions with an infinite number of terms. The complex fraction in the exercise, if continued indefinitely according to the law which its form suggests, is called an infinite con- tinued fraction. Brouncker was the first to study such expressions. CHAPTER XXVI LINEAR EQUATIONS IN ONE UNKNOWN (In Part Review) 122. Use of the axioms in solving equations. Two or more equations, even if of very different form, are equivalent if all are satisfied by eviery value of the unknown which satisfies any one. ;' Of the four axipms, or assumptions (see page 33), we shall make constant use. If the " same number " referred to in each is expressed arithmetically, the result is always an equation equivalent to the 'original one. Further, if identical expressions involving the unknown be added to or subtracted from each member of an equation, the resulting equation is equivalent to the first. If, however, both members of an equation be multi- plied by or divided by identical expressions containing the unknown, the resulting equation nnay not be equivalent to the original one. In other words, under the condition just stated, roots may be introduced or lost by the use of Axiom III or IV respectively. The examples which follow illustrate the impor- tant facts concerning the introduction or loss of roots which enter or disappear even though a proper use of the axioms is made. Example 1. Let a; - 2 = 7. (1) Multiplying by a; - -3, a:2 - 5 X -t- 6 = 7 a; - 21. (2) From (2), a;2 - 12 X -1-27 = 0. (3) Whence Qc - 3) (x - 9) = 0. (4) Therefore a; = 3 or 9. Since (1) has the root 9 only, and (3) has the two roots 3 and 9, (3) is not equivalent to (1), that is, a root was introduced by the use of the multiplication axiom. 308 Example 2. Let a:^ - 4 = ^ + 2. Dividing by a: + 2, x-2 = l. Solving (2), x = 'i. But from (1), i;2 -a; - 6 = 0. Whence (x - - 3) (a; + 2) = 0. Therefore a; = 8 or - LINEAR EQUATIONS IN ONE UNKNOWN 309 (1) (2) (3) (4) (5) ■2. Here (5) shows that (1) has the two roots 3 and — 2, and since (2) has but one root, 3, it is evident that a root was lost by the use of the division axiom. The student shoiild note the preceding illustration carefully, as the possibility of dividing each member of an equation by a common factor involving the unknown frequently arises. A very common' type is the following : Example 3. Let x^-2x = 0. (1) Dividing by a;, a; — 2 = 0. (2) Whence x = 2. (3) But (1) has the root also, which is lost by dividing both members of (1) by x. If proper methods of solution are applied to an equation (or to a false statement in the form of an equation), and one or more values of the unknown which are thus obtained do not satisfy the original statement, such values are called extraneous (or extra) roots. An extraneous root is a root of an equation which is not equivalent to the original statement, but of one which is derived from the original statement in the process of solution. 123. Principles. The preceding discussion may be summed up thus : Principle I. If identical expressions (which may or may not contain the unknown') be added to or subtracted from, each mem- ber of an equation, the resulting equation is equivalent to the original one. Pbinciple II. Sxtraneous roots m,ay be introduced into a solution by multiplying both members of an equation by an integral expression containing the unknown. 310 COMPLETE SCHOOL ALGEBRA Principle III. If both Tnembers of an equation he divided hy an integral expression containirig the unknown, one or more roots will usually he lost hy such a division. It can be seen from Example 1 that the root introduced is the value of the unknown obtained by setting the multiplier, x — S, equal to zero and solving the resulting equation. Similarly, it can be seen from Example 2 that the root lost is that value of the unknown obtained by setting the divisor, x + 2, equal to zero and solving the resulting equation. Sometimes a statement in the form of an equation has no root; yet the ordinary method of solution appears to give one. For ex- 4a;-l a; + 8 ample, consider the statement = 1- 5. X — S X — S If one multiplies each member by a; — 3 and solves as usual, he obtains x = 3. This answer cannot be verified because x — S, the denominator, becomes zero for x = S. Here the multiplication by X — S introduced the value 3 for x. Checking will always discover the falsity of such a result (see page 161). Extra roots usually occur in the solution of equations (or in attempts to solve false statements in equation form) which involve fractions or radicals. For solving equations in one unknown -which, may or may not involve fractions we have the Rule. Free the equation of any parentheses it may contain except such as inclose factors of the denominators. Where polynomial denominators occur, factor them if possihle. Find the lowest common WAiltiple of the denominators of the fractions and multiply each fraction and each integral term of the equation hy it, using cancellation wherever possible. Transpose, solve as usual, and reject all values for the unknown which do not satisfy the original equation. Checking the solution of an equation is often called testing, or verifying, the result. For this we have the EuLE. Substitute the value of the unknown obtained from the solution in place of the letter which represents the unknown in the original equation. Then simplify the result until the two members are seeri to he identical. LINEAR EQUATIONS IN ONE UNKNOWN 311 EXERCISES 1. Define and illustrate : equation, identity, equation of con- dition, linear equation in one unknown, satisfy, root, extraneous root, substitute, verify, check, and axiom. 2. G-ive an example of (n) a numerical identity ; (6) a literal identity ; (c) a conditional equation ; (d) two equivalent equations ; (e) a statement in the form of an equation, but which has no root. 3. Define and illustrate transposition, (a) On what axiom does transposition depend ? (b) If one equation is obtained from another by transposition, are the two equivalent ? Explain. 4. What extraneous roots, if any, are introduced if both members of the following equations are multiplied by the expression on the right? (a) x + 3 = T x + i (b) x + S = x + 5 (c) X + c = x — c (d) X -\- a = X (e) a; = 5 4 (f) x-1 =0 a;= + 3a; + 2 5. What roots, if any, are lost by dividing both members of the following equations by the expression on the right ? (a) x'-4: = x-2 (b) x^-ix + 3 = x-3 x-3 (c) a;^ + a; — 12 = X + 4 x + i (d) x^ — 2x = ix X (e) x*-16 = x^-4. x'-i (f) (x-af = {x-af x-a Solve the following for the unknowns involved, considering a, b, c, and d as known numbers : Bx a _ X 3 3 7_ m— 2 _ 17 2r~6~i4' 2^~20~6a;' m-3~18' 312 COMPLETE SCHOOL ALGEBRA 2.r + 5 3(2a; + l) _ **■ 10a; 2a; *' 5 a; — 7 5 /4 — a;\ 15 a; — 22 10. 6 2\ 10 / 6 1-8.T 2(1- 6a:) _ 1- 24a; ^^- 6 24x-3 ~ 15 j^g 3a; — 9 O — ^^ ~^ - 14 =^ , a; + m ^ "3a; — 6 8 — 3a; '« J ~' 13. i + i-i = i. 15. |^-c^ = 2^^^-c_= a 6 c a; 2a c „ 2a;-3& 2rt-3x 9S 4a ^ lb. 1 = — 5. a a b 17. 3 ^ 1 _ -3(&+3) 6 (a — a;) a. — a; 2 aft 18. _3 l + 2y 4j 2y+l 2y-l 1-4/ -- 3& + 9x a — 2x _ 9a + 6a;~2a; + 3a"~ ' 2j^ 8 2 -6a; 27 22. a;— 7 a;^— 6x — 7 a; + l X 5 bx 2(a + b) b-a b^-d' 24. 82.4 - 13 a; = 32 a; -52.6. 25. .01 (2 a; + .205) - .0125 (1.5 a; - .5) = .01955. 26. X - 1 -.33a + 1 5 2x 4 a 3 — .22 a 27. ax c 3 c — 3a 1 a + ex LINEAR EQUATIONS IN ONE UNKNOWN 313 PROBLEMS 1. At what time between 4 and 5 o'clock will tlie hands of a clock be together ? Solution : First, the minute hand moves twelve times as fast as the hour hand. Second, at i o'clock the hour hand is 20 minute spaces ahead of the minute hand. Now let x equal the number of minute spaces that the minute hand travels from its position at 4 o'clock until it overtakes the hour hand. Obviously the hour hand must travel a; — 20 spaces before it is overtaken by the minute hand. Therefore x = (x — 20) 12. Whence x = 21^9^. Hence the hands are together at 21J^ minutes after 4 o'clock. 2. At what time between 7 and 8 o'clock are the hands of a clock together ? 3. At what time between 2 and 3 o'clock are the hands of a clock in a straight line ? 4. At what time between 6 and 7 o'clock is the minute hand (a) 10 minute spaces ahead of the hour hand ? (h) 10 minute spaces behind the hour hand ? 5. At what times between 5 and 6 o'clock are the hands of a clock at right angles ? 6. If the earth is between a planet and the sun and in a line with them, the planet is said to be in opposition. The earth and Mars revolve about the sun in (approximately) 365 days and 687 days respectively. Mars was in opposition Septem-^ ber 24, 1909. What is the approximate date of the next opposition ? Solution : For the sake of simplicity we will suppose in this and in similar problems that the planets move in the same plane and in circular paths, of which the sun is the center. Let X = the required number of days. Now in X days the earth will make — - revolutions about the sun. X And in x days Mars will make — — revolutions about the sun. 687 But to be in opposition the earth must in x days go round the sun once more than Mars does. 314 COMPLETE SCHOOL ALCEBRA Therefore = 1- 1. 365 687 Clearing, 687 x = 365 x + 250755. Whence i = 779+. Therefore the required date is November 11, 1911. 7. If a planet is between the earth and the sun and in a line with them, it is said to be in conjunction. Venus was in (superior) conjunction April 28, 1909. If Venus revolves about the sun once in 225 days, find the approximate date of the next conjunction. 8. Jupiter revolves about the sun once in 4332 days. On February 28, 1909, the planet was in opposition. Find the approximate date of the next opposition. 9. Two men travel in the same direction around an island, one making the circuit every 2^ hours and the other every 3 hours. If they start together, after how many hours will they be together again ? 10. Three automobiles travel in the same direction around a circular road. They make the circuit in 2f hours, 3-J- hours, and 4§ hours respectively. If they start at the same time, after how many hours are the three together again ? 11. Is the answer to Exercise 9 an integral multiple' of 2^ and 3 ? Is it the least integral multiple ? 12. Is the answer to Exercise 10 an integral multiple of 2|, « 3^, and 4| ? Is it the least integral multiple ? 13. Eeduce 2|, 3J, and 4| to improper fractions and divide the L.C.M. of the numerators by the G.C.D. of the denomi- nators. Compare the result with the answer to Exercise 10. 14. The method of finding the L.C.M. of two or more frac- tions or mixed numbers is hinted at in Exercise 13. State a rule therefor. Find by the rule the L.C.M. of 1^, 2^, and 3^. 15. Find by the same rule the L.C.M. (a) of 7 and ^ ; (5) of a c . e b d J, -, and - ■ b d f LINEAR EQUATIONS IN ONE UNKNOWN 315 16. How many ounces of alloy must be added to 56 ounces of silver to make a composition 70% silver? 17. Gun metal of a certain grade is composed of 16% tin and 84% copper. How much tin must be added to 410 pounds of this gun metal to make a composition JL8% tin ? Hint. Since the composition is 16% tin, then ^^^-410 = the number of pounds of tin in the first composition. Let X = the number of pounds of tin to be added. 16-410 Then h x = the number of pounds of tin in the second composition, and 410 + a; = the number of pounds of both metals in the second composition. 16-410 Therefore 100 "^ ^ 18 — -: = ' etc. 410 -f X 100 18. A 30-gallon mixture of milk and water tests 16% cream. How many gallons of -srater has been added if the milk is known to test 20% cream ? 19. How many gallons of alcohol 90% pure must be mixed with 10 gallons of alcohol 95 % pure so as to make a mixture 92% pure? 20. The diameter of the earth is 3§ times that of the moon, and the difference of the two diameters is 5760 miles. Find each diameter in miles. 21. The diameter of the sun is 3220 miles greater than 109 times the diameter of the earth, and the sum of the two diameters is 874,420 miles. Find each diameter in miles. 22. The distance of the earth from the sun is 387^ times the earth's distance from the moon. Light traveling 186,000 miles per second would require 8 minutes 18f f seconds longer to go from the earth to the sun than from the earth to the moon. Find each distance in miles. 23. The diameter of Jupiter is 10^^ times the diameter of the earth, and the sum of their diameters is 94,320 miles. Find each diameter in miles. 316 COMPLETE SCHOOL ALGEBRA 24. A can do a piece of work in 15 days and B in 25 days. After they have worked together 3 days, how many days will B require to finish the work ? 25. A can do a piece of work in a days, B in i days, and C in a + & days. How many days will it take them all working together to do the work ? 26. A cistern has two pipes. By one it can be filled in 2 m hours ; by the other it can be emptied in — - — hours. Assume w + 1 2 m less than — - — and find the number of hours required to o fill the cistern if both pipes are opened. 27. Discuss Problem 26 thus : What is the relation between m and n if (a) the water could run out more slowly than it comes in ; (S) the water could run out as fast as it comes in ; (c) the water could run out faster than it comes in ? 28. If both pipes in Problem 26 had been intake pipes, how many hours would have been required to fill the cistern one-xth full? 29. If the radius of a circle is increased 7 inches, the area is increased 440 square inches. Pind the radius of the first circle (tt = \2- approximately). Facts from Geometry. The area of a circle is the square of the radius multiplied by tt (;r = 3.1416 approximately). This is expressed by the formula A = ivR''. The circumference of a circle equals the diameter times tt. The usual formula is C = 2 irR. 30. Imagine that a circular hoop one foot longer than the circumference of the earth is placed about the earth so- that it is everywhere equidistant from the equator and lies in its plane. How far from the equator will the hoop be ? 31. Compare the result of Exercise 30 with the one obtained when a similar process is carried out on a flagpole 6 inches in diameter, instead of the earth. LINEAR EQUATIONS IN ONE UNKNOWN 317 32. A passenger train whose rate is 42 miles per hour leaves a certain station a hours and b minutes after a freight train. The passenger train overtakes the freight in b hours and a minutes. Find the rate of the freight train in miles per hour. 33. The arms of a lever are 3 feet and 4 feet in length respectively. What weight on the shorter arm will balance 100 pounds on the longer? 34. A beam 12 feet long supported at each end carries a load of 3 tons at a point 5 feet from one end. Find the load in tons (excluding the weight of the beam itself) on each support. 35. The arms of a balanced lever are 8 feet and 12 feet respectively, the shorter arm carrying a load of 24 pounds. If the load on the longer arm be reduced 4 pounds, how many feet from the fulcrum must an 8-pound weight be placed on the longer arm to restore the balance ? 36. A horizontal beam 12 feet long of imiform cross section is hinged at one end and rests on a support which is 4 feet from the other. The free end carries a load of 130 pounds. Excluding the weight of the beam itself, what is the weight in pounds on the support ? Hint. The products of the upward and downward pressures by their respective arms are equal. 37. A 14-foot horizontal beam of uniform cross section weigh- ing 200 pounds is hinged at one end and rests on a support at the other end. (a) What is the weight in pounds on the support ? (5) If the support is moved in 3 feet from the end of the beam, find the pressure in pounds on the support. 38. A 16-foot horizontal beam of uniform cross section weighs 300 pounds. It rests on two supports, one at one end and the other 4 feet from the other end. Find the weight in pounds on each support. CHAPTER XXVII DETERMINANTS AND REVIEW OF LINEAR SYSTEMS 124. Graphical solution of linear systems. The graph of a linear equation in two variables is a straight line. Therefore it is necessary in constructing the graph of such an equation to locate only two points whose coordinates satisfy the equation and then to draw through the two points a straight line. It is usually most convenient to locate the two points where the line cuts the axes. If these two points are very close together, however, the direction of the line will not be accurately determined. This error can be avoided by selecting two points at a greater distance apart. The graphical solution of a linear system in two variables consists in plotting the two equations to the sajne scale and on the same axes, and obtaining from the graph the values of x and y at the point of intersection. Two straight lines can intersect in but one point. Hence but one pair of values of x and y satisfies a system of two independent linear equations in two variables. Through the graphical study of equations we unite the subjects of geometry and algebra, which have hitherto seemed quite separate, and learn to interpret problems of the one in the language of the other. The student should make such a review of the definitions, illus- trations, and theory on pages 187-200 as will enable him to solve the following exercises. EXERCISES ^ 2a; + 2/ = 8, ^ a; + 5 = _3y^ Solve graphically : 2a; + 2/ = 8, a: + 2y = 7. "' 6y + 2a; - 8 = 0. a; -2/ =6, 2,x + 4^ = 20, 3x + 4y = 4. 22/-10 = -x. g a; + 2y + ll = 0, ^ x + y = 6, x+2y = 0, „ + 4 = 0, 82/ + 2x = 3. ■ 2-a; = 0. 318 LINEAR SYSTEMS 319 In Exercise 9 graph each equation. Then add or subtract the cor- responding members of the two equations and graph the resulting equation to the same scale on the same axes. Note the position of the third graph with reference to the other two. Proceed in like manner with Exercises 10 and 11. x + y = 4:, x-y = 5, 3a;-4y = 12, ' x + 2i/ = T. ■3a; + 2y = 5. 4x + 3y = -G. 12. In each of the last three exercises will the values of the X- and y-coordinates of the point of intersection of the two lines, as obtained from the graph, verify in the third equation obtained by adding the two given equations ? Why ? 13. Graph the equation a; — 2 ?/ = 2. Then multiply both mem- bers by 2 or 3 and graph the resulting equation. Compare the two graphs. Then try — 2 or — 3 as a multiplier and graph the resulting equation. Compare the three graphs. What con- clusion seems warranted ? 14. What are the coordinates of the origin ? 15. Is a graphical solution of a linear system in two varia- bles ever impossible ? Explain. 16. In the example on page 198 could different scales have been used on the two axes ? Could the two lines have been plotted to different scales ? Explain. 17.- What is the form of the equation of a line parallel (a) to the a;-axis ? (b) to the y-axis ? 18. What is the form of the equation of a line through the origin ? 19. Give ,an example of a system in two unknowns which has (a) no graphical solution ; (b) an infinite number of sets of roots. 20. The boiling point of water on a Centigrade thermometer is marked 100°, and on a Fahrenheit 212°. The freezing point on the Centigrade is zero and on the Fahrenheit is 32°. Conse- quently a degree on one is not equal to a degree on the other, nor does a temperature of 60° Fahrenheit mean 60° Centigrade. Show that the correct relation is expressed by the equation 320 COMPLETE SCHOOL ALGEBRA C = |(F — 32), where C. represents degrees Centigrade and F degrees Fahrenlieit. Construct a graph of this equation. Can you, by means of this graph, express a Centigrade reading in degrees Fahrenheit, and vice versa ? 21. By means of the graph drawn in Exercise 20 express the following Centigrade readings in Fahrenheit readings, and vice versa. (a)60°C; (J) 160°F; (c)-20°C; (d)-30°¥. 22. What reading means the same temperature on both scales ? 23. A boy starts at the southwest corner of a field and walks 20 rods, keeping twice as far from the south fence as from the west fence. He then walks east until he is three times as far from the west fence as he is from the south fence. Lastly he walks north until he is as far from one fence as he is from the other. Construct a graph of his path. Find (by measurement) the length of each portion of it and his distance from the starting point. 125. Elimination. The process of deriving from a system of n equations a system of »i — 1 equations, contaiaing one variable less than the original system, is called elimination. If one equation of a system can be obtained from one or more of the other equations of the system by the direct ap- plication of one or more of the axioms, it is called a derived equation ; if it cannot be so obtained, it is called independent. Only two methods of elimination will be considered, -^ that of addition or subtraction, and that of substitution. The student should review the definitions, examples, and rules on pages 203-209. EXERCISES 1. What is a constant ? a variable ? 2. Define and give examples involving two unknowns of (a) a linear equation ; (b) a system of linear equations ; (c) a simultaneous linear system ; (d) equivalent equations ; (e) a de- terminate system ; (/) an indeterminate equation ; (g) an inde- terminate system ; (A) an incompatible or inconsistent system. 3. What is the graph in each case in Exercise 2 ? LINEAR SYSTEMS 321 4. What is the general form of a linear equation in two variables ? 5. To what general form may any incompatible linear sys- tem in two unknowns be reduced ? 6. What is the general form of a linear system in two unknowns which has an infinite number of sets of roots ? Solve by addition or subtraction : * 2x + 5tj = 8, 9t-2n = 18, a;-102/ = 9. ■20^ = 771 + 63. 5x + 38 = 122/, llm-10 = -18ra, 3a; + 82/=0. 9m + 12n = -15. 11. 3a;-2y = 18, 30 + 87/ = 5a;. Solve by substitution : 3r-8s = 13, r _i_14 .„ n 12. , „ rt 1- 6a; + ^-12 2/ = 0, r + 6 s = 0. 15. o j3 2 (a: + 2/) + 32/ = 4, 7y-3x-i = 0. ■5 = x + y. 8^L^ + 6,.=-9, 16a; + 7 = 15y, 16. 2 ' ix + Sy = 0. 4m — 1 = 3 w. Solve by either of the preceding methods : 1 1__1 17. 3 ^^--■' x + 2/~ 6' ?_§ = _! a; y 3 18. 4 2 12 2w + 3w-2 _4 m + n + 6 3 1.1 9 19. 2x^ 3 +^- ^- s 5a; — 3?/ = 105. 3/- 7 4 2~ s 12' r- + 8 = - -2s. 32/ + 1 « + 20 13 12 z 2y 8 9 1. 22. = - 3, - + - = ■ .25 7i. + 8 + .1 7c _ 7c - 10 + /t 23 5x + .7y = -1.25, '^ ,_20_^ ^"- .12a;-.082/ = 3f. .8 7i - 2.2 ^ 36 - 5.5 7c ' 322 COMPLETE SCHOOL ALGEBRA f (2 /i + 5 k) = 39, 3(x + y) 24. 2h 26. 3A; 4 f 5 25. 3(/c-7i). m — 6 in — § (3 t)i - + ^^^ = 0, t -V- 2 a; + 2/ = 7. 1 2n)-^ n M — 10 Solve for X and y : 2 ex — 2/ = 5 ac, 27. 2 a; ?/ 2w-l + 3 = *^- ay+{a + b)x ■■ 30. ft + ^' 4_ il _ 4 "3" :a;y, y 28. « 3 a - + — = 1, X y 3 (' a 1 X y 2 a: + ?/ 29. 10 a x = y. = 2 4a 31. 32. 33. (c + i)y + aa; = 1, ay = 1 — (c + 6) a;. aa,' -\- by = c, dx -\- ey ^ f. 126. Determinants of the second order. The arrangement of 4 2 has been given the meaning 4 • 3 — 5 • 2. numbers 5 3 Such an arrangement is called a determinant. The value of any such determinant is easily found since means ad — he. Accordingly Similarly, And 3 ^. — 4 The preceding operations can be reversed and the difference (or sum) of two products written as a determinant. Thus mn — rs can be written number of other ways. Similarly, ah — h = ab — 1-k = Accordingly, 5 6 2 3 Similarly, 6-2 8 3 And 3 3 -4 9 -5 ~ = 5-3-2.6 = 15-12 = 3. = 6 ■ 3 - 8 • (- 2) = 18 + 16 = 34. : 3[_ 15 _ (_ 36)] = 3[_ 16 + 36] = 63. , and in a , etc. m s in r r n , or s n a k b 1 ~ 1 b k a LINEAR SYSTEMS 323 EXERCISES Find the value of the determinants : 1. 4 1 3 6 2. 6 - 8 2 1 4. 5. 3 2 a —10 b Sa 3h _5^ 2a 3. 3 6. 4 2 3 4 -3 Write as a determinant : 10. ax — cr. 12. f — ar. 11. mz — 3d. 13. hk — c. Find the value of the fractions : 16. »I 3e 9d 2a -14e= 14. aS + cd. 15. a-i. 5 3 1 -1 - 17. 3 2 4 1 . 18. c b f e ■ 19. c 7c (l-c) (7 -12 c) 1 1 1 -1 3 2 4 3 a b d e c 3 7 c 36 Write as the quotient of two determinants : 20. 21. st — cd 3cx — 5r ■ 1_2 e 22. a a' 12 23. ax - -6 2r- -5t m 3 " -7 24. 25. 2m-l 6 a; — hm 3a -3r' x^' + O a^ 15' 4 "^ a 127. Solution by determinants. For the general linear system in two unknowns ¥. bd b f e a h d e ax + by = c, dx + ey =/, (3), and y = a c af — cd d f ae — bd a b d e (1) (2) (4) The determinant expressions for x and y in (3) and (4) can be used as formulas to solve any pair of linear equations in two unknowns. This method is particularly useful in the 324 COMPLETE SCHOOL ALGEBRA solution of linear equations with literal coefficients. The de- terminant forms can be easily remembered and written down at once if we observe carefully the following points : I. The determinants in the denominators are identical, and each is formed by the coefficients of x and y as they stand in the original equations (1) and (^). II. The determinant in the numerator of the value of x is formed from the denoTninator by replacing the eoefficientsr of x, a c d, by the constant terms f. III. The determinant in the num,erator of the value of y is formed from the denom,inator by replacing the coefficients of b c y, e, by the constant terms f. Biographical Note. Gottfried Wilhelm Leibnitz. For the last few hundred years the study of the higher mathematics has been carried on almost entirely by professors in the universities. It is rather exceptional for a man not connected with any educational institution to achieve dis- tinction in this field. Before this was the case, hovyever, scholars were accustomed to devote themselves to any or all branches of learning which attracted them, and many men of wide erudition in various walks of life flourished at different times during the two or three hundred years fol- lowing the fifteenth century. But of them all, the man who perhaps most clearly deserves the title of universal genius is Leibnitz (1646-1716). He was born in Leipzig, Germany, and on account of the poor instruction in the school to which he was sent, he was obliged to learn Latin by himself, which he did at the age of eight. By the time he was twelve he read Latin with ease, and had begun Greek. Not until the age of twenty-six, when he was sent to Paris on a political errand, did he become deeply interested in mathe- matics. Prom 1676, for nearly forty years, he held the well-paid posi- tion of librarian in the ducal palace of Brunswick, serving under three princes, the last of whom became George I of England in 1714. This post afforded him time for the deep study of mathematics, philosophy, theology, law, politics, and languages, in all of which he distinguished himself. An incomplete edition of his mathematical works has been published in seven volumes. It is in his writings that we find the first mention of determinants. He also discovered the calculus independently of Sir Isaac Newton, and the last years of both men were embittered by a most unfortunate wrangle in which the friends of Newton accused Leibnitz of publishing as his own, results which really belonged to Newton. '».■■'■. , --./'^^i*^: /^.i:;' ^ Iffli 1 ^;t:|v ^ ■■ '" *li "'^ 1, "'.'*•■ "'■- ''r ;■,. GOTTFRIED WILHELM LEIBNITZ LINEAR SYSTEMS 325 Personally Leibnitz was quick of temper, impatient of contradiction, overfond of money, and one of the few really great men wlio have been ofEensively conceited. Example : Solve by determinants i „ ^ „ ~ ' ^ •' . l5a; = 2y + ll. Solution : Writing the equations in the standard form, we have 5x-2y = \l. 7 2 -14-22 -36 Then 11 -2 2 -2 - 2 - 10 - 12 = 3. In solving for y the denominator is the same as before ; hence 1 7 11 11-35 - 24 y = ■12 - 12 - 12 = 2. 3=7, : 4 + 11. 6. Check: {Jg+ EXERCISES Solve by determinants and check results 2a; + 32/ = 7, 3«-22/ = 4. 5 2/ + 6 = 3 a. 5x + 4.y = 10 a + 4:, X — 2 ay = 0. .3 a; +,02 y = 185, .5a; + .04> = 335. 4 a; + 3 y = 6, 1. 2. 4. 2 3' 3aJ 2y_ 7 a; + 6 2/ = 21 c, 7. x_y_ 2c' 3. 3a; 32/_ 8. X y _a ■ a b I y- ah 128. Indeterminate equations. If numerical values are given to any two variables in the equation m + n -\- p = &, a value for the third variable can be found, which, taken with the values assigned to the other variables, satisfies the equation. 326 COMPLETE SCHOOL ALGEBRA For example, let in = 1 and n = 2. Then m + n + p = 6 becomes 1 + 2 + /) = 6, whence p = 3. Obviously m = 1, n = 2, and p = 3 satisfy the equation. Other values may be given to m and n (or m and p, or n and p), and the foregoing process repeated, thus obtain- ing set after set of roots. A few sets of roots are tabulated here. in 1 2 1 6 i -4 10 n 2 2 3 1 2 -1 P 3 2 6 2 3 8 -3 It can easily be shown that the above table can be indefinitely extended ; that is, that every linear equation in three variables has an infinite number of sets of roots. It can also be shown that a system of two independent equations of the first degree in three variables has an infinite (unlimited) num- ber of sets of roots. A system of three independent equations of the first degree in three variables, no two' equations being incom2)alible, has one set of roots and only one. A system of four independent linear equations in three variables has no set of roots. Note. It is not a little remarkable that the writings of the first great algebraist, Diophantos of Alexandria (about 300 a.d.), are devoted almost entirely to the solution of indeterminate equations ; that is, to finding the sets of related values which satisfy an equa- tion in two variables, or perhaps two equations in three variables. We know practically nothing of Diophantos himself, excepting the information contained in his epitaph, which reads z^ follows : "Dio- phantos passed one sixth of his life in childhood, one twelfth in youth, one seventh more as a bachelor ; five years after his marriage a son was born who died four years before his father, at half his father's age." From this statement the reader was supposed to be able to find at what age Diophantos died. As a mathematician Dio- phantos stood alone, without any prominent forerunner, or disciple, so fax as we know. His solutions of the indeterminate equations were exceedingly skillful, but the methods which he used were so obscure that his work had comparatively little influenoe upon that of later times. LINEAR SYSTEMS 327 129. Determinate systems. The method of obtaining the set of roots of a determinate system is illustrated in the example on page 222. If necessary, the student should refer to that example and the rule and explanations on page 223. EXERCISES 1. Mnd five sets of roots for x— 2y + s = 6. 2. Find three sets of roots for the system TO + w — J? = 8, 3to^2to + 4j)=6. Solve the following systems : 2x + 3i/ = -U-4:Z, Ar + .3s-8t = i, 3. a;-2/ + 3» = 0, 5. .5r + t + .Ss = 1.2, 5x + s = 14:-2y. 2.6t + .3-r=+.5s. x + 22/ + 3z = U, .25a3 + .062/ = -l + .103, 4. 4a;-5y+6s = 12, 6. .60 a; - .30 y = 0, x + 15tj + 9z = 58. .052/ + .04« = 3. In Exercises 7 and 8 consider a, b, c as known numbers. r,^_J_^Q h + 2k-l = 3b + c, a 2a 3a ' 5h-ik- il = a + b-8o, 7. Zr + it = 6s, '±- 3tt + ^' , ^ _ 2c 9 6 In Exercises 9-11 solve for x, y, and z. Solve Exercise 15 for x only. a-.a^y.a' = a-\ «, 5,c_o 9 g2:r.g-;,.g» + l_g-19^ X y »~ ' e-.e-v.e-^' = e^^ ^^ a_b 2c^ &'".6!'+.2 = 5i2, ' X y » ' 10. c^.) we replace the coefficients of z,f, by the constant P i terms q, we obtain a determinant equivalent to the numerator of the fraction (6) whose value is z. (The student should perform the work outlined in the last two sentences.) Therefore we may write the values of x, y, and z for the given system in determinant form as follows : p h c q e f r h t a b c d e f 9 h I (') 2/ = a P c d q f Vll. 18. Ve, V3. 20. VS, a/6. 19. Vi9, V7. 21. 2V6, ^89. 346 COMPLETE SCHOOL ALGEBRA 22. 3 V3, 2 a/10. 24. Vs, 4^6, -?/l25. 23. a/48, -v/64. 25. 4a^,3^25, 4^64. Reduce to respectively equivalent surds of the same order : 26. VS, -^/is^. 28. 2x-V5xy,?>x VSxy. 29. 27. Va + i, Va — &. Square : 30. a/3. 31. 2-^1. Cube: 36. 3V5. xi/, ^xij', 7 2 Vs. 32. VS- 33. 4 \/3 - Vs. 34. ^^4 - 4 V3. 35. v'e - 3 V2. 37. 3 V2 - 2 V3. 38. ( Vl - V2)'. Simplify : 39. (5V5 + 9V3- V7 + 2Vi06)(V3 + V5- Vr). 40. (Va — Vac +'Vc)(Va + Vac + Vc)(a + c + Vac). 41. (V2a;-l-V5)(2V2x-l+V45)(4a--Vl0x-5-17). 42. l.^R^ f^-f 44. ^R^ R V4- V2 ■{e'-2 e-=) (e^ - 2 e"'') + e^^ + e" 140. Division of real radicals. Division of one real radical by another may often be performed as in Examples 1-3, page 251. Direct division of radical expressions in which the divisor is a polynomial is very difBcult. Where division by a polynomial divisor is necessary we use the rule of pages 252-253. This rule applies in all cases, while the rule for direct division fails when dividing a real radical by a radical of even order whose radicaad is negative. ROOTS AND RADICALS 347 EXERCISES Find a simple rationalizing factor for : 1. 3V7. 4. V^. 7. V3^-V2^. 2. 5 -v'i. 5. VS - 7. 8. Va;-c - Va. 3. 7- where a is the altitude and b is the area of the base. 9. The base of a pyramid is a square, each side of which is 10 feet. The other four edges are each 20 feet. Find the altitude and the volume of the pyramid. 10. The base of a pyramid is a rectangle 8 by 18. The other four edges are each 16. Find the altitude of the pyramid. ROOTS AND RADICALS 351 11. The side of an equilateral triangle is 18. Find the two parts into which each altitude is divided by the other altitudes. Fact from Geometry. The altitudes of an equilateral triangle inter- sect at a point -which divides each altitude into two parts whose ratio is 2 to 1. The altitude of a regular tetrahedron {BK in the adjacent figure) meets the base at . the ^ point where the altitudes of the base intersect. 12. ABCD is a regular tetra- hedron. If each edge is 12, find CR, CK, and lastly the alti- tude DK. Fact from Geometry. A regular tetrahedron is a pyramid whose four sides are equal equilateral triangles. 13. Find the altitude and vol- ume of a regular tetrahedron whose edge is 15. 14. Show that the altitude and the volume of a regular tet- rahedron whose edge is e are respectively ^ VG and q-r V2. 15. The base of a pyramid is a regular hexagon each side of which is 10 inches. The other edges of the pyramid are each 16 inches. Find the altitude and the volume of the pyramid. CHAPTEE XXIX EXPONENTS 143. Fundamental laws of exponents. The laws of exponents may be stated as follows : I. Law of Multiplication, Law I may be stated more completely thus : jfU , X^ ' X^ ' ' • y g + & + C This follows directly from the definition of an exponent and from the Associative Law. For instance, xx = x^, and xxx = x^, and xxxxx = x^ by definition. Hence xx ■ xxx = xxxxx, or x'^-x^ = x^. 11. Law of Division,