Cornell University Library The original of tliis book is in tlie Cornell University Library. There are no known copyright restrictions in the United States on the use of the text. http://www.archive.org/details/cu31 924031 263795 Cornell University Library arV19285 Bradbury's Elementary algebra : 3 1924 031 263 795 olin.anx (Bntan anir ^rairbwrg's Piat^cmatital S'&tm. EATON'S ELEMENTARY ALGEBRA DESIGNED FOR THE USE OP HIGH SCHOOLS MD ACADEIIES. WILLIAM F. BEADBUEY, A. M., nOPKIXS MASTER IN THE CAMBRIDGE HIGH SCHOOL ; AOTaOR OP A TREATISE 01 TRIGONOMBTM AND SIIKTEYING, AND OF AH ELEMENTARY GEOMETRY. BOSTON: THOMPSON, BROWN, AND COMPANY, 25 & 29 COBNHILL. 1875. EATON AND BRADBURY'S TJSED WITH TOfEXAMPLED SUCCESS IN THE BEST SCHOOLS AUD ACADEMIES OF THE COUNTRY. Eaton's Prima by Arithmetic Eaton's Intellectual Arithmetic. ^^—,~, — ^ .^.^2Jh^Z2^. Eaton's Common School Arithiu^|> C\ t-^ \ Eaton's High School Arithmi !?«. ■ «. . . t / f— r-s<-% i -b-^ ;UNfVERSiT Eaton's Elements of AEiTHMETSbx, I I R f ' Eaton's Grammar School ApttttmWtt!":: -y Bradbury's Eaton's Elementary Algebra. Bradbury's Elementary Geometry. Bradbury's Elementary Trigonometry. Bradbury's Geometry and Trigonometry, in one volume. Bradbury's Trigonometry and Surveying. Keys of Solutions to Common School and High School Arithmetics, to Elementary Algebra, Geom- etry, AND Trigonometry, and Trigonometry and Sur- veying, /or the use of Teachers. Entered according to Act of Congress, in the year 1868, ET WILLIAM P. BRADBURY AND JAMES H. EATON, in the Clerk's Office of the District Court of the District of Massachusetts. University Press : Welch, Bigelow, & Co., Cambridge. PREFACE. It was tlie intention of the author of Eaton's Arith- metics to add to the series an Algebra, and he liad com- menced the preparation of such a work. Although its completion has devolved upon another, the author, as far as practicable in a work of this character, has followed the same general plan that has made the Arithmetics so popular, and spared no labor to adapt the book to the wants of pupils commencing this branch of mathematics. A few problems have been introduced in Section II., to awaken the pupil's interest in Algebraic operations, and thus prepare him for the more abstract principles which must be mastered before the more difficult problems can be solved. Special attention is invited to the arrangement of the equations in Elimination ; to the Second Method of Completing the Square in Affected Quadratics ; and to the number and variety of the examples given in the body of the work and in the closing section. The Theory of Equations, the Explanation of Negative Results, of Zero and Infinity, and of Imaginary Quantities, are omitted, as topics not appropriate to an Elementary Algebra. It may also be better for the younger pupils to IV PREFACE. pass over the two theorems in Art. 74, until they become more famiHar with algebraic reasoning. While the book has not been made simple by avoiding the legitimate use of the negative sign before a parenthesis or a fraction, the difficulty which is caused to beginners by the introduction of negative indices in simple division has been obviated by deferring their introduction to the section on Powers and Roots, where they are fully ex- plained. The utmost conciseness consistent with perspicuity has been studied throughout the work. It is hoped the book will commend itself to both teachers and pupils. W. F. B. Cambkidge, Mass., May 17, 18S8. CONTENTS. SECTION I. Definitions and Notation The Signs 7 | Axioms , Page . 7 . 9 AU3£BBAI0 OfEEATIONS . SECTION II. 10 SECTION III. Defdotions and Notation (Continued from Section I.) 14 SECTION IV. Addition 19 SECTION V. Sobteaotion 25 SECTION VI. MULTIPLrCATION 31 SECTION VII. Division 37 Demonstration of Theorei«is SECTION VIII. 43 Faotorino Greatest Common Ditisor . SECTION IX. SECTION X. 47 54 Least Comsion Multiple , SECTION XI. Fractions SECTION XII. General Principles 63 Signs of Fractions 64 To reduce a Fraction to its lowest terms 65 To reduce Fraetions to equivalent Frac- tions having a Common Denominator 66 To add Fractions 68 To subtract Fractions 69 To reduce a Mixed Quantity to an Im- proper Fraction 70 To reduce an Improper Fraction to an Integral or Mixed Quantity ... 72 To multiply a Fraction by an Integral Quantity 73 To multiply an Integral Quantity by a Fraction 75 To divide a Fraction by an Integral Quantity 76 To divide an Integral Quantity by a Fraction 77 To multiply a Fraction by a Fraction . 77 To divide a Fraction by a Fraction . , 79 SECTION XIII. Bquations of the First Degree containing dot one Unknown QuANTirr Definitions 82 Transposition . 83 Clearing of Fractious 85 Reduction of Equations 86 Problems 91 VI CONTENTS. SECTION XIV. Equatiohs op the TntsT Degree containing two Unknown QtJANTinES . - . • 104 Elimination by Sntstitution .... 105 I Elimination by Combination .... 108 Elnnination by Comparison .... 106 | Problems ^^ SECTION XV. Equations op the Fibst Degree containing more than two Unknown Quan- tities * ' SECTION XYI POWKRS anh Koots . .... N^atiTe Indices 125 118 125 Multiplication and Division of Powers of Monomials 126 Transferring Factors from Numerator to Denominator, or Denominator to Numerator of a Fraction .... 127 Involution of Monomials 129 Involution of Fractions 130 Involution of Binomials • ... 131 Square Root of Numbers .... 139 Cube Root of Numbers 142 Evolution of MonomiaL? 147 Square Root of Polynomials .... 148 To find any Root of a Polynomial . 152 SECTION XVII. Radicals 154 Definitions 154 To reduce a Radical to its Simplest Form 154 To reduce a Rational Quantity to the form of a Radical 157 To reduce Radicals having different In- dices to equivjUent ones having a Common Index . 1^ To add Radicals 160 To subtract Radicals . .... 361 To multiply Radicals 162 To divide Radicals . 163 To involve Radicals 165 To evolve Radicals 16S Polynomials having Radical Terms . 167 SECTION XTIII. Pdbb Equations which require in their Reduction eit3er Involution or Eto- 170 SECTION XIX. Affected Quadratic Equations . . . 178 Completing the Square 178 I Third Method of Completing the Square 185 Second Method ofCampleting the Square 182 I Problems 193 SECTION XX. Quadratic Equations containing two Unknown Quantitibs . ... 196 SECTION XXI. Ratio and Proportion . ... , ... 207 SECTION XXII. Arithmetical Progression ,, ^ 215 Geometrical Progression Miscellaneous Examples SECTION XXIII. SECTION XXIV. SECTION XXV. 225 236 253 ELEMENTARY ALGEBRA SECTION I. . DEFINITIONS. 1. Mathematics is the rscience of quantity. 2i Quantity is that which can be measured ; as distance, time, weight. 3i Arithmetic is the science of numbers. In Arithmetic quantities are represented by figures. 4.i Algebra is Universal Arithmetic. In Algebra quan- tities are represented by either letters or figures, and their relations by signs. NOTATION. 5. Addition is denoted by the sign +, called plus; thus, 3 + 2, i. e. 3 plus 2, signifies that 2 is to be added to 3. 6« Subtraction is denoted by the sign — , called minus; thus, 1 — 4, i. e. t minus 4, signifies that 4 is to be sub- tracted from t. 7, Multiplication is denoted by the sign X i thus, 6X5 signifies that 6 and 5 are to be multiplied together. Be- tween a figure and a letter, or between letters, the sign X is generally omitted ; thus, 6 a & is the same as 6 X a X &• Multiplication is sometimes denoted by the period ; thus, 8 . 6 . 4 is the same as 8 X 6 X 4. 8 ELEMENTARY ALGEBRA. 8. Division is denoted by the sign -^ ; thus, 9-^3 sig- nifies that 9 is to be divided by 3. Division is also indi- cated by the fractional form ; thus, f is the same as 9 -^ 3. 9. Equality is denoted by the sign = ; thus, $1 = 100 cents, signifies that 1 dollar is equal to 100 cents. An ex- 'pression in vs^hich the sign = occurs is called an equa- tion, and that portion which precedes the sign = is called the first member, and that which follows, the second mem- ber. 10. Inequality is denoted by the sign > or <, the smaller quantity always standing at the vertex ; thus, 8 > 6 or 6 < 8 signifies that 8 is greater than 6. 11. Three dots .•. are sometimes used, meaning hence, (here/ore. ( 12. A Parenthesis ( ), or a Vinculum , indicates /that all the quantities included, or connected, are to be considered as a single quantity, or to be subjected to the \same operation ; thus, (8 -j- 4) X 3 = 12 X 3, or = 24 .+ 12 = 36 ; 21 — 6 -r- 3 = 15 -=- 3, or = 7 — 2 = 5. Without the parenthesis, these examples would stand thus :8 + 4X3 = 8-|-12 = 20;21— 6-^-3 = 21 — 2 = 19 ; the sign X, in the former, not affecting 8 ; nor the sign -j-, in the latter, 21. Examples. 1. 9 + T — 3 + 4 = how many? 2. (9 + 15) -^ 3 = how many? 3. — Yq — X 14 :^ how many ? 4. (14 + 13) X (5 — 2) = how many? 5. 10 + (7 — 4) -- 3 X 4 = how many ? 6. 25 — (6 + 7) = how many? 7. 150 — (18 — 11) = how many? DEFINITIONS. 9 8. Prove that ITS + 8 — 49 = 14 + 190 — 54 — 16. 9. Prove that 216 — 44 -f 14 > 144 + 13 — 75. 10. Place the proper sign (^, >, or <) between these two expressions, (247 + 104) and (546 — 195). 11. Place the proper sign (^, >, or <) between these two expressions, (119 — 47 + 16) and (317 — 104). 12. Place the proper sign (:=:, >, or <) between these two expressions, (417 + 31) — (187 — 72) and (127 + 179). AXIOMS. IB, All operations in Algebra are based npon certain self-evident truths called Axioms, of which the following are the most common : — 1. If equals are added to equals the sums are equal. 2. If equals are subtracted from equals the remainders are equal. 8. If equals are multiplied by equals the products are equal. 4. If equals are divided by equals the quotients are equal. 6. Like powers and like roots of equals are equal. 6. The whole of a quantity is greater than any of its parts. 7. The whole of a quantity is equal to the sum of all its parts. 8. Quantities respectively equal to the same quantity are equal to each other. 1* 10 ELEMENTARY ALGEBEA. SECTION II. ALGEBRAIC OPEEATIONS. II. A Theorem is something to be proved. 15, A Problem is something to be done. 16. The Solution of a Problem in Algebra consists, — 1st. In reducing the statement to the form of an equa- tion ; 2d. In reducing the equation so as to find the value of the unknown quantities. Examples for Practice. 1. The sum of the ages of a father and his son is 60 years, and the age of the father is double that of the son ; what is the age of each .'' It is evident that if we knew the age of the son, by doubling it we should know the age of the father. Sup- pose we let X equal the age of the son; then 2x equals the age of the father ; and then, by the conditions of the problem, x, the son's age, plus 2 a;, the father's age, equals 60 years; or 3a; equals 60, and (Axiom 4) x, the son's age, is i of 60, or 20, and 2x, the father's age, is 40. Expressed algebraically, the process is as follows : — Let X = son's age, then 2 a; =^ father's age. a: -|- 2 a; == 60, 3 a; = 60, ^ = 20, the son's age. 2 3:- = 40, the father's ag-e. DEFINITIONS. 11 2. A horse and carriage are together worth $450 ; but the horse is worth twice as much as the carriage ; what is each worth? Ans. Carriage, $150; horse, $300. All problems should be verified to see if the answers obtained fulfil the given conditions. In each of the pre- ceding problems there are^^two conditions, or statements. For example, in Prob. 2 it is stated (1st) that the horse and carriage are together worth $450, and (2d) that the horse is worth twice as much as the carriage ; both these statements are fulfilled by the numbers 150 and 300. 3. The sum of two numbers is 72, and the greater is seven times the less ; what are the numbers ? / 6 J i 4. A drover being asked how many sheep he had, said that if he had ten times as many more, he should have 440 ; how many had he ? Y J 5. A father and son have property of the value of $8015, and the father's share is four times the son's; what is the share of each? Ans. Father's, $6412; son's, $1603. 6. A farmer has a horse, a cow, and a sheep ; the horse is worth twice as much as the cow, and the cow twice as much as the sheep, and all together are worth $490 ; how much is each worth ? OPEBATION. Let- X =: the price of the sheep, then 2 a: = " " " " cow, and 43; = " " " " horse ; and their sum "T a; = 490, X = TO, the price of the sheep, and 2 a; = 140, " " " " cow, and 4 a; = 280, " " " " horse. 12 ELEMENTARY ALGEBRA. 7. A man has three horses which are together worth $540, and their values are as the numbers 1, 2, and 3; what are the respective values? Let X, 2x, and Sx represent the respective values. Ans. $90, $180, and $2Y0. 8. A man has three pastures, containing 360 sheep, and the numbers in each are as the numbers 1, 3, and 5 ; how many are there in each ? T ^: 9. Divide 63 into three parts, in the proportion of 2, 3, and 4. Let 2 a;, 3 a;, and 4 x represent tlie parts. 10. A man sold an equal number of oxen, cows, and sheep for $ 1500 ; for an ox he received twice as much as for a cow, and for a cow eight times as much as for a sheep, and for each sheep $ 6 ; how many of each did he sell, and what did he receive for all the oxen ? Ans. 10 of each, and for the oxen, $960. 11. Three orchards bore 8Y2 bushels of apples; the first bore three times as many as the second, and the third bore as many as the other two ; how many bushels did each bear ? / [ 9 12. A boy spent $4 in oranges, pears, and apples; he bought twice as many pears and five times as many apples as oranges ; he paid 4 cents for each pear, 3 for each orange, and 1 for each apple ; how many of each did he buy, and how much did he spend for oranges ? how much for pears, and how much for apples ? . (25 oranges, 50 pears, and 125 apples. (Spent for oranges, $0.'?5; pears, $2; apples, $1.25. 13. A farmer hired a man and two boys to do a piece of work ; to the man he paid $ 12, to one boy $ 6, and to the other $ 4 per week ; they all worked the same time, and received $ 264 ; how many weeks did they work ? Ans. 12 weeks. DEFINITIONS. 13 14. Three men, A, B, and C, agreed to build a piece of wall for $ 99 ; A could build 7 rods, and B 6, while C could build 5; how much should each receive? -i' , i' 15. Four boj^s, A, B, C, and D, in counting their money, found they had together $1.98, and that B had twice as much as A, C as much as A and B, and D as much as B and C ; how much had each ? Ans. A 18 cents, B 36, C 54, and D 90. 16. It is required to divide a quantity, represented by a, into two parts, one of which is double the other. OPERATION. Let X = one part, then 2x = the other part. 3 a; = a. o one part. 2x = — , the other part. 17. If in the preceding example a = 24, what are the required parts ? ^?^"- 3 = T = ^' ^""^ T = T = ^^- 18. It is required to divide c into three parts so that the first shall be one half of the second and one fifth of the third. )L 1-3. y ^ry - C^ Ans. '-. , -^ and tl. ,8 8 8 19. Divide n into three parts, so that the first part shall be one third the second and one seventh of the third. t7 ■'^;/^ '^' 20. A is one half as old as B, and B is one third as old as C, and the sum of their ages is p ; what is the age of each ? ^„^_ j^,^ E ^,^ ^_ ^^^ C's '-^. 14 ELEMENTARY ALGEBRA. SECTION III. DEFINITIONS AND NOTATION. [Continued from Section I ] 17. The last letters of the alphabet, x, y, z, &c., are used in algebraic processes to represent unknown quanti- ties, and the first letters, a, b, c, &c., are often used to represent known quantities. NuMEEicAL Quantities are those expressed by figures, as 4, 6, 9. Literal Quantities are those expressed by letters, as a, X, y. Mixed Quantities are those expressed by both figures and letters, as 3 a, 4 a;. 18. The sign plus, -f-, is called the positive or affirm- ative sign, and the quantity before which it stands a pos- itive or affirmative quantity. If no sign stands before a quantity, -\- is always understood. 19. The sign minus, — , is called the negative sign, and the quantity before which it stands, a negative quantity. 20. Sometimes both -\- and — are prefixed to a quan- tity, and the sign and quantity are both said to be am- biguous ; thus, 8 ± 3 = 11 or 5, and a ± b ^= a -\- b, or a — b, according to circumstances. 21. The words -plus and minus, positive and negative, and the signs -|- and — , have a merely relative signifi-' cation ; thus, the navigator and the surveyor always rep- resent their northward and eastward progress by the sign -[-, and their southward and westward progress by the sign — , though, in the nature of things, there is nothing to prevent representing northings and eastings by — , and, southings and westings by -|--- So if a man's pi-op- DEFINITIONS. 1, erty is considered positive, his gains should also be cor sidered positive, while his debts and his losses should b considered negative; thus, suppose that I have a farr worth $5000 and other property worth $3000 and tha I owe $1000, then the net value of my estate is $500' + $3000 — $1000 = $1000. Again, suppose my farr is worth $6000 and my other property $3000, whil I 'owe $12000, then my net estate is worth $500i + $3000 — $12000 =— $4000, i. e. I am wortl — $4000, or, in other words, I owe $4000 more than can pay. From this last illustration we see that the sigj — may be placed before a quantity standing alone, am it then merely signifies that the quantity is negative without determining what it is to be subtracted from. 22. The Terms of an algebraic expression are the quan titles which are separated from each other by the signi 4- or — ; thus, in the equation 4a — 6 =3a;-)-c — 1 y the first member consists of the two terms 4 a and — h and the second of the three terms 3 x, c, and — 1 y. 23. A Coefficient is a number or letter prefixed to i quantity to show how many times that quantity is to b( taken ; thus, in the expression 4 x, which equals x -\- a -\- X -\- X, the 4 is the coefficient of a; ; so in Sab, whicl equals ab -\- ab -\- ab, 3 is the coefficient of a 6; in 4 a 6 4 a may be considered the coefficient of 6, or 46 the co efficient of a, or a the coefficient of 4 b. Coefficients maybe numerical or literal or mixed; thus in 4c ab, 4 is the numerical coefficient of ab, a is the lit eral coefficient of 4 6, 4 a is the mixed coefficient of b. If no numerical coefficient is expressed, a unit is un derstood ; thus, x is the same as Ix, be as 16c. 24. An Index or Exponent is a number or letter placec after and a little above a quantity to show how many times that quantity is to be taken as a factor; thus, in the ex- 16 ELEMENTARY ALGEBRA. pression ¥, which equals b X b X b, the 3 \s the index or exponent of the power to which b is to be raised, and it indicates that 6 is to be used as a factor 3 times. An exponent, like a coefficient, may be numerical, lit- eral, or mixed ; thus, of, a;", x^^, &c. If no exponent is written, a unit is understood ; thus b =2 b^, a ^ a}, &c. Coefficients and Exponents must be carefully distin- guished from each other. A Coefficient shows the num- ber of times a quantity is taken to make up a given sum ; an Exponent shows how many times a quantity is taken as a factor to make up a given product ; thus A,x=^x-\-x-\-x-\-x, and x^ ^^ x X x X x X x. 25. The product obtained by taking a quantity as a factor a given number of times is called a •power, and the exponent shows the number of times the quantity is taken. 26. A Boot of any quantity is a quantity which, taken as a factor a given number of times, will produce the given quantity. A EooT is indicated by the radical sign, \/ , or by a fractional exponent. When the radical sign, \/, is used, the index of the root is written at the top of the sign, though the index denoting the second or square root is generally omitted ; thus, js/ X, or xh, means the second root of x ; ^ X, or a;i, " " third " " x, &c. Every quantity is considered to be both the first power and the first root of itself 27. The Recipeocal of a quantity is a unit divided by that quantity. Thus, the reciprocal of 6 is -, and of a: - 5 X DEFINITIONS. 1' 28. A Monomial is a single term; as a, or Sx, o bhxy. 29. A Polynomial is a number of terms connectec with each other by the signs plus or minus ; &s x -\- y or 3a -j- 4x — *lahy. 30. A Binomial is a polynomial of two terms ; ai 3 a; -|- 3 y, or a; — y. 31. A Eesidual is a binomial in which the two termi are connected by the minus sign, as a; — y. 32. Similar Teems are those which have the name power, of the same letters, as x and 8 a;, or baa? and — 2 a a? But X and a?, or 5 a and 5 h, are dissimilar. 33. The Degree of a term is denoted by the sum o the exponents of all the literal factors. Thus, 2 a is oi the 'first degree ; 3 a^ and 4 a 6 are of the second de gree ; and 6 a' x* is of the seventh degree. 34. Homogeneous Terms are those of the same degree Thus, 4a^a:, 3 a 6c, 3?y, are homogeneous with each other 35. To find the numerical value of an algebraic expres sion when the literal quantities are known, we must sub stitute the given values for the letters, and perform the operations indicated by the signs. The numerical value of t o — 6* -|- c^ when o ^ 4 6 = 2, and c =: 5 is 7 X 4 — 2* + 5^ = 28 — H + 25 = 37. Examples. Find the numerical values of the following expressions; when a ^ 2, S ^ 13, c ^ 4, c? = 15, m = 5, and n=.1. 1. a-\-b — G-\-2d. Ans. 41. 2. a'' -{- 3bc — 2cd. Ans. 40. 3. ^"' + "'' Ans. 219. 18 ELEMENTARY ALGEBRA. 4. m' — 2mn + n". 5. J - (^ - n'). 6. (a^ — c -{- b) {m + n). 7. ^' ~ '' X (rf - m + n). n — c Ans. 867. 8. Vc + 's/i.d + c — s/bm. Ans. 1. 9. 3a\/b — c X 4:n\/'2bm. 10. 5 m — 6n + Zd d — 2c Ans. 4. 11. V<^ — n + \/7n. 12. (6 — a) {d — c) — m. Ans. 116. 13. 13 (4«i) + 4(Z — 7a. 14. 4a 6 + v/lOOc ^ v^d — n. Ans. 122. 15. 4a VeO^ + 5a^6^ 16. b — a — {d — n). Ans. 3. lY. I — a —d — n. Ans. — 11. 18. (6 — a) (d — n). Ans. 88. 19. (6 — a) d — n. Ans. 158. 20. a + 6 \/10 (d — m) + \l\fc. S6. Write in algebraic form : — 1. The sum of a and 6 minus the difference of m and n. (m 3> n.) 2. Tour times the square root of the sum of a, b, and c. 3. Six times the product of the sum and diflFerence of c and d. (c >• d.) 4. Five times the cube root of the sum of a, m, and n. 5. The sum of m and n divided by their difference. 6. The fourth power of the difference between a and m. ADDITION. li SECTION lY. ADDITION. 37i Addition in Algebra is the process of finding th( aggregate or sum of several quantities. For convenience, the subject is presented under threi cases. CASE I. 38. When the terms are similar and have like signs. 1. Charles has 6 apples, James 4 apples, and Willian 5 apples ; how many apples have they all ? OPERATION. It is evident that just a 6 a 6 apples and 4 apples an( or, letting a 4 a 5 apples added togethe represent 5 a make 15 apples, so 6 a an( one apple, 4 a and 5 a added togeth 15 a er make 15 a. 6 apples, 4 apples, 6 apples, 15 apples. In the same way — 6 a and — 4 a and — 5 a ar( equal together to — 15 a. Therefore, when the terms are similar and have lik( signs : RULE. Add the coefficients, and to their sum annex the commoi letter or letters, and prefix the common sign. (2.) (3.) (4) (5.) (6.) a-) 5 ace Za" 4.x 6y — 3x8 — bhy Sax 4a^ X lOy — 2ar' — 2hy iax 7a2 5x y — tr' - hy 2ax 3a^ 3x •^y — 4ar' - hy 19 a a; 13a; ■16ar' 20 ELEMENTARY ALGEBEA. 8. What is the sum of a x% 3aa^, 2ax', and 4.ax'? Ans. 10 a x^. 9. What is the sum of 35a;, 45a;, 65x, and 5a:? 10. What is the sum of 2a;y, 6xy, 10 a; y, and 8xy? 11. What is the sum oi—lxz, —xz, —ixz, and ^2? Ans. — 13 xz. 12. What is the sum of — 2 5, — 3 5, — 6 5, and — 35? 13. What is the sum of — abc, — 3a5c, — 4o6c, and — a 5 c ? CASE II. 39t When the terms are similar and have unlike signs. 1. A man earns t dollars one week, and the next week earns nothing and spends 4 dollars, and the next week earns 6 dollars, and the fourth week earns nothing and spends 3 dollars ; how much money has he left at the end of the fourth week ? If what he earns is indicated by -f-, then what he spends will be indicated by — , and the example will appear as follows : — OPERATION. -|- 7 dollars, — 4 dollars, -}- 6 dollars, — 3 dollars, -}- 6 dollars, or, letting d represent one dollar, -\-1d — Id + 6d ~3d + 6d Earning 7 dollars and then spending 4 dollars, the man would have 3 dollars left; then earn- ing 6 dollars, he would have 9 dollars; then spending 3 dollars, he would have left 6 dol- lars ; or he earns in all 7 dollars -f 6 dollars ^13 dollars ; and spends 4 dollars -|- 3 dollars = 7 dollars; and therefore has left the differ- ence between 13 dollars and 7 dollars = 6 dollars; hence the sum of 4- 7d, — Ad, + ed, and — 3 d is -|- 6 d. Therefore, when the terms are similar, and have unlike signs : ADDITION. 2: EULE. Mnd the difference between the sum of the coefficients oj the positive terms, and the sum of the coefficients of the neg ative terms, and to this difference annex the common lette. or letters, and prefix the sign of the greater sum. (2.) (3.) (4.) (5.) 3xy if 13 a 6 c= Ux^'y xy -2f 6o6c^ — ■ \33?y — bxy If — a be'' lOa^y 1 xy -3f — Sab(f — x' y — "^xy 14j/= 4 a 6.0= 2ix'y ixy Uabc' (6.) (^•) 2bxy z 8 {x + y) — - bOxyz — 4 (a; + 2/) lOxyz 1{x + y) — -eixyz — 3{x + 2/) 8xyz - (^ + y) — lixyz ^ (.^ -\- y) 8. Find the sum of 8 x^y\ — 14 x'f, 11 x^y'', and —x^f 9. Find the sum of 1 {x -\- y), 8 (a; + y), — (x -\- y) and 4 (a; -|- y). Ans. 18 (x -\- y). 10. Find the sum of — ax^, -{-a x^, — 10 a t?, -|- 25 a a;= and — 13 a x^. 11. Find the sum of 27 a J, — 34 a J, — 150 ah, 27 a J and — 13 a 5. Ans. — 143 a h. 12. Find the sum of an?, — 14 a a:', 11 ax?, — ax^ 44 a a^, and — aa^. 13. Find the sum of 17 (a + b), — (a + h), (a + J) and — 13 (a 4- J). Ans. 4 (a + b). 22 ELEMENTARY ALGEBRA. CASE III. 40. To find the sum of amj algebraic quantities. The sum of 5 a and 6 6 is neither 11a, nor lib, and can only be expressed in the form of 5 a + 6 S, or 6 J + 5a; and the sum of 5a and —45 is 5a — 4&; but in finding the sum of 5 a, 6 J, 5 a, and — 4 S, the a's can be added together by Case I., and the b's by Case II., and the two results connected by the proper sign ; thus, 5 a + 6J + 5a — 4J=10a + 2 6. 1. Find the sum of 6 d, — 2 S, x, By, 5 x, — 3 J, 3 5c -{-Id, 5x, T 5 + 2 X, and — 3 5 c. OPERATiosr. Poj. convenience, simi- Qd—2b-\- a; + 3y + 35c lar terms are written un- 4,d—3b]-5x —3 be der each othej ; then by -\-1b-\-bx -\-2x Case I. the first column at the left is added; the second by Case II., and 10^+25+ 18 as + S^^ soon;+35cand — 36c cancel. This case includes the two preceding cases, and hence to find the sum of any algebraic quantities : EULE. Write similar terms under each other, find the sum of each column, and connect the several sums with their proper (2.) 4a; — Ta-|-3y — 45-l-3z 6a— y -\. ib — 2z 4a — 2y-|-85— z — 3a —85 —10 c ix — 10 c ADDITION. 23 (3.) 10 + 45— 3c + It^ X — 36+ %c — *\l/x-\-y — 10 c + 8\/^— y 3v/¥ •70+ 6— lOc + 6 \/^ 4. Add together *l sf x, — 8 a;, la;'', — ^ s/ x, 4 a;'', — 8 a;, 4 a;, and 1 a;^, — 8 %/ ^• Ans. 18 a;" — 12 a; — T v/x. 5. Add together 3aa; — ^ah -\-1xy, *\ ah -\- b x — 4a, *l xy — 3 a + 4a;, and -\- ahc — ax-\- Qxy. 6. Add together *l x — 3ay — 5aJ + 4c, 3aa; + 4a; + 5 a 6 — 5 c, and 3c — 3aa;+7^ + c. Ans. 11a; — Say + Sc + ty. t. Add together 6a — 32 + '7a;+4aa; — 3a 5, bab — 5a + 22 — 4aa; + 4, and 6 — 2a6 + 3a; + 4^/ + 4oa;. 8. Add together 6 a;y + 6 xz — 6mn-\-4:n, imn — 3xy + 2n — 8m n, — 6a;z + 4n — 3a;y + 6, and 10 mii — 10n + 3 — 9. Ans. 0. 9. Add together 8a»i+19«a; — 55 6 + c, — 19i; + 14 J — 16 c + y, and 18 nx — 44a?re+ 15v — iy. 10. Add together It aa:" + 19aa;' — 14aa;* + 16 aa;^ 13 a a;' — 5 a a;* + 6 a a;^ — 10 a a;', and 14 a a;'' + i? a a;'' — 3aa;'+15aa;^ Ans. Tl aa;^ + 19aa;^ — Saa;*. 11. Add together m + n — 4a + 6c — 1y, 8c — 4m + 3n — 5a + 3c, t a— 17c + 1y — 10m — 6n, and 14ra — 8a — 7c+10y — 8 m. 12. Add together 8axy-\-l1bxy — 16ea;y — 9axy, 16bxy — 18c a;y + 10a a;^ — 14aa;2, 16 ca;y + 2baxy — Ibxy + 2bcxy, and 10aa;2 + Sbxy — lOca;^ + 4aa;z. Ans. 34aa;y + 29 Ja;y — Sexy. 24 ELEMENTARY ALGEBRA. 13. Add together B {x -\- y) , — 4. (x -\- y), and 7 {x + y). Ans. 6 (a; + y). 14. Add together 5 (2 a; + y — 3 2), and — 2 (2 a: + y — 3 2). Note. — If several terms have a common letter or letters, the Slim of their coefficients may be placed in parenthesis, and the com- mon letter or letters annexed ; thus, 6a;4-8a; — 5a;=(e + 8 — 5)x; ax-\-Sbx — 2ca;=(a + 36 — 2c) a;; icxy -\- adxy — acxy = (b c -\- a d — a c) x y. 15. Add together ax — hx -|- 3x, and lax -\- A:hx — x. Ans. (3 a + 3 5 + 2) a;. 16. Add together hy — Bey -\-bay, and cy -j-4iy — 2 ay. It. Add together 2xy — axy, and Qxy — 3axy. Ans. (8 — 4 a) xy. 18. Add together 1 (3a; + 5j^) + 3a — 6a; + 8 ab, 3x + 5(3a; + by) -\- 1 a —bab, and 8a; +2(3a; + 5y) — Ta — 3ai. Ans. 47 a; + YOy -|- 3 a. 41 • From what has gone before, it will be seen that addition in Algebra differs from addition in Arithmetic. In Arithmetic the quantities to be added are always con- sidered positive ; while in Algebra both positive and neg- ative quantities are introduced. In Arithmetic addition always implies augmentation ; while in Algebra the sum may be numerically less than any of the quantities added ; thus, the sum of 10 a: and — 8 a; is 2 a;, which is the numerical difference of the two quantities. SUBTRACTION. 25 SECTION V. SUBTRACTION. 42. SuBTKACTioN in Algebra is finding the difference between two quantities. 1. John has 6 apples and James has 2 apples ; how many more has John than James? Let a represent one apple, and we have 6 a 2a 4a ., or 6a — 2a = 4a. 2. During a certain day A made 9 dollars and B lost 6 dollars ; what was the difference in the profits of A and B for the day ? If gain is considered -\-, then foss must be considered — , and letting d represent one dol- lar, it is required to take — Qd from 9 d. OPERATION. 9d — Sd It is evident that the difference be- tween A's and B's profits for the day- is 9d -\- 6d == 15d; that is, 9d — 15 <^ (— 6d) = 9rf-|- 6d= 15d. Hence it appears that, as addition does not always im- ply augmentation, so subtraction does not always imply diminution. Subtracting a positive quantity is equivalent to adding an equal negative quantity; and subtracting a negative quan- tity is equivalent to adding an equal positive quantity. Suppose I am worth $1000; it matters not whether a thief steals $400 from me, or a rogue having the author- ity involves me in debt $400 for a worthless article; for 2 26 ELEMENTARY ALGEBRA. in either case I shall be worth only $600. The thief sub- tracts a positive quantity ; the rogue adds a negative quan- tity. Again, suppose I have $1000 in my possession, but owe $400: it is immaterial to me whether a friend pays the debt of $400 or gives me $400; for in either case I shall be worth $1000. In the former case the friend subtracts a negative quantity ; in the latter, he adds a pos- itive. Or, to make the proof general : 1st. Suppose + i to be taken from a -{- b the result will be a ; and adding — J to a -\- b we have a -\- b — b, which is, as before, equal to a. 2d. Suppose — J to be taken from a — b the result will be a ; and adding -\- b to a — b we have a — b -\- b, which is, as before, equal to a. 3. Subtract 6 4" c from a. OPERATION. 6 subtracted from a gives a — (b -\- c) = a — b — e a — 6; but in subtracting b we have subtracted too small a quantity by c, and therefore the remainder is too great .by c, and the remainder sought is a — b — c. 4. Subtract b — c from a. OPERATION. I„ subtracting b from a we a — (5 — c) = a — b -\- c subtract a quantity too great by c ; therefore the remainder (a — J) would be just so much too small, and the remainder sought is a — 6 -|- c. 43. By examining the examples just given it will be seen that in every case the sign of each term of the subtrahend is changed, and that the subsequent process is precisely the same as in addition ; hence, for subtrac- tion in Algebra we have the following SUBTRACTION. 27 RULE. ' Change the sign of each term of the subtrahend from -\- lo — , or — to -\-, or suppose each to be changed, and then proceed as in addition. (1.) (2.) (3.) (4.) (5.) (6.) (^0 Min. 9 9 9 9 9 9 .9 Sub. 9 6 3 — 3 — 6 — 9 Rem. 3 6 9 12 15 18 In examples 1-7, the minuend remaining the same while thie subtrahend becomes in each 3 less, the remainder in each is 3 greater than in the preceding. (8.) (9.) (10.) (11.) (12) (13.) (14.) Min. 9 6 3 0—3—6—9 Sub. 9 9 9 9 9 9 9 (19.) (20.) (210 — 3 — 6 — 9 — 3 — 6 — 9 Rem. 0- ^3 — 6 — 9 — 12 — 15 — 18 In examples 8-14, the minuend in each becoming 3 less while the subtrahend remains the same, the remainder in each is 3 less than in the preceding. (15.) .(16.) (11.) (18.) Min. 9 6 3 Sub. 9 6 3 JO Rem. ,0 .0 In .examples 15-21, both minuend and subtrahend de- creasing by 3, the remainder remains the same. (1.) (2.) (3.) (4.) (5.) (6.) Min. 26a; 21 axi/ —I3ab —18c i9xy — 438& Sub. 10a; — 4oa;^ 4a5 — , 6c — 25a;2^ 21b Rem. 16x Slaxy —llab — .12e 28 ELEMENTAEY ALGEBRA. (1.) (8.) (9.) (10.) (11.) (12.) Min. 10a: — 4=axi/ iab — 6c — 25,a;y 21b Sub. 26a; 21 axi/ —13a& —18c i9xy —438* Eem. —16a: — 31aa;y llab 12c (13.) (U.) Min. 6x—Uy-\-3z 1 « + 18 & — 10 c Sub. 3x-\- 3j/-\- z _255+ 6c — 8^~x take J a;'' -|- c a; — d t^Tx. 8. From a;/ -f a;= — x^y'^ take y"^ ■\- x" y -^ a^y-". MULTIPLICATION. 31 SECTION VI. MULTIPLICATION. 46i Multiplication is a short method of finding- the sum of the repetitions of a quantity. ' 47i The multiplier must always be an abstract num- ber, and the- product is always of the same nature as the multiplicand. The cost of 4 pounds of sugar at 17 cents a pound is 11 cents taken, not 4 pounds times, but 4 times ; and the product is of the same denomination, as the multi- plicand 17, viz. cents. In Algebra the sign of the multiplier shows whether the repetitions are to be added or subtracted. 1. . (+«)X(+4)= + 4a; i. e. -|- " added 4 times is -\-a-\-a-\-a-{-a = -\-4:a. 2. (+ a)X(— 4) = — 4a; i. e.-\- a subtracted 4 times is — a — a — ' a — a-^= — 4 a. 3. (— a) X (+4)= — 4a; ,i. e. — a added 4 times is — a — a — a — a = — 4 a. 4. (_a,),x(— 4) = + 4a; i. e. — a subtracted 4 times is -|- a -|- a -j- a 4- a=^-\- ia. In the first and second examples the nature of the product is -\- ; in the first, the + sign of 4 shows that the product is to be added, and -f- 4 a added is -f- 4 a'; in the second, the — sign of 4 shows that the product is to be subtracted, and -|- 4 a subtracted is — 4 a. In the third and fourth examples the nature of the product is — ; in the third, the -f- sign of 4 shows that the prod- uct is to be added, and — 4a added is — 4a; in the 32 ELEMENTARY ALGEBRA. fourth, the — sign of 4 shows that the product is to be subtracted, and — ia subtracted is + * «• 48. Hence in multiplication we have for the sign of the product the following BTJLE. I/ike signs give -{- ; unlike, — . Hence the products of an even number of negative fac- tors is positive, of an odd number, negative. 49. Multiplication in Algebra can be presented best under three cases. CASE I. 50. When both factors are monomials. 1. Multiply 3 a by 2S. OPERATION. 3oX25=3XaX2X*=3X2XaXJ=6a5. As the product is the same in whatever order the factors are arranged, we have simply changed their order and united in one product the numerical coefficients. Hence, when both factors are monomials, RULE- Annex the product of the literal factors to the product of their coefficients, remembering that like signs give -\- and unlike, — . 2. Multiply a' by a". OPEKATION. o' X a^= (o X a X a) X (o X a) = o X a X « X o X a = «" As the exponent of a quantity show's how many times it is taken as a factor, a* = a X « X a ; and a« = a X a ; and a' X a" = a X o X a X a X a, and this is equal to aK (Art. 24.) Hence, Powers of the same quantity are multiplied together by adding their exponents. MULTIPLICATION. 33 (3.) (4.) (5.) (6.) (r.) kxy bx^j^ *l ah < — 14»»M* ■— a^b* Zah 3xf — 8 a" J Qan* — 4o=i I2abxy 15a^y — SCa'J" — 84a»»«' ia*^ 8. Multiply x^ by jc». 9. Multiply a^ by «'. 10. Multiply a* I by ^ a. 11. Multiply — o* by a". 12. Multiply — c' by — c*. 13. Multiply 8a;/ by lax. 14. Multiply 504 o^ J^ by — 8 e^b. 15. "Multiply — '2bxyz by ixyz. 16. Multiply — 417 a J c^ by — 3 a J" c. 17. Multiply tOj^ether 444 a; 2^, 3 a;''/, and — 2«. (AflS. — 2664a;'/«. 18. 'Multiply ia'ir'cd hy —iabc^ d^. 19. Mliltiply — 5-ir by — 6af. AriB. 30*"+". 20. Multiply together 14a6c^ — 5a^Jc, and —iaF. 21. Multiply 25Vay'l5y 3 Ja;. Atts. *lbhxs/ay. 22. Multiply '4 (a; -{- y) by 3 (a; 4- y). Ans. 12 (a; + 2^)". Note. — Any number of terms ■ enclosed in a parenthesis may be treated as a monomial. 23. Multiply — 12 (a^ ^ JS) by — 4 (a= — P). Ans. 48 {a" — W)\ 24. Multiply (a — x)* by (a — xf. 25. Multiply 4 (a + S)"* by 2. (a + J)»- Ans. 8 (a + i)-"*"- 26. Multiply a' (a; + ef by a J^ (a: + 2;). 2* c 34 ELEMENTARY ALGEBKA. CASE II. 51. When only one factor is a monomial. 1. Multiply 8 + 5 by 3. OPERATION. 8+51 84- 5 8+ 5 24+15 or 8 + 5 = 13 -3 3 24 + 15 = 39 In this example, not the sum of 3 repetitions of 8 only, but of 8 and 5, is re- quired ; the sum of 3 repeti- tions of 8 = 24 ; of 3 repeti- ticms of 5 = 15. Hence, the sum of 3 repetitions of 8 + 5 = 24 + 15. 2. Multiply 8 — 5 by 3. OPERATION. 8 — 5] 8 — 5 8 — 5 24 — 15, 24 — -15 =9 The sum of 3 repetitions of 8 = 24 ; but it is not the sum of 3 repetitions of 8 that is required, but of a num- ber 5 units less than 8 ; 24, therefore, will have in it the sum of 5 units repeated 3 times, or 15, too much ; the product required, therefore, is 24 — 15. , Therefore, The product of the sum is equal to the sum of the prod- ucts, and the product of the difference to the difference of the products. 3. Multiply X -\-y — z by a. OPERATION. X -\- 11 z ^^ ^""^ °^ ^^^ repetitions of X a times, of y a times, and of — z a times is ax -\- ay ax -\- ay — az — az. RULE. MuUiply each term of the multiplicand by the multiplier, and connect the several results by their proper signs. MULTIPLICATION. 35 (4.) (5.) Sax — 4a; 12 aa? — 24aa:^4-42aa;y — iw'x-j-Sabx — ll/'x. 6. Multiply 5 m w + 4 m" — 6 w"^ by 4 « *■ 7. Multiply IQa' X — S xz -{- ii/ by — 3a: y. 8. Multiply J a? — c a;" + rf a; by — ^ a;'. 9. Multiply — 63 a;y — 14 a: — 6 z by — 4 ». Ans. 252 a; 2^ 2 + 56 a; « + 24 «^ 10. Multiply 14 a* _ 13 a' + 12 a^ — 11 a by 4 a\ 11. Multiply x — 2 a + 14 by ax. 12. Multiply 17 ax — lAby -\- 11 c z by -^iahcxyz. 13. Multiply 210=4" — 3a;y^^4Jc by —Qaxy. CASE III. 52i When both factors are polynomials. 1. Multiply 7 + 4 by 5 — 3.' OPBKATION. Multiplying 7 -|- 4 by 5 is tt \A =11 taking the multiplicand 3 too r q 9 many times; therefore,' the _^ true product, will be found by 35 + 20 — 21 — 12 =22 subtracting 3 (7 + 4) from 5(7+4). 2. Multiply X — y by a + 6. OPERATION. a times x -^ y =^ ax — a^i but X — ^ is to be taken, pot a times only, but a + 6 times; therefore, '^-^ — [ a(x-^,y) is- too small, by 6 (a; — j/) ; a.x — ay -\-bx — hy ^nd the product required is a a; — • ' ay-\r'h.x — by., Hericfej ' ' ,' X — y a + 'j RULE. Multiply each term of the multiplicand by each term of the multiplier, and find: the sum of the several products. : ?36 ELEMENTARY ALGEBBA. (3.) (4.) a'' -\- 2 a X -\- 3^ xy-\-ab a — X xy — ah a?-\-2a^x-^ av? x^y^-{-abxy — a^x — 2a a? — a? ■ — ahxy — a^W a«+ a^x— a7? — v? x^y^ — a^¥ 5. Multiply a^-^-V" — (? by a^ + A Ans. a* 4- a" S^ ^ 52 ^a _ ^4 6. Multiply 3? — 2xy-\-y^ by -x^ '-\- y"^. Ans. 7^ — 2x^y-\-23?y'^ — 2xf-\-y'^. 1. Multiply 4 a* — 2 a* S + 3 a^ J'' by 2a^ — 2h\ Ans. 8 a« — 4 a" 5 — 2 a* J^ + 4 „s js _ g „a j4 8. Multiply a;^+2a^ + 3a^-f 2a;+l bya;" — 2x+l. 9. Multiply ar' + 2^2l|.^2_3.j^_3,^_y^|jy^_|_^_|_^ Ans. s^-\-f-\-^ — ^xyz. 10. Multiply 4 a:^ _ Y a;* + 10 a^ — 13 «= by 3 a; — 2. 11. Multiply a' + a'' — 1 by a^ — 1. 12. Multiply a;2 + 7 a ar — 14 a' by a; — T a. Ans. a:' — 49 a'' a; — 14 a' a; + 98 a*. 13. Multiply x-\-y — a by x — y -\- a. 14. Multiply o" -f- If hj a'" — i™. Ans. a "■+" -]- a" J" — a" J™ J^ + n^ 15. Multiply '7 a:y — 14 a:^^ + 21 ar'/ by 6 a;y — 3. Ans. —2lxy-{-84=a^f—U'Ja?f^ 126 x* y*. 16. Multiply 6aH—9a¥—12aH^hj2ab—3b\ Ans. 12a'i2_3g„2j3_24„3j3_^27a6* + 36a»i*, IT. Multiply x* — 7?-\-v^ — a:+l bya;+l. Ans. a:* 4- 1. 18. Multiply a!< — 3:^ + :b^ _ ^ + 1 by 3. _ j, Ans. a^— 2a;*4-2x' — 2a;2+2a;— 1. 19. Multiply x* + a!> + x'-\.x+ lby»+l. 20. Multiply af* + a^ + ^2 ^ ^ -^ 1 by a; — 1. DIVISION. 37 SECTION YII. DIVISION. 53i Division is finding a quotient which, multiplied by the divisor, will produce the dividend. In accordance with this definition ahd the Eule in Art. 48, the sigh (Jf the quotient must be -f- when the divisor and the dividend have like signs ; — when the divisor and the dividend have unlike signs ; i. e. in di- vision as in multiplication we have for the signs the fol- lowing EULE. Like signs give -\- ; unlike, — . CASE 1. 54. When the divisor and dividend are both monomials. 1. Divide 6 a J by 2 J. OPERATION. The coefficient of the quotient must 6a6-7-2J = 3o be a number which, multiplied by 2, the coefficient of the divisor, will give 6, the coefficient of the dividend ; i. e. 3 : and the literal jpart of the quotient must be a quantity which, multiplied by 6, will give a b ; i. e. a: the quotient required, therefore, is 3 a. Hence, for division of monomials, RULE. Annex the quotient of ffie literal quantities to the quotient of their co^cienls, remembering that like signs give -\- and unlike, — . S§ ELEMENTAEY ALGEBBA. 2. Divide a" by c?. OPERATION. a? ^ a^ — a' For a? X a' = «'• (Art. 50.) Hence, Poiuers of the same quantity are divided by each other by subtracting the exponent of the divisor from that of the dividend. (3.) 9 xy = Zxy (4-) 48 a^a;y — IQ ax .— Say (5.) 216 x'y = — Qx (6.) — 16 a^ x^ y — 4 axy z — =4aa: Y. Div: 8. Div 9. Div: 10. Div 11. Div 12. Div 13. Div 14. Div 15. Div 16. Div \n. Div; 18. Div 19. Div 20. Div 21. Div 22. Div de 34 a hy'^ by 2 ay. de 297 xV ty — 99 a; y''. Ans. — 3 xy. de — *l4ixy^ z by 2xy. de — 144 a'b^x by — 24 a J x. Ans. 6 a J. de a a;* by a 3?. de 8 a;' by — 8 x\ de — 210a;'y by 42 a;' y. de — 270 a J a; by — 135 a & a;. de — 474 a' W c" by 158 aWc. de ai" by a;"- Ans. a:""". de 14 a"" a:" by — 7 a" a;. Ans. — 2o"'-"a;"-^ de — 747 a^ b" e^d' by 83 a^b c d\ de 12 {x + yf by 4 (a; +y). Ans. 3 (x + y). de _ 27 (a — bf by 9 (a — i)'^. de (6 — cf l)y (J — cf. Ans. (6 — c)\ de 14 (a; — yy by 7 (a: — r/)^. DIVISION. 39 .CASE II. 55i When the divisor only is a tnohomial: I. Divide ax-\-ai/-\-az hy a. OPERATION. In the multiplication of a poly- a)ax-\-ay-\-az nomial by a monomial, each term 2; _i_ „ _r_ g of the multiplicand is multiplied by the multiplier ; and therefore we divide each term of the dividend ax -\- ay -\- az by the divisor a, and connect the partial quotients by their proper signs. Hence, RULE. Divide each term of the dividend by (he divisor, and con- nect the several results by their proper signs. (2.) (3.) 3 a) 6 a x^ — 24.0^ — 5a:°y )— 15 a:»y — 25 a:'y 2 a;'' — 8a;' 3a; + 5 4. Divide 12 a a;* — 24: a x^ -\- 4:2 a x y hy Sax. 5. Divide — ^a'x-^-Sahx — 4¥x by — 4x. 6. Divide 6 a" x* — 12 a^ a? -\- 15 a* a^ hy 3 a^ x\ Ans. 2x'^ — 4:ax -\- ba^a^. 1. Divide 12 a*f — 16 a^f -\- 20 a'y* — 28 aV by 4aV- 8. Divide — 5 ce' + 10 cc'' — 15 a; by — 5 a;. 9. Divide 273 (a + a:)? — 91 (a + x) by 91 (a + x). Ans. 3(a + x) — l=3a + 3a; — 1. 10. Divide 20a6c — 4ac -f- 8acd — I2a^c^hy —4:ac. II. Divide 16 a= a;'' — 32 a= a;* + 48 a* x* by 16 a^ x\ 12. Divide 72 a;'' 2?'' — 36 a;» / — 54 a;» / a» by — 18 a;'' y\ Ans. — 4 + 2a;2/ + 3a;3^ 13. Divide ISaa;*^/ — 54a;=2/2 + 108 ca;^/ by Q7?y. 14. Divide 40 a* i'l + 8 a= 5= — 96 a» ¥ a? by 8 a'' b\ 15. Divide 39 a:* 3* — 65 o a;' a« + 13 a:» z^ by — 13 a;' 3=. , 10 ELEMENTARY ALGEBKA. CASE II'I, 56. When the divisor and dividend are both polyno- mials. 1. Divide ^ —Zx^y -^-Zxy" — ]/' hj x"" -^ li x y -\- y\ OPERATION. 3? ^2xy -\.y'^)3? — Bx^y + Zxy^^f (ar-— y 3?—2x^y-\- xy^ — x'^y -\-2xy'^-^y — x'^y-^lxy-' — f The divisor and dividend Are arranged in the order of the powers of X, beginning with the highest power, i", the highest power of a; in the dividend, must be the product of the highest, power of x in of the quotient and ^ in the divisor; therefore, ■^ = ^ must' be the highest power of x in the quotient. The divisor :j? — ixy -\-y'- mul- tiplied by X must give several of the partial products which would be produced were the divisor multiplied by the whole quotient. When (jf — I X y -\- ■tp) X = x^ — 'i, x^ y ■\- x if is subtracted from the dividend, the remainder must be the product of the divisor and the remaining terms of the quotient; therefore we treat the retnain- der as a new dividend, and so continue until the dividend is ex- hausted. Hence, for the division of polynomials we have the f61- lowing EULE. Arrnnge (he divisor and dividend in the order of the powers of one of the letters. Divide the first term of the dividend by the first term of the divisor ; the result will be the first term of the quotient. Multiply (he whole divisor by this quotient, and subtract the product from the dividend. Consider the remainder as a new dividend, and proceed as before until the dividend is exhausted. DIVISION. 41 Note. — If the dividend is not exactly divisible by the divisor, the remainder must be placed over the divisor in the form of a fraction and connected with the quotient by the proper sign. 2. Divide as* -\-4.f hy a? — 2xy -{-2f. p^-2xy + 2/)a:* + 4y' {a? ^ 2xy + 2f a:*— 2a;»y + 2a^/ 2a:«y- 2a?y- -2:^f-\-^f -A.3?f + 4.xf 27?f — 4.xf + 4:y* Note. — By multiplying the quotient and divisor together all the terms which appear in the process of dividing will be found in the partial products. 3. Divide a^ — 1 by x — 1. X - -1) 3*- -1 (a:' ' + ^ + a: + 1 a?- a? — 1 ■x' ar"- a?- — 1 — X X - -1 X — -1 4. Divide o x — ay -\-bx — by -\- z by x — y. X — y) ax — ay -\- hx — hy -{- z (a + b -{- -^-^ ax — ay bx — by bx-^- by 5. Divide 2by — 2¥y — 3l^yz + 6Py + byz—yz by 2i — «. Ana. 3b'y—by-\-y. i2 ELEMENTAEY ALGEBRA. 6. Divide c' + a;= by c + a;. Ans. ' ~-^ ^ a^ — an -{- d'V — a^W + ah* —¥, and so on. It follows from the two preceding statements that The difference of the same even powers of two quantities is divisible by either the sum or the difference of the quart- titles. III. The sum of the same odd powers of two quantities is divisible by the sum of the quantities. ?-±-f — a* ^ c^ b + aH^ — a¥ 4- ¥, a -{-b ' ' ' ^^ = a^ — an + a*S'—a'¥ + an* — al^+¥, and so on. 52 ELEMENTARY ALGEBRA. 1. Find the factors of a;* — /. OPERATION. (a? — y^) -h (a; — y)=x* + a^y + a^/ + a;s^ + / By I. of this article, the difference of the same powers of two quantities is divisible by the difference of the quantities; therefore x — y must be a, factor of a;* — ^; and dividing af — y^ hyx — y gives the other factor i^ -\- 3? y -'r ^ f' -\- x f -{■ f. 2. Find two factors of c° — cZ". OPERATION. (c' — d') -r- (c + d) =c^ — c^d-\-(?d^ — c^d^ + cd^ — d^ By 11. the difference of the same even powers of two quantities is divisible by the sum of the quantities ; therefore c -\- d must be a factor of c° — c?'; and dividing c" — rf° by c ■\- d gives the other factor c= — c* rf + c' tZ= — c'' rf» -f c rf' — dK 3. Find the factors of w? -\- n^. OPERATION. {m^ 4- n^) -V- (m -\- n) =z m^ — m^n -\-m^n^ — mrc^ -\- w* By III. the sura of the same odd powers of two quantities is divisible by the sum of the quantities ; therefo*e m -\- n must be a factor of m' "1- n' ; and dividing m^ -j- n' by m -j- re gives the other factor m* — m' n ■-\- m? n? — mn' -\- n*. 4. Find the factors of a* — x^. Ans. {a — x) (a^ -{- ax -\- x^). 5. Find the factors of a^ -\- x^. Note. — In Example 2, the factors of c» — rf» there obtained are not the only factors; for by I. c' — d' is divisible by c — d; and dividing c' — d" hy c — d gives another factor, c' + d" d -{- (f d' -{- c^ d' + c d* + d'; or by Art. 70, C — d« = (c= + d') (c» — d'). FACTORING. 53 But (fi — (^ d + &• d' — c' d' + c d* — d'; (J' + c* d -\- d' d' + c' d' + c d* + d^, c» + d', c» — d', are not prime quantities ; for the first can be divided by c — d, and the quotient thus arising can be divided hy c' ± cd-\- d^; the second can be divided by c -\- d, and the quotient thus arising ■will be the same as after the division of the first quantity by c — d, and can be divided hy <^ ± c d -{- d'; the third can be divided by c -|- d, and the fourth by c — d. Performing these divisions, by each method we shall find the prime factors of c° — d" to be c -{- d, c — d, (^ -\- c d -\- d', and c' — cd-\- d'. In finding the prime factors, it is better to apply first the princi- ple of Art. 70 Jis ftr as possible. 6. Find the prime factors of a:^" — y". ,: . 3.10 _ 2,10 = (a.6 _j_ 2,5) (a.6 _ ^6). ajS _ 3,5 = (a: _ J,) (as* + a^y + a^f -\-xf-\- y*). Ans. (x-\-y){x—y){sii^ — a?y-\-x^y^ — xf -\-y^) {x*-\-x'y-\-x='f-{-xi/'-\-y'). 1. Find the prime factors of a° — 1. Ans. (fl + 1) (a— 1) (a' + a + l) (a= — a+1). 8. Find the prime factors of a^ — 2 a^ a^ + **• Ans. (a -\-x) (a + x) (a — a:) (a — x). 9. Find the prime factors of a;° + 2 a^ ^^^ + /■ Ans. {x-\-y) {x-\-y) {7? — xy-^f) (3? — xy-\rf). 10. Find the prime factors of 1 — a*. Ans. (l+a) (I — a) (!+«'). 11. Find the prime factors of 8 — (?. Ans. (2 — c) (4 + 2c-f c^. ELEMENl'AEY ALGEBRA. SECTION X. GEEATEST COMMON DIVISOR.* 72. A -GosTMON ©ivisoE of two or more quantities is any quantity that -will divide each of them without remainder. 73. The Greatest Common Divisor of ,two or more quantities is the greatest quantity that will divide each of them without remainder. 74. To deduce a rule for. finding the greatest common divisor of two or more quantities, we demonstrate the two following theorems : — ^ Theorem I. A common divisor of two quantifies is also a common divisor of the sum or the difference of any multiples of each. Let A and B be two quantities, and let d he their common di- visor; d is also a common divisor of mj4 :t n B. Suppose A -^ d = p; i. e. A ^ dp, and m A== dmp, and B -^ d^ q; i. e. B = dq, and nB = dnq; then mA±nB = dmp±dnq = d (mp ± n q). That is, d is contained in m A -\- n B, mp -^ n q times, and in m A — n B, mp — n q times,; i. e. rf is a common divisor of the sum or the difference olf any multiples of A and B. Theorem II. The greatest common 'divisor of two quan- tities is also the greatest common -divisor of the less and the remainder after dividing the greater by the less. Let A and B be two quantities, and A '^ B; and let the process of dividing be as appears in B) A {q the margin. Then, as the dividend is equal to qB the product of the divisor by the quotient plus the remainder, A^r+qB. (1) * See Pre^e. GEEATEST COMMON DIVISOR. 55 And, as the remainder is equEil to the dividend minus the product of the divisor by the quotient, r==A — qB. (2) Therefore, according to the preceding theorem, from (1) any divisor of r and B must be a divisor of A ; and from (2) any divisor of A and B, a divisor of r ; i. e. the divisors of A and B and B and r are identical, and therefore the greatest common divisor of A and B must also be the greatest common divisor of B and r. In the same way the greatest common divisor of B and r is the greatest common divisor of r and the remainder after dividing B by r. Hence, to find the greatest common divisor of any two quantities, RULE. Divide the greater by the less, and the less by the remain- der, and so continue tin the remainder is zero ; the last di- visor is the divisor sought. Note 1. — The division by each divisor should be continued until the remainder will contain it no longer. Note 2. — If the greatest common divisor of more than two quan- tities is required, find the greatest common divisor of two of them, then of this divisor and a third, and so on ; the last divisor will be the divisor sought. "•*^ Note 3. — The common divisor of xy and a; z is a; ; a: is also the common divisor of x and x z, or of ax y and xz\ i. e. the common divisor of two quantities is not changed by rejecting or introducing into either any factor which contains no factor of the other. ^ Note 4. — It is evident that the greatest common divisor of two quantities contains all the factors common to the quantities. CASE I. 75t To find the greatest common divisor of monomials. 1. Find the greatest common divisor of i a^ I? c d, 16 o» 6= c^ and 28 a* b" c. The greatest common divisor of the coefiicients found by the gen- eral rule is 4; it is evident that no higher power of a than a\ of 66 ELEMENTARY ALGEBRA. b than 6', of c than itself, will divide the quantities ; and that d will not divide them ; therefore, the divisor sought is 4 a' J' c. Hence, KULE. Annex to the greatest common divisor of the coefficients those letters which are common to all the quantities, giving to each letter the least eacponent it has in any of the quantities. 2. Find the greatest common divisor of GSa^^c^d', 27 a* b' c5, and 45 a' A" c» d. Ans. 9 a^ V c'. 3. Find the greatest common divisor of 1bx''y^z^ and 4. Find the greatest common divisor of 99 a P c* d^ x'^ i^ and 22aH*c''d^x\ Ans. Ualr'c^d^x^. 5. Find the greatest common divisor of Yl x^y^, 19 a;^^', and 2\2bx'y^!!?. CASE II. 76t To find the greatest common divisor of polynomials. 1. Find the greatest common divisor oi a^ — ^ and 3?—2xy-\-f. !^x' — 2xy+ y^(l - f — 2xy^2f BqecBng the actor 2y X- y):^-fix^y s? — xy xy — f ^y — y^ Ans. X — V- RULE. Arrange ike terms of both quantities in the order of the powers of some letter, and then proceed according to the general rule in Art. 74. Note 1. — If the leading term of the dividend is not divisible by the leading term of the divisor, it can be made so by introducing GREATEST COMMON DIVISOR. 57 in the dividend a factor which contains no factor of the divisor; or either quantity may be simplified by rejecting any factor which contains no factor of the other. (Art. 74, Note 3.) Note 2. — Since any quantity which will divide a will divide — a, and vice versa, and any quantity divisible by a is divisible by — a, and vice versa, therefore all the signs of either divisor or dividend, or of both, may be changed from -\- to — , or — to -[-, without changing the common divisor. Note 3. — When one of the quantities is a monomial, and the other a polynomial, either of the given rules can be applied, although gen- erally the greatest common divisor wiU be at once apparent. 2. Find the greatest common divisor of ax' — a'^a;" — 8 a^x'^ ani '2cx* — 2acx^ -{-4:0,^0x^ — 6 a' ex — 20 a*e. az' — a'a^ ■ DiTiding bj a a? 3^ — aa? - 8o* 2a'i? — 5ci'x — 2a* Dividing by fl S 2x'—3ax—2ci' 2c3i^—2aca?-\-ia'c3? — e(^ex — 20a*c Dividing by 2 C x' — ai?-\-2a'x^ — Sa'x — 10a* (1 3*~ai^ —8 a* 2a'a^ — 3(fx — 2a' istKem. X*— aif- 8a* Multiplying by 2a 2x*—2a3? — 16a*(x' ey 23f — Za3? — 2cfi^o£ a3?-\-2d?x' — l6a* A^ Multiplying by 2 2aj? + 4a^a? — 32a* {ax 2aa^ — Sa'ar"— 2a^x 7a"a^+ 2a'a; — 32a* Multiplying by 2 14a''a?+ 4 0=3; — 64a* (7a' Ua^a:'— 21 a' a:— 14 a* 25 a' a; — 50 a* zdBem. 25 a* a: — 50 a* " Dividing by 25 a „ , , a;— 2a) 2^*— 3aa;— 2a^ (2a; + a 2a^ — 4aa; ax — 2 a' ax — 2 a' Ans. X — 2 a. 3* 58 ELEMENTARY ALGEBRA. ,3. Find the greatest common dmsor of a^ — «* and a'-{.d^x — ax' — x^ Ans. a" — x^ 4. Find the greatest common divisor of a* — x* and q6 jj8 ^_ Ans. a^ — x^. 5. Find the greatest common divisor of 2aa:^ — a^x — a' and 2x''-^3ax-\- a\ 6. Find the greatest common divisor of & ax — 8 a and 6 a *' + a a;2 — 12 o x. Ans. 3 a a; — 4 a. t. Find the greatest common divisor of x* — ^ and x' + f. y-^j^ 8. Find the greatest common divisor of 3 a;' — '24 x — 9 and 2a;'— 16a; — 6. 9. Find the greatest comnaon divisor of a:*^— ^' and a;2 — y'^. Ans. x — y. 10. Find the greatest common divisor of 10 a;* — 20^a;^y + 30 / and a;' + 2x''j/ -\- 2 x.f + f. Ans. x -\- y. 11. Find the greatest common divisor of a^ -\- a' -{- a^ + a — 4 and o^ + 2 a= + 3 a^ + 4 a — 10. Ans. a — 1. 12. Find the greatest common (divisor of T a a;* -(-'21 « a;' -|- 14 a and a;' -(- a;^ -j- a;' — x. Ans. a: -|- 1. 13. Find the greatest common divisor of 27 a'^ — 8 a^y and 3 / — 2 a / -|- 3 a^ 3^« — 2 a^y . Ans. 3^" — 2 ay. 14. Find "the greatest common divisor of a' -|- a — 10 and a* — 16. Ans. a — 2. Note 5. — The greatest common divisor of polynomials can also be found by factoring the polynomials, and finding the product of the factors common to the polynomials, taking each factor ^e^least num- ber of times it occurs in any of the quantities. (Art. 74, Note 4.) 15. Find the greatest common divisor of3aa;^ — 4aa;-}- Saxy — 4zay and a' x — x -\- a^y — y. 3 ax^ — 4:ax -{- 3axy — iay = a{x -\- y) (3a; — 4) a^a; — a; -^ a*!/ — 2/ = (a;-f2/) (a — 1) (a^-l-a-f 1) Ans. a; -|- 2/. LKAST COMMON MULTIPLE. 69 SECTION XI. LEAST COMMON MULTIPLE. 77t A Multiple of any quantity is a quantity that pan be divided by it without remainder. ,78i A Common Multiplet of two or more quantities is any quantity that can be divided by each of them with- out remainder. 79, The Least Common Multiple of two or more quan- tities is the least quantity that can be divided by each of them without remainder. 80i :It -is -eyident that a mpltiple pf any quantity must contain the factors of that quantity ; and, vice versa, any quantity that coijtg,in8 the factors of another quantity is a multiple of it : and a common multiple of two or more quantities must contain the factors of these quantities ; and the Zeosf common multiple of two 'or more Quantities must contain only the factors of these quantities. CASE I. To find;the least common multiple of monomials. 1 . Find the least common multiple of 6 a^ 6^ c, ■ 8 a' ¥ c' d, and 12 a*, box. The least common multiple of the coefficients, found by inspection or the rule in Arithmetic, is 24 ; it is evident that no quantity which contains a power of a less than a*, of 6 less than 5^ of c less than e\ and which does not contain d and x, can be divided by each of these quantities ; therefore the multiple sought is 24 a* 6' c" d x. Hence, in the case of monomials. 60 ELEMENTAEY ALGEBRA. RULE. Annex to the least common multiple of the coefficients all the letters which appear in the several quantities, giving to each letter the greatest exponent it has in any of the quan- tities. 2. Find the least common multiple of 3 a* h^ c®, Q cU h*c d^, and IQabcu^. Ans. ^Qa'hU^d^xK 3. Find the least common multiple of \Qahx, 9i{iah*x^, and 35 a' 6 a;*. Ans. bQQd¥a*. 4. Find the least common multiple of 9a'i', \ba*hx^, and l^axif Ans. 90a*b^x^i/^. 5. Find the least common multiple of 18 a' bc*x, '2.4.a¥cx^y, and SOa^J^xz. 6. Find the least common multiple of IOQxt/z, 4:5abc, and 25 m n. Y. Find the least common multiple of 10 a'^by^, 13 a*Pe, and lla^b'c^ 8. Find the least common multiple of 14 aP b^ c^, 20 a' b c*, 25 a'ic^ and 28abcd. CASE II. 81. To find the least common multiple of any two quantities. Since the greatest common divisor of two quantities contains all the factors common to these quantities (Art. 74, Note 4) ; and since the least common multiple of two quantities must contain only the factors of these quantities (Art. 80) ; if the product of two quanti- ties is divided by their greatest common divisor, the quotient will be their least common multiple. Hence, to find the least common multiple of any two (JUantities, LEAST COMMON MULTIPLE. 61 RULE. Divide one of the quantities by their greatest common di- visor, and multiply this quotient by the other quantity, and the product will be their least com/mon multiple. Note 1 . — If the least common multiple of more than two quanti- ties is required, find the least common multiple of two of them, then of this common multiple and a third, and so on ; the last com- mon multiple will be the multiple sought. Note 2. — In case the least common multiple of several monomials and polynomials is required, it may be better to find the least com- mon multiple of the monomials by the Rule in Case I., and of the polynomials by the Rule in Case II., and then the least common multiple of these two multiples by the latter Rule. 1. Find the least common multiple of ar' — y'' and x'—2xy-\-y^. OPERATION. Their greatest common X — y) x^ — 2xy -{-y' divisor is x — y, with which we divide one of ^ y the quantities ; and mul- (a^ —f) {x— y), Ans. tiplying the other quan- tity by this quotient, we have the least common multiple (i? — ^) (x — y). 2. Find the least common multiple of 2 a^ n?, 4 a^ y, a* — a;*, and o^ — a' oi?. The least common multiple of the monomials is icPx'y; and the least common multiple of the polynomials is a' (a* — a;*). The greatest common divisor of these two multiples is o? ; and dividing one of these multiples by a', and multiplying the quotient by the other, we have ic^x'y (a* — a^) as the least common mul- tiple. 3. Find the least common multiple of 3 a^ J', 6 a' by, a' — 8, and a^ — ia-^-i. Ans. Gan^y (a» — 8) (a — 2). 62 ELEMENTARY ALGEBKA. 4. Find the least commoD multiple of 3 a;' — 24 a: — 9 and 2 ar« — 16 a; — 6. (See 8th Example, Art. 76.) 5. Find the least common multiple of a^ — a;* and a" — X*. 6. Find the least common multiple of a;* — 1, a;^-|-2a;-(- 1, and (x — 1)^ Ans. a;° — a;* — a;^ -(- 1 . T. Find the least common multiple of a;* — y^ and x^ -\-^. 8. Find the least common multiple of a' -)- a — 10 and a* — 16. Note 3. — The least common multiple of any quantities can also be found by factoring the quailtities, and finding the product of all the factors of the quantities, taking each factor the greatest number of times it occurs in any of the quantities. (Art 80.) 9. Find the least common multiple of x^ — 1xy-\-y^, x* — y*, and {x + yf- x^—2xy-\-y^=(x — y){x — y) x* — y*= (x^ + y'') (x + 2/) (« — 2/) (x + yy = (x -\- y) (x -\- y) Hence L, G.M. = (x — y) {x — y) {x^ + y^) {x-{-y) {x^y) ^ x^ — x^y^ — x^y* -\- y^. 10 Find the least common multiple of 3ax^ — iax -{- 3axy — 4: ay and a" x — x -\- a'y — y. (See 15th Example, Art. 76.) Ans. a{x-\-y) (3 a; — 4) (a^+a+.l) (a — 1) =:3a^a;'' — 4a^a; + ^a^xy — 4.a^y — Bax^ + 4aa; — 3axy + 4.ay. FRACTIONS. 63 SECTION XII. FEACTIONS. 82. When division is expressed by writing the dividend over the divisor with a line between, the expression is called a Fraction. As a fraction, the dividend is called the numerator, and the divisor the denominator. Hence, the value of a fraction is the quotient arising from dividing the numerator by the denominator. Thus, — is a fraction whose numerator is xy and denominator y, and whose value is x. 83. The principles upon which the operations in frac- tions are carried on are included in the following THEOREM. Any multiplication or division of the numerator causes a like change in the value of the fraction, and any multiplica- tion or division of the denominator causes an opposite change in the value of the fraction. Let — ^ be any fraction ; its value =xy. y 1st. Changing the numerator. Multiplying the numerator by y, which is y times the value of the given fraction. Dividing the numerator by y, xy y which is — of the value of the given fraction. y 64 ELEMENTARY ALGEBEA. 2d. Changing the denominator. Multiplying the denominator by y, — . - X, t which is - of the value of the given fraction. y Dividing the denominator by y, which is y times the value of the given fraction. CoROLLABT. — Multiplying or dividing both numerator and denominator by the same quantity does not ctiange the value of the fraction. For if any quantity is both multiplied and divided by the same quantity its value is not changed. _, xy cxy X Thus, -^ = — - = - = r. y cy 1 84. Every fraction has three signs: one for the numer- ator, one for the denominator, and one for the fraction as a whole. Thus, +^- If an even number of these signs is changed from -\- to — , or — to -{-, the value of the fraction is not changed ; but if an odd number is changed, the value of the fraction is changed from -f- to — , or — to -f-. Thus, changing an even number, _i_ + ^y__ —xy __ +'^-'y = I "^y — I a;- -\-y +2/ —y "*" —y "•" ' but, taking 1 ±xy _ ■ ^ FRACTIONS. 65 and changing an odd number, -{•y -\-y —y —y The various operations in fractions are presented under the following cases. CASE I. 85i To reduce a fraction to its lowest terms. Note. — A fraction is in its lowest terms when its terms are mu- tually prime. 1 6 Q^ X V 1. Eeduce „, „ " . to its lowest terms. 24 a'xy' OPERATION. Since dividing both terms Wa*xy _ixy _ 2 °^ ^ *"''^'='^°" ^^ *^® "^""^ 2id'xf — 6^7 — Vy quantity does not change its value (Art. 83, Cor.), we divide both terms by any factor common to them, as 4 a^ ; . and both terms of the resulting fraction by any factor common to them, as 2xy; or we can divide both terms of the given fraction by their 2 greatest common divisor Bcpxy; the resulting fraction - — is the fraction sought. Hence, RULE. Divide both terms of the fraction by any factor common to them, ; then divide these quotients by any factor common to them ; and so proceed till the terms are mutually prime. Or, Divide both terms by their greatest common divisor. 2. Keduce —.-„ to its lowest terms. Ans. — ■■ i?f xy 3. Eeduce txtt-i — —, to its lowest terms. Ans. -—5 — 408 a^x f Sa'y 24 X V z 23 4. Reduce — ;: — ^— to its lowest terms. Ans. — 12axy a 5. Eeduce i^::—,:-,- to its lowest terms. 51 ffoxy 66 ELEMENTARY ALGEBRA. 6. Reduce Hl'^ff. to its lowest terms. 1. Reduce „ ."S^^. , to its lowest terms. Ans. — r— ^• 8. Reduce -. „ , . ,a to its lowest terms. 9. Reduce ^ — -. s — °^ i~." » — a tt — to its lowest ■ 2 c a;' — 2 a c ar" -[- 4 a^ c a;' — 4a'ca; terms. a a^ ^^^- c(2a? + 4a») 10. Reduce g_iri^ ^ " ^ — to its lowest terms. a — X* CASE II. 86t To reduce fractions to equivalent fractions having a common denominator. 1. Reduce j— and =— to equivalent fractions having a common denominator. OPERATION. "We multiply the numerator and a a b X denominator of each fraction by the by ¥xy denominator of the other (Art. 83, 1 Cor.). This must reduce them to c bey / i— — 12 equivalent fractions having a common denominator, as the new denominator of each fraction is the product of the same factors. 0B> In the second operation we find the o ax least common multiple, bxy, of tlie iy bxy denominators by and 6x; as each df- c cy nominator is contained in this multi- b X bxy pie, each fraction can be reduced to a fraction with this multiple as a de- nominator, by multiplying its numerator and denominator by the quotient arising from dividing this multiple by its denominator. Hence, FRACTIONS. 67 Multiply all the denominators together for a common de- nominator, and m/uUiply each numerator into the continued product of all the denondnators, except its mm, for new numerators. Or, Find the least common multiple of the denominators for the least common denominator. For new numerators, mul- tiply each numerator by the quotient arising from dividing this multiple by its denominator. 2. Eeduce — , — r. and -^— to equivalent fractions xy ab ahy ^ having the least common denominator. , ahm nxy - a;'' Ans. ~= — f , > and ahxy abxy abxy 3. Eeduce — -=. ttti-' and ^ to equivalent frac- 15 10 oc 25 acd ^ tions having the least common denominator. , 80 d' c d 45 adxy , 12 6a: Ans. -— — = — ,, -^„ — ,— H- and 150 a 6c d 150 abed 150 abed 4. Reduce - — > ' ■. and z— ; to equivalent fractions m nxy 5 d ^ having the least common denominator. 5. Eeduce . , and y- to equivalent fractions hav- ing the least common denominator. . a" — Sab-i-b' . a^ m -\- ab m Ans. 5 TT — and ~ — tz — a^ — W a" — 6^ 6. Eeduce . and —'^-— to equivalent fractions x — 4 x ^ 1 having the least common denominator. Y. Reduce -^ => — -. — -i and to equivalent frac- ,ar — ■f X -{■ y x — y ^ tions having .the least common denominator. 68 ELEMENTARY ALGEBRA. CASE III. 87. To add fractions. b , c 1. Find the sum of - and -• OPBKATiON. If anything is divided into equal J g S -I- c parts, a number of these parts rep- ■^ I" ^ ^^^ ^ resented by 6, added to a number represented by c, gives b -\- c of these parts. In the example given, a unit is divided into x equal parts, and it is required to find the sum of 6 and c of these parts ; i. e. 6 1^ c b -\- c X ' X X It is evident, therefore, that fractions that have a common denom- inator can be added by adding their numerators. But fractions that do not have a common denominator can be reduced to equivalent fractions having a common denominator. Hence, RULE. lieduce the fractions, if necessary, to equivalent fractions having a coTnmon denominator; then write the sum of the numerators over the common denominator. n Ajj*" 2T , a . bmv-\-bnx-\-any 2. Add -, -, and x- Ans. , ^^ -• n y b b ny 3. Add -—, —,, and --^• 7 46 2d 4. Add — , — ^, and -= — ^^ ixy bab Sabxy ■^°s- Z,, J. iOabxy ^- ^^^ ^'-' -97d' ^'^•l 27^- 6. Add^ and j-L_. Ans.- ^, 1. Add \+^ and i^. Ans. ^+^'. 1 — a 1+a °-l_a2 FRACTIONS. 69 8. Add ^— and ^ l + a;' 9. Add ^fy, and -^. 7 (a; + 2^) a; — y 10. Add ^=i'. ?^^^ and "-^ + ^^. Ans 1 11. Add -= 3 and 2xy Ans. ^+i^ 12. Add jwa; and 18 a mx Note. — Consider ma;= — , and then proceed as before. , ISamx 4- 7 X Ans. —-^ 18 a 13. Add a: + y and — j—r- ' " a-\-b 14. Add «» + 2 a;y + / and — i- • a; y Ans ^ + ^y-^y'-y'+l a; — 2^ CASE IV. 88. To subtract one fraction from another. 1. Subtract - from -• X X OPERATION. If anything is divided into x J c 6 — c equal parts, a number of these X X X parts represented by c, subtracted from a number represented by 6, leaves 6 — c of these parts ; i, e. = — Hence, '^ X X X RULE. Reduce the fractions, if necessary, to equivalent fractions having a common denominator ; then subtract the numerator of the subtrafwnd from that, of the minuend, and write the result over the common denominator. 70 ELEMENTARY AtGEBBA. J, ah — 14ca; 2. Subtract ^ from g^- ^ns. gc 3 a 7 *^ 3. Subtract ^^ from g^^- liab . 15 c 4. Subtract -jg^ from j^- 99ac 39 a; . „ 273 a;' — 116 a eg 5. Subtract ^ from — • Ans. ^^^.^ 1.1 Ana ^° 6. Subtract ^ _ ^ from ^-:^- -"-ns. ^, _ ^ 7. Subtract |i| from^-_j_-^- a6_|_6c f ah — be 8. Subtract „,^^-^ from ^.^-ZTw^' 9. Subtract ^^Tj ^'■°™ ^"+6' ''^°^' ^^^^'" 10. Subtract -^^—^ from ^^zrj- ■^'^^- a? + 1 11. Subtract 16 from'j-q^- Note.— Consider 16 = y, and then proceed as before. a? — 16 a;— 23 Ans. ^^7^ ^ 3 12. Subtract , from xy. a — Ja; 4- 46 V -\-ix 13. Subtract 03 + 6 from ^T^ ■• Ans. j-^T' CASE V. 89. To reduce a mixed quantity to an improper fraction. 1. Reduce a; + | to an improper fraction. o OPERATION. As eight eighths make .a_8a:.a_ 8 a: + d a unit, there will be in ^"T~8 ~8~~r"8 8 X units eight times « 8a; ,8a;, a 8a; + a „ eighths; i. e. a; = -^; and — + ^ = — ^ — . Hence, FRACTIONS. 71 ETJLE. Multiply the integral part by the denominator of the frac- tion; to the product add the numerator if the sign of the fraction is plus, and subtract it if the sign is minus, and under the result write the denominator. Note. — By a change of the language, Exalnples 12 -14 in Art. 87, and 11-13 in Art. 88, become examples under this case. Thus, Example 12, Art. 87, might be expressed as follows: Reduce mx-{- y^ — to an improper fraction. 7 2. Reduce s? -\- ^ to an improper fraction. . a:^ V -t- 4 « — 7 y 3. Reduce 25 a — 25 ai -) ^^ to an improper frac- tion. 4. Reduce a — 1 -|- T to an improper fraction. . a' — a Ans. — j—r • a-\-l 5. Reduce y -\ ^-^ — to an improper fraction. 6. Reduce — T {a-\-h) to an improper fraction. . a" — ah Ans. — r T. Reduce x — 1 J— to an improper fraction. Note. — It must be remembered that the sign before the divi('' line belongs to the fraction as a whole. X — 1 ^-^ p- = — j— r'or 7-— > Ans. x-^l x-{-l a-f-l K+l 8. Reduce x -\- 1 ^^ to an improper fraction. Ans. , 1 — X 72 ELEMENTARY ALGEBRA. 9. Eeduce x' — 2ax -\-a' — - ^ ~^ to an improper fraction. Note. — According to the same principle an integral quantity can be reduced to a fraction having any given denominator, by multiplying the quantity by the proposed denominator, and under the product writing the denominator. 10. Eeduce x -\- 1 to a fraction whose denominator is ^ ■'^- Ans. -■ X — 1 11. Eeduce x — 1 to a fraction whose denominator is a — ■ b. 12. Eeduce iax to a fraction whose denominator is a' — z. CASE VI. 90. To reduce an improper fraction to an integral or mixed quantity. /*•* ^__ A, fl rp \ 5 fit 1. Eeduce — to an integral or mixed quan- tity. OPERATION. X — 2 a) x-^ — ^ax -\- 5 a" (x — 2 a A \- x^ — 2ax — 2ax -\- he? — 2ax-\-^a^ As the value of a fraction is the quotient arising from dividing the numerator by the denominator (Art. 82), we perform the indicated division. Hence, RULE. Divide the numerator by the denominator ; if there is any remainder, place it over the divisor, and annex the fraction so formed vyith its proper sign to the quotient. FRACTIONS. 73 2. Reduce — to an integral or mixed quantity. Ans. a — 4 J. 3. Reduce r — ^^^ to an integral or mixed quantity. 4 j;S 4 aS 4. Reduce ^ to an integral or mixed quantity. r T. J 24y— 12ow'+7 , • . 1 .J 5. Reduce — =! „ „ — ■ — to an integral or mixed l^^^i^^y- Ans. 4y-2a + gL. 6. Reduce — — — — to an integral or mixed quan- X tity, 2x ^ _ . Sax — 106a; — 5cx . . , , . , 7. Reduce — — ^ to an integral or mixed quantity. 8. Reduce s ^-s-e to an integral or mixed - Ans -'- • ^•^ 2iy "^ 20" — 8a* *" 3 -f ■ 6 « FRACTIONS. 79 10. Multiply ,^^_, by ,3„./_14ar 11. Multiply ^^i^ by ^. Ans. ^^. 12. Multiply -j—hy-^. 13. Multiply y + -^ by y ^ "2/ a — y •' " a-f-i Ans. ^°/-/. 14. Multiply together ^, -^. ^nd |±|., Ans. 1. 15. Multiply together ^, ^,_^> and — • 16. Multiply together a -^ — i 5 -(- - . and y — ^- Ans. ajy + 6^_g-|. CASE XII. 96i To divide a fraction by a fraction. 1. Divide - by -r • OPERATION. X T --^a-- (Art. 93); but the a; _ X y fly y ' ay divisor is not a, jbut 5-: we have used _^ y. J ^ a divisor Jf limes too great, *nd t^ere- " * fore the quotient — is 6 times too ay X h X small, and the quotient sought is — x 6 =s= - (Art. 91). It will he noticed that the denominator of the dividend is multiplied by the numerator of the divisor, and the nuinerator of th^ diF,id.6^d by the denominator of the divisor. Hence, EULE. Invert the divisor, and then propel as in multiplicqiion of a fraction by afra^ion. 80 ELEMENTARY ALGEBRA. Note 1. — All cases in division of fractions can be brought under this rule, by ■writing integral quantities as fractions with a unit for the denominator. Note 2. — After the divisor is inverted, common factors can be cancelled, as in multiplication of fractions. Note 3. — Mixed quantities should be reduced to improper frac- tions before division. 2. Divide - by — ■ Ans. — ■ c •' n cm 3. Divide - by - • Ans. ~ ■ y " i x'y 4. Divide -^ by -3^. 5. Divide ^tL^ by 1^. Ans. l^^l+l^, 6. Divide ^^f+^ hy^-. 1. Divide ^ by '^±^y. Ans ^"^ . ^ + y ■' x—y • 3?-\-xy 8. Divide t±^ by tizl. a — 6 ^ 4 9. Divide 3^ by -^ . Ans ^^ . 10. Divide ^^fl! by ^=i. x-\-\ "' 4 11. Divide ^+^ by i££ilf^. 12. Divide 3"''"-3»' by "■' + "'" + »• Ans. 3 (»»» + »'). 13. Divide 1 + '-±1 by — £ 1 Ans. '^(^ + y) + (^ + .y)' c (c — X — y) 14. Divide x-^-y—^ ^y 5 + * + y- FRACTIONS. 81 1+a; , 1—x ^ 1 — 1? 15. Divide ii^ + "—^ by ^-^=^ . Ans. 10+^'. (1 — x'y Note. — The division of fractions is sometimes expressed by writ- a ing the divisor under the dividend. Thus, i-. Such an expression is called a Complex Fraction. A Complex Fraction can be reduced to a simple one by performing the division indicated. 16. Reduce -=- to a simple fraction. Ans. — • _ 7 X 5 X 11. Reduce j- to a simple fraction. T t cx-\-ac e Ans. ~ '■ — J—. lex — ox x+1 X —-\ 18. Reduce ^ ^ to a simple fraction. x-\-i 19. Reduce — , . to a simple fraction. \—x a^ — V Ans. 20. Reduce — ,\ to a simple fraction. a-f-o '^ Ana X -\- ) 21. Reduce — to a simple fraction. a + 6 Ans. "(a: -\- y) (« + V). Note. — A Complex Fraction can also be reduced by multiplying its numerator and denominator by the least common multiple of the denominators of the fractional parts. Thus, if both terms of the fraction in Ex. 16 be multiplied by 5x, or both in Ex. 17 by ex, the result ivill be the same as above. 4* F 82 ELEMENTARY ALGEBRA. SECTION XIII. EQUATIONS OF THE FIRST DEGREE CONTAINING BUT ONE UNKNOWN QUANTITY. 97i An Equation is an expression of equality between two quantities (Art. 9). That portion of the equation which precedes the sign = is called the first member, and that which follows, the second member. 98i The Degree of an equation containing but one un- known quantity is denoted by the exponent of the highest power of the unknown quantity in the equation. An equation of the first degree, or a simple equation, is one that contains only the first power of the unknown quantity. For example, 2x — ax = 21. An equation of the second degree, or a quadratic equation, is one in which the highest power of the unknown quantity is the second power. For example, x^ — ax^b -]- c, or ax'^ — 5=17. An equation of the third degree, or a cubic equation, is one in which the highest power of the unknown quantity is the third power, and so on. 99. The Eeduction of an Equation consists in finding the value of the unknown quantity, and the processes involved depend upon the Axioms given in Art. 13. The processes can be best understood by considering an equation as a pair of scales which balance as long as an equal weight remains in both sides : whenever on one side any additional weight is put in or taken out, an equal weight must be put in or EQUATIONS OF THE FIRST BEGEEE. 83 taken out on the other side, in order that the eqiaMibrium may remain. So, in an equation, whatever is donp to one side must be done to the other, in order t|iat the equality may rem am. 1. If anything is added to one member, an equal quantity must be added to the other. 2. If anything is subtracted from one member, an equal quantity must be subtracted from the other. 3. If one member is multiplied by any quantity, the other member must be multiplied by an equal quantity. 4. If one member is divided by any quantity, thp other member must be divided by an equal quantity. 5. If one member is involved or evolved, the other must be involved or evolved to the same degree. TEANSP'OSITION. lOOt Transposition is the changing of terms from one member of an equation to the other, ijFithout destro^ying the equality. The object of transposition is to tiring oall -the nnknosssn terms into one member and all the known into the other, so that tbe unknown may become known. I. Find the value of x in the equation x -\- 1^ ,:;=: 24. Subtracting 16 ftom the first OPERATION. member leaves a; ; but if 1 6 is sub- *■ T" l" -— "* tracted froiii the first menljjer, it X = '24: — 16 = 8 must also be subtracted from the second. 2 Find the value of x in the equation x — ,h ^ a. OPERATION. Adding 6 to the first member J. J = a gives x ; but , if h is , ^dfje^d ito the „ „ 1 I first member it must also be added X = a -\- o ...... to the second. 84 ELEMENTARY ALGEBEA. 3. Find the value of x in the equation 2 a; = a: -j- 16. OPERATION. ~ i_ 1 « Subtracting x from both mem- „ , „ bers, we have 2 x — a; = 16, or 2x — X = 16 ,„ x= 16. a; =16 It appears from these examples that any term which dis- appears from one member of an equation reappears in the other with the opposite sign. Hence, RULE. Any term may he transposed from one member of an equation to the other, jrromded its sign is changed. 4. Find the value of x in the equation 8x — 15 = 4a:-|-5. OPERATION. 8x — 15= 4a;+ 5 Transposing, 8a; — :4a;^ 5 +15 Uniting terms, 4 a; =20 Dividing both members by 4, a; = 5 5. Find the value of a: in 4 a; -)- 46 = 5 a; -|- 23. Note. — Reducing, we have — x ^ — 23. If each member of this equation is transposed, we shall have 23 = a; ; i. e. 23 equals X, or X equals 23. Dividing both members by — 1 will give the same result. Hence, the signs of all the terms of an equation may be changed without destroying the equality. 6. Find the value of a; in 1^3:+ IT = 19 a; + 13. Ans. a; = 2. 1. Find the value of a; in 8 a; — 14 =: 13 a; — 29. 8. Find the value of a: in 5 a; -|- 25 = 10 a; — 25. Ans. X = 10.- 9. Find the value of a; in 24 x — 17 = 11 a; + T4. 10. Find the value of x in 37 x — (4 -f 7) = 41 x — 23. EQUATIONS OF THE FIRST DEGKEE. 85 CLEARING OF FRACTIONS. 101 • To clear an equation of fractions. 1. Find the value of a; in the equation | — 2 = ^4-1- OPBEATION. If the given equation is mul- E 2 -^ ^ I 1 tiplied by 6, the least common * ^ multiple of 6 and 3, it will give 2x — 12=a;-j-6 2x — 12 = a;-|-6, an equation a; = 18 without a fractional term. Hence, RULE. Multiply each term of the equaMon by the least common multiple of the denominators. Note 1. — In multiplying a fractional term, divide the multiplier by the denominator of the fraction and multiply the numerator by the quotient. Note 2. — An equation may be cleared effractions by multiplying it first by one denominator, and the resulting equation by another, and so on, till all the denominators disappear; but multiplying by the least common multiple is generally the more expeditious method. Note 3. — Before clearing effractions it is better to unite terms which can readily be united ; for instance, the equation in Ex. 1, by transposing — 2, can be written - = - -|- 3. Note 4. — When the sign — is before a fraction and the de- nominator is removed, the sign of each term that was in the nu- merator must be changed. 2. Given j — ^-|-25 = 33- """ 2 OPBEATION. X X „ X ■ Transposing 25, 7 — 5 ^^ ^ ' 2 Multiplying by 20, 5 a; — 4 a; = 160 — 10 a; -]- 60 Transposing and uniting, 11 x = 220 Dividing by 11, a; =20 86 ELEMENTARY ALGEBEA. Note. — The sign of the numerator of — - is +> ^^^ "'"s* ^^ changed to — when the denominator is removed ; for — (+ 4 J:) = ix; and so the sign of each term of the numerator of the fraction ^ must be changed when the denominator 2 is removed ; for — (+ 10 X — 60) = — 10 x -|- 60. 102. To reduce an equation of the first degree contain- ing but one unknown quantity, we deduce from the preced- ing examples the following RULE. Clear the equation of fractions, if necessary. 2'ranspose the known terms to one member and the un- known to the other, and reduce each member to its simplest form. Divide both members by the coefficient of (lie unknown quantity. Note 1. — To verify an equation, we have only to substitute in the equation the value of the unknown quantity fouad by reducing the equation. Por instance, in Ex. 2, Art. 101, by substituting 20 XX X ■ 6 for a:, in - — - ^ 25 = 33 — — - — , we have ?2_20+25 = 33-^^:^, 4 5' 2 ' 5 — 4 + 25 = 33 — 7, 26 = 26. Note 2. — AVhen answers are not' given, the work should be veri- fied. 103. Since the relations between quantities in Algebra are often expressed in the form of a proportion, we intro- duce here the necessary definitions. EQUATIONS OF THE FIRST DEGREE. 87 104i Eatio is the relation of one quantity to another of the same kind ; or, it is the quotient which arises from di- viding one quantity by another of the same kind. Batio is indicated by writing the two quantities after one another with two dots between, or by expressing the division in the form of a fraction. Thus, the ratio of a tp h is written, a -.h, ov j] read, a is to 6, or a divided by h. 105i Pkoportion is an equality of ratios. Four quan- tities are proportional when the ratio of the first to the second is equal to the ratio of the third to the fourth. The equality of two ratios is indicated by the sign of equality (=) or by four dots (::). Thus, a : J = c : rf, or o : J : : c : rf, or j ^ ^ ; read, a to & equals c to d, or a is to J as c is to d, or a divided by h equals c divided by d. The first and fourth terms of a proportion are called the extremes, and the second and third the means. 106. In a proportion the product of the means is equal to the product of the extremes. Let a : h = e : d i. e. Clearing of fractions, ad = bc a c b~d A proportion is an equation ; and making the product of the means equal to the product of the extremes is merely clearing the equation of fractions. Examples. 1. Eeduce ^ -f 10 := ^ + 13. Ans. x = 30. 2. Eeduce IT a; — 14 = 12 a; — 4. Ans. a: = 2. 88 ELEMENTARY ALGEBRA. 3. Reduce 6 a; — 25 + a; = 135 — 3 a; — 10. Ans. X = 15. 4. Reduce 3a; + 5— a; = 38 — 2a;. Ans. x — 8^. 5. Reduce ^^ + | = 30 — ^-J^. Ans. x = 12. 6. Reduce x — T^ = — — • Ans. x = llxV- T. Reduce | + f-J-^4-?= 154. Ans. x — 120. 8. Reduce | + |= 16 + |. Ans. x = 24. 9. Reduce ^ + o = f — ^ + rf. OPERATION. Multiplying bj J c A, chx-\-abc h = bcx—bhx-\-bcdh Transposing, chx — bcx-{-bhx=bcd/i — abch Factonnglstmem.,(ch~bc-\-bh)x = bcdh — abch Dividing by coefficient of a;, x = bcdh — abch 10. Reduce x -\- mx = c. 11. Reduce ^^=^ — 3 = 7. 12. Reduce l-f-^^a;. a c 13. Reduce — [- - = c 14. Reduce 8 = -^^ U fi. a; — 2 ' 15. Reduce - — - = - _ c s a; a; ch — bc + 6A Ans. X = c T+m' Ans t. x = 6 — c " 10 Ans. X = a J -|- c a c Ans . X = _a + b c Ans. ar = 9. Ans. X = .« + l KQUATIONS OF THE FIRST DEGREE. 89 16. Reduce ^ + f -f- f = 39. 2 o 4 IT. Eeduce — — 7 + c = «f. a ' X 1 18. Reduce (a — b) x -\ — = - ■ C a 19. Reduce x — /| — ^^=h. Ans. a; = 6. 20. Reduce 6 — ^^ii = 3. _ 4, 01 T> J o 9a; — 29 ,„ 6a;+ll 21. Reduce 2 a: ; =18 ^ 4 5 Ans. a; = 9. 22. Reduce ^^~-~~ = 3 a; + - ~ ^^. , 23. Reduce 2 a: — ?^-=i = 14 — ''-^^-^. Ans. a; = 5. 24. Reduce 6 a: + TJ — | = 9^ — ^ + ^''. Note. — Before clearing of fractions, transpose 7^ and unite it with 9J; also transpose — -, and unite it with —^. 25. Reduce 4 a: + ^±^ = 5 + li±-^-l5. o o 26. Reduce ^^ — ^^ = 21 — -t-^ Ans. x = 39. 2 o b CM, T, I X , X , X J , nhcd 27. Reduce — h t. H — = 0. Ans. a; ^ a'^b'^c ' ' be -\-'ac-\- ab 28. Reduce ^~^ — 6 = ^=-^ + 7. » 4 29. Reduce ^^ = 6 — ---— — ii^- Ans. a; = 7. 6 a a 90 ELEMENTARY ALGEBRA. 2 3; — 22 3 a;— 75 , 284 — 4 a: 12 30. Reduce 19 + „, ^ , 4a: + 5 5x — 5 _ x-\-l 31. Reduce — y j — — — g ^• Ans. X = f). „ , 18 — 5x 3a;+3 . ,^ . 5a;-|-3 32. Reduce g =!— = 4=x—11 -] J-~ ■ ^ . i a; — 12 , . 20a: + 21 1 33. Reduce 4a: ^ 1- 5 = ^ ^■■ 34. Reduce = - • Ans. x == t j ; • :j c m om-\-cd X ax 35. Reduce t = 1 — 3a c. 6 c „« T. J 5a;+3 , . 3a:+15 . , 6a;4-10 86. Reduce — ^ + 6 ^ — = * H ^ 2 ' 4 '4 „^. „ , „ 3x — 19 „ 23 — X , 5x — 38 , ,„ 31. Reduce 3 a; 8= — (-10. Ans. X = 19. „„ ^ , 13 — 3a; 8x+2 ^ „ , 8x— 13 38. Reduce — ^^ ^— = 1 — 6x-\ ^ „„ „ , 4(a:— 7) , 3(a: + l) 7a:— 17 39. Reduce ^ ^ ^ + -\f-^ = ^q X 21 4a: — 6 19— 4a: 5a; — 6 . 7x + 8 40. Reduces [-3 = — 5 ^ \- o 7 a; , a; , 3; , a: 41. Reduce t^±^f^±: a ' b ' c ' d i«T>j 7x+5,6a; — 30 ,, 42. Reduce — ^— + -y^^y- = a: + 1. y ^ V Note. — Multiply by 7, transpose, and unite. 43. Reduce 2 (3 + a;) : 6 a: — 9 = 2 : 3. Ans. x=6. 44. Reduce ^ + ^ : -^il- = 11:^- 6 ' 5 2a;-|-14 2 45. Reduce b -. c -\- d ^^ — : n. X ' X EQUATIONS OF THE FIRST DEGREE. 91 PROBLEMS PRODUCING EQUATIONS OF THE FIRST DEGREE CON- TAINING BUT ONE UNKNOWN QUANTITY. 107. The problems given in this Section must either con- tain but one unknown quantity, or the unknown quanti- ties must be so related to one another that if one be- comes known the others also become known. 108. With beginners the chief diflSculty in solving a problem is in translating the statements or conditions of the problem from common to algebraic language ; i. e. in preparing the data, and forming an equation in accord- ance with the given conditions. 1. If three times a certain number is added to one half and one third of itself, the sum is 115. What is the number ? SOLUTION. Let X =. the number ; then by the conditions of the problem. Clearing of fractions, \?>x-\-Zx-\-2x^ 690 Uniting terms, 23 a' ^ 690 Dividing by 23, a; = 30 VERIFICATION. 3X30 + ^ + ^=115 115 = 115 In this problem there is but one unknown quantity, which we rep- resent by X. 2. There are three numbers of which the first is 6 more than the second, and 11 less than the third ; and their sum is 101. What are the numbers? 92 ELEMENTARY ALGEBRA. SOLUTION. Let X = the first, In this problem then X — 6 ^ the second, there are three un- and a; + 1 1= the third. ^"0^° quantities; . ; ' ZZT but they are so re- Their sum, 3 a; + 5 = 101 ; ' ' lated to one an- ^^= ^^ other that, if any X = 32, the first, ^^^ becomes known, X — 6 ^ 26, the second, the other two will a; + 11 = 43, the third. be known. VERIFICATION. 32 + 26 + 48 = 101 101 = 101 From these examples we deduce the following GENERAL RULE. Lei X {or some one of the latter letters of the alphabet) represent the unknown quantity ; or, if there is more than one unknown quantity, let x represent one, and find the others by expressing in algebraic form their given relations to the one represented by x. With the data thus prepared form an equation in accord- ance with the conditions given in the problem. Solve the equation. The three steps may be briefly expressed thus : — 1 St. Preparing the Data ; 2d. Forming the Equation ; 3d. Solving the Equation. 3. The sum of three numbers is 960 ; the first is one half of the second and one third of the third. What are the numbers ? Ans. 160, 320, and 480. 4. Find two numbers whose diflference is 18 and whose sum 112. Ans. 41 and £5. EQUATIONS OF THE FIRST DEGREE. 93 5. A man being asked how much he gave for his horse said, that if he had given $ TO more than three times as much as it cost, he would have given $ 445. How much did his horse cost him ? , ~ 6. A man being asked how many sheep he had, replied that if he had as many more, and two thirds as many, and three fifths as many, he should have 8 more than three times as many as he had. How many sheep had he ? 1. Divide $515 between A and B in such a manner that B may have two thirds as much as A. Ans. A's share, $ 345 ; B's " $230. 8. A father divided his estate among his three children so that the eldest had $ 1440 less than one half of the whole, the second $500 more than one third of the whole, and the youngest $250 more than one fourth of the whole. What was the value of the estate ? SOLUTION. Let X = whole estate. Then X 2~" 1440 = share of the eldest, f + 500= " " " second. sum 1+ 250= " " " youngest, Their 13a; 12 690 = X, whole estate. 5=690 a; = 8280, whole estate. 9. A gentleman meeting five poor persons, distributed $7.50, giving to the second twice, to the third three times, to the fourth four times, and to the fifth five times as much as to the first. How much did he give to each ? , , ^ ' 94 ELEMENTARY ALGEBRA. 10. Divide '795 into two such parts that the greater di- vided by 3 shall be equal to the less divided by 2. Note. — To avoid fractionB, let 3a;=the greater and 22;=the less. Ans. 417 and 318. 11. Divide a into two such parts that the greater di- vided by b shall be equal to the less divided by c. Let then actions, bx + 0. a — SOLUTION. X = the greater, - X = the less. And Clearing of fr Transposing, Dividing by b •■ + X b~ c X ■=■ ex ^ a — X c : ab — bx : a b cb ., , , ' theg ab ac ,11 a — a; =: a — .—t — = j— j — , the less. o-j-c 0-f-C 12. What number is that which, if multiplied by 7, and the product increased by eleven times the number, and this sum divided by 9, will give the quotient 6 ? 13. If to a certain number 55 is added, and the sum divided by 9, the quotient will be 5 less than one fifth of the number. What is the number? Ans. 125. 14. As A and B are talking of their ages, A says to B, "If one third, one fourth, and seven twelfths of my age are added to my age, the sum will be 8 more than twice my age." What was A's age ? 15. A farmer having bought a horse kept him six weeks at an expense of $20, and then sold him for four fifths of the original cost, losing thereby $ 50. How much did he pay for the horse? Ans. $150. 16. A man left $ 18204, to be divided among his widow, three sons, and two daughters, in such a manner that the widow should have twice as much as a son, and each son as much as both daughters. What was the share of each ? EQUATIONS OF THE FIRST DEGKEE. 95 It. If a certain number is divided by 9, the sum of the divisor; dividend, and quotient will be 89. What is the number? Ans. 12. 18. if a certain quantity is divided by a, the sum of the divisor, dividend, and quotient will be b. What is the quantity ? . ab — a^ Ans. j — — • a -j- 1 19. Verify the answer to the preceding problem. 20. A farmer mixed together corn, barley, and oats. In all there were 80 bushels, and the mixture contained two thirds as much corn as barley and one fifth as much bar- ley as oats. How many bushels of each were there ? 21. Three men. A, B, and C, built 512 rods of fence. A built 8 rods per day, B 7, and 5. A worked one half as many days as B, and B one third as many as C. How many days did each work ? 22. What number is as much greater than 340 as its third part is greater than 34 ? Ans. 459. 23. A man meeting some beggars gave 3 cents to each, and had 4 cents left. If he had undertaken to give 5 cents to each, he would have needed 6 cents to complete the dis- tribution. How many beggars were there, and how much money did he have ? SOLITTION. Let X = the number of beggars ; then, according to the first statement, Sx -\- 4: = the number of cents he had, and, according to the second statement, 5 a; — 6 ^ the number of cents he had. Therefore, 5a; — 6^3a: + 4 2 a; =10 X = 5, the number of beggars, and 3 a: -|- 4 = 19, the number of cents he had. 96 ELKMENTAEY ALGEBRA. 24. A boy wishing to distribute all his money among his companions gave to each 2 cents, and had 3 cents left ; therefore, collecting it again, he began to give 3 cents to each, but found that in this case there was one who had received none, and another who had only 2 cents. How many companions, and how much money had he ? Ans. Y companions, and 17 cents. 25. What two numbers whose difference is 35 are to each other as 4:5? 26. A man being asked the hour, answered that three times the number of hours before noon was equal to three fifths of the number since midnight. What was the time of day ? SOLUTION Let X = the number of hours since midnight, i. e. the time ; then 12 — a: = the number of hours before noon. Then 36— 3x=^-^ a Clearing of fractions, 180 — 15 a; = 3 a; Whence 18 a; = 180 X = 10. Ans. 10 o'clock, 27. A gains in trade $ 300 ; B gains one half as much as A, plus one third as much as C ; and C gains as much as A and B. What is the gain of B and C ? Ans. B's, $375; C's, $675. 28. What number is to 28 increased by one third of the number as 2 : 3 ? Ans. 24 29. What number is that whose fifth part exceeds its sixth by 15 ? 30. Divide $3740 into two parts which shall be in the ratio of 10 : 7. 31. Divide a into two parts which shall be in the ratio oi b : c. ab , ac Ans. 5— i — and b + c b + c EQUATIONS OF THE FIRST DEGREE. 97 32. What number is that the sum of whose fourth part, fifth part, and sixth part is 37 ? 33. What quantity is that the sum of whose third part, fifth part, and seventh part is o ? . 105 a 34. A farmer sold IT bushels of oats at a certain price, and afterward 12 bushels at the same rate ; the second time he received 65 shillings less than the first. What was the price per bushel ? 35. A certain number consists of two figures whose sum is 9 ; and if 21 is added to the number, the order of the figures will be inverted. What is the number ? SOLUTION. Let X = the left-hand figure ; then 9 — x=. the right-hand figure. As figures increase from right to left in a tenfold ratio, 10a;-|-(9 — a;) = 9a;-|-9= the number ; and when the order of the figures is inverted, 10 (9 — a:) -j- a; = 90 — 9 a; = the resulting number. Therefore 9a: + 9-}-27=:90 — 9a: Or 18x = 54 Whence a; = 3, the left-hand figure, and 9 — a: = 6, the right-hand figure. Ans. 36. 36. A certain number consists of three figures whose sum is 6, and the middle figure is double the left-hartd figure ; and if 198 is added to the number, the order of the figures will be inverted. What is the number ? Ans. 123. 3T. Two men 90 miles apart travel towards each other till they meet. The first travels 5 miles an hour and the second 4. How many miles does each travel before they meet? j G 98 ELEMENTARY ALGEBRA. 38. A man hired six laborers, to the first of whom he paid 75 cents a week more than to the second; to the second, 80 cents more than to the third ; to the third, 60 cents more than to the fourth ; to the fourth, 50 cents more than to the fifth ; to the fifth, 40 cents more than to the sixth; and to all he paid $68.15 a week. What did he pay to each a week ? 39. What number is that to which if 20 is added two thirds of the sum will be 80 ? 40. What number is that to which if a is added - of the sum will be c?? ^^^ ^ ^ b 41. A man spent one fourth of his life in Ireland, one fifth in England, and the rest, which was 38 years, in the United States. To what age did he live? 42. A post is one fifth in the mud, two sevenths in the water, and 18 feet above the water. How long is the post ? 43. What number is that whose half is as much less than 40 as three times the number is greater than 156? Ans. 56. 44. Two workmen received the same sum for their la- bor ; but if one had received $15 less and the other $15 more, one would have received just four times as much as the other. What did each receive ? * 45. Of the trees on a certain lot of land five sevenths are oak, one fifth are chestnut, and there are 32 less wal- nut trees than chestnut. How many trees are there ? 46. Divide 4*74 into two parts such that, if the greater part is divided by 7 and the less by 3, the first quo- tient shall be greater than tjje second by 12. Ans. 35'7 and 117. EQUATIONS OF THE FIRST DEGREE. 99 4:1. Two persons, A and B, have each an annual income of $1500. A spends every year $400 more than B, and at the end of five years the amount of their savings is $6000. "What does each spend annually? Ans. A $1100, and B $'700. 48. In a skirmish the number of men captured was 41 more, and the number killed 26 less than the number wounded ; 45 men ran away ; and the whole number en- gaged was four times the number wounded. How many men belonged to the skirmishing party ? Ans. 240. 49. A and B have the same salary. A runs into debt every year a sum equal to one sixth of his salary, while B spends only three fourths of his ; at the end of five years B has saved $ 1000 more than enough to pay A's debt. What is the salary of each ? Ans. $ 2400. 50. A man lived single one third of his life : after hav- ing been married two years more than one eighth of his life, he had a daughter who died ten years after him, and whose age at her death was one year less than two thirds the age of her father at his death. What was the Other's age at his death ? SOLUTION. Let X = his age ; then - ^ his age at marriage, - -[- - 4" 2 := his age at daughter's birth, o o and X — / ^ -^ ^ -|- 2 j = her age at his death. Then 0.-1-1-2+10 = ^-1 Transposing and uniting, — -.= — 9 X = '72, the father's age. 100 ELEMENTARY ALGEBRA. 51. Divide $ 864 among three persons so that A shall have as much as B and C together, and B $5 as often as C $ 11. 52. A father and son are aged respectively 32 and 8. How long will it be before the son will be just ' one half the age of the father ? 53. A man's age wag to that of his wife at the time of their marriage as 4:3, and seven years after, their ages were as 5:4. What was the age of each at the time of their marriage ? 54. One fifth of a certain number minus one fourth of a number 20 less is 2. What is the number? Ans. 60. 55. There are two numbers which are to each other as J : J ; but if 9 is added to each, they will be as ^ : ^. What are the numbecs ? Ans. 9 and 6. 56. A person having spent $ 150 more than one third of his income had $ 50 more than one half of it left. What was his income ? 5T. A merchant sold from a piece of cloth a number of yards, such that the number sold was to the number left as 4 : 5 ; then he cut off for his own. use 15 yards, and found that the number of yards left in the piece was to the number sold as 1:2. How many yards did the piece originally contain ? Ans. 45. 58. Four places, A, B, C, and D, are in a straight line, and the distance from A to D is 126 miles. The distance from A to B is to the distance from B to C as 3 : 4, and one third the distance from A to B added to three fourths the distance from B to C is twice the distance from C to D. What is the distance from A to B, from B to C, and from C to D ? 59. A laborer was hired for 40 days ; for each day he wrought he was to receive $2.50, and for each day he was idle he was to forfeit $1.25. At the end of the time he received $68.15. How many days did he work? Ans. 29. EQUATIONS OF THE FIRST DEGREE. 101 60. A cask which held 44 gallons was filled with a mixture of brandy, wine, and water. There were 10 gal- lons more than one half as much wine as brandy, and as much water as brandy and wine. How many gallons were there of each ? 61. Two persons, A and B, travelling each with $80, meet with robbers who take from A $5 more than twice what they take from B ; then B finds he has $ 26 more than twice what A has. How much is taken from each ? Ans. From A, $69 ; from B, $32. 62. Four persons. A, B, C, and D, entered into part- nership with a capital of $84816; of which B put in twice as much as A, C as much as A and B, and D as much as A, B, and 0. How much did each put in ? 63. In three cities, A, B, and C, 1188 soldiers are to be raised. The number of enrolled men in A is to that in B as 3 : 5 ; and the number in B to that in C as 8 : Y. How many soldiers ought each city to furnish ? Ans. A, 288 ; B, 480 ; C, 420. 64. Divide $65 among five boys, so that the fourth may have $2 more than the fifth and $3 less than the third, and the second $4 more than the third and $5 less than the first. 65. A merchant bought two pieces of cloth, one at the rate of $ 5 for T yards, and the other $ 2 for 3 yards ; the second piece contained as many times 3 yards as the first times 4 yards. He sold each piece at the rate of $6 for 7 yards, and gained $24 by the bargain. How many yards were there in each piece ? Ans. First, 84 ; second, 63. 66. A drover had the same number of cows and sheep. Having sold IT Cows and one third of his sheep, he finds he has three and a half times as maiiy sheep as cows left. How many of each did he have at first? 102 ELEMENTARY ALGEBEA. 61 . A flour dealer sold one fourth of all the flour he had and one fourth of a barrel ; afterward he sold one third of what he had left and one third of a barrel ; and then one half of the remainder and one half of a barrel; and had 15 barrels left. How many had he at first ? SOLUTION. Let X = number at first; S X 1 then -; 7 = number after first sale, gi-^ i) — 3^^2 — 2^ number after second sale, 1 /x 1\ 1 a: 3 and gig — gl — 2^^4 — 4^^ number after third sale. Then ? _ ! = 15 4 4 Clearing of fractions, x — 3 = 60 Whence x = 63, number at first. 68. A merchant bought a barrel of oil for $50; at the same rate per gallon as he paid, he sold to one man 15 gallons ; then to another at the same rate two fifths of the remainder for $ 14. How many gallons did he buy in the barrel ? 69. Two pieces of cloth of the same length but dif- ferent prices per yard were sold, one for $ 5 and the other for $7.50. If there had been 5 more yards in each, at the same rate per yard as before, they would have come to $ 15.4'?^^. How many yards were there in each? Ans. 21. "70. A and B began trade with equal sums of money. The first year A lost one third of his money, and B gained $760. The second year A doubled what he had at the end of the first year, and B lost $150, when the two had again an equal sum What did each have at first ? EQUATIONS OF THE FIRST DEGREE. 103 Vl. A man distributed among his laborers $2.50 apiece, and had $25 left. If he had given each $3 as long as his money lasted, three would have received nothing. How many laborers were there, and how much money did behave? Ans. 68 laborers, and $195. 72. A man who owned two horses bought a saddle for $ 35. When the saddle was put on one horse, their value together was double the value of the other horse ; but when the saddle was put on the other horse, their value together was four fifths of the value of the first horse- What was the value of each horse ? • V3. From a cask two thirds full 18 gallons were taken, when it was found to be five ninths full. How many gallons will the cask hold ? 74. A farmer had two flocks of sheep, and sold one flock for $60. Now a sheep of the flock sold was worth 4 of those left, and the whole value of those left was $8 more than the price of 8 sheep of those sold, and the flock left contained 40 sheep. How many sheep did the farmer sell, and what was the value of a sheep of each flock ? Ans. Number sold, 15; value, $4 and $1. 16. A man has seven sons with 2 years between the ages of any two successive ones, and the sum of all their ages is ten times the age of the youngest. What is the age of each ? 76. Divide 75 into two parts such that the greater in- creased by 9 shall be to the less diminished by 4 as 3 : 1. 77. Divide a into two parts such that the greater in- creased by b shall be to the less diminished by c a,s m:n. 78. What two numbers are as 3:4, while if 8 be added to each the sums will be as 5 : 6 ? 79. Divide 127 into two parts, such that the difierence between -the greater and 130 shall be equal to five times the difference between the less and 63. 104 ELEMENTAKY ALGEBBA. SECTION XIY. EQUATIONS OF THE FIRST DEGREE CONTAINING TWO UNKNOWN QUANTITIES. 109. Independent Equations are such as cannot be de- rived from one another, or reduced to the same form. Thus, ar + y= 10, 1+1 = 5, and 4x+3y = 40— y are not independent equations, since any one of the three can be derived from any other one ; or they can all be reduced to the form x -\-tf= 10. But a:-f-y=10 and 4a:^y are independent equations^ 110. To find the value of several unknown quantities, there must be as many independent equations in which the unknown quantities occur as there are unknown quantities. From the equation x -|- y = 10 we cannot determine the value of either a; or 3^ in known terms. If y is transposed, we hare a; = 10 — y; but since y is unknown, we have not determined the value of X. We may suppose y equal to any number whatever, and then x would equal the remainder obtained by subtracting y i'rom 10. It is only required by the equation that the sum of two niimbcTs shall equal 10; but there is an infinite number of pairs of numbers whose sum is equal to 10. But if we have also the equation 4 a: = y, we may put this value of y in the first equation, X -\- y = 10, and obtain a;-^4x=10, orT=2; then 4x=S = y, and we have the value of each of the unknown quantities. ELIMINATION. 111. Elimination is the method of deriving from the given equations a new equation, or equations, containing one (or more) less unknown quantity. The unknown quantity thus excluded is said to be eliminated. EQUATIONS OF THE FIBST DEGREE. 105 There are three methods of elimination : — I. By substitution. II. By comparison. III. By combination. CASE I. 113i Elimination by substitution. 1. Given j^^ + ^2^ — 23| t^gnj ^ and y. OPERATIOIT. 4x4-53^ = 23 (1) 5a;-f 4y= 22 (2) 23— 4 a; (3) 5. + 4(^^7^^)=22 (4) 25a; + 92 — 16a: = 110 (5) 23 — 8 y=—i—= :3 (T) a;= 2 (6) Transposing 4 a; in (1) and dividing by 5, we have (3), which gives an expression for the value of y. Substituting this value of y in (2), we have (4), which contitins but one unknown quantity ; i. e. y has been eliminated. Beducing (4) we obtain (6), or a; = 2. Substituting this value of x in (3), we obtain (7), or y =s 8. .Hence, RULE. Find, an expression for the value of one of the unknown quantities in one of the equations, and substitute this value for the same unknown quantity in the other equation. Note. — After eliminating, the resulting equation is reduced by the rule in Art. 102. The value of the unknown quantity thus found must be substituted in one of the equations containing the two unknown quantities, and this reduced by the rule in Art. 102. Find the values of x and y in the following equations : — 2. Given \- + y=^n. Ans. {-=!«• 106 ELEMENTARY ALGEBRA. f?_|-|=12] (a; =14 3. Given <^ ^ ^ • ^ns. { ^ ^^ L 7 ^^ 5 J 5. Given 6. Given {3x — y-. CASE II. 113. Elimination by comparison. 1. Given {l~J.l'Z 2l} ' ^ ^^^ ^ *°^ ^• OPEKATION. x — 2y = 6 (1) 2a; — y = 27 (2) a; = 6 + 2y (3) a: = ^Hif (4) 6 + 2y = HI±i! (5) 12 + 42/ = 2'7+y (6) y= 5 (t) a; = 6 + 10 = 16 <8) Finding an expression for the value of x from both (1) and (2), we have (3) and (4). Placing these two values of x equal to each other (Art. 13, Ax. 8), we form (5), which contains but one unknown quantity. Keducing (5) we obtain (7), or y = 5. Sub- stituting this value of y in (3), we have (8), or a; = 16. Hence, EQUATIONS OF THE FIRST DEGREE. RULE. 107 Find an expression for the value of the same unknown quantity from each equation, and put these expressions equal to each other. By this method of elimination find the values of x and y in the following equations : — 2. Given Ans (a;= 1. 3. Given i + y = 12 ,1-1= « 5. Given g + y ^ — y — A 3 6 3 "■ 5 Ans. { x = 24. ^^= 0. 4. Given j^^f ^^l" ^^'- I""!" Ans. {.:=: 2. 6. Given (3(:r-y)-9 = 0| t. Given !':-'^ = ;i|- i 2 a; — « = 13 > 8. Given - 4 5 4 ^^ 5 108 ELEMENTAEY ALGEBRA. CASE III. 114. Elimination by combination. 1. Given P^~^^=H,to find a: andy. (2x — 3i/= 3 J OPERATIOS . 3x — 2y= 1 (1) 2x — 3y = 3 (2) Gx — 4y= 14 (3) 6x — 9y= 9 (4) 5y= 5 (5) 2a; — 3 =3 (T) tl= 1 (6) x = 3 (8) If we multiply (1) by 2, and (2) hj 3, we have (3) and (4), In which the coefficients of x are equal; subtracting (4) from (S), we have (5), which contains but one unknown quantity. Redu- cing (5), we have (6), or y = 1 ; substituting this value of y in (2), we obtain (7), which reduced gives (8), or a; = 3. 2. Given ■ ' X 2'_ 4 2 6 12 ■ , to find X and y. DPI ilKATION. 2-4= ^ (1) I+I='2 (2) (6) .-1=12 (3) 9-1= 6 if = ^ W y = 12 (^) a; =18 (5) If we multiply (1) by 2, we have (3), an equation m which y has the same coefficient as in (2) ; since the signs of y are different in (2) and (3), if we add these two equations together, we have (4), which contains but one unknown quantity. Reducing (4), we have (5), or a; = 18. Substituting this value of x in (1), we have (6), which reduced gives (7), or ^=12. Hence, EQUATIONS OF THE FIRST DEGREE. RULE. 109 Multiply or divide the equations so that the coefficients of the quantity to be eliminated shall become equal ; then, if the signs of this quantify are alike in both, subtract one equa- tion from the other ; if unlike, add tJie two equations to- gether. Note. — The least multiplier for each, equation will be that which will make the coefficient of the quantity to be eliminated the least common multiple of the two coefficients of this quantity in the given equations. It is always best to eliminate that quantity whose coefficients can most easily be made equaL By this method of elimination find the values of x and y in the following equations : — 3. Giv«. \i^±i'-'=in- i.2x-\-1y = 34t) 4. Given | 8-+62/ = 6|. (lOa; — 32^ = 4J Ans. |^=^- Ans. 6. Given Ans. \ ^ 21. 12. 6. Given ■ ^ + 1 6 x — y 2 3 = 1 = 1. Given 3 2 110 ELEMENTARY ALGEBRA. 115i Find the values of x and y in the following Examples. Note. — Which of the three methods of elimination should be used depends upon the relations of the coefficients to each other. That one which will eliminate the quantity desired with the least work is the best. 1. Given j2- + 32/ = 25). "5a; — y = 2. Given \ o 7a:4--y=31 Ans = 2. T. ix= 3. 1^=15. Ans. 4 3. Given • 4. Given 2y -l/ = 2 a;— 1 17 ) 2a: — 4^=17 ) ly= 1. Ans (a; = 9. 6. Given '2a:- f-3y 1 x+_2j-\-3' 3 3 2 Ans. (a; = ll. Xu= 5. 6. Given 7 ~ 3 a; . y 8 + 16 "= ^2 t. Given I 3 3-18 11 7 — °T1 EQUATIONS OF THE FIRST DEGREE. Ill 'Sx,5y 8 8. Given - 5 1" 8 —15 L 4' ■I/ = 40. Ans.-!^ = : 5a: + |= 54 9. Given ■{ ^ 1 + 52, = 102 Ans, = 10. 20. 10. Given "^4-' + | = ^^ + 24f ,-8=1+4 11. Given -j 1 + 1= ^[ • ( 4 a; — 3 V = 25 ) >■• Ans (y = = 48. 28. f.= 4. (.3r = — 3. Ans.-^^ = 12. Given L o Ans.j^ = « + ' 13, Given '-'-i^=» !-«=l Ans.j* — (y = = 10. 20, 14, Given < -^ = y_2 4y — 4 2 = 15. Given r'~^"^'~' 5-i=i-« 112 16. Given 17. Given 18. Given 19. Given ELEMENTARY ALGEBRA. 4x — 1 y ■ x — 4 Sx-\-y _ = 2, + 3 PEOBLEMS PEODTJCING EQUATIONS OF THE EIRST DEGREE CON- TAINING TWO UNKNOWN QUANTITIES. 116. Many of the problems given in Section XIII. con- tain two or more unknown quantities ; but in every case these are so related to each other that, if one becomes known, the others become known also ; and therefore the problems can be solved by the use of a single let- ter. But many problems, on account of the complicated conditions, cannot be performed by the use of a single letter. No problem can be solved unless the conditions- given are suflScient to form as many independent equa- tions as there are unknown quantities. 1. A grocer sold to one man 7 apples and 5 pears for 41 cents; to another at the same rate 11 apples and 3 pears for 45 cents. What was tue price of each ? EQUATIONS OF THE FIRST DEGREE. 113 SOLUTION. Let X = the price of an apple, and y^" " "a pear. Then, by the conditions. 7a; + 5y = 41 (1) and 11 a; + 3y= 45 (21 55x4-15^ = 225 (3) 21a;+153r=123 (4) 2l + 5y = 41 a) 34 a; =102 (5) y= 4 (8) x= 3 (6) We multiply (2) by 5 and (1) by 3, and obtain (3) and (4) ; subtracting (4) from (3) we have (5), which reduced gives (6), or z = 3. Substituting this value of x in (1), we have (7), which re- duced ^ves (8), or y = 4. 2. There is a fraction such that if 2 is added to the Jumerator the fraction will be equal to ^ ; but if 3 is added to the denominator the fraction ■will be equal to J-. What is the fraction ? SOLUTION. Let - = the fraction. Then, by the conditions, 4 '+-'-i (1) J, 2 ^ ' (2) 3a; = y-l- 3 (3) 2x + 4=y (4) a; — 4=3 (5) a;='7 (6) t-l-4 = 18=^ (T) X 7 V~" 18 (8) Clearing (1) and (2) of fractions, we obtain (3) and (4) ; sub- t.-acting (4) from (3), we obtain (5), which reduced gives (6), or 1=7. Substituting this value of x in (4), we have (7), ov y= 18. , X 7 hence, - = - H 114 ELEMENTARY ALGEBRA. 3. There are two numbers whose sum is 28, and one fourth of the first is 3 less than one fourth of the second. What are the numbers ? Ans. 8 and 20. 4. The ages of two persons, A and B, are such that 5 years ago B's age was three times A's ; but 15 years hence B's age will be double A's. What is the age of each ? Ans. A's, 25 ; B's, 65. 5. There are two numbers such that one third of the first added to one eighth of the second gives 39 ; and four times the first minus five times the second is zero. What are the numbers ? 6. Find a fraction such that if 6 is added to the nu- Tierator its value will be ^, but if 3 be added to the de- nominator its value will be ^ ? Ans. ^. Y. What are the two numbers, whose difference is to their sum as 1:2, and whose sum is to their product as 4 : 3 ? SOLUTION. Let X = the greater and y = the less. Thena;— y:a;+y=l:2 (1) a: + y:a;y = 4:3 (2) 2x — 2i/ = x-\-y (3) 3x-\-3y = ixy (4) x = 3y (5) 9y + Sy=12f (6) x = 3 (1) l=y (8) Having written (1) and (2) in accordance with the statement In the problem, we form from them (3) and (4) by Art. 106. Re- ducing (3), we obtain (5) ; substituting this value of x in (4), we have (6), which, though an equation of the second degree, can be at once reduced to an equation of the first degree by dividing each term by y ; performing this division and reducing, we obtain (8) or 2^ = 1 ; substituting this value of y in (5) we obtain (7), or a;= 3. EQUATIONS OF THE FIRST DEGREE. 115 8. What are the two numbers whose difiference is to their sum as 3 : 20, and three times the greater minus twice the less is 35 ? 9. There is a number consisting of two figures, which is seven times the sum of its figures ; and if 36 is sub- tracted from it, the order of the figures will be inverted. What is the number ? Ans. 84. 10. There is a number consisting of two figures, the first of which is the greater ; and if it is divided by the sum of its figures, the quotient is 6 ; and if the order of the figures is inverted, and the resulting number divided by the difiference of its figures plus 4, the quotient will be 9. What is the number? Ans. 54. 11. As John and James were talking of their monej'-, John said to James, " Give me 15 cents, and I shall have four tiaies as much as you will have left." James said to John, " Give me 7^ cents, and I shall have as much as you will have left." How many cents did each have ? Ans. John, 45 cents ; James, 30 cents. 12. The height of two trees is such that one third of the height of the shorter added to three times that of the taller is 360 feet ; and if three times the height of the shorter is subtracted from four times that of the taller, and the remainder divided by 10, the quotient is 17. Ee- quired the height of each tree. Ans. 90 and 110 feet. 13. A farmer who had $41 in his purse gave to each man among his laborers $2.50, to each boy $1, and had $15 left. If he had given each man $4 and then each boy $ 3 as long as his money lasted, 3 boys would have received nothing. How many men and how many boys did he hire ? 116 ELEMENT AEY ALGEBEA. 14. A man worked 10 days and his son 6, and they "•eceived $ 31 ; at another time he worked 9 days and ais son 1, and they received $29.50. What were thfi wages of each ? 15. A said to B, " Lend me one fourth of your money, and 1 can pay my debts." B replied, "Lend me $100 less than one half of yours, and I can pay mine." Now A owed $1200 and B $1900. How much money did each have in his possession ? Ans. A, $800 ; B, $1600. 16. If a is added to the difference of two quantities, the sum is 6 ; and if the greater is divided fey the less, the quotient will be c. What are the qiiantities ? . be — ac , b — a Ans. ,— and -■ c — 1 c — 1 It. A man owns two pieces of land. Three fourths of the area of the first piece minus two fifths of the area Df the second is 12 acres ; and five eighths of the area of the first is equal to four ninths of the area of the second. How many acres are there in each ? Ans. 1st, 64 acres ; 2d, 90 acres. 18. A and B begin business with different sums of money ; A gains the first year $350, and B loses $500, and then A's stock is to B's as 9 : 10 If A had lost $500 and B gained $350, A's stock would have been to B's as 1:3. With what sum did each begin ? Ans. A, $1450; B, $2500. 19. If a certain rectangular field were 4 feet longer and 6 feet broader, it would contain 168 square feet more ; but if it were 6 feet longer and 4 feet broader, it would contain 160 square feet more. Eequired its length and breadth. EQUATIONS OF THE I'TKST DEGREE. 117 20. A market-man bought eggs, some at 3 for t cents id some at 2 for 5 cents, and paid for the whole $2.62 ; 3 afterward sold them at 36 cents a dozen, clearing 0.62. How many of each kind did he buy ? 21. A and B can perform a piece of work together in i days. They work together 7 days, and then A fin- hes the woisk alone in 15 days. How long would it ike each to do the work ? Ans. A 36 and B 18 days. 22. " I was ten times as old as you 12 years ago," lid a father to his son ; " but 8 years hence I shall be nly two and one half times as old as you." What as the age of each ? 28. If 8 is added to the numerator of a certain frac- on, its value will be § ; and if 4 is subtracted from the enominator, its value will be ■^. What is the fraction ? 24. A farmer sold to one man 7 bushels of oats and 5 ushels of corn for $12.76, and to another, at the same ite, 5 bushels of oats and 7 bushels of corn for $13.40. 7hat was the price of each ? 25. Find two quantities such that one third of the first linus one half the second shall equal one sixth of a ; nd one fourth of the first plus one fifth of the second hall equal one half of a. . 34 a j ^^ a Ans. -^ ana ---• 26. A person had a certain quantity of wine in two asks. In order to obtain an equal quantity in each, he cured from the first into the second as much as the Bcond already contained; then he poured^ from the sec- nd into the first as much as the first then contained ; nd, lastly, he poured from the first into the second as luch as the second still contained ; and then he had 16 allons in each cask. How many gallons did each origi- ally contain ? Ans. 1st, 22 ; 2d, 10 gallons. 118 ELEMENTARY ALGEBRA. SECTION XV. EQUATIONS OF THE riEST DEGREE CONTAINESTG MORE THAN TWO UNKNOWN QUANTITIES. 117. The methods of elimination given for solving equa- tions containing two unknown quantities apply equally well to those containing more than two unknown quantities. 1. Given X -\- y — s = 4 ■ 2a;-|- 3y 4- 4z: %x — 2y-\- bz: = 5) to find X, y, and z. OPERATION. xJfy — z = i (1) 2a; + 3y + 4z = 17 (2) 3a: — 23/+ 5«= 5 (3) 2a; + 2^ — 2z= 8 (4) Zx + Sy— 3g=12 (5) y + 6«= 9 (6) 5jr- 8j= 7 (7' 5j + 30z = 45 (8; 1 + 3 — 1=4 (13) 3^ + 6 =9 (II) 382 = 38(9) x = 2 (14) a =3 (12) 2= 1(10) Multiplying equation (1) by 2 gives equation (4), -which we sub- tract from (2), and obtain (6) ; multiplying (1) by 3 gives (5), and subtracting (5) from (3) gives (7). We have now obtained two equations, (6) and (7), containing but two unknown quantities. Mul- tiplying (6) by 5, we obtain (8), and subtracting (7) from (8), we obtain (9), which reduced gives z = 1. Substituting this value of z in (6),_ and reducing, we obtain y = 3. Substituting these values of y and z in (1), and reducing, we obtain a; ^ 2. 2. Given X -\-y z=2Q y + z = 29 z -\- w ^ bQ w -j- M = 81 u -|- a; = 46 ■ , to find u, w, X, y, and z. EQUATIONS OF THE FIRST DEGREE. 119 OPERATION. x + y = 2S a) ¥ + 2 = 29 (2) z + w = 56 (3) m. + « = 81 (4) u-\-x^4S (5! » + ^ = 26 z— a:= 3 w + a:=53 u—x = 2S 2—1= 3 (6) w + a:=63 (7) k — 1=28 (8) 2x = 18 (9) » = 17 (11) 2 = 12(12) to = M (13) » = 37 (14) 1=9(10) Here we subtract (1) from (2), and obtain (6) ; then (6) from (3), and obtain (7) ; then (7) from (4), and obtain (8) ; then (8) from (5), and obtain (9), which reduced gives (10), or a; = 9. Sub- stituting this value of x in (1), (6), (7), and (8), and reducing, we obtain (11), (12), (13), and (14), or y = 17, z = 12, m; = 44, and «=37. Hence, for solving equations containing any number of unknown quantities, RULE. From the given equations deduce equations one less in number, containing one less unknown quantity; and con- tinue thus to eliminate one unknown quantity after an- other, until one equation is obtained containing but one unknown quantity. Reduce this last equation so as to find (ke value of this unknown quantity ; then substitute this value in an equation containing this and but one other unknown quantity, and reducing the resulting equation, find the value of this second unknown quantity ; substitute again these values in an equation containing no more than these two and one other unknown quantity, and reduce as before ; and so con- tinue, till the value of each unknown quantity is found. Note. — The process can often be very much abridged by the exercise of judgment in selecting the quantity to be eliminated, the equations from which the other equations are to be deduced, the method of elimination which shall be used, and the simplest equa- tions in which to substitute the values of the quantities which have been found. 120 ELEMENTAEY ALGEBEA. Find the values of the unknown quantities in the fol- lowing equations : — 3. Given X -\- t/ -\- z -\- w ::= 16 y-j-z -^w-\- u ^ 18 X -\- Z -\-W-\-U=::^ 11 x-\-y-\-w-\-u= 14 X-{-1/-\-Z-{-U=:l6 Note. — If these equations are added together and the sum di- vided by 4, we shall have x-\-y-{-z-\-w-\-u = 20; and if from this the given equations are successively subtracted, the values of the unknown quantities become known. r n y = B. Ans.< z=: 6. M= 4. w= 5. 4l. Given x-\-3y-\-2z = 2Q 3^ + iy + iz=lQ Ans rx='. ■X=:2. :4. 6. 5. Given (-2a; + 3y + 42: = 6T 4 25 J 9 ' 9 ' 2 2y + ■x= I. Ans. -j y ^ 7. 2=11. X — • y — z= 1 6. Given -l a; + 2y — 10z:=l .2a; — 4y+ 33=1 Ans, / a; = i \z = . ■a; = 5. 3. 1. T. Given -a , 1 + \^ = 22 jX + 1. + \' = 24 \^ + h + \' = 10 Ans. .. = : ]y = (« = ; ■ a: = 20. 12. :32. EQUATIONS OF THE FIRST DEGREE. 121 8. Given 1,1 5 x~^ y 6 - + - = - y^ z 12 1.1 3 a; "•" z 4 Note. — The best method for this example is that used in Ex- ample 3, without clearing of fractions. ( a;+ y-\- z= Qs rx = •|2a; + 32^4-43 = 20[. . Ans. -|y = (3x4- 1« + 5e = 32; (2: = ■x= 1. 2. 3. {^ + iy 10. Given -{y -\- + 1! »a; = 29) 11. Given rx + y = a^ fx - y +« = *[- • Ana. j y '.X -\- z = c) Iz = ^(« + = H5 + b + c) b — c) c — a) 12. Given fabx-\-aby = a-\-b\ ^aex-{-aez = a-\-G>-- \bcy-\-bez = b -\- c) Ans. " r 1 x=- a 1 _i . 13. Given (3/ + i^ + *^=18) \x+2n=iy+28[. ( 9ar = 2y ) 122 ELEMENTAEY ALGEBRA. PEOBLEMS PEODtrCING EQUATIONS OF THE FIRST DEGREE CON- TAINING MORE THAN TWO UNKNOWN QUAIJTITIES. 118. 1. A merchant has three kinds of flour. He can sell 1 bbl. of the first, 2 of the second, and 3 of the third for $85; 2 of the first, 1 of the secopd, and J bbl. of the third for $45.50 ; and 1 of each kind for $41. What is the price per bbl. of each? Ans. 1st, $ 12 ; 2d, $ 14 ; 3d, $ 15. 2. Three boys, A, B, and C, divided a sum of money among themselves in such a manner that A and B re- ceived 18 cents, B and C 14 cents, and A and C 16. How much did each receive? Ans. A, 10 ; B, 8 ; 0, 6 cents. 3. As three persons. A, B, and C, were talking of their ages, it was found that the sum of one half of A's age, one third of B's, and one fourth of C's was 33 ; that the sum of A's and B's was 13 more than C's age ; while the sum of B's and C's was 3 less than twice A's age. What was the age of each ? Ans. A's, 32 ; B's, 21 ; C's, 40. 4. As three drovers were talking of their sheep, says A to B, "If you will give me 10 of yours, and one fourth of his, I shall have 6 more than C now has." Says B to 0, "If you will give me 25 of yours, and A one fifth of his, I shall have 8 more than both of you will have left." Says C to A and B, "If one of you will give me 10, and the other 9, I shall have just as many as both of you will have left." How many did each have ? 5. Divide 32 into four such parts that if the first part is increased by 3, the second diminished by 3, the third multiplied by 3, and the fourth divided by 3, the sum, difference, product, and quotient shall all be equal. 4.ns. 3, 9, 2, and 18. EQUATIONS OF THE FIRST DEGREE. 123 6. If A and B can perform & piece of work together in 8^ days, B and in 9^^ days, and A and C in 8J days, in how many days can each do it alone? Ans. A in 15, B in 18, and C in 21 days. *l. Find three numbers such that one half of the first, one third of the second, and one fourth of the third shall together be 56 ; one third of the first, one fourth of thfe second, and one fifth of the third, 43 ; one fourth of the first, one fifth of the second, and one sixth of the third, 35. 8. The sum of the three figures of a certain number is 12 ; the sum of the last two figures is double the first ; and if 29*1 is added to the number, the order of its fig- ures will be inverted. What is the number? Ane. 417. 9. A man sold his horse, carriage, and harness for $450. For the horse he received $25 less than five times what lie received for the harness ; and one third of what he received for the horse was equal to what he received for the harness plus one seventh of what he received for the carriage. What did he receive for each ? Ans. Horse, $225; carriage, $175; harness, $50. 10. A man owned three horses, and a saddle which was worth $45. If the saddle ia put on the first horse, the value of both will be $ 30 less than the value of the second ; if the saddle is put on the second horse, the value of both will be $ 55 less than the value of the third ; and if the saddle is put on the third horse, the value of both will be equal to twice the value of the second minus $10 more than one fifth of the value of the first. What is the value of each horse ? Ans. 1st, $100; 2d, $175; 3d, $275. 11. The sum of the numerators of two fractions is 7, and the sum of their denominators 16 ; moreover the sura of the numerator and denominator of the first is equal 124 ELEMENTARY ALGEBRA. to the denominator of the second ; and the denominator of the second, minus twice the numerator of the first, is equal to the numerator of the second. What are the fractions ? Ans. f and |. 12. A man bought a horse, a wagon, and a harness, for $180. The horse and harness coat three times as much as the wagon, and the wagon and harness one half as much as the horse. What was the cost of each ? 13. A gentleman gives $600 to be divided among three classes in such a way that each one of the best class is to receive $10, and the remainder to be divided equally among those of the other two classes. If the first class proves to be the best, each one of the other two classes will receive $5 ; if the second class proves to be the best, each one of the other two classes will receive $4f- ; but if the third class proves to be the best, each one of the other two classes will receive $2. What is the number in each class? 14. A cistern has 3 pipes opening into it. If the first should be closed, the cistern would be filled in 20 min- utes ; if the second, in 25 minutes ; and if the third, in 30 minutes. How long would it take each pipe alone to fill the cistern, and how long would it take the three together ? Ans. 1st, 85f minutes ; 2d, iQ^^ minutes ; 3d, 35-jAf minutes. The three together, 16tjV minutes, 15. Three men. A, B, and C, had together $24. Now if A gives to B and as much as they already have. and then B gives to A and C as much as they have after the first distribution, and again C gives to A and B as much as they have after the second distribution, they will all have the same sum. How much did each have at first? Ans. A, $13 ; B, $7, and C, $4. EQUATIONS OF THK FIRST DEGREE. 125 SECTION XYI. POWEKS AND EOOTS. 119f A PowEK of any quantity is the product obtained Y taking that quantity any number of times as a factor ; id the exponent shows how many times the quantity is iken (Art. 24). Thus, a^^a^ is the first power of a ; a a = a^ " second power, or square, of a ; a a a^a* " third power, or cube, of a ; a a a az= a^ " fourth power of a ; ad BO on. 120. In order to explain the use of negative indices, '6 form, by the rules of division, the following series: — ^, «', «s a, 1, 1 a' 1 1 1 a*' 1 i\ «», a\ a\ a« a-\ «-^ a-'. a-*. a- We form the first series as follows: a* divided by a gives a*; ' by a, gives a' ; a' by a, gives a' ; a' by a, gives a; a by a, gives 11, -111, -111, -1 J ; 1 by % gives -; - by a, gives -j; -, by a, gives -j, and so on. le second series is formed in tbe same way from a' to a; but if 'e follow the same rule of division from a toward the right as from ' to a, viz. subtracting the, index of the divisor from that of the div- Imd, a divided by a, gives a" ; o° by a, gives a-' ; read a, with the egafive index one; a-^ by a, gives o""^; o~^ by a, gives o"*; and } on. From this we learn, 1st. That the power of every quantity is 1 ; 2d. Thai cr^, a~^, cr*, &c., are only different ways of ...111. 126 ELEMENT AEY ALGEBEA. Any two quantities at eqxial distances on opposite sides of a°, or 1, are reciprocals of each other. 121. The rules given for the multiplication and divis- ion of powers of the same quantity (Arts. 50 and 54) apply equally well whether the exponents are positive or negative. For a« X «-" = «' X ^3 = j! = «' «« H- 3. Transfer the denominator of L. to the numerator. 4. Transfer the numerator of °f ^^ to the denominator. oca 128 ELEMENTAKT ALGEBRA. 4 a-' V c-° 5. Free from negative exponents „ ,_7— -— u- Ans. ^= — i 1 ac xy 6. Free from negative exponents ^ ,^ - a (x -\- v)""' 7. Free from negative exponents — ^ Ans x' — f (x — y) (x -f- ^)'~' 8. Free from negative exponents , . a (^\y\ • */. or by a fractional exponent. Thus, mJ~x, or a;^ indicates the square root of x. ^'x, or a;i " " cube ^Ic, or a;« " " mth It €t ft <> << II 134 1 A root and a power may be indicated at the same time. Thus, \/^, or a:*, indicates the cube root of the fourth power of x, or the fourth power of the cube root of X ; for a power of a root of a quantity is equal to the same root of the same power of the quantity. a^^S" or 8^ is the square of the cube root of 8, or the cube root of the square of 8, i. e. 4. 135i A perfect power is a quantity whose root can be found. A perfect square is one whose square root can be found ; a perfect cube is one whose cube root can be found ; and so on. 136> Since Evolution is the reverse of Involution, the rules for Evolution are derived at once from those of Involution. And therefore, as according to Art. 125 an odd power of any quantity has the same sign as the quantity itself, and an even power is always positive, we have for the signs in evolution the following RULE. An odd root of a quantity has the same sign as the quan- tity itself. An even root of a positive quantity is either positive or negative. An even root of a negative quantity is impossible, or im- aginary. EVOLUTION. 139 SQTTAEE BOOT OF NtJMBEES. 137t The Square Root of a number is a number which, taken twice as a factor, will produce the given number. 138 1 The square of a number has twice as many figures as the root, or one less than twice as many. Thus, Eoots, 1, 10, 100, 1000. Squares, 1, 100, 10000, 1000000. The square of any niimber less than 10 must be less than 100 ; but any number less than 10 is expressed by one figure, and any number less than 100 by less than three %ures ; i. e. the square of a number consisting of one figure is a number of either one or two figures. The square of any number between 10 and 100 must be between 100 and 10000; i. e. must contain more than two figures and less than five. And the square of any number between 100 and 1000 must contain more than four figures and less than seven. Hence, to ascertain the number of figures in the square root of a given number. Beginning at units, point off the number into periods of two figures each; there mil be as many figures in the root as there are periods, and for the incomplete period at the left, if any, one more. 139, To extract the square root of a number. 1. Find the square root of 5329. From the preceding explanation, it is evident that the square root of 5329 is a number of two figures, and that the tens figure of the root is the square root of the greatest perfect square in 53 ; i. e. v'49, or 7. Now, if we represent the tens of the root by a and the units by 6, a + 6 will represent the root; and the given number will be (a + by = a? + 2 ab -{- ¥. Now o' = 70« = 4900 ; therefore, 2ai + V= 5329 — 4900 = 429. But 206 + 6"= (2a + 6)6; 140 ELEMENTARY ALGEBEA. If therefore 429 is divided by 2 a + S, it will give b the units of the root. But b is unknown, and is small compared with 2a; we can therefore use 2 a = 140 as a trial divisor. 429 ~ 140, or 42 -^ 14 == 3, a number that cannot be too small but may be too great, because we have divided by 2 a instead of 2 a -|- J. Then 6=3, and 2 a -\-b = liO -\- 3 = 143, the true divisor ; and (2 (I + J) 6 = 143 X 3 = 429 ; and therefore 3 is the unit figure of the root, and 73 is the required root. The work will appear as follows : — OPEEATION. 5329 (13 a = 10 49 6=3 2a + 6 = 14 3)429 (2a4-6)&== 429 Hence, to extract the square root of a number, RULE. Separate the given number into periods of two figures each, by placing a dot over units, hundreds, &g. Find the greatest square in the left-hand period, and place its root at the right. Subtract the square of this root figure from the left-hand period, and to the remainder annex the next period for a dividend. Double the root already found for a trial divisor, and, omitting the right-hand figure of the dividend, divide, and place the quotient as the next figure of the root, and also at the right of the trial divisor for the true divisor. Multiply the true divisor by this new root figure, subtract the product from the dividend, and to the remainder annex the next period, for a new dividend. Double the part of the root already found for a trial di- visor, and proceed as before, until all the periods have been employed. EVOriUTlON. 141 Note 1. — Wh6n a root fjgure is 0, annex also to the trial di- visor, and bring down the next period to complete the new dividend. Note 2. — If there is a remainder, after using all the periods in the given example, the operation may be continued at pleasure by annexing successive periods of ciphers as decimals. Note 3. — In extracting the root of any number, integral or deci- .mal, place the first point over unit's place ; and in extracting the square root, over every second figure from this. If the last period in the decimal periods is not full, annex 0. 2. Find the square root of 46225. operation. 41)62 41 We suppose at first that a rep- resents the hundreds of the root, 46225 (2 15 and 6 the tens ; proceeding as in 4 Ex. 1, we have 21 in the root. Then letting a represent the hundreds and tens together, i. e. 21 tens, and 6 the units, we have 4 2 5) 2 1 2 5 2 a, the 2d trial divisor, = 42 212 5 tens ; and therefore h =^ b\ and 2 a -|- 6 ^ 425 ; and 215 is the required root. 3. Find the square root of 5013.4. operation. 50 13.4 0(10.80 5+ 49 14 0.8)113.4 112.6 4 141.605) .760000 .708025 4. Find the square root of 288369. Ans. 537. 5. Find the square root of 42849. Ans. 207. 6. Find the square root of 173.261. Ans. 18.16+. 7. Find the square root of .9. Ans. .948+. 142 ELEMENTAKY ALGEBRA. 8. Find the square root of 2. Ans. 1.4142-}-. 9. Find the square root of 484. 10. Find the square root of 48.4. 11. Find the square root of .064. 12. Find the square root of .00016. Note. — As a fraction is involved by involving both numerator and denominator (Art. 127), the square root of a fraction ii the square root of the numerator divided by the square root of the denominator. 13. What is the square root of f ? AnS. |. 14. What is the square root of ^| ? 16. What is the square root of -^j^ ? -^^j = ^. Ans. |. Note. — If both terms of the fraction are not perfect squares, and cannot be made so, reduce the fraction to a decimal, and then find the square root of the decimal. A mixed number must be re- duced to an improper fraction, or the fractional part to a decimal, before its root can be found. 16. What is the square root of f ? Ans. .53-|-. 17. What is the square root of ^-fy ? 18. What is the square root of -^-j- 1 19. What is the square root of 7f ? CUBE EOOT OF NUMBERS. 140t The Cube Eoot of a number is a number which, taken three times as a factor, will produce the given number. 141 • The cube of a number consists of three times as many figures as the root, or of one or two less than three times as many. Roots, 1, 10, -100, 1000. Cubes, 1, 1000, 1000000, 1000000000. The cube of any number less than 10 must be less than 1000; but any number less than 10- is expressed by one figure, and any EVOLUTION. 143 number less than 1000 by less than four figures; i. e. the cube of a number consisting of one figure is a number of less than four figures. The cube of any number between 10 and 100 must be be- tween 1000 and 1000000; i. e. must contain more than three figures and less than seven. And in the same way we see that the cube of any number between 100 and 1000 must contain more than six figures and less than ten. Hence, to ascertain the number of figures in the cube root of a given number, Beginning at units, point off the number into periods of three figures each ; there uiill be as many figures in the root as there are periods, and for the incomplete period at the left, if any, one more. 142. To extract the cube root of a number. 1. Find the cube root of 428T5. From the preceding explanation, it is evident that the cube root of 42875 is a number of two figures, and that the tens figure of the root is the cube root of the greatest perfect cube in 42 ; i. e. ^27, or '3. Now, if we represent the tens of the root by a and the units by 6, a-\-b will represent the root, and the given number will be (a -(- 6)» = a" + 3 a= J + 3 a 5= + 6'. Now a' = 30' = 27000; therefore, 3n'b -\- 3aV-{- V = 42875 — 27000 = 15875. But 3 a'b -\- 3 aV' -\- b" = (3 a' -\- 3 ab -\- V) b. If therefore 15875 is divided by 3 a= + 3 a 6 -f- 6^ it will give 6, the units of the root. But b, and therefore 3 ab -^V,a. part of the divisor, is unknown, and we must use 3 a' = 2700 as a trial divisor. 15875 -f- 2700, or 158 -^ 27 = 5, a number that cannot be too small but may be too great, because we have divided by 3 a? instead of the true divisor, 3 a' -^ 3 ab -{- }^. Then 6 = 5, and 3a'-)- 3064-6"= 2700 -|- 450 -[- 25 = 3175, the true divisor; and (3 a' + 3 a 5 -f 6") 6 = 3175 X 5 = 15875, and therefore 5. is the unit's %ure of the root, and 35 is the required root. The work will appear as follows : — 144 ELEMENTARY ALGEBRA. OPEEATION. 4 2 8 '7 5 (35 Eoot. 21 Trial divisor, 3a'^ = 2 7 00 3ab= 450 b'== 2 5 True divisor, So" -{- 3ab + l^ = 31 1 5 15 8 7 5 Dividend. 15875 Hence, to extract the cube root of a number, EULE. Separate the number into periods of three figures each, by placing a dot over units, thousands, &g. Find the greatest cube in the left-hand period, and place its root at the right. Subtract this cube from the leflrhand period, and to the re- mainder annex the next period for a dividend. Square the root figure, annex two ciphers, and multiply this result by three for a trial divisor ; divide the dividend by the trial divisor, and place the quotient as the neost figure of the root. Multiply this root figure by the part of the root previously obtained, annex one cipher and multiply this result by three; add the last product and the square of the last root figure to the trial divisor, and the sum will be the true divisor. Multiply the true divisor by the last root figure, subtract the product from the dividend, and to the remainder annex the next period for a dividend. Find a new trial divisor, and proceed as before, until all the periods have been employed. Note 1. — The notes under the rule in square root (Art. 139) apply also to the extraction of the cube root, except that 00 must be annexed to the trial divisor when the root figure is 0, and after placing the first point over units the point must be placed over every third figure from this. Note 2. — As the trial divisor may be much less than the true EVOLUTION. 145 divisor, the quotient is frequently too great, and a less number must be placed in the root. 2. Find the cube root of 1819144T. 1st Trial Divisor, OPEUATION. 3 a' =12 Sab = 360 b^= 3 6 18191447 (2 63 8 10191 1st Dividend. 1st True Divisor, 3a' + 3o6 + 6' = 1 5 9 6j 95 76 2d Trial Divisor, 3 a' =202800 Sab = 2340 ¥= 9 6 1544 7 2d Div. 2d True Divisor, 3a' -{-3ab-\-t^ = 205 14 9j 6 1 544 7 We suppose at first that a represents the hundreds of the root and b the tens: proceeding as in Ex. 1, we have 26 in the root. Then letting a represent the hundreds and tens together, i. e. 26 tens, and b the units, we have 3 a\ the 2d trial divisor, = 202800 ; and therefore 6 = 3; and Sa' -{- Sab -\-I^, the 2d true divisor, = 205149; and 263 is the required root. Note. — Though the 1st trial divisor is contained more than 8 times in the dividend, yet the root figure is only 6. 3. Find the cube root of 68n6.47. OPBEATION. 6 8 '7 16.4 '7 6 (4 0.9 5+ 64 48 0.0 10 8.0 .8 1 47 16.470 2 9 8.5410 00 4908.81 4=4.11.929 5 1 8.4 3 6.1350 .0 02 5 5024.56751 251.22837 5 47.312625 146 ELEMENTARY ALGEBRA. 4. Find the cube root of 2924207. Ans. 143. 5. Find the cube root of 8120601. Ans. 201. 6. Find the cube root of 36926037. 1. Find the cube root of 67917.312. 8. Find the cube root of 46417.8. 9. Find the cube root of .8. Ans. .928-|-. 10. Find the cube root of .17164. 11. Find the cube root of .0064. 12. Find the cube root of 25.00017. 13. Find the cube root of 2.7. Note. — As a fraction is involved by involving both numerator and denominator (Art. 127), the cube root of a fraction is the cube root of the numerator divided hy the cube root of the denominator. 14. What is the cube root /y? Ans. |. 15. What is the cube rOot of //g ? 16. What is the cube root of f||? |M = HI- Ans. f. Note. — If both terms of the fraction are not perfect cubes, and cannot be made so, reduce the fraction to a decimal, and then find the cube root of the decimal. A mixed number must be reduced to an improper fraction, or the fractional part to a decimal, be- fore its root can be found. 17. What is the cube root of t\ ? Ans. .899+. 18. What is the cube root of /x? 19. What is the cube root of ^ ? 20. What is the cube root of 117f ? EVOLUTIOJI. 147 EVOLUTION OP MONOMIALS. 143i As Evolution is the reverse of Involution, and since to involve a monomial (Art. 126) we multiply the exponent of each letter by the index of the required power, and prefix ihe required power pf the numei'ipal coefiScient, Hence, to find the root of a monomial, RULE. Divide the exponent of each letter by ffie index of the re- quired root, and prefix the required root of the numerical cvefficient. Note 1. — The rule for the signs is given in Art. 136. As an even root of a positive quantity may be either positive or negative, we prefix to such a root the sign ± ; read, plus or minus. Note 2. — It follows from this rule that the root of the product of several factors is equal to the vroduct of the roots. Thus, 1. Find the cube root of 8a^y'. Ans. 2xy'. 2. Find the square root of 4 a:^. Ans. ±2x. 3. Find the third root of — 125,o«a;. Ans. ^—5 a" a;*. 4. Find the fourth root of 81 a"^ 5. Ans. ± 3 ffl-i b^. 5. Find the fifth root of 32.ai'',6'. Ans. 2 a^ 6*. 6. Find the cube root of — 729a;V- Ans. — 9xi^. 1. Find the fourth root of 256 a;^/. 8. Find ±he cube root of — 512 a"' b". 9. Find the fifth rooi of 243a^/°, 148 ELEMENTARY ALGEBEA. Note. — As a fraction is involved by involving both numerator and denominator (Art. 127), a fraction must be evolved by evolving both numerator and denominator. - 4 a" . ,2a 10. Find the square root of g-j- Ans. ± —,- Perform the operations indicated in the following ex- pressions : — 11. ^ — 729 a= 6« c». 12. (ida'x^y^)^. 14. /^a^a;™". 15. (256a*a;i°j^")^. 16. >i/ 81 a" ¥. 11. S/o^S^^c™". SQUAEE ROOT OF POLYNOMIALS. 144i In order to discover a method for extracting the square root of a polynomial, we will consider the rela- tion of a + J to its square, a^ -]- 2 a i -{- U'. The first term of the square contains the square of the first term of the root ; therefore the square root of the first terra of the square will be the first term of the root. The second term of the square contains twice the product of the two terms of the root ; therefore, if the second term of the square, 2 a b, is divided by twice the first term of the root, 2 a, we shall have the second term of the root J. Now, 2ab-{-l^={2 a-\-h)i; therefore, if to the trial divisor 2 a we add b, when it has been found, and then EVOLUTION. 149 multiply the corrected divisor by h, the product will be equal to the remaining terms of the power after a^ has been subtracted. The process will appear as follows : — OPEKATION. Having written a, the square a^ -\- 2 ah -\- h^ {a -\- b root of a\ in the root, we sub- o^ tract its square (if) from the 2o4-J)2a64-62 given polynomial, and have 2 ah A-b"^ 2a6-l-6^ left. Dividing the first term of this remainder, 2 a J, hy 2 a, which is double the term of the root already found, we obtain 6, the second term of the root, which we add both to the root and to the divisor. If the product of this corrected divisor and the last term of the root is subtracted from 2 a 6 -j- 6", nothing remains. 145i Since a polynomial can always be written and involved like a binomial, as shown in Art. 131, we can apply the process explained in the preceding Article to finding the root, when this root consists of any number of terms. 1. Find the square rootof a^-|-2a6-l-^^ — 2ac — 2hc-\-c\ OPERATION. 2a + i)2Bi + 62 2 a S 4- *^ 2a + 2J — c)— 2ac — 25c + c'' — 2ac — 2bc-^c^ Proceeding as before, we find the first two terms of the root a -j- 5. Considering a -\- b as a single quantity, we divide the remainder — 2ac — 2bc -\- c' by twice this root, and obtain — c, which we write both in the root and in the divisor. If this corrected divisor is multiplied by — c, and the product subtracted from the dividend, nothing remains. 150 ELEMENTARY ALGEBRA. Hence, to extract the square root of a polynomial, RULE. Arrange the terms according to the powers of some letter. Find the square root of the first term, and write it as the first term of the root, and subtract its square from the given polynomial. Divide the remainder by double the root already found, arid annex the result both to the root and to the divisor. Multiply the corrected divisor by this last term of the root, and subtract the product from the last remainder. Proceed as before with the remainder, if there is any. 2. Find the square root of 4 x^ — 4 xi/^ -\- y*. Ans. 2x — y^. 3. Find the square root ofo^-|-2aJ-|-J^-)-4ac + 4 J c + 4 c^, Ans. a 4- J -|- 2 c. 4. Find the Bquave root of 9 a;* — 12 s^ -\- i x^ -\- 6 a x^ — 4:ax-\-a\ Ans. 3 a;^ — 2a; + a. 5. Find the square root oi ia' -\- 8 ah — 4a-|-4i' — ib-\-l Ans. 2a + 26 — 1. 6. Find the square root of 25 a;* — 10 a:' 4- 6 a;^ — a: + i- Ans. 5 ay' — ^ -\- h 1. Find the square root of a;« 4- 2 a;^ — a;< — 2x'-{- x'. 8. Find the square root of 4=n^ — 4aJ 4- V — 4ac — 4.ad-\-2hcJ^2hd-\-v^-{-2cd-\-d\ Ans. 2 a — h — c — d. 9. Find the square root of a;« — 4a;5 4- 6 a;* — 6 a;' 4- 5a;'' — 2a; 4- 1. 10. Find the square root of 4 a* -|- 8aH — Sa^i' — 12 a J= 4- 9 h\ EVOLUTION. 151 Note 1. — According to the principles of A(i;. 136, the signs of the answers given above may all be changed, and still be correct. Note 2. — No binomial can be a perfect square. For the square of a monomial is a monomial, and the square of the polynomial with the least number of terms, that is, of a binomial, is a trinomial. Note 3. — A trinomial is a perfect square when two of its terms are perfect squares and the. remaining term is equal to twice the product of their square roots. For, (a + by = a' + 2ab-{-¥ (a — by = a^—2ab + l^ Therefore the square root of a' ± 2 a 6 -f- 6^ is a ± 6. Hence, to obtain the square root of a trinomial which is a perfect square. Omitting the term that is equal to twice the product of the square roots of the other two, connect the square roots of the other two by the sign of the term omitted. 11. Find the square root of - — ^ + r* A- 1-1- 12. Find the square root of a;" -|- 2 ar -j- 1. Ans. X -{- 1. 13. Find the square root of 4a;'' — 8xy -{- iy'^. a' 14. Find the square root of r 2 a 6 + 9 6^ 15. Find the square root of 16/ + 40^2;^ + 252*. Note. — By the rule for extracting the square root, any root whose index is any power of 2 can be obtained by successive extractions of the square root. Thus, the fourth root is the square root of the square root ; the eighth root is the square root of the square root of the square root; and so on. 16. Find the fourth root of a» — 12a»S + 54 a* 6^ — 108 a^ 6» + 81 J*. Ans. a' — 3 b. 152 ELEMENTARY ALGEBRA. 11. Find the fourth root of -« + ^ + j^ + ^ + ^. Ans. - -I X ' y 18. Find the fourth root of a^ — da;' -f 10a;» — IGa;* + 19x*— 16a:*+10a^ — 4a:+l. Kas. 7? — x -\- I. 146t To find any root of a polynomial. Since, according to the Binomial Theorem, -when the terms of a power are arranged according to the power of some letter begin- ning with its highest power, the first term contains the first term of the root raised to the given power, therefore, if we take the re- quired root of the first terra, we shall have the first term of the root. And since the second term of the power contains the second terra of the root multiplied by the next inferior power of the first term of the root with a coefiicient equal to the index of the root, therefore if we divide the second term of the power by the first term of the root raised to the next inferior power with a coefficient equal to the index of the root, we shall have the second term of the root. In accordance with these principles, to find any root of a polynomial we have the following ETJLE. Arrange the terms according to the powers of some letter. Find the required root of the first term,, and write it as the first term of the root. Divide the second term of the polynomial by the first term of the root raised to the next inferior power and multiplied by the index of the root. Involve the whole of the root thus found to the given power, and subtract it from the polynomial. If there is any remainder, divide its first term by the di- visor first found, and the quotient mil be the third term of the root. Proceed in this manner till the power obtained by involv- ing the root is equal to the given polynomial. EVOLUTION. 153 Note 1. — This rule verifies itself. For the root, whenever a new term is added to it, is involved to the given power, and whenever the root thus involved is equal to the given polynomial, it is evident that the required «)0t is found. Note 2. — As powers and roots are correlative words, we have used the phrase given power, meaning the power whose index is equal to the index of the required root, and the phrase next inferior power meaning that power whose index is one less than the index of the required root. 1. Find the cube root of a' — 3a^-\-5a^ — 3 a — 1. OPEEATION. Constant divisor, 3 a*) a^ — 3 a^ -\- 5 a,^ — 3 a — 1 (a^ — a — 1 a° — 3 a^ + 3 a* — g' — 3 a*, 1st term of remainder. a6_3a5-|-5a= — 3a— 1 The first term of the root is a", the cube root of a", a' raised to the next inferior power, i. e. to the second power, with the co- efficient 3, the index of the root, gives 3 a', which is the constant divisor. — 3 a^, the second term of the polynomial, divided by 3 a*, gives — a, the second terra of the root. (a° — a)' ^ a° — 3 a' -If-Sa* — a'; and subtracting this from the polynomial, we have — 3a* as the first term of the remainder. — 3 0* divided by 3 a* gives — 1, the third term of the root, (a^ — a — l)'=the given poly- nomial, and therefore the correct root has been found. 2. Find the fourth root of 16 x* — 32 a:' f + 24 a:V — Sxi/'+f. OPERATION. 4X (2a;)' = 32a?) 16 a;*— S2 3? y' -{- 24 sf y* — 8 xf -}- f (2 x — f 16 3*—S2ii?f-}-2ix'f—8xf-\-f 3. Find the cube root oi a' + 3 aH + 3 ab'' -\- b^ —3 a^ c — 6abc — 3b^c + 3ac^-{-3bc^ — c\ 4. Find the fourth root of 16 a* c* — 32 a'c'x+ 24 a^ c^ x^ — 8aca;^ -\- x*. 7* 154 ELEMENTARY ALGEBRA. SECTION XVII. EADICALS. 147. A Eadical is the indicated root &f any quantity, as V"^, a*, \/"2, 3^, &c. 148. In distinction from radicals, other quantities are called rational quantities. 149. The factor standing before the radical is the co- efficient of the radical. Thus, 2 is the coefficient of \/2 in the expression 2 \/ 2. 150. Similar EadicAls are those ■which have the same quantity under the same radical sign. Thus, \/a, 2 v'a, and X \/a are similar radicals; but 2 /\/a and 2is/b, or 2 x^ and 2 x are dissimilar radicals. 151. A Sued is a quantity whose indicated root cannot be found. Thus, \/2 is a surd. The various operations in radicals are presented under the following cases. CASE I. 152. To reduce a radical to its simplest form. Note. — A radical is in its simplest form when it contains no factor whose indicated root can be found. 1. Eeduce v"r5a^i to its simplest form. OPERATION. \/15an = \/25a^X3b — \/25a^XV3b = 5a^/3b We first resolve 75 a' 6 into two factors, one of which, 25 a% is the greatest perfect square which it contains ; then, as the root of RADICALS. 155 the product is equal to the product of the roots (Art. 148, Note 2), we extract the square root of the perfect square 25 a?, and annex to tihis root the factor remaining under the radical. Hence, RULE. Resolve the quantity under ibe radical sign into two fac- tors, one of which is the greatest perfect power of the same name as the root. Extract the root of the perfect power, multiply it by the coefficient of the radical, if it has any, and annex to the result the other factor, with the radical sign between them. Reduce the following expressions to their simplest form : — 2. v'lSa;, 3. y/4:9x'. -4 x'y\ 9 a' 64a;» ^Ty-- Ans. 2 v/3a;. Ans. 1 3? y/ X. 4. ,^na^lr'. Ans. 2a /^QH". 5. 5 4^ 64 a 6*. Ans. 10 5 1^4 a. 6. SA^UIa^bK 1. 25 ^56 a;. Ans. 21 a 6= is/ 3. Ans. bdt^lx. 8. 4v'128ar'y. 9. ^343a;». in / ^^''''' / 2Ta?c _ y 128z»2^~ 12. A/l6x^if—32. 3a /3c . = 8^V2,' ^'^^• ^°^- 3 V 2^ ^/ 16 x^f — 32 x^f = x/ 16 x'yWj^^ ^^f z=4.xy»^l — 2x^y\ Ans. 156 ELEMENTARY ALGEBRA. 13. 4 ^ 81 o»c 4- 270'. Ans. 12o4j^3c + l. 14. (a + b) ^3a' — 6ab + 3b\ An8. {a^ — ly') -s/l. 15. 7 ^aSOa;"/— 125 a;«/. 16. (ar — y) (a'a:— 0^)^ 17. («'+ a^J^)^. 18. \/— 16. V— 16 = ^/16^/— 1=4=\/— 1, Ans. 19. ^— 1250. 20. \/19a^ — 4.P. 153t When a fraction is under the radical sign, it can be transformed so as to have only an integral quantity under the radical sign, by multiplying both terms of the fraction by that quantity which will make its denominator a perfect power of the same name as the root, and then re- moving a factor according to the Bule in Art. 152. 1. Reduce 4'/- to its simplest form. OPERATION. Transform each of the following expressions so as to have only an integral quantity under the radical sign. 2. l^l- Ans. 1^-6. 3.4^1- Ans. VV^. 4. g^-^- Ans. -^!^343. 6. j-y 18^- Ans. gjV34o. EADICALS. 157 /^7^ 3 _ y "50"" Ans. -V6 ^)* u(S)*=u(i)'(?)*.A...6V3 / 294 7 y 845' ^°S- 65^30. 1. 2 Y o a: 8. 14 9 X. X. 10. 11. 432 1875' (« + *) y ^-ZfTj • Ans. V a'' — 5^. CASE II. 154. To reduce a rational quantity to the form of a radical. 1. Keduce Sar* to the form of the cube root. OPBKATiON. Since 3x' is to be placed 3 a^ = t^ 21 3? under the form of the cube root without changing its value, we cube it and then place the radical sign, &, over it. It is evident that ^27a» = Z3?. Hence, EULE. Involve the quantity to the power denoted by the index of the root required, and place the corresponding radical sign over the power thus produced. 2. Eeduce 4 a' J to the form of the square root. Aus. VlSa^J^. 3. Reduce 2a5^c~* to the form of the fifth root. Ans. ^32a*5"c- 4. Reduce 5 c? <^ to the form of the cube root- 158 ELEMENTAKY ALGEBRA. 5. Heduce -^ to the form of the fourth root, 3xyi 6. Keduce x — 2y to the form of the square root. Ans. is/ x' — 4:X^ -\- 4y^. 155. On the same principle the rational coefficient of a radical can be placed under the radical sign, by involv- ing the coefficient to a power of the same name as the root indicated by the radical sign, multiplying it by the radical quantity, and placing the given radical sign over the product. 1. Place the coefficient of hi^ly under the radical sign. OPERATION. h 1^ ly = 1^ \2b 1^ 1y = ^ 250y In the following examples, place the coefficient under the radical sign. 2. Zii/^x^y. Ans. ^324x'y. 3. 2xy^2a?y. Ans. -^16a^y*. 5. i^m. 6. (a — 5) ^~i- Ans. ,»^ a« — 2 a^ fi -j- o 6^. '7. 4:xy-index, we reduce the fractional indices to equivalent ones hav- ing a common denominator. It is evident that we have not changed the values of the given radicals by the process. Hence, RULE. Reduce the fractional indices to equivalent ones having a common denominator ; involve each qusmtity to the power de- noted by the numerator of the reduced index, and indicate ihe root denoted by the denominator. 2. Eeduce t^2 and ^3 to equivalent radicals having a common index. '- Ans. 3. Eeduce \/T ^.nd 4^^ to equivalent radicals having a common index. Ans. i^-^^s and f^ ■^■ 4. Eeduce 1/ 1 and 1/ | to equivalent radicals having a common index. 5. Eeduce n^ a, /^ a — h, and V « + * t° equivalent radicals having a common index. Ans. ^"5*, i^ (H^=^)S and ^ (^T+T)'. 6. Eeduce V^, ^"4, and ^ 3 to equivalent radicals having a common index. T. Eeduce y/li: and X^'y to equivalent radicals having a common index. Ans. •S7"^,and V7- 160 ELEMENTARY ALGEBRA. CASE IV. 157t To add radical quantities. 1. Add V^ and i^ y. Ans. s/ x -\- r^ y. It is evident that the addition can only be expressed. 2. Add Z\/x and bs/x. Ans. 8 \/a;. It is evident that 3 times the ^ x and 5 times the ^ x make 8 times the ^ x. 3. Add */S and \/50 together. OPERATION. In this case we make the radi- ^~g —- 2 \/~2 ^^ parts similar by reducing them /Xn K /^ *° ^''^'^ simplest form (Art. 152), ^ Zm — and then add their coefficients as Sum = 7 \/ 2 in Example 2. Hence, RULE. Make the radical parts similar when they are not, and prefix the sum of the coefficients to the common radical. If the radical parts are not and cannot be made similar, con- nect the quantities with their proper signs. 4. Add 2\^50ax and B\/98ax. Ans. 31\/2ax. 5. Add 4^24a;» and x/^Sl. Ans. 11x^3. 6. Add \/2i and s/Ms. Ans. 14 \/^. 7. Add ^5l2^* and ^162/. Ans. (4a; + 32^)^2. 8. Add \/~5 and Vi- Vi = \/SV5 = iV5; /v/5 + iV5 = fV5, Ans. 9. Add J^ e^nd^^-f. Ans. ^ ^l2. RADICALS. 161 lO: Add Vl, lOVlv. and 6 a/20. Ans. lav's. 11. Add V^ and V^- CASE V. 158t To subtract one radical from another. 1. From VTS take ik/¥l. OPERATION. We make the radical parts sim- x"hT r r'S ilar by reducing them to their sim- .-^ . plest form (Art. 152). And 3 y/l taken from 5 y/ 3 evidently leaves 2 v'^. Hence, 2^/3 EULE. Make the radical parts similar when they are not, sub- tract the coefficient of the subtrahend from that of the min- uend, and prefix the difference to the common radical. If {he radical parts are not and cannot be viade similar, indi- cate the subtraction by connecting them with the proper sign- 2. From Vsl take ^3. Ans. 2 ^^J. 3. From 9\^a^xi/^ take 3a\^xy^. Ans. ^ays/x. 4. From 7/\/20a; take ^s/^bx. Ans. 2/>/bx. 5. From ^500 take ,^l08. Ans. 2^4. 6. From 2 a/ A take \/^. Ans. y\ V 5- 1. From /v/f take \/ ijV- -^°s. -i^ \^ l(). 8. Prom 2Vl-'?6a:5 take V891a;^ 9. From a i^ aP' take 7 i^ a^s?. 10. From VlHi take ^1892. 162 ELEMENTAKY ALGEBRA. CASE VI. 159. To multiply radicals. 1. Multiply 3 V a by 5 y/l. OPERATION. 3 ^/'a X 5 s/l = 3 X 5 X V a X "/h = 15 K/~ab As it makes no difference in what order the factors are taken, we unite in one product the numerical coefficients ; and ^ ay, ^b = \/~ab (Art. 143, Note 2), 2. Multiply 4\/2a6 by b/^Zay. OPERATION. We reduce the radical parts to equivalent radicals 4ulti- Product == 20 ^12^^^ P'y ^' '"" ^^^ preceding ex- ample. 3. Multiply i^~a by ^a^x. Ans. 2/>/a~^. 5. Divide 4 x/a^ — 5^ by 2/s/a — h. Ans. 2 \/« + *• 6. Divide Gv'ST by 3^3. Ans. 6 1. Divide 'i/ x by >/« — 2v'*- 6. Subtract a^"32 — \/"2i2 from — 3 ^1 — V /v/"3. T. Multiply i>/a — \/6 by ^a — ^/x. OPERATION. ; — /\/ ab — /i^ ax -\- \/h: 8. Multiply a; 3/4" v'"* ^y ^ — -v/afi. Ans. 4a;y -}- (4 — xy) nJ ah — ah. 9. Multiply t + \/T0 by 6 — V" 10. _ Ans. 32 — V 10. 10. Multiply sfa + V* by \/a — s/h. Ans. a — 6. 11. Multiply V''5 — 4^3 by ^45 -f ^"9. 12. Multiply iVi + T V3 by i VI — T V3. 18. Divide -v/'oS + V ay 4-^ + ^2; y by v'« + V»'- OPERATION. \/a + V^) Vaa; + ^Z ay -\- x -\- \^xy {s/ x -\- sf y s/ ax -|- X sfay -\-Vxy Is/ ay ■\-'>/'xy p^- EADICALS. 169 14. Divide Vac — Va^ — \/ Jc + ^6rf by Vc—Vrf. Ans. V'« — -v/S. 15. Divide a^ x-\-b^x-}-aii/i-\-b^yi hy x-{-yi. 16. Divide a; — y by \/^ — V^- Ans. V^ + Vy- IT. Divide4a;2?-l-4v'«^ — a;y\/a^— a&by4 — \/fflJ. 18. Expand (/i/x + \/V)^- Ans. a; + 2 \/^ -}- y. 19. Expand {J — J~^)^ Ans. a — 2 v/?+F 20. Expand (v/a — V*)*- Ans. a^ — 4a^6^-}-6a5 — 4a^6^ + 52_ 21. Expand (4-/1/3)*- Ans. 100 — 51 ^"3. 22. Expand (a-i — x-y. Ans. a-t — 3 a"^ a;-i + 3 a-i a;"" — a;"'. 23. Expand (i-^iy, Ans. 24. Expand (^l-y'iy. . J 1 ^J ^4-1 a? 2 x^yi j^ 4 a;^ j/^ . ^ Ans. - — ■ — =? — \- xy ye " 3^6 9 25. Find the square root of a — 2 a^ 5^ + 5*. Ans. /\/ a — -^ 6^ 26. Find the cube root of a;* — 3 a;^y* + ^^y — 2/- Ans. x — y*. 21. Find the fourth root of 16 o —32 a*/ + 24 a*/ — 8 ffli j^" _j_ yl Ans. 2 a* — ^*. 8 170 ELEMENTAEY ALGEBRA. SECTION XVIII. PUEE EQUATIONS "WHICH EEQTJIEE IN THEIR EEDUCTION EITHER INVO- LUTION OR EVOLUTION. 164, A Pure Equation is one that contains but one power of the unknown quantity ; as, V^ -\-acz=b, 4:X^-\-3 = 1, or 14 a;" = a J. 165, A Puke Quadratic Equation is one that contains only the second power of the unknown quantity ; as, 6x^—Ua=51b, a f =13 cd, or acz^= 14. 166i Eadical Equations, i. e. equations containing the unknown quantity under the radical sign, require Invo- lution in their reduction. 167i To reduce radical equations. 1. Eeduce v'a; — 3n=H. operation. V"i — 3 = 8 Transposing, a^ x =^ 11 Squaring, • x = 121 2. Eeduce ^x _ 4 -f 7 = 10. operation. ^^=^ + 7 = 10 Transposing, /^ a; — 4 = 3 Cubing, 9= — ' 4 = 27 Transposing, -c = 31 RADICAL EQUATIONS. 171 3. Eeduce V^_V^ = ^-^. ^ a OPERATION. V a Clearing of fractions, \^ d^ -\- />,/ x = a Squaring, d^ -^ \/ x ^ a' Transposing, n^ x=. c? — d"^ Squaring, x-={d? — d'^Y Hence, to reduce radical equations, we deduce from these examples the following general RULE. Transpose the terms so that a radical part shall stand by itself; then involve each member of the equation to a power of the same name as the root ; if the unknown quantity is still under the radical sign, transpose and involve as before ; finally reduce as usual. L Eeduce 4 + f + 3 V^ = ^/-. Ans. a; = 16. 5. Reduce 7 «/- = -. Ans. a: = 2. 4 y a: 2 6. Reduce (v'^ + 4)^ = 2. Ans. a: = 144. T. Eeduce v'll + a? = -N^^ + 1- Ans. a; = 25. 8 Reduce \/aJ — 7 = \/a;4-18 — V^. Ans. a; = 21. 9. Reduce ■ ^^ = -4= . Ans. x = • X — ex ^ X 1 — "^ 10. Reduce V^_^-2 ^ V^_^-l . ^^^ ^ ^ 9_ 172 ELEMENTARY ALGEBRA. 11. Eeduce f>J x -\- s/ x — a = \j X ■ 12. Reduce v'a; — 30 + \/a; + 21 = Va; — 19. 13. Reduce ^^9 2; + 13 = 3\/a;+ 1- Ans. a; = 4. 14. Reduce \ ' — = '^, ' Ans. x = 5. 15. Reduce Va:^ — 32 = a; — ^V 32. 168i Equations containing the unknown quantity in- volved to any power require Evolution in their reduction. 169i To reduce pure equations containing the unknown quantity involved to any power. 1. Reduce-^ - = -. OPERATION. \^_ 3 97 Clearing the given equation of 5 7 35 fractions, transposing, and divid- 28a;^ — 15= 9T ing, we have :/■ = i-^ extract- 28 a;'^ = 112 ing the square root of each mem- a;2^^ 4 ber of this equation, we have x=±2 ^=±2. (Art. 136.) 2. Reduce T a;= — 89 = 100. OPEEATION. ij ^3 _ 89 _ jQo Transposing and dividing, we 3 have ^3 = 27; extracting the 189 g^jjg j,pgj. ^^ gg^gj^ member of — ■^' this equation, we have a; = 3. X ^^ 3 Hence, RULE. Reduce the equation so as to have as one member the un- known quantity involved to any degree, and then extract that root of each member which is of the same name as the power of the unknown quantity. jhcd- ■ acd -V <=- ■d Ans. X = ±4. Ans. X 1 ~3' Ans. X = — T. PUEE EQUATIONS ABOVE THE FIRST DEGREE. 178 Note. — It appears from the solution of Example 1 that every pure quadratic equation has two roots numerically the same, but with op- posite signs. 3. Eeduce Ix'^ -^1 = lar^ 4- 3. Ans. x=±&. 4. Eeduce a = J — ^ ■ c a 5. Eeduce 5 =^ 10. 1 4 6. Eeduce 3a;' + - = -- 1. Eeduce ^ + 50 = 1. 8. Eeduce f ~ , = ^ i~. ■ Ans. a!= ± V — 5. 2.2; -|- 1 a; -J" 4 9. Eeduce 4 a:' — 4 a;= = 0. 10. Eeduce Sa:"— 3 a; = 3 a;'' — 3 a; +50. 1 7 11. Eeduce a; -J- - ^ - ^ '^ — 1- 12. Eeduce 2 a; + 2 = (a: + 1)^ 13. Eeduce 1 + lia;-^ = 2 — 23?-^ 14. Eeduce 3 a;-i — 5 ar" = 2 a;^^ + 3 a;-i — ^. 15. Eeduce (c + x)' — 6 c^* = (c — a:)" + 16 C. ^ , 85 3a; — 3 3a;+3 16. Eeduce - — 3^^ = g^^Ts' 17. Eeduce -^qiT- + 'J-^i" - ^- 174 ELEMENTARY ALGEBRA. no. Equations containing radical quantities may re- quire in their reduction both Involution and Evolution; and in this case the rule in Art. 16'7, as well as that in Art. 169, must be applied. Which rule is first to be ap- plied depends upon whether the expression containing the unknown quantity is evolved or involved. 1. Keduce 17 — \/a;= — 2 = 12. OPERATION. 11 — ^x"- Transposing, &c., \/^' - Squaring, a:' - Transposing and uniting. Extracting the cube root. -2 = 12 -2= 5 -2 = 25 x= = 27 X = S Ans. Ans. X = 2. Keduce (Va;= — 4 + 3)»= 125. a; = 2. 3. Eeduce 4 / — s = \/x. a 1 6 = ^\- 4. Keduce \/x -\- a — ',"'"• ya; — a Ans. a; = ± \/2 a^ + 2 a i 4- 6^ 5. Keduce f- ^3 (.^ + 11)=^^. 6. Keduce /^ 2 x^ -|- 8 a;'' -f- 24 x'^ + 32 a: = a; + 2. 1. Keduce 4^9 {x* -{■ 19) + 100 — ^ = ^ . 171i Equations which contain two or more unknown quantities may require for their reduction involution, or evolution, or both. In these equations the elimination is eifected by the same principles as in simple equations. (Arts. 112-114.) 1. Given ^5 4 >-, to find x and y. (2x^ + y = 64) PURE EQUATIONS ABOVE THE FIRST DEGREE. 175 OPBRATION. ¥-!=" (1) 2a;=4-y= 54 (2) iF-,= ae (3) r-"» (4) 15-?= 14 4 (^) x'= 25 (5) y= 4 (8) a; =±5 (6) Subtracting four times (1) from (2), we obtain (4), which re- duced gives (6), or a: = ± 5 ; substituting this value of x in (I), we obtain (7), which reduced gives (8), or y = 4. Find the value of the unknown quantities in the follow- ing equations: — 2. Given ( a:y = 15 ) ^^ 3. Given }3.-4y=2y|. ^^^ ja: = }2xz = C3yz = 5. Given JA?'^-^l=n. Ans. ^ 6. Given|-^ + ^ = ^^ , }• (.a; —2^ =y — 2a;) ix^ — 2y^=l4: ) ±5. ±3. = ±6. x^ + 6y^ = 90!) " (.y=±3. 20 \ ra;=±l. 4. Given ^2xz=10y- Ans.-jy=±3. ■ 3yz = 45) (z=±5. = 21. = 8. 1. Given 1T6 ELEMENTAEY ALGEBEA. PKOBLEMS PEODTJCING PTJEE EQUATIONS ABOVE THE FIRST DEGREE. 172. Though the numerical negative values obtained in solving the following Problems satisiy the equations formed in accordance with the given conditions, they are practically inadmissible, and are therefore not given in the answers. 1. A gentleman being asked how many dollars he had in his purse, replied, " If you add 21 to the number and subtract 4 from the square root of the sum, the remainder will be 6." How many had he? SOLUTION. Let X r= number of dollars. Then, */x-\-21—4l= 6 Transposing, */x-\-21^ 10 Squaring, a; + 21 = 100 Transposing, a; = 79, number of dollars. 2. Divide 20 into two parts whose cubes shall be in the proportion of 27 to 8. Ans. 12 and 8. 3. What two numbers are those whose sum is to the less as 8 : 3, and the sum of whose squares is 136 ? Ans. 10 and 6. 4. What number is that whose half multiplied by its third gives 54 ? 5. What number is that whose fourth and seventh multiplied together gives 46f ? Ans. 36. 6. There is a rectangular field containing 4 acres whose length is to its breadth as 8:5. What is its length and breadth ? PURE EQUATIONS ABOVE THE FIRST DEGREE. 177 7. There are two numbers whose sum is IT, and the less divided by the greater is to the greater divided by the less as 64 : 81, What are the numbers ? Ans. 8 and 9. 8. The sum of the squares of two numbers is 65, and the difference of their squares 33. What are the numbers ? 9. The sum of the squares of two quantities is a, and the difference of their squares b. What are the quantities ? Ans. ± s/ ^ (« + V) and ± V^ (« — *)• 10. A gentleman sold two fields which together con- tained 240 acres. For each he received as many dollars an acre as there were acres in the field, and what he received for the larger was to what he received for the smaller as 49:25. What are the contents of each? Ans. Larger, 140 ; smaller, 100 acres. 11. What are the two quantities whose product _is a and quotient hi , , — - , la Ans. ± V a 6 and ± i / r • 12. What two numbers are as m:n, the sum of whose squares is a? , , nai/a" • , , nUli Ans. ± -; i and ± -. V (»»' + "') VO'i' + 'O 13. What two numbers are as m:n, the difference of whose squares is a ? * , mv/a J , re\/ Ans. ± -. and ± '^ a 14. Several gentlemen made an excursion, each taking $484. Each had as many servants as there were gentle- men, and the number of dollars which each had was four times the number of all the servants. How many gen- tlemen were there? Ans. 11. 15. Find three numbers such that the product of the first and second is 12 ; of the second and third, 20 ; and the sum of the squares of the first and third, 34. 8* ^ 178 ELEMENTARY ALGEBRA. SECTION XIX. AFFECTED QUADRATIC EQUATIONS. 173. An Affected Quadratic Equation is one that con- tains both the first and second powers of the unknown quantity ; as, 3a;^ — 4a; = 16 ; or ax — hx^ = c. 174. Every affected quadratic equation can be reduced to the form x^ -\- bx = c, in which b and c represent any quantities whatever, posi- tive or negative, integral or fractional. For all the terms containing a^ can be collected into one term whose coefficient we will represent by a ; all the terms containing x can be collected into one term whose coefficient we will represent by d\ and all the other terms can be united, whose aggregate we will represent by e. Therefore every affected quadratic equation can be reduced to the form ax'-\-dx = e (1) Dividing (1) by a, ' a? + -x=~ (2) d e Letting - ^ J, and - = c, we have a!' -\- bx ^ c (3) 175. The first member of the equation x' -j- bx = c cannot be a perfect square. (Art. 145, Note 2.) But we know that the square of a binomial is the square of the first term plus or minus twice the product of the two terms plus the square of the last term; and if we can find the third term which will make a;^ -}- bx a perfect EQUATIONS OF THE SECOND DEGEEE. 179 square of a binomial, we can then reduce the equation ic' -\- bx = c. Since 6 x has in it as a factor the square root of 3?, 3? can be the first term of the square of a binomial, and 6 x the second term of the same square; and since the second term of the square is twice the product of the two terms of the binomial, the last term of the binomial must be the quotient arising from dividing the second term of the square by twice the square root of the first term of the square of the bino- mial ; i. e. the last OPERATION. jg^^^ ^f (,,g ,,;„„. a;^ 4- J a; = c (1) . , . 6 a; h mial IS z— = ;; ; J2 j3 2 a; 2 a;2 + 6 a; + - = - -f c (2) ^^^ therefore the third term of the X -\- — =^ ±. \ T ~\' '^ (^) square must be a;=-5±y/^ + c (4) V2/ 4 — to each raem- 4 ber, we have (2), an equation whose first member is a perfect square. Extracting the square root of each mouiber of (2), and transposing, we obtain (4), ot x = — -^± \l -^ ■}- c, which is a general expression for the value of x in any equation in the form oi 3? -\-hx = c. Hence, as every affected quadratic equation can be re- duced to the form a-'-^hx—c, in which 6 and c repre- sent any quantities whatever, positive or negative, integral or fractional, every affected quadratic equaltion can be re- duced by the following RULE. Reduce the equation to the form x'-^bx = c, and add to ea^h member the square of fialf the coefficient of x. Extract the square root of each member, and then reduce as in simple equations. 180 ELEMENTARY ALGEBRA. 1. Eeduce •? a;= — 28 a; + 14 = 238. OPEKATION. T ar* — 28 a; + 14 = 238 Transposing, T ar" — 28 a; = 224 Dividing by T, a;^— 4a;= 32 Completing the square, x'' — 4a;-|-4^ 36 Evolving, X — 2 = ±6 Transposing, a; = 2±6 = 8, or — 4 Note. — Since in reducing the general equation a? -|- 6 a; = c we find x^ — oiv/j "i""' ^^^ry affected quadratic equation must have two roots; one obtained by considering the expression V 2 4" c positive, the other by considering this expression nega- tive. Whenever 4 / — -j- c ^ these two roots will be equal. 2. Eeduce I?- ^^ + i-3 = J + |. OPBEATION. ^ _K I 13 ^_|_£ 5 10 "T" 20 2"' 4 Clearing of fractions, 4a:^ — 2 a: -|- 13 ^ lOa:^ + 5ar Transposing, — ^a? — ta; = — 13 Dividing by — 6, a:^ + Ix 6 ~ 13 ■ 6 Completing the square. ^^+0 + 49' 144 ~ 49 144 : + 13 _ 6 361 144 Evolving, X + 7 12 ± 19 12 Transposing, t = -l^± 19 12 ~" 1, or- -H EQUATIONS OP THE SECOND DEGREE. 181 TfoTE. — In completing the square, as the second term disappears when the root is extracted, we have written ( ) in place of it. 3. Reduce 3 a:^ — 25 + 6 a; = 80. Ans. a; = 5, or — 7. 24 X 4. Reduce x = 3. Ans. x = 6, or — 4, 5. Reduce 2 a; -1 ^^ = Y. Ans. a: = 2. ' X — 1 Note. — In this example both roots are 2. 6. Reduce 1 x ^~r = 5 x — 1. X — 4 Ans. a; = 8, or — 1. •?. Reduce 17 — ii^ = ^^^ + 10 8 —X Ans. a; = t, or — 2T. X _[_ 1 x_ a; +~5 "■" 3 lO' 8. Reduce , ^ -!-- = —• Ans. x = 10, or — 1§. 9. Reduce 1 5 = 4. X ' x^ ,„ „ , 16 100 — 9x „ 10. Reduce 7-5 — = 3. x 4ar 176. Whenever an equation has been reduced to the form x^ -\- bx^ c, its roots can be written at once ; for this equation reduced (Art. 1T5) gives x=: — - ± t/- -j- c. Hence, The roots of an equation reduced to the form a^ -\- hx = c are equal to one half the coefficient of x with the opposite sign, plus or minus the square root of the sum of the square of one half this coefficient and the second member of the equation. 182 ELEMENTARY ALGEBRA. In accordance with this, find the roots of x in the fol- lowing equations : — 1. Reduce x'' + 8 x = 65. a; = — 4 ± V i6 + 65 = 5, or — 13, Ans. 2. Reduce a;^ — 10 a; = — 24. a; = 5 ± ^25 — 24 = 6, or 4, Ans. 3. Reduce i^ — 6 a: = — 5. Ans. a; =: 5, or 1. 4. Reduce x" + 1 a; = ITO. ar = — ^ ± 4/ J + no = 10, or— 17, Ans. 5. Reduce 3?' -\- -a; = -. 1, Ans. 1 9 6. Reduce a;'' + H^=s- Ans. ar=l, or — 1^. 1 o SI 7. Reduce x^ — -;x=. — ;r;- Ans. x^-> or r- 5 25 00 8. Reduce - = --)- 5|^. Ans. a; = 7, or — 5^. SECOND METHOD OF COMPLETING THE SQUARE. 177i The method already given for completing the square can be used in all cases ; but it often leads to in- convenient fractions. The more difficult fractions are in- troduced by dividing the equation by the coefficient of x^, to reduce it to the form x^-\-hx-^c. To present a method of completing the square without introducing these fractions, we will reduce equation (1) in Art. 174. EQUATIONS OF THE SECOND DEGREE. 183 1. Eeduce ax''-\-dx = e. OPERATION. ax^ -\- dx^ e (1) a'3y'-\-adx = ae (2) a^a? -{- adx -{- ~-^ — -\- ae (3) ax-[. 5 = ± y/j + «« (4) Multiplying (1) by a, the coefficient of n?, we obtain (2), in which the first term must be a perfect square. Since adx, the second term, has in it as a factor the square root of c? n?, a^ ^ can be the first term of the square of a binomial, and adx the second term ; and since the second term of the square is twice the product of the two terms of the binomial, the last term of the binomial must be the second term of the square divided by twice the square root of the first term of the square of the binomial, or = - ; and therefore the term required to complete the square is — , which is the square of d^ one half of the coefficient of x in (1). Adding — to both members of (2), we obtain (3), whose first member is the square of a binomial. Extracting the square root of (3) and reducing, we obtain (5), or i(-f*v/r+"'> Hence, to reduce an affected quadratic equation, we have this second RULE. Beduce the equation to the form ax^ -\- dx^e; then muir tiply the equation by the coefficient of x^, and add to each member the square of half the coefficient of x. Extract the square root of each raember, and then reduce as in simple equations. 184 ELEMENTARY ALGEBRA. Note 1. — This method does not introduce fractions into the equa- tion when the numerical part of the coefficient of x is even. When the coefficient of 3? is unity, this method becomes the same as the first method. Note 2. — If the coefficient of s? is already a perfect square the square can be completed without multiplying the equation, by add- ing to both members the square of the quotient arising from dioiding the second term by twice the square root of the first. This method also becomes the same as the first method when the coefficient of a? is unity. Note 3. — As an even root of a negative quantity is impossible or imaginary, the sign of the first term, if it is not positive, must be made so by changing the signs of all the terms of the equation. 2. Eeduce 3 a;^ + 8 a; = 28. OPERATION. 3 a;^ + 8 a; = 28 Completing the square, 9a;2+()H-16=:164-84 = 100 Extracting square root, 3 a; + 4 = ± 10 Whence, 3a: = — 4 ±10 = 6, or— 14 -A-nd a; = 2, or — 4f 3. Reduce 25a;''— 10a; = 195. OPERATION. 25a;«— 10a;=195 Completing sq. by Note 2, 25a;^— () + 1 = 1 -f 195 = 196 Extracting square root, 5 a; 1 = ± 14 Whence, 5a;=l ± 14 = 15, or — 13 ^^^ x= 3, or — 2f 4. Eeduce 6 3;==— 20 a; = —15. Ans. a; = 3, or 1. 6. Reduce 7a:2— 8a;=12f. Ans. a; = f ± f ^26. 6. Reduce Ta;^ — 4a:c=^'. Ans. a; = ^, or — -■ 7. Reduce 4a; X — 1 14 = a; — 3. Ans. a; =10, or 31 EQUATIONS OF THE SECOND DEGREE. 185 8. Reduce x^-4- — = — ——4: 4 4 9. Reduce — = — = THIRD METHOD OF COMPLETING THE SQUARE. 178. The method of the preceding Article introduces fractions whenever the numerical coefficient of x is not even. To present a method of completing the square without introducing any fraction, we vi^ill again reduce equation (1) in Art. 174. 1. Reduce ax^ -\- dx = e. OPERATION. a3? -\- dx = e (1) ar" -1 — X =- ' a a (2) d . cP

or 3, or — 1. 190 ELEMENTARY ALGEBRA. 4. Keduce a; + 7 — T Va; + ^ = 8 — 5V^ + T- Ans. a? = 9, or — 3. 40 5. Reduce {x — 5)^ — 3 /s/x — 5= ^ _ ^ Ans. a; := 9, or 5 -{- v'25! 6. Eeduce x^ + Bx + a^x^ + 3a; + 6 = 14. Note. — Add 6 to both members. Ans. a; = 2, or — 5, or — f ± i V 85. i. Eeduce 4, + x^ — 2x — 2 ^/6 — 2x -\- x^= 1. Ans. a; = 3, or — 1, or 1 ± 2 \/ — 1. 8. Eeduce \^ x'^ -\- x -\- 6 60 y'ri 4- a- + 6 Ans. a; := 5, or — 6, or — ^ ± ^ /y/SIT. 181 • Of the methods given for completing the square, the first is the best when the coefficient of the less power of the unknown quantity is even, and the coefficient of the higher power is unity, or when these become so by reduc- tion ; the second method is better than the third when- ever the coefficient of the less power of the unknown quan- tity is even. When the equation cannot be reduced by the first method without introducing fractions, if the co- efficient of the higher power of the unknown quantity is a perfect square, and the coefficient of the less power is divisible without remainder by twice the square root of the coefficient of the higher power, the method given in Note 2, Art. Ill, is the best. Let each of the follow- ing equations be reduced by the method best adapted to it. 1. Eeduce 4a;2 — 14=: 3a;^ — 12a; — 1. Ans. a; = 1, or — 13. 2. Eeduce 36a;2-j- 24a; = 1020. Ans. a; = 6, or — 5§. EQUATIONS OF THE SECOND DEGREE. 191 3. Reduce a; — - -f- ^ = 21. Ans. x = 21, or }. 4. Eeduce X 8 — X X — 2 2x — H X — 3 5. Eeduce J_|_± = 23. 6. Beduceg^^:"^^-^). a X Ans. as = 3 (c — rf), or 3 «?. t. Reduce ^ '" „ = ^ "7 „ + 2|. 8. Reduce X — 7 a;-[- 3 a; — 4 a; — 2 X — 4 2 ' 9 Reduce - — - — li^^l^ X 9 a? 10. Reduce - + - = -• Ans. a; = 1 ± \/ 1 — a", 11. Reduce 3a; + 3 = 13 + ^. 12. Reduce 5 a; — = 2a;-| ^^. a; — 3 ' 2 io -D J a; + 4 4x4-7 , 7 — x 13. Reduce -^ ^ + 1 = ^^3. 14. Reduce 2\/a; — v'^ — 7 = 5. Ans. x= 16, or 7|^. 15. Reduce 2 V^"^^^+ 3 ^2^= 7 a + 5x _ \ X — a Ans. a; = 9 a, or — a. 16. Reduce 4 V'S — V2a; + 1 =-=i£^. ^ ^ ^ V2Z + 1 Ans. a; = 4, or — 2f . n. Reduce 5 /v/25 — as = 6 v'25 — a; + a; — 13. Ans. X = 16, or 9. 192 ELEMENTARY ALGEBRA. 18. Keduce ^-ti — a/ i +^^23^ + xK Ans. X = 12, or 4. 19. Eeduce 6 + 4 a:"* — 12 a;-^ = 100 x-^ 20. Eeduce ^ -^^^ + 6 \/"^— y = y- 21. Eeduce 3 a;* — 24 a;^ — 80 = 304. 22. Eeduce ^ + 10 = 1 + 4a;^ Ans. a; = ± 4, or ± 2 V— 2. Ans. X = ^'9, or 4^- 23. Eeduce 5 a;* — 3 a;^ + — = 27. Ans. a;=±Jv'6, oriSV — -j^. 24. Eeduce 2 a;^ — 5 a;^ ■+ 4 = 2. 25. Eeduce 6 a;* + 1184 = 5 a;*. 26. Eeduce V'a;+ 3 — ^a:4- 3 = 2. Ans. a; =: 13, or — 2. 27. Eeduce x^ — is/ x" -\-x — b = 25 — x. Note. — By transposing — x and subtracting 5 from each mem. ber, make the expression without the radical in the first member like that under the radical; then complete the square, &c. 28. Eeduce a;^ — 2a;4-3V2a;^ — 6x— ll=x + 33. Ans. a; = 6, or — 3, or f ± ^ \/2'78. 29. Eeduce 21 x* + 11 a:^ — 69 = 321 — (11 x^ -\- 5 x'^). Ans. x= ± i \/T3, or ± i a^—15. 30. Eeduce -, — = - -] 31. Eeduce (a;'' — 4 a;)2 = 12 a; — 3 x^ 32. Eeduce a; + (a;^ — a;)^ = a;2-|- 5112. EQUATIONS OF THE SECOND DEGREE. 193 PROBLEMS PRODUCING AFFECTED QUADRATIC EQUATIONS WITH BUT ONE UNItNOWN QUANTITY. 182. Though the numerical negative values obtained in solving the foUoviring Problems satisfy the equations formed in accordance with the given conditions, they are prac- tically inadmissible, and are therefore not given in the answers. 1. Divide 40 into two parts such that the sum of their squares shall be 1042. SOLUTION. Let a; = one part ; then 40 — a; = other part. Then, 3? + (40 — xf = 1042 Expanding, a? -|- 1 600 — 80 a; -(- r* = 1042 Transposing and unfting, a? — 40 a; = — 279 Whence, a; = 20 ± 11 = 31, or 9 And, 40 — a; = 9, or 31 2. Divide 20 into two parts such that their product will be 99|. Ans. 9J and 10^^. 3. The ages of two brothers are such that the age of the elder plus the square root of the age of the younger is 22 years, and the sum of their ages is 34 years. What is the age of each? Ans. Elder, 18; younger, 16. Note. — The other answers found by reducing the equation, viz. 25 and 9, satisfy the conditions of the equation only upon consid- ering v/y= — 3. To make the problem correspond to these an- swers, the word "plus" must be changed to "minus." 4. A merchant had two pieces of cloth measuring to- gether 96 yards. The square of the number of yards in the 194 ELEMENTARY ALGEBRA. longer is equal to one hundred times the number of yards in the shorter. How many yards are there in each piece ? Ans. 60 and 36. 5. Find two numbers whose difference is 3, and the sum of whose squares is 117. Ans. 9 and 6. 6. A merchant having sold a piece of cloth that cost him $42, found that if the price for which he sold it were multiplied by his loss, the product would be equal to the cube of the loss. What was his loss ? Note. — If the word " loss " were changed to gain, the other an- swer, — 7, or as it would then become, -|- 7, would be correct. Ans. $6. I. Find two numbers whose difference is 5, and prod- uct 176. Ans. 11 and 16. 8. There is a square piece of land whose perimeter in rods is 96 less than the number of square rods in the field. What is the length of one side ? Ans. 12 rods. 9. Find two numbers whose sum is 8, and the sum of whose cubes is 152. 10. A man bought a number of sheep for $240, and sold them again for $6.75 apiece, gaining by the bargain as much as 5 sheep cost him. How many sheep did he buy? Ans. 40. II. Find two numbers whose difference is 4, and the sum of whose fourth powers is 1312. Note. — Let x — ,2 and a; -|- 2 be the numbers. Ans. 2 and 6. 12. A man sold a horse for $312.50, and gained one tenth as much per cent as the horse cost him. How much did the horse cost him? Ans. $250. 13. The difference of two numbers is 5, and the less minus the square root of the greater is 7. What are the numbers? Ans. 11 and 16. EQUATIONS OF THE SECOND DEGRKE. 195 14. A and B started together for a place 300 miles dis- tant. A arrived at the place 1 hours and 30 minutes be- fore B, who travelled 2 miles less per hour than A. How many miles did each travel per hour ? Ans. A, 10 ; B, 8 miles. 15. A gentleman distributed among some boys $15; if he had commenced by giving each 10 cents more, 5 of the boys would have received nothing. How many boys were there ? Ans. 30. 16. Find two numbers whose sum is a, and product b. . a ± ^ a' — 4i , a 'X \/ a' — 46 Ans. ^—^ and ' 11. A merchant bought a piece of cloth for $45, and s6ld it for 15 cents more per yard than he paid. Though he gave away 5 yards, he gained $4.50 on the piece. How many yards did he buy, and at what price per yard ? Ans. 60 yards, at "15 cents per yard. 18. A certain number consists of two figures whose sum is 12 ; and the product of the two figures plus 16 is equal to the number expressed by the figures in inverse order. What is the number? Ans. 84. 19. From a cask containing 60 gallons of pure wine a man drew enough to fill a small keg, and then put into the cask the same quantity of water. Afterward he drew from the cask enough to fill the same keg, and then there were 41^ gallons of pure wine in the cask. How much did the keg hold ? Ans. 10 gallons. 20. There is a rectangular piece of land '75 rods long and 65 rods wide, and just within the boundaries there is a ditch of uniform breadth running entirely round the land. The land within the ditch contains 29 acres and 96 square rods. What is the width of the ditch ? Ai:s. .5 of a rod. 196 ELEMENTARY ALGEBRA. SECTION XX. QUADRATIC EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 183i The Degree of any equation is shown by the sum of the indices of the unknown quantities in that term in which this sum is the greatest. Thus, ^xy — 2a: = 1 is an equation of the second degree, lx^y^-\- xy^^a^c " " " fourth " 5y*_ 14cc=;a;2/ " " " fifth " Note. — Before deciding what degree an equation is, it must be cleared of fractions, if th6 unknown quantities appear both in the denominators and in the numerators or integral terms ; and also from negative and fractional exponents. 184i A Homogeneous Equation is one in which the sum of the exponents of the unknown quantities in each term containing unknown quantities is the same. Thus, 4a;^ — 4ixy -\-y^=:\Q or «'+ ^xy^-^-Zx'y-^-y^ — il or a;* — 4 a;' 2^ + 6 x'^y'^ — 4 x/ + / = 256 is a homogeneous equation. 185. Two quantities enter Symmetrically into an equa- tion when, whatever their values, they can exchange places without destroying the equation. Thus, a;^ — 2 a; y -|- y^ ^ 25 or a:'+3a;2y-4-3a;y2 4- /= 8 or x'^-\-2xy-\-y^-\- 2x-\-2y = 2l QUADRATIC EQUATIONS. 197 186i Quadratic equations containing two unknown quan- tities can generally be solved by the rules already given, if they come under one of the three following cases : — I. When one of the equations is simple and the other quadratic. II. When the unknown quantities enter symmetrically into each equation, III. When each equatiott is quadratic and homogeneous. CASE I. 187i When one of the equations is simple and the other quadratic. 1. Given \ v V >• , to find x and «• OPERATION. 2x + 23/ = 22 (1) 3a;»+2^«=lll (2) 2^=11 — a: (3) 32r'-|-121 — 222; + a;==lIl (4) 4a«— 22r = — 10 (5) 42a;8_()_|_ll2=121— 40 = 81 (6) 4a:=ll±9 = 20, or2 (7) y = 6, or 10^ (9) a;= 5, or ^ (8) From (1> we obtain (3), or y = 11 — s. SnbstitUting this value i;f y in (2), we obtain (4), an affected quadratic equation, which reduced gives (8); and substituting these values of x in (3), we obtain (9). In this Case the values of thd nnkfiown quantities can generally be found by substituting in the quadratic equation the value of one unknown quantity found by reducing the simple equation. L98 ELEMENT AEY ALGEBRA. 2. Given 1 (.X ■ '^2' = 28 I to find X and 2?. — y— 3) OPERATION. xy = 2% (1) X — y = 3 (2) x^ — 2xy-{-y''= 9 (3) 4:xy =112 W ^ x^-\-2xy+f=l2l (5) x + y =±11 (6) 2 a; = 14, or — 8 0) 2y = 8, or — 14 (8) a: = t, or — 4 (9) y = 4, or — 7 (10) Adding four times (1) to the square of (2), we obtain (5) ; ex- tracting the square root of each member of (5), we obtain (6) ; adding (2) to (6), we obtain (7) ; subtracting (2) from (6), we ob- tain (8) ; .and reducing (7) and (8), we obtain (9) and (10). Note. — Though Example 2 can be solved by the same method as Example 1, the method given is preferable. By this method find the values of x and y in the fol- lowing equations : — 3. Given j^ "2'= H- Ans. 4 ^ = ^• (a:^ — 2/2=32) iy = 2. 4. Given |^ + 2/ = 13) . ^^^^ ( a; = 7, or 6. 5. Given j ^y = 20) I 5 a; + y = 29 ) 6. Given } ^2^-247 (.3a; — 2v= 10) 3xy =45 xy = 15 a:^- -2x^4-/= 4 x — y = ±2 QUADEATIC EQUATIONS. 199 CASE II. 188. Wheu the unknown quantities enter symmetrically into each equation. 1- Given {", + J,Zi52}'*°fi^da:andy. OPERATION. ^+3'= 8 (1) a:» + /=152 (2) x' + 2xy-\-y^ = U (3) a;2— xy +y^= 1 9 (4) (5) (6) 0) (8) 2 a; = 10, or 6 (9) 2y = 6, or 10 (10) X = 5, or 3 (11) y = 3, or 5 (12) Squaring (1), we obtain (3) ; dividing (2) by (1), we obtain (4) ; subtracting (4) from (3), we obtain (5), from which we obtain (6) ; subtracting (6) from (4), we obtain (7) ; extracting the square root of each member of (7), we obtain (8) ; adding (8) to (1), we ob- tain (9); subtracting (8) from (1), we obtain (10) ; and reducing (9) and (10), we obtain (11) and (12). Note 1. — It must not be inferred that x and y are equal to each other in these equations ; for when a; = 5, y = 3 ; and when X = 3, y = 5. In all the equations under this Case the values of the two unknown quantities are interchangeable. Note 2. — Although i^ -^ y" = 152 is not a quadratic equation, yet as we can combine the two given equations in such a manner as to produce at once a quadratic equation, we introduce it here. 200 ELEMENTAKY ALGEBRA. 2 Given I ^2/ = 6 ) .^^ g^^^ ^ ^^^ lx^-\-f — 2x — 2y = By OPERATION. xy=(i (1) a?-{-3^_2a: — 2y= 3 (2) 2a:y =12 (3) (.^+yy-2(F+y) = -^^ (4) (^+y)°- ()+ 1-16 (5) x-{-y=l±4 = 5,OT — 3 (6) I = 3, or 2, or J^ (7) y = 2,or3,or:=i5_V'-J.^(8) Adding twice (1) to (2), we obtain (4) \ completing the sqinare in (4), we obtain (5) ; extracting the square root of each member of (5), and transposing, we obtain (6) ; and combining (1) and (6) as the sum and product are combined in the preceding example, we obtain (7) and (8). In Case II. the process varies as the given equations vary. In general the equations are reduced by a proper combination of the sum of the squares, or the square of the sum or of the difference, voith multiples of the product of the two unknown quantities; and finally, of the sum, with ilie difference of the two unknown quantities^ l^OTE 3l — When the unknown quantities enter into each equation symmetrically in all respects except their signs, the equations can be reduced by this same method ; e. g. i — y ^ 7, and a? — t/* = 511. In such equations the values of the unknown quantities are not inter- changeable Note 4. — The signs ± if standing before any quantity taken in- dependently are equivalent to each other ; but when one of two quan- tities is equal to ± o while the other is equal to q: b, the meaning is that the first is equal to -|- a, when the second is equal to— b; and the first to — a, when the second is equal to -j- 6. QUADRATIC EQUATIONS. 201 By this method find the values of x and y in the follow- ing equations : — a: = 4, or 3. y = 3, or 4. 3. Given |2. +2^= IM . ^^, | Ua:»+3/ = 2t3) t 4. Given 1^-2' = H. Ans. |^= ^' "'^ - 1- (x^ — /=m> 1y=l, or — 9. 5. Given |^ +^/^+^ =14|. UoTE. — Divide the second equation by the first. 6. Given j^-V^ + S' = U. CASE III. I89i When each equation is quadratic and homogeneous. 1. Given \ ^2' ~r y (■ , to find x and y. tSa;" — a;3r=: 10) OPERATION. 2a;y+3/' = 5 (1) 3a;= — a;y==10 (2) Let x^vy 2vy-'-\-f = i (3) Zv''f — vy'=\^ (4) 2 5 ,^. „ 10 -. («) 5 _ 10 ?„ _|1, 1 — 3 (;« — !, ^''' 15 ?72 — 5 « = 20 j; + 10 (8) Zv' — ^v = '2. (9) 17 = 2, or — A (10) / = z4t' --~hr, (11) 2/ = ± 1, or ± V 15 (12) .T = r ^ = ± 2, or T ^ VTS^ (13> 9* 202 ELEMENTARY ALGEBEA. Substituting vy for x in (1) and (2), we obtain (3) and (4) ; from (3) and (4) we obtain (5) and (6) ; putting these two values of y' equal to each other, we obtain (7), which reduced gives (10); sub- stituting this value of v in (5), we obtain (11), which reduced gives (12) ; and substituting in a; = w!/ the values of v and y from (10) and (12), we obtain (13). Examples under Case III. can generally be reduced best by substituting for one of the unknown quantities the product of the other by some unknown quantity, and then finding the value of this third unknown quantity. When the value of this third quantity becomes known, the values of the given unknown quantities can be readily found by substitution. Note. — Whenever, as in the example above, the square root is taken twice, each unknown quantity has four values ; but these values must be taken in the same order, i. e. in the example above, when y = -\- 1, X = -\- 2 ; when y = — 1, x = — 2; when y = -\- ^TE, x^= — ^ y/ 15 ; and when y = — ^ 15, x = -\- ^ ^ 15. By this method find the values of x and i/ in the follow- ing equations : — • 2. Given f ^' " ^^ = ^H ■ t^y — 2y Ans. < ± 5, or ± 11 /i/ — i 3. Given j-^+3-y = 2Y|. Ans. |-=±T.or± 4^-^ Ans. \-=±3,.ov±9V-i. Li h= ±2, or rpSV— i 4. Given i-' + i-y = U-2f} (a;2/ — 3/= 3 — x^ ) ^„g_ ( a; = ± 2, or ± 24 V— -,-V- ^y = ± 1, or T 11 -(/^^ 5. Given | 32 - 3xy = 2.^ -y^} <. x^ — 3=y^ + 2 ) QUADRATIC EQUATIONS. 203 190. Find the values of x and y in the following Examples. Note. — Some of the examples given below belong at the same time to two Cases. Thus in Example 1 both the equations are symmetrical, and both are quadratic and homogeneous, and there- fore it belongs both to Case II. and Case III. Example 3 belongs both to Case I. and Case II. 1. Given j ^ ,"^ = 20|. ^^^ |.= ±5, 2. Given \ ^2'= ^ |. 3. Given p + 2^ = ^ I. Ans. 1^ = *' ly = 3, 4. Given } ^y=l^ \. or ± 4. or ± 5. J J-IZ^ — ^^l. Ans. \' (3a;y— 7 = 56) ' 5. Given -^ x—y '""^- Ans. \^ J °^ J l^ == 3, or t. 6. Given | "'""^^.o^?!- Note. — Considering zy a single quantity, find its value in the second equation. r. Given \-'y--f=^n \ ar' — «» = 98 ) Note. — Subtract from the second equation three times the first, and extract the cube root of each member of the resulting equation. a P,. (2a;'^y + 2a:/=168) 8. Given \ * a , , „, f • ( a;« + /= 91) (a; = 4, ^- 1=3, . (a; = 4, or 3. 2/ = 3, or 4. 204 ELEMENTARY ALGEBRA. 9. Given }3«.^^ _ 3a:/ = 18> !a^/ — 2a:2/ 10. Given ( , . 13 (x — yY) 11. Given n^-2^y=88> Ans. j^=±5. iy=±2. Ans |^=±4, or± 66V^. (2^= ± 3, or T 175 V 5^5- 12. Given |^- 2a. + 2y = 30 -/) ( 4x« = 60 > 4x2^ = 60 /ea:*- 13. Given> i . a.?/ = 4/> ■ — 2y^= 10 ) Ans. f^=1000,or8. ly= 625, orl. 14. Given ( 3xy=18) (a;* — y = 65i Ans. = ± 3, or ± 2^ — 1. Ly = ± 2, or ±3^— 1. 15. Given P (^ -y) = 3 (V^+ V7) ]. . ( a:«=36 ) i:y= 36 Ans, ±V^— 23 — 11 X = — '^ 5 , or 9, or 4. T l/— 23 — 11 y — ^^ s , or 4, or 9. 16. Given x-\~y = 41 ) 17. Given |»^*-/= U. (a; — y =19) QUADRATIC EQUATIONS. 205 18. Given ^^^^+'^1/= 5) 19. Given i^~' " 3^" = sfI . ix-^ — tr^ = i ; 20. Given |»^'+ 2^V + 2^2,- + / = 95} 21. Given I ^2'= H. («* + / = 272) 22. Given (a; +y = 6) (.a;* + /=626i PKOBLEMS PRODUCING QUADRATIC EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 191. Though the numerical negative values obtained in solving the following Problems satisfy the equations formed in accordance with the given conditions, they are practically inadmissible, and, except in Example 4, are not given in the answers. 1. The sum of the squares of two numbers plus the sum of the two numbers is 98 ; and the product of the two numbers is 42. What are the numbers ? Ans. 7 and 6. 2. If a certain number is divided by the product of its figures the quotient will be 3 ; and if 18 is added to the number, the order of the figures will be inverted. What is the number '' Ans. 24. 3. A certain number consists of two figures whose product is 21 ; and if 22 is subtracted from the number. 206. ELEMENTARY ALGEBRA. and the sum of the squares of its figures added to the remaitider, the order of the figures will be inverted. What is the number? Ans. 37. 4. Find two numbers such that their sum, their prod- uct, and the difierence of their squares shall be equal to one another. Ans. § ± ^ \/~S and J ± i \/ 5. 5. There are two pieces of cloth of diflerent lengths ; and the sum of the squares of the number of yards in each is 145 ; and one half the product of their lengths plus the square of the length of the shorter is 100. What is the length of each ? Ans. Shorter, 8 ; longer, 9 yards. 6. Find two numbers such that the greater shall be to the less as the less is to 2f, and the difierence of their squares shall be 33. Y. The area of a rectangular field is 1575 square rods ; and if the length and breadth were each lessened 5 rods, its area would be 1200 square rods. What are the length and breadth ? 8. Find two numbers such that their sum shall be to 6 as 9 is to the greater, and the sum of their squares shall be 45. Ans. 9 VT and 3 VT, or 6 and 3. 9. The fore wheels of a carriage make 2 revolutions more than the hind wheels in going 90 yards ; but if the ch'cumference of each wheel is increased 3 feet, the car- riage must pass over 132 yards in order that the fore wheels may make 2 revolutions more than the hind wheels. What is the circumference of each wheel ? Ans. Fore wheels, 13^ feet ; hind wheels, 15 feet. 10. Find two numbers such that five times the square of the -greater plus three times their product shall be 104, and three times the square of the less minus their prod- uct shall be 4. KATIO AND PROPORTION. 207 SECTION XXI. EATIO AND PROPORTION. 192i Ratio is the relation of one quantity to another of the same kind ; or, it is the quotient which arises from dividing one quantity by another of the same kind. Ratio is indicated by writing the two quantities after one another with two dots between, or by expressing the division in the form of a fraction. Thus, the ratio of a to I is written, a : h, or t ; read, a is to b, or a divided by h. 193i The Teems of a ratio are the quantities compared, whether simple or compound. The first term of a ratio is called the antecedent, and the other the consequent ; and the two terms together are called a couplet. 194i An Inverse, or Reciprocal Ratio, of any two quan- tities is the ratio of their reciprocals. Thus, the direct ratio of a to S is a : J, i. e. r; and the inverse ratio of a to 6 is 11.116 , - : ri 1. e. - -i- 1 = -t or o : a. a aba 195i Proportion is an equality of ratios. Four quan- tities are proportional when the ratio of the first to the second is equal to the ratio of the third to the fourth. The equality of two ratios is indicated by the sign of equality {=) or by four dots (: :). Thus, a : b = c : d, or a : b : : c : d, or J :=^: read, a to 5 equals c to d, or a is to 6 as c is to d, or a divided by S- equals c divided by d. 208 ELEMENTARY ALGEBRA. 198. In a proportion the antecedents and consequents of the two ratios are respectively the antecedents and con- sequents of the proportion. The tirst and fourth terms are called the extremes, and the second and third the means. 197. When three quantities are in proportion, e. g. a : b ^ b : c, the second is called a mean proportional be- tween the other two ; and the third, a third proportional to the first and second. 198. A proportion is transformed by Alternation when antecedent is compared with antecedent, and consequent with consequent. 199. A proportion is transformed by Inversion when the antecedents are made consequents, and the conse- quents antecedents. 200. A proportion is transformed by Composition whem in each couplet the sum of the antecedent and consequent is compared with the antecedent or with the consequent. 201. A proportion is transformed by Division when in each couplet the difference of the antecedent and conse- quent is compared with the antecedent or with the con- sequent. THEOREM I. 202. In a proportion the product of the extremes is equal to the product of the means. Let a c Clearing of fractions, ad=^bi EATIO AND PROPOETION. 209 THEOREM II. 203. If the product of tvjo quantities is equal to the prod- uct of two others, the factors of either product may be made the extremes, and the factors of the other the means of a proportion. Let ad^h e Dividing hj bd, I —~d 1. e. 6 = , THEOREM III. 204. If four quantities are in proportion, they will he in proportion by alternation. Let a : b = e : d By Theorem L ad=bc By Theorem II. a : c = 6 : cf THEOREM IV. 205. If four quantities are in proportion, they will he in proportion by inversion. Let a:b — c:d By Theorem I. ad^bc By Theorem II. b : a=zd:c THEOREM V. 206. ^ three quantities are in proportion, the product of the extremes is equal to the square of the mean. Let a:b=.l:c By Theorem I. ac=^V^ THEOREM VI. 207. If four quantities are in proportion, they will he in proportion by composition. 210 ELEMENTARY ALGEBRA. Let a:6 = : C :d i. e. a b~ C d Adding 1 to each member, !+'= c a + 1 a-\-b c -\- d or ~~b~~ IT je. a-\-h: b=::^c-\-d: d THEOREM VII. 208. If four quantities are in proportion, they will be in proportion by division. Let a : b = c : d a c i-«- l = d d c Subtracting 1 from each member, t — ^ ^^ ^ — ^ a — 6 c — d or 1. e. 6 d a — b : b = c — did THEOREM VIII. 200. Two ratios respectively equal to a third are equal to each other. Let i. e. Hence (Art. 13, Ax. 8), i. e. . a : b :^ c : d THEOREM IX. 210i ^ four quantities are in proportion, the sum and difference of the terms of each couplet will be in proportion. a : b ^ m ; : n and c : d^m : ; n t a m. b~n and a c h~d c m d~~n RATIO AND PBOPOETION. 211 Let a : h = c : d By Theorem YI. a -^ b : b = c -\- d : d (l) and by Theorem VII. a — 5:i = c — d : d (2) From (1), by Theorem III. a-\-b: c-]-d=b: d From (2), by Theorem III. a — b : c — d^= b : d By Theorem VIII. a-\-b:c-\-d^a — b : c — d Hence, by Theorem III. a -\- b : a — b ^= e -\- d : c — d THEOREM X. 211. Equimultiples of two quantities have the same ratio as the quantities themselves. For by Art. 83, i = ^ •' ' 6 mb i.e. a : b :=^ ma : mb Cor. It follows that either couplet of a proportion may be multiplied or divided by any quantity, and the result- ing quantities will be in proportion. And since by Theo- rem III. ii a : b = m a : m b, a : ma = b : mb, or ma : a := mb : b, it follows that both consequents, or both ante- cedents, may be multiplied or divided by any quantity, and the resulting quantities will be in proportion. THEOREM XI 212. If four quantities are in proportion, like powers or like roots of these quantities will be in proportion. Let a : b = c : d a c 1- e. r = J b d Hence, F = d^ i. e. o" : J" = c" : «f Since n may be either integral or fractional, the theorem is proved. 212 ELEMENTARY ALGEBRA. THEOREM XII. 213. If any number of quantities are proportional, any antecedent is to its consequent as the sum of all the antece- dents is to the sum of all the consequents. Let a : b:= c : d:=e :f Now ah = ab (1) and by Theorem I. ad^=bc (2) and also af=ibe (3) Adding (1), (2), (3), a{b^d+f)=:b{a + c^e) Hence, by Theorem II. a: b=^a-\-c-\-e:b-\-d -\-f THEOREM XIII. 214. If there are two sets of quantities in proportion, their products, or quotients, term by term, will be in proportion. Let a : J =: c : rf and e:f=g : h By Theorem I. ad := b c 0) and efi^fff (2) Multiplying (1) by (2), adeh^:^h cfg ;, (3) Dividing (1) by (2), ad he e It ~ fg ' W From (3), by Theorem 11. ae : bf^ eg : dh and from (4), ah c d e ' J~ g ' h' PEOBLEMS IN PROPORTION. 215. By means of the principles just demonstrated, a proportion may often be very much simplified before making the product of the means equal to the product of the extremes ; and a proportion which could not oth- erwise be reduced by the ordinary rules of Algebra may often be so simplified as to produce a simple equation. RATIO AND PROPORTION. 213 1. The cube of the smaller of two numbers multiplied by four times the greater is 96 ; and the sum of their cabes is to the difference of their cubes as 210 : 114. What are the numbers ? SOLUTION. Ijet a; = the greater and y = the less. Then 4:xf=96 (1) a^ -\-f:x^ — f = 210 -.lU (2) From (2), by Theo. X., Cor. a;» + / : a:' — / = 35 : 19 By Theorem IX. 2 a:' : 2/ = 54 : 16 By Theorem X., Cor. x^ : f = 21 : 8 By Theorem XI. a; : y = 3 : 2 By Theorem I. 2x = 3y (3) From (1) and (3) we find x ^ 3 and y = 2. 2. The product of two numbers is 78 ; and the differ- once of their cubes is to the cube of their difference as 283 : 49. What are the numbers ? SOLUTION. Let X = the greater and y = the less. Thena;y=78 (1) 3? — j^: 3i^ — 3x'y-\-Sxf — f = 28S -.49 (2) From (2), by division, Sx'y — 3xf : (x — yy =234 : 49 Dividing 1st couplet by a; — y, 3xy : {x — yY = 234 : 49 Dividing antecedents by 3, xy : (x — yy = 78 : 49 Substituting the value of xy, 78 : (a; — yy = 78 : 49 Dividing antecedents by 78, 1 '• (x — y)* = 1 : 49 E.xtraetlng the square root, 1 : x — y = 1 -.7 Whence, x — y= 7 (3) From (1) and (3) we find a; = 13 and y ^ S. B. The sum of the cubes of two numbers is to the cube of their sum as 13 : 25 ; and 4 is a mean proportional be- tween them. What are the numbers? 214 ELEMENTARY ALGEBRA. 4. The difference of two numbers is 10 ; and their prod- uct is to the sum of their squares as 6 : St. What are the numbers ? SOLUTION. Let X = the greater and y = the less. Then X— 2^=10 (1) xy:o?-\-f= 6:37(2) From (2), by Theorem X., Cor. ^xy::^ -\- f =\1 ■.Zl By Theorem IX. x* + 2 a;y + 3^« : a;« — 2 a;?^ + ys = 49 : 25 By Theorem XI. x -\- y : x — y= 7:5 By Theorem IX. 2a;:2y =12:2 By Theorem X., Cor. x:y =6:1 By Theorem I. a: = 6 y (3) From (1) and (3) we find a; = 12 and ?/ = 2. 5. The product of two numbers is 136 ; and the dif- ference of their squares is to the square of their differ- ence as 25 : 9. What are the numbers ? Ans. 8 and 17. 6. As two boys were talking of their ages, they dis- covered that the product of the numbers representing their ages in years was 320, and the sum of the cubes of these same numbers was to the cube of their sum as 7 : 27. What was the age of each ? Ans. Younger, 16; elder, 20 years. 7. As two companies of soldiers were returning from the war, it was found that the number in the iirst multi- plied by that in the second was 486, and the sum of the squares of their numbers was to the square of the sum as 13 : 25. How many soldiers were there in each company? Ans. In 1st, 27 ; in 2d, 18. 8. The difference of two numbers is to the less as 100 is to the greater ; and the same difference is to the greater as 4 is to the less. What are the numbers ? J^ToTE. — Multiply the two proportions together. (Theorem XIII.) PEOGRESSION. 215 SECTION XXII. PEOGEESSION. 216. A Progression is a series in which the terms in- crease or decrease according to some fixed law. 217. The Teems of a series are the several quantities, whether simple or compound, that form the series. The first and last terms are called the extremes, and the others the means. AEITHMETICAL PEOGEESSION. 218. An Arithmetical Progression is a series in which each term, except the first, is derived from the preced- ing hy the addition of a constant quantity called the com- mon difference. 219. When the common difference is positive, the series is called an ascending series, or an ascending progression ; when the common difference is negative, a descending se- ries. Thus, a, a -\- d, a -{- 2d, a-\- 3d, &c. is an ascending arithmetical series in which the common difference is d ; and a, a — d, a — 2d, a — 3d, &c. is a descending arithmetical series in which the common difference is — d. 220. In Arithmetical Progression there are five elements, any three of which being given, the other two can be found : — 1. The first term. 2. The last term. 216 ELEMENTAKY AL.GEBKA. 3. The common difference. 4. The number of terms. 5. The sum of all the terms. 221 1 Twenty cases may arise in Arithmetical Progres- sion. In discussing this subject we shall let a = the first term, I = the last term, d = the common difierence, n = the number of terms, (S':= the sum. of all the terms. CASE I. 222. The first term, common difference, and number of terms given, to find the last term. In this Case a, d, and n are given, and I is required. The suc- cessive terms of the series are a, a-j- d, a-\- 2d, a-{- 3d, a -\- id, &c. ; that is, the coefficient of d in each term is one less than the number of that term, counting from the left ; therefore the last or nth term in the series is a-\-(n — l)d or 1= a -\- (n — 1) d in which the series is ascending or descending according as d is posi- tive or negative. Hence, RULE. To the first term add the product formed by multiplying the common difference by the number of terms less one. 1. Given a = 4, rf:= 2, and re = 9, to find I. l=a-\-{n — \) rf = 4 + (9 — 1) 2 = 20, Ans. 2. Given a = T, tf = 3, and n = 19, to find I. Ans. ^=61. PROGRESSION. 217 3. Given a = 29, d = rrrr 2, and n = 14, to find I. Ans. ^ = 3. 4. Given a = 4.0, d =z 10, and n = 100, to find /. 5. Given a = 1, rf = i, and n = It, to find I. 6. Given a = J, d = — ^i^, and n = 13, to find I. 1. Given a = .01, d = — .001, and « = 10, to find I. CASE II. 223i The extremes and the number of terms given, to find the sum of the series. In this Case a, I, and n are given, and S Is required. Now S=a + {a + d)+ia + 2d) +{a. + 3d) + + 1 or, inTerting the series, S = I + {l — d) + (l — 2d) +ll — 3d) + +(i Adding these together, 2 S = (a + «) + (a + i) + (a + i) + (a + i) + . . . . . . -f- (a + Z) And since (a -j- Z) is to be taken as many times as there are terms, hence 2S = n (a -{- I) or S = - (a -[- /). Hence, RULE. Find one half the product of the sum of the extremes and the number of terms. Note. — If in place of the last term the common difference is given, the last term must first be found by the Kule in Case I. 1. Given a = 3, Z = 141, and « = 26, to find S. 5^ = ^ (a + Z) = y (3 + 141) = 1872, Ans. 2. Given a = i, Z ^ 25, and n = 63, to find S. Ans. ^=793f. 3. Given a = 4, S'IkI '' = — ^< to find S. Ans. S=^£^. CASE III. 235. The first term, ratio, and number of terms given, to find the sum of the series. In this Case a, r, and n are given, and 5 required. The last term can be found by Case I., and then the sum of the series by Case II. Or better, since Ir = ar^ Substituting this value of Z r in the formula in Case II. we have r" 1 S = X "■ Hence, r — 1 228 ELEMENTARY ALGEBRA. RULE. From the ratio raised to a power whose index is equal to the number of terms subtract one, divide the remainder by the ratio less one, and multiply the quotient by the first term. 1. Given a = 4, r :^ 7, and n = 5, to find S. S='^^^X a= ^^ X 4=11204, Ans. 2. Given a = ^, r = 5, and n = 6, to find S. Ans. S^= 558. 3. Given a = ^, »• = |-, and w ^ 1, to find »S^. Ans. S=:iU. 4. Given a ^^ — 5, r ^ — 4, and w = 4, to find S. Ans. ;S'=255. 6. Given a ^ — -J, r ^ 6, and n = 5, to find S. 6. Given a ^ §, r = — 3, and n = Q, to find S. t. Given a ^ — f , r := 2, and ra = 8, to find S. 236t In a geometrical series whose ratio is a proper frac- tion the greater the number of terms, the less, numeri- cally, the last term. If the number of terms is infinite, the last term must be infinitesimal ; and in finding the sum of such a series the last term may be considered as noth- ing. Therefore, vsrhen the number of terms is infinite, the formula »S^= becomes S = Hence, to find the sum of a geometrical series whose ra- tio is a proper fraction and number of terms infinite, RULE. Divide the first term by one minus the ratio. PEOGRESSION. 229 1. Find the sum of the series 1, |, J, &c. to infinity. '5 = 1^ = 1^ = 2, Ans. 2. Find the sum of the series %, I, j\, &c. to infinity. 3. Find the sum of the series -, -^, -, &c. to infinity. Ans. -^%. Ans. c — 1 4. Find the sum of the series 6, 4, 2§, &c. to infinity. Ans. 18. 5. Find the value of the decimal .4444, &c. to infinity. Note. — This decimal can be written -jty -|- ^^ -|- -^^^, &c. Ans. |. 6. Find the value of .324324, &c. to infinity. t. Find the value of .32143214, &c. to infinity. CASE IV. 237i The extremes and number of terms given, to find the ratio. In this Case a, I, and « are given, and r is required. From Case I. l=ar'"-'^ Whence, r= »/-. Hence, RULE. Divide the last term hy the first, and extract that root of the quotient whose index is one less than the number of terms. I. Given a =: Y, 1= 56Y, and n = 5, to find r. »-VT »-V2 4/567 „ . 2. Given a = 6f , l = i, and n — 6, to find r. Ans. r ^ ^. 230 ELEMENTARY ALGEBRA. 3. Given a = — I, I ^ 31^, and n :^ A, to find c. Ans. r =: — 5. Note. — This rule enables us to insert any number of geometri- cal means between two numbers ; for the number of terms is two greater than the number of means. Hence, if m ^ the number of m+l/J means, vi -\- 2 ^ n, or m -\- 1 = n — 1 ; and r = i /-. Having found the ratio, the means are found by multiplying the first term by the ratio, by its square, its cube, &c. 4. Find three geometrical means between 2 and 512. Ans. 8, 32, 128. 5. Find four geometrical means between 3 and 30'72. Ans. 12, 48, 192, t68. 6. Find three geometrical means between 1 and y\. Ans. i, i, f Note. — When m = 1, the formula becomes Multiplying by a, ar^a . /- = . /a I. But ar is the second term of a series whase first term is a and ratio r ; or the geometrical mean of the series a, a r, a r^. Hence, the geometrical mean between two quantities is the square root of their product. T. Find the geometrical mean between 8 and 18. Ans. 12. 8. Find the geometrical mean between f and 343. Ans. 7. 9. Find the geometrical mean between | and -j-^-g. 10. Find the geometrical mean between — f and — ^j\j. PROGRESSION. 231 238. From the formulas established in Arts. 233 and 234, l=:ar"-^ (1) can be derived formulas for all the Cases in Geometrical Progression. From (1) we can obtain the value of any one of the four teriiis, /, a, n, or r, when the other three are given; from (2), the value of 5, I, r, or a, when the other three are given. Formulas for the remaining twelve Cases which may arise are derived by combining the formulas (1) and (2) so as to eliminate that one of the two unknown terms whose value is not sought. 1. Find the formula for the value of S, when I, n, and r are given. From (1). ^1 = « r — l Substituting this value of a in (2), »S = Note. — The four formulas for the value of n cannot be derived or used without a knowledge of logarithms ; and four others, when n ex- ceeds 2, cannot be reduced without a knowledge of equations that can- not be reduced by any rules given in this book. 239. To find any one of the five elements when three others are given. RULE. Substitute in that one of the formulas (1) or (2) that con- tains the four elements, viz. the three given and the one re- quired, the given values, and reduce the resulting equation. If neither formula contains the four elements, derive a for- mula that will contain them, then substitute and reduce the resulting equation ; or substitute the given values before deriv- ing the formula, then eliminate the superfluous element and reduce the resulting equation. 232 ELEMENTARY ALGEBRA. 1. Given r = 3, ra == 5, and ^=726, to find I. l = ar''- (1) *=^ (2) l=Sla (3) m = "7° i^) I 81=" (5) a = 3 Z — 1452 (6) 3 1 — 1452 = I '81 0) 242? = ■ 1452 X 81 (8) 1 = :486 (9) Substituting the given values of r, n, and S in (1) and (2), we obtain (3) and (4) ; finding the value of a, the superfluous element, from (3) and (4), and putting these values equal to each other, we form (7), an equation containing but one unknown quantity. Ke- ducing (7) we obtain (9), oi I = 486. 2. Given a = 4, r = 5, and S= 15624, to find I. Ans. Z= 12500. 3. Given a = 2, w = 5, and ?= 512, to find S. ' Ans. /S— 682. 4. Find the formula for the value of a, when r, n, and -S are given. Ans. a = ^P'V*- j-n — 1 5. A gentleman purchased a house, agreeing to pay one dollar if there vs^as but one window, two dollars if there were two windows, four if there were three, and so on> doubling the price for every window. There were 14 windows. How much must he pay? Ans. $8192. 6. A man found that a grain of wheat that he had sown had produced 10 grains. Now if he sows the 10 grains the next year, and continues each year to sow all that is produced, and it increases each year in tenfold ratio, how many grains will there be in the seventh harvest, and how many in all? . f In Tth harvest, 10000000 grains. tin all, 11111111 grains. PEOGEESSION. 233 PEOBLEMS TO WHICH THE FORMULAS DO NOT DIRECTLY APPLY. 240i In solving Problems in Geometrical Progression, if we let X = the first term and y = the ratio, the series will be X, xy, xf, xf, &c. It will often be found more convenient to represent the series in one of the following methods : — ' 1st. When the number of terms is odd, - in or n ' for three terms: ^\ xy. t ) A xy, f. t X for five terms. 2d. When the number of terms is even, 3? X, y. t X for four terms; 3? X-, y> t x' -„ for six terms. ST* Which method is most convenient in any case will de- pend upon the conditions that are given in the problem. 1. There are three numbers in geometrical progression, the greatest of which exceeds the least by 32 ; and the difference pf the squares of the greatest and least is to the sum of the squares of the three as 80 : 91. What are the numbers ? SOLUTION. Let X, xy, and xy^ represent the series. Then a;/ — 4=32(1) j?3/*— a^:i= + 2^/ + ar'y=80:91 (2) 3/»— 1:14-/ + / = 80:91 (3) 91j^_91 = 80 + 80j/=+802^ (4) ll2,*_80 2f'=171 (5) x=i (7) 3/ =3 (6) 234 ELEMENTARY ALGEBRA. Dividing the first couplet of (2) by x', we obtain (3) ; from (3) we form (4), which reduced gives (6), or y =^ S. Substituting the value of y in (1), we obtain (7), or x = 4:. Ans. 4, 12, 36. 2. The sum of three numbers in geometrical progres- sion is 39, and the sum of their squares 819. What are the numbers ? SOLUTION. Let x,s/xy, and y represent the series. Then :r + V^+y=39 (1) x^ + xy + ^= = 819 (2) a: — >v/^ + y = 21 (3) Ix-^-ly — m (4) a^+ y = 30 (5) 2\/'iy=18 (6) xy = %\ a) Dividing (2) by (1), we obtain (3) ; adding (3) to (1), we ob- tain (4), which reduced gives (5) ; subtracting (3) from (1), we obtain (6), which reduced gives (7). Combining (5) and (7) as the sum and product are combined in Example 1, Art. 188, we obtain a; = 27 and y = 3. Ans. 3, 9, 2t. 3. Of four numbers in geometrical progression the dif- ference between the fourth and second is 60 ; and the sum of the extremes is to the sum of the means as 13 : 4. What are the numbers ? SOLUTION. Let X, xy, xy^, and xy" represent the series. Then xf- 64a;- xy = QQ (1) a;^' -|- X : xy^ -[- xy ^ 13 : 4 (2) 2/^ — y + 1 : y = 13 : 4 (3) 4y^ — 42/ + 4 = 13y (4) 4a; = 60 (Y) 4y^ — ny = — 4 (5) 0^=1 (8) y = 4. (6) PEOGKESSION. 235 Dividing the first couplet of (2) by a; 2; -|- x, we obtain (3) ; from (3) we form (4), which reduced gives (6), or y = i. Substituting this value of y in (1) and reducing, we obtain (8), or a; = 1. Ans. 1, 4, 16, 64. 4. Of four numbers in geometrical progression the sum of the first two is 10 and of the last two 160. What are the numbers ? Ans. 2, 8, 32, 128. 5. A man paid a debt of $310 at three payments. The several amounts paid formed a geometrical series, and the last payment exceeded the first by $240. What were the several payments? Ans. $10, $50, $250. 6. In the series x, a/x^, and y what is the ratio? A... ^|. I. In the series -, x, y, and — what is the ratio ? y ^ 8. There are four numbers in geometrical progression whose continued product is 64 ; and the sum of the series is to the sum of the means as 6 : 2. What are the num- bers? Ans. 1, 2,4, 8. 9. There are five numbers in geometrical progression; the sum of the first four is 156, and the sum of the last four T80. What are the numbers? 10. There are three numbers in geometrical progression whose sum is 126 ; and the sum of the extremes is to the mean as 17 : 4. What are the numbers ? II. The sum of the squares of three numbers in geo- metrical progression is 2215 ; and the sum of the ex- tremes is 35 more than the mean. What are the numbers ? 12. Of four numbers in geometrical progression the sum of the first and third is 52 ; and the difference of the means is to the difiference of the extremes as 5 : 31. What are the numbers ? 236 ELEMENTARY ALGEBRA.' SECTION XXIV. MISCELLANEOUS EXAMPLES. 1. From 6ac~ 5ab-\-c^ take Sac — [Sab — (c — c^) + Tc}. Ans. 3ac — 2a6-4- 2c2_]_6c. 2. Eeduce x^i/^ — ( — xi/^-^-x" ) ^y — ^ \ — jy' ■ — y (xy — a;^) } ) to its simplest form. Ans. Ix^'i^ -\- x^. 3. Eeduce {a — b -^ cY — (a (c — a — h) — {6 (a + J -\- c) — c {a — b — c) } j to its simplest form. Ans. 2 (a= + J'' + c=). 4. Eeduce (x -\- a) a -[- y — { (y + *) {^ -{• V) — y {x ~\- a — 1) — {x -\- y) (6 — a) I to its simplest form. Ans. a^ — 5^ 5. Eeduce {a^ — W) c — (a —■ b) [a (b-\-c) —b (a — c)} to its simplest form. Ans. 0. 6. Eeduce (a -{- b) x— (b — c) c— \{b — x) b— {b — c) {b-{-c)\ — ax to its simplest form. Ans. 2 6a; — be. 1. Multiply o' + 2 a^ i — 3 a 4^ by — (— 3 a' 6 + a^ J2). 8. Multiply a*-\-6a^-\-9 by a* — 6 a'^ + 9. 9. Multiply a -{- b — c by a — b -{- c. 10. Divide 28a2 _ 6a» — 6a« — 4a^ — 96 a + 264 by 3a2 — 4a + ll. 11. Divide 1 — 18x2 _^ 81 a:* by 1 + 6a: + 9 a;'^. 12 Divide 9 a^ + 1 _ 4 a* — 6 a by 1 -f 2 a^ — 3 a. 13. Divide 9x'>—1x''f-^2y^hy Sx* + 2x^y— y\ MISCELLANEOUS EXAMPLES. 237 14. Divide 23 a — 30 — T a' + 6 a^ by 3 o — 2 a^ — 5. 15. Find the prime factors of a* — ¥. 16. Find the prime factors of 4 m' m^ — 49 m* n^". VI. Find the prime factors of a;'^ — ^xy-i^y^. 18. Find the prime factors of a;' — j^. 19. Find the greatest common divisor of 5 a;° — XQx^y -{-Ibff and 4 a;' -|- 8 a;^ y -|- 8 a; j/^ -|- 4 /. Ans. x -\-y. 20. Find the greatest common divisor of 8 a 6^ -(- 24 a J* -f 16 a J and 1 5= + 7 i^ + 7 6* — U^ Ans. ¥ + 6. 21. Find the greatest common divisor of 6 ai" -)- T a; y — 3y-^ and 12 a;^ + 22 xy + 6 y''. 22. Find the greatest common divisor oi ^x -\- 4ca? — 40 and Zx^y — ^^y. Ans. x — 2. 23. Reduce -. — ... . , „ ■ , .^s to its lowest terms. (a — 6) (a^-^ 2ab -\-lr) 24. Reduce , „ to its lowest terms. a — o 25. Reduce 7-1-, ,x / 7" %. j— rC to its lowest terms. 26. Find the least common denominator and reduce ^ — 5 to a single fraction. I — a 1 + a 1 + 0" 1 — a" ® ^ Ans. 21. Find the least commou denominator and reduce -- '- . — - — i — ; to a single fraction. 1 — m' \ -{-■ne ° 28. Find the least common denominator and reduce " Tl — ,- — 1 ^ ° ^ o .^ to a single fraction. 4a^ — 3a6 16a*— 9a'6^ ^ 29. Reduce to one fraction with the least possible de- a l^ — a^-\-ah 3b — a , c nominator ;- 7 — r j r aT; ' 6 bed cd ' bd 238 ELEMENTARY ALGKBEA. 30. Eeduce to one fraction with the least possible de- . ^ a4-b i-f-c a-\-c (6 — c) (c — a) (a ^ — b) 31. Find the least common denominator and reduce 3 + 2a; /2 — 3x (16— a;)a;\ , . , » ,. — -! I — — i ^^— = J— ) to a Single fraction. . 1 Ans. 2 + a; 32. Eeduce to one fraction with the least possible de- . ^ lA-x 4x 1 — X . 2a; + 6x' nominator ^^—^-^—^-^5-^-^,. Ans. -^3^,. 33. Eeduce a — c ^^ — to its simplest form. 34. Eeduce ^^ to its simplest form. 35. Eeduce -^„ \- x and — ^ 2y each to a X — 2^' X -\- 2y " single fraction and find their product. Ans. 4^ — 3? 36. Subtract -^^— from X y — X X +y 37. Subtract ZxA-^ from x^ ' o Multiply (-i^y by ^-j: a^ar* 39. Divide by -^ -5, and multiply the result by a^. 40. Divide ^TZTg-iT+l^ ^y -^^• 41. Divide ^ by f^-, + J^\. m-\-n •' \a-\-b ' a — 0/ MISCELLANEOUS EXAMPLES. 239 42. Divide V°'-° ^> by -^^. 43. Divide — ■ -. — by — ■ ; — , and eive a — X ' a-\-x ^ a — x a -\- x' ^ • ■ 1 C? I 3^ the answer in its lowest terms. Ans. ■ — — — 2ax 44. Eeduce ^—r = " , . What is the value of a a -\- a — b X, if o = — 2 and 5 = 3? 45. Eeduce x ^^ = 7 H s 46. Reduce (a -[- a;) (5 — a;) — a{h — c) — b What is the value of a;, if a = 2, J = — 3, and c ^^ ^ 1. 47. Eeduce — =■ — = • "What is the value of a X, if a = 2, 6 = — 1, and c = 3 ? 48. Reduce a — !-i^ = 0. 49. Find the value of x in the equation x = — r^ — ~ — - in its simplest form. a — i> a + b 50. A man spends $2. He then borrows as much money as he has left, and again spends $2. Then borrowing again as much money as he has left, he again spends $2, and then has nothing left. How much money did he have at first ? 51. If 5 is subtracted from a certain number, two thirds of the remainder will be 40. WTiat is the number ? 52. Having a certain sum of money in my pocket, I lost c dollars, and then spent one ath part of what re- mained and had left one Sth part of what I had at first. What was the original sum ? What does the answer be- come if o ^ 3, J ^ 9, and c ^ 5 ? 240 ELEMENTARY ALGEBRA. 53. If I buy a certain number of pouads of beef at $0.25 a pound, I shall have $0.25 left; but if 1 buy the same number of pounds of lard at $0.15 a pound, I shall have $1.25 left. How much money have I? 64. Divide 84 into three parts so that one third of the first, one fourth of the second, and one fifth of the third shall be equal. 55. In a certain orchard 25 more than one fourth of the trees are apple trees, 2 less than one fifth are pear trees, and the rest, one sixth of the whole, are peach trees. How many trees are there in the orchard ? 56. A merchant spent each year for three years one third of the stock which he had at the beginning of the year; during the first year he gained $600, the second $500, and the third $400. At the end of the three years he had but two thirds of his original stock. What was his original stock ? 5T. From a cask of wine out of which a third part had leaked, 84 liters were drawn, and then the cask was half full. What is the capacity of the cask ? 58. A gentleman has two horses and a chaise. The chaise is worth n dollars more than the first horse and b dollars more than thesecond. Three fifths of the value of the first horse subtracted from the value of the chaise is the same as seven thirds of the value of the second horse subtracted from twice the value of the chaise. What is the value of the chaise and of each horse ? What are the answers if a ^ — 50 and J = 50 ? 59. A had twice as much money as B. A gained $30 and B lost $40. Then A gave B three tenths as much as B had left, and had left himself 20 per cent more than he had at first. How much did each have at first? MISCELLANEOUS EXAMPLES. 241 60. A mimber of men had done one third of a piece of work in 9 days, when 18 men were added and the work completed in 12 days. What was the original number of men? 61. A boatman can row down the middle of a river 14 miles in 2 hours and 20 minutes ; but though he keeps near the shore where the current is one half as swift as in the middle, it takes him 4 hours and 40 minutes to row back. What is the velocity of the water in the middle of the river ? Ans. 2 miles an hour. 62. A had three fifths as much money as B. A paid away $80 more than one third of his, and B f 50 less than four ninths of his, when A had left one third as much as B. What sum had each at first? 63. A farmer hired a man and his son for 20 days, agreeing to pay the man $3.50 a day and the son $1.25 for every day the son worked ; but if the son was idle, the farmer was to receive $0.50 a day for the son's board. For the 20 days' labor the man received $61. How many days did the son work ? 64. I purchased a square piece of land and a lot of three- inch pickets to fence it. I found that if I placed the pick- ets 3 inches apart, I should have 50 pickets left ; but if I placed the pickets 2^ inches apart, I must purchase 50 more. How much land and how many pickets did I pur- chase ? Ans. 18906:1 square feet and 1150 pickets. 65. A criminal having escaped from prison travelled 10 hours before his escape was discovered. He was then pursued and gained upon 3 miles an hour. When his pursuers had been on the way 8 hours, they met an ex- pressman going at the same rate as themselves, who had met the criminal 2 hours and 24 minutes before. In what time from the commencement of the pursuit will the crimi- nal be overtaken ? Ans. 20 hours. U P 242 ELEMENTAEY ALGEBRA. 66. In February, 1868, a man being asked the time, answered that the number of hours before the close of the month was exactly one sixth of 10 less than the number that had passed in the month. What was the exact time ? Ans. February 25th, 10 o'clock, p. m. 61. A and B owned adjoining lots of land whose areas were as 3 : 4. A sold to B 100 hectares of his, and after- ward purchased of B two fifths of B's entire lot ; and then the original ratio of their quantities of land had been re- versed. How much land did each own at first ? Ans. A, 300 ; B, 400 hectares. 68. A laborer was hired for 70 days ; for each daj'^ he wrought he was to receive $2.25, and for each day lie was idle he was to forfeit $0.75. At the end of the time he received $118.50. How many days did he work? 69. A sum of money was divided equally among a num- ber of persons by giving to the first $100 and one sixth of the remainder, then to the second $200 and one sixth of the remainder, then to the third $300 and one sixth of the remainder ; and so on. What was the sum divided and what the number of persons ? 70. A besieged garrison had a quantity of bread which would last 9 days if each man received two hectograms a day. At the end of the first day 800 men were lost in a sally, and it was found that each man could receive 2| hectograms a day for the remainder of the time. What was the original number of men ? 71. Find a fraction such that if 1 is added to the de- nominator its value will be J ; but if the denominator is divided by 3 and the numerator diminished by 3, its value will be f. 72. If 7 years are added to A's age, he will be twice as old as B ; but if 9 years are subtracted from B's age, he will be one third as old as A. What is the age of each ? MISCELLANEOUS EXAMPLES. 213 13. A, B, and C compare their fortunes. A says to B, "Give me $100 of your money, and I shall have twice as much as you retain." B says to 0, "Give me $1400, and I shall have three times as much as you retain." C says to A, " Give me $420, and I shall have five times as much as you retain." How much has each? 14. An artillery regiment had 39 soldiers to every 5 guns, arid 4 over, and the whole number of soldiers and officers was six times the number of guns and officers. But after a battle in which the disabled were one half of those left fit for duty, there lacked 4 of being 22 men to every 4 guns. How many guns, how many officers, and how many soldiers were there ? Ans. 120 guns, 44 officers, 940 soldiers. 15. A car containing 5 more cows than oxen was started from Springfield to Boston. The freight for 4 oxen was $2 more than the freight for 5 cows, and the freight for the whole would have amounted to $30; but at the end of half the journey 2 more oxen and 3 more cows were taken into the car, in consequence of which the freight of the whole was increased in the proportion of 6 to 5. What was the original number of cows and oxen, and what was the freight for each ? . j 9 cows and 4 oxen. I Freight for a cow, $2 ; for an ox, $3. 16. Two sums of money amounting together to $1600 were put at interest, the less sum at 2 per cent more than the other. If the interest of the greater sum had been in- creased 1 per cent, and the less diminished 1 per cent, the interest of the whole would have been increased one fif- teenth ; but if the interest of the greater had been increased 1 per cent while the interest of the other remained the same, the interest of the whole would have been increased one tenth. What were the sums, and the rates of interest? Ans. $1200 at 1 per cent ; $400 at 9 per cent. 244 ELEMENTARY ALGEBRA. 77. A and B can perform a piece of work together in 17f days. They work together 10 days, and then B fin- ishes the work alone in 16| days. How long would it take each to do the work? 78. The Emancipation Proclamation of President Lincoln was promulgated on the 1st day of January in a year rep- resented by a number that has the following properties: the second (hundred's) figure is equal to the sum of the third and fourth minus the first ; or to twice the sum of the first and fourth ; the third is a third part of the sum of the four ; and if 1818 is added to the number, the order of the figures will be inverted. What was the year ? 79. A and B can do a piece of work in a days ; A and C in h days ; B and C in c days. In how many days can each do it? 80. A can do a piece of work in a days, B in J days, and C in e days. In how many days can A and B together do it ? B and together ? A and C together ? All three together ? 81. A market-man bought some eggs for $0.28 a dozen, and sold some of them at 3 for 8 cents and some at 5 for 12 cents, receiving for the whole $6.24, and clearing $0.64. How many did he sell at each rate ? 82. One cask contains 56 liters of wine and 40 of water, and another 96 of wine and 16 of water. How many liters taken from each cask will make a mixture containing 52 liters of wine and 24 of water ? 83. A and B are travelling on roads which cross each other. When B is at the point of crossing, A has 720 me- ters to go before he arrives at this point, and in 4 minutes they are equally distant from this point ; and in 32 minutes more they are again equally distant from it. What is the rate of each? Ans. A's, 100; B's, 80 meters a minute. MISCELLANEOUS EXAMPLES. 245 84. Multiply a;"* by a^. 85. Multiply a:' by a;-*. 86. Divide f^ by y-''. 87. Divide cT^ by a=^. 88. Transfer the denominator of —f^i to the numerator. 89. Free _„ ^^; from negative exponents. 90. Expand (— 2 a')*. 91. Expand (a= J)"*. 92. Expand {—Zx-^fY. 93. Expand (a:" X a:^)'. 94. Expand {x — \/y)^ 95. Expand {a^ — IVf. 96. Expand (2a;— y^. 9T. Expand (2 a: — 3)*. 98. Expand (3 a — 2 J)'. 99. Find five terms of {x — y)» LOO. Expand (2 — a; — 2^)»- 101. Expand (3 — a — 5 + cf. 102. Find ^'IP. 103. Find /om + n Y a"*-" 104. Find s/ — 16a;«. 105. Find the square root of t — yla-^^x — 4j t/v + a^ — 4ffl-|-2aa;-f-4 — 4a:-l-a;''. 106. Eeduce 4^ 256 a* J' — 768 a° 6' c' to its simplest form 246 ELEMENTARY ALGEBRA. 107. Keduce ^1 ^ ^°^ V I *° equivalent radicals hav- ing a common index. 108. Add V^. ^/^> and Vil- 109. From ^HM take ^lO. 110. Multiply i>^f by i^J. 111. Divide — SV^O by v''5. 112. Divide -^"6 by ^^. 113. Find the cube of Z/s/Yx. 114. Find the square root of 6^3. 115. Multiply 6 + v/S by 3 — ^/'Z. 116. Expand (x^ — 2 \/x)^ in. Expand (y/^-J'-'y- 118. Expa..d(y/^-fy. 119. The area of a rectangular iield is 4 acres and 35 square rods ; and the sum of its length and breadth is equal to twice their difference. What are the length and breadth ? 120. Two travellers, A and B, set out to meet each other. They started at the same time and travelled on the direct road toward each other. On meeting it appeared that A had travelled 18 miles more than B, and that A could have travelled B's distance in 9 days, while it would have taken B 16 days to travel A's distance. How far did each travel ? Ans. A, 72 miles ; B, 54 miles. 121. Find three quantities such that the product of the first and second is a ; of the second and third, b ; and of the first and third, c. A I /<^c , I ah I b c Ans. ±^-^, ±^_, ±^_. MISCELLANEOUS EXAMPLES. 247 122. A and B invest in stocks. At the end of the year A sells his stocks for $108, gaining as much per bent as B invested ; B sold his for $49 more than he paid, gaining one fourth as much per cent as A. What sum did each invest?, Ans. A, $45; B, $140. 123. Eeduce ISa;^ — 33 a: — 40 = 0. X X -\-l 1 124. Eeduce X — I X 6 2 125. Eeduce g - y)" -(l-y) = ^-^. Ans.. y = — 2.1 ±3^1-4:9, or 5, or — f . 126. Eeduce {x* — a?-{- If -\- a? — 5704 + x^. Ans. a;— -±3, or ±2^—2, or ±w'l±''V— 3- ''1 12T. Podnno ^/^f-L.^-i-'>. i^/r — - Jteuuoe /^ ^J, -X- s. -^ ^i\ ^ d 2x + l 128. Eeduce f — 2 V/ — 3y + 5 = 3^? — 2. 129. _ , 2v''^— 5 /x — 2 Eeduce -^ = - — = — ^ x-\-2 - , to find x and «. (z* + / = 192n * 141. A drover sold a number of sheep that cost him $29t for ft each, gaining $3 more than 36 sheep cost him. How many sheep did he sell ? 142. A merchant sold a piece of cloth for ftS, gaining as much per cent as the piece cost him. What did it cost him 1 143. A drover bought 12 oxen and 20 cows for $920, buying one ox more for $160 than cows for $66. What did he pay a head for each ? 144. A started from C towards D and travelled 4 miles an hour. After A had been on the road %\ hours, B started from D towards C, and travelled every hour one fourteenth of the whole distance, and after he had been on the road as many hours as he travelled miles an hour, he met A. What was the distance from to D ? MISCELLANEOUS EXAMPLES. 249 145. A person bought a number of horses for $1404. If there had been 3 less, each would have cost him $39 more. What was the number of horses and the cost of each? 146. Find a number of. four figures which increase from left to right by a common difference 2, while the product of these figures is 384. Ans. 2468. 147. A rectangular garden 24 rods in length and 16 in breadth is surrounded by a walk of uniform breadth which contains 3996 square feet. What is the breadth of the walk? Ans. 3 feet. 148. A square field containing 144 ares has just within its borders a ditch of uniform breadth running entirely round the field and covering 381.44 centares of the area. What is the breadth of the ditch ? Ans. 0.8 meter. 149. A and B hired a pasture into which-A put 5 horses, and B as many as cost him $5.50 a week. If B had put in 4 more horses, he ought to have paid $6 a week. What was the price of the pasture a week ? Ans. $8. 150. A father dying left $3294 to be divided equally among his children. Had there been 3 children less, each would have received $183 more. Ilow many children were there ? 151. A merchant bought a quantity of tea for $66. If he had invested the same sum in coffee at a price $0.71 less a pound, he would have received 140 pounds more. How many pounds of tea did he buy ? 152. Find two quantities such that their sum, product, and the sum of their squares shall be equal to one an- other. Ans. i (3 ± V^=^) and i (3 T V— 3). 153. Find two numbers such that their product shall be 6, and the sum of their squares 13. 11* 250 ELEMENTARY ALGEBRA. 154. A and B talking of their ages find that the square of A's age plus twice the product of the ages of both is 3864 ; and four times this product, minus the square of B's age, is 3516. What is the age of each? Ans. A's, 42 ; B's, 25. 155. Find two numbers such that five times the square of the less minus the square of the greater shall be 20 ; and five times their product minus twice the square of the greater shall be 25. 156. A and B purchased a wood-lot containing 600 acres, each agreeing to pay $17500. Before paying for the lot, A ofiered to pay $20 an acre more than B, if B would consent to a division and give A his choice of situ- ation. How many acres should each receive, and at what price an acre ? Ans. A, 250 acres at $ TO an acre ; B, 350 at $50. 15t. A merchant bought two pieces of cloth for $115. For the first piece he paid as many dollars a yard as there were yards in both pieces ; for the second, as many dol- lars a yard as there were yards in the first more than in the second ; and the first piece cost six times as much as the second. What was the number of yards in each piece ? Ans. In 1st, 10 yards ; in 2d, 5. 158. Two sums of money amounting to $14300 were lent at such a rate of interest that the income from each was the same. But if the first part had been at the same rate as the second, the income from it would have been $532.90; and if the second part had been at the same rate as the first, the income from it would have been $490. What was the rate of interest of each? Ans. First, 1 per cent ; second, Yytr per cent. 159. Divide 29 into two such parts that their product will be to the sum of their squares as 198 : 445. MISCELLANEOUS EXAMPLKS. 251 160. What is the length and breadth of a rectangular field whose perimeter is 10 rods greater than a square field whose side is 50 rods, while its area is 250 square rods less than the area of the square field ? Ans. Length, 75 rods ; breadth, 30. 161. A rectangular piece of land was sold for $5 for every rod in its perimeter. If the same area had been in the form of a square, and sold in the same way, it would have brought $90 less; and a square field of the same perimeter would have contained 272|- square rods more. What were the length and breadth of the field ? Ans. Length, 49 ; breadth, 16 rods. 162. A starts from Springfield to Boston at the same time that B starts from Boston to Springfield. When they met, A had travelled 30 miles more than B, having gone as far in 1} days as B had during the whole time ; and at the same rate as before B would reach Springfield in 5f days. How far firom Boston did they meet ? Ans. 42 miles. 163. The product of two numbers is 90 ; and the dif- ference of their cubes is to the cube of their difierence as 13 : 3. What are the numbers ? 164. A and B start together from the same place and travel in the same direction. A travels the first day 25 kilometers, the second 22, and so on, travelling each day 3 kilometers less than on the preceding day, while B travels 14^ kilometers each day. In what time will the two be together again ? Ans. 8 days. 165. A starts from a certain point and travels 5 miles the first day, 1 the second, and so on, travelling each day 2 miles more than on the preceding day. B starts from the same point 3 days later and follows A at the rate of 20 miles a day. If they keep on in the same line, when will they be together ? Ans. S or 1 days after B starts. 252 KLEMENTARY ALGEBRA. 166. A gentleman offered his daughter on the day of her marriage $1000; or $1 on that day, $2 on the next, $3 on the next, and so on, for 60 days. The lady chose the first offer. How much did she gain, or lose, by her choice ? 167. The arithmetical mean of two numbers exceeds the geometrical mean by 2 ; and their product divided by their sum is 3|. What are the numbers ? 168. A father divided $130 among his four children in arithmetical progression. If he had given the eldest $25 more and the youngest but one $5 less, their shares would have been in geometrical progression. What was the share of each ? 169. The sum of the squares plus the product of two numbers is 133 ; and twice the arithmetical mean plus the geometrical mean is 19. What are the numbers ? 170. The sum of three numbers in geometrical progres- sion is 117 ; and the difference of the second and third minus the difference of the first and second is 36. What are the numbers ? 171. There are four numbers in geometrical progression, and the sum of the second and fourth is 60 ; and the sum of the extremes is to the sum of the means as 7 : 3. What are the numbers ? LOGARITHMS. 253 SECTION XXV. LOGAEITHMS. 241. Logarithms are exponents of the powers of some num- ber which is taken as a base. In the tables of logarithms in common use the number 10 is taken as the base, and aU. numbers are considered as powers of 10. By Arts. 119, 120, 10°^ 1, that is, the logarithm of 1 is 10^=10, " " 10 " 1 10^=100, " " 100 " 2 10»=1000, " " 1000 " 3 &c., &c., &c. Therefore, the logarithm of any number between 1 and 10 is between and 1, that is, is a fraction ; the logarithm of any number between 10 and 100 is between 1 and 2, that is, is 1 plus a fraction ; and the logarithm of any number between 100 and 1000 is 2 plus a fraction; and so on. By Art. 120, 10° = 1, that is, the logarithm of 1. is 10-1 = 0.1, " " 0.1 "—1 10-2=0.01, " " 0.01 " —2 10-»= 0.001, " " 0.001 " —3 &c., &c., ikc. Therefore, the logarithm of any number between 1 and 0.1 is between and — 1, that is, is — 1 plus a fraction ; the logarithm of any number between 0.1 and 0.01 is between — 1 and — 2, that is, is — 2 plus a fraction ; and so on. The logarithm of a number, therefore, is either an integer (which may be 0) positive or negative, or an integer positive or negative and a fraction, which is always positive. 254 ELEMENTARY ALGEBRA. The representation of the logarithms of all numbers less than a unit by a negative integer and a positive fraction is merely a matter of convenience. The integral part of a logarithm is called the characteristic, and the decimal part the mantissa. Thus, the characteristic of the logarithm 3.1784 is 3, and the mantissa .1784. 242« The characteristic of the logarithm of a number is not given in the tables, but can be supplied by the following RULE. The characteristic of the logarithm of any number is equal to the number of places by which its first significant figure cm the left is removed from units' place, the characteristic being posi- tive when this figure is to the left and negative when it is to the right of units' place. Thus, the logarithm of 59 is 1 plus a fraction ; that is, the characteristic of the logarithm of 59 is 1. The logarithm of 5417.7 is 3 plus a fraction ; that is, the characteristic of the logarithm of 5417.7 is 3. The logarithm of 0.3 is — 1 plus a fraction ; that is, the characteristic of the logarithm of 0.3 is — 1. The logarithm of 0.00017 is — 4 plus a fraction; that is, the characteristic of the logarithm of 0.00017 is — 4. 243. Since the base of this system of logarithms is 10, if any number is multiplied by 10, its logarithm will be in- creased by a unit (Art. 50) ; if divided by 10, diminished by a unit (Art. 54). That is, the log of 5549 " " 554.9 " 55.49 " " 5.549 " " .5649 " " .05549 .005549 being 3.7442 is 2.7442 1.7442 0.7442 1.7442 2.7442 3.7442 LOGARITHMS. 255 Hence, the mantissa of the logarithm of any set of figures is the same, wherever the decimal paint mxiy he. As only the characteristic is negative, the minus sign is ■written over the characteristic. TABLE OF LOGARITHMS. 244 • To find the logarithm of a number of two figures. Disregarding the decimal point, find the given number in the column N (pp. 268, 269), and directly opposite, in the column 0, is the mantissa of the logarithm, to -which must be prefixed the characteristic, according to the Rule in Art. 242. Thus, the log of 85 is 1.9294 « " 26 " 1.4150 The first figure of the mantissa, remaining the same for sev- eral successive numbers, is not repeated, but left to be supplied. Thus, the log of 83 is 1.9191 As, according to Art. 243, multiplying a number by 10 increases its logarithm by a unit, therefore, to find the loga- rithm of any number containing only two significant figures with one or more ciphers annexed, we use the same rule as above. Thus, the log of 850 is 2.9294 " 750000 " 5.8751 The principle just stated is applicable also in the cases that follow. 245. To find the logarithm of a number of three figures. Disregarding the decimal point, find the first two figures in the column N, and the third figure at the top of one of the columns. Opposite the first two figures, and in the column under the third figure, will be the last three figures of the deci- mal part of the logarithm, to which the first figure in the 256 ELEMENTARY ALGEBRA. column O is to be prefixed, and the characteristic, according to the Eule in Art. 242. Thus, the log of 295 is 2.4698 " 549 " 2.7396 In the columns 1, 2, 3, &c., a small cipher (o) or figure d) is sometimes placed below the first figure, to show that the figure which is to be prefixed from the column O has changed to the next larger number, and is to be found in the horizontal line directly below. Thus, the log of 7960 is 3.90Q9 " " 25900 " 4.4133 246i To find the logarithm of a number of more than three figures. On the right half of pages 268 and 269 are tables of Propor- tional Parts. The figures in any column of these tables are as many tenths of the average difierence of the ten logarithms in the same horizontal line as is denoted by the number at the top of the column. The decimal point in these differences is placed as though the mantissas were integral. 1. To find the logarithm of a number of four figures, find as before the logarithm of the first three figures ; to this, from the table of Proportional Parts, add the number stand- ing on the same horizontal line and directly under the fourth figure of the given number. Thus, to find the log of 5743. The log of 5740 is 3.7589 In " Proportional Parts," in the same line, under 3, " 2.3 Therefore, the log of 6743 " 3.7591 It is always best to find the logarithm of the nearest tabu- lated, number, and add or subtract, as the case may be, the correction from the table of Proportional Parts. LOGARITHMS. 257 Thus, to find the log of 6377. 6377 = 6380 — 3 The log of 6380 is 3.8048 correction for 3 " 2 Therefore, the log of 6377 " 3.8046 Whenever the fractional part omitted is larger than half the unit in the next place to the left, one is added to that figure. 2. For a fifth or sixth figure the correction is made in the same manner, only the point must be moved one place to the left for the fifth, two for the sixth, figure. Thus, to find the log of 3.6825. The log of 3.68 is 0.5658 correction for 2 " 2.4 " " 5 " .59 Therefore, the log of 3.6825 " 0.5661 To find the log of 112.82. 112.82 = 113 — 0.18 The log of 113 is 2.0531 correction for .18 is (3.8 + 3.02) " 6.82 Therefore, the log of 112.82 " 2.0524 The logarithm of a common fraction may best be found by reducing the fraction to a decimal, and then proceeding as above. 247. To find the number corresponding to a given log- arithm. Find, if possible, in the table the mantissa of the given logarithm. The three figures opposite in the column N, with the number at the head of the column in which the log- arithm is found, affixed, and the decimal point so placed as to make the number of integral figures correspond to the 258 ELEMENTARY ALGEBEA. characteristic of the given logarithm, as taught in Art. 242, will be the number required. Thus, The number corresponding to log 5.6378 is 345000 " 1.8745 " 74.9 If the mantissa of the logarithm cannot be exactly found, take the number corresponding to the mantissa nearest the given mantissa ; in the same horizontal line in the table of proportional parts find the figures which express the difference between this and the given mantissa ; at the top of the page, in the same vertical column, is the correction that belongs one place to the right of the number already taken, — to be added if the given mantissa is less, subtracted if greater. Thus, 1. To find the number corresponding to log 2.7660 next less log, 2.7657, and number corresponding, 583. difference, 3 correction, 0.4 Number required, 583.4 2. To find the number corresponding to log 3.8052 next greater log, 3.8055, and number corresponding, 0.00639 difference, 3 correction, 44 Number required, 0.0063856 The nearest number in the taWe of Proportional Parts to 3 is 2.7 ; corresponding to this at the top is 4, which belongs as a correction one place to the right of the number (0.00639) already taken, but 3 — 2.7 = 0.3 ; this, in like manner, gives a still further correction of 4, one place farther still to the right. The whole correction, there- fore, is 44, to be deducted as shown in the operation above. 3. Find the log of 3764. 4. Find the log of 2576000. 6. Find the log of 7.546. 6. Find the log of 0.0017. LOGARITHMS. 259 7. Find the log of ^. 8. Find the number to log 3.807873. 9. Find the number to log 1.820004:. 10. Find the Tiumber to log 2.982197. 11. Find the number to log 2.910037. 12. Find the number to log 4.850054. 248i The great utility of logarithms in arithmetical opera- tions is that addition takes the place of multiplication, .and subtraction of division, multiplication of involution, and di- vision of evolution. That is, to multiply numbers, we add their logarithms ; to divide, we subtract the logarithm of the divisor from that of the dividend ; to raise a number to any power, we multiply its logarithm by the exponent of that power; and to extract the root of any number, we divide its logarithm by the number expressing the root to be found. This is the same as multiplication and division of different powers of the same letter by each other, and involving and evolving powers of a single letter or quantity ; the number 10 takes the place of the gi-een letter, and the logarithms are cthe -exponents of 10. MULTTPLICATION BY LOGARITHMS. B, U L E . 249i Add the 'logarithms of the factors, and the svm wUl he the logarithm of the product (Art. 50). 1. Multiply 347.6 by 0.04752. Ans. 16.517. 2. Find the product of 0.5«8, 0.7496, 0.0846, and 1.728. Ans. 0.06224. (It must be carefully borne in mind that the mantissa of the logarithm is alwaifs positive.) 260 ELEMENTARY ALGEBIiA. 3. Multiply 0.00756 by 17.5. 0.00756 log 3.8785 17.5 " 1.2430 Product, 0.1323 " 1.1215 4. Multiply 0.0004756 by 1355. Although negative quantities have no logarithms (Art. 262), yet, since the numerical product is the same whether the fac- tors are positive or negative, we can use logarithms in multi- plying when one or more of the factors are negative, taking care to prefix to the product the proper sign according to Art. 48. When a factor is negative, to the logarithm which is used n is appended. 5. Multiply —0.7546 by 54.5. —0.7546 log 1.8777 n 0.00545 " 3.7364 Product, —0.004113 « 3.6141 n 6. Find the product of —0.017, 25, and —165.4. 7. Fmd the product of —14, -^7.643, and —0.004, Ans. —.428. DIVISION BY LOGAEITHMS. EULE. 250i From the logarithm of the dividend subtract the log- arithm of the divisor, and the remainder will he the logarithm, of the quotient (Art. 64). 1. Divide 78.46 by 0.00147. 78.46 log 1.8946 0.00147 " 3.1673 Quotient, 53374. « 4.7273 LOGAEITHMS. 261 2. Divide O.OOU by 756. 0.0014 log 3.1461 756 « 2.8785 Quotient, 0.000001852. " 6.2676 Negative numbers can be divided in the same manner as positive, taking care to prefix to the quotient the proper sign, according to Art. 53. 3. Divide 0.7478 by 0.00456. Ans. 164. 4. Divide 6000 by 0.00149. 5. Divide 0.00997 by 64.16. Ans. 0.0001554. 6. Divide —14.55 by 543. Ans. —0.0268. 7. Divide —465 by —19.45. Ans. 23.9. 251. Instead of subtracting one logarithm from another, it is sometimes more convenient to add what it lacks of 10, and from the sum rejept 10. The result is fividently the same. For a; — y =