*^^r%-' Myy\ ^'1' ^^ it; • .SC ■; S.V • (jx, I • r.^-^ LIBRARY OF THE University of California, GIFT OF Class ■ lff\ ) « «■? -. v^, ..- S* H ^hv ""^t- -i^/) ^4. ■ ii •/'•^^ fii.*^ Astronomy (Newcomb and Holden), v • '2 -50 The same, Briefer Course, ... i 40 //£yi//?K HQLl & CO., Publishers, New York. NEWGOMB'8 MATHEMATICAL COURSE A SCHOOL ALGEBRA BY SIMON E-EWOOMB Professor of Mathematics in the Johns Hopkins University NEW YORK HENRY HOLT AND COMPANY 1887 Copyright, 1882, HENRY HOLT & OO. PEEFACE. The guiding rule in the preparation of the present work has been to present the pupil with but one new idea at a time, and, by examples and exercises, to insure its assimilation be- fore passing to another. This system is carried out by a minute subdivision of all the algebraic processes, and the end kept in view is that in no case where it can be avoided shall the pupil have to go through a process of which he has not previously learned all the separate steps. As a part of the same plan, definitions are so far as practicable brought in only as they are wanted, the first exercises in indicated opera- tions are performed with numbers, and the pupil is set to work on exercises from the start. Correlative with, the system of subdivision is that of ex- tending the scope of the exercises so as to include not only the elementary operations of algebra, but their combinations and applications. It is hoped that the wider range of thought and expression in which the pupil is thus practised will be found to tell in his subsequent studies. As another part of the same general plan the subject has been divided into three separate courses. The First Course, which extends to Simple Equations, is intended to drill the student in all the fundamental processes by exercises which are, for the most part, of the simplest character. The varied exercises in algebraic expression which are scattered through this course form a feature to which the attention of instructors is especially solicited. In the Second Course the processes are combined and the whole subject is treated on a higher plane. The general arrangement of this course is the same as that of the Elemen- tary Course in the author's College Algebra. But the exer- cises are all different, and greater simplicity of treatment is aimed at. This course terminates witli Quadratic Equations. 183977 iv PREFACE. The Third Course consists of three supplementary chapters which, however, should be mastered before entering college. An attempt has been made to treat quadratic equations with such fulness as to avoid the usual necessity of reviewing that subject after entering college. In the preparation and use of such a work no question is more difficult than that of the extent to which rigorous de- monstrations of the rules and processes shall be introduced. At one extreme we have the old method, in which the teach- ing is purely mechanical; at the other, the modern demand that nothing be taught of which the reason is not fully ex- plained to and understood by the pupil. The latter method should of course be preferred, but we meet the insuperable difficulty that we are dealing with a subject of which the reasoning cannot be understood until the pupil is familiar with it. The rule adopted in the present case has been to present a proof, reason or explanation wherever it was thought it could be clearly mastered. Most teachers will, however, admit that long explanations of any kind weary the pupil more than they instruct him, and that the best course is to present the examples and exercises in such a form that their logical correctness shall gradually become evident without much further help. Were the author to make a suggestion respecting the system of teaching to be adopted, it would be to commence the study of algebra at least one year earlier than usual, and to devote the whole of that year to the first course, taking two or perhaps three short lessons a week. The habit of using algebraic symbols in working and thinking would thus V)ecome established before taking up the subject in its more difficult forms. TABLE OF CONTENTS, FIEST COURSE. ELEMENTARY PROCESSES OF ALGEBRA. Chapter I. Algebraic Language. PAOB Reference List of Algebraic Signs and Symbols 2 General Definitions 3 Algebraic Signs 4 Parentheses 6 Sign of Multiplication 7 " " Division 9 Symbols of Quantity 11 Powers and Exponents 13 Coefficients 15 Exercises in Algebraic Expression 16 Chapter II. Algebraic Operations. Section I. Definitions 19 11. Addition and Subtraction 19 III. Multiplication 27 IV. Division 34 Miscellaneous Exercises in Expression 41 Memoranda for Review 42 Chapter III. Algebraic Fractions. Section I. Multiplication and Division of Fractions 43 II. Reduction of Fractions 49 III. Aggregation and Dissection of Fractions 53 IV. Multiplication and Division by Fractions 57 Miscellaneous Exercises in Fractions 60 Memoranda for Review 62 Vi CONTENTS. Chapter IV. Simple Equations. PAGE Definitions .63 Axioms 64 Transposing Terms 65 Division of Equations ... 68 Multiplication of Equations 71 Problems leading to Simple Equations 73 Memoranda for Review 80 SECOND COUESE. ALGEBRA TO QUADRATIC EQUATIONS. Chapter I. Theory op Algebraic Signs. Use of Positive and Negative Signs, 83 Algebraic Addition 87 Subtraction 89 Rule of Signs in Multiplication 90 " " " '* Division 91 w «« « '* Fractions 92^ Chapter II. Operations with Compound Expressions. Section I. Preliminary Definitions and Principles 94 Principles of Algebraic Language 95 II. Clearing of Compound Parentheses 96 III. Multiplication 97 IV. Division and Factoring 106 V. Fractions 116 VI. Substitution 124 VII. Highest Common Divisor 125 Chapter III. Equations of the First Degree. Section I. Equations with One Unknown Quantity 131 II. " " Two Unknown Quantities 134 III. " " Three or More Unknown Quantities 139 Problems leading to Equations of the First Degree 143 Interpretation of Negative Results 149 Memoranda for Review 152 Chapter IV. Ratio and Proportion. Section I. Ratio 153 II. Proportion 158 Problems in Ratio and Proportion 163 Memoranda for Review 167 CONTENTS. Vii Chapter V. Powers and Roots. PAGE Section I. Powers and Roots of Monomials 168 Fractional Exponents 173 Negative Exponents 176 II. Powers and Roots of Polynomials 179 Square Root of a Polynomial 184 Square Roots of Numbers . . 188 III. Operations upon Irrational Expressions 192 To Complete the Square 207 Memoranda for Review 209 Chapter VI. Quadratic Equations. Section I. Pure Quadratic and Other Equations 211 II. Complete Quadratic Equations 216 III. Equations which may be solved like Quadratics 224 IV. Factoring Quadratic Expressions 227 V. Solution of Irrational Equations 232 VI. Quadratic Equations with Two Unknown Quantities. . . 235 VII. Imaginary Roots 240 Memoranda for Review 243 THIRD COURSE. PROGRESSIONS, VARIATION AND LOGARITHMS. Chapter I. Progressions. Section I. Arithmetical Progression 247 II. Geometrical Progression 253 Limit of the Sum of a Progression 258 Chapter II. Variation 263 Chapter III. Logarithms 271 Table of Four-Place Logarithms 278 APPENDIX. Supplementary Exercises. Factoring 281 Problems in Ratio and Proportion 284 Quadratic Equations 288 Progressions t 291 FIRST COURSE. THE ELEMENTARY PROCESSES OF ALGEBRA. EEFEEE^CE LIST OF ALGEBEAIC SIGNS AND SYMBOLS. r= RepresentSy sign that -f- Plus, sign of addition. — Minus, sign of subtraction. X Times, sign of multiplication. -ir Divided by, sign of division. Is to, or divided hy,' sign of ratio. : : So is, sign of equality of ratios. := Equals, sign of equality. ' two expressions are identically equal, a symbol stands for an expres- sion (§ 109). > Greater than. A ^ B means A is greater than B. < Less than, A < B means A is less than B. signs of aggregation, indicate that the included expression is to be treated as a single quantity. |/ Radical, indicates a root. "|/ nth. root. Continued, stand for any number of unwritten letters or terms. Because. Therefore, or hence. oc Infinity, represents a quantity increasing beyond all limits. ( ) Parentheses, — Vinculwn, SCHOOL ALGEBRA. CHAPTER I. THE ALGEBRAIC UNGUAGE. P\^' General Definitions, 1. Definition. Mathematics is the science which treats of the relations of magnitudes. 2. Def. A magnitude is that which can be divided into any number of separate parts. Example. Length, breadth, time and weight are magni- tudes. 3. Def. A quantity is a definite portion of any magnitude. Example. Any number of feet, miles, degrees, bushels, years or dollars is a quantity. 4. In arithmetic and algebra quantities are ex- pressed by means of aarotos? . /l ) ( i v vj ' ,^ ^ To^ To express a quantity by a Tminhrr tt take a cer- tain portion of the magnitude as a unit, and state how many of the units the quantity contains. Example. We may express the length of a string by say- ing that it is 7 feet long. This means that we take a certain length called a foot as a unit, and that the string is ec^ual to 7 of these units. 4 . ALGEBRAIC LANGUAGE. Algebraic Signs. 5, Sign of equality. — read equals.^ or is equal to., is the sign of equal- ity. It indicates that the quantity which precedes it is equal to the quantity which follows it. G. Signs of addition and subtraction. + read plus, is the sign of addition. It indicates that the quantity which foUows it is to be added to the quantity which precedes it. Def, The number resulting from the addition is called the sum. — read minus., is the sign of subtraction. It indi- cates that the quantity which follows it is to be sub- tracted from the quantity which precedes it. Def. The quantity from which we subtract i15 the minuend. The quantity subtracted is the subtrahend. The result is the remainder. Examples. The expression 9 + 4 = 13 means 4 added to 9 makes 13. 9-4 = 5 means 4 subtracted from 9 leaves 5. Remark. Numbers to be added or subtracted may be written in any order. Thus —5 + 7 is the same as 7 — 5, so that - 5 -f- 7 = 7 - 5 = 2. EXERCISES. 1. 15 + 7 = what? 2. 15 - 7 = what? 3. - 7 + 15 = what? 4. — 8 + 8 = what? 7. Positive and negative quantities. Def The signs 4- and — are called the algebraic signs. The sign + is called the positive sign. The sign — is called the negative sign. ALGEBRAIC SIGNS, 5 Def. Positive quantities are those which have the sign + before them. Negative quantities are those which are preceded by the sign — . 8. We may have any number of quantities con- nected by the signs + and — . Example. The expression 9-2-4+5 means 2 to be subtracted from 9, leaving 7; then 4 more to be subtracted, leaving 3; then 5 to be added, making 8. Hence we may write 9-2-4 + 5 + 12 -2 = 18. But when we have several quantities connected by the signs + and — it is generally easiest to add all the positive (Quanti- ties and all the negative quantities separately, and then subtract the sum of the negative from the sum of the positive quanti- ties. Example. The preceding expression is calculated thus: 9-2 5-4 26 12-2 - 8 26-8 18 "We add 9, 5, and 12, making 26. Then we add -2,-4, and making — 8. Then . 36 - 8 = 18. EXERCISES. Compute the values of the following expressions*. 1. 14 _ 3 _ 8 + 22 + 17. 2. 41 - 9 - 10 - 11 + 1. 3. 12 - 13 + 14 - 15 + 16. 4 1+2+3+4- 9. 6. 1 + 3 + 6 + 10 + 15. 6. 14- 4 — 9 _ 16 + 25. 7. 1 + 8 + 27 + 64 + 125. 8. 1 - 8 + 27 - 64 + 125. 9. 24 - 31 + 1 - 2 + 50. 10. 9 -|_ 19 _|_ 29 - 39 + 40. 11. _ 24 - 13 + 7 - 4 + 101. ALGEBRAIC LANGUAGE. 12. 9- 5+ 5- 9 + 14. 13. - 32 - 23 + 50 + 32 + 23. 14. 117 _ 13 - 31 - 17 - 56. 15. 1008 - 1008 - 500 + 650. 16. 1205 + 1336 - 296 - 1694. 17. 439 + 940 - 631 - 142. Parentheses, 9. When combinations of numbers are enclosed in parentheses the results to which they lead are to be treated as if they were single numbers or quantities. Example. 19 — (9 — 4) means that the quantity 9 — 4, that is 5, is to be subtracted from 19. The remainder is 14. Therefore 19 - (9 - 4) = 19 - 5 = 14. The expression 12 - (7 - 3) + (8 -2) - (9 - 6) means that 7 — 3, that is 4, is to be subtracted from 12; then 8 — 2, that is 6, is to be added; then 9 — 6, that is 3, is to be sub- tracted. We may write the expression thus: 12 - (7 - 3) + (8 - 2) - (9 - 6) = 12 - 4+ 6 - 3 = 11. EXERCISES. Compute the values of the following expressions: 1. 12 -(11 -10). 2. 17 - (13 - 4). 3. 17 -(13 + 4). 4. 46 + (19 -4). 6. 46 -(19 -4). 6. 13 + (14 + 15). 7. 27- (16-9-4). 8. 41 -(31 + 1 -7). 9. (101 - 99) + (1 + 8 + 27). 10. (34 + 13 - 9) - (34 - 13 - 9). 11. 1 + 3 + 6 - (1 - 3 + 6). 12. (1 + 3 + 6) - 1 - 3 + 6. 13. (9 - 8) - (8 - 7) + (7 - 6). 14. _ (9 _ 8) + 18 - (33 - 17). 15. (74 - 69) - (8 + 3 - 7) + 145. SIGI^'JS OF MVLTIFLICATION 7 16. (27 + 8) - (27 - 8) 4- (1 + 2 - 3) - (1 - 2 -h 3). 17. -(ll-7)-l-(134-T-19-0)-27+(lH-2-3+4-5). 18. (136 - 48) -f- (49 - 1 - 35) - (10 - 7 + 13). 19. (7»-ll+13)-(7 -11-4-13) +(13-11) - (11-13+4). 20. (746 - 614) -- (42 - 18 - 17) + 973-39-(973 - 39). 21. 8 + 1- (8- 1). 22. 8 + 2 - (8 - 2). 10. Signs of Multiplication. X read multiplied by, or times^ indicates tliat the numbers between which it stands are to be multiplied together. Example 1. 12 x 5 = 60 means 12 multiplied by 5 make 60. Ex. 2. (11 - - 3) ;< (4 + 2) means that the remamder, when 3 is subtracted from 11, is to be multiplied by the sum 4 + 2, that is 6. Therefore (11 - 3) X (4 + 2) = 8 X 6 = 48. Def. The quantity to be multiplied is called the multiplicand. The number by which it is multiplied is the multi- plier. The result is called the product. Multiplier and multiplicand are called factors. 11. Omission of the sign x. The sign X is gen- erally omitted, and when two quantities are written after each other without any sign between them it in- dicates that they are to be multiplied together. Between numbers a dot may be inserted. The reason for inserting the dot is that such a product as 3 X 5 would be mistaken for 25 (twenty-fire) if nothing were inserted. Examples. 2(5 + 3) means the sum of 5 and 3 multi- plied by 2. Therefore 2(5 + 3) =r 2 . 8 = 16. 3.7.2 means 3 times 7 times 2, which makes 42. 12. We may have any number of quantities mul- tiplied together. Any two of them are then to be S ALGEBtiAIC LAMJL'AGE. multiplied together ; this product multiplied by the third, this product multiplied by the fourth, this pro- duct again by the fifth, etc. Examples. 2. '3. 4=2x3x4 = 6x4 = 24. 3 . 5 . 2(1 + 3) (1 4- 5) = 3 . 5 . 2 .4 . 6 = 720. We may get this result thus : 3 . 5 = 15; 15 . 2 = 30 ; 30 X 4 = 120; 120 X 6 = 720. The multiphcations may be performed in any order with- out changing the result. EXERCISES. Calculate the values of the following expressions: 1. 4(9-2 + 3). Ans.40. 2. (9 - 6) (6 - 3). Ans. 9. 3. (5 -2) (10- 3). Ans. 21. 4. (7 - 4) (17 - 4). Ans. 39. 5. 3(13 - 7) (19- 14). Ans. 90. 6. (8 - 4) 4(9 - 5). Ans. 64. 7. (1-2+3) (4-5 + 6). 8. (1 + 2) (2 + 3) (3+4) (4+5). 9. 10(13 - 9) (5 - 1). 10. 17(17-1). 11. (13 + 7) (13 - 7) = 13 . 13 - 7 . 7. 12. (8 + 5) (8 - 5) = 8 . 8 - 5 . 5. 13. (19 - 3) (19 - 3). 14. (14 + 4) (14 - 4). 15. (1 + 9) (1+10). 16. (3 + 4 + 5) (3 + 4 - 5). 17. (3 + 4 + 5) (3-4+5). 18. (3 - 4 + 5) (3 + 4 - 5). 19. (25 + 10 + 4) (5 - 2) = 5 . 5 . 5 - 2 . 2 . 2.- 20. 9(9 - 3) - 8(8 - 3) + 7(7 - 3). 21. (8 - 2) (7 - 3) - (6 - 3) (5 - 2). 22. (6.8-5.9) (2.3. 4-4. 5). 23. (7.8- 2.12) (5.6-4.7). 24. (2 . 3 . 4 - 20) (4 . 5 . 6 - 7 . 5 . 3). 25. 3(2.6-3.1) (4. 5 -2. 8). 26. 2(2 . 3 - 3 . 1) (4 . 5 - 2 . 7)5. SIOJVS OF DIVISION. 9 13. Signs of Division. -i- read divided by, indicates that the number which precedes it is to be divided by that which follows it. Examples. 15 -j- 3 = 5; 19 -f- 7 = ^ = 2f . Def. The quantity to be divided is called the dividend. That by which it is divided is the divisor. The result is called the quotient. 14. Division in algebra is commonly expressed by writing the divisor below the dividend with a line be- tween them, thus forming a common fraction, which fraction will be the quotient. Illustration. Let us show that the quotient 2 -i- 3 = f . This means that if we take two units and divide the sum of them into three equal parts, each of these parts will be f of the unit. I unit I unit 1 lilililililil Let us draw a hne as above, two inches in length, the inch being the unit. Divide each unit into three equal parts. Each of these parts will then be \. Since the two units have 6 parts in all, one third of the line will be 2 parts, that is f . EXERCISES. 1. Show in the same way that if we divide a line 3 units long into 5 equal parts, each part will be f of the unit. We first divide each unit into 5 parts, making 15 fifths. Dividing these 15 parts by 5 we shall have 3 parts, that is |, for the quotient. 2. Divide a line 5 units long into 3 parts, and show that each part is | of the unit. We divide each unit into thirds, making 15 thirds in all. 3. Divide a line 7 units long into 4 parts, and show that each part is J of the unit. . 9 + 20-2_27_ 7-4 ~ "3 ~ • 10 ALGUBBAIG LANGUAGE. 2.3(6 + 8) 2.3.14 _ (3 + 4) (5 - 3) 7.3 3 + 12 - (9 - 7) _ 15 - 2 _ 13 _ „, ^- 2"." 3 ~6~ - T - ^• Note, We obtain this result by multiplying the numerator of the fraction by the multiplier 7. It is explained in arithmetic that we multiply a fraction by multiplying its numerator. 13-4 + 9 8. 10. 9 X 11 -(1 + 2)2 • 14 - (14 - 4) + 17 1+2+4 • (2 + 1) (2 + 1) (1 + 2) (1 + 2-3 + 4)- (8 + 1) (1 + 8)8 + 1 8-1 12 2 l + (l + 2)2 + (l + 2 + 3)3 ^-^^ '^ • 2 (4 - 3 + 2 - 1) + 3 (3 - 2 + 1)* j3 1.3.5.7 14. 15. 16. 17. 18. 19. 2. 4.6.8* l + 7.4(7 + 4)(7- 4) ^ (4 + 7) + 7 8. 7.6.5' 8.7.6 .5.4 1. 2.3.4 • 1.2.3 .4.5' (2 . 2 + 3) (3 . 3 - 2) 7x7 13 .13 + 31 2, .5.5.2' (1 . 3) + (3 . 5) + (5 . 7) (2 . 4) + (4 . 6) + (6 . 8)- (1 . 1 . 1) + (2 . 2 . 2) + (3 . 3 . ;3) + (4 . 4 . 4) (1 + 2 + 3 + 4) X (1 + 2 + 3 + 4) 2.2.4.4.6.6 ^^' ^^1.3.3.5.5.7* REDUCTION TO NUMBERS. 11 Symbols of Quantity. 15. In algebra quantities are represented by the letters of the alj^habet. Def. The letters used to represent quantities are called symbols. 16. A quantity may be represented by any symbol we choose. The symbol may then be regarded as a 7imne for the quantity. Example. We may draw a straight line and call it a. The letter a is then the name of that line. But we might also have called it h, c, Xy or any other name we choose. The only restriction is that two quantities must not have the same symbol or name in the same question. 17. Def. The value of a symbol is the number or quantity which it represents. Example. In the preceding example the length of the line is the value of the symbol a. If I have the number 275 and call it n, then 275 is the value of n. Reduction of Algebraic Expressions to Numbers. 18. Def. An algebraic expression is any com- bination of algebraic symbols. Rule. To reduce an algebraic expression to mimbers we put, 171 place of the symbols, the numbers tvhich they represent, and perform the operations indicated. Example. Find the value of the following expression, en — ab supposmg 3n- -6c* a = 3, g = i^> 6 = 6, n = 15, 6- =7, m = 15. 12 ALGEBBAIC LANQUAOE. We do this as follows: c/i = 7 . 15 := 105; a^> = 3 . G = 18; hence en — ab — 87: 3;i = 3 . 15 = 45 66' = 6 . 7 = 42 hence 3?^ — 6c = 3 and then V- = ^9- Ans. EXERCISES. Using the above values of the symbols «, h, c, g and n, compute the values of: 2. a -\-h -\- c — g. 4. gc + «&. 6. an —{a -{-h-{- c). 8. ^7j -\- eg — be -\- ab, 10. 2aJc - 3^^ + 6a. bn -|- c^ 1. rt + ^ + 6'. 3. n — g — a. 5. ^c — «Z>. 7. abe — c(/. 9. 3«^ — 4:be -\- bcgn, en + ab 11. 13. 15. en — ab' aaa + bb 9<^ 2ng a^ b If a = 1, ^ values of: 17. ab\ abe; abed; abede. 19. ab [1 + c + cd (1 + e)] b — a a — e 23. ^« + ^>(c + ^)(e+/). 25. {a-\-b-{-e-^ d)e^f. 12. 14. 16. Z'/j — eg' Saaaa — 2bbb + aabb aa -\- bb — 9 _gn_ n a -\-b a 2, c = 3^ ^ = 4, e = 5, etc., then find the 18. ab + abc + abed + «Jcc?e. 22. (« -{-b)(e + d) {e -f/). 24. {a-^b)e-\-d(e-{-f). 26. ( 29. «ce 4^ - 7« 5^ - Qe 33. ^{1 + ^ [2 4-^(3+'/)] 32. 2<^ + « hdf \\^c{\-\-d)\. POWERS AND EXPONENTS. 13 34. Find the values of the following expression, for w = 0, then for n = 1, then for n = 2, etc., to n = 7: n{n -{-! ) { n ^ 2) 6" For w = 0, the expr. =0; for w = 1, the expr. =1; iov n = 2, the expr. = 4, etc. When n = 0, the expression = 0, " n = 1, " " = 1. " n = 2, " '' = 4. " n = 3, " " = 10. " ?i = 4, " " = 20. - w = 5, " " = 35. " n = 6, " " = 56. " n = 7, " " = 84. In like manner 'find the values of: 35. in{7i + 1), for 7i ^ 0, 1, 2, 3, 4, and 5. 36. ^n(n + 1) (j^''>i +1)^ foi" same values. 37. iw . 7i('/i -|- 1) (u 4- 1), foi' same values. 38. IT — ; — -, for same values. Zn-\- 1 39. Can you show that the quantity S -\- n-{- (S ~ n) is the same for all values of ^ ? Powers and Exponents. 19. Bef. A power of a quantity is the product obtained by taking the quantity as a factor any num- ber of times. Def. The degree of a power means the number of times the quantity is taken as a factor. If a number is to be raised to a power, the result may, in accordance with the rule for multiplication, be expressed by writing the number the required number of times. Examples. 5.5 = 25 is the second power of 5. bbi is the third power of b. xxxxx is the fifth power of x. To save repetition, the number whose power is to be ex- pressed is written only once, and the number of times it is taken as a factor is indicated by small figures written after and above it. 14 ALQEBBAIC LANGUAGE. Examples. Instead of 5.5 we write 5'. aa '' '' a\ " " UlU '' " v. Also, (1 -f 3)^ means (1 + 3)(1 + 3)(1 + 3) = 4 . 4 . 4 = 64. (2.3y '' 2.3.2.3 = 6.6 = 36. 20. Bef. The number wMcli indicates a power is called the exponent of that power. EXERCISES. Find the values of: 1. 2^; 4^; 2^ 3'; 5^^; 4^ ,3. 2^3^; (1-1-2)^(2 + 3); (l+2)(2 + 3)'[(l + 2)(2 + 3)p. 4. l-h4^: (1 + 4)^; (7-5)'-^; r_5^-*; (10 - 7) (9 - 6) (8 - 5) = 3^ (10+ 5)^ 10 + 5^ 10' + 2 . 5 . 10 + 5' ^- 5 ' 5 ' 5 • 6. 1 + 2 + 3^ 1 + (2 + 3)=^; (1 + 2 + 3)^ ^ r + 2^ 4-3^4- 4^ + 5^ 8, 3^-1 10. — „ ^^r -; 7.2^; (7.2)^; 7\ 2. (4 + 7)' + 11(7 + 4)- -(13 -2) (19 - 9)=^ + 2^ - 5^ - 7 12. (6 - 5)^ + (6 - 4)^ + (6 - 3)« + (6 -2)^ + (6-1)^=^^^)-'. 13. 7(8-1) (9 -2) (10 -3) (11 -4) ^ U. 87 - 3 . 5'; (87 - 3 . 5)'; 87 - 3\ 5. (1 + 2 + 3 + 4 + 5)' 2 • 1 + 2 + 2'^ + 2^ + 2^ ' + 3' 2^-1 1+3 + 3^^ + 3^ + 3* + 3' COEFFICIENTS. 15 Write the following expressions with exponents: 15. mmm. 16. ppppp. 17. (a + b) {a -\-h){a^ h), 18. {a - h) (a ^ b) {a + b). 19. xxyyzzyzz. 20. nn{7n + w) (m + ^). 21. (jo + ^) (g -\-p)rrr. 22. jfif(:?; - ?/) {^ -y){^- y)- 23. a^a^a; (a -\- x) (a -\- x). 24. ^/^y^ (/ + ^) (^ +»> 25. (a + J + c) (Z> + c + flj) (c + fl^ + b). Find the values of the following expressions when « = 5, b = S, m = 2, 71 = S: 26. {b - aym\ 27. (b - «)- 28. (b - my-\ 29. (b - nr- 30. (a - ny\ 31. (771 + 7l)^ (a + 7n)"' 32. (b - my 33. (a-m)^' ' b"" Coefficients. 21. Bef. The coefficient of an algebraic symbol is any number which multiplies it. Example. In the expression 3x -4:y -\- \%z, 3 is the coefficient of a:; 4 is the coefficient of y; 12 is the co- efficient of z. Use of Coefficients. If we call this line a, a then this line = 2«, — go this line = 3ff, ■ — 3^^^ and this line = 4^. Aa Remark. Compare this definition of coefficient with that of expo- nent of a power. a-\-a-\-a-\-a — ^a, and 4 is here the coefficient; also, a X a X fl^ X « = a\ and 4 is here the exponent. 22. A coefficient may be an algebraic symbol or a product of several symbols. Any quantity may be supposed to have the coeffi- cient 1. 16 ALGEBRAIC LANGUAGE. Examples. In the expression mx, rn is the coefficient of x. In the expression %7nax, 2ma is the coefficient of x; 2m is the coefficient of ax; and 2 is the coefficient of max, EXERCISES. In the expression 2amx + bcyz + mpqr. What is the coefficient of a;? What is the coefficient of mx^ What is the coefficient of 2;? What is the coefficient of yz? What is the coefficient ot mp? 23. Exercises in Algebraic Expression. Note. The object of the following exercises is to practise the student in algebraic expression. The answers are therefore to be expressed or indicated in the manner of algebra, instead of being calculated 1. How many dollars will 7 knives cost at S2 each? To get the cost we must multiply 2 by 7; this is done algebraically by writing 7.2; therefore the required cost is 7 . 2 dollars. 2. What will 7 pounds of tea cost at $2 per pound? 3. If I buy 9 sheep at 15 each, and 15 turkeys at $2 eacli, what is the total cost? 4. A man had $15, and spent $3; how many dollars had he left? 5. If a boy has a cents in one pocket and b cents in another, how many has he in both? Ans. a -{- h. 6. A man had x dollars, and paid out y dollars; how many had he left? 7. If I buy 7 cows at $45 each, and 3 horses at $175, express the total cost? Ans. $(7 . 45 + 3 . 175). 8. A huckster sold a chickens at x cents each, and lost m cents on his way home; how many cents had he left? 9. A train having to run a distance of m miles, went for 2 hours at the rate of 30 miles an hour; how much farther had it to go? EXERCISES m ALGEBRAIC EXPRESSION, 17 10. A man bought two pounds of tea for x cents; how- much would one pound cost? , x Ans. - cents. 11. If 8 pounds of tea cost %x, what is the cost per pound? Ans. %-. u 12. Three men took dinner at a hotel at a cost of m dol- lars for all three; how much had each to pay? 13. One boy has w cents and another n cents; if they make an equal division of the money, how much will each have? 14. A boy had h cents, and out of them gave c cents to one person and d cents to another; how many had he left? 15. A quantity m is to be multiplied by another quantity /^; express the product. 16. What will a -\- h sheep cost at x dollars apiece? 17. What will m -f- 'n pounds of beef cost at x -\- y cents a pound? 18. A man bought j(; dollars' worth of flour at q dollars a barrel; how many barrels were there? 19. How many cents are there in x dollars? How many dollars in x cents? 20. A man had x dollars in one pocket and x cents in another; how many dollars had he in both? How many cents? 21. A man went out with k dollars in his pockets, and paid out m cents; how many cents had he left? 22. What will be the value of 7i houses at x dollars apiece? 23. How many dollars will m pounds of beef cost at .^ cents per pound? 24. A man bought from his grocer a pounds of tea at .r cents per pound, and b pounds of sugar at y cents i)er pound, and c pounds of coffee at z cents per pound; how many cents would the bill amount to? How many dollars? 25. A man bought/ pounds of flour at m cents per pound, and handed the grocer an x dollar bill to be changed; how many cents ought he -to receive in change? How many dollars? 26. Of 3 boys one had n cents in each of his two pockets, another had p cents in each of his three i)0('kots, and the 1 8 AL a EBUA J L ' L. I ^'G UA GE. third had q cents ; if they make an equal division of the money, how much would each have? 27. A huckster had 3 large baskets, each containing p apples, and 7 small baskets, each containing q apples. He divided his ai)ples equally among the 10 baskets; how many were there in each basket? 28. The sum of the quantities 17 and 29 is to be divided by the sum of the quantities 5 and 3; what will be the quotient? 29. Tlie sum of the quantities a and h is to be divided by the sum of the quantities m and n-, what will be the quotient? 30. A man left to his children a bo'nds worth x dollars each and s acres of land worth y dollars an acre; but he owed m dollars to each of q creditors; what was the value of the estate? 31. Two numbers x and y are to be added together, their sum multiplied by 5, and the product divided by « + ^; ex- s(x + y) press the (quotient. Ans. , . . 32. The sum of the numbers p and q is to be divided by the sum of the numbers a and h, forming one quotient, and the difference of the numbers jt? and q is to be divided by the difference of the numbers a and b, forming another quotient; express the sum of the two quotients. 33. In the preceding exercise express the product result- ing from multiplying the sum of the quotients by m. 34. The quotient of x divided by a is to be subtracted from the quotient of y divided by h, and the remainder mul- tiplied by the sum of x and y divided by the difference be- tween /'and _?/; express these operations in algebraic language. 35. 'I'he number x is to be increased by n, the sum is to multiplied by a + h, and q is to be added to the product, and the sum is to be divided by rs', express the result. 36. The quotient of x divided by y is to be divided by the <|Uotient of a divided by h. 37. The quotient of x divided by y is to be added to the quotient of a divided by h, and the sum is to be divided by the sum of m and n. 38. X -\- y houses each had a -\- h rooms, and each room w -\- n pieces of furniture; how many pieces were there in all? ALGEBRAIC OPERATIONS. 1^ CHAPTER II. ALGEBRAIC OPERATIONS. Section I. Definitions. 34. Term. The terms of an expression are the parts connected by the signs + and — . Example. In the expression a-\-2b — 3c the parts a, 2b and 3c are the terms, there being three terms m all. Algebraic expressions are divided into monomials and polynomials. A monomial consists of a single term. A polynomial consists of more than one term. A binomial is a polynomial of two terms. A trinomial is a polynomial of three terms. Examples. The expression l^^abx is a monomial. %ab + omxy is a binomial. I -\- m -\- n and x^ -j- 2axy + y^ are trinomials. Factor. When any number of quantities are multi- plied to form a product, each of them is called a factor of the product. Equation. An equation is composed of two equal expressions with the symbol = between them. Example. 7 + 5 = 15 — 3 is an equation. Lilie Terms. Like terms are those which contain identical symbols and differ only in their numerical coefficients. Example, a, 2a and Ha are like terms; like; 9a:-}- 12a; are also like; but l^pq and 6qr are nnlike. Sectiot^ IL Addition and Subteaction. Addition of Positive Quantities. 25. The addition of any number of terms may be indicated by writing them down with the sign -f- be- tween them. *20 ALQEBBAIC OPERATIONS. Example. The sum of the quantities 2x, A:ax, 6b and bbx may be written 2x + 4:ax + 5^ + 6bx. If tlie terms are all U7ilike, this is the only way of express- ing addition of algebraic quantities. 26. Rule for Like Positive Terms. If like positive terms are to be added, take the swn of all the coefficients of the symbol and affix the symbol to their sum. In this addition a symbol without a coefficient must be considered as having the coefficient 1. Example 1. If we have to add together the quantities x, HXy 4tXy %x, 12x, we proceed thus: Ix 2x 4:X Sx Sum = 27:*; Ex. 2. x-{-x-^x-\-x-{-x = 5x. Ex. 3. am -\- 2a7n + 3am = 6am. EXERCISES. Add up the quantities: 1. 9a-{-7a-\-6a-\-Sa. 2. x -\- 2x + Sx -}- ^x. 3. 2y -\-dy + by. 4. "dab + ^ab + ab. 5. ab -\- ab -\- ab -\- ab. When there are several sets of similar terms, add each set, and connect the sums by the sign +• Example. Add a + 2x + 3« 4- lab -\- 6x -{- hab -\- x -\- 12a. Work: We pick out all the symbols a and write them under each other with their coeflBcients; then all the symbols «; then all the products a6. We then add the coefficients, thus formmg the s^"°- Sum = lea + 92: -f- 12rh -la-{-2x-{- tab 3a-{- 6x -\- bah 12a + Ix ADDITION. 21 6. Add lU + 4^ + l^abc 7. ah + 'de a-\-%b -\- mabc 22ab -\- I2e -^ d Sa -{- 6b -\- 12abc 29ab + e 8. {x + !/) + n^ + y) + H^ + y)' Ans. 16(0^ + y). 9. (a -' ^>) + 2{a - b) + 'S(a - b), 10. (rn — j») + O^i — P) + 3(m — jo) + 24:(m — p). 11. 3(m — /i)^ -j- 5(w — ^)^ + 7(/?i — ^O^'- 12. If Thomas has x dollars, John 3 times as many as Thomas, and James as many as John and Thomas together, express the number all three have in the language of algebra. 13. If a man owes his grocer b dollars, his tailor c dollars, his shoemaker 4 times as much as his grocer, and his butcher twice as much as his shoemaker, what is his total indebted- ness Addition of Negative Quantities. 27. Let us have to add T — 4, that is 3, to 10. This is the same thing as adding 7 and subtracting 4, because 10 + 7 — 4 is the same as 10 + 3, namely, 13. Whatsoever numbers a, b and e represent, if we add b — c to a the sum will be a -\- b — c. Hence Rule. //" negative qtianfities are among those to be added, they are to be subtracted from the sum of the others. Algebraic addition therefore means something more than addition in arithmetic or arithmetical addition, because it may include subtraction. 28, Def. Algebraic addition means the combi- nation of quantities according to their algebraic signs, the positive ones being added, and the negative ones subtracted. Bef. The result of algebraic addition is called the algebraic sum. Example. The algebraic sum of ~ 2—3 + 15 — 6 is 4. Def. Numerical addition and numerical sub- traction mean addition and subtraction as in arith- metic, without regard to the algebraic signs of the quantities. /^^^5>v 22 ALGEBBAIC OPERATIONS. 29. KuLE FOR Algebraic Additioi^. Take the mm of all the positive terms and the sum of all the negative terms. Subtract the less siom fro7n the greater and, if the negative Slim is the greater, lurite the sign — before the difference. Remark. If the positive and negative sums are equal, the total sum will be zero. Example 1. 4 - 2 + 3 - 8 = + 7 - 10 = - 3. Here the sum of the positive quantities is 7 and of the negative ones 10. In arithmetic we cannot subtract 10 from 7, but in algebra we ex- press this subtraction by taking 7 from 10 and writing — before the difference to indicate that tlie subtractive quantities are the greater. Ex. 2. Add 4.T -h %ay - 3z Ex. 3. Add 7x - 5y + 4 3x — 4:a?j — z — 2x — dg — S 2x — dag -{-4:Z — 5a; + 8^ + 4 Sum, 9x — 6ag Sum, EXERCISES. Add: 1. dx-4:X-^a-\- Tx + 5x -i-2b-\-3a- 7.?; - 6a. 2. 26?^ — 2y -^ (jg -\- Sag — Ig -\- lOxg + 6ag — 5xg. 3. 8^- + 9 - 3// + :x -12-\-Sg - 2x - 3- 7g. 4. 7?/ + 6?/ + 3rt - 9// - 2g ~ ba -{- 2g - 4^ + 2a. 5. 9x -\- axg -\- 3// — 6x — 2xg -}- 6g. 6. 9x — 3xg — l^x + ^xg — \x — Qx. 7. 4a - 2b, Oft - bb, 8a — lib, a + 75. 8. 4.^• — dg, 3x — bg, — x -\- g, — ^x + 4?/. 9. 5a + 35 + c, 3a + 35 + 3c, a + 35 + be. 10. 3X + 2g — z, 2x—2g-\- 2z, — x -]- 2g -i- 3z. 11. 7^f _ 4j _|_ c, 6a + 35 — be, — 12a + 4c. 12. .t- — 4a + 5, 32;-f25, a — .t — 55. 13. a + 5 — c, 5 + 6' — a, c + a — 5, a + 5 — c. 14. a 4- 25 + 3c, 2a — b — 2c, b — a — c, c — a — h. 15. ^_25+3c-4^, 35-4c+5fZ-2a, 5c-6^+3a--45. 16. a - 25 + c, 2a - 35 - 4c, 3c - 4^. 17. 2x + 3, 5^; + 7, 13:c +1, ^ - 4. 18. ax-\- 5, ca; + d, ex -\-f 19. m«/ -\^n, bag — n, 2n — lag. 20. a+54-c, a + 5 — c, a — 5 + c, — a+5 + «» 21. x^^2xg^g\ x'-2xg^y\ 2x^-2y\ 22. x^'-y^ g''-z\ z^-x\ SUBTRACTION. 23 23. llax — 19hy 13a; - c, lib — a-\- c, Hb - lOax. 24. 7« - 85 + 9c, 9b -7c- Sa, 9a - n ^ Sc. 26. a'+l, a'-\-l, a + 1, 1 - a, 1 - a\ 26. a -\-m, n -\-%a, p— 3a, Aa — q. 27. axi/ 4- P^^ + '^y + ^> 4/?2:* — 3«a;?/ + 9X; — Smy, l^axy — 5^.r^ — k. 28. m — 2/i, — 2m + 3/i — 4/?, 3w — 4?^ + ^j^ — 6§'? — 4m +5^ — 6jo + 7^ — 8r. 29. «:?;'+ %'— a'5% ax" — by"" + tt'Z*', — d^x -\- m — n, 30. « — 3^ + 2;^, 4« + 7y — 3;^, «/ + ^ — 5fl^. Subtraction. 30. Def. Subtraction consists in expressing the difference between two algebraic quantities. Example 1. Let us have to subtract 5 — 2 from 11. Since 5—2 =3, we must subtract 3 from 11, leaving 8. But this is the same thing as first subtracting 5 and adding 2. Hence the result is 11-5+2, and the sign of 2 is changed from — to -\-, Ex. 2. Take a — b from y. It is plain that if we subtract a from y we shall subtract b too much, because a — b i^ less than a. But if after sub- tracting a we add b, we shall make the answer right. Hence the answer is y -a-^b. In this answer the positive symbol a has the sign — in the remainder and the negative symbol b the sign +. We thus derive the following rule for subtraction: 31. Rule. Change the signs of all the terms to be sub- tracted, or imagine them to be changed, and then proceed as in addition, Numerical Examples. The learner should first practice on the purely numerical exercises until he sees how the rule leads to correct results. Each result should be proved by showing that the answer is right. From 21 + 10 Take 9-5 Rem. 12 + 15. 24 ALOEBBAIG OPERATIONS. Operation. Imagine 9 to have the sign minus. We must by § 29 subtract it from 21, leaving 12. Imagining 5 to have the sign +, we must add it to 10, making + 15. So the answer is 12 + 15. Proof. Minuend = 21 + 10 = 31 Subtrahend = 9 — 5 = 4 Remainder = 12 + 15 = which is right. From 10 + 6 10 + 6 10 + Take 9 9-2 9 - 37, 6 4 10+6 9- 6 Rem. 1 + 6 1 + From 25-3 Take 16 + 1 8 1 + 10 1 + 12 27-5 29-7 15 + 2 14 + 3 Rem. 9-4 12-7 31-9 13 + 4 33-4 12 + 5 35- 13 11 + 6 Algebraic Examples. From 3a; — ^ay + 55 + c Subtract x — lay + 85 + ^. We write the minuend, Zx — ^ay + 55 + c and subtrahend, with signs changed, — x -\- lay — Sb — d Result, applying Rule § 29, 2x + Say — 3b -\- c — d From a -\-b subtract b + y. « + 5 a-y EXERCISES. 1. From Ix — ^bxy — 12cy + 85 + Sac Take 2x + 75^:^ - Scy - 5b - 2d Difference, 6x — llbxy — 4:cy + 135 + Sac + 2d Note. After the beginner is able to operate by simply imagining the signs changed, he need not actually change them. 2. From 8« + 95 - 12c - 18c? — 45a; + Sexy Take 19a - 75 - 86- - 25^^ + 3a; - 4?/ SUBTBAGTION. 25 3. From 267z + 201;^' + d2y + S6ax - 6 Take UOz — S2z'' -\- 202/ -j- 92ax -{- U 4. From 8^^ + lU subtract 6a + 10^. 5. From 7a — 3b — c subtract 2a — 3b — 3c. 6. From 8^^ — 2^ + 3c subtract 4:a — 6b — c — 2d. 7. From 2x'' -8^-1 subtract 5x' - 6x -{- 3. 8. From 4a;* - 3x' - 2x' - 7x -{- 9 subtract x* - 2x' - 2x' + 7a; - 9. 9. From 2a;'' — 2ax -\- 3a^ subtract x^ — ax -\- a^. 10. From x"" — 3xy — y^ -\- yz — 2z^ subtract x" + 2xy + hxz - 3^' - 2z\ 11. From 5a;' + 6xy — 12xz — ^y'' — lyz — 6z' subtract 20;=^ - 'Txy + 4a;;2 - 3y' + 6yz - 6z\ 12. From a' - 3a'b + 3ab'' - b' subtract - 3ab'' + b\ 13. From 7a;'-- 2a;''+ 2a; + 2 subtract 4a;'- 2a;'- 2a; - 14. 14. From 6a'^b—6mn-\- 19xy subtract Ha^b-^-Smn-^-lOxy—c. 15. From 8{a-\-b)-i-12{m-^n) subtract 3(a+^)+5(m+70- 16. From 9{m^n) — 6{p -f- q) subtract 5(m + n) — 8(p-\-q) + 3 (a; + y). Ans. 4(m + ^) + 2{p + ^) — 3(a; + y). 17. From 9a + 12y + 16{a + y) subtract 12(a -f y) — 6y f 10^. 18. From 19^ + 23^ _ 18- + ^ subtract 19^ - 23^ b d y b d y 19. From a-\-b subtract a — b. 20. From x + 2y subtract x —. 2y. 21. From 2a; — 2y subtract x -\-2y. 22. From a -\- b subtract b -{- a. 23. From 2jo + a; subtract x -\- 2p. 24. From 7n -{- n -{- x subtract m — n -\- x. 25. From a — b subtract b — a. 26. From a -\- b — c take a — b -\- c. 27. From a — b -\- c take a-\-b — c. 28. From 1 -m-^-ni' take 1 -\- m -\- m\ 29. From a' - ab -\- V take b"" - 3ab + a\ 30. From 2' + 2; + m^i take z — z^ -\- nm. 31. From ab take Ja; from 4 . 7 take 7 . 4. 26 ALGEBBAIG OPERATIONS, Clearing of Parentheses. 33. Plus Sign before Parentheses, If a poly- nomial is enclosed between parentheses and preceded by the sign +, the parentheses may be removed and all the terms added without change. Example. 22 + (8 — 15 + 12) is the same as 22 + 8 — 15 + 12. Proof. 22 + (8 - 15 + 12) = 22 + 5 = 27 and 22 + 8-15 + 12 = 42 - 15 = 27. EXERCISES. Clear of parentheses and add: 1. 5^ - 3^ + 2c + (- 2a + 9^ - 3c) + {la - lU). 2. 36m +i? - g + (24m - 5^) + (- 42m + Sp). 3. a-^l)-\-{a+b-c)-\-{a-i-{-c). 4. a-\-'b-c-\-{a-l)-Yc)^{-a^h-\-c), 5. a-h-^{l)-c)-]-{c-a). 6. 2:^ - 4^/ + 2^ + (- ^x + 2?/ + %z). 7. m + ^ - 2j(? + (m - 2w -^p) + {- 2m + ^ -\-p). m a n b 9. m-\-^n-{- {7n - 2n) + {p -{- 2q) + (p - 2q), 10. x-{-y-z-^(y-\-z-x)-\-(z-^x-y). -fH-14-3' m ^ I f ^ _ ^ ] I ( ^ _ !!^ ' ~n b \b yJ\y ti 12. ax-by-\- (2by - 2ax) + {3by - 3fl^a;). 13. 277171 — 6xy -\- {577171 — 2xy). 33. Minus Sign before Parentheses. If the paren- theses are preceded by the sign — we remove them and change the sign of each enclosed term (§ 31). Note- Remember that if a term has no sign before it, then it is positive, and must by this rule be changed to negative. EXERCISES. 1. a -3b — {2b - 5a) -\- {m -\- Sa - b) — {a - m). Solution. Changing the signs by the rule, where the parentheses are preceded by the sign — , the result is TO MULTIPLY A MONOMIAL BY A NUMBER 27 a — db--2b-\-5a-{-m-\-da~b — a-{-7n. The coefficients of a are + 1 + 5 + 3 - 1 = +8. " 6 " -- 3 - 2 - 1 = - 6. « " " m " +l-]-l = +2. Therefore Ans. = 8a - 6^* + 2m. 2. x-3y - (27/ - 6x) -}- {z -]- 3x - y) ~ (x - z). 3. Qm + 9/i — (m — h) + [h — m) — (2m + 2h), 4. a - (- « + ^) - (- fl^ + c) - (- ^ + ^). 5. X — y — {y — x) — X -{- ^y- 6. « + Z> - 26? - (c - 2^> + fl^) - (^ - 2fl^ + c). 7. 2z — X — y — \z — 2y ^- x) — {y — 2x -{- z). 8. «a; — (dax -\- 2by) + 9<^^- 9. m — n — {a — h) — {c -[- d) — {e -\- g), 10. 2am + {3am — ?i) — (n — 3am) + 2n. 11. 77i + "Ih - (7A - 7/1') + {^k - ^^0- 12. 13px — qy — {~ qy — rz) — {rz + 2px). 13. 2«jo - 3bq — {Sap + SJ^^) + (Qap - c - d). 14. fl^ — (:r + ?/ — ;2; + ^^ — ^ + ^)* 15. i?; — (a; — ?/ + ?^ — w). Sectiot^ III. Mtjltiplicatiois". To Multiply a Monomial l>y a Number. 34. EuLE. Multiply the coefficient ^ and to the product affix the algebraic symbols of the multiplicand. Example 1. Multiply 3ax by 7. 3ax 7 Ans. 21ax Ex. 2. Multiply ab by 5. Here the coefficient of ab is 1 and the product is ^ab, EXERCISES. Multiply: 1. "laby by 8. 2. hmx by 6. 3. ^pqr by 7. 4. 22J(« + ^) by 2. Ans. 44J(a + x). 5. 3(6' + .v)m by 9. 6. 2(« + ^) by 3. 7. a + J by 7. Ans. 7(« + b). 8, iz; — ^ by 8. 9. 2(5 - c) by 12. 28 ALQEBKAIG OPERATIONS. To Multiply one Monomial by Another. 35. Rule. Multiply the coefficients, and affix to the pro- duct all the symbols loth of the multiplier and 7nultiplicand, Example. Multiply 12amx by lacy. Multiplicand, 12amx Multiplier, Hacy Product, S4:aacmxy = S4:a^cmxy Solution. 7 X 12 = 84 is the product of the coeflQcients, to which we affix the symbols. The symbol a being taken twice as a factor, we write a\ Multiply: 1. dab by 12mx. 2. 5amy by 11^5. 3. '7ahj Qamx. 4. 6{m -\- n) by 3w. 6. %(x-{-y)hj — c) by a. 15. a;?/ by 3(a; + y). 17. 5m/zjo by 4:mnq. 19. mp{a-\-b) by ^^'(a+J). 20. 6fl^^(a; + 1/) by 7«c(a; + y). 36. Z7se 0/ Exponents. Let us have to multiply By definition. Therefore Therefore a' = aaa; a^ = aa. a^ y a^ = aaa X «« = aaaaa = a' MULTIPLICATION OF MONOMIALS. 29 Hence EuLE. The exponents of like symbols in the factors must he added to form the exponent in the product. Note. In the following exercises the pupil may advantageously go through the above process in each case, until he fully understands reason for it. EXERCISES. Multiply: 1. x'y.x\ ^. (^+^rx(^+2/r. Atih. (x\ 3. a' X a\ 4. V X l\ 6. {a + hY X (« + l)\ 6. (m + nY X (m + n)\ 7. 2A* X 3/i\ Ans. Qh\ 8. 3m X 6m' Ans. 18m*. Where there is no exponent, the exponent 1 must be understood. 9. Wh' X %a'h. Ans. lQa'b\ 10. brn^n X 5mw^ 11. ^ahx X ba'lfxy, 12. Sa'^^V X ^a'hxy. 13. 86j'(m + 7i) X 9a'(m + t^)'. Ans. 72«X^ + ^)'« 14. 2(2; + ?/) X 3(.c + ^)'^«^. 15. 3(^ + ^r2:/x3(j9 + ^)^^. 16. 6(A + ^)%»?^ X {h + Jcym7i'x, 17. 2^c(^ + c) X ^V'(^ + c)^. 18. M'x(d + a;)'' X 2^2;'(^ + x)\ 19. 3(m + ^) (i? + qY X 2(m + ^^)' (p + ^). 20. 4(« + bY iff + ^)' X 6{a + J)* (^ + h). 37. C'flj^e 0/ more than Tzvo Factors. When there are three or more factors, multiply two of them, then multiply that product by the third, etc. , until all are multiplied. All the preceding rules may be combined in the following Rule for the Multiplicatiok of Moj^omials. Fonti the product of the numerical coefficients and to it affix all the symbolic factors, giving each the sum of its exponents in the several factors. 30 ALGEBRAIC OPERATIONS. EXERCISES. Multiply: 1. ab X ic X ca. Solution, a, h, c are the factors, and the sum of the ex- ponents of each is 2. Hence ah X he X ca = o^V&. Ans. 2. xy X yz X zx. 3. mn X np X pm. 4. ah X ac X ad. Ans. a^hcd. 5. 2mx X Smy X mz. Ans. Qm^xyz, 6. 2mx X dmy X Qmz. 7. a X ah X ahx X ahxy. Ans. a^h^x'y. 8. 2« X ^ab X 4:ahx X hahxy. 9. "^a X {m -\- n) X h. Ans. %ah{m + w). Kemark. The pupil will treat quantities in parentheses as mo nomials. 10. 2m X {x-[-y) X a. 11. '2mxlx-\-y) X 3a. 12. ^aX {a-\-h) X 6ax. 13. 46j'm' X (fl^ + ^) X 2amY Ans. 8«'m*(a + h)y. 14. Stt''^' X h'c' X c'a\ 15. mVi" X ny X p'm\ 16. a" X a^ X a, 17. a^ X a'^'a; X a\ Ans. a'^'^r. 18. am X a'n X a'p, 19. h'd X d'c X c'h. 20. h^ X 2^' X U\ 21. a^'^ X "lax^ X 2b'x\ 22. m X nf Xtri" X m\ 23. wa; X wa;' x mx^, 24. 2«a; X 2ay X 2az X 2xyz. 25. 2a'x' X Say. To Multiply a Polynomial by a Monomial. 38. Rule. Multiply each term of the polynomial hy the monomial and take the algehraic su?n of the products. Numerical Example. Multiply 3 + 4 + 5 by 3. 3+4+5 = 12 3 3 9 + 12 + 15 = "36 We see by this example that the product of the sum of several numbers, as 3, 4 and 5, is equal to the sum of the pro- TO MULTIPLY A POLYNOMIAL BY A MONOMIAL. 31 ducts found by multiplying each number separately, because we get the same result, 36, in either case. Algebraic Example. Multiply 2a -\- dixy + ^^^ ^J '^am. Multiplicand, 2a + Sbxy + ^^(^ Multiplier, 7am Product, 14a^m -|- ^lahmxy + Boa^mc, Operation. Each term is multipHed separately, by the last rule, and the sum taken. EXERCISES. Multiply: 1, a -\- b hj x\ 2, m-\- n by a, 3. a-{-chj2x, 4. 2a + 3c by 32;. 5. a -\- 2b -{-de by 2ai/, 6. 2a-{-3b-{- 4:C by abc. 7. Sab + 4«c by babe. 8. 4:ax + 5by + 7cz by Sxyz, 9. «m + Z>;? + eg by 2abc, 10. 6fl!m + 8^/^ + 9c/^ by 7«a;. 11. 6x-{-6xy-\-'7xyz by «a:?/;2;. 12. 2a: + ^2/ + ^^ + ^^ ^J ^^^• 13. m-{-7nn-\-mnp by m?z^. 14. ^ + 2g + 3r by 2apq. 15. ProYB the equations 9(8 + 7) =908 + 9.7. 6(5-3) = 6 o 5 - 6 . 3. 2(1 + 1) = 2 + 2. 2(1 + 1 + 1) = 2 + 2 + 2. 7(9-5) =7o9-7.5. 6(6 - 3) = 6' - 6 , 3. q\q -5) =6. 39. If any terms of the polynomial are negative their product by a positive multiplier is negative. EXERCISES. Multiply; 1. a — bhj m, Ans. ma — nib, 2. m — n by a. 3. a — nhj b. 4. 2« — Sn by c. 5. —2a-\- 3n by 2J. Ans. — 4a5 + 6Jw. 6. a — b-{-c — dhyx. 7. a — 2Z> + 3c - 4c? by ab. 8. 2a - 3^ + 46- by abc. 9. — a-\- b — dc by aa;?/. 82 ALGEBRAIC OPERATIONS. 10. '7ab -\- 2bx — ax by 2abx, 11. — 6am -\- 7bn — Sep by mnp, 12. — a^x + b'^y -\- &z by xyz. 13. %rrei - Zn^j + 4j9'X; by ijK 14. 7«''m — %lfn — l^c^jt? by ^mnp, 15. 12^2;'' - 10Z>^=^ - Icz" by 8a5c. 16. - 8«m" + 9^?^' + lOc/ by 2abc. 17. 3«J + 2^c — cfl^ by a^c. 18. ^a'b - Ub' by 12a;2^. 19. — l^ab — Sbc - 9ca by '7abc. 20. — xy — 2yz — 32;a; by 5xyz, 21. «^'* — ^'^o; by axy. Ans. a'*a:^y — a^x*y. 22. - 2mb -i- 2m'b' hj 3mbr. 23. — 3«'m — 2Z>'?^ by 3a2»mw. 24. Sfl^rc — 36?'^?/ ^y ^f^^y* 40. Indicated Multiplications. Tlie multiplica- tion of a polynomial may be indicated by enclosing it in parentheses and affixing or prefixing tlie multi- plier. EXERCISES. Execute the following indicated multiplications: 1. 3(2fl^ — hbx — c). Ans. 6a — lobx — 3c. 2. 2(a—2b + 3cy). 3. 4:(mx — 2ny + 3pz). 4. 4:a{ab -\- ac — ad). Ans. 4:a^b -\- ^a^c — ^a^d, 5. 2m{m^ + '^'^^ + ^*)' 6. 3Z''^(Z''^^'' - cY). Ans. 3Z>*a;'^ - 3^>Vy. 7. c^c'x'' - c?/''). 8. cy(c' + 2/'). 9. 2«"^^(3aa; - 2%). 10. 2a(« + ^)a2'. ' Ans. 2a'b -\- 20^^. We first multiply the two factors 2a and db which are without the parentheses, making 2a'^b. Then we multiply each term within the parentheses by this product. 11. 2m{m + n)mn. 12. 3h{h — hyik. 13. 2m{m - n)mn. 14. 4:a\a - ¥)b\ 15. 2^'^(2: - i/)a:.v. 16. Vi\2m - 3n)k. 17. Sb{b' + A)^^ 18. 3a'^^^(2a - 3h)h. 19. 2mn\m. + 27^)3m. 20. 2a^(«'' — ^'')2rt'^. TO MULTIPLY A POLYNOMIAL BY A MONOMIAL. 33 41. Negative Multipliers. Rule. When the 7nuUipUer is negative, change the signs of all the terms of the product. The reason for this rule will be given in Course II. EXERCISES. 1. _ 3(^ _ 2§ + 3c - 4:d). Ans. - 3fl^ + 6J - 9c + 12d, 2. — 2{m — 2n + 3p — 4:q). 3. -a{x-{-y-z). 4. — a^{ax — hy -{- cz). 5. —2al){—^ax + 21)y). Ans. Mix — ^a¥y, 6. - %al{a^ - V)a. Ans. - ^a'l + %a^h\ 7. - 2mn{7)i^ - n')n, 8. - 3mx{x' - y')xy, 9. — 8p{x — y)xy. 10. — 2a(ax — a'y + a^z). 11. ^rn{x-y-z)m\ 12. - U'h\h - 2h')h. 13. - 4/m^(« - 7i - 2m)2am. 14. - 4a^^>V(a^ - ^'^ + c''). 43. Comlination of Multiplication and Addition, EXERCISES. Execute the following indicated multiplications, and simplify the results by addition: 1. 2a{^x — 4.y) - a{x + 2y) — 3«(— x — Sy), Work: 2a{'Sx — Ay) = Qax — Say — a{x -\-2y) = — ax — 2ay —da{—x—3y) = Sax + 9ay Sum = Sax — ay Ans. 2. 3(fl^ + b)- 2{h + c) - (c - a). 3. 2«(o^ + ^) + 2^(^ + a). Ans. 2«'' + ^ah -\- 2h\ 4. 37i(^ + m) + 3m(7/i + h). 6. 2«(a + ^) - 2J(^' + a). 6. - ^72(a -^l-c) - 2a{m - V). 7. a{l) -c) - l(c -a) - c(a + i), 8. a:(,?/ - 2;) + y{z - x) + ;^(2; - y). 9. m'^(a'' - Z*) - a'ifn' - b) - 2h{m' + «'). 10. - 3(3rc - 2y) + 2x - 5?/ + 4(8a: - 2?/). 11. 2a;(a:'' + 2x) - Sx^x - 2) - 4(6 - x'), 12. 2a''(aj'' - a') - 2x\a'' - x'). 13. 2x{x ~y)-j- 2y{x - y). 34 ALQBBnjLlG OPERATIONS. 14. ^h' + U{Vi - 7) - U{U' -n- 2). 15. 4?i' — bn{n — 3) — n{n — 2). 16. 2x{x — «/)?/ 4- ^^(y — ^)^' 17. 2a;(a - Z>) - 2a{x - h) -\- %bx. Section IY. Divisiot^". 43. Def. The quotient is that quantity which multiplied by the divisor will produce the dividend. The dividend is said to be exactly divisible by the divisor when it contains the divisor as a factor. 44. First Principle of Divisiojs". A product may he divided hy any of its factors by simply removing them. Example 1. 3.4, which is 12, may be divided by 4 by removing the 4 leaving 3 as the quotient. Ex. 2. The product abc may be divided by h by remov- ing b leaving ac as a quotient. Remark. When all the factors of the dividend are re- moved the quotient is not but 1. Ex. 3. 2.3, which is 6, divided by 2 . 3, which is 6, is 1; ah divided by ab is 1. Hence, to divide one expression by another, which is an exact divisor of it: Rule. Remove from the dividend those factors whose pro- duct forms the divisor. The product of the remaining factors will he the quotient. If all the factors are removed the quotient is 1. EXERCISES. 1. Divide 3 . 4 . 7 by 3 . 4. 2. Divide 6 . 8 . 9 by 8. Ans. 6 . 9 = 54. 3. Divide ab by a. 4. Divide ^amn by 2m. Ans. %an. 5. Divide l%nxy by 4?/. 6. Divide l^bpr by ^br. 7. Divide l{a -\.b)hja^b. Ans. 7. 8. Divide 5m{m + w) by 5m. 9. Divide 6p{r + s) by 3(r + s). 10. Divide Su{h + g) by 4:u{h + g). 11. Divide 12d{x + y)fhj 4/. DIVISION. 35 12. Divide a{x + y) {m -\- n) by a{m -\- n), 13. Divide 15{f -{- g)2x{h + g) by 6x. 14. Divide 16(r + s)4:mn{u + ^) by 8m(?^ + ^)- 45. i^w^e o/ Exponents, Let us have to divide a"" by a". We have a^ = aaaaa, a^ = aaa. Hence by the preceding rule a^ ~ d z=. aa^= a^. Here 2, the exponent m the quotient, is obtained by sub- tracting the exponent of the divisor from that of the dividend. Hence EuLE. Tlie exponent of any symbol in the divisor is to be mbtr acted from the exponent of the lihe symbol in the dividend. Remark. Iq applying this rule remember that a quantity without any exponent is to be considered as having the exponent 1. EXERCISES. Divide: 1. m^ by m". Ans. m^. 2. m^ by m, 3. m' by m^, 4. (a + b^ by o^ + 5. 5. am^ by am. 6. a^m^ by am. Ans. am^. 7. p^(i' \iypq\ 8. 12/7i' by 4^. Ans. ^(jh\ 9. l^g^h by 5^. 10. ^km'^n^ by ^7imk. Ans. dmn\ 11. 8AFm^by2mk 12. 67^X« + ^)' ^J ^^K^ + ^)- 13. 'ir\x + ^)' by t{x + ^)V. 14. 14?^ V(^^ -h ?;) by 2^^. 15. lb{a + Z*)^ (a; + vY by 5(«^ + 2*) (a; + ^). 16. lQab''c\x-\-yyhj ^{x-\-y)abc. 17. 3a'm'(jy - qY by da'm'{p - qY. 18. 5pq^{m — 7i) hj pq\m — n). 19. pa^xy^z bv aa:?/;?. 20. 12(« - Z<)'3^^(r - sY by 9(fl^ - J) (r - s)q. 21. 9(« - Z») (r - 5) by 9(r - s) (a - 3). 22. x"" by a:". Ans. a;"*-r^ 23. fhjy^. } 24. (:r + hYr^ hyr^{x + A)". 25. a'^x''' by a^z". 26. {m + yi)2'^2;^ by (m + nYx"", 27. (a- J)^^by(a-Z')" + ^ 36 ALGEBRAIC OPERATIONS. To Divide a Polynomial. 46. Rule. Divide each term of the polynomial separately. The sum of the separate quotients will he the quotient required. Example. Divide m^a + ma^ by m. We first divide m^a, leaving the quotient ma. Then we divide ma^, leaving the quotient a^ Hence the answer is m^a + ma^ -^ m = ma + a^, EXERCISES. 1. iri^x + mx^ -^- m = what ? 2. a^y + ay" -^ a. 3. 9bY + 12by' ^ 3by. 4. 8p'x -{- 4:p'y + 16p*z -^4:p\ 6. 9abx' + I'Zab'x + ISa'^a; ^ 3aJa;. 6. 5/i'(m + w) + 10y\m -j- n) -^ m -\- n. 7. (a; + 2/) (:?; - 2/) - (x -\- yY -^ x -\- y. 8. (« + Z*) (« - by + (« + ^)' («^ - ^') 4- (a 4- ^) {a - h). 9. 3^; - 6a;' + 3^' -^ 'dx. 10. 6m' (a; + y) - 12m' (a; — y) -^ 6m. 4*7. /S^^'^7^ o/" the Quotient. When the divisor is positive, the quotients of negative dividends must be negative. Example. If we have to divide 12 — 8, that is 4, by 2, the quotient from the separate terms will be 6 — 4. It is evident that the true answer is 6 — 4, and not 6 -f- 4. Hence 12 -8^2 = 6-4 = 2. This principle may be expressed in the following form: Like signs give -\- ; unlike signs give — . EXERCISES. 1. 9a"x^ — 12ax" -f- 3ax. Ans. Sax' — 4ic. 2. 8m'a; — 4:m^y — lQm*z -^ 4m'. 3. 9pqx" — 12pq"x + 15pq^x ~- Spqx. 4. r"{m — nY — 5r^(m — w)' -=- r{m — ny, 5. V(g - hy - 2b{ff - ny -=- l{g - li)\ 6. bx + cx^ — dx^ — gx* + hx^ -^ x. 7. hax — lOa'^y + ba^z -^ 6a. 8. dmu — Qm'^w — 9mw" -r- 3m. FACTORS AND MULTIPLES. 37 9. 5m' (ic — y) — l^m{x — yY -^ bm{x — y), 10. 2(m + x)y - Qy\m + xf -^ %y. 11. Zx\x + A) — 9^;^^ + lif -^ 3(ic + }i)x. Factors and Multiples. 48. Def. A prime expression is one wMch has no factors except itself and unity. Def, A composite expression is one wMch can be expressed as a product of two or more factors. Examples. The number 7 is prime because no two numbers multiphed together will make 7. 21 is composite because it is equal to 7 X 3. a-\-x\% prime. a" + ax is composite because it is equal to (a-\-x)y^a which is a product. 49. Bef. The degree of a monomial is the number of literal factors which it contains. Example. The monomial Zax^ is of the fourth degree because it contains the literal factors a, x, x, x, EXERCISES. "What is the degree of these expressions? 1. hx. 2. hx\ 3. ^'bx\ 4. ^c{a-^x). 5. 1c{a^x)\ 6. mV. 50. Def. To factor an algebraic expression means to express it as a product of several factors. Problem. To factor a number or algebraic expression. EuLE. Find hy trial what prime number or expression tvill divide it ; then find what number or prime expression will divide the quotient. Continue the process until a prime quotient is reached. The divisors and last quotient will be the factors required. Example. Factor the number 420. 420 -^ 2 = 210; 210 ^ 2 = 105; 105 -^ 3 = 35; 35 -^ 5 = 7. Hence 420 = 2\ 3 . 5 . 7, which are the factors required. 38 ALQEBEAIO OPERATIONS. EXERCISES. Express the following numbers as products of prime factors: 1. 36. Ans. 2' . 3^ 2. 12. 3. 28. 4. 81. Ans. 3*. 6. 256. 6. 324. 7. 72. 8. 140. 9. 56. 10. 82. 11. 100. 12. 52. 51. Factoring Algebraic Expressions, Example 1. To factor ¥ -\- hy. We see that h will divide both terms. Dividing by 1) the quotient is ^ -|- ^ which is prime; therefore V -^hy = h{h + y), Ex. 2. «' - a'y = a\a - y). EXERCISES. Factor: 1. ni^ -\- mn. 2. lx-\-hy, 3. ix-^ly -\- hz. Ans. l{x -\- y -\- z). 4. ex — 2cy -\- Scz. Ans. c{x — 2y -{- 3z). 5. hp — dhg -^ bhr. 6. hp — ahg + hhr. 7. 2/i> - 2a/iV + 4:Z>AV. 8. n-^n" -\- 7i\ 9. n- dn'' + 5n\ 10. Ux' - 8i'x + 12^>a;. 11. Shy + 6/^V + 12^Xv^- 1^- 4^' - ^«^' - l^^'- 13. 4:mn' + 8m' w' + ^m'n. 14. (« + ^•)a; -\-{a-\- b)y. Ans. (a + Z>) (a: + y), 15. (w + w)a; + {m + w)y. 16. {g + >^)?^ - (^ + ^)v- 17. a(x -y)-{- h(x - y), 18. a\x -^ y) - h{x + y). Highest Common Divisor. 53. Def. A common divisor of several quantities is any quantity which will divide them all without a remainder. Def. The highest common divisor of several quan- tities is their common divisor of highest degree. LOWEST COMMON MULTIPLE. 89 Def. When two or more quantities liave no com- mon divisor but unity, they are said to be prime to each other. Notation, The highest common divisor is written, for shortness, H. CD. 53, Problem. To find the H.O.D. of several quantities. Rule. Factor each of the quantities. The continued product of the factors common to all, each with its lowest expo- 7ienty is their H. 0. D. Example. Find the H. C. D. of 24, 36 and 48. 24 = 2^ 3; 36 = 2». 3="; 48 = 2*. 3. The common prime factors are 2 and 3; the highest ex- ponent of 2 is 2; . •. 2^ 3 = 12 is the H. 0. D. EXERCISES. Find the H. 0. D. of the following: 1. 54; 90; 144. 2. 14; 56; 63; 84. 3. 72; 108; 132. 4. ax\ bx; dcx\ Ans. x, 5. bni; hn\ Ih. 6. 2mn^x; 6mn^x*, 7. Sx'yz; dxYz; 15a;>. Ans. 3x'y, 8. VZm'n*; 20mn'', 2^m'n\ 9. I){x — h); c{x — A). Ans. x — h. 10. m''{m-\- n)', mn{m -{- n), 11. 3m{g-h); Qm\ 12. 6m{g ^ h); 12{g - h). Lowest Common Multiple. 54. Def. The multiples of any quantity are all quantities which contain it as a factor. Example. The multiples of 3 are 3, 6, 9, 12, etc., and 3a, 6i, 15c, etc., and in general all quantities which contain 3 as a factor. Def. A common multiple of several quantities is any expression which contains all the quantities as factors. Example 1. 24 is a common multiple of 3, 4, 6, 8, be- cause 24 contains 3 as a factor, 4 as a factor, 6 as a factor and 8 as a factor. Ex. 2. ab^{x 4- ^) is a common multiple of a, i, db, ai*, X -\- y, etc., because it contains each of them as a factor. 40 ALGEBRAIC OPERATIONS. Def. The lowest common multiple of several quantities is their common multiple of lowest degree. The lowest common multiple is written, for shortness, L. C. M. ^^. Problem. To find the lowest common multiple. Rule. Factor each of the quantities. The product of the different factors, each aff'ected with the highest exponent it has in any one of the quantities, is the L. C. M. Example. Find the L. 0. M. of "^xif, 3yz and 6^'^. The different factors are 2, 3, x, y, z. The highest exponents, 1, 1, 2, 2, 1. Lowest common multiple, 2.3. x'^y'^z = Qx'^y^z, Quotients, dxz, 2x''y, y^. EXERCISES. Pind the L. C. M. of the following expressions and the quotients formed by dividing the multiple by the several quantities: 1. 4:a'h'c; ISa'h'c; Sadc\ 2. 2mn^; 3np; 6m>. 3. qr; rs; st; qt. 4. 4m?^; 8mj9; 4,np, 5. Qgh""', ^''h\ ^g'h^', ZghK 6. mw': nr"^', rs; srri^. 7. 2u^vw; 4:uv^tu; 6u^v^w; 6u. 8. 3Z'(m + w); 6h' (m -{- n) ; 3(m + ^)'. Ans. L. 0. M. = 6b'{m + n)'. Quotients := 21f{m -\- n), m -{- n and 25'. 9. c^{m-\-n); c^{m-^ny; c{'m -{- n)'. 10. pq\g - h); p'q{g-h); p\\g - h). 11. x\ xy, xyz', xyz{x -\- y\ 12. 12A; 4^; 3/^U\ 13. {a-^-l) {a-iy, {a-^l)\ 14. \a - xy-, U\a - 1); W{a - 1). 15. %x - yY; l(x + y) {x - y)', 14.{x + y)\ 16. 24(« - hyx; 36(« - l))x\ 17. m; ni^n; m^n'^p; 'm*n^p^{m -\-p). 18. ax; 2d^x; 3a^x; 4:a*x. 19. a{x — y); a{x -]- y); a{x — y) {x -{- y). 20. abc{2J — q); abed; ah{p — q). 21. (m-^n) (p — q); (p — q){m — ?i); {m — n) {m-\- n). MISCELLANEOUS EXERCISES. 41 MISCELLANEOUS EXERCISES. 1. In the division of an estate A got y dollars, B got c dollars more than A, and C got 3 times as much as A and B together. How much did they all get? 2. A pedestrian on the first day walked h hours at the rate of q: miles an hour; the second day he walked the same length of time, but one mile an hour faster. How far did he go? 3. If the distance of the earth from the sun is x diameters of the earth, the distance of Jupiter 5 times that of the earth, the distance of Saturn twice that of Jupiter, and the distance of Uranus twice that of Saturn, what is the sum of all these distances? 4. A charitable association working six consecutive days collected x dollars on the first day, and on each of the remain- ing five days it collected a dollars more than on the day pre- ceding. It then divided the amount equally among ^x white and ha colored families. How much did each family get? 5. A library contains ^x books of history, 2y books less of biography than of history, as much of poetry as of biog- raphy and history together, and Zy fewer novels than books of poetry. The books were divided equally among 9 alcoves. How many were in each alcove? 6. In winding up a company each stockholder got x dollars, each ordinary creditor got {h + x) dollars more than each stockholder, and each preferred creditor got 2a; dollars more than each ordinary creditor. There were in all a stockholders, h creditors and c preferred creditors. What was the total amount divided? 7. A railway train having to make a journey of 600 miles ran x hours at the rate of 30 miles an hour and y hours at the rate of 45 miles an hour; it then became disabled and had to make the remaining distance at the rate of 15 miles an hour. How long did it require to make the entire journey? Metliod of Solution. We must subtract from the totaJ distance the distances it ran during the x hours and the y hours. The remainder is the distance it had to run at the rate of 15 miles an hour. Dividing this remainder by 15 will give the time required for the last stage of the journey. Adding the times of the first two stages, which are given, we shall have the whole time required, which we shall find to be (40 — ic — 2y) hours. 42 ALGEBRAIC OPERATIONS. Addition. < Memorakda for Review. Define and Explain: Use of Parentheses; Value oi Symbol; Power; Exponent; Degree; Expression; Ooejfficient; Term; Monomial; Binomial; Trinomial; Factor; Equation. Addition and Subtraction. Define: Like Terms; Algebraic Addition and Subtrac- tion; Numerical Addition and Subtraction; Minuend; Subtra- hend; Kemainder. When the terms are unlike. Rule for like terms. Case of negative terms, i General rule for positive and negative terms. Subtraction. Rule; Reason for the rule. Clearing of J Sign -|- before parentheses; Rule. Parentheses. ( Sign — before parentheses; Rule; Reason. Multiplication. Define: Multiplier; Multiplicand; Product. A monomial by a number; Rule. One monomial by another; Rule. Like symbols, using exponents; Rule. More than two factors; Method; Rule. A polynomial by a monomial; Rule. A negative term by a positive multiplier; Principle. Any term by a negative multiplier; Rule. Divisiion. Dividend; Divisor; Quotient; Exactly Divisible; Prime; Composite; Common Divisor; Highest Common Di- visor; Multiple; Common Multiple; Lowest Common Multiple. First principle of division; Rule deduced. Rule of exponents. \ Division of polynomials; Rule. Product of Define ; Rules and Principles. Rule of signs. Rule for factorinsf. Divisors and j Highest Common Divisor; Rule. IMultiples. ( Lowest Common Multiple; Rule. CHAPTER III. ALGEBRAIC FRACTIONS. 56. Def. A fraction is the expression of an indi- cated division formed by writing the divisor under the dividend with a line between them. Example. The quotient oip ^ q\% the fraction — . The numerator of the fraction is the dividend. The denominator is the divisor. Numerator and Denominator are called terms of the fraction. SECTIOiq" I. MULTIPLICATIO]^ AIS^D DlVISIOlS^ OF Fractions. 57. Theorem. Multiplying the numerator multiplies the fraction. 71% Reason. If we call a fraction — , it will be the same as n m wths. If we multiply it by any factor, as 3, the result will be 3m ;^ths, that is ~^-. n EXERCISES. Multiply: 1. - by a. 3. - by m. n ^ n ^ 3. —^ by aV, 4. — by nm. ^ ^ \ 5. — by^?^ 6. i-^ by 5' by 9«J. ax — by 12. ' 3 , ^ by — 3w. Ans. 3 , » . 13. -::^by-i. 14. -^; — — ^ by — ahc. a-\-b — c ^ Perform the indicated multiplications: ,„ ^ (a am\ . da" da^m 15. Sai-r ). Ans. — c J bo \y X 16. J^-^-l). 2 I 'in. 'r 17. ^a^b \n^ mj' 18. da'(--i-'^-^-\b'c\ fa — b , Z> — c , (?— «\ „ .3. -4^-^). V c^ ' a' ' b' . am — an , cm — en Ans. • ; = . 58. Theokem. Dividing the numerator divides the fraction. The reason is nearly the same as in the case of multipli- cation. EXERCISES. Divide: , 2ax' , , 2x' 1. by ax, Ans. — . mn '' mn 2. __5_by2^y. 3. -^^^J^n^m. MULTIPLICATION AND DI VISION OF FRACTIONS. 45 . aia — bY, , ^ a(a — b)\ ., . ^' 77 — rrf by « — &. 5. 77 — — rf by — (^ — a). Note, First free the divisor from the parentheses. Coinp. §33, which shows that ~(b — d) is the same ^^—h-\- a ox a — h. 6. ^^^!£Lby7cW. ^ 3 . 4 . 5(« + J) (c + ^) , ^ ., , ,, 7. 17^/ by^.4(g + ^). 8. — ^- by - «m%^ 9. ^(^qr^- by ^' - y'. 10. -T by Sa, ^^ aim — n) c(m — n) , 11. -^ — 5 ^^ — 7 — - m m — n. b d ^ 13. !^ + l^by3.'. ? i' j3_ {a + h)(a-l) _ (.-^)(« + i) . _ (J _ „)^ m n '' ^ ' 14. — + _- + _ by «fc. 15. (l+<_(l + .)(l-«) ( )^ mn nr j \ * / 16. ^ilzi.'!!) + Oi+IIO?Lzl)by«'-l. m n ^ ^^ abjx^ + y' + z') Hf + ^^ + ^•^)^ . (^^ + ^' + /)^^ c a b by ^' + 2/' + ^'. 18. -^__---^-^---by7«Jm^ 19./^+ ~^^^~^n y/-^. « b ^ c ^ ^^- r - 273 +273747 5 ^y-*^- 46 ALGEBBAIC FRACTIONS. 59. Theoeem. Multiplying the denominator divides the fraction. Reason, Suppose a line first divided into 4 parts. | ' ' 1 ' ' | ' ' 1 ' '1 One of the parts will then be - of the line and n of them will then be - of it. 4 4 Now multiply the denominator 4 by 3. This will be dividing the line into 3 . 4 = 12 parts, so that each fourth will be divided into 3 parts. Therefore j "^ ^ = — , and n _ n 4""^-i2- EXERCISES. ■1 ^' o A ^' ^ 3« 1. _^3. Ans.^. 2. -^-^5. 3. ^^ ~- 5/^. 4. i -^ 2n. 5k n 5. 3-^-4(l-a). 6. 3^-2(1 + ^). 11. -\ - ^ -^ «Jc. 12. J- - -i- ^ 3 . 4 . 5. ab be 3.44.5 13. x' -\-l-\-\~ x\ Ans. 1 + -, 4- -,. ' a;' ^ x^ X 14. ^,+ --^3F. 15. «_-+-_ J ^a'. 16. -2 -^ aa:^. 2/ ^ 17 _1 ^ _ ^ . ^^ •*•'• 55 9 /-kQ "»' aXt ax ax 2ax MULTIPLICATION AND DIVISION OF FRACTIONS. 47 60. Theorem. Dividing the denominator multiplies the fraction. Reason. If the fraction is ^, and we divide the denomi- nator 13 by 3 we have j, which is 3 times as much ^ j-^. ^ , n So also £ is 3 times as much as ^^, or EXERCISES. 1. |X2. 2. |X3. 6 a 3.|X6. 4.^jXl2. 5. (f^^).x (« + *)• 6.^x-^™'«- Note, n — m = — (m — n). ^•I(3^W)^'- ^'•^^(*-"^- 20. -^^i;X(:^ + 2/)- 21. ;;.^Xa + J. 23 ^ + g X («' + 1). 23. ^-^ X *. ■* a'm -\-m ^ ^ ' ax-bx 48 ALGEBRAIC FRACTIONS. 61. If we multiply a fraction — by its denominator q we haye by the preceding rule^, that is jt?, as the product. Therefore A fraction is multiplied by its denominator ty simply re- moving it. Examples. -x2 = 1; -x«^ = 2; ^x(l— F)=m. it a \. — Ic 63. When the multiplier and the denominator contain common factors, we may multiply by these factors by removing them from the denominator, and then multiply the remaining factors into the numer- ator. Example. Multiply — by mx. ^ '' mn "^ We multiply by m by removing it from the denominator, and then multiply the numerator by x. Therefore a ax — . X mx = — . mn n EXERCISES. 6. j-3- X (« - 1) {a + 1). 7. ^,- X U^'2/'. 10- (,_^) (,!»)(,_,) X s{s - a). REDUCTION. 49 "• ( l_^)(l_^ -)X-a(g-l)(l-^). Section II. Reductiois- of Fractions. 63. Def. Reduction means changing the form of an expression without changing the value. 64. Reduction to lowest terms. Theorem. // loth terms of a fraction le multiplied or divided hy the same quantity, tlie value of the fraction will not he altered. Reason. By §§ 57, 58, 59 and 60 the multiplications of the two terms have opposite effects, which cancel each other, and so have the divisions. Hence all coiannon factors m the two terms may be cancelled. Note. Observe that only factors can be cancelled. G5. Def. When all the common factors are cancelled the fraction is said to be reduced to its lowest terms. To reduce a fraction to its lowest terms: EuLE. Factor each term when necessary, and cancel all factors common to both, ^ , mnp^x^ mn Example 1. — ^7- = — . rsp X rs The factors p^x^, common to both terms, are cancelled. x^y _x xY y the factors x'^y being cancelled. mx — nx _(m — n)x __ x Ex. 3. my — ny {m — n)y y EXERCISES. Reduce to lowest terms: a¥c' 2 ah + ac* •^- a'Vc c'a - aV 50 ALGEBRAIC FRACTIONS. ' n'x{i - a) ' ' i?Y+i^V+i?V* (l-xYJl + ^y . (a-i){a-2) (1 -x) {x -{-!)' (1 -a) (2- a)' Note, a — 1 = — {1 — a). ISa^b^c* amx — amy TZa^h^cxy^ ' ' m^c^x -|- ma^y rs - rV /V+/y 11. (!-«' + a')^y i3_ 7.9 + 3 _ a:?/ + a*xy ' '7.9 — 2.3* 3a^{k — q)xy ' abc abc abc a;?/- "^ y;2- "^ ;2a;-* 3. 4. 5 "^ 3. 4. 5 "^ 3 . 4. 5* 66. Reduction to Given Denominator. An entire quantity may be expressed as a fraction with any re- quired denominator, Z>, by supposing it to have the denominator 1 and then multiplying both terms by D. For if a is any entire quantity, we have a aD '' = i='B- Example. If we wish to express the quantity ai as a fraction having xy for its denominator^ we write adxy xy * 67. If the quantity is fractional, both terms of the fraction must be multiplied by that factor which will produce the required denominator. Example. To express j- with the denominator nd^ we multiply both members by nb^ -^ h = nb^. Thus, a _ anb^ b~''W This process is the reverse of reducing to lowest terms. REDUCTION. 51 EXERCISES. Express the quantity: 1. m with denominator a. Ans. — . 2. m with denominator m. Ans. — . Z. a — b with denominator x. a m 4. - with denominator or, Q 5. — with denominator 12. 4 6. a^ with denominator a -\- b. 7. X — y with denominator x -\- y. 1 — k^ 8. with denominator abc. a 9. ^ — Jl — with denominator lOaa;^ (F — 1), 10. — -^^ with denominator 20m^ — 4m 1 — ^' 11. ^ with denominator 1 — a. 9 -^ Note. 1 — ^^ = — (^ — 1). 12. :r-, — and — : with denominator abc» be ca ab 13. — , — and — with denominator rst, r s t X^ |/2 14. —J, j-^ and 1 with denominator a^b^. s s s 15. , i and with den. {s—a)(s—b)(s—c). s — a s — b s — c ^ ^^ ^^ ' 16. with denominator + a. — a ct X 17. T- and - with denominator abxy, by ^ 52 ALGEBBAIG FRACTIONS. 68. To reduce fractions to a common denominator: Rule. Choose a common muUijjle of the denominators. Multiply both terms of each fraction by the multiplier 7iecessary to change its denominator to the chosen multiple. Note 1. Any common multiple of the denominators may be taken as the common denominator, but the L. 0. M. is the simplest. Note 2. The required multipliers v/ill be the quotients found by dividing the chosen multiple by the denominator of each separate fraction. Note 3. When the denominators have no common factors, the multiplier for each fraction will be the product of the denominators of all the other fractions. Note 4. An entire quantity must be regarded as having the denominator 1. (§ 66.) Example 1. Eeduce to common denominator \ T) , m w, — , -^ and — . m mn np The given denominators are 1, m, mn, np. Their L. 0. M. is mnp. The multipliers are (Note 2) mnp, np, p and m. Multiplying by these quantities, the fractions become m^np np p^ ^ m^ -, —^—, -^-— and . mnp mnp mnp Ex.3. \ I i. a b c By Note 3 the multipliers are be, ac and ab. Multiplying by them, the fractions become bcx acy _ abz —Ti —r and -^. abc abc dbo EXERCISES. Reduce to common denominator: 1 1 - - 2 b c a abc 5. X 3 1 1 1 a' y T m" a'' m ~, m. a 2x a' 3^ 42 1 - «' a' AGGREGATION. 53 q ^ ^ ? 10 i ^ 1 ^•3' V 5* 3? 4/ bz M ^ ^ £1 19 A m 5n_ ■f -r ,. . -a' 13. /; — , ^^^^ . 14. «^, ^-^, ^345. 15. -, , — . Id. c m—V 1 — 7n' m ' 1 — x' 2a;(l — x)' Sx"^' Section III. Aggregation^ and Dissection of Fractions. 69. I>ef, Aggregation is the expression of the algebraic sum of several fractions as a single fraction. Def. Dissection is the separation of a fraction into an algebraic sum of fractions. 70. Case I. When the fractions to be aggregated have the same denominator. EuLE. 1. Unclose each numerator detween parentheses, or suppose it so enclosed. 2. Prefix to each numerator so enclosed the algebraic sign of the fraction. 3. Form the algebraic sum of the expressions thus found and write the cofnmon denominator under them. Reason. 1. By the definition of a fraction, the numerator expresses a number of fractional units. 2. Hence the sum of several fractions with a common de- nominator is the sum of all the fractional units indicated by the several numerators. 3. If the fractions are preceded by the minus sign, it in- dicates that the fractional units in its numerator are to be algebraically subtracted. This subtraction is indicated by enclosing the numerator between parentheses and prefixing the minus sign. 4. Hence the algebraic sum of all the fractions is formed in the manner directed in the rule. 54 ALGEBRAIC FRACTIONS. Example 1. ;/ ~ ^ + j means l fractional units to be subtracted from a fractional units, and c fractional units to be added. Hence a___i c _a — 'b-\-c Ex. 2. c^^x - 3g — % x __ 4a; -8c _ c-^hx — {^c — 2a;) — (4a; - 8c) D ~ D D D _ c 4- 5a; — 3c + 2a; — 4a; + 8c . _ _ _ 6c + 3a; ~" D • Aggregate: EXERCISES. ^ a — X ^ a-\-x . %a 1. ' — . Ans. — . mm m _« + a; a — X . a-\-x—{a — x\ 2x 2. — ' . Ans. — ■ -5^ ^ = — . mm mm « + 2a; a ~ X 2a — 3x a -\-'Sx m m ' ' m m ^ a -\- h 2a — dd ^ y x 5. ■ . 6. - ^ m — n m — n ^ — y ^ — V Sx 3a a — 2b-]-Sc a-{-2b~ Sc ' X -{- a X -\- a' ' 1 — x^ 1 — x^ ' s s s ' ' m m ' When a+ 5 + c = 2s, what does the answer to Ex. 9 reduce to? 4:xy^ 4:xy^ Axy^ ' ,_ a^-{-2aI)-\-i' a"" - 2ah -\- V 12. mn mn ^_ m{\ - a) ■ m{a - 1) 2(1 - F ) 1-F If ^ ¥ ' 2.3.4 2.3.4' 2a -^l 2a ^b 5a -2b a-}- 2b ■^^' D D '^ D ' D ' AGOBEQATION. 55 71. Case II. If an entire quantity and a fraction are to be aggregated, we reduce the entire quantity to a fraction having the same denominator as the fraction. Example. Aggregate a-\ — . Til CLYh Eeducing by § 66, we have a = — . So the sum is an m _an -\- m n n~ n We now see that the result is obtained by the following rule: Multiply the entire quantity hy the denominator, and add the product with the numerators. EXERCISES. Aggregate: ^ ^ ^ A 1 — x — a 1. 1 — . Ans. — . 1 — x 1 — x. 2. l+T-^-. 3. l-:rT-- ' 1 — a; l-\-x 6./+}. 7. . + 3,-1-^:^^. 8. 1 — r-r — . 9. Amn — - ^ a , . 1 -\- a 4rm^n* 10. x-j-^^. 11. A ^ a: + 2* 4 + 5a'' 12. m-h + -, 13. 4:Aa ^ m' ' a-\- A' 14. . + 1^. 15. . + i. 16. x^a-^-. 17. l^x--^-=^ 18. a-* a' 4- 2a^ + ^' _ 4:ah a'-2ab-^b* Samx' 3amx' Zamx' 5e ALGEBUAIC FRACTIONS. *72. Case III. When the fractions have different denominators we reduce them to a common denomina- tor and proceed as in § 70. EXERCISES. Aggregate: ■• ,-+r Ans. a-\-b ab ' 2. 1 1 b a 3 !?.l_!?L 4. m n 'a I' m — n n 5 "^ -4- n 6. a 3b O, — r- m-\-n m a — b a' ^ a h c x' x^ '• ^-+3 + 4* 8. ^ + 7 + ^- 1 1 1 a — b b — c c — a ab be +-.«• 10. c ' a ' b ' n.i + i + c ' a 12. ^^\^W,- 13. l-f + r 14. . /' , /' 2.3. 4' •' 2.3 ' a. 3. 4. 5" 1 1 1 15. 1-4- 16. ^'-W + '- .^ 1 1 1 1 i^- — 2e' 18. rrf wr m' m^''' 73. Dissection of Fractions. If the numerator is a polynomial, each of its terms may be divided separately by the denominator, and the several frac- tions connected by the signs + or — . The principle is that on which the division of polynomials ^ founded (§ 46). The general form is ±i:^iL£±^ = £ + ^ + £ + etc. (1) m m m m The separate fractions may then be reduced to their lowest terms. Example. Dissect the fraction 12^^g + Qab — 3g UabG ' MULTIPLICA TION AND DIVISION B T FBA CTI0N8. 57 The general form (1) gives for the separate fractions 12abc VZabc 12abc Reducing each fraction to its lowest terms the sum becomes 1 + 1— i-. EXERCISES. a* — a;' + 2ax x^ — 2xy -f- y* ab-\-bc-{-ca . am^ — 2amn — «' J' O. 7 » 4:. 2 a • abc aw; Note, f—g and h— j are to be considered as single quantities; for example, like A and B in Ex. 5. aY + ^'^' - ^'^' o 1 + ?i?/^ * 5^/i * ' \b ~^ c I ' c' 11. (4_^«]^?!|!. 13. (_!!!_ _l+£)^l±« W y J « Vl — a n J 1 — a 13. (^ + ^)^ {s — a) (s — b)' 14. (i _ 1) ^ -J_. 15. (ir + ri + il) -. Jl. \ a J 1—a \s a sqj grs l'7a'b'xy"' ^ 61m''n'x 13 , 14 * 19m/i2; * 19a%^* a— 1 * 1 — a ^^- 9 • 9a ^^' af^'^a^' 20. fl--lU4. 21. 1-^ Note. The answers to Exercises 5 and 21 show that the quotient of unity divided by any fraction is equal to the fraction inverted. MISCELLANEOUS EXERCISES. Express and reduce: 1. I of I of Hx. 2- T ^f I of ^^- 3 5 o z 3. I of I of {5x - a). 4. J- of (|x - «). 5. lof(^*+nA 6. lof'^of^ofx. m \n ^ j m n p 7. From a sum of a: dollars - of the amount and a dollar o more were taken. Express — the remainder. o MEMORANDA FOR REVIEW. 61 8. Increase the quantity xhj — part of itself, and express the result as a fraction, with ic as a factor. 9. Diminish the quantity x by the ni\i part of itself, and express the result in the same way. 10. From a cistern containing g gallons of water - the o water was taken one day and ^ of what was left the next day. Express the remainder. 11. A father left h thousand dollars to each of a children. Each of these children had h other children between whom the money was equally divided. How much did each grand- child get? 12. A charitable association collected x dollars from each of a people and y dimes from each of h people. It divided the amount equally among a almoners, and each of these almoners divided his share among h poor. How much did each poor person get? 13. A man having to make a journey of m miles went — the distance and a miles more on the first day, and — the re- maining distance -f ^ miles on the second day. How far had he still to go? 14. A huckster with t turkeys sold half of them at — dollars 1 m" each, and - the remainder at —^ dollars each, and what were 3 n m' left at —5 dollars each. How much did he realize? n Memokakda for Review. The Two Conceptions of a Fraction. 1. As parts of a unit. 2. As the quotient of an indicated division. Explain significance of numerator and denominator in each mode of conception. ALGEBRAIC FBACTIONS. Multiplica tion Division Multiplication and Division of Fractions, " By multiplying numerator,- Reason. By dividing denominator; Eeason. By removing denominator; Reason. When multiplier and denominator have a common factor. By dividing numerator; Explain. By multiplying denominator; Explain. When divisor and numerator have a common factor. Define Reduction Aggrega- tion Dissection. Reduction of Fractions, Reduction; Lowest Terms. ' By multiplying both terms by the same fac- tor; Explain. By dividing both terms by the same di- visor; Explain. To lowest terms. ( Entire quan- To given denominator; Reason. •< tity. ( Fraction. To common denominator. Aggregation and Dissection, Of fractions having the same denominator; Give rule and reasons. Of an entire quantity and a fraction. Of fractions having different denominators. Show when applicable; Give rule; reason. Multiplication and Division iy Fractions, Multiplica- tion. Division. Define multiplication by a fraction. Give general rule; Explain reason. Show how general rule is applied when the multiplicand is also a fraction. Define division by a fraction. Deduce general rule. CHAPTER IV. SIMPLE EQUATIONS. Definitions. 77. Bef. An equation is a statement, in the lan- guage of algebra, that two expressions are equal. Def. The two equal expressions are called members of the equation. 78. Def. An identical equation is one which is true for all values of the algebraic symbols which enter into it, or which has numbers only for its members. Examples. The equations 14 + 9 = 29 - 6, (5 + 13) - (3 X 4) - 6 = 0, which contain no algebraic symbols, are identical equations. So also are the equations X = Xy X — X = 0, ^x -f y) = 72; + ly, because they are necessarily true, whatever values we assign to X and y. Remark. All the equations used in the preceding chapters to express the relations of algebraic quantities are identical ones, because they are true for all values of these quantities. 79. Def. An equation of condition is one which can be true only when the algebraic symbols are equal to certain quantities, or have certain relations among themselves. Example. The equation a: + 6 = 22 can be true only when x is equal to 16, and is therefore an equation of condition. Remark. In an equation of condition, pome of the quantities may be supposed to be known and others to he unknown. 64 SIMPLE EQUATIONS. 80. Def. To solve an equation means to find such numbers or algebraic expressions as, being substi- tuted for the unknown quantity, will render the equa- tion identically true. Any such value of the unknown quantity is called a root of the equation. Example 1. The number 3 is a root of the equation W - 18 = 0, because when we put 3 in place of x the equation is satisfied identically. Prove this. The number — 3 is also a root. An algebraic equation is solved by performing such similar operations upon its two members that the unknown quantity shall finally stand alone as one member of an equation. Remark. It is common in Elementary Algebra to represent unknown quantities by the last letters of tlie alphabet, and quantities supposed to be known by the first letters. But this is not at all neces- sary, and the student should accustom himself to regard any symbol as an unknown quantity. Axioms. 81. Def. An axiom is a proposition which is taken for granted, in order that we may, by it, prove some other proposition. Equations are solved by operations founded upon the fol- lowing axioms, which are self-evident, and so need no proof. Ax. I. If equal quantities be added to the two members of an equation, the members will still be equal. Ax. II. If equal quantities be subtracted from the two members of an equation, they will still be equal. Ax. III. If the two members be multiplied by equal factors, they will still be equal. Ax. IV. If the two members be divided by equal divisors (the divisors being different from zero), they will still be equal. Ax. Y. Similar roots of the two members are equal. These axioms may be summed up in the single one, Similar operations upo7i equal quantities give equal results. TRANSPOSING TERMS. 66 Transposing Terms. 82. Theorem. Any term may he transposed from one member of an equation to the other member if its sign be changed. Example 1. From the equation 7 + 18 = 25 we obtain, by transposing 18, 7 = 25 - 18; by transposing 7, 18 = 25 - 7. Ex. 2. From the equation 9 = 12 - 3 we obtain, by transposing 3, 9 + 3 = 12, and from this last equation, by transposing 9, 3 = 12 - 9. EXERCISES. Form two equations from each of the following by trans- position: 1. 16 = 9 + 7. 2. 8 + 5=13. 3. 15 = 6 + 9. 4. 23 - 10 = 13. 5. 14 = 20 - 6. Ans. 14 + 6 = 20; 6 = 20 - 14. 6. 14 = 21 - 7. 7. 17 - 8 = 9. Form as many equations as you can from: 8. 8 - 5 = 9 - 6. 9. 19 - 7 = 15 - 3. 83. General Proof of Transposition. Let us have the equation a-\-t = b. Now subtract t from both members, a-\-t — t = b — t; whence, by reduction, a = b — t. This equation is the same as the one from which we started, except that t has been transposed to the second member, with its sign changed from + to — QQ SIMPLE EQUATIONS. If the equation is i — t =K a, we may add t to both members, which would giye EXERCISES IN SOLVINQ EQUATIONS BY TRANSPOSITION, 1. Find that number which, when 9 is added to it, will make 25. Solution. Let us call the required number n. The prob- lem says 9 must be added to it. Adding 9 the sum is w + 9. The problem says this sum must make 25. Therefore ^-f-9 = 25. Transposing 9, w = 25 — 9 = 16. Ans. Note. This and some of the following problems are so simple that the pupil can answer them mentally; but he should do them by algebra in order to learn methods which may be applied to more diflBcult prob- lems. 2. Find that number which, when 17 is added to it, will make 30. 3. Find that number which, when 12 is subtracted from it, will leave the remainder 11. 4. Find a number which, subtracted from 16, will give 9 as the remainder. Solution. Let x be the number. When subtracted from 16, the remainder is 16 — a:. By the conditions of the ques- tion, 16 - a; = 9. Transposing x, we have 16 — 9 -f a;. Transposing 9, U-9=x, whence x = 7. Ans. 5. Find a number which being subtracted from 15, the remainder shall be 6. 6. If a number be diminished by 6, and the remainder multiplied by 3, the product shall be double the number. What is the number? Solution. Let r be the number. Diminishing it by 6 the remainder is r — Q. Multiplying this remainder by 3 EQUATIONS SOLVED BT TRANSPOSITION, 67 the product is 3r — 18 (§38). By the condition of the problem this product is equal to 2r. Therefore 3r - 18 = 2r. Transposing 18, 3r = 2r + 18. Transposing 2r, 3r — 2r = 18, , or, by reduction, r = 18. Ans. ^ Proof, 18 - 6 = 12; 12 X 3 = 36, which is twice 18. Note. The student should prove all his answers by showing that they fulfil the conditions of the problem. 7. If 4 be subtracted from a number, and the remainder multiplied by 4, the product will be three times the number. Find the number. 8. A baker started out with x loaves of bread. He sold all but 8 of them at '5 cents each, and then had as much money as if he had sold them all at 4 cents each. What was his number ic? 9. One baker started out with y loaves, and another with 9 loaves less. The first sold all his at 5 cents each, and the other sold all his at 6 cents each and realized an equal amount. How many loaves had each? 10. A huckster bought a lot of turkeys at $1 each. Nine were spoilt, and he sold the remainder at $2 each and made a profit of $7. How many did he buy? Solution. Let x be the number he bought. At $1 each they cost him x dollars. 9 being spoilt he had ic — 9 to sell. At $2 each the amount was 2a: — 18 dollars. Because he made a profit of $7 this amount must, by the conditions, be 17 more than the cost; that is $7 more than %x. Therefore 2a; - 18 = a; + 7. Transposing x and 18, 22; - a; = 7 + 18, or x = 25. Ans. Proof. 25 — 9 = 16; 16 X 2 = 32, which is the same as 25 + 7. 68 SIMPLE EQUATIONS. Division of Equations. 84. From the equation 12 - 8 = 4 we obtain, by dividing by 2, 6-4 = 2, and, by dividing by 2, 3-2 = 1. EXERCISES. Form as many more equations as you can from the follow- ing by division, and see if the results are true: 1. 24 - 16 = 8. 2. 36 - 24 = 12. 3. 72-45 = 27. 4. 84-70 = 35-21. 85. From the equation ^x = 24 we obtain, by dividing by 3, a; = 8. Proof, Putting 8 for x m the given equation, we have 3 . 8 = 24, which equation is identically true. If we have the equation ax = m, we find, by dividing by a, m x= —, a Hence, to solve an equation in which one member is known and the other member is the unknown quantity mul- tiplied by a coefficient: EuLE. Divide loth members by the coefficient of the un- known quantity, EXERCISES. Solve the following equations by reduction and transposi- tion: 1. a-\-x = b —2x, Solution, Transposing a and 2x, x-\-2x — b — a, ox 3x = b — a, DIVISION OF EQUATIONS. Dividing by 3, x = 5- 3 ■ a 2. 7 + 3:2^ = 37-20;. 3. 5(a;- 3)=a; + 12. 4. 6(2; + 5) = 10a:. 5. 3(8- a;) = 23 + 2a;. Solution of 5. 24- -dx = : 23 + 2a;. Transposing 3a;, 24 = : 23 + 2a; + 3a; = 23 \ + 5x. Transposing 23, 24- 23: = 62 •9 or 1 = 5a ;. Dividing by 5, 5 = x, or X _1 ~5' . Ans. 6. 7(9 -x) = 5(11 -a;). 7. 8(2- n) = b{n - 2) 8. 4(1 -x) = :5(2- -X). 9. x-{-a = 2a; + J. 10. 3a; -21 + 4ic- 7 = 0. 11. ax — b = 0. 12. bx = c. 13. ax = a-{-l. 14. {a-\-b)x = : a — 5. 15. mx-\- 1 = wa; + 2. 16. A man divided 42 apples between two boys, giving A twice as many as B. How many did each get? Solution. Let us call n the number of apples B got. Then, because A got twice as many, he got 2n. So, by adding, B's apples = n A's apples = 2n Both together got 3n The condition is that this number shall be 42. Hence 3n = 42; and dividing by 3, ^ = 14. Therefore A's share = 2^ = 28; B's share = n = 14=. 17. A, B and C had between them $78. B had twice as much as A, and C had three times as much as A. How much had each? 18. A drover bought a flock of sheep at $6 per head. After losing 40 he sold the remainder at $5 per head and gained 11050. How many did he buy? 70 SIMPLE EQUATIONS. 19. A man made a journey of 244 miles in 3 days, going 10 miles less on the second day than on the first, and 12 miles less on the third day than on the second. How far did he go each day? Note, If we call x the distance made on the first day, that on the second will be a? — 10 and that on the third a? — 10 — 12 = « — 23. 20. In dividing a profit of $2500 between two partners, A got twice as much as B and 1100 more. What was the share of each? 21. A huckster had 50 apples, some good and others bad. He sold the good at 5 cents each and the bad at 2 cents each, realizing $1. 78 in all. How many apples of each kind had he? Method of Solution. Let us call g the number of good apples. Then, because the good and bad together numbered 50, the bad alone num- bered 50 — g. So the sums realized from sales were: g good apples at 5 cents each 5^ 50 — ^ bad ones at 2 cents each 100 — 2^ Total amount received ... 100 + ^9 This expression is to be equated to the amount realized, namely, 178 cents. By solving the equation thus formed, we shall find ^ — 26 = number of good apples; 50 — ^ = 24 = number of bad apples. 26 X 5 4- 24 X 2 = 130 + 48 = 178. Proof. 22. A man had 45 pieces of coin, some 3-cent pieces and the remainder 10-cent pieces, amounting in all to $1.84. How many pieces of each kind were there? 23. A train made a journey of 374 miles in 13 hours, going part of the time at the rate of 25 miles an hour and the re- mainder of the time at the rate of 32 miles an hour. How many hours did it run at each rate of speed ? Note. If we call x the number of hours it ran at one rate, 13 — a; will be the number of hours it ran at the other rate. 24. A man made a journey of 112 miles in 4 days, going 4 miles farther each day than he did the day before. How far did he go on each day? Note. Take the distance he went on the first day as the unknown quantity. 25. Divide 100 into two parts such that three times the one part shall be equal to twice the other. CLEARING OF FRACTIONS. 71 Multiplication of Equations. 86. Clearing of Fractions. The operation of multipli- cation is usually performed in order to clear the equation of fractions. To clear an equation of fractions: FiKST Method. Multiply its memhers hy the least com- mon multiple of all its denominators. Second Method. Multiply its memhers hy each of the denominators m succession. Remark 1. Sometimes the one and sometimes the other of these methods is the more convenient. Rem. 2. The operation of clearing of fractions is similar to that of reducing fractions to a common denominator. Example of First Method. Clear of fractions the equation 1 + 1 + 1 = 26. Here 24 is the least common multiple of the denominators. Multiplying each term by it, we have (§ 68) 6x-{-^x-{-Sx = 624, or 13a; = 624. Example op Secoi^d Method. XX 1 a'^h~'^' Multiplying by a, . ax 1 * + T = F- Multiplying by b, or hx -]- ax = 1, (a 4- h)x = 1. EXERCISES. Clear of fractions and solve the following equations 1 ^-7 = -. •^•2 ^4 2. 1 - 2^ = Sx. a:-l_a; + l ^' 2^3- ^ X 4 a; X ' X ' X X ^ X ' X 72 . SIMPLE EQUATIONS. » 9. ?4^-+-^=7.. 10. 1+1-1=4. llJ^^%-^ = m. 12. 2 . 3;r - 7 = 0. ' a r h c ^ li\ apx - ^^ = 0. 14. 1 _ ?^ r. 3. / . X mx 15. r— = 3. lo. — ^ -. = d.4. < 17.-^^ j-=«J. 18.^ 1 = ^ + 1. 19. «- ^^~^ =^y + 7. 30. 4k + 13.'!; = 4. 12. o 7 Problems Leading to Simple Equations. To solve a problem, we hnYe first to state the conditions of the problem in algebraic language, and secoyid to solve the equation resulting from such statement. We have already shown how to solve the equation, but for the statement no general rule can be laid down. The follow- ing precepts will, however, serve as a guide to the beginner, who must trust to practice to acquire skill in solving problems. 1. Study the problem carefully to see what is the unknown quantity required to be found. Sometimes there are several, only one of which need be taken. 2. Represent this quantity by x, y, or any other letter or symbol whatever.* 3. Perform on and with this symbol the operations de- scribed in the problem (as in Chapter I., § 23). 4. Express the conditions, stated or implied, in the prob- lem by means of an equation. 5. Solve the equation by the methods already explained in the last four sections. * Any symbol may be used. In the early history of algebra the un- known quantity was called the thing — in Italian cosa, which for brevity was written co. UNIVERSITY OP CAffPg^^a^ LEADING TO SIMPLE EQUATIONS. 73 I *• PROBLEMS FOR PRACTICE. 1. A man asked a shepherd how many sheep he had. He replied : If you add 32 to the number of my sheep and multi- ply the sum by 9, the product will be 27 times the number of my sheep. 2. Another shepherd answered: If you subtract 32 from my sheep and multiply the remainder by 6, you will have double the number. 3. A baker said: If you divide the number of my loave$ by 6 and add 60 to the quotient, the sum will be the number^' of my loaves. 4. A man sets out upon a journey from one city to an- other. The "first day he travels one half the distance between the cities; the next day he travels one third the distance be- tween the cities, and then finds he has still 12 miles to go. Find the distance between the cities. Solution, Let d = the distance between the cities; then - = the distance travelled the first day, - = ** " " " second day, o and these two amounts plus 12 miles must, by the conditions of the problem, equal the whole distance d; that is, in alge- braic language the conditions of the problem are stated in the following equation: f + f + 12 = .. To solve this equation multiply each member by 6, whence 3c? + 2g? + 72 = 6d. Transposing, 72 = Gc? — 3^7 — 2d; and uniting terms, 72 = d. Hence 72 miles is the required distance between the cities. 72 Froof. Distance travelled the first day = -- = 36 miles; 72 " " second *' = — = 24 miles; " " third '' = 12 miles. Whole distance travelled = 72 miles. 4 74 SIMPLE EQUATIONS. 5. Another said: If you add 9 to three times the number of my loaves and divide the sum by 9, the quotient will be 38. 6. A third baker said: If you add 136 to twice the number of my loaves and divide the sum by the number of my loaves, the quotient will be 10. 7. If you add 41 to a number and divide the sum by the number minus 15, the quotient will be 8. What is the mim- ber? Ans. 23. 8. If you add 96 to 6 times a number and divide the sum by twice the number minus 52, the quotient will be 17. Find the number. Ans. 35. 9. If you subtract 3 times a certain number from 480 and divide the difference by the number minus 18, the quotient will be 139. Find the number. 'Ans. 21. 10. Find a number such that if we divide it by 6, add 7 to the quotient, and multiply the sum by two, the product will be the number itself. Ans. 21. 11. From the following condition find the age at which Sir Isaac Newton died: If you take one third his age, one fourth his age, and one sixth his age, then add them together, and add 105 to the sum, you will get just double his age. Ans. 84. 12. If you add 5 to the year in which Sir Isaac Newton was born, then divide the sum by 9 and add 638 to the quo- tient, the sum will be just half the number of the year. What was the year? Ans. 1642. 13. If you take the year in which John Hancock died, add 16 to it, then take ^, -, and ^ of the sum, the sum of these three quotients will be 871. Find the year. Ans. 1793. 14. If you subtract the age of John Hancock from 100, divide the difference by 4, and add the quotient to one half his age, the sum will be 17 years less than his age. Find liis age. Ans. 56. 15. If you take the population of the District of Colum- bia in 1870, divide it by 300, subtract 400 from the quotient, and multiply the remainder by 7, the product will be 273. What was the population? Ans. 131,700. 16. If you divide my age 10 years hence by my age 21 years ago, the quotient will be 2. What is my present age? PBOBLEMS LEAUij-sG TO SIMPLE EQUATIONS. 75 17. Divide 1200 among three persons, A, B and C, so that B shall have $25 more than A, and C $18 less than A and B together. Solution. This question differs from tliose preceding in having three known quantities. We therefore show how to solve it. Let us put y for A's share. Then B's share will hey-\- 25, and the share of A and B together will be 2/ + y + 35; that is, 2y -\- 25. C's share is said to be $18 less than this sum; it is therefore 2y -f 7. We now add up the shares : A's share is y B's - - y + 25 C's " " 2y+ 7 Sum of all, 4y + 32 Kow by the conditions of the problem this sum must be $200. Hence we put it equal to 200 and solve the equation. We shall thus find y = 42. Therefore we have for the answer: A's share = y =42 B's " = y -f 25 = 67 C's •' = 109 - 18 = 91 Total, $200. Proof. / 18. A father left $7050 to be divided among five children, directing that the eldest should have $200 more than the second, the second $200 more than the third, and so on to the youngest. What was the share of each? Ans. $1810, $1610, $1410, $1210, $1010. 19. A is 15 years older than B, and in 18 years A will be just twice as old as B is now. What are their ages? Ans. 48, 33. 20. Of three brothers the youngest is 4 years younger than the second, and the eldest is as old as the other two together. In 10 years from now the sum of their ages would be 98. What are their ages now? Ans. 15, 19, 34. 21. An uncle left his property to two nephews, the elder to receive $100 more than the younger. But if a certain cousin was still living, the elder was to give her one fourth of his share, and the younger one sixth of his. The cousin was living, and got $1275. How much did the uncle leave, and what was the share of each heir? Ans. $3000, $3100. 22. The head of a fish is 9 inches long, the tail is as long as the head and half the body, and the body is as long as the 76 SIMPLE EQUATIONS. head and tail together. What is the whole length of the fish? Ans. 6 feet. 23. Divide the number 104 into two such parts that one eighth of the greater part shall be equal to one fifth of the lesser. Ans. 64, 40. Note. If one part be x, the other will be 104 — x. 24. Divide 188 into two such parts that the fourth of one part may exceed the eighth of the other by 18. 25. Two gamblers, A and B, engage in play. When they begin A has $200 and B has $152. When they finish A has three times as much as B. How much did he win? 26. A father has five sons each of whom is 5 years older than his next yo anger brother, and the eldest is five times as old as the youngest. What are their ages? 27. A father is now 4 times as old as his son, but in four years he will only be 3 times as old. How old is each? 28. The sum of $345 was raised by A, B and C together. B contributed twice as much as A and $15 more, and three times as much as B and $15 more. How much did each raise? 29. Divide the number 97 into two parts such that twice the one part added to three times the other shall be 254. Ans. 37, 60. 30. A father divided his estate among four sons, directing that the youngest should receive -J- of the whole; the next, $1000 more; the next, as much as these two together; and the eldest, what was left. The share of the eldest was $5000. What was the value of the estate, and the shares of the three younger? Ans. $14,000; $1750, $2750, $4500. 31. An almoner divided $86 among 59 people, giving the barefoot ones $1 each and the others $1. 75 each. How many barefoot ones were there? 32. An almoner divided $75 among a crowd of people, giving one third of them 50 cents each, one fourth $1 each, and the remainder $1.50 each. How many people were there? 33. A father, in making his will, directs that the second of his three sons shall have half as much again as the young- PROBLEMS LBADINa TO SIMPLE EQUATIONS. 77 est, and the eldest one third more than the second. The eldest receives $2250 more than the youngest. What was the share of each? 34. Divide the number 144 into four parts such that the first part divided by 3, the second multiplied by 3, the third diminished by 3, and the fourth increased by 3 shall all four be equal to each other. Call X tlie quantity to which these four results are equal. Then the part which divided by 3 will make x must be ^x, the second part must X be =, the third x-\-Z, and the fourth a? — 8. To form the equation note o the sum of all the parts is 144, which gives the equation to be solved. Ans. 81, 9, 30, 24. 35. Divide the number 100 into four parts such that the first being multiplied by 4, the second divided by 4, the third increased by 4, and the fourth diminished by 4 shall give equal results. 36. A man making a journey went on the first day one third of the distance and 36 miles more; on the second, one third the remaining distance and 36 miles more, which jast brought him to his journey's end. What was the length of the journey? 37. What number of apples was divided among three people when the first was given half the apples and half an apple more, the second half of what remained and half an apple more, and the third half of what then remained and half an apple more, which emptied the basket? 38. In the division of an estate between A, B, C and D, A got one tenth of the whole, B got half as much as A and $2505 more, got half as much as B and $2505 more, and D got $4050 dollars. What was the amount of the estate? 39. A person journeying to a distant town travelled on the first day half way, wanting 20 miles; on the second day half the remaining distance, wanting 5 miles; and on the third day half the distance still remaining and 10 miles more, when he still had 12 miles to travel on the fourth day. How long was his whole journey? 78 SIMPLE EQUATIONS. 40. A trader having made a profit of r per cent on his capital found that capital and profits together amounted to a dollars. What was his capital? Note. By the definition of percentage, r per cent of a sum means r hundredths of that sum. Hence r per cent is found by multiplying by T^. If the capital is c, r per cent of it is :7^, and when this is add- ed to the capital the sum will be 6 + r^. 41. A trader having increased his capital by 8 per cent found that it amounted to $4320. How much was it at first? 42. A merchant increased his capital by 8 per cent the first year and increased that increased capital by 12 per cent the second year, when the total amounted to m dollars. How much had he at first? 43. Of three casks the second contained 15 per cent more than the first, and the third 20 per cent more than the second. The first and third together contained 119 gallons. How much did the second contain? 44. A man investing a sum of money got 5 per cent inte- rest on it the first year, which he added to his principal, and then got 4 per cent on the amount for the second year. At the end of the second year he found interest for the two years to amount to $2457. How much did he invest? 45. Of two men A and B, A had in money 25 per cent more than B. B having received $225 then had 25 per cent more than A. How much had each at first? 46. The perimeter of a triangle measures 320 yards. The first side is 40 yards longer than the base, and the second side is half the sum of the base and first side. What is the length of each side? Note, By the perimeter of a triangle is meant the sum of its three sides. The base is any one of the three sides. 47. The perimeter of a triangle measures 'p feet. The first side is m feet longer than the base, and the remaining side is m feet longer than the first side. What are the lengths of the three sides? 48. A pedestrian made a journey of 125 miles in 5 days, going 3 miles less on each day than he did on the day preced- PROBLEMS LEADING TO SIMPLE EQUATIONS 79 ing. How far did lie go on the first day? How far on the last day? 49. A man bought 3 works each containing a volumes, and 2 works each containing h yolumes, for x dollars. What was the price of each volume? 50. A man bought a work of m volumes for x dollars. How much would h volumes cost at the same price per volume? 51. A line 64 feet long is divided into three parts such that the second part is one fourth longer than the first, and the third part one fourth shorter than the first and second together. What is the length of each part? 52. A factory employed men at $2 per day, 38 more women than men, paying them $1 per day, and 35 more boys than women, paying the boys 60 cents per day. The daily wages amounted to $197. How many operatives were there of each class? 53. In another factory there were one third more women than men, and one third more boys than men and women together. The wages were: men $1.50, women $1, boys 50 cents, — making a sum total of $276.50. How many operatives were there of each class? 54. Four men having received a sum of D dollars to be divided between them, the first 2:ot — of the whole, and the second half as much as the first and x dollars more; the remainder being divided equally between the third and fourth. How much did each get? 55. If from a sum of x dollars I take one third, then one half of what is left, and then one third of what is still left, how much remains? 56. Having an equation ax = 5, I want to put for a and h two numbers whose sum shall be 20, and which shall give f for the value of x. What numbers shall I use for a and h? 57. In another equation of the same form I want h to be greater than a by 12, and the value of x to be 2. What num- bers shall I use for a and J? 80 SIMPLE EQUATIONS. 58. In the fraction m-\-x dm + X I want to put for x such a quantity that the value of the fraction shall be f. What value of x shall I use? 59. What must be the value of m in order that the sum of the two fractions — -— and — --^ may be — ? m + 1 m-{-l ''■ /J 60. Out of a bin of wheat one man took one fourth of the wheat and 3 bushels more, and a second took one third of what was left and 5 bushels more. What he left was 5 bushels more than the first man took. How much wheat was in the bin, and how much did each take? Memoranda aitd Exercises for Eeyiew. Define : Equation; Identical equation; Equation of con- dition ; Solving an equation ; Root; Axiom. Axioms of addition and subtraction. Rule for transposition; G-ive examples. Prove the general rule. Write an equation which may be solved by simple transposition and aggregation of terms. Addition and Sub- traction. Division. Multipli- cation. Axiom of division. Solution of an equation by division; Rule. Write an equation with four terms which can be solved by transposition and division. C Axiom of multiplication. ^ When multiplication is required. 1^ Two methods; Explain both. SEOOKD OOUESE. ALGEBEA TO QTJADRATIO EQUATIONS. CHAPTER I. THtORY OF ALGEBRAIC SIGNS. Use of Positive and Negative Signs. 88. Opposite Directions of Measurement. Most of the quantities we have to express in algebraic lan- guage may be measured in two opposite directions. Examples. Ti7ne may be either time before or time after. Distance on a straight road may be measured from any point in two opposite directions. If the road is east and west, one direction will be west and the other east. The scale of a thermometer is divided so as to measure from zero in two opposite directions. 89. Positive and Negative Measures. In order to measure such quantities on a uniform system, the numbers of algebra are considered to increase from in two opposite directions. Those in one direction are called positive; those in the other direction, nega- tive. Positive numbers are distinguished by the sign +, negative ones by the sign — . If a positive number measures years after Christ, a nega- tive one will mean years before Christ. If a positive number is used to measure toward the right, a negative one will measure toward the left. If a positive number measures weight, the negative one will imply lightness, or tendency to rise from the earth. If a positive number measures property, or credit, the negative one will imply debt. 84 THEORY OF ALGEBRAIC SIGNS. 90. The series of algebraic numbers will therefore be considered as arranged in the following way, the series going out to infinity in both directions : .^ Negative Sirootioa. Positive Bireotion. ^ Before. After. Downward. Upward. Debt. Credit. etc. etc. 5, -4, -3, -2, - -1, 0, +1, , +2, +3, +4, +5, etc. etc. 91. Clioice of Direction. It matters not which direction we take as the positive one, so long as we take the opposite one as negative. If we take time lefore as positive, time after will be nega- tive; if we take west as the positive direction, east will be the negative; if we take debt as positive, credit will be nega- tive. 93. The Scale of Numbers. Positive and nega- tive numbers may be conceived to njeasure distances from a fixed point on a straight line, extending in- definitely in both directions, the distances on one side being positive, on the other side negative, as in the following diagram : * etc. - 7, -6, -5, -4, -3. -2, -1, 0,-f 1, +2, +3 , -|-4, 4-5. -j-6. +7, etc. I — i — I — I — \ — \ — ! — [— i — \ — I — \ — I — \ — I Positive direction. ■>■ >. Negative direction. < In the above scale of numbers the positive direc- tion is from left to right and the negative direction from right to left^ as shown by the arrows ; and the. distance between any two consecutive numbers is con- sidered a unit or unit step. * The student should copy this scale of numbers, and have it before him in studying the present chapter. MEANING OF MINUS. 85 93. Def. In the scale of numbers any number which lies in a positive direction from another is called algebraically greater than that other. Thus, — 2 is algebraically greater than — 7 ; " " " " -2', 5 " " " " -5. Def, The absolute value of a quantity is its value without regard to its algebraic sign. Example. The absolute vahie of — 3 is the same as the absolute value of + 3; namely, 3 without any sign. 94. Meaning of Minus. We now have the following new definition of the minus sign: Minus means opposite. A minus sign shows that the quantity before which it is placed must be taken in the opposite sense from that in which it would be taken if the sign were not there. Example 1. — x is the opposite of x. Ex. 2. — (— a;) is the opposite of — x; that is, it is x. Therefore — {— x) = x. Ex. 3. — [— (— ^)] is the opposite of — (— x), or of x. Hence — [— {— x)] = — x. etc. etc. EXERCISES. What is the meaning of: 1. — n years after Christ? 2. — (—n) years after Christ? S. —n years before Christ? 4. — {— n) years before Christ? 5. The thermometer is — 15° above zero? 6. The thermometer has risen — 20°? 7. The thermometer has fallen — (— 20°)? 8. To-day is — 25° colder than yesterday? ^ 9. To-day is — 25° warmer than yesterday? 10. To-day is — (— 25°) warmer than yesterday? 11. John lives — 2 miles east of William? Ans. John lives 2 miles west of William. 12. James lives — 4 miles south of John? 86 THEORY OF ALGEBRAIC SIGNS. 13. Smith is — 6 years older than Jones? 14. Jones is — 5 years younger than Brown? 15. John owes the grocer — $5? 16. Thomas weighs — 12 pounds more than Jane? 17. Mr. Weston is — $1000 richer than Mr. Brown? Answer the following questions in algebraic language: 18. James owed William $12 and paid him $7. How much did he still owe? How much did William owe James? 19. If James had owed $12 and paid $15, how much would he still owe? 20. If he owed William a dollars and paid him b dollars, how much would he still owe? How much would William owe him? 21. The thermometer was 15° on Sunday, and next day it was 22° lower. What was it then? 22. What day is the 0th of February? The - 1st? The -4th? The -31st? Note that the 0th day is the day before the 1st day. 95. Def. When two quantities are numerically equal but with opposite signs, each is said to be the negative of the other. Examples. — a is the negatiye of a. a is the negative of — a. X — y is the negatiye oi y — x, EXERCISES. What is the negative of: 1. 3a? 2. —25? 3. x—a — h'^ 4. -2«/ + 3a-5? 5. -(«-5)? 96. Lines to represent Nwrnbers. A number may be represented to the eye by drawing a line long enough to reach from the zero point to the number on any fixed scale. To fix the scale a length must be assumed as a unit, and a direction must be chosen as positive. The line must then have as many units as the number to be represented. A negative number is drawn in the opposite direc- tion from a positive one. ALGEBIIAIC ADDITION. 87 Examples. If we take this length as the unit, -f 3 is this line o — 2 this hne o and — 1 this line o The zero point is always the beginning of the line. V ♦ EXERCISES. < Taking any unit you please, draw, by the eye, lines to represent: 1. +2. 2. +5. 3. -4. 4. -3. 5. -1. Algebraic Addition. 97. Algebraic addition is the operation of com- bining quantities according to their algebraic signs, and is performed by taking the difference between the sums of the positive and negative quantities and prefixing the sign of the greater sum. Def. The result of algebraic addition is called the algebraic sum. Example. The algebraic sum of 4 -j- 5 — 7— 8 + 2 is + 11 -15 = -4. 98. Algebraic addition is represented on the scale of numbers by measuring off the positive quantities in the positive direction, and the negative quantities in the opposite direction, commencing each measure at the end of the one preceding. Example. To represent the algebraic sum 4-6 + 3-5-1, we start from and measure off 4 unit steps to the right, which takes us to + 4. Then we measure 6 from + 4 toward the left, which brings us to — 2. Then 3 to the right brings us to + 1. Then 5 to the left brings us to — 4. Then 1 to the left brings us to — 5, which is the algebraic sum. Thus 4-6 + 3-5-1 = - 5. 99. Formation of Algebraic Sums. We see from the above that the algebraic sum of several numbers may be 88 THEORY OF ALGEBRAIC SIGNS. formed by putting the lines which represent them end to end in their proper directions, the beginning, or zero point, of each one being at the end of the preceding one. The preceding example (+ 4 — 6 -f 3 — 5 — 1) is then represented in this way: Sum = o '■ Examples of Algebraic Sums. Three traders agree to divide their profits and losses equally. The first year A gained 13000, B $4000 and C $8000. The algebraic sum of the profits to be divided was therefore $3000 + 14000 + 18000 =: $15,000. Each man's share = $5000. Next year A gained $4000 and B $5000, while C lost $3000. The algebraic sum to be divided was therefore $4000 + $5000 - $3000 = $6000. Each man's share = + $2000. The third year A gained $2000, B lost $1000 and C lost $4000. The algebraic sum of their profits was therefore $2000 - $1000 - $4000 == - $3000. Therefore each man got — $1000; that is, he suffered a loss of $1000. EXERCISES. Draw a scale of numbers (§ 92) from — 8 to -j- 8, and then form the following algebraic sums by lines under the scale: 1.-4-^6-3+5 + 1. 2. 1-2 + 3-4. 3. 5-3-3-3-3. 4.-4-4 + 3 + 4 + 5. 5. The temperature at 9 points scattered equally over the country is +15°, +20% -8°, -13°, -2°, +7°, -15°, — 10° and — 12°. What is the mean temperature? Note. The mean of any series of numbers is obtained by dividing their algebraic sum by the number of terms. 6. A merchant in 5 successive years lost $3000, made $2000, made $8000, lost $1000 and made $3000, What was bis average aiiiiuul ])ryfit? ALGEBRAIC SUBTRACTION. 89 Algebraic Subtraction. 100. Subtraction in algebra consists in expressing tbe algebraic difference between two quantities. Def. The algebraic difference of two quantities is their number of units ajjart on the scale of numbers. EXERCISES. What is the algebraic difference between: 1.-3 and +5? Ans. 8. 2. +3 and +5? 3. -7 and -3? 4. + 5 and - 3? 5.-9 and + 9? 101. Sign of Remainder. Tlie sign of the alge- braic difference is shown by the direction from the subtrahend to the minuend. Examples. From —8 to — 3is+c-. — 3 — (— 8) = +5. From -3 to -8 k ~.'.-~S — {-3) = -b. Remark, When we subtract a lesser positive quantity from a greater one, this rule gives a positive remainder, so that the result is the same as in arithmetic. But the algebraic process takes account of cases whicli arithmetic does not; for example, tliose where the subtra- hend is greater than the minuend. This reversal of the case is indi- cated by the negative sign of the difference. EXERCISES. Give the values of the three quantities a-\-If, a —b and b — a in the following cases: 1. When a = + 13 and h = -{-^. 2. When a = -f 13 and ^ = — 8. 3. When a = - 13 and ^ = -f- 8. 4. When a = - 13 and b = ~ 8. 5. Of two bankrupts, A and B, A owes 15000 more than all he possesses, and B owes $3000 more. Which is the richer, and. how much richer is he? 6. In this case, what is the sum of their possessions? 7. Of two merchants, A and B, A made $2000 and then lost $3000; B lost $2000 and then made $7000. How much more was A worth than B? How much more was B worth than A? 90 THEORY OF ALGEBRAIC SIGNS. 8. During two years A gained a dollars and afterward lost b dollars; B first gained x dollars and then lost y dollars. Express how much A was worth more than B. Rule of Signs in Multiplication. 102. I. In multiplying a line by a positive factor, we leave its direction unchanged. II. In multiplying by a negative factor, we change the direction of the line. Illustration. Suppose the quantity a to represent a length of one centimetre from the zero point toward the right on the scale of § 92. Then we shall have a = this line i i The products of this line by the factors from -]- 2 to — 3 will be: rt X 2 I I I = + 2a. a X 1 . a X . a X - 1 a X — 2 a X - 3 zzz a. = 0. = — a. = — 2a, 3:: 3a. We shall also have — a = this line | | The products of this line by the factors from -f- 3 to — 2 will be: — a X3 . . — a X 2 . . — a X 1 . . — a X . . — a X - 1 . — a X - 2 . = - 3a. = - 2a. = — a. = 0. — + «. = + 2a. The preceding result may be expressed thus: In muUipUcation like siffus give plus. IIJILE OF SIGNS IN DIVISION. 91 103. Wlien there are tliree or more negative factors, the product of two of them will be -f^ the third will change the sign to — , the fourth will change this — to -|- again, and so on. Hence: The proihict of an evei^ number of negative factors is posi- tive. The product of an odd number of negative factors is nega- tive, EXERCISES. Form the following products by the method of § 37, and assign the proper signs: 1. - 2 X 3 X - 4. ' 2. 3 X - 5 X - G X - «. 3. — a X — a'^x X — a^x^. 4. m X cm'' X — c^x. 5. - 2Z' X 3Z* X ^bc. 6. 3^ X - hab X Ic 7. - 5 X - 6 X - 7 X - 8. 8. - 1 x - 2 X - 3 X «^ 9. -lX-lX-1. 10. -lX-lX-lX-1. 11. — ahc X — ah X —aXx, 12. m X — 1 X m X nf. 13. (-l)^ 14. {-ly. 15. (-ly. le. {-ly. 17. (-l)^ 18. --X-'' n m a b' . a 1 0. ^- X - -^3 X - ab X y 20. (- 1)^^ Rule of Sigfiis in Division. 104. The rule of signs in division corresponds to that in multiplication, namely: If dividend and divisor have the same sign, the quotient is positive. If they have opposite signs^^ the quotient is negative. Proof -\-mx -^ {-\-m) — -[-X, because -\- x x (4-^) = -\-mx. -f- 7nx -^ (— 7n) = — x, " — X X (— m) = + mx. — 7nx -h (+ w) = — X, " — X X {-\-m) = — 7nx, — 7nx -^ (— w) = -f- ^- *^ -\-x X {— 7n) — • — mx. The condition to be fulfilled in all four of these cases is that the product, quotieiit X divisor, shall have the same algebraic sign as the dividend. 02 THEORY OF ALGEBUAIC SIGNS. EXERCISES. Express the following diyisioiis, reducing fractions to their lowest terms: 1. a'—a''b-\-a¥— ab^-^ + ab. 2. «' — «V ^ ax" -, ax. 3. — ab — be — ca-\- abc. 4. d'b — Vc — &a -. rr^V^ 105. Rule of Signs in Fraetions. Since a fraction is an indicated quotient, the rule of signs corresponds to that for division. The following theorems follow from the laws of multiplication and division: 1. If tJie terms are of the same sign^ the fraction is positive; if of opposite signs., it is negative. 2. Changing the sign of either term changes the sign of the fraction. 3. Changing the signs of hotli terms leaves the fraction with its original sign. 4. The sign of the fraction maij he changed by changing the sign written before it. m, — n TT.v A .MP m — m — 1)1 T T? 1 — n — n n Ex. 2. m — 7)1 — 1)1 m, n —n n — ri Ex. 3. m — n 11 — m in — n a — b b — a b — a EXERCISES. n — in Express each of the following fractions in three other ways with respect to signs: m — n Q m — n c J. . /C. . o. c c )n — n 4. ^~y . 5. ^±-^. 6. i'^i-tr. ' m — ri ' p— (f ' p-\- ^1 — '' Write the folloAving fractions so tliat the symbols x and y ,hall be positive in both terms: x-a g a~x Q a-x- h ' c — y ' b — y ' m ^y — c 10. _ l:z±, 11. "^. 12. - i^£r i o—y b-^ hy m + y-c RULES OF SIGNS IJV DIVISION. ya Aggregate the following fractional expressions: X 2x 3x X — 2a a ni — m — 7H — m rn ' c- y c+y c-2y 2c - y ^** h - /i "^ - h h ' m — 71 — m — m — 2a . 2m — a lo. :: + a — n a — 71 n — a n — a 106. General Remark. It is necessary to the inter- pretation of an algebraic result that the positive sense or direction be defined in the case of each symbol which admits of either sign. The understanding is that the sense in which the symbol is defined is positive. Example. If we say, ^^Let t be the number of days he- fore,^^ we understand that days before are positive and days after negative. But if we say, ^^ Let t be the number of days after y^^ the reverse is understood. 107. Exercises in changing Algebraic Expressions into Nmnbers. Compute the values of the following expressions when a = 2, p = — S, c = 5, r = — S. 1. a -j-p. 2. a —p. 3. b -^ q. 4:. b — q. 6. a-\-r. e, a — r. 7. 2r — 3b. 8. qr — cp. 9. (pq-ac)''. 10. {cr-abp)^. 11. (A] 12. (- -Y. 13. ^' + ^~ \ 14. c"r-p'q ^^ pqr-abc a" — q-P' ' q-\-r ' ' pqr-\-abc 16. -^. 17 ^^P ~ ^^^ 18 P'-^'"' q — p aqp — bcr p ^ c 19. Compute the several valncs of the expression ax"^ -\-qx-\-r for iC=:-4, -3, -2, -1, 0, 1, 2, 3, 4. CHAPTER II. OPERATIONS WITH COMPOUND EXPRESSIONS. Sectiot^ I. Preliminary Defiis^itions and . Principles. 108. Aggregate. A polynomial enclosed between parentheses in order to be operated upon as a single symbol is called an aggregate. Entire. An entire quantity is one wliicli is ex- pressed without any denominator or divisor, as 2, 3, 4, etc. ; a, ^, x^ etc. ; 2aZ>, 2m^, ah {x — y\ etc. Formula. A formula is an algebraic expression used to show how a quantity is to be calculated. Reciprocal: The reciprocal of a number is unity divided by that number. In the language of algebra, Reciprocal of N = -^. Function. An algebraic expression containing any symbol is called a function of the quantity repre- sented by that symbol. Example 1. The expression dx'^ is a function of x. Ex. 2. The expi-ession — ^^— is a function of x. It is ^ a — X also a function of a. Degree. The degree of a term in one or more symbols is the number of times it contains such sym- bols as factors. Examples. The expression abx^y"^ is of the fourth degree in a^ l and x, because it contains these symbols four times as factors. It is also of the third degree in x, of the fifth degree in X and ijy and of the seventh degree in a, b, x and y. PIUJS'GIPLKS OF ALGEDllAIG LANGUAGE. 95 The expression al/x^ is of the fifth degree in b and :r, be- cause it contains h twice and x three times as a factor. When an expression consists of several terms, its degree is tliat of its liighest term. Principles of Algebraic Language. 109. First Principle. We may tnvploy a single symbol to represent any algebraic expression what- ever. The fact that a symbol is meant to represent an expression is indicated by the sign =. Example. The statement p ^ ax -\- hy means, we wa'ite the symbol p to represent the expression ax, + lij. Second Principle. Any algebraic expression may be operated with as if it were a single symbol. An expression thus operated on is enclosed in parentheses wlien necessary to avoid ambiguity. As a consequence of this principle, an algebraic expres- sion between parentheses may be enclosed between other j^a- i-entlieses, and these between others to any extent. Each order of parentheses must then be made thicker or different in form to distinguish them. Third Principle. An identical equation will re- main true wlien any expression or number is written in place of each symbol. EXERCISES. Let us put P ^ a -\- h', Q =a - h; and R = ah. It is then required to substitute in the following expressions a and b in place of P, Q and JL 1. PQ. Ans. {a-^b){a-b). 2. rQR\ Ans. {a -\- b){a - b){ab)\ 3. P(P-Q). Ans. {a-^b){a-^b- {a-b)\. 96 OPERATIONS WITH COMPOUND EXPRESSIONS. 4. P{R - P), 5. P(7^ _ 0. 6. Q{P - QR), Ans. {a - h) [a -^ h - {a - h)ah}. 7. Q(Q-FE). 8. P{QR~P), 110. Exercises in Compound Expressions. Compute :lie value of the following expressions when rt = 6, m = — 4, i — ^, n = — 5. 1. {(f« - b)m 4- 3w{^. Ans. (- 3/m + 3?i)9 = — 27, 2. {vm(^- ^0 4- ^H^ + ^0 1(^-^0- 3. u\b{m -f ?^) — rt(m — n)]. 4. Jrr + a{h - 7n)Y — \m + ni^m - u)Y. 5. \{a-hY-\-{in-nY\{in-^n). 0. (^/, — ^ 4~ ^^' ~ ^0^ (^'^ + ^0- 7. {(a - m){b - ny - {a - n){lj - mY}\ Sectio:n^ II. Cleaeit^g of Compound Pakentheses. 111. When expressions in parentheses are enclosed be- tween others, they may be removed by applying the rules of §§ 32 and 33 to one pair at a time. We may either begin with the outer ones and go inward, or begin with the inner ones and go outward. It IS common to begin with the inner ones. EXERCISES. 1. Clear of parentheses P — {a— \in — (x — y)] }. Solution, Beginning with the inner parentheses and ap- l)lying the rule of § 33 to each i3air of parentheses in succes- sion, the expression takes the following forms: P-\a- {m-x-{-y)] — P — \a — nfi -\- X — y] = P— a -\- m — X -\- y. Ans. 2. x—{'ila — x—{a-hx)-{-Za-^j)\. Solution, Eemoving the inner parentheses, we have x —\^a — X — a -\-bx-\-^a — x\ =^ X — '^a -{- X -\- a — ^x — 'da -\- X := — 2x -- 4ca. Ans. GENERAL LAWS OF MULTIPLICATION. 97 3. a + { - [a - 2b) + (2a - Z/) - (- ^a - 3b)}. 4. -(x- a) - {2x - 3a) + {6x + ^a). 5. |w- [^y? - (m-2m)]}. G. 2?/2.2: - J3m2; -\-p!/ - (omx - 2py) - {- ^py -r t^x)]. 7. _ {{]]jy _j- '^mu) +i - 2% - {mu - 3/;?/) + 5/^^ -f 2?/iw j. 9. a - X -{a-x-{a-\- x) - (x - a) J . 10. 3ax - 2hy - {ax - by - [ax -by - {by + ax)]}. 11. X — {2m -\- (5a- — a — b) — (7rt + 2b) }. 12. - (367/ + 2mz) - {2r/y - {3mz - 2cy)]. 13. p-q - {2p - 3q - {^p + ^(j) + (G^^ - ^) }• 14. p-^qJr{-3p - 4r/ - {2p - Try) + (3;^+ 4^)}. • Seotiox hi. Multiplk ation. General Laws of Multiplication. 112. Law of Commutation. Multiplier and wnltiplicand may he interchanged witliout altering the product. This law is proved for whole numbers in the folloAving way: Form several rows of quantities, each represented by the letter a, with an equal number in eacli row, tlius: a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a Let m be the number of rows, and n the number in eacli row. Then counting by rows there Avill be n X tn quantities. Counting by columns there will be m X n quantities. Therefore m X n = 7i X m, or nyn = 7mi. Remark. This is called the kw of eoinmntation, from commuto, I cxchaiiffe. 98 0PEnAT102s'S WITH COM POUND EXl'RESSlOyS. 113. Law of Association. When there are three factors, m, n and a, 7n{na) = {vin)a. Example. 3 x (5 X 8) :== 3 x 40 = 120; (3 X 5) X 8 = 15 X 8 = 120. Proof for Whole Numbers. If a in the above scheme rei)resents ii number, the sum of each row will be na. Be- cause there are m rows, the whole sum will be m(na). But the whole number of as is mn. Therefore 7ti{na) = {mn)a. Remark. This is called the law of association, because the middle factor, n, is associated first witli a and then with m. • 114. The Distributive Law. The product of a polynomial hy a factor is equal to the sum of the products of each of the terms hy the same factor. That is, m{p -[- q -\- r -X- etc. ) = mp -\- mq -\- mr -{- etc. (1) Proof for Wliole Numhers. Let us write each of the quantities j9, q, r, etc., m times in a horizontal line, thus: p -{■ p -\- p -\- etc. , m times = mp. q -\- q -{- q -\- etc., m times = mq. r -\- r -\- r -\- etc., m times = mr. etc. etc. etc. If we add up each Ycrtical column on the left-hand side, the sum of each will ha p -{- q -\- r -\- etc., the columns being all alike. Therefore the sum of the 7n columns, or of all the quanti- ties, will be m{p -\- q -\- r -^ etc.)t The first horizontal line of 7^'s being 7np, the second mq, etc., the sum of the right-hand column will be Trip + mq -\- mr + etc. Since these two expressions are the sums of the same quantities, they are equal, as asserted in equation (1). Remark. This is called the distnbutive law, because the multiplier m is distributed to the several terms jo, q, r, etc., of the polynomial. FOiiMATioy OF riionucTS. 99 Forinatioii of Products. 115. Products of Aggregates. Expressions between parentheses may be multiplied and the parentheses removed by successive application of the principles of § 38, and of the distributive law. EXERCISES. 1. a\m — n{p — q)]. Solutmi. Apj)lying the rule of § 38, we have a{m — np -\- nq] = am — anp -\- anq. Ans. 2. m{ax — h(y — z)]. 3. p{qm- l)\a-h)]. 4. PxP + qip- q)\- 5. m^\ax — '6{iix — hy) -\- 5{a.'^x — Ify)]. 6. n{p-\-q[x-Yy(a-b)\\. 7. r{t-m[x-y{p-q)Y^. ' 8. r{l-\-r[l^r(l + r)]]. 9. rjl- r[l - r(l - r)]}. 10. a-[-x\h^ x[c + x{cl -f x)'\ ], 11. d-x{c- q:\1) - x{a - .r)] }. 12. a^\m{b — a)'}i — n{a — b)m}ax^. 13. pq{a{b - c) + b{c - a) + c{a - b)]pr. 14. pq\a{b - c)-\- 2b{c - a) + 3r(a - b)\pr. 15.^]-^^- n [ m w I n n \ 'ill 116. Arrangements according to Powers of a SyinboL We may collect all the coefficients of each power of some one symbol, and affix the power to their aggregate, by the process of § 26. Def. When a polynomial is arranged according to powers of a symbol it is called an entire function of that symbol. Example. Arrange according to powers of x, x{iax - 2b) - b(3x'- F) + (a - 4b)x - nax'-\- ic. Solution. Clearing of i)arentheses, 4:ax' - 2bx - 3^.^-' + ^' + ax - ibx - dax'' -\- 4.c. 100 OPERATIONS WITH COMPOUND EXPRESSIONS. Taking the coefficients of the several powers of x, we find them to be: Coefficients of x^ = 4:a — 3b — 3a = a — 3b; Coefficients ot x = — 2b -{- a — 4:b = a — Qb; Term without 2; = b^ -\- 4c. Therefore the expression is equal to (a - 3b)x' + (« - (jb)x + Z*' + 4c. EXERCISES. Arrange the following expressions according to the powers of x or other leading symbol: 1. (a + x)x'' — {b — x)x^ -\- ex — d. 2. {a -\- bx)x + (c + dx)x'^ — a. 3. i^nx^ — n — 2))x -\- [2]) — q)x — ax^ -j- ^^^ — ^^' 117. By combining the operations of the two j)receding articles entire functions of one symbol may be expressed as entire functions of another symbol. EXERCISES. Arrange the following polynomials according to poAvers of x\ 1. {2x - 3x')f -{x^ 2x')y' + {3x' - 2x' - a)y. 2. (ax^ + bx^)y -\- y^'lax' — (3a + y)x} —2xy\ 3. {a + a^xy + x^)a'^ -\- {b -{- ex -{- mx^)a -\- 2cx — nx^. 4. aimix' -y)- n {x' -y')\-{- b{y' + y'x + yx' + x'). 5. (a' + 2b'x + 2cx' + x')xy - 7n{b' + 2c'x - Mx"), Multiplication of Polynomials by Polynomials. 118. Let us consider the product {a^b){p^q + r). liegarding n^ + ^ 'is ^ single symbol, the product by §114 is (a + b)p + (^ + b)q + (a + b)r. But {a -f- b)p z= ap -\- bp\ (a + b)q = aq -j- bq; (a + b)r = ar -\- br. Therefore the product is ^1^ + ^P + ^(/ ~\- M ~V ^^^ + ^^« MULTIPLICATION OF POLYNOMIALS. 101 It would have been shorter to first clear the parentheses from {a -{-h), putting the product into the form

12. (tf^l)(y^J^^)(y-4.)---f+y-^2. 13. (rt'^+3)(^-,1)K-5)-f-«-l. 14. (^^_2)(V-4)(r?,-8) -^r?'^+2«-fl. 15. a{a'-\~ 1)K+ a){a -1) -^ a''-2a\ ^^ 108 OPmATtONS WITH COMPOUND KXPUESSlONf^. 128. Use of Detached Coefficients. In dividing by the preceding method, there is no need of repeating the symbol after each coefficient. Example, x'— 4:x'-\- Zdx — 11 ~ 2^'+ 2^-1. X* X^ X^ X^ X^ X -fl_4 + 35-111 + 1 + 2-1 + 1+2-1 + 1^' - 62; + 13 . - 6 + 1 + 35 - 6 - 12 + 6 + 13 + 39 - + 13 + 26 - 11 13 + 3+ 2 Here the power of x which is written above each column of coeffi- cients is to be understood as if repeated after each coefficient below it . Result: Quotient, x^ — Qx -{- 13. Remainder, Zx + 2. EXERCISES. 1. a^- 2a'+ 4« - 8 -^ ft - 2. 2. 2a*+ 6«'+ 12a - 3 -^ «'- 2ft + 2. 3. 2ft*- 2ft'+ 6ft - 6 -^ a'- 1. 4. 3a:'+ Qx'- lix'- x'- 11a: + 15 -^ x'+ 2x ~ 5. 5. x'- 2.t'+ 4:x'^2x -b~- x'- 2x + 5. 129. Case II. When there are several algebraic sym- bols in the dividend. Rule. Arrange the terms of the dividend and divisor ac- cording to the poicers of some one symhol, preferring that sym- bol which has the greatest number of poivers. Then proceed as in Case /. Example 1. Divide a:' + Zax^ + '6a^x + ft' by a: + a. OPERATION. x^ Ar Sfto;' + 3ft'^a; + ft" \x -\- a x' + ax^ x' + 2ax + a' 2fta;' + 3ft'a: ^ax" + ^a'x ft^r + ft® a^x + ft' ~V DIVISION OB' ONE POLYNOMIAL BY ANOTHER. 109 Ex. 2. Divide x"" — ax^ -f a{l) -\- c)x — ahc — hx"— cx^— hex hj X — a. Arranging according to § 116, we liave the dividend as follows: x^ — (« + J + c)x^ -\- {ab -\- be -{- ea)x — abc \x — a x^ — ax"^ x'— {b-\-c)x -j- be (b -f c)x'' -f (ab -\- be -\- ea)x (b -\- c)x^ + (^^ + ^^)^ bex — abc bcx — abc 4. x' + o\ 7. x' - c\ 10. x" - e\ EXERCISES. Divide the following binomials hj x-{- c and hj x — c\ 1. x"" + e\ 2. x' + e\ 3. x' + e\ 5. a;" + e\ 6. a:^ + e\ 8. ^' - c^ 9. x' - c\ 11. x' - c«. 12. :z:^ - c\ 13. Divide a;' + (^ + c)^:' + ^>c^ + aix"" -\- bx -^ ex -{- be) by .T + a, then by a; + Z>, then hy x -{- c. 14. Divide a' + J* + c* - 2{a'b' + ^V' + cV) by (a-^b-{- e)(a -{- b — e){a — 5 + c){a — b — c). Divide by the four factors of the divisor in succession. 15. Divide {ax + bf -[-{a- bx)' by x' + 1. 16. Divide x* + Ua* by x"" — 4«.r + 8r?'. 17. Divide 4.a' + Sa'^Z* -^ b' - da' -\-3ab - Ihja + b -I. 18. Divide x'-\- y'-\. 2x\f by (a; + iff. 19. Divide 2:"+ if— 2.*+ Z>« by (a"- ab + Z'')(«''+ «5 + V), 22. Divide (« + 2Z>) «'- (Z> + 2«)^' by r? - Z». 23. Divide {x^-\- a')(a:'+ w^ + a') bya;*+ aV+ «♦. 110 OPrntATIO^S WITH COMPOLWJJ i-:x rUK.s.SlU^^S. Division into Prime Factors. In the following exercises we apply the definitions and processes of §§ 50 and 51 to compound expressions. 130. Difference of Tivo Squares. By § 123 the difference of two squares is equal to the product of the sum and differ- ence of their square roots, which sum and difference are there- fore the factors. EXERCISES. Factor: 1. a^-h\ Ans. {a J^h)[a -h). 2. 7}f- n\ 3. ^nf- n\ ■ 4. ^m'- 9^^ 5. IG/- 25f/'\ 6. x^-l. 7. x'-l, 8. a;'— rt'. Ans. {x^-^ a''){x''— a"). 9. x'- 1. • 10. ^x'- 4. 11. x'-a\ Ans. {x''-^ce){x'-a'') = {x''-^a:'){x-^a){x-({). 12. x'- 1. 13. Ua'-V. 14. m^x — n'x. Ans. {iif— n^)x = (w + n)i(m — n)x. 15. {a'- lf)y. IG. {nf- 7i'){x-]-y). 17. {m''-n'')\x''-if). 18. x^—xif. Ans. x{x -\- y){x — y)- 19. a'-4:ax\ 20. 9m'- 4mV. 1 14 21. a'-~ 22. -;,- - jj-. a^ a h 131. Perfect Squares. By §§ 121 and 122 a trinomial consisting of the sum of two squares j9??^5 or minus twice the product of their square roots is the square of a binomial, and may therefore be divided into two equal factors. EXERCISES. Factor: 1. x:'-2x-\- 1. Ans. {x - ly or (1 - x)\ 2. x" -4:X-^ 4. 3. a" + 4.ax + 4.x\ 4. a" + ^a 4- 9. 5. w' - 2w'.'r + x\ 6. 2:^ + lOo;^ + 25. 7. 49F+14A: + 1. Divisions L\T(j VlUMh: FAuTOltS. m 8. x^ + 'Hax' + era: Ans. a'{;A' + a)\ 0. a' — 4:a'w + ^am\ 10. 2;^ + 2«a; + a' - b\ Ans. {x J^ay-F = (x-}-a -f- Z»)(rr + ^ - b). Note. Here we combine the methods of this and the pieceding sec- tious, 11. x"- -2ax-{-a'-h\ 12. m' — 4w - x' + 4. 13. 3a' - 6«m + 3wi'. Ans. ^ce-2am-^m'')=^a-m)\ 14. Sft'^ - 6aZ> + U\ 15. 3w' - 6«& + 3^^' - 4^1 IG. 5m'+5?i'^-20//-10ww. 17. 2A' + 8F- 8//1- - 18AU-\ Ans. 2(7/. + 2h + 37/y[;)(7i + 21c - 37/y^). 18. 3/-12p^+12^'^-27r?'y. 19. ax'-\-2(i'x-\-a' -^a\ 20. wijt?' + 4m> + 4^//' - 1G?»>*. Ans. ^6' + ly. 21. '^ + ^ + \. 22. c^-^ + l- 24. "'\ 4 + 4't 20. a' 2ac & 111^ mil n^' 23. 25. 27. 132. Oilier Trinomials. If we perform the multiplica- tion (^ + ci){x + ^'), we find as the result x" + (a + 2')^ -y cib = {x + rO(:?^' + ^). We see that in the trinomial (1) The coefficient of x is the sum of a and ^, and (2) The term independent of x is their p^^oduct. Hence if in a trinomial of the form 0^ -\- px -\- q we can find two numbers whose sum is p and whose product is q, the trinomial may he factored. 112 0PML4.TION8 WITH COMPOUND EXPRESSIOI^S. Example. x^ + bx + 6. Here we find by trial that 2 and 3 are sucli numbers that 2 + 3 = 5 and 2 X 3 == 6. Hence x" -]-bx -\- Q = [x-]- 2)(^ + 3) EXERCISES. Factor: 1. x"" -\-^x-ir 2. 2. x''-\-^x^ 3. 3. x'-\-1x-\- 10. 4. c'^ + 6c + 8. 5. x' + W + 12. 6. c' + Sc' + 12. 7. m" + 9m' + 14. 8. m' + 8m' + 15. 9. ax" + "iax + 12«. Ans. a{x + 4)(a; + 3). 10. rri'x'' -\- 12mx + 35. 11. !!^ + i3!!^ + 40. 12. ^; + 11^ + 24. 71 71 C C 133. If the second term of the trinomial to be factored is negative and the third positive, both the quantities a and h must be negative. For the product being positive, the signs must be like, and the sum being negative, one at least must be negative. The fornf is {x — a){x — b) = x"^ — (a -\- h)x -{-ab. EXERCISES. Factor: 1. x"" - bx-{- 6. Ans. (X- 2)(^ - 3). 2. x" - ix-^ 3. 3. x^- ^x^ 12. 4. x" -^x-{- 18. 5. m* - 5m' + 4. 6. b' -Qb' -\- 8. 7. c' - ^& +20. 8.. ^x' - %nx + 60. 9. 5^^^ - 25^ + 20. 10. W - %W + 21. 11. -2 h 35. a a 12. ^'\ - 11^- + 24. n n 13. i'-\+- 134. If the last term of the trinomial is negative, one of the quantities a and Z* must be negative and the other positive, and the coefficient of the second term must be their algebraic sum. DIVmON INTO PRIME FAOTORS. 113 Hence, in this case, we must find two numbers whose product is the third term and whose numerical difference is the coefficient of a-. Then prefixing to the greater number the sign of the coefficient of x, and to the lesser number the opposite sign, we have the quantities to be added to x to form the factors. EXERCISES. Factor: 1. x' -\-^x- 8. Ans. {x -\-4:){x- 2). 2. x' -2x- 8. 3. a' -^U- 3. 4. a' -^a- 3. 6. m' — m - 9. 6. r>i* -{- m — Q. 7. 5a;' - lOx - 15. Ans. ^{x - d){x + 1). 8. 5^^^ + 10:c - 15. 9. -[ + 2- - 3. W' ' 7i 1 3 5 . By combining the above forms others may be found . For example, the factors (a' -{-ab-{- h'){a' - ab -}- h') (1) are respectively the sum and difference of the quantities a^ + ^' a^d ^^• Hence the product (1) is equal to the difference of the squares of these quantities, or to (a' + by - a'b' = a*-\- cfb' + b\ Hence the latter quantity can be factored as follows: a' 4- a'b' + b* = {a' + ab + b'){a' - ab -\- b'), EXERCISES. Factor: 1. m* + m'7i' + n\ 3. a* + 9a'b' + 81b\ 5. 3x* + 12a'x' + A8a\ 7. 16 + 4rt' + a\ 9. ^:+i+-t a* a'm'' m*_ C C 71 71 2. a' 4- la'b' + Wb\ 4. Ua' + 4.a''b'' + b\ 6. U* + 18^''^ + 162. 8. ?+4+''- 10. ^^■+>- 12. i'+.^- Teachers requiring additional exercises in factoring will find them in the Appendix. 114 OPEnATlONS WITH COMPOUND EXPTtE88I0N8.. Factors of Binomials. 136. Let us multiply OPERATION. X — a a;" _^ ^^ » - 1 _|_ ^,2^ n - 2 _^ ^^3^n-3_|_ . _ , _|_ « " - l^g — ax""-^ — (ex""-^— a'x''-^— .... — n^'-'^x— a'' Prod. , a? 0^~-^". Putting — a in j)lacc of b, and supposing n — 2, 3, etc., in succession, we have X — b = X -\- a (because b — — a), x^ — b^ = x"" — (f (because b'' — a"), x^ — b'^ = x^ -\- CI? (because b"" — — we conclude: Theorem. When n is odd, the binomial x"" + «" is divis- ible by X -\- a. When n is even, x^ — «" is divisible by x + ct. EXERCISES. Divide the following expressions by x -{- a, and thus find their factors: 1. x' + a\ 2. x' - a\ 3. x' + a\ 4. x' - a\ When n is even, x^ — «"can, by § 136, be divided hy x — a as well as by :r + ^« Therefore both of these quantities are factors, and the binomial may be divided by their product, x"" — a". EXERCISES. Divide the following by x^ — a^, and thus factor them: 1. x' - a\ 2. x' - a\ 3. x' - a\ 4. x'' - a'\ Lowest Common Multiple. 138. The L.C.M. of any polynomials maybe found by factoring them and applying the rule of § 55. Example. Find the L.C.M. of ^ct" - 2b\ a' + 2ab + b% a' - 2ab + b\ 3a* - 3b\ Factoring, we find these four expressions to be ^a -f b){a - b), (a + b)', {a - b)\ 3{a' + b'){a + b){a - b). Applvmg the rule, avo find the L.C M. to be 2 . 3(a -[- by{a - bfia' -f b'). 116 OPERATIONS WITH COMPOUND EXPRESSIONS. EXERCISES. Find the L. CM. of: 1. a' - b\ a' - b\ 2a + 2b. 2. a'-l, a' -]-a-2, a-1, a-\- 2. 3. 2a - 1, 4a*' - 1, 4ta' + 1. 4. x' - X, x:' -1, x' -^1, x-\- 1. b. X, X — a, x^ — a^f X -\- a, 6. X — a, x^ — ax, x^ — ax"^. 7. iy' - W, 4:y' + U% 2y + 2b. 8. x' — a\ x^ — a\ 9. Cy a, a — 6', a-\- c. 10. 21a' - ^c\ 9a*^ - U\ da + 2c. 11. 2a - b, 4:a' - b\ 4a* + b\ 12. a^ 4- 4a^ + 46% a^ - ^ab + 46=^, a» - 4^>'. 13. % + z), y{x - z), xyz. 14. m — n, m^ — n", m' — w^ 15. 7?i — n, m^ — 2mn -\- n^, m^ — n\ 16. a-\-b, a' - b\ a' + b\ 17. 46-' - ^cn + 7i% 4c2 - n". 18. a + 1, a'' + 1, a^ - 1. 19. a; — 4, a;'' - 16, a;' - 8, a; - 2. 20. a — 6, b — a, a-{-b, b -\- a. 21. a: - «/, ?/» - ic% x-\- y. 22. it? - 2^, ^ - 2p. 23. ^>* — a^c% ac — b, a'^c' — 2abc + b\ 24. aa: + ay, x -\- y, a. 25. ex + c?/, cV - 6'Y, c'. 26. 2fla:, 3a'^a:% 4:a'x\ Sectiot^^ y. Fractions. 139. Fractions having compound expressions for the numerators or denominators may be aggregated, multiphed, divided or reduced by the methods of §§ 57 to 76. The general rule to be followed is: 1. Observe what part of the expression constitutes the numerator, and what the denominator. 2. Operate o?i these expressions accordmrj to the rules prcr scribed for fract\on9% FRACTIONS. 117 EXERCISES. 140. Multiplicatio7i by Entire Quantities, Execute the following multiplications: ^ a-\- h ' , , . a^ — b"^ 1. — - — Xia — b). Alls. . x-y x-y 2. ^-±^- X (^ + 2b), 3. ?^ X {2x - y). x — y ^ ' a-\-b ^ ^' . ax -{-by , , . „ 1 + a; + 2a;^ .. ^ . 4. — -V--- X (ay - bx). 5. , , ^r-, X (1 - 2iz:). „ d^—ab-\-¥ , ,, ^ a^b , , ^ '■ ^+I~ X (" - *)• ^- ?^^ X (" + 2')- Here, because x^ — y^ = (x — y){x + ^), the multiplier is a factor of the denominator; so we operate by § 60, getting a-\- b ,, as the answer. x-y 8- :^4? ^ (^ - ^^)- '•*• ;7^^+A^ >< (» - *)■ Here denominator and multiplier have the common fac- tor a — 2x, which we suppose cancelled. Multiplying by the remaining factor, we get {a-^x){a-{- 2 x) _ a'-^ Sax + 2x'' - — _T •. Ans. a — 2x a — Zx 11. , ^^ ' -—-. X K-^O- 1^. -T^A X {a-\-b). ■ 7ri —2m7i-\-n^ ^ ^ a -\- b ^ ^ ^ fl^ + 2^ a — 2b , -. * 13. ^ri:^iX(a+M. 14. -^^---,x(a-b). 1^- j-r^-T^i X (^ + .). 16. ^^:^^:^^ X (b - c). i^-^xK-5^). Here the denominator is a factor of the multiplier, so thq product is an entire quantity, namely, a X (a — b) = a"" — ab. 18. ^^- X {m'- 2}nn^n'). 10. "-^\ x ia' - b'). 118 OPEltATIOSS WirU COMPOUND EXPUliti^IOA-S. 3S. i!i+|^ X (..' - 8/). 23. ^iil^- X (.. - V). x — 'Zy ^ ^ ' ax — ay ^ '' ' 141. Dividing hy dividing the Numerator or 7nuUi]jly- ing the De7iominator (§§ 58, 59). a -\-b ^ ^' Ans. 4. a-b '' ("+^)- 5. .+ 1^(^+^)- 6. xXx''^^ !)• 7. .-2-^ ^(^ + ^^)- 1 x' -\-"Zx-{-X 8. ^4-K- ^')- a — 9. — ^ 4- (a - 2^). 10. -^^-- -^ (?yi^ - ^r). 11. ^^- - (2^ + c). 12. -^^|- -^ (7/i + n). b -{-2c m — 2n ^ ' / 13. 5£_^^^ _^ (:^;« _ y^). 14 — ^ - (a.-^ - f). X -\- y ^ ^ ' x — y ^ ^ ' 15. - ^ "^ '\ , -- (2;^ - ?/^). 142. Reduction to Lowest Terms (§65). 2. .. 2. „ — „ ic — rt X — a a;^ ~r 3. X -\-a x' + «" ^ x' - 2^y + // t). ^^-2/^ x^ — a' a — b -\- c (f - 2ab + b' -(f FRACTIONS. 11^ am + hn + ^mx h^ + GZ>g + 9^' a'x' - 2abxy + by a' - 2a'b -\- a'b' ^' '^-by * «^-a'^^'^ • c'-[-^c^x-^4:d'x'' 1 — X ~^V^=^1^^''~' 1 - o;^* bx + by ' «'— a^ + b^' ^2_ 5;;^ + 6* * x'-x- 6* 143. Reduction to Given Denominator (§67). 1. Express a -{- b with denominator a -^ b. (a^bY a' + 2ab-}-b' Ans. ^ ' ^ — a -{- b a -\r b with denomi 2. Express , with denominator a^ — J^ 3. Express a-{- 2x 4. Express 7- " 5. Express — ** 6. Express—^— " 7. Express ^^ X — i 8. Express ,^ ^ '' ^ a -^ 2x 9. Express a" — ^x\ J:r + %. W^2 ^ + ^^ m^ -n\ a;* -1. 46? »^' - «*. x' -a\ a — X 10. Express — . ^ . — with den. <5^' + 2ac — c' + J'» a-\-b -\- c 11. Express — i-r '' *' ^^ - 5a; + 6. ■^ a; — 2 12. Express ^\ '^ ^\'\ 1' " " a* - ^'^'^ + ^*. 13. Express ^ " " m' - 2w/^ + w' - a;'. m ^ n — X 120 OPERATIONS WITH COMPOUND EXPRESSIONS. 144, Reduction to L.C.D. and Aggregation. The fol- lowing algebraic sums of fractions are to be reduced to their L.C.D., and then aggregated into single fractions by the methods of §§ 68 to 72. a-b^a-\-h' a'-¥~~ a' -b'' a -{-b a — b' 3 ^ + 2y x-^ ^ 1_ 1__ ' X — %y x-\-%y' 'a a ^ b' K _i L_ f. a , ^ a-\-b a — b' ' h — c~ c — b' 1.1.1 ^ a a' . a' 9. a ' a-[-b ' a — b' 'a — b a"" — b"" ^ a' — b*' a^b a-b AF a-b a-\-b a^ -¥' 10. — — '—-. 11. a + -7 . i?;+la; — 1 b — a 11 n^ 12. 1 - :; :r-^^. 13. a -\- b ^ 1 — a 1 — a a — b 14. « + i + -_^. 15. ;. - 2, - -^^i^. 16. ^ ^ 1 - c 1 + c 1 1 4- 10c c''- -1 * 3 ^'^' a-l o + 2 a'+ ia + 4' 18. i±l-^^_(^ + y) 1 + 1). 19. (w + 7i) — (m — ^i) . ^ ^ \n m) ^ ' \n m j 20. (a-*)(i-i) + (« + J)(l+y. 33. 5^-+-!- + ^. 24. i + ^'-*^y+%- . FRACTIONS. 121 145. MuUipUcation of one Fraction hy Another (§ 75). Note. In these exercises, see whether common factors can be cancelled before multiplying. a x' -¥ X ' X — b b We see that the first denominator and the second numerator have the common factor x — b. So we cancel it, and only multiply x + b_a{x + b) ''^~~r~ b ' 2 ^ + ^ V ^ + y Ans L ^- ^-^-.y^^ a" - b'' ^' (a -b){x-' y)' „ a -\-b — c a — b — c a -{-b -\- c a — b -\- c -G-??)(|-^)(' + r^)- ^ a^b I a' , b' a — b \ a — b ' a-\-b. 6. ^^f-l+ -^ x-\- Qy \ X — ^y X -{- 2yJ ^ a^ -\-ab a^ — b^ a' + ^' «'^ + «^'' _ «' — '^ax -\- x"^ X a^ A- X* 8. . ! X X — -T— . a -\- X a — X X 1 «- + y «+a a — b a — b a — b -•^ ^ + V x"^ — y^ 1 10. — ^-^ X 3 , 3 X 11 _^lzZ_v ^-^ a'-Ub^b''' a'-\-ah' 12. fl + ^ + ^Vi-!i + ^). V X ' aj\ X ^ aJ 146. Division by inverti7ig the Divisor (§ 76). The dividend and divisor are first to be aggregated and reduced if necessary, 1. n ^-^ n . ^ n n*-l ^ n^—l _ n'-l n _ n^-\- 1 n n n n — 1 n 122 OPERATIONS WITH COMPOUND EXPRESSIONS. a ^ h a — h ^ _j — _ _i_ ^ ' a -{- b a — b a-\-b 71 — 1 71 -\- 1 71 -{-1 ' a-\-b a — b ' a-\-b' a X ^ 2x X -\- a X — a ' X — a a -\-b a — b \ 2a ' a — b a -{- b ' a -\- b' 9, ' ' ' ' x-1 x-^1 x' ' x\x' - !)• a -}- b m — n ^ 1 a — b 7n-{- 71 ' {a — b) {in -\- 71)' 13. l + ^!_ + ^+ • . a — X a — X X -f- a 147. deduction of Complex Fractions. Def. A complex fraction is one either or both of "vrhose terms is fractional. The minor fractions are those which enter into the terms of the complex fraction. Their terms are called minor terms. Problem. To reduce a complex fraction to a simple one. EuLE. Multiply both terms by the L.C.M. of the minor denominators. Reason. 1. The value of the fraction remains unchanged (§64)- 2. The minor denominators are removed. m n Example 1. Reduce to a simple fraction. m n n 7n FRACTIONS. 123 The minor denominators are m and n, and their L.C.M. is fan. Then: Numerator, 1 X mn = m^ + n*. 71 m Denominator, X mn = m^ — n^. n m Therefore the given fraction is equal to —2—^^ — ^. m m m — —Xnq Ex. 2. n _ n " _ mq p ~ p ^ np' -- — X nq ^ q q EXERCISES. Beduce the following complex fractions to simple ones: m 3. — -% m m y a 5. J. 7. 9. 1 n — \ ^- 1 1 1 + - n. ^ 1 - 9 '+r « 1 — ^ h a b 4. X y a-\- b a-b 6. a — b' a-\- b • a b 8. a+b ' a-b a a a-\-b ' a—b 10. ' 1. a 1 X X 1+1-. VZ. 1 + x • 124 0PEUATI0N8 WITH COMPOUND EXPBESISIONS, 1 1 1 14. ~ . 13. - l-x X 15. ^ 1 + x m — n 1 1 ^-^ 17. 10 ' m + n - + -+- x^y^ z x^ y '^ z 1 x-\-y X -\- a X — a X — a X -\- a X -\- a X — a X — a X -\- a a b 18. "-* "+^. a a -\- b a — b m m — n 111- 20. KZ»±». — h T H — 1^' , M 4- n a b c ' — 71 m — n Section YI. Substitutioi^. 148. In algebra we often have to substitute an expres- sion for a single letter in some other expression. This is done by making the substitution as in Chapter II., § 109, and then reducing. EXERCISES. Make the following substitutions: 1. In a;^ + -5- substitute x = -, Ans. -- 4- cC X a a^ ^ 2. In substitute x =' —. 1 — X a X 3. In :; substitute x = 1 — a. 1 — X 4. In -— — substitute x = l-x I- a 5. In 1 substitute x := 1 . X a G. In x^ + 2 + -2- substitute x = y. THE G.G.D. OF TWO J^UMBEM8, 125 7. In ax" -j- Ifx substitute x = a — I), 8. In -n- substitute x = -r, X a 0^ x^ 9. In — substitute x =^ a, x a ft I QC 10. In substitute x = am, a — X 11. In — ■ — substitute x = . X -{- a a — c 12. lix = T and b = , find x in terms of c, 1 — b 1 — c Section VII. The Highest Common Divisor. The G.C.D. of Two Numbers. 149. Theorem I. If ttvo numbers have a common divisor, their su7n will have that same divisor. Proof, Let <^ be a common divisor; m the quotient of one number divided by r/; n the quotient of the other number divided byrf. Then the two numbers will be dm and dn\ and their sum is d{rn + n). This sum is evidently divisible by ^; and the quotient w + w is a whole number because m and n are whole num- bers; hence follows the theorem as enunciated. Theorem II. If two numbers have a common dirisor, their difference will have that same divisor. Proof. Almost the same as in the last theorem. Cor. If one number divides another exactly, it will divide all multiples of that other exactly. Remark. The preceding theorems may be expressed as follows: A common divisor of two numbers is a common divisor of their sum, difference and multiples. 126 OPERATIONS WITH COMPOUND EXPRESSIONS. Eemark. If one number is not exactly divisible by another, a remainder less than the divisor will be left over. If we put D = the dividend; d = the divisor; q = the quotient; r = the remainder; we shall have I) = dq -\- r, or D — dq = r. Example. 7 goes into %Q 9 times and 3 over. Hence this means 66 = 7 . 9 + 3, or 66 - 7 . 9 = 3. 150. Pkoblem. To find the greatest common divisor of two numbers. Let J[f and N be any two numbers, and let if be the greater. 1. Divide if by N. If the remainder is zero, iV^will be the common divisor required, because every number divides itself. If there is a remainder, let q be the quotient and R the remainder. Then M - Nq = R. Let d be the common divisor required. Because M and iV^ are each divisible hj d, M — Nq must also be divisible by d (Theorem II.). Therefore R is divisible by d. Hence: a. Every common divisor of M and N is also a common divisor of JV^and R. Conversely, because M= Nq-{- R, /?. Every common divisor of iV^and R is also a divisor of M, and therefore a common divisor of ilf and Nf. Comparing a and ^ we see that whatever common divisors M and iV^ may have, those same common divisors and no others have aYand R. Therefore the greatest common divisor of M and N is the same as the greatest common divisor of N and R, and we proceed with these last two numbers as we did with Jf and N, 2. Let R go into N q' limes with the remainder R\ Then X = Rq' + R', or NT - Ra' = J?\ THE Q.a.D. OF TWO NUMBERS. 127 Then it can be shown, as before, that d is a divisor of R'y and therefore the greatest common divisor of R and R'. 3. Dividnig R by R\ and continuing the process, one of two results must follow. Either — a. We at length reach a remainder 1, in which case the two numbers are prime to each other; or, /?. AYe have a remainder which exactly divides the pre-' ceding divisor, in which case this remainder is the common divisor required. To clearly exhibit the process, we express the numbers J/, i\^and the successive remainders in the following form: M = JV. q + R, (R < J^); jsr = R.q^-i-R\ {R' If now we find the H.C!D. of these expressions (2) and call it D, then Dx will be the H.C.D. of the given polyno- mials (1). We shall find, by going through the process, D = x^l. Therefore the H.C.D. of (1) is x"" -\- X. 154. Throwing out Factors. If we carry on the pre- ceding process without modification, we shall commonly find that numerical fractions enter into the remainders. These may be avoided by applying the following principle: If a divisor contains any factor prime to the dividend, if may he rejected before dividing. The reason of this is that we are seeking, in the final re- sult, only for the product of all those factors which are common to both divisor and dividend. Therefore a factor contained in one but not in the other is not a factor of the H,C.D. sought, and hence may be rejected. For a similar reason, we may multiply any divide7id by any factor prime to the divisor. Example. Find the H.C.D. of x' - 4:x' + 12a;' + 4:x' — 13:r and X* - 2x' + 4:x' -{- 2x - 5. First Division. x' - 4.x* -f- 12a;'' 4- 4:x' - 13a: | x* - 2x' -}- 4a;' + 2a: - 5 x' - 2a:^ + 4:x' + 2a:' - 5x x -2 _ 2x'~-{- Sx' + 2a;' - 8a: - 2a;* -j- 4a:' - 8a;' - 4a; + 10 4a;' + 10a:' — 4a; — 10 = first remainder. This remainder contains the factor 2, which is not con- tained in the dividend. So we divide by it. But then the first term of the next divisor, 2a:', will still not go into ar* 130 OPERATIONS WITH COMPOUND EXPRESSIONS. without a fractional quotient. So we multiply the new divi- -dend by 2. Second Division. '^x' - 4.x' + ^x"" + 4:x - 10 I 2x' -\-5x' -2x-6 2x' + 5:^:=' - 2x' - 5x ^' - | - 9x' + 10^:' + \^x — 10 - dx' - ^x' H- 9^ + -¥ -^^x"" — -%^- = second remainder, or \H^^ ~" 1) = second remainder. To have avoided all fractions, we should have multiplied the dividend by 4. But we could not know this until after we had begun the division, and the failure to multiply does no harm. We now reject the factor %S- from the remainder, leaving ic^ — 1 as the next divisor. Third Division. 2x' -\-Dx' -2x- d\ x' -1 2x^ — 2x 2x +5 5x' - 5 6x' - 5 = third remainder. Hence ^^ — 1 is the H.C.D. sought. EXERCISES. Pind the H.C.D. of: 1. x' + x' and x' - 2x' + 2x' - 2a; + 1. 2. 2x' + x' - 5a; + 2 and 4:x' - ^x' - 5a; + 3. 3. a;' + 1 and x' + ax"" -\- ax -{- 1. 4. x' - ^x' + 2a;' + a; - 1 and x' - x' - 2x + 2. 5. 2a;' - llo;' - 9 and 4a;' + 11a;' + 81. 6. 6x' - Hax^ - 20a'x and 3a;'^ + «a; - 4«l 7. 12a;* + 4a;^ + 17a;' - 3a; and 24a;' - 52a;^ + 14a;' 8. a* - a' -{-20" -\- a -\- 3 and a' -{- 2a' - a — 2, 9. ea* + a' — « and 4:a' — 6a' — 4« + 3. CHAPTER III. EQUATIONS OF THE FIRST DEGREE. Section I. Equations with One Unknown- Quantity. 155. Bef. An equation of the first degree is one which, when cleared of fractions, contains only the first power of the unknown quantity. All equations of the first degree may be solved by the pro- cesses of multiplication, transposition and division explained in §§ 82 to 86. These processes are embodied in the follow^ ing KuLE. 1. Clear the equation of fractions, 2. Transpose the terms which are multiplied hy the un- known quantity to one meml)er and those which do not con- tain it to the other. 3. Aggregate the coefficients of the unknown quantity; and 4. Divide loth members by the coefficient of the unknown quantity. Example 1. Let us take the equation x — 1 _ 2a; — 6 2^ +• 10 ~" Ix +2* Clearing of fractions, we have ^x" - ^Qx - 14 = ^x"" + 8a; - 60. Transposing and reducing, 46 = 34a;. Dividing both members by 34, _46_23 ^ ~ 34 ~ 17* This result should now be proved by computing the values of both 23 members of the original equation when — is substituted for x. 132 EQUATIONS OF THE FIB8T DEGSEE. Ex.3. -^ ^ = -.-P- X — a X -\- a x^ ^ d^ Here the L.C.M. of the denominators is 7^ — a'. Multi- plying each term by this factor, the equation becomes m{x -\- a) — n{x — a) = p^ or (m — n)x -{- {m -\- n)a =p. Transposing, {m — n)x z= p — am — an; , p — am — an whence x = ^ m — n EXERCISES. Solve the following equations, regarding x, y or u as the unknown quantity: 1. i + I = 1. 2. -^— + -^ = 2. a X — a X — u — d u — 1 u -\-a u -\-^a ^ 5^-1 9t/-6 ^ 9y-7 7 "^ 11 5 7. {x -2) {x-5) = {x- S)x. .0 ^ ?y__o 9 ^ ,1^ 6 i/-l 2-y 2y- 6~^ y -3 Sy - 1' 10. Note. When common factors appear, divide them out. 1111 X — 2 X — 3 X — ^ X — a _x — b a-\-b_ a b 11. ; — 7 — ; . 1/5. — 7. X -{- X -f- c X — c X — a X — X — a X ^ b X — c _x — {a -\- b -i- c) ±0, — — . c a abc 14. ax-\-b = -^\, a b ^ „ 7n(x + a) . n{x + b) , 15. — ^^ — r^-^ -\ — ^^ — r — - = m-\-n. x-\-b X -^ a 16. (x - a)' + (a; - b)' + {x - c)' = S(x - a){x - b)(x - c). EQUATIONS WITH ONE UNKNOWN qUANTITY. 133 17. «.Il^-^^=0. 18. l = a. u -\-m u -\- n X 19. l = i. 20. l = i. X b X b a; — fl5^4~^ ^ X — a X — mx , nx . X — m X — 71 \yt n't Find the value of - from each of the following equations X without clearing of fractions: 26. - = 4. Ans. - = 2. 27. - = 15. XX X 38. ^- = a«. 29. ^L±_l = ,«. X X 30. ■ — =zm — n, 31. — = m' — rf. X XX 32. TT- + o - = «• 33. \-— = a — b, 2x 2x ax bx Find - and z from: z S4, —--=z- ^^ m-\-n _m-\-n z a b z m — n 36. ^^- = ^. 37. " + " - ^ W2 ^ mz — nz m — n 38.^^^ = "-. ' 39. ^ + l + f=^. (a + c)2; c 2 2;z 3;? In the following equations find the value of each symbol in terms of the others: 40. 2x-^a=zbb. . '^x-bb ^ 2x -da 3a + 6b Ans. a = —^-; b = — ^-; x = — ^— . 41. 5a- U = 2u, 42. Ha - Ub = 21y. 43. ax = by. 4:4:. a(x— y) = b{x -\- y). 45.^ = ^. 46.^ = 2. b — c 2a — c by 134 EQUATIONS OF THE FIRST DEGREE. Section II. Equations of the First Degree WITH Two Unknown Quantities. 156. Def. An equation of the first degree witli two unknown quantities is one which admits of being reduced to the form ax-\-hy= (?, in which x and y are the unknown quantities and a, b and c represent any numbers or algebraic expressions^ which do not contain either of the unknown quanti- ties. Def. A set of several equations, each containing the same unknown quantities, is called a system of simultaneous equations. 157. To solve two or more simultaneous equa- tions, it is necessary to combine them in such a way as to form one equation containing only one unknown quantity. Def, Elimination is the process of combining equations so that one or more of the unknown quanti- ties shall disappear. The term "elimination" is used because the unknown quantities which disappear are eliminated. There are three methods of eliminating an unknown quantity from two simultaneous equations. Elimination by Comparison. 158. EuLE. Solve each of the equations with respect to one of the unknown quantities and put the tioo values of the unknown quantity thus obtained equal to each other. This will give a new equation with only one unknoivn quan- Hty, of which the value can he found from the equation. The value of the other unknown quantity is then found hy substitution. Example. Solve the following set of equations: x-^y = 28,) Sx -2y = 29. ) ELIMINATION BY SUBSTITUTION. 135 From the first equation we find a; = 28 - «/, _29_+2^. ~ "3 ' _ 29 + 2^ and from the second from which we have 2S — y = whence i/ = 11. Substituting this talue in the first equation in x, it becomes a; = 28 - 11 = 17. If we substitute it in the second, it becomes 29 + 22 51 " ^=-3— = -3- = 17, the same value^ thus proving the correctness of the work. EXERCISES. Find the values of x and y from the following equations: x-3y= 7. 1. x-i-2y = ll! '^, %x— y = 8 3. X — 2y = b 4. 2x -Sy= 7 5. 2x — 3y = 7)1 7 2 X y' 3x-i-2y = 40. 3y — x= 8. 2x-{- y = 35. 2x -\- y = n. 9 -3 3 2 5 X // + 3 5 y X-\-4: 1 • %-y x-\-y = 42. 2y-\-x = 4A. 2x-\-y = 30. Elimination by Substitution. 159. Rule. Find the value of one of the unknown quantities in terms of the other from one equation, and sub- stitute this value in the other equation. The latter will then have hut one u7iknown quantity. Example. We take the same example as in the preced- ing method, namely, x-{- y = 28, 3x - 2?/= 29. 136 EQUATIONS OF THE FIRST DEGREE. From the first equation we have xzzz'il^ - y. Substituting this value in the second equation, it becomes 84 - 3.?/ -2y = 29, from which we obtain as before 84-29 ,, y = —^— = ll. This method may be applied to any pair of equations in four ways: 1. Find X from the first equation and substitute its value in the second. 2. Find x from the second equation and substitute its value in the first. 3. Find y from the first equation and substitute its value in the second. 4. Find y from the second equation and substitute its value in the first. EXERCISES. Solve the following equations in four ways: 1. x-}-2y = 18 2. X- y = 1 3. X -{- 2y = m 2x — y = 6. 2x + 3y = 14. 2x + y = n. Elimination by Addition or Subtraction. ' 160. Rule. Multiply each equation ly such a factor that the coefficients of one of the unknown quantities shall hecome 7iumerically equal in the two equations. Then hy adding or subtracting the equations we shall have an equation loith hut one unknow?i quantity. Remark. We may take for the factor of each equation the coefficient of the unknown quantity to be eliminated in the other equation, unless we see that simpler multipliers will answer the purpose. Example. Taking again the same equations as before, x-\- y = 2S, 3x — 2y = 29, we multiply the first equation by 3, obtaining 3x + Sy = 84. ELIMINATIONS BY ADDITION OR SUBTRACTION. 187 The coefficient of x is now the same as in the second given equation. Subtracting the second, we have by = 55, whence ^ = 11. Again, if we multiply the first equation by 2 and add it to the second, we have 6x = 85, whence x = 17. Eemark. We always obtain the same result, whatever method of elimination we use. But, as a general rule, the method of addition or subtraction is the simplest and most elegant. Very often little or no multiplication is necessary. Take the following case, for instance: 161. Peoblem of the Sum aj;d Differekce. The sum and difference of two numbers being given, to Ji?id the numbers. Let the numbers be x and y. Let « be their sum and d their difference. Then, by the conditions of the problem, x^ y = s, X — y = d. Adding the two equations, we have 2x = s -\- d. Subtracting the second from the first, 2y = s — d. Dividing these equations by 2, s -{- d s , d _ s — d _ s d y ~ 2 ~ 2 ~" 2 * We may therefore state these conclusions in the form of the following Theorem. Half the sum of any two numbers plus half their difference is equal to the greater number; and Half the sum of any tivo numbers minus half their differ- ence is equal to the lesser number. 138 EQUATIONS OF THE FIR8T DEGREE. EXERCISES. Solve the following pairs of equations: 1. X + 2?y =: 36 2. 2a;-f i/ = 8 3. dx- by = 17 4. x-\-2y = 20 5. dx — 4ty = c 6. ax -\- by =z m '7. ax -\- by = c 8. ax -\- by = p X- 6 _ y-\-7 2 ~ 3 9. 10. 11. 12. X —2y = 24. 2x — y = 8. 3x -i- 6y = 37. 2i2; + ?/ =r 25, 2x -]-'7y= e. ax — by = n, mx + ny = p. mx — ny = q. 2a; - 4 by - ax -\-by = c, 1; a~^ b ^ I y a -\- b a — b 13. - 14. - 15. x-j-y a y b 2 771 \ y-a x-y _ 2 a'x + b'y •1 _ ^ = 1 « ^ 2 a; — ^/ii/ + a X -\- my h. y -. = 1, 3 16^. Sometimes we may advantageously treat expres- sions containing the unknown quantities as if they were single symbols, in accordance with Priilciple II. of the algebraic language. EXERCISES. 1. 5{x + y)-2{x-y) = U; 5(a: + ^) + 2(2: - ^z) = 76. Solution. Taking the sum and difference of the two equa- tions as they stand, we have 10(:?; + y) =z 44 + 76 = 120, whence x -{- y = 12; 4(0; — i/) = 76 — 44 = 32, whence x — y =■ 8. Finally, adding and subtracting the last pair, we have 2x = 20, 2y = 4; whence x = 10, y = 2. 2. 3(x + 2y) + 2{x - 2y) = 65; 3{x + 2y) - 2(x - 2y) = 17. 3. 2(2a;-?/) + 3(2a: + ?/) = 28; 4(2a; - i/) + 3ferr + ?/) := 42. 4. 2(5a:-3i/) + (4a;-2/) = 40; 2 ^x-3jf) i^x—y) --20. EqUATIONS WITH THREE UNKNOWN QUANTITIES. 139 ^' x~^y 24' X y 24* Solutioyi. Adding and subtracting the equations as they stand, we find - = — = -, whence - — 77 and a: = 8; a: 24 4' X ^ = -, whence - = kt and «/ = 24. . = 3. Section III. Equations of the First Degree WITH Three or Mo^ Unknown Quantities. 163. When the values W scA^eral unknown quantities are to be found, it is necessary to have as many equations as un- known quantities. 164. Method of Elimmati^. When the number of un- known quantities exceeds two, the most convenient method of elimination is generally that by addition or subtraction. The unknown quantities are to be eliminated one at a time by the following • Rule. 1. Select an unknowit quantity to he first elimi- nated. It is best to hegin ivith the quantity which appears in the fewest equations or has the simplest coefficients. 2. Select one of the equations containing this unknown auantify as an elfll^altfki envafu)}^ (a) 140 EQUATIONS OF THE FIRST DEGREE. 3. Eliminate the quantity letween this equation and each of the others in successi07i. We shall then have a second system of equations less by one in number than the original system, and containing u number of unknown quantities one less. 4. Repeat the process on the neiu system of equations, and continue the repetition until only one equation u^ith one un- known quantity is left. 5. Having found the value of this last unknoivn quantity, the values of the others can he found hy successive substitution in 07ie equation of each system. Example. Solve the equations (1) ^x — dy — z -\- u = 62 (2) X - y -^ 2z -^ 2u = 20 (3) 2x -{- 2y - z- 2u = 3 (4) X -\-2y -j- z -^ u = 2. We shall select x as the first quantity to be eliminated, and take the last equation as the eliminating one. We first mul- tiply this equation by three such factors that the coefficient of X shall become equal to the coefficient of x in each of the other equations. These factors are 4, 1 and 2. We write the products under each of the other equations, thus: (1), 4:X — Sy — z -\- u = 62, (4) X 4, 4x -]- Sy -\- 4:z -{- 4.U = 8. (4) X 1, ^4) X 2, By subtracting one of each pair from the other, we ob- tain the equations (1') lly -^ 6z -i-Su = - U,) (2') dy - z - u= - 18, y (b) (3') 2y ^3z-\-4:U = 1. ) The unknown quantity x is here eliminated, and we have three equations with only three unknown quantities. Next we may eliminate y by using the last equation as the elimi- nating one. We proceed as follows: X - y -\-2z -\-2u = ^ + ^.y + ^ + ^ = 20, 2. 2x -i- 2y — z — 2u = 2x + 4i/ -{- 2z -\- 2u =: 3, 4. EQUATIONS WITH THREE UNKNOWN QUANTITIES. 141 (1') X 2, (?/) X 11, 22^/ + 10^ + 6w = 22?/ + 332; 4- 44?^ = - 88, 11. (2') X 2, (3') X 3, Qy - 2z - 2tc = 6y -\- dz -\- 12u = - 36, 3. Subtracting, we have (1") 232 + 3Su = 99, ) . . (2") 11^ + 14w. = 39, f ^"^^ a system of two equations with only two unknown quantities. From these equations we find: (1") X 11, 253;^ + 418?^ = 1089, (2") X 23, 253;^ + 322w = 897. 96?^ = 192, whence u = 2. Having thus obtained the value of one unknown quantity, we find the values of tlie remaining ones as follows: From (2") we have 11;^ = 39 - 14?^ = 39 - 28 = 11, whence z = 1. From (1') we have Uy = - u - 6z - du = - 44 - 5 - 6 = - 55, whence ^ = ~ ^• From (1) we have 4.x = 62 -\- Sy -{- z — u • = 52 - 15 + 1 - 2 = 36, whence a; = 9. Note. The student should now verify these results by substituting the values of x, y, z and u in the four original equations {a) and see wliether they are all satisfied. EXERCISES. 1. One of the best exercises for the student will be that of resolving the previous equations (a) by taking the last equation as the eliminating one, and performing the elimina- tion in different orders; that is, begin by eliminating u, then repeat the whole process beginning with z, etc. The final results will always be the same. 142 EQUATIONS OF THE FIRST DEGREE. 2. Find the values of x, y, z and ti from the equations x-\-y-^z-\-u = 4:a, X -\- y — z — u = 4:b, x — y -\- z — u = 4:C, X — y — z -\- u = 4:cl. Remark. This exercise requires no multiplication, but only addi- tion and subtraction of the different equations. 3. %x — by -\-^z = 11, 4. ax -\- hy -\- cz = A, dx -\-2y - z = 31, a'x + % + c'z = A'\ 5x-\-3y - 2z = 52. a'x + b'y + c'z = A\ Many of the following equations can be simplified by add- ing and subtracting, so as to reach a solution more expedi- tiously than by following the general rule: 5. X ^ y = a, 6. X -\r y = a, x^ y -^ z = h y ^ z = i, X -\- y -\- z -{- u = c, z -\- X = c. X - y — d. 7. X — 7iy = a, y — nz = b, z — nx ^= c. 8. y ' z Z X 9. 1 _1^_1 X y c' 1 i _ 1 - y z ~ a' 1 + 1 = 1 Z ^ X h' 10. X ^ y '- + ^ = 1. Z ' X 11. X — 2y = 2, 12. ax -\- hy = c, X - y -\- z = 5, cy -\-hz = a. X + 2u = 7, az -{- ex = h. 2x — z = 6. 13. 3u — X = m, dx — y = n, Sy - z =p, 3z — y = q. 14. mx — y = Of my- z =0, mz — X •= a. EQUATIONS WITH THREE UNKNOWN QUANTITIES. 143 15. x^y -^z = 0, (m + n)x + (71 -{-p)y -\- (p + m)z = 0, 7)inx -j- npy -\- pmz = 1. 16. X -\- y = 2x — y -{- IS = Sx — y — (5. Note. Equations of this kind, which often trouble the beginner, art; easily solved by equating separately different pairs of the equal Micnibers, For instance, the second and third members are J 2x - y -^ IS = dx - y - Q. i By transposing the unknown quantities to the second member, and 6 to the first, we have at once 24 = x or X = 24. (1) The first two members alone are x-{.y = 2x — y-\-18, which gives by transposition 2y - x = 18, from which y is found by (1). It is to be noted that in all such cases tliere are as many equations as signs of equality. 17. 2x -dy -'d = dx-2y -22 = X- y^ 1. 18. xy — X — y -\- 24: = xy -j- x -{- y = xy -{- x — 3y -{- 6. 19. dx = bx — y =: 7x — y — 4: = z. 20. xy -}- X = xy -{- Sx -^ y — 28 = xy — 3x + 44. 21. 2x = - 3y = 5{x + ?/' + 1). 22. ax = by = (a - b) (x + y + 1). 23. xy = (x + 5) (y -2)= (x + 9) (^ - 3). 165. Problems leading to Equations of the First Degree. Ill the solution of the following problems the student will sometimes find it convenient to use but one unknown quan- tity, and sometimes to use two or more. He must always state as many equations as there are unknown quantities, but the method to be followed in the solution must be left to his own ingenuity. 1. Divide 11.25 between two boys, giving A 43 cents more than B. Svggestion. Call x A's share and y B's share. 2. Divide a line 85 feet long into two parts of which one shall be 22 feet longer than the other. 3. Half the sum of two numbers is 95 and half their differ- ence is 55. What are the numbers? 4. The total vote for two candidates for Congress was 144 KQUATIONti OF THE FlRiST DEGREE. 20,185 and the Republican majority was 1093. How many votes were cast for each party? 5. A line being divided into two parts, the whole line/;//^s the greater part is 81 feet, and the whole line plus the lesser part 54 feet. What is the length of the line? 6. Divide $455 between two men, so that ^ the share of one shall be equal to -J the share of the other. 7. There were subscribed 125,000 to a college fund. A subscribed 15000 less than B and together, and B subscribed 1^2000 more than 0. What did each subscribe? 8. A man has two horses, with a saddle Avorth $50. The saddle and first horse are together worth as much as the sec- ond horse; the saddle and second horse are together worth twice as much as the first horse. What is the value of each? 9. A man has a buggy and two horses worth in all $800. When the best horse is harnessed to the buggy the team is Avorth $400 more than the other horse. When the poorest horse is harnessed the team is worth $100 more than the good horse. What is the value of the bnggy and of each horse? 10. Two men have together 225 acres of land worth $15,000. A's land is worth $50 per acre and B's worth $100 per acre. How much land has each? 11. A sum of $15.60 was divided among 72 children, each boy getting 25 cents and each girl 20 cents. How many Avere boys and how many Avere girls? 12. The first of tAvo cisterns has twice as much Avater as the second. If 120 gallons be poured from the first into the second, the latter will then have twice as much as the first. How much has each? 13. What fraction is that which becomes equal to ^ Avhen its numerator is increased by 1, and to ^ when its numerator is diminished by 1? x Suggestion. Call — the fraction. 14. What fraction becomes equal to J both when 1 is sub- tracted from its numerator and when 2 is added to its de- nominator? 15. A and B each had a certain sum of money. A got $25 more and thus had twice as much as B. Then B got PI10BLEM8. 145 ^IO'k) more and had twice as much as A. How much had each at first ? 16. In a Congressional election 17,346 votes were cast^ the Democrat getting 2432 more than the National, and the RepubUcan 878 more than the Democrat. What Avas the vote of each? 17. A man is 7 years older than his wife, and 10 years hence his age will be double what his wife's was 10 years ago. What are their present ages? Suggestion. If we call x the wife's present age, what is the man's present age? What will be his age 10 years hence? What were tlieir respective ages 10 years ago? 18. A boy is now half the age of his elder brother, but in 24 years he will be f his age. What are their present ages? 19. The combined ages of a man and his wife are now 62 years, and in 11 years he will be older than she will by ^ of her age. What are their present ages? 20. Two men having engaged in gambling, A won 14 from B and then had twice as much money as B. Then B won $25 and their shares were equal. How much had each at first ? 21. A sum of money being equally divided between A and B, A got $75 more than ^ of it. What was the sum divided? 22. What is the length of that line which being divided into three equal parts, each part is 2 inches longer than one fourth of the line? 23. A sum of $520 being divided between A, B and C,. B's share was f of A's share and $40 more than C's share. What was the share of each? 24. Another sum being divided, A got $40 less than halL B $40 more than one fourth, and $40 more than one sixth. What was the amount and the share of each? 25. Two men had between them 96 dollars. A paid B one fourth of his (A's) money, but B lost $6 and then had twice as much as A. How much had each at first? 26. Two cisterns being each partly full of water, ^ of the water in the first was poured into the second, which then had 120 gallons. Then \ of what was left was poured from the first into the second, when the latter had twice as much as the first. How much did each contain at first? 146 EQUATIONS OF THE FIRST DEGREE. 27. At a city election Jones had a majority of 244 votes over Smith. But it was found that ^ of Jones's votes and yig- of Smith's votes were illegal, and on correcting this Smith had a majority of 77. What was the legal vote of each? Ans. Smith, 10,458; Jones, 10,381. Suggestion. Take the whole numbers of votes first cast as the un- known quantities. 28. An apple- woman bought a lot of apples at 5 for 2 cents. She sold half at 2 for a cent, and half at 3 for a cent, thus gain- ing 10 cents. How many apples were there? Ans. 600. 29. A train ran half the distance between two cities at the rate of 30 miles an hour and half the distance at 50 miles an hour. It performed the return journey at the uniform speed of 40 miles an hour, thus gaining half an hour on its time in going. What was the distance? Ans. 300 miles. Remark. In all questions involving time, constant mlocity and dis- tance we have the fundamental relation: Distance = 'oelocity X time. 30. A man bought 29 oranges for a dollar, giving 3 cents a piece for poor ones and 4 cents for good ones. How many of each kind had he? 31. A grocer bought a lot of sugar at 8 cents a pound and of coffee at 12 cents a pound, paying $8.60 for the whole. He sold the sugar at 10 cents a pound and the coffee at 14 'Cents, realizing $10.40. How much of each did he buy? 32. A huckster bought a lot of oranges at 1 cent each and lemons at 2 cents each, paying $1.45 for the lot. 5 of the oranges and 10 of the lemons were bad, but he sold the good fruit at 2 cents each for oranges and 3 cents for lemons, real- izing $1.90. How many of each kind did he buy? 33. The sum of the ages of two brothers is now 4 times tlie difference, but in 6 years it will be 5 times the difference. What are their present ages? 34. For $6 I can buy either 4 pounds of tea and 20 pounds of coffee or 2 pounds of tea and 25 of coffee. What is the price per pound of each? 35. An almoner had 3 equal sums of money to divide be- tween 3 families. The second family had 5 more than the first and so got $2 apiece less, and the third had 4 more than the second and got $1 a|)iece less. What were the numbers of the families .-md tlic equal ;nTir)unts divirlorl? riWBLEMS. 147 36. If A can do a piece of work in 3 drys and B in *: days, in what time can they do it if both work together? Suggestion. In one day A can do \ and B can do |. Hence both together can do ^ + i- But if x be the time in which both can do it, they can do — of it in a day. Hence - = - + - 37. If one pipe can lill a cistern in 12 minntcs and ar.other in 18 minutes, in what time can they both fill it? 38. A cistern can be emptied by two faucets in 12 minutes and by one of them in 36 minutes. In what time can it be emptied by the other? 39. A cistern can be filled by a pipe in 25 minutes when the faucet is left running, and in 15 minutes when the faucet is closed. In what time will the faucet empty it? 40. Three men can together perform a piece of work in 12 days. A can do twice as much as B, and B twice as much as 0. In what time could each one separately do the work? 41. A and B can together do a job of work in 8 days, B and in 9 days, and and A in 12 days. In what time can each one alone do it? In what time can they all do it? Ans. A, 20f ; B, Id^ ; 0, 28f ; all, 6^V 42. A train performs the journey from Washington to Chicago in a certain time at a certain speed. By going 16 miles an hour faster it gains 9 hours, and by going 8 miles an hour slower it takes 9 hours longer. What is the original time and speed and what the distance? 43. A privateer sights an enemy's ship 9 miles away, flee- ing at the speed of 6 miles an hour. If the privateer chases her at the rate of 8 miles an hour, in what time and at what distance will she overtake her? Method of Solution. Let t be the time and d the distance. Because the privateer sails 8 miles an hour, we have d = 8t. The ship having 9 miles less to go will, when overtaken, have sailed {d — 9) miles. Therefore d - 9 = 6t. From these two equations we find d and t. 44. A man gets into a stage-coach driving 7 miles an hour for a pleasure-ride; but he must walk home at the rate of 3 miles an hour, and be gone only 5 hours. How far can he ride? 148 EQUATIONS OF THE FIliST DEGREE. 45. A train performed a journey in 6 hours, going one third the way at the rate of 30 miles an hour and two thirds at the rate oi' 40 miles an hour. What was the distance? 46. A man had two casks of wine containing together 75 gallons. He poured one fourth the contents of the first into the second, and then poured one third the increased contents of the second cask into the first. The first then contained 'Z^ gallons more than the second. How much did each contain at first? 47. Two casks contain between them s gallons of wine, and one had d gallons more than the other. How much wine had each? 48. A board is divided into two parts. The whole board plus the greater part is m feet and the Avhole board plus the lesser part is n feet. What is the length of the board? 49. Divide a board I feet long so that two thirds of one part shall be equal to one half the other part. 50. A grocer mixed k pounds of tea worth j(? cents a pound with Ji pounds worth n cents a pound. How much per pound was the mixture w^orth? Note. The solution of this question does not really require any equation. 51. A grocer has t pounds of tea worth r cents a pound, which is formed by mixing two kinds, one worth p cents a pound and the other q cents a pound. How much of each kind did he mix? 52. A boy's age is now one third that of his elder brother, but in t years it will be one half. What are their present ages? 53. Three casks contain altogether on gallons of wine. By pouring a gallons from the first into the second and then h gallons from the second into the third, the quantities in the three casks become equal. How much did each cask con- tain at first? 54. A man who must be back in t hours starts in a coach going m miles an hour and walks back at the rate of 7i miles an hour. How far can he go and get back in time? 55. If one man can do a piece of work in a days and another in l days, in what time can they both do it? INTERPRETATION OF NEGATIVE RESULTS. 149 56. If A can do the work in a days, B in days and C in c days, in what time can they all do it if work together? 57. If A and B together can do a piece of work in 7n days and A alone in a days, in what time can B alone do it? 58. A man bonglit tea and coffee, p pounds in all, for r cents, giving m cents a pound for tea and n cents a pound for coffee. How much of each did he buy? To prove the results add the two amounts and see wlicther tliey make p pounds. 59. Divide m dollars among 3 men, giving B a dollars more than 0, and A b dollars more than B. GO. If a train makes half its journey at the rate of vi miles an hour and the other half at the rate of n miles an hour, what is its average speed? Interpretation of Negative Results. 166. An answer to a problem may sometimes come out negative. This shows that the answer must be reckoned in the opposite direction from that assumed as positive in the enunciation of the problem. Example. A man is 30 years old and his wife is 24. In how many years will he be half as old again as she is? Solution. Let us put t for the required number of years. In t years his age will be 30 + if and hers will be 24 + t. Because the conditions of the problem require his age to be half as much again as hers, we have 30 + ?f = |(24 + 0- Solving this equation, we find ^ = — 12. This negative result shows that the time when the re- quired condition was fulfilled is not m the future but in the past, when the wife was 12 and the man 18. Had we stated the problem. How many years ago was his age half as much again as hers? t would have had the opposite sign all the way through and would have come out +12 years, because in the enunciation years past would then be regarded as positive. Therefore whether time in the future J5() F.qUATJO^'S OF THE FIRST DEOBEE. shall be positive or negative depends on what we assume as positive in the enunciation of the problem. Ex. 2. A man is 4 years older than his wife, and 6 times his age is equal to 7 times hers. How many years ago was 7 times his age equal to 8 times hers? Soluho7i. Let us j^ut x for lier present age and t for the required number of years ago. Then his age will be a; -j- 4 vears, and by the conditions of the problem iS{x + 4) r= 7x, But t years ago his age must have been x -{- 4: — t. and hers must have been x — t. By the conditions of the problem 7(^ + 4 - t) = 8{x - t). Reducing and solving these two equations, we obtain x= 24, t= ~ L Here t comes out negative as before; but it does not mean years ago as it did in the first example, but years in the future, because in stating the problem we assumed years in the past to be positive. 16T. Problem, of the Couriers. A courier starts from his station riding 8 miles an hour. 4 hours afterwards he is fol- lowed by another riding 10 miles an hour. How long will it require for the second to overtake the first, and wliat will be the distance travelled? If t be the number of hours required, the first will have travelled t -\- ^ hours and the second t hours when they come together. Because one goes 8 miles an hour and the other 10, the whole distance travelled by the one will be 8(^ -j- 4), and that travelled by the other will be 10^. Because these distances are equal, we have 8if + 32 = l^t Solving this equation, we have t = 16, whence distance = 160. Now let us change the problem thus: The first courier still riding 8 miles an hour, the second starts out after him in 2 hours and travels 6 miles an hour. In what time and at what distance will they be together? Treating the equation in the same way as before, we find the equation to be 8(t + 2) = 6/ INTEUPHKTATION OF NEGATIVE BESifLTti. l|)i Solving this equation, the result is t = - S, d= - 48. These answers being negative show that there is no point on the road in the positive direction, and no time in the future when the couriers would be together. Thus far the algebraic result agrees with common-sense; but common-sense seems to say that the couriers would never be together, whereas the algebraic answer gives a negative time and a negative distance on the road when they were together. The explanation of the difficulty is this. Supposp, 8 to be the point from which the couriers started, and AB ^he road along which they travelled from S toward B. Suppose also that i — ^^ the first courier started out from S at 8 o'clock and the second at 10 o'clock. By the rule of positive and negative quantities, distances towaids A are negative. Now, because algebraic quantities do not com- mence at 0, but extend in both the negative and positive directions, the algebraic problem does not suppose the cour- iers to have really commenced their journey at S, but to have come from the direction of A, so that the first one passes ^S', without stopping, at 8 o'clock, and the second at 10. It is plain that if the first courier is travelling the faster, he must have passed the other before reaching S\ that is, the time and distance are both negative, just as the problem gives them. The general principle here involved may be expressed thus: In algebra, roads and journeys, like time, have no hegi7i- ning and no end. 168, In the following problems, when negative results are obtained, the pupil should show hoAv they are to be in- terpreted. 1. Albany is 6 miles below Troy on the Hudson River. Let us suppose the river to flow at the rate of 4 miles an hour. How fast must a man starting from Troy roAV his boat through the water in order that he may reach Albany in 1 hour? 152 EQUATIONS OF THE FIRST DEGREE. 2. The same thing being supposed, how fast must he row in order to reach Albany in 2 hours? Interpret the negative result. 3. If, starting from Albany, he rows up the river at the rate of 6 miles an hour, how long will it take him to reach Troy? 4. If he rows at the rate of 3 miles an hour, how long will it take him to reach Troy? Explain the negative result. Apply the principle laid down by tlie diagram in tiie preceding section. Memoeanda for Review. Define : Equation of the first degree; System of simul- taneous equations; Elimination. Equations of the First Degree luith One Unhioivn Quantity. Give rule for solution. Two Unknoivn Quantities, [Comparison; Rule. \ Substitution; Rule. Ellmma- ( Comparison; Rule. ^ ( Addition and subtraction; Rule. Theorem of the Sum and Difference. Explain theorem. Three or More Unknown Quantities. Method of elimination; Rule, CHAPTER IV. RATIO AN D PROPORTION Section I. Of Ratio. 169. Def. When a greater quantity contains a lesser one an exact number of times, the greater is said to be a multiple of the lesser, and the lesser is said to measure the greater. Example. If the line a ^ ' ' contains the line b exactly three times, « is a multiple 6 of l, and b measures a. 170. Def. When two quantities may each be measured by the same lesser quantity, they are said to be commensurable, and the lesser quantity is said to be their common measure. Example. If the line A , ^ , 5 i ^> i & i b i contains the measure b five times, and B contains it three \ h \ h \ h \ times, A and B are cornmen- ^ " .surable, and b is their common measure. 171. Def. The ratio of two niafrnitudes is the number by which one must be multiplied to j)roduce the other. Example. In the first fio^ure ahove the ratio of a to b is 3, because b X S = a. 173. Belation of Batio to Quotient. To divide a ((imn- tity by a number means to separate it into that number of erpial parts. The divisor is then a number, and the (|U()tient is a quantity of the same kind as the dividend. For instance, if we divide a line by 3, the result will be a line one third as long; if we divide weight by 4, the result will be weight one fourth as heavy. 154 RATIO AND PROPORTION. But when we measure one line by another or one weight by another, the result is neither a line nor a weight, but a number; namely, the number by which one must be multi- plied to produce the other. This number is the ratio. Hence another definition: 173. A ratio is the quotient of two quantities of the same kind. A ratio is expressed by writing the divisor after the divi- dend with the sign : between. Thus means the ratio of « to ^ =: 3, or Z> X 3 produces a, or a divided by J = 3. Hence from the equation we have both a : h = ^ and ^ = b. 174. Def. The antecedent of the ratio is the quantity divided. The consequent is the divisor. Example. In the ratio a : b, a is the antecedent and b the consequent. Def. Antecedent and consequent are called terms of the ratio. 175. When the consequent measures the ante- cedent, the ratio is an integer. When the consequent and antecedent are com- mensurable, the ratio is a vulgar fraction. Example. In the example of § lo9, because the ante- cedent contains the consequent 3 times the ratio is the in- teger 3. But in ti)e exjiniple of g 170, to find the ratio A : B, wo hn^'o to divide U into )> parts and take 5 of these parts to make A, Hence the ratio of ^ to 7^ is '—, or o RATIO. 155 The fractional ratio — has for its numerator the number o of times the common measure is contained in the antecedent, and for its denominator the number of times the common measure is contained in the consequent. Hence if we divide botli terms into equal parts, we have the theorenj: A ratio is equal to the quotient of the number of parts i?i the a7itecedent divided hy the nmnber of equal parts i7i tlie consequent. 176. When the two terms of the ratio have no common measure they are said to be incommensur- able. An incommensurable ratio cannot be exactly expressed as a vulgar fraction, but we can always find a vulgar fraction which shall be as near as we please to the true value of the ratio, though never exactly equal to it. Reaso7i. Let us divide the consequent into any number n of equal parts. Measuring the antecedent with one of these l)arts we shall, when the ratio is incommensurable, always have a piece of the part left over. By dropping this piece from the antecedent it Avill become commensurable. Xow by making the number 7i of parts great enough, we may make each part, and therefore the piece left over, as small as we l)lease. For example: When n > 100, piece over < ^\^ of consequent; When n > 1000, piece, over < y-J^^ of consequent: When n > 1000000, piece over < xo-o^o-^ ^^ consequent; etc. etc. EXERCISES. Here are four Hues of different leugths, a, h, c and d. Find as nearly as you can, by dividing up the lines, the following ratios: 1. a :h. Ans. 2. a : c. Ans. 3. a : d. Ans. 4. b : c. Ans. 5. b : d. Ans. 6. c \ d. Ans. 156 RATIO AND PROPORTION. Proi)erties of Ratios. 177. Def. If we intercliange the terms of a ratio, the result is called the inverse ratio. That is^ ^ : ^ is the inverse of A : B. If B:A = -, n then B = — A, n fix and we have, by dividing by — , fi 7)1 or A : B = —. m Because — is the reciprocal of — , we conclude: m n Theorem I. The inverse ratio is the reciprocal of the direct ratio. 178. Theorem IT. If hoth terms of a ratio he midti- 2)lie(l hy the same factor or iliridcd Inj the same divisor, tite ratio is not altered. Proof. Ratio of B to A = B : A = --. A If m be the factor, then Ratio of 7nB to mA = mB : mA = — j = -;-, mA A the same as the ratio of ^ to ^. Again, if both terms are divided by tlie same divisor, this operation amounts to dividing both terms of the frac- tion which expresses tlie ratio, and so leaves the value of the fraction unaltered. 179. Theorem III. If both terms of a ratio be increased by the same quantity, the ratio will be increased if it is Jess than 1, ajid diminished if it is greater than 1; that is, it will be brought nearer to unity. BATIO. 157 Example. Let the original ratio be 2 : 5 = |. If we repeatedly add 1 to both numerator and denominator of the fraction, we shall have the series of fractions 2 3 4 5 pfp 5' 6» 7» 8> ^^^'i each of which is greater than the preceding, because I - I = a'a; whence | > |. I - I = A; whence | > |. 8 — V ■-- 5^; whence f > f, etc. etc. General Proof. Let a : b ha the original ratio, and let both terms be increased by the quantity u, making the new ratio a-\-2i : h -\- n. The new ratio mimis the old one will be a-\- u ci _{b — '^)?^ h -[-u ~ b" b' + bu ' If b is greater than a, this quantity wll be positive, show- ing that the ratio is increased ])y adding u. If b is less than a, tlie quantity will be negative, showing that the ratio is dimin- isbed by adding u. EXERCISES. 1. What is the ratio (a) of 27 inches to 3 feet? Ans. f. (b) of 1122 feet to half a mile? (c) of 3 miles to 2244 yards? 2. What is the ratio (a) of I of an inch to -f^ of a foot? (/;) of I of an inch to f of an inch? (r) of I of a yard to f of a foot? 3. Which ratio is the greater, (a) 3 : 5 or 5 : 8? (b) 5 : 3 or 8 : 5? (c) G : 7 or 7 : 8? (d) 7 : 4 or 14:8? 4. The ratio of two numbers is f, and if each of them is increased by 4 their ratio wdll be f . Find the numbers. Note. If we call x and y the numbers, the fact that their ratio is I is expressed by tlie equation X _d 2/ ~ 5' x-\- 4 When each number is increased by 4 tlie ratio will be equal to — - .. ^ ^ yi-4 158 RAllO AND PROPORTION. 5. If 1 metre = 39.37 inches, what is the ratio of 10 metres to 1270 feet? 6. The ratio of two numbers is — , and if eacli of them 71 be increased by a their ratio will become — . Find the num])ers. . . . ^ 7. If the ratio of two quantities is |, what is the ratio of tAvice the antecedent to three times the consequent? What is the ratio of m times the antecedent to n times the conse- quent? Sectiois- II. Proportion^. 180. Bef. Proportion is an equality of two or more ratios whose terms are written. Since each ratio has two terms, a proportion must liave at least four terms. Bef. The terms which enter into the two equal ratios are called terms of the proportion.. If « : Z> be one of the ratios and p : q the other, the pro- portion will be a:h=p'.q. (1) A proportion is sometimes written a :h :: p : q, which is read, " As a^ is to h, so is p to 9." The first form is to be pre- ferred, because no other sign than that of equality is necessar}^ betM^een the ratios; but the equation may be read, " As ^ is to b, so is p to q," whenever that expression is the clearer. Def. The first and fourth terms of a proportion are called the extremes; the second and third are called the means. Theorems of Proportion. 181. Theorem I. In a proportion the product of the extremes is equal to the product of the means. Proof. Let us write the ratios in the proportion (1) in the form of fractions. It will give the equation :- = f ^'} PBOPORTION, 159 Multiplying both members of this equation by Iq, we shall have aq = hp. (3) 183. Def. A proportion is said to be inverted when antecedents and consequents are interchanged. Theorem IL A proportion remains true after incersion. Proof, If a \l) — p \ q, ,, a p then — =z =!-' q and by taking the reciprocal of each member, 1 = 1 a p^ whence h : a — q : p. 183. Theorem III. If the means in a proportion he in- terchanged, the proportion will still he trite. Proof Divide the equation (3) by p^q. We shall then have, instead of, the proportion (1), a _h p~ q' or a : p = h '. q. Def. The proportion in which the means are in- terchanged is called the alternate of the original pro- l^ortion. The following examples of alternate proportions should be studied, and the proof of the equations proved by calculation: 1:2= 4:8; alternate, 1:4 =2:8. 2:3=6:9; '' 2:6=3:9. 5 : 2 = 25 : 10; " 5 : 25 = 2 : 10. 184. Theorem IV. If in a proportion we increase or diminish each antecedent hy its consequent, or each consequent by its own antecedent, the proportion will still he true. Example. In the proportion 5 : 2 = 25 : 10 the antecedents are 5 and 25, the consequents 2 and 10 (§ 174). Increas- ing each antecedent by its own consequent, \\\e proportion will be 5 + 2 : 2 = 25 -4- 10 : 10, or 7 : 2 = 35 : 10. 160 BATIO AND PROPORTION. Diminishing each antecedent by its consequent, tlie proportion will become 5 - 2 : 2 = 35 - 10 : 10, or 3 : 2 = 15 : 10. Increasing each consequent by its antecedent, the proportion will be 5 : 2 + 5 = 25 : 10 + 25, or 5 : 7 := 25 : 35. These equations are all to be proved numerically. General Proof. Let us put the proportion in the form If we add 1 to each member of this equation and reduce, it will become a + h _ p^q^ b ~~T"' ^^ that is, a -\- J) \ b = p -\- q '. q. (6) In the same way, by subtracting 1 from each side, we have h - q ' ^^^ or a — h \ h = p — q \ q. (8) If we inv^ert the fractions in equation (4), the Latter will become a p ' By adding or subtracting 1 from each member of this equation, and then again inverting the terms of the reduced fractions, we shall find a -.h ^ a = p :q ^p\ (9) a : h — a =^ p '. q — p. (10) The forms (7) and (9), in which each pair of terms is added, are said to be formed by composition. The forms (8) and (10) are said to be formed by division. 185. Composition and Division. Taking the quotient of (5) by (7), we have a — h p — q^ that is, a -\- J) : a — 1) = p -\- q \ p — q. Def. This form is said to be formed from a ; h = p : q by composition and division. PROPORTION. 161 EXERCISES. 1. From the proportion 2 :7 = 6 :21 form as many more proport ions as you can by alternation, inversion, composition and d i vision: Ans. Alt. 2:6=7 : 21. Inv. 7:2 = 21 : 6; a 6 :2 = 21 : 7. Comp. 9 : 7 = 27 :21; a 2:9=6 :27. Div. 5 : 7 = 15 :21; a 2:5= 6 : 15. Comp. and div. 9 : 5 = 27 : 15. 2. Do the same with tlie proportion 24 : 9 = 8 : 3. 186. 1'heorem \. If each term of a proportion he raised to the same poiver, the proportion will still be true. Proof. If a \ h = p : qy a p or r- = ^, b q then, by multiplying each member by itself repeatedly, we shall have - t -Pi ~ q' etc. etc. Hence, in general. Cor. If then, from Th. IV. ^ and rt" : b"" = 7J" : q\ a :b = p : q, : a" ± b"" = p"" : p" ± (f ± //' : b" = jy" ± f : q'\ 187. Theorem VI. When any three terms of n pro- portion are given, the fourth can always be found from the theorem that the product of the means is equal to that of the extremes. We have shown that whenever a :b ■:;=■ p : q, then aq = bp. 162 RATIO AND PROPORTIOK Considering the different terms in succession as unknown quantities, we find h aq J' ' aq p = -p Ip ^ a 188, Theoeem VII. If the ratio of ttvo quantities is given, the relation bettveen them may he expressed hy an equa- tion of the first degree. Proof. Let x and y be the quantities, and let their given m ratio be — . By the definition of ratio this will mean that if n -^ nz we multiply y by — the result will be x. That is, 711 X = — y. Multiplying by n, my =nx, an equation of the first degree. Cor. If we know that the ratio of two unknown quanti- m ties is — , we may call one of them mx and the other nx. n For mx : nx = ~. (§ 178) The Mean Proportional. 1 89. Def. When tlie middle terms of a propor- tion are equal, either of them is called the mean pro- portional between the extremes. The fact that h is the mean proportional between a and c is expressed in the form a : i = i : c. Theorem T. then gives ¥ = ac. PltOPOllTION. 163 Extracting the square root of both members, we have I =z Voir. Hence Theorem VIII. The iuean proportional of two quanti- ties is equal to the square root of their product. Def. Three quantities are said to be in proportion when the second is a mean proportional between tlie first and third, and the third is then called the third proportional. EXERCISES. What is the mean i)roportional between: 1. 16 and 36? 2. 2 and 8? 3. a' and Z>'? 4. a -\- h '' Explain and Distinguish the same quantity. Result when both terms are | J^^J^ihiTshed ^^^' the same quantity. Proportion. Define : Proportion; Terms; Extremes; Means; Inver- sion; Alternation; Composition; Division; Mean propor- tional; Third proportional. Products of Extremes and Means, Theorem of; Prove. Inversion, Theorem of; Prove. Alternation, Theorem of; Prove. Theorems Composition; Prove. of Pro- ■{ Division ; Prove. portion. Composition and division; Explain. Each term raised to the same power; Th. Express any term in terms of the three others. Express that two quantities have a given ratio. Mean proportional; Tlieorem of» CHAPTER V. POWERS AND ROOTS. Section I. Powers and Roots of Monomials. 190. Bef, The result of taking a quantity A n times as a factor is called the nth power of A^ and, as already known, may be written either AAA, etc., n times, or A^. Bef, The exponent n is called the index of the power. Bef, Involution is the operation of finding the powers of algebraic expressions. The operation of involution may always be indicated by the application of the proper exponent, the expression to be involved being enclosed in parentheses. Examples. The ^^th power oi a -\- h \^ (a -\- Vf, The wth power of abc is (abcy. Involution of Monomials. 191. Involution of Products. The nth. power of the product of several factors, a, I, c, may be expressed without exponents as follows: ahc abc abc, etc., each factor being repeated n times. Here there will be altogether n a\, n J's and n c's, so that, using exponents, the whole power will be a"Z>"c" Hence (abcY = aVc\ That is, Theorem. The potver of a product is equal to the pro- duct of the poivers of the several factors. INVOLUTION OF MONOMIALS. EXERCISES. Form the squares, the cubes and the nt\i powers of: 1. mn. 2. %hp, 3. ^xy, 4. 'Zan. 5. ^r. 6. bpq. 192, Involution of Fractions. Applying the methods to fractions, we find that the nila. power of x"- r For XXX . . . , etc., n times y y y xxXy etc., n times yyyy etc., n times a;" same X . — IS (§^5) EXERCISES. am hn ' ]^orm the squares, the cubes and the wth powers of: 2cpq dpx abf/z 4:mqy' 2. dxyz ^fiinx 193. Involutiofi of Powers. Let it be required to raise the quantity a^ to the ;^th power. Solution. The nth. power of r/"' is, by definition, a"^ X or X a^, etc., n times. By § 36, the exponents of a are all to be added, and as tlie exponent m is repeated n times, the sum m + m + m + etc., n times is mn. Hence the result is fl^"*", or, in the language of algebra, (ary —a '"^ Hence Theorem. // any poioer of a quantity is itself to he raised to a poiver, the indices of the potvers must he multiplied together. Examples. (a")' — a'a^a'' — r/"; 170 POWERS AND ROOTS. EXBRCISES. Write the squares and the cubes of the ■ following expres- sions: 1. al)\ 4. aVx\ 2. 2m'n. 5. 'doVi'x. 3. 6. fq\ bg'^m'^x. J. ^,. 10. '"''". 8 ^^ 11. ^^•'^^ 9. 12. Form the wth powers of: 13. /^\ 14. 2m'x. 15. 18. 3pY 2cr 19. ^T. ». -r. 21. Wx^' Keduce: 22. (/r)^ 23. (jy''')^ 24. {D^n'xy. 25. (2^^=')^ 26. {%iil)y. 27. m- 194. Algebraic Signs of Potvers. Since the continued product of any number of j)ositive factors is positive, all the l^owers of a positive quantity are positive. By § 103, the product of an odd number of negative fac- tors is negative, and the product of an even number is positive. Hence Theorem. The even poivers of negative quantities are positive, and the odd poivers are negative. Examples. (— af = a""; (— ay= — a^; (— aY= a*; etc. EXERCISES. Find the values of: 1. (-3)\ 2. (- 3)^ 3. (-3)\ 4. {-ay. 5. {-ay. 6. {-ay. 7. {-aby. 8. (-mV)^ 9. {- cn'xy, 10. (-l)^ 11. (-1)1 12. {-ly. 13. {-hf^. 14. (-A)2« + i. 15. {-hf^-K 16. (-1)2". 17. (-1)^"-^ 18. (-1)2"+^ ROOTS OF MONOMIALS. 171 Roots of Monomials. 195. Def. The nt\i root of a quantity q is sucli a number as, being raised to the iit\\ power, will pro- duce q. The number n is called the index of the root. The second root is called the square root. The third root is called the cube root. Examples. 3 is the 4th root of 81, because 3. 3.3. 3 = 3' = 81. I)ef. Evolution is the process of extracting roots. 196. Sign of Evolution^ or the Radical Sign. |/, called root^ indicates the square root of the quan- tity to which it is prefixed. When any other than the square root is expressed, the index of the root is prefixed to the sign \/. Thus y, ^^, "|/ indicate the cube, the fourth and the wth roots respectively. The radical sign is commonly followed by a vin- culum extended over the expression whose root is indicated. Examples. In Va-^-h -\- x the root applies to a only. In i/a -{- b -\-c the root indicated is that of a -\- h. In Va-\- b -\- c the root indicated is that oi a -\- b -{- r. But we may use parentheses in place of the vin- culum, writing |/(a -{-h) -\- c^ \/{a -\-h -\-c\ etc. EXERCISES. What are the values of the following expressions? 1. i/7 + 9. Ans. 4. 2. 1^20 + 5. 3. i/(3'+ 4'^). 4. |/(l7 4-8'). 5. Ve . 15 . 1^. 6. Va\ 7. x^ia'-^-^ab-^b'), 8. Vs! 9- Vl^'+^'+S). 10. V81. 11. WW^^^ 12. Vl4^- 13^ 17^ POWERS ANt) MOOTS, 19*7. Division of Exponents. Let us extract the square root of a\ We must find such a quantity as, being multiplied by itself, will produce a\ It is evident that the required quan- tity is a^f because, by the rule for involution (§191), n Tlie square root of dP- will be d\ because j n M !! + ?! n In the same way, the cube root of cC" is a^, because n n n a'X a'x (("'= (("". The following theorem will now be evident: Theorem. The square 7'oot of a poiver may he expressed hy dividing its exponent ly 2, tlie cube root ly dividing it by 3, and the 7ith root by dividiiig it by n. EXERCISES. Find the cube roots of: 1. p\ 2. p\ 3. p\ 4. p^. 198. Since the even powers of negative quantities are positive, it follows that an even root of a positive quantity may be either positive or negative. This is expressed by the s ign ± ; read, plus or minus. Examples. Va"" = ± a; V{a — by = a — b or b — a. 199. If the quantity of which the root is to be ex- tracted is a product of several factors, we extract the root of each factor and take the product of these roots. Example. The square foot of a^ni^p^ is ± a^mp^, because (a^mpy = a'nep\ by §§ 191 and 193. 200. If the quantity is a fraction, we extract the roots of both members. The root of the fraction is then the quotient of the roots. Examples, y -,- = -3; y -, = ± ^. ' y y ' c c FMACTIONAL EXPONENTS, 175 EXERCISES. Extract the square roots of: 1. a'm\ 2. c'x\ 3. 4^2n^4n^ '■'?■ 5 ^^^ 6. 7. (a+bY(a. -b)'. Express the cube roots of: 8aW 11. 23h ^ 36n Express the nth roots of: 12. x\ 13. x^\ 14. cc*"". 15. x^\ 16. :c^'+^ 17. ^n^ + 2n^ 18. 1". 20. Fractional Exponents. 201. If we apply the rule of § 197 for division of ex- ponents so as to express the square root of x% we shall have Vx' = xl Because x = x\ we also have i^x = x^ . Reversing the process and applying the general form, we have ici X :r^ = a:^+i = x^ = x, x^ X x^ = x^ + i = x\ Hence A fractional exponent mdtcates the extraction of a root. If the denominator is 2, a square root is indicated; if 3, a cube root; if n, an nth root. A fractional exponent has therefore the same meaning as the radical sign |/, and may be used in place of it. It is a mere question of convenience which method of indicating the root we shall use. 174 POWEMS AM) MOOTS. EXERCISES. Express the following roots by exponents only, 1. V^. 2. ^{a - 1)). 3. V{a - b)\ 4. V{a - bf. 5. V^. 6. Va^. 7. VI^^\ 8. y^v. 9. y^. cy 10. 'l/?^. 11. ^V7^, 12. V^^iV. 13. ;/'^-. 14 y/EEs 15 v^^ * ^ (a-b)y'' ^' ^ {a-x)y^'' ^ {b-\-x) {b-x)' Express the nth roots of: 19. X. 20. x\ 21. a:'. 22. bmx''. 23. 7c'a;'". 24. 10c'x'\ Powers of Expressions with Fractional Exponents. ^02. Theorem. The pth poicer of the nth root is equal to the nth root of the pth poiuer. Example. CViy = 2' = 4, W = V64 = 4; or, in other words, the square of the cube root of 8 (that is, the square of 2) is the cube root of the square of 8 (that is, of 64). General Proof Let us put x = the nth. root of a. The pth power of this root x wall then be x^. (1) Because x is the nth root of a, x"" = a. Raising both sides of this equation to the j^th power, w^e have ^np _ ^p _ p^Yi power of a. The nth root of the first member is found by diyiding the exponent by n, which gives wth root of j9th power = x^, FRACTIONAL EXPONENTS. 175 the same expression (1) just found fortlie pi\\ power of the nth Yooi.. 303, This theorem leads to the following corollaries: CoK. 1. The expresmm p a"" 1 mat/ mean either the pth poiver of a" or the nth root of a^\ these quantities being equivalent. Cor. 2. The numerator of a fractional exponent ix the index of a power. The deno^ninator is the index of a roof. Cor. 3. The potvers of expressions having fractio)iaI ex- ponents ?nay be for^ned hy multiplying the exponents by the index of the poiver. EXERCISES. Express the squares, the cubes and the nt\\ powers of the following expressions : 1. c^. 2. ch. 3. cl 1 4. ach 5. aid 6. aic''. 7. a'y". 8. c'l/". 9. c'^y^ 10. ™*? ah 11. "*f-. 1 7inyi ^^' a^b- 13. (''+*).:. 14. (™ + 1 71)'' 2' 15 {m-i-71)^ {a — by {m — n)^ {m — 7i)» 204. Multiplication of Fractional Exponents. One fractional exponent may be multiplied by another wlien a root is to be extracted, according to the rules for the multiplication of fractions. The process is the same as that of § 193 except that the exponent is fractional instead of entire, and roots are indicated as well as powers. Example 1. Square root of «* = («*)^ = ak Ex. 2. Cube root of m§ = {ml)\ — ml. Ex. 3. Square root of {a-\-xYbl = \{a-^xYbl\^ —{a-\-x)\U. 176 POWERS AND ROOTS. EXERCISES. Express the square roots of: 1. xK 2. ahxi. 3. {a^h)^x% xh (a + b)^ ^m'xi (v^ (a — ^)3 /i Express the following indicated roots: 7. {a^M)l Ans. WB^i 8. (aibi)l Ans. «iZ»i 9. (7/^3^S)i. 10. (mhi'^)l i H i\IL 11. \(a^x)^ynlm, 12. (f/i^)M)i 13. ((7"/?")^ 14. (c'^§:r§)i. 15. (jo^'^a;'')*. 16. \m"w"J2 1 n m Negative Exponents. 205. The meaning of a negative exponent may be de- fined by the formula of § 45 : rim -n = «'"-^ (1) Let us suppose m = 3 and n = 5. The equation then be- comes _ = «»-=«- or, reducing, -^ = «-«. (2) We hence conclude: A negative exponent indicates the reciprocal of the corre- ifponding quantity ivith a, positive exponent. Multiplying (2) by «' gives a" Xa-'= 1. Dividing by a "" ', we conclude NEGATIVE EXP0NE2^'TS. 17t Hence A factor may he transferred from one term of a fraction to the other if the sign of its exjponent he changed. 206, If, in the formula (1), we suppose m = n, it be- comes a" = 1. Heuco, because a may be any quantity whatever. Any quantity with the exponent is equal to unity. This result may be made more clear by succes- sive divisions of a power of a by a. Every time we effect tliis division, we subtract 1 from the ex- ponent, and we may suppose this subtraction to continue to negative values of the exponent. In the left-hand members of the equations in the mar- gin, the division is effected symbolically by dimin- ishing the exponents; in the right-hand members tlie result is written out. " EXERCISES. Transform the following expressions by introducing nega- tive exponents: 1. ^-. Ans. a'h-\ 2 ^' a' = aaa a' = aa a' = a a" = 1 a~^ = 1 a a-'' = 1 aa etc. etc. ¥' ^"" "" • "• hh' mix"^' 3. i^. 4. i^y Ans. a'b'm-'n-'. 5. (^y. 6. 0^- \pqy \mnj 7. ~. 8. ^. Ans. {x-^h){x~h)-'. 1 (^^'(^- + n)\V ' f ah' y \ n(m — n) }' ' Kxy'^zj ' Express the following with positive exponents, reducing to fractions where necessary: 13. a-\ 14. — ,. 15. a a-'' 178 POWERS ajsu Hoots. IG. 1 17. ab-\ 18. c 10. m' m-'' 20. a-'h. 21. 22. ce'b-''. 23. a-'xh-\ 24. 7}1~'^X'^V~\ •>■» {a-\h)-\ 2G. {a-^x) (a-^x)-' . 27. a-^x {a^x)-'' 28. {a-\-x)-^ {ii-x)-'' 29. a-' 30. 1 ^07. Tlie rules of involution and evolution by ex- ponents apply to negative exj)onents, regard being paid to algebraic signs. -17 11, Examples. — ^ = -— =i a , a~ 1 IF or («-')-' = {;?)"'= (^' -i-y-"'- Because — 2x — 3 = + 6^ the sign of tlie final exponent corresponds to the rnle of signs in multiplication. EXERCISES. Free the following exponential expressions from paren- theses and negative exponents: 1. {m~y. Ans. -«. 2. (m-y, 3. (m-')-'- 4. (a¥)-' 5. {fj-'h)\ 6. (g-'k)-\ 7. (/i-^F)-\ 8. (h-'x-')-'- 9- {h-^x-y. 10. (rkj-i)\ 11. {p-kj^)-\ 12. (;^V)-*. 13. (;rlvO-*- 14. (;>-'.r^)'. 15. {mhj-^)-\ THE BINOMIAL THEOREM. 179 Section II. Involution and Evolution of Polynomials. The Binomial Tlieoreni. 208. Let us form the successive powers of the binomial 1 _^ X. We multiply according to the method of § 121; Multipliei, l-\-x l + x + X +3? (i+^r -1+%Z + X^ 1 +x l + -ix-\-x' x + 2x' + x' (i+^r = 1 + 3x + Zx' + x' 1 + x l + dx + dx" + a;' X + 3x' + 3x' + x' Multiplier, Multiplier, (1 + a-)* = 1 + 4^: + 6x' + 4x' + x\ It will be seen that whenever we multiply one of these powers by 1 4- x, the coefficients of x, x^, etc., which we add to form the next higher power, are the same as those of the given power, only those in the lower line go one place toward the right. Thus, to form (1 -\- x)\ we took the coefficients of (1 + ^)^ and wrote and added them thus: Coef. of (1 + x)\ 1, 3, 3, 1 1 , 3, 3, 1 Coef. of (1 + x)\ 1, 4, 6, 4, 1 1, 4, 6, 4, 1 Again adding, 1, 5, 10, 10, 5, 1 And so on to any extent. It is not necessary to write tbe numbers under each other to add them in this way; we have only to add each number to the one on tlic left in the same line to form the corresponding number of the line below. Thus we can form the coefficients of the successive powers of X at sight as in the following table. The tirst figure in each line is 1; the next is the coefficient of ^i the third the coefficient of ic"^, etc. 180 POWERS AND ROOTS. For practice let the pupil continue this table to n — 10. n = 1; coefficients, 1. n = 2', (< 2, 1. 71 = 3; it 3, 3, 1. w = 4; it 4, 6, 4, 1. n = 5; a 5, 10, 10, 5, 1. n = 6, (< 6, 15, 20, 15, 6, etc. etc. 1. It is evident that the first quantity is always 1, and that the next coefficient in each line, or the coefficient of x, is n. The third is not evident, but is really equal to ^^>, («) as will be readily found by trial; because, beginning with w=3, _ 3.2 _ 4.3 ,_ 5.4 , 3 = — , 6 = -y-, 10 = -^, etc. The fourth number on each line is 7i[n - l){n - 2) ^ ,^. Z . o Thus, beginning as before with the third line, where """ ^_3_^^1 4_iJ^2 10-^-^ etc (c) ^- 2.3' ^~ 2.3' ^^- 2.3' ^^''' ^""^ 209. The numbers given by the f-ormulae {a), (b), (c), etc., are called binomial coefficients. In writing them, we may multiply all the denominators by the factor 1 with- out changing them, so that there will be as many factors in the denominator as in the numerator. The fourth column of coefficients (6*) will then be written 3.2.1 4.3.2 5.4.3 1.2.3' 1.2.3' 1.2.3' • We can find all the binomial coefficients of any powei when we know the value of 7i. In every binomial coefficient the first factor in the nu- merator is 71, the second w — 1, etc., each factor being less by unity than the preceding one, THE BINOMIAL THEOREM, 181 The first factor in the denominator is 1, the second 2, the third 3, etc. The numerator and denominator of the second coefficient will contain two factors, as in {a) ; of the third, three factors, as in {h) and (c); of the fourth, four factors, etc. EXERCISES. 1. Form all the binomial coefficients for the fiftli power of 1 + X. Ans. -:^-5; ^=10; ^-^ = — = 10; 5.4.3.2 _ 5 _ 5.4.3.2. 1 _ 1.2.3.4" 1 ~ ' 1.2.3.4.5 ~ It will be seen that after we reach the middle of the series of coefficients the last factors begin to cancel the preceding ones, so that the numbers repeat themselves in reverse order. 2. Form all the binomial coefficients for n = 6; 71 = 7; n — 8; n = 9. 310. The conclusion reached in the preceding section is embodied in the following equation, which should be per- fectly memorized: -, , , ^iO^ — 1) o , n(n — l)(n — 2) „ (l + xr = l-}-7ix-\- \,^ \ t-4--^ ^^-^3 ^-x' "+" 1.2.3.4 " ^ -h. . . -t-^. 211. Def. When a single expression is changed into the snm of a series of terms, it is said to be developed, and the series is called its development. ^ { f2 1 ) Example. The preceding series 1 -|- nx -j — ^j— r — - x"" -\- etc., is the development of (1 + xy, EXERCISES. Develop the following expressions: 1. (l + ff)*. 2. (\+ax)'. 3. (l+^)'- 4. (1 + hj. 5. (1 + Vy. 6. {l+ah')\ 182 POWERS AJSD ROOTS. 212. If the second term of tlie binomial is nega- tive, its odd powers will be negative (§194). Hence the signs of the alternate terms of the develojjment will then be changed as in the following form: (1 - .)» = 1 - .. + 'iif-^' r - ^M .' + ec. EXERCISES. Develop: 1. (1 - h)\ 2. (1 - by. 3. (1 - by. 4. (1 - ahy. 5. (1 - ahy. G. [l - "'X 213. The developnient of any bmomial may be reduced to the preceding form by a transformation. Let us put a = the first term of the binomial; b~ the secoiid tei'ni; WE the index of tlie power. The binomial will then be a -\- b. which we may write in either of the forms .{, + t) ,. .(i + ^A). The nth i)o\ver of tlie iirsi form will be By § 210, / b Y_ b n(, - 1) b-^ n{n-l) {n - 2) b^ V^aJ -^^\'^'T72~a-'~^ rr2:3 a'^ Multiplying by a", we luivc etc. a'-hna „ ,, , n(u — l) „ .,,., , n{n — l)(7i — 2) ,,3,3, , '^ + — h — ^ a''-^b' + — ^— j — ^---7, Uc'' ^b -{- etc. We see from this that, in the develo})ment of {a -\- by\ The first term is a^'b^ (because ^"=1). The second term is r/"~^^\ aiul its coefficient is the iudvx of the power. In each successive tcrui the exponent of r/ is diniiiiislM d THE BINOMIAL THEOREM. 183 and that of b increased by unity, while the coefficients are the same as in the case of (1 + ^') • Remark. The work of developing a power of a binomial is facilitated by the following arrangement: 1. Write in one line all the powers of the first term, be- ginning wich the highest and ending with the zero power, or unity. 2. Write under them the corresponding powers of the fiecond term, beginning Avith the zero power, or unity, and ending with the highest. 3. Under this, in a third line, write the binomial coef- ficients. 4. Form the continued product of each column of three factors, and by connecting with the proper signs we shall have the required power. Example. Form the sixth power of 2a — 3a;^ Powers of 2a, 64a«+ 32a« " " - 3a;2, 1 - 3^a Binom. coef., 1+6 (2a - 3x2)6 z= + 16a4 + 8a3 + 4a^ -j- 2a +1 4- 9x* - 27a:« + 81a;8 _ 243a;io + 729x1 » + 15 +20 +15 +6 +1 64a8- 576a6x2+ 2l60a*x*- 4320a3x«+ 48ma^x»- 2916axio+ 729x»». EXERCISES. Develop : 1. {a + by. + .7)' 7. {a-i-2xy. «. {a-\-bY. 5. (a-a-'y. 8. (c-'dh'y. 10. (c'-x'y- 13. (a-' + b-y. 14. Write the first five terms of (w -f- n 3. {a + by. 9. (»-!)'. 12. KB 15. - (a + bY\ 16. - (a - b)l^ 17. '' {a- 2xy. 18. " (a + 3^;)^. 19. "(■+a-- 184 POWERS AND BOOTS. Square Root of a Polynomial. 214. Sometimes the square root of a polynomial can be extracted. To extract a root, the polynomial must be ar- ranged according to the powers of some one symbol. Sup- pose X to be that symbol. Then supposing the roof arranged according to the powers of x, calling 71 the index of the high- est power of X, and calling a, h, c, etc., the coefficients of the powers of x, we shall have Eoot = ax'' 4- Z>:i-"-i + . . . -f /. Squaring, we find that will be the highest term of the square, and r the lowest term. Hence A polynGmial can have no root unless both its highest and mvest terms are i)erfect squares. 315. If the preceding condition is fulfilled, ^er/^ajos the polynomial has a root. If it has, the root is found by the following KuLE. 1. Arrange the polynomial according to ]^owers of that syinhol of which the powers are most numerous, 2. Extract the square root of the highest term, and subtract its square from the polynomial. 3. Fi7id the next term of the root hy dividing the second term of the polynomial by double the tenn of the root already found. 4. Multiply twice the first term plus the second term by the second term, and subtract the product. 5. Find each term i?i succession by dividing the first term of the remainder by the first divisor. 6. Multiply every new term by tivice the part of the root already found plus this term, and subtract the produ^ct. If there is any root, the last remainder will come out zero. 216. Reason of the Rule. To show the truth of this rule let us take the polynomial tr* -\- 2(m + ^)^' + (^^ + n'^)x'^ — 2mn{m -\- n)x -\- m^n^. (a) The square root of the first term is x^. Let us put P for the sum of the remaining terms, so that the root is x^-\-P. (b) SQUARE HOOT OF A POLYNOMIAL. 185 Then, by squaring, or, dropping x^ fron each member, 2Pa;' -\-P' = 2{m + 7i)x' + (ju' + n')x' + etc. (c) Since these two members are to be identically equal, the highest terms must be equal; that is, the highest term of 2Px^ must be 2{m + n)x\ or, dividing by 2x^, Highest term ot P = {m-{- n)x. That is, we find the second term of the root hy dividing the second term of the polynomial by twice the first term of the root. This is No. 3 of the rule. Next, by the rule, we multiply twice the first term plus the second term by the second term, and subtract the pro- duct. That is, if we put, for shortness, a = x^, the first term; Z* = (m -|- n)x, the second term; the quantity the rule tells us to subtract is {2a + l))h = 2aT) + h\ But, by No. 2 of the rule, we have already subtracted a^ = X*, the square of the first term of the root. Hence we have subtracted in all a' + 2al} -\-h'={a-\- h)\ which is the square of the sum of the first two terms of the root. Reasoning in the same way, we find that, at each step, the total quantity subtracted from the polynomial is the square of the sum of all the terms of the root already found. Hence if at any step we have zero for a remainder, it will show that the square of the sum of the terms of the root is equal to the polynomial, and therefore that the root is really the root of the polynomial. The arrangement of the work is shown in the following examples, which the student should perform, and by which he should show that the sum of the quantities subtracted at each step is really the square of that part of the root already found. 186 POWERS AND ROOTS. ErAMPLE 1. I x^-\-{m-\-n)x—mn x*-\-%{m-\-n)x^-\-{m''-\-n'^)x^—2mri{m-\-n)x-\-m^7i'' X* 2x^ -]- {m-\-n)x Mult. {m-}-n)x 2x''-{-2{m-{-7i)x—mn Multiplier, —fun 2(m-^7i)x'-{-{7n'-\-n')x' 2{m^'n)x'-{-(m'^2m7i-{-?i')x' —2mnx'' —2mn{'m-^7i)x-\-nfn^ —2m)ix^ —2rnn(m-\-?i)x-{-m'^7i'^ The order of operations is this: We find the term x* of the root by extracting the root of the highest term. We write this term three tiaies: first on the right, as a part of the root, and then on the left, as a trial divisor, and again as a multiplier of this divisor. Multiplying x"" X x' we have x* to be subtracted from the given expression. We then bring down two more terms. Forming x^ ^ x^ = 2x\ the latter is a trial divisor, which, divided into the first term brought down, gives {m + 7i)x as the second term of the root. We write this term three times: first on the right, then after 2x\ and then under it- self as a multiplier. Multiplying the two terms already found by (m + n)x, we subtract the product from the terms brought down, and thus find a second remainder, to which we bring down the remain- ing terms of the expression. Adding {7n-\-7i)x to 2x''-{-(m-\-7i)x, we have 2a;"+ 2{m-{-n)x as the third trial divisor. Dividing the first term by 2x\ we have the term — rim of the root, with which we proceed as before. Example 2. X* - 2ax' + {a' -f- W)x' - 4:ab'x + 4J* \x'-ax + 2b' 2x^— ax — ax - 2ax' -f (a'' -f W)x'' — 2ax^ -{- a^x^ 2x'- 2ax-^2b' 2b' 4:b'x' - 4.ab'x + U* SQUARE BOOT OF A POLYNOMIAL. 187 EXERCISES. Extract the square roots of: 1. a* -W^^a" -2a-\~l, 2. a* - 4«' + 8« + 4. :j. 4:x* - I2ax' + 26a'x' - 2Aa'x -\- 16a\ 4. x' - Gax' + 16a'x* - 20a'x' + 15a'x' - 6a'x + a\ b. 2bx* - 30ax' + 49«V - 24:a'x + 16«\ 6. rt* + (4^ - 2c)a' + {W - Uc + 3c')a^ + {Uc' - 2c')a + c\ 7. «'^>' + «V'^ + b'c' + 2«^Z>c - 2ab'c - 2abc\ 8. 9:^;^ + la^c^* + 10a;' + 4a; + 1. 3» n 9. 9^:^^"+ 12a;' + lOa.^ + 4x^ + 1. Remark. The root may be extracted by beginning with the lowest term as well as with the highest. The pupil should perform the above exercises in both ways. The fol- lowing is the solution of Ex. 6, when we begin with the low- est term in a: I c'-{-{2b-c)a-{-a' e c'+ (4^r - 2c')a + (4Z'' - 4J)c + 3c')a' + {4tb ^ 2c)a'-\- a^ te^i^Zb- c)a {2b— c)a (4^c' - 2c')a + (4^' - ^bc + 3c')a' (4.bc^- 2c^)a + {W - ^hc + c')a' 2c' -f (U - 2c)a + a' 2c'a'-\-{4.b-2c)a'-\-a* 2c'a'-\-(Ab-2c)a' + a* 217. It is only in special cases that the root of a poly- nomial can be extracted . But when it cannot be extracted, we may continue the process to any extent, though we shall never get zero for a remainder. If the polynomial is arranged according to ascending powers of the leading symbol, the degree of the root in re- spect to that symbol will go on increasing by unity Avith every term added. In the contrary case the degree will go on diminishing, tinally becoming negative. 188 POWMiS ASD MOOTS. Example. Extract the square root of 1 1 1 -[- :, I 1 + 1., _ x^x' + A^x" X. 5 la's :«:* + "r5T^ etc. 1 1 2 + i^ X + ¥ X + ix' i+ X - K ' - K -K -K - ¥' + tV*' a+ X - i-*^-" + iV^" + i^' - ijx' + iV-'" + ¥■' + tV^' - A^^ etc. ;i+ X - i-^' - -ii^" + liV*:' etc. - ii^" - ihx' etc. EXERCISE. Extract the square root of 1 — x by the aboye process, carrying the result to x\ Square Roots of Numbers. 218. The square root of a number may be extracted by a process similar to that of extracting the square root of a ])olynomial. But there are some points peculiar to a number to be understood. 219. Niimler of Figures in the Root. By studying the equations, 3^ = 9, or |/9 =3 9' = 81, or |/8L = 9 10^ = 100, or i^'ioO" = 10 50' = 2500, or 4/2500 = 50; Ave see that If a number has 1 or 2 digits, its square root has 1 digit; If a number has 3 or 4 digits, its square root has 2 digits: If a number has 5 or 6 digits, its square root has 3 digits; etc. etc. ; or, in general: 1. The sqtiare root has as niany dif/its as the numher itself has pairs of digits, a sijiglc (tUjit left over heiiig coioifp'l fs a pair. SQUARE ItUOTS OF 2s UMBERS. 189 2. The first Jigure (ff tlte rout will be the square root of the greatest perfect square contained in the first figure or pair of figures of the numler. 220. Case of Decimals. If we study the equations 0.9^ - 0.81, or |/078l == 0.9; 0.3' = 0.09, or f/(U)9 =0.3; 0.09^ =. 0.0081, or V0.0()81 = 0.09; 0.03'^ = 0.0009, or l/'OOOOO = 0.03; we see that: 1. The first figure of the square root of a decimal is the square root of the greatest square contained in the first pair of figures of the decimal. 2. // the given decimal has two or more zeros after the decimal point, the decimal of the root will hegin with as many zeros as the number begins ivith pairs of zeros. EXAMPLES AND EXERCISES. How do the square roots of the following decimal frac- tions commence? 1. 1/0.223. Ans. 0.4. l)ccause 0.4 is the largest single digit whose square is contained in 0.22. 2. VO.033. Ans. 0.1, because 0.1 is the largest single digit whose square is not greater than 0.03. 3. Va729. 4. /a053. 5. ^0.0092. 6. i/0.0005. 7. 1^000032. 8. i^'. 00008. 221. Extraction of the Square Root. Let us put N = tlie number whose root is to be extracted. n = the value of the first figure of the root. This value may be 7i units, n tens, 7i hundreds, etc., ac- cording to the place it occupies. P = the value of all the figures after the first. Q E the value of all after the second, etc. Then we have, by supposition, Root = Vy = ^? -f /'. etc. etc. 190 POWERS AN2J HOOTS. Squaring, we have iV^ = {n + Py = n' + 2/iP + P'; whence ^nP -^ P' = N - n\ or {%n -{- P) P = N ~ n\ BiYiding by P, we have P = ^^ ~ '^' 2n + P' From this equation we have to find, by trial, the first figure of P, when we shall have the first two figures of the root. Kepeating the process, we shall find a third figure, etc. The general rule will be: KuLE. 1. Separate the figures of the member into pairs, starting fro7n the decimal point or unifs place. 2. The first figure of the root is the greatest number whose square is contaified in the first pair or figure of the number. 3. Subtract the square of the first figure of the root from the first pair or figure of the number, and bring down the next pair. 4. Divide this result by twice the first figure of the root as a trial divisor; add the quotie7it to the trial divisor, multiply the sum by the qiiotient and suUract from the dividend. For the quotient must be tahen the largest numher which will give a product less than the dividend. 5. After subtracting this product, bring dowii another pair of figures, double the part of the root already found, and repeat the process until as many figures of the root as are wanted are found. The reason of this rule is nearly the same as that of the rule for polynomials. If we call n, p, q, etc., the values of the successive figures of the root, the sum of the qiiantities we subtracted is the square Q>i n -\- p ■\- q -\- etc. We first point the given number off into pairs of digits from the unit's place, as is commonly explained in arith- metic. We may then liave, on the left, either one figure or two. The largest square which it contains is the first figure of the root. We subtract the square of this figure, and use double the root figure as a trial divisor. This gives 1 as the SQUARE ROOTS OF NUMBERS. 191 second figure, which we affix to the trial divisor and also use as a multiplier. Subtracting the product again, we use double the part of the root as a second trial divisor, and so on. After getting a certain number of figures, we may find nearly as many more without changing the divisor. We may tlien cut off one figure from the divisor for every one we add to the root. As an example, let us find the square root of 26890348. 26890348 | 5185.5904196 25 101) 189 1 101 1028) 8803 8 8224 10365) 57948 5 51825 10370.5) 6123.00 5 5185.25 10371.09) 937.7500 9 933.3981 10i3|7!l.|l|8) 4.3519 4.1485 2034 1037 997 933 ~64 62 EXERCISES. Extract the square roots of: 1. 2193. 2. 46084. 3. 293462. 4. 89189219. 5. 82.6292. 6. 41693. 192 POWERIS AND BOOTS. Section III. Operations upon Irrational Expressions. Definitions. 222. Def. A rational expression is one in which the only indicated operations are those of addition, subtraction, multiplication or division. All the operations we have hitherto considered, except the extraction of roots, have led to rational expressions. Def. A perfect square^ cube or jith power is an expression of which the square root, cube root or nth. root can be extracted. 223. Def. An expression which requires the ex- traction of a root is called irrational. Example. Irratiomil expressions are or, in the language of exponents, a^, {a + h)h, 27i Def. Expressions irrational in form are called ir- reducible when incapable of being expressed without the radical sign. Remark. Only irreducible expressions are ])roperly called irra- tional. Example. The expressions though irrational in form are not so in reality, because they are equal to « -f ^ and 6 respectively, which are rational. Def. A surd is an indicated root which cannot l)e extracted exactly. Def. A quadratic surd is a surd with an indi- cated square root. Example. The expression a -j- ^ \^ is irrational, and tlie Burd is VXi which is a quadratic surd. AG0RE0A2I0N OF SIMILAR 8U1W8. 191-^ DeJ\ Irrational terms are similar when they con^ tain identical surds. Examples. The terms -/SO, 7 l/30, {x + y) V^O are similar, because the quantity under the radical sign is 30 in each. The terms {a + i) Vx -\- y, ^ Vx -\- y, m Vx -j- y are similar. Agj^regation of Similar Surds. 234. Irrational terms may be aggregated by the rales of §§ 28 and 30, the surds being treated as if they were single symbols. Hence When similar irrational terms are connected hy the sig7is -\- or — , the coefficients of the similar surds may be aggre- gated, and the surd itself affixed to their sum. Example. The sum ax/{^-^y)-i y (^ + ^) + 3 V{^ + y) may be transformed into {a — b -\- ^) ]/(x -\- y). EXERCISES. Reduce the following expressions to the smallest number of terms: 1. dVa- 5ab V2 + 6 |/« + Hab V2. 2. 6«f - 7b'^ + 8ah + 9bk 3. (a'by + {b'cc)h - aa\ - bbl Ans. {a - b)bh -{-{b- a)aK 4. {c'x)h + {b'yY^ - bxh - cy^. 5. {a + b) Vx-\- (a - b) Vx. 6. a(xh + yh) + b{xh - yi). 7. a(xi - yh) + b{yi - zh) + c{zh - x\). 8. {d + c) sT^^J^y - [cl - c) V^~^y. 9. 7 x/x -f y -f 8 Vx - y - 5 Vx -}- y - 7 Vx^^. 10. i Vm -j- n — ^ Vm — u -\- ^ Vm -f- 7i — ^ V'm — n, 11. i{a + b){ Vx-\- Vy) - i{a - b){ V^ - Vy). 12. (m 4- n){ai - M) -f {m. - n){ah + M). 13. (m-{-n-2p)(ai^bh- c^)^{7n -2n-^ p){ai -M~-hd) -f- (- 2m + n + p)(-- ai + M -f c^). 194 fowehs and hoots. Multiplication of Surds. 225. Irrational exjoressions may sometimes be trans- formed so as to have different expressions under the radical sign by the method of § 197, applying the following theorem: Theorem. A root of the product of several factors is ■equal to the product of their roots. In the language of algebra, \ahcd, etc. = \^ Wb Vc" V^, etc. 1111 = aH^'d^d^', etc. Proof. By raising the members of this equation to the jiQi power, we shall get the same leeult, namely, a X b X c K d, etc. Example. V3Q = 1^4 l/9, ivhich, by extracting the roots, becomes 6 = 2.3, ii true equation.. EXERCISES Prove the truth of the following equations by extracting the roots in both members: VI 1^49 = VTM. VI V2b --= VIM. VI VM = VIM. 1/94/25 = VdM. 226. Ap^j)lication of the Theorem to the Multiplicatiori of Surds. The preceding theorem gives the following rule for finding the product of two or more surds with equal indices: Rule. Form the product of the quantities under the radi- cal sign, and ivrite the similar indicated root of the product. If this root can he exactly extracted, the product is rational, EXERCISES. Express each of the following products by a single surd, or without surds. 1. Vx X \^. Ans. Vxy. 2. VbX\^. 3. \7ixVbX\'c. 4. \^i X Vp X V"/. MULTIPLICATION OF 8U1WS. 195 6. {a + by {a - b)K Ans. {a' - by, 7. {x + yy{x^ y)K 8. '/ic 4- «2/ X Vx — ay. 9. |/a''+ 1 X Va-\-l X l/a- 1. 10. Vm'x V^^J X V'm +*7i. 11. /5 X '/S X V«+^. 12. Va X Vb X Va + b X V^ - b. 13. i/«J X /^ X Vck Ans. ^?^>c. 14. {2m7iyx(27npYx{:npy. 15. 2%i^ X 2^?i^ X 2imK 16. 2K« + by X 2i{a -^ J)* X (a -{- b)K 17. (aJc)^ X (bcdy X (cda)i X {dab)K 18. a Vx Vy X b Vx Vy X c Vx V~y, 19. a(« + by X{a- by 4- b(a + &)* X (tt - J)i Ans. {a-{-b){a''-by. 20. a Vx~^ X Vx — y ^b Vx -]- y X Vx — y. 21. V^niTi X V2mx X VnxX Vy- 22. |/6 X VT^ X I^IO. 23. «*^1 + -J- 24. ;^iif^ - ~-]\ 25. |/(^+^X i/(m+^i)t 26. V2'x V2 X 'VTe. 27. V(«+ J) X W{a - b). 28. V'^ X |/-^. 29. l/^ X V- ^ n in ^ aA- b ^ 1 m ^ a-\- b {a-\- by 30. r -y- X r yj. 31. r -—^ x y — — y. A A^ a — b a-{- b 32. /;4^ X /f -^ X j/y-fr X / A -|- ^ li — k h -\-k h — k Factoring Irrational Expressions. 227. Conmrsely, if the quantity under the radical sign can be factored, we may express its root as the product of the roots of its factors. (Comp. 191, 199.) t96 POWERS A^n JWOTS. EXERCISES. Factor: 1. {aljc}i. Ads. ciiMc^. 2. {mnp)^. 3. Va"" — ah. Alls. Va \'a — b. 4. \^m^ — mx. 5. \d'—'b'' 6. ^a' - b\ 7. Va' - a^Z>^ 8. 1/55. 9. V66. 10. (?//^' - 4m V)^. 11. (a'b-aby. 12. (a:^;^' + «V + «':r)i Ans. r^ia;%^+ «:?; -f r;^)^. 13. {a'm'-a'm'+a'niy. 14. (P'^ _ .^^^-^^^ 228. When the quantity under the radical sign is factored, one of the factors may be a perfect power. We may then extract the root of this power, and affix the surd root of the other factor to it. In the problems of this and the next three articles the ob- ject is to remove as many factors as possible from under the radical sign. EXERCISES. Factor: 1. V¥b. Ans. Va' Vb = aVb. 2. Vrnri^. 3. Vnfn^. Ans. mn Vn. 4. Vm'n'x\ 5. Va'bc\ 6, (g'rY)l 7. {4.a'x)h. 8. 4/56. Ans. 2 VU. 9. |^'27. 10. V36^. 11. V2 Vn. 12. Vx\a-\-b). 13. Va'b-a'b\ 14. |/Zt4- Ubx + b'^x, Ans. {a -\-b)Vx. Here the quantity under the radical sign is equal to (^2 4. 2ah + W)x = {a-[- hfx. In questions of this class the beginner is apt to divide an expression like ^^a -\- b -\- c into Va -\- Vb -\- Vc, wliich is wrong. The square root of tlie sum of several quantities cannot be reduced except by factoring. Hence such an expression as the square root of a -{-b-\-c is not reduci- ble, but must stand as it is. FACTOlUJSiJ IRRATIONAL EXPRhJSlSlOJSS. 197 15. (m^ — 'im^n + m7i^)h. 16. ym* — 2m^n -\- m^n^ 17. VAa^ - Hacy + 4:c'y. 18. Va' - a'b\ 19. V(fb'-b\ 20. 4/4wV-T6?. n. (m'-tmy. 22. (^^^ + w^)i 23. (a' - 2a' + r?)i 24. (a?^'' + 4«;^' + 4:any. 25. (w^ + 4^i^ + 47z^)i 229. When a quantity to be factored by the pre- ceding method is affected by a fractional exjionent, this exponent may be divided into an integer (positive or negative) and a fraction, and the quantity factored accordingly. Exj^mples. ai = a^+i= a^ . ah a-i = a-^+i = a^^ . a^ = — . a EXERCISES. Factor: 1. ml ■ 2. chnh 3. chul • 4. 8i Ans. 16 V2, 5. 24i 6. t'24^. 7. ax^ -h hx^ -f cxl Ans. (a + bx + cx'^yxi. 8. 2mi + 3«//?i — 4.7i^ml 9. x-i-{-xK Ans. fl + -j xK 10. «-^a:i-«*a;--*. 11. a* + 2«§ + 3^§. 12. aixi-\-aixl 13. 8i^ + 8^a;i 230. The preceding method may be applied to any other indicated roots as well as to square roots. Example 1. Factor 'VT6 = 16i We have 16^ = 8^ X 2^ = 2 . 21. Ex. 2. Factor W~a^ + ^Va^. Solutio7i. a~l-{- a^= a~^. a\ -{- aal = (— ■+-V^» ....=: {a''W(f . . . .)". 200 POWERS ANJJ Room. EXERCISES. Express with single surds the products; 1. V2xW'd. Ans. 72i 2. 3*2i 3. am. 5. a^Uxk 7. mxk 9. 2^{x-y)i. 11. 2m. 1 1 4. xp^y''. 6. 2ai Ans. (4a)i. 8. 32*«i^i 10. a^(x -^ y)\ (:c - y)h, 12. 3H^5i 13. t 14 (^^' + 2/)^ {x - yY 234t. Co7'ollary. Because a rational quantity may be considered as affected with the exponent — , the preceding method can be applied to reduce the product o<* q rational quantity and a surd to the form of a surd. Example 1. Eeduce 2h^ to a single index. 2m = 2m = (2T)^ = {8by =Wsb\ Ex. 2. dabkr:h = dia^x^ = (SWb'xy = 'VU^a'b'x Ex. 3. a^h Vm = a + VWi. EXERCISES. Reduce to single surds : 1. 2Vb. 2. 3 Vx. 3. bax^. 4. m Vn. ■ 5. (m + n) Vm — n, 6. a Va. 7. {a + b) Va-{-b. 8. (m - n)(m ~ n)-i. 9. ^^ + ^ 10. «(a + J)i 11. m{m — 7i)'i. 1 m ^^^ «^^ ^g xy'z^ h fk^m^ X x^ • ax^y\ y' y' ' cp^q^' MULTIPLICATION OF lliii ALTON AL Ji]XL^TiE88L0NS. 201 235. Irrational expressions may be multiplied by multi- plying each term of the multiplier into each term of the mul- tiplicand and taking the sum of the products. Example. Develop the product {a -[- b Vx){g -\-h Vy). a + bV^ ff + ^i'^y ag + bg Vx ah Vy + ^^^ ^^y Product = ag -{- bg Vx -\- ah Vy + bh Vxy, EXERCISES. Multiply: 1. {a-c V^){a + b Vy). 2. (a + Vy){a - Vy), 3. {a + n Vx){b — m Vy). 4. ai{aib — a^x). 5. (am -\- n Va — x){n — m Va — x). Ans. mnx -j- {rC — am"^) Va — x. 6. {:m-nVy){m + nVy). 7. {a -\- c Vx){a - c Vy). 8. {a' -\- c V^^^^a){c ^ a V^^'^^a). 9. {a + Va' - x'){a - Va' - x'). 10. (m - ^m'-%n'){m + Vm'- 2n'). 11. (m + V7n' - l)(m - |/m^ - 1). 12. (V^-{.Vb)(Va- Vb). 13. (c+ i^+ |/^)(c - i/^- Vy). 14. (a* + M + c*)(«^* -bi+ ci){ai -\-bi - ci)(ai - §i - c*). 15. ai[l-{a-l)i][l-]-2(a-l)i]. 16. ( I^M^ + Vx - a){ Vx-{-a — Vx — a). 17. {m + l/^+ l/^)(m - Vx^ Vy). 18. (a: + «/ - |/J- Vy){l + l/^+ i/^). 202 POWERS AND ROOTS. \Vx-a J\ Vx -a ) 21. \{i)t + 1)^ + '^A [(^^^' + 1)^ ~ H- 236. Rationalizing Fractions. The quotient of two surds may be expressed as a fraction with a rational numer- ator or a rational denominator by multiplying both terms by the proper multiplier. Example. By multiplying both terms of —^ by 4^6, we have 1/5" _ 6 and the numerator is rational. In the same way, multiplying by Vl , we have ^ VJ_ _ v^ Vl~ 7* EXERCISES. Express the following fractions with rational denominators: 1 -^. 2. -^ ' i^d* '2 1^2 3. i^ 4 -l^^_±i ^ Vm ^ Vx'-l ^ \m — n] ' d c 9. "4. 10. i. ni al RATIONALIZING FRACTIONS. 203 237. If one term of tlie fraction contains a quad- ratic surd, it may be rationalized by multiplying by tlie term itself with the sign of the surd changed. Example. To rationahze the denominator in Vr we multiply both terms hj P -\- Q V R. The numerator becomes QR-{-P Vr. The denominator becomes {P-Q VR){P -\-QVR) = P'- Q'R. So we have _ V~R ^ QR^P \^R P-QVR~ P'~g'R ' EXERCISES. Eeduce the following fractions to others with rational denominators: ^ a-^-h Vx ^ a — c Vy a — h Vx a -\~ c Vy ^ Vh , Vx Vb a -\- c Vx f, Vm -\- n c! ci-\- x^ a — Vm — n a — x^ 7 ^+ ^ 3 Wn-\- \/{m-{-n ) VA- Vb ' Vm- \/{m + 7i) ^ V^- V(a-x) ^ ^^^ 1 Vx-\- \/{a — x) m -\- Vm^ — a^ 11 1 12 (^^^ + nY ~ {m - ny ' m — Vrrf — a'' ' {m + n)^ + (?^ — n^ 13. ^-y_ . 14. ^ x-\-y-2Vxy ^-^'^'^ 204 POWERS AND ROOTS. Irrational Factors. 338. By introducing surds many expressions can be factored which are prime Avhen only rational factors are con- sidered. The following theorem may be applied for tliis purpose: Theorem. The differeyice of any ttvo quayitities is equal to the product of the sum and difference of their square roots. Proof. If a and h are the quantities, we shall have a - h = {ai - l)^){a^ ^ h^), as can be proved by multiplying, or by § 123. Factor: EXERCISEa 1. a — X. 2. a -1. 3. a" - bx. 4. 16 - 3. 5. 1 a _ 1 c ' 6. x' -{a + by 7. {x - cy X 8. x'- ■ {P + q)' 9. {x-cf- -i{p- -q)' 10. X a .1 c ' 11. 1 - ia + ^). 12. m"- m - - n^ m — n ■ n m -{- n' 13. x'-^.ax + 4.a' -I. 14. x"- -2x-6. 15. x'~2ax-i- a'- {p-\-q). 16. 2a'- 4:{p - q). 239. Irrational Square Roots. When a trinomial con- sists of two positive terms with twice the square root of their product, its square root may be found by the method of § 131. EXERCISES. Find the irrational square roots of the following expres- sions: Note. A few of the roots are rational. 1. a — 2 Vab + b. Ans. l/« — Vb. 2. a-^2Vab-^b. 3. « + - + 2. Ans. ai + \, a a* IRRATIOJS'AL F AC TOMS. 205 ' a 6. a' - 2 + 4. 7. «^" - 2 + i. 8. 9 + 5 - 6 y^. ^* ^ + 1^4+3- + ^- 11- ^ + 2^ + 16- 12. ?^-?i + !:i. 13. ^-2 + ^' 16 10^25' ' c a 14. 4 H . 15. 6-4 . c a n m 16. i» + y-2H r—. 17. (x^y)-^{x-y)-^2V¥^^\ 18. ic-l" 2:^+1. 19. ai-'Za^-^rl. 20. «i — 2^ + «i 21. 4 2 "^ ~4 * 22. a;§ + 2^: + .-ci 23. x^ — x -\- -x%. 24. mi-2 + m-i 25. (a; + «) + 2(:c + a)i+l. 26. 4(^ — «) + 4wi(a; - ^)* + m\ 27. (a: - <^) - 2«i(:?; -- a^ + a. 28. l-^+2c|/l-- + c». 29. «'Z>' + 4 + 4^-''^-*. 340. Square Roots of Irrational Binomials. If we have such a binomial as « + Vh, its square root may be expressed in the form ^a-^Vh, ' in which a surd is itself under the radical sign. If this root can be so reduced that no snrd shall be under the radical sign, it is said to be reducible; otherwise it is irreducible. 206 i"0 WEllS AMJ HOOTS. Problem. To lind whether the square root of an irra- tional binomial is reducible, and when it is to express it in the reduced form. Solutioii. Let us suppose Squaring, a-\-Vb = x-\-y-\-^ Vxy. This equation may be satisfied by putting x-\- y = a, h xy = j. By putting ^a — Vb = Vx — Vy, we get the same ex- pressions for X and y. Therefore, if we can find tivo rational quantities, x and y, such that their su7n shall he equal to the ratiojial terrn of the binomial and their product to one fourth the square of the surd term,, Then the root is the sum or difference of the square roots of these quantities. Example 1. Vb^%V<6 = ^5+ VYl We have to find two numbers of which the sum shall be 5 and the product 24 -t- 4 = 6. Such numbers are 2 and 3. Therefore ^5 + 2 1/6= |/3 + \% and ^5-2 V^ :^ Vd- 1^2. ' Remark. We see that when we express the binomial in the form a ± 2 V'c, a will be the sum and c the product of the required numbers EXERCISES. Express the square roots of ; 1. 7 + 2 V^, 2. 7 - 4 |/3. 3. 7 _ 2 1/6. 4. 9 - 1/80. 5. 3 _^ |/8. 6. 6 + 4 i^. 7. 6 + V^, 8. 7 + ^^■ 9. 8 + |/60. 10. 14-8 \^. TO COMPLETE TEE SQUARE. 207 11. 15 - 6 y^. , 12. 14. 17 + 4 Vl5. 13. 2m + 2 V'm'* - 1. m -1- \/m' — 1. 15. 2 - 2 |/1 - m\ 16. 18. 1 - fl - m\ 17. 2« + 2 Vcc" - ^^ a - Va" - b\ 19. ^'^ + 2a Vb' - d\ 20. 4.a-^2V^a'-b\ 21. 2a - b- 2 Va'-ab. 22. 4:X-2i/4:X'-a.\ I To Complete the Square. 241. If one term of a binomial is a perfe'ct square, such a term can always be added to the binomial that the trinomial thus formed shall be a perfect square. This operation is called completing the square. Proof. Call a the square root of the term which is a perfect scjuare, which term we suppose the first, and call w the other term-, so that the given binomial shall be a^ -\- m. Add to this binomial the term — ^, and it will become 4rr This is a perfect square, namely, the square of , m H that is, „= + ,„ + g==(„ + |Ly. Hence the following Rule. Add to the binomial the square of the second term divided by four times the first term. Example. Wluit term must be added to the expression x^ — 4:ax to make it a perfect square? The rule gives for the term to be added {-4:axy . , 4:X^ Therefore the required perfect square is x" - 4:ax + 4^" = (x - 2ay, 208 F0WER8 AND ROOTS. EXERCISES. Complete the square in each of the following expressions, and extract the root of the completed square: 1. a\- 2ab. 2. «' + 4:ax. 3. 4«' - Sax. 4. 4:a' + 4:a'x. 6. a' -\-b. 6. a' - h. 7. a' - 4tt^ 8. a'x' + a'x. 342. In the preceding article, by adding 4«* to the binomial x^ — Aax, we formed the equation x"" — 4:ax + 4a* = {x — 2a) ^ By transposing the added term, we have x^ — 4:ax = (x — 2ay — 4a^ The original binomial is now expressed as the difference of two squares. Therefore the above process is a solution of the problem Having a Imomial of wliich one Irr^'^ is a perfect square, to express it as a differ ejice of two sqtKn^ -. EXERCISES. Express the following binomials as differences of two squares: Ans. {a -hY - l\ 3. x^ -^^ax. 1. a^ - lah. 2. a" - 4a5. 4. rr' + 2ax. 6. x' _ 2x a^ a' 8. Aa'x' - Ah'x, 10. mV + 2, 12. ;..-■ 14. :c' X 4a' + «' 5. a\v^ — a^x. 7. a' ' ■^* 9. w'a;' - 1. 1. 1+. 13. a'x' - Wx. 15. .'+;^. MEMOBAJ^BA FOE REVIEW. Memoranda for Eeview. '2m Involution. - Fotuers cmd Roots of Mono7nials. Define: Power; Square; Cube; nih. Power; Index of power; 7ii\\ Koot; Index of root; Square root; Cube root; Evolution. Of products; Theorem; Proof. Of fractions; Theorem; Proof. Of powers; Theorem; Proof. Algebraic sign of powers; Theorem. Sign of evolution; Index; Vinculum. Division of exponents; Theorem. Sign of even root. Of products; Eule; Proof. Of fractions; Rule; Proof. Indicate root; Explain. Theorem of power and root (§ 202). Significance of terms of the fractional expo- nent; Explain. Powers ) „ . -. . ^ , . , P , j- of expressions havnig fractional ex- ponents; How formed. Meaning; Explain application of preceding rules to negative exponents. Evolution. Fractional Exponents Negative ( Exponents. ] Binomial Theorem. Square Root of Polynomial or Number. Powers and Roots of Polynomials. Powers of 1 + :r; How formed. Expression for coefficients. Define binomial coefficients. When one term is negative. Powers of a + &• Arrangement of terms. Criterion that a root is possible. ^ Eule for polynomial; Explain. When exact root cannot be extracted. ' Square root of numbers; Rule; Explain. 210 POWERS AND IIOOIS. Surd Fac- tors and Products. Irrational Expressiotis. Define : Rational; Irrational; Perfect power; Irreducible; Surd; Quadratic surd; Similar surds. Aggregation of Similar surds; Rule. Theorem of roots; Proof. Products ) „ -, „ T -p ; I surds 01 same degree. When one factor is a perfect power; Rule. Case of fractions. Reduction of surds to common degree; Rule. Product of surds of same degree; Rule. Of irrational polynomials. Rationalizing Irrational fractions; Rule. To factor any binomial; Theorem. Irrational square roots. Square roots of binomial surds; When re- ducible; Expression for root. To complete the square; Rule; Result. Irrational Factors and Roots. CHAPTER V. QUADRATIC EQUATIONS. Section I. Purp: Quadratic and Other Equations. 243. Def. A quadratic equation is one which, when cleared of fractions, contains the second and no higher power of the unknown quantity. Remark. A quadratic equation is also called an equa- tion of the second degree. Def. A pure quadratic equation is one which contains no power of the unknown quantity except the second. Def. A complete quadratic equation is one which contains both the first and second powers of the un- known quantity. Pure Quadratic Equations. 244. If, on clearing of fractions, arranging according to powers of x and transposing, we put xi = the coefficient of x^^ B = the terms not containing x, a pure quadratic equation will reduce to Ax' = B. Dividing by ^, we have . B x=--. Extracting the square root of both members,, 'B _ & T~ "^ Ar 212 QUADRATIC EQUATIONS, !<545. Positive and Negative Roots. Since the square root of a quantity may be either jjositive or negative, it fol- lows that when we have an equation such as x^ = a and extract the square root, we may have either X = -\- ai or X = — a^. Hence there are two roots to every such equation, the one positive and the other negative. We express this pair of roots by writing X = ± aif the expression ± # meaning either -\- a^ or — aK Remark, It miglit seem that shice the square root of x^ is either -\-xoT — X, we should write ± X = ± a^, having the four equations x = a^, X = — a^, — X = -\- a^, — a; = — a^. But the first and fourth of these equations give identical values of x by simply changing their signs, and so do the second and third; hence more than two of the equations are unnecessary. Example. Solve the equation X — na _ nx — h X — a X — b' Clearing of fractions, we ha^e {x — b)(x — na) = (a: — a)(nx — b)y or x^ — bx — nax + ^«^ = ^i^^ — i^ct^ — bx-\- ah. Removing equal terms, x^ + nah = nx"^ + ab. Transposing, nx"^ — x^ = nab — ab, or (n — l)x'' — (n — l)ab. Hence x"^ = ab. Extracting root, x = ± Vab. FUUE QUADHATIC EQUATIONS. 213 246. Studying the preceding process, we see that a pure equation is solved by the following Rule. 1. Clear the equation of fractions. 2. Transpose all terms containing x^ as a factor to one member; tJiose not containiyig x to the other. 3. Divide hy the aggregated coefficient of the square of the imknown quantity. 4. Extract the square root of loth memhers. EXERCISES. Solve the following equations, regarding x, y or z as the unknown quantity: 1. x"" = a. 2. x' = c\ 3. dx" = 27. 4. 2x' - 98 = 0. 5. 3x' - 36 = 0. 6. 7x' - 4: = 0. 7. (a - h)x' = c. 8. {a - b)a'' = a' - h\ 9. mx'- m''^ nx'-^ n^= 0. 10. (x + 5){x - 5) = 11. 11. (x- a'){x - c') + (a' + c')x = 3a' c\ 12. ax^ — b-i-c = 0. 13. (x - a){x - b) + (:r + a){x + b) = a\ ^. a ff ~ x' 7,2 16. (x' + ay 10 _ 8^ 1 x'-i-2- 6x^ - 32- ^^' '"^ + ^ = ^''^- m 18. ^.r 4- _- :::= ^^:^,. jg^ x{a-x) - x(a + :z,-) = 0. 22. a : .^' = a; : ^. 23. « : ^.-c = c^ : «. 24. x'-^-a -.x" — a=p-[-q :p — q, 25. (:^- + « + ^) (^ - a + ^) + (a; 4- « - J) (:^ _ « _ J) = 0. 27. ^^a-^2b :x^a-2b = b-2a-^2x:b-{-2a--2x. Reduce by composition and division (§ 185). 214 QVABliAl'IC EqUATIOJSS. Pure Equations of any Degree. 347. Def. An equation wMcli, when cleared of fractions, contains only the Tith power of the un- known quantity is called a pure equation of the nth degree. A pure equation of the third degree is called a pure cubic equation. One of the fourth degree is called a pure biqua- dratic equation. Def. A pure equation reduced to the form x"" ^ a (1) is called a binomial equation. 24:8. Solution of a Binomial Equation. 1. When the exponent is a wliole nuniber. If we extract the ^^th root of both members of the equation (1), these roots will, by Axiom V. , still be equal. The /^th root of x"^ 1 being x, and that of a being a'\ we have X — a"", and the equation is polved. 2. When tlie exponent is fractional. Let the equation be x:^ = a. Raising both members to the /?th power, we have x"^ — a^. Extracting the mth root, X = a"^. If the numerator of the exponent is unity, the equation will take the form i x"" = a. By raising it to the ^th power, X = a^. Hence the binomial equation always admits of solution by forming powers, extracting roots, or both. PURE EQUATIONS OF ANY DEGREE. 215 EXERCISES. Find the values of x in the following equations: 2. -r = a, X 1. a ¥ = "• 3. Xi 5. 8 X' X 27* 7. x' ~ X ' 9. x^ ei 11. V7nx' - ¥ = X. Square both members. 12. {x - h)^ = mK -iA /^2 1 7,2\1 _ ^ 4. x^ — c x^ — a 6. ^ = »K X^ x^ m O. — —r. a oc^ 10. \/{x^ - a") = c. 13. {(x^ - d)^ = qh. (x' - ny 249. Problems leading to Binomial Equations, 1. Find two numbers one of which is three times the other, and the sum of whose squares is 90. 2. Find two numbers of which one is twice the other, and of which the product is 48. 3. Two numbers are required whose ratio is 2 : 3 and the difference of whose squares is 61:|^. Note § 188, Cor. 4. Two numbers are required whose ratio is 3:5 and the difference of whose cubes is 264G. 5. Find three numbers of which the sccoiul i,s i \vi(;c the first, the third tAvice the second, and tlie sum of the ,<(]iiarc& 378. 6. Find a number such that if 4 l)e added lo i{, mikI sub- tracted from it the product of the sum uud difference shall be 273. 7. Of what two numbers is one twice the olher, and the difference of the squares 60? 216 QUADRATIC EQUATIONS. 8. Find two numbers wliose ratio is 2 : 3 and the sum of whose cubes is 945. 9. Find two numbers whose ratio is 3 : 4 and the square of whose sum is 392. 10. What number multiplied by its own square produces 1331? 11. What number multiplied by its own square root pro- duces 729? 12. Find tw^o numbers in the ratio 2 : 5 tlie difference of whose squares is greater than the square of their difference by 216. 13. Find two numbers in the ratio m : n whose product is rt^ 14. Find two numbers in the ratio m : n the difference of whose squares is greater by m^ — 7i^ than the square of their difference. 15. What number multiplied by its own square root pro- duces 48 V^? 16. Of what number is the product of the square and cube roots 243? 17. Find three numbers in the ratio 1:2:3 the sum of whose squares shall be 350. Section II. Complete Quadratic Equations. 250. Normal Form. Every complete quadratic equation can, by clearing of fractions and transpos- ing, be reduced to the form ax" + J^ -f c = 0, in which a, h and c may represent any numbers or algebraic expressions which do not contain the un- known quantity x. Def. The form ax' + hx-{-c = (1) is called the normal form of the quadratic equation. The quantities a^ h and c are called coefficients of the equation. SOLUTION OF qUADRATIC EQUATIONS. 217 251. General Form. If we divide all the terms by a, the equation (1) will become x' -\--x-\-- = 0. a a Kow let us put, for brevity, h c ^ a ^ a The last equation will then be written x' ^px^q = 0. (2) Def. The form x^ -\-px-\- q = is called the general form of a quadratic equation, because it is a form to which every such equation may l)e reduced. Solution of a Complete Quadratic Equation. 252. The Equation m its Ge7ieral Form. AVe first transpose q, obtaining from the general equation (2) x^ -\- px — — q. By § 241 the first member may be made a perfect square by adding ~. Adding this quantity to both members, we have ^ -{-P^-h-^-^'j-Q- The first member of the equation is now a perfect square. Extracting the square roots of both members, we have Transposing ^, we obtain a value of x which may be put in either of the several forms, p Vf- -i, 2-^ 3 f or x = i{ — p ± yp" — ^q), and the equation is solved. 218 QUADRATIC EQUATIONS. ^53. The Two Roots. Since the square root in the ex- pression for X may be either positive or negative, there will be two roots to every quadratic equation, the one formed from the positive and the other from the negative surd. If we distinguish these roots as x^ and x^, their values will be _ -p- 4/(7/ - 4g) ^« - ^z Example 1. Solve the equation X ^^ _ 9 X — S X -\- '6 Clearing of fractions, x' + Sx- 2a;'+ 6x = - 2(x' - 9) = -2x' -{- 18. Transposing and arranging, x' + 9x = 18. Completing the square, 2 . n 1 81 1Q 1 81 153 ^ +9:f-f- = 18 + -^- = -j- 9 . V153 (3) Extracting the root, x -{- — = ± whence x = 2 " -" 2 ' - 9 ± VTEs 2 Ex. 2. -x' -\-Sx-l = 0. The coefficient of x"^ being negative, we must, in ord^r to form a perfect square, change the sign of each term. The equation then becomes x'-3x-\-l = 0, or Completing x'- square, x'- -3x -3x-^ 9 4 _ 9 ~ 4 1. - 1 = 5 Extracting root, X — 3 2 = ± 2 ' whence X _ 3 ± 1/5 2 SOLUTION OF QUADRATIC EQUATIONS. 219 EXERCISES. Solve: 1. x' - ix = 3. 2. x" -^x = 0. x — ^_'^x — ^ x — ^_x-\-l x-\-% x-\-ll a; + 2 2a; -8 5. x" - 2x = 3. 6. 2;' - 5a: = 14. 7. 2{x - 2){x -S) = (x- l){x + 2) - 16. 8. {2x + 2){x 4- 3) - (a; - l)(a; + 2) = 5. 9. x' -j-ax-{-b = 0. 10. a;' + «a: - ^' = 0. 11. a:' - «a: + Z> = 0. 12. x' - ax - b = 0. 254. TAe Equation in its Normal Form. If in the equation (1), ax? -^hx -\- c =■ ^, we transpose c and multiply the equation by «, we obtain the equation d'x^ + abx = — ac. To make the first member a perfect square, we add — to each member (§ 241), giving a^x^ + alx 4- — = ac. Extracting the square root of both members, we have ax-\- - — - |/(&^ — 4«c). We then obtain, by transposing and dividing, -'b± \/{y - 4:ac) - = -£±i^(*'-^«^) 2a Hence the two roots are _-b-j-{b' - 4.ac)i x^ — and x„ = — 2a {If - ^acy 2a (4) We can always find the roots of a niven quadratic equation by sub- stituting tlie coefficients in tlie preceding expression for x. But the student is advised to solve eacli separate equation by the process just iriven, wliich is embodied iu the followina: rule. 220 Q UADRA TIV KQ L A TWyS. 255. Kulp:. 1. Reduce the equation to its normal or its general form, as may be most cofivenient. 2. Transpo.se the terms which do not coidain x to the sec- ond member. 3. If the coefficient of x^ is unity, add one fourth the square of the coefficient of x to both members of the equation and extract the square root. 4. If the coefficient of x^ is not unity, either divide by if so as to reduce it to unity, or multiply or divide all the terms by such a factor or divisor that the term in x" shall become a perfect square. 5. But if the term in x^ is already a perfect square, no multiplication or division need be performed. 6. Complete the square by the rule of § 241, and extract the square root. Example. Solve the equation 7 = 2x. X — 4: Clearing of fractions and transposing, we find the equation to become 2x' - 9x + 1 = 0. X 2, ' 4:X^— 18:k = — 2. 92 81 Adding x ~ IT ^^ ®^^^^ member of the equation, we have 4^ - I8.7; + -- ^ - - 2 = --. ll.itracting the square root, 2x 9 _ V73 _ 1/73^ 2 ~ ^ 4' ~ 2 ' whence we find x 9 ± V73 ~ 4 So the two roots are ^. = i(9 + 4/73), .?•„ = i(9 - 1 73). SOL UTJOS 01^ Q VA DBA TIC EQ UA TIONS. 2^1 EXERCISES. Solve the equations: 1. x''^Jix = ilc\ Steps of Solutio7i. h _ {h^ + k'^f ^ + 3" - 2 ' _ -h± jh'' + A;^)* X - 2 ' or . »i = ^-y-^ , -Ti- {M + ^2)^ x,= — ^ 2. a:'' + 2Aa; = F. 3. ^'^ + 2^:r + ¥ = 0. 4. x'' + 4m^ - 7z = 0. 5. 2;' - 2ax = 3a\ 6. a;' - 4:ax + 4«' = 0. 7. m'2;' + 2/irz; = p. 8. wa;' - m'^ = 4m'. 9. «a;' - 2bx = W, 10. 4 _ 2- - 3 = 0. 11. |-^ + 6- =: 3m. x^ 12. — ,- — mx = n. 13. m'a;^ — 2m'ic + m* = 1. 14. 4:cx' - {a + b)x-\-c = 0. 15. (a + b)x' + (« - ^')a; = a' - b\ X — a m — 16 ^~ ^ I ^ — g _Q 17 _^^_±_^ = _ 1 *^' — C.T — a 2* * .:?; + « m H ' — X — a 18 _1- = 1+1 + 1 19 ^_?. = i!4.i a; + 5 a^S^x" 3 X S^ x' 30. i - - =. i + «. 31. i+ 1 =L+L. m X n X X X -\- a a 2a 22 _^4_-J_=_15_ as ^ + 1 , ■^-1^4a.--l a; + 2 "^ it- + 4 a; + 6' a; + 2 "^ a; - 2 2x - V 24. a' + 5' + cc' = l-2aJa;. 25. {a - xy + (.'T - hy = (a - hY. 222 QUADRATIC EQUATIONB. 26. (^ -\-¥ -\- x' = 2{ax -i-bx-i- ah). ^^_ {x - ay + (:, - ^^ _ a^ -f h' (.^ _ af -{x- by ~ a' - ¥' 28. ^ + - = ^. 29. a; + - = 3. X 'Z ^ X 30. X = — -. 31. ic =1. X 4 X 32. a; + - = — . 33. X — c. X n X In the following equations find the value of y in terms of X, and of x in terms of y\ 34. x"" + xy + I/' - c' = 0. -y ± Vic' - Zy' -x± VI^^^' Ans. X - — ^ =^; y = ^ . 35. x"" - xy -^ if - h' = 0. 36. y''~dxy-\-'2x''-x-y = c. 37. a;^ - 4^y + 4^^ = 0. 38. x" + 7ia;?/ + ny' = 0. 39. alf^ -h «-Ja:?/ + i'y' = 0. 40. x' - Q—xy + 9-^.y'^ = 0. 256. Problems leadmg to Quadratic Equations. 1. Find two numbers of which the sum is 32 and the pro- duct 231. 2. Find two numbers of which the sum is ]) and the product q. 3. Find two numbers of which the sum is 40 and the pro- duct 48 times the difference. 4. Of what two numbers is the difference 10 and the pro- duct 375? 5. Find those two numbers whose sum is 38 and the sum of whose squares is 724 6. Find those two numbers whose sum is m and the sum of whose squares is n". 7. Find those two numbers whose difference is m and the difference of whose squares is n^. 8. Divide the number 26 into two parts whose product added to the sum of their squares shall be 556. 9. Divide 20 into two such parts that the square of their (difference shall be equal to their product. PROBLEMS. 223 10. Divide 12 into two such parts that the sum of their squares shall be 4 times their product. 11. Can there be two unequal numbers the square of whose difference is equal to the difference of their squares? 12. A drover bought a certain number of sheep for $663. Had he bought them for II apiece less he would have got 48 more sheep. How many did he buy? Note. If he bought x sheep, each sheep must have cost him — dollars, 13. When tea costs 50 cents a pound more than coffee you '-Q) \ (§341) = {x + ip + iv{p''-ig)\{^ + ip~ii^(p'-ii)\- (§238) 230 qUADMATIC EQUATIONS. If we now seek the roots of the equation formed by equat- ing these expressions to zero, we have the general principle : In order that a product may be equal to zero at least one of the factors must be zero. Hence we must have either whence x = — ^p — ^ ^{^p^ — 4g), or ^-i-ip-i V{P' - ^q) = 0, ^ whence x = — ip -\- ^ ^(p^ — 4^). Hence we may find the roots of any quadratic equation in the general form by factoring the expression lohicJi the equation states to be zero. EXERCISES. Find the roots of the following equations by factoring: I. x" -^x = (i. 3. x' - (a + b)x = 0. 5. (x-iy = a(x'-l). II. x"* -\- 6x -\- S = 0. 9. x"" - X - I = 0. 11. x'-ix-l = 0. 13. x"" - ax -i-b = 0. 15. x''-{-ax + b = 0. 17. x^ + mxy -ir f- = 0. 264. The principle of § 261 enables us to determine the signs of the two roots of a quadratic equation by inspection. 1. Because in the equation ^^+pxJrq = the term q is the product of the roots, q is positive when tlie roofsi Juire like signs, and negative when they have unlihe signs. 2. Because^ is the nrgalire of the sum of the roots, At least one of the roofs )nnst be negative when p is posi- tive, and positive when p is negative. Hence in such an equation as x^ + mx + '?i = 2. x"" — ax = 0. 4. x"" -2x-l= 0. 6. x"" ^'^x- ^ = 0. 8. x' -\-x-l = 0. 10. x' -{-ix~2 = 0. 12. x' -ax-b = 0. 14. .t' -{- ax — b = 0. 16. x' + xy-f^ 0. 18. x'^ — mxy -j- y'^ = 0. RELATIONS OF BOOTS AND COEFFICIENTS. 231 the roots, being like in signs and one at least negative, must both be negative. In such an equation as x^ — mx -|- ^ = both roots must, for a similar reason, be positive. In such an equation as x"^ ± 7nx — 71 = one root must be positive and the other negative. 265. Other Relations between the Roots and the Coeffi- cients of a Quadratic Equation. The preceding theory will enable us to express many functions of the roots in terms of the coefficients without solving the equation. Example 1. Find the sum of the squares of the roots of the equation x'-\-bx-^'2 = 0. {a) Solution. If we put a = the one root, (3 = the other root, we have, by § 261, « + /?=- 5, (h) ^/^ = +^. {c) Squaring the first of these equations, doubling the second and subtracting, we have a' + 2a/3 -f- y5" .= 25 "Zafi = 4 a' -f p' = 21. Ans. Note, The student may verify this result by solving the equation. Ex. 2. Find the sum of the cubes of the roots of the above equation. Solution. Cubing the equation (b), we have a' + 3a' /3 + Sa/3' -^ /3' = -. 125. The two middle terms of the first member may be put into the form da'j3 + 3aj3' = 3a/3(a + /3). Substituting the values of the factors from (b) and (c), we have 3a/3(a -\. jS) = - 30. «'-30 + /J^= - 125, a'-^r ^"= - 95. Ans. 232 QUADRATIC EQUATIONS. EXERCISES. Find the values of the following functions of the roots of equation {a): 1. a''-\-aft-{- fi\ 2. a" - a^-{- /3\ 3. «' + a'ft'' + p. 4. a' - daft + /?'. Prove that if x^ and ic, are the roots of the equation x^ -\- px -\- q = 0^ we shall have: 5. x^^ -\- x^' = p"" — 2q. 6. x^^ + x^x^-{- x^^ = 2^' — q- 7. x^^ — x^x^ + x^ = p"" — dq. 8. x; + x^' = 3pq -p\ Find the values of: 9. x^' + x;x^-\-xX'-i-x^ 10. a;/ — iCjX — ^X' + ^,'- Section Y. Solutiois- of Irrational Equations. 366. An irrational equation is one in which the unknown qnantity appears under the radical sign. An irrational equation may be cleared of fractions in the same way as if it were rational. Example. Clear from fractions the equation a h ah . . ■ — — = — - . (a) Vx-{- c Vx — c Vx" — c^ The L.C.M. of the denominators is V{x -\- c){x — c) = ^x' — c'. Multiplying all the terms by this factor, the equa- tion becomes a Vx — c — b Vx-\-c = db, (5) and the equation is cleared of fractions. 267. To reduce an irrational equation to a rational one. EuLE. 1. Clear the equation of fractions, 2. Transpose terms so that a surd expression shall he a factor of one member of the equation. SOLUTION OF IBRATWNAL EQUATIONS. 23:3 3. Square both memiers so as to rationalize the surd m em- ber of the equation. 4. Repeat (2) and (3) until no surds containing the un- known quantity are left. Example. To rationalize the equation (Z>). Transposing, a ^x — c ~h \/x -\- c -\- ah. Squaring, d\x — c) = ¥{x -\- c) + a'h'' + 2a J' I^F+T. Transposing, a^{x — c) — Jf{x -\- c) -\- a^W = 2ab^ \/x -\-~c, or ya"- b')x - a'c - Wc + a^h^ = 2ab' V^^. Squaring, (a"" - ^)V + 2(a'b'' - a'c - b'c)(a' - b')x + {a'b' - a'e - b'cy = 4:a'b\x + c), a rational equation. EXERCISES. Reduce and solve the following equations: 1.-4=+ ' ' Vx-{-4: Vx — 4: S/x^ — 16 Principal Steps. 9{x-4)= 40 + « - 12 Vx^fl, 9aj + 36=361 -'7Qx-^4xK 65 X = o or -7-. 4 2. {x-{- 4)i + (a; - Sy = 7. 3. (a; + 2)* + 2(x + 7)* = 8. 4. - _ ^ J- 4/^» J, ^. Vx''-\-c 6. Vx-{-a — Vx — a=i Vx. 6. a -}- Vx"" — ax __ ^ a— Vx"" — ax 7. \-v;^+' + i-o. 1+1/^^=1 o. — ==z = n. Vx^ + a' - « 234 qUAJJHATlC EQUATIONS. 9. Vx' + a' + Vx' -a' = 2a. 10. Va' -{-x'+ Va' - x' = a. 11 ^ 1 ^ _ 1 .^^-^a'-a ' V:^-^-«^ + «^ a 12. 1/^^=^ - Vx-'d = Vx. 13. 4/(^ + ^) - l^(^ + I) "" ^^• 14. 4/(4:r + 21) + ^{x + 3) = |/(:?: + 8). 15. 2 |^= 1 + |/4M-V{?2;+T)^ 16. V(a + a;) + |/(a — a;) = 2 Va. 17 i/(^ + ^) ■ ^ yi"^ - ^) Va 4- ^{a + 2:) Va - \/{a - x) 18. \'x\ y{a- x) - i/(a + .^•) } = Va { ^{a' — x^)—a\, 19. Va' - X 4- V^' + X = a-{-i. 20. |/«~^^^ + Vb — X = Va-{-b. 21. Va-x-{- Vx — b= Va-[-h — 2x. 22. Va — x^ VI = |/^ i^a — a; — Vx — b X — b 23. Vl^-x\^ Vl-x' _a Vi + ^-'^ + Vi the first member 24. |/4« + Z> - 4a: - 2 l/^'+T^^^ = l/^. Express the first member as a ratio and reduce by composition and division. 25. a Vm-^x— b Vm — x= Vm(a'' + J"). 26. V{a + x){x-\-b) + V(a - x)(x -b) = 2 Va^, 8IMULTAJVB0UJS QUADRATIC EQUATIONS. 235 Skctio]^ YI. Simultaneous Quadkatic and Other Equations. 268. Def. Two equations, one or both of the sec- ond degree, between two unknown quantities, are called a pair of simultaneous quadratic equations. The most geiiei'Ml form of an equation of the second de- gree hetvveen the unknown quantities x and y is ax' -f bxy + cy' J^ dx -{- ey +/ = 0. If we eliminate one unknown quantity between two such (.'({nations, the resulting equation, when reduced, will be of the fourth degree with respect to the other iinknown quan- tity, and therefore cannot in general be solved as a quadratic. Hut there are several cases in which a solution of two equations, one of which is of the second or some higher degree, maybe effected, owing to some special relations among tlie coefficients of the unknown quantities in one or both equations. The following are the most common: 269. Case I. Whe7i one of the equations is of the first degree only. Tin's case may be solved thus: Rule. Find the value of one of the unhnoimi quantities in terms of the other from the equation of the first degree. This ralup heing substituted in the other equation, we shall have a Huadratic equation from lohich the other unhnoiun quantity may he found. Example. Solve x-^ - 2/ = a, ) . '^x -^ =h.\ ^""^ From the second equation we find . _ % + h^ (*) whence x^ = — -~^ — 4 Substituting this value in the first equation (a) and re- if + Qhy -f 9b^ = ^a + 2h'). ducing, we find 236 QUADRATIC EQUATIONS. Solving this quadratic equation, y = -U ±^ Va + %h\ This value of y being substituted in the equation {h) gives a; = - 4Z» ± 3 Va + W. The same problem may be solved in the reverse order by eliminat- ing y instead of x. The second equation {a) gives %x -h If we substitute this value of y in the first equation, we shall have a quadratic equation in x, from which the value of the latter quantity can be found. The result should be pioved by substituting the values of x and y in the first equation. EXERCISES. Solve 1. x-\-y = a', xy = b\ 2. X — y = cr, X 3. .x + ^/=:3; x' -f y' = 29. 4. X - y = 3; x' - / = 51. • 5. x-\-y = c; a^-f=f. 6. X — y = 2c; x' -\-if = 2b\ 7. x-^ = l; x' - 3?/^ = 121. 8. '"'? x' -~ = 21. Consider - as an unknown quantity. 9. .-1 = 10; .^ + i, = 58. 10. ^-^ = n; f + 1 = 145. 11. 2/' + i, = 39; ^ X 12. 1 ^ xy = c; ^ = m. simulta:neou8 quadratic equations. 237 13. ^±^' = 1; x'^-y' = o\ 14. ?-±^ = ^; ax' + (a - ^>)a;v - ^>2/' = c\ ax - hy ^ ^._^ .^^..^ ex — dy ' ' -^ 16. -^—7 —T-^- = -?.-; X — y = m. (a-x)y 6' 17. x-\- y — a; t = o • ' ^ ' h — y X 2 18. X -\- y = c'j x' -{- y' = mxy. 270. Case II. If each equation contains only one term of the second degree, and these terms are similar, they may be eliminated by the method of addition and subtraction, lead- ing to an equation of the first degree. We may then proceed as in Case I. Example. Solve 2xy — X — y — 4, '-hxy-\-%xArZy= — 6. Multiplying the first equation by 5 and the second by 3, and adding, we have V)xy — hx — by = 20 — lOxy + 4a; + 6?/ = — 12 ~ a;+ 2/= 8 Hence y = % -\-x, (a) Substituting this value of y in the first of the given equa- tions and reducing, we find 2x' + 14:r - 8 = 4, or x' + '7x = 6. Solving this quadratic, we find x= - I ± i V73; whence, from (a), y = I ± i ^'«'3. 238 QUADRATIC EqUATI0N8. EXERCISES. Solve: 1. x" ^y = 1', x" -^-'dy -x = 2. 2. a; + 2/ + ^' = 22; x — by -{- x' = 10. 3. X — y -{- xy — a; cc -^ y — xy = 3a. 4. x' + 3xy + Sy = 43; x' -f- 3:?-^ - 3x = 25. 6, X = a i^x -\- y; y ■=■!) Vx -\- y. 2i*ll\. Case III. When the functions of the unknown quantities in the two equations contain one or more common factors, we may sometimes form an equation of lower degree by taking the quotient of the two equations and dividing out the common factors. Example. x"^ -{-xy = «; y"^ + xy = h. Factoring, we have x(x-\- y) = a, y(x •\-y) = 'b; whence, by taking the quotient, X a h y V ^ a The last equation is of the first degree, and by substitu- tion we readily find __ a _ h EXERCISES. Solve: 1. x^ + xy"^ = a\ y^ + ^"^V ~ ^' 2. x^y + xy^ = a; x''y — xy"* = h, 3. X Vx -\- y = a^; y Vx -\- y ==■ 5'. 4. x-\-y =. m(x^ + «/'); x — y = n{x* + y*), 6. x-\-y = m(x^ — y""); x — y =. n{x* — y'). G. x^ + xy"" = a; y* -{- x^y = 5. 7. a; + ^ = 7; x^ -\- y^ = 133. 8. xy + ;/ = «; rr'^ — .^/'^ = h. SIMULTANEOUS QUADRATIC EQUATIONS 239 272, Case IV. The solution may often be simplified by combining the two equations so as to get others simpler in form or admitting of being factored. The following is an elegant example: x^ -j- y^ = a, xy = 1), Take twice the second equation^ and add and subtract it from the first. Then x^ 4- '^^y + 2/" = ^ + ^^^ or (^ "h yY = « + ^h; X* ~ 2xy -\- y^ = a — %h, or {x — y)- = a — 2h. Ext. root, X-]- y = Va + 'Zb X — y = Va — 21j 2x = Va -\- 2b -^ Va - 2b 2y = Va-{-2b~ Va — 2b Dividing by 2, we have the values of x and y. As another example, take the equations x^ -\- y = nx, if J^x=z ny. If we take their difference we shall find it divisible by cc — y, giving an equation which leads to x-\-y = ri-\-l, or y = 71 -{- 1 — X. Substituting this value of y in the first equation, we have a quadratic in x. EXERCISES. Solve: 1. x^ -{- 4;?/* = 4c*; xy = ?i, 2. x* — xy = a^; y"^ — xy = b^, 3. x' -i- y' -\- X + y = 62; x^ + y"" - x - y = 44. 4. x' -\-y' -x-Jf-y = 49; x' + y^ -{- x - y = 33. 5. x"" - Sxy = 12; xy - Ay' = 8. 6. x'^ — y"^ -\- X — y = 2m; x^ — y* — x -\- y = 2n. The following equations may be solved by various com- binations of the preceding methods: ^ X . y 10 2,2 Ar. 7. - -f- - = ^; ^ -\-y = 40. V X 3 ' ' -^ 240 QUADRATIC EQUATIONS. 8. x-" + ^' = 74; x^y-\- xy = 47. To solve this form a quadratic with « + y as the unknown quantity. 9. x"" + / = 85; x-y -{-xy=%h, 10. x'-\-f-\-x-\-y = 48; 2(x + ^) = Zxy. 11. x^ — xy -\- y"^ = 2a'; x^ -}- xy -]- y^ = 2c^. 12. X -\- y^ = ax; x' -\- y = ly. 13. x" -{-y Vxy = 36; y"" ^ x Vxy = 72. J^" Teachers requiring problems leading to equations with two or more un- known quantities will find them in the Appendix. Section YII. Imaginary Roots. 373. Since the squares of both negative and posi- tive quantities are positive, the square root of a nega- tive quantity cannot be any quantity which we have hitherto considered. Def. The indicated square root of a negative quantity is called an imaginary quantity. The term imaginary is applied because in this case we have to imagine or suppose a quantity of which the square shall be negative. Def. The positive and negative quantities of al- gebra are called real. 2*74. Theokem. An imaginary quafitity can always be expressed as the product of a real quantity into the square root of-l. Proof. Let — « be the negative quantity whose root is to be expressed. Because — a = a X — 1, we have, by § 227, V - a = Vax V — 1. Because a is positive, Va is real, so that the theorem is proved. Notation. It is common to use the symbol i for V — 1, so that i = V^^l. This is because it is easier to write i than V — \. IMAGIJS'AHY ROOTS. 241 EXERCISES. Express the square roots: 1. V— 4. Ans. 2 l/— 1 or %L 2. v^To. 3. i/rg, 4. 1/-F. 5. V-F. 6. |/- 4A;. Alls. 2 l/^ V^ or 2kH. 7. l/- 9/i. 8. V- 367? 9. V"- 20. Alls. 2 . 5i V- 1 or 2 . 5H*. 10. V"- 12F. Extract the square roots of: 11. — a"" -[- 2ab — F. Ans. (a — b)i. 12. - a' + 4«Z> - 4<^'. 13. - w' - 2?/?/^ - n\ 14. _4 + 2^-l. 15. --!^-2^-l. 275. Since the roots of the general quadratic equation x' '-f- ]}X -^ q — ^ contain the expression \^^f — 4^/, it follows that if tlie quan- tity under the radical sign is negative there is no real root. In such cases the algebraic solution may be expressed by imaginary quantities. EXERCISES. Solve the following quadratic equations: 1. x^ _ 2a: = - 5. Ans. x=\±2 V^^ =p 1 ± 2L 2. .r' -4:x= -6. 3. x"^ — 2ax = — 2«^ 4. x' + 4:ax + Sa' = 0. 21 Q, When the solution of a problem contains an imaginary quantity it shows that the conditions of the prob- lem are impossible. Example. To find two numbers of which the sum shall be 6 and the product 10. Solving in the usual way, we may put X = one number; 6 — x^ the other. 242 QUADRATIC EQUATIONS. Then x{Q - :r) = 10, x' - Ga- = - 10, x" — Qx + 9 = - 1, X — 'd ± V^^ ='d ± i. The answer being imaginary, there are no numbers which fulfil the conditions. Indeed, by assigning to x different values fi'om to G we see that 9 is the greatest possible product of two numbers whose sum is G. If one number is less than or more than G the product will be negative. EXERCISES. 1. Find two numbers of which tlie sum shall be 8 and the 1 and the second when r <1. By this formula we are enabled to compute the sum of the terms of a geometrical progression without actually forming all the terms and adding them. EXERCISES. 1. A farrier having told a coachman that he would charge him $3 for shoeing his horse, the latter objected to the price. The farrier then offered to take 1 cent for the first nail, 2 for the second, 4 for the third, and so on, doubling the amount for each nail, which offer the coachman accepted. There were 32 nails. Find how much the coachman had to pay for the last nail, and how much in all. 2. Having the progression a -\- ar -{- ar^ -{- ar^ + ^^% what is the common ratio of the new progression formed by taking {a) Every alternate term of this progression? {])) Every mth term? 3. Insert two geometrical means between the extremes a and «^^ Note. See Problem TIL and note that when two means are in- serted we shall have a progression of four terms, and, in general, that wlien n means are inserted between two extremes the whole forms a G. P. of 71 -|- 2 terms. 4. Insert three geometrical means between the extremes m. and ?/^^ 5. Insert two geometrical means between the extremes ah and a'F. 6. What is the geometrical mean of a and «i? Of a and ah'^ Oia-^h and a — M 7. Insert two geometrical means between ab and a^h^. 8. The arithmetical mean of two numbers is 6^ and their geometrical mean is 6. Find the numbers. PROBLEMS IN OEOMETRIGAL PROQRESStON. ^57 9. Find two numbers whose sum is 30 and the sum of whose arithmetical and geometrical means is 24. 10. In the following series of numbers and expressions state which are geometrical progressions and which are not such: (a) 1, 3, 6, 10, etc. \b) 2, 4, 8, 16, etc. (c) 4, 2, i, h etc. (d) 2, - 3, +4, - 6, etc. (e) a, a% a^, a", etc. (/') a, 2a% 3a% 5a% etc. (g) 3m\ - 6m% 12m% - Um'\ etc. 11. If we take the geometrical progression a, ar, ar"^, ar^, ar*, etc., (a) and form a new series of terms by adding each term to the one next following, thus: a(l + r), a(r + 0, a(r^ + O. «(^ + O. etc., is this new series or is it not a geometrical progression? and if it is such, what is its common ratio? 12. If, in the preceding example, we form a new series by subtracting each term of the progression from the term next following, will the new series be a geometrical progression or will it not. 13. If we multiply all the terms of a G.P. by the same constant factor, will the products form a G.P. ? 14. If we add the same constant quantity to all the terms, will the sums form a G. P. 15. Do the reciprocals of a G.P. form another G.P. ? If so, what is the common ratio? 16. What is the continued product of the first three terms of the progression {h), Ex. 10? Of the first four terms? Of the first n terms? 17. One number exceeds another by 15, and the arith- metical mean of the two is greater than the geometrical mean by |. What are the numbers? 18. Find a progression of three terms of which the first term is 1 and the continued product of the three terms 343. 19. Find the common ratio of that progression of which 258 phogressions, the sum of the third and fourth terms is one fourth the sum of the first and second. 20. What is the common ratio when the sum of the first and fourth terms is to the sum of the second and third as 13 to 5? Note that 1 + ^'^ is divisible by 1 + ^• i^^ Teachers requiring additional problems in geometrical progression will find tiiem in the Appendix. Limit of tlie Sum of a Progression. 393. Theorem. If the absolute value of the common ratio of a geometrical pi^ogi^ession is less than U7iity, then there always will he a certain quantity which the sum of all the terms can never exceed, no ^natter how many terms we take. Example 1. The sum of the progression i + i + i -h etc., in which the common ratio is i, can never amount to 1, no matter how many terms we take. To show this, suppose that one person owed another a dollar and proceeded to pay him a series of fractions of a dollar in geometrical progression, namely, i. h h -^-^y etc. When he paid him the i he would still owe another i, when he paid the i he would still owe another i, and so on. That is, at every payment he would discharge one half the remaining debt. Now there are two propositions to be under- stood in reference to this subject: I. The entire debt can never he discharged by such pay- For, since the debt is halved at every payment, if there was any payment which discharged the whole remaining debt, the half of a thing would be equal to the whole of it, which is impossible. II. The debt can be reduced belong any assignable limit by continuing to pay half of it. For, however small the debt may be made, another pay- ment will make it smaller by one half; hence there is no smallest amount below which it cannot be reduced. LIMIT OF THE 8UM OF A PROGBESSION. 259 These two propositions, which seem to oppose each other, hold the truth between them, as it were. They constantly enter into the higher mathematics, and should be w^ell understood. We therefore present another illustration of the same subject. A B I \ . \ LJ-J i i i tV Ex. 2. Suppose AB to be a line of given length. Let us go one half the distance from ^ to ^ at one step, one fourth at the second, one eighth at the third, etc. It is evident that at each step we go half the distance which remains. Hence the two principles just cited apply to this case. That is: (1) We can never reach ^ by a series of such steps, be- cause we shall always have a distance equal to the last step left. (2) But we can come as near B as we please, because every step carries us over half the remaining distance. This result is often expressed by saying that we should reach B by taking an infinite number of steps. This is a convenient form of ex- pression, and we may sometimes use it; but it is not logically exact, because no conceivable number can be really infinite. The assumption that infinity is an algebraic quantity often leads to ambiguities and ditfi- culties in the application of mathematics. 294. Def. The limit of the sum ^ of a geomet- rical progression is a quantity which ^ may approach so that its difference shall be less than any quantity we choose to assign, but which ^ can never reach. Examples. 1. The limit of the sum i + i + i H- tV + etc. is 1, because this sum fulfils the two conditions {a) That it can never be as great as 1 ; {b) That it can be brought as near to 1 as we please i y increasing the number of terms added. 2. The point B in the preceding figure is the limit of all t\\Q steps that can be taken in the manner described. The following i)rinciple will enable us to find the limit of tlie sum of a progression: 395. Principle. If r < 1, the power r"" can be made as small as we ])1ense l)y increasing the value of ;?, but can never be made equal to 0. 260 PROGRESSIONS. Suppose, for instance, that Then every time we multiply by r we diminish r"" by J of its former value; that is, r' = lr = (1 - i)r = r - Jr, r^ ~ f r^ = r^ — i/-^, r* = fr' = r' — ;^r', etc. etc. etc. 296. Problem. To find the limit of the sum of a geo- metrical progression. To solve this problem we must take the expression for the sum of 71 terms, and see to what limit it approaches when we suppose n to increase indefinitely. The required expression, as found in § 292, is 1 — r which we may put into the form 1 — r 1 — r ^ ' This expression for "2, the sum of n terms, is identically the same as (4), but different in form. If r is greater than unity, the quantity r"" will increase indefinitely when n increases indefinitely, and the expression will have no limit. If r is less than unity, then, by § 295, when n increases indefinitely r'^ will approach as its limit, and therefore the expression r'^ will also approach 0, so that we shall have Limit of ^ == --^. (7) Example 1. If ^ = 1 and r = ^, then = 2; and When n = 2, l + i = 2-i; When n = ^, 1 + | + |- = 2 - }; When n = 4., 1 + ^ + ^ + 1 = 2- i; When 7^ = 5, 1 + ^ + J + i -f ^ = 2 - -^; etc. etc. etc. LIMIT OF THE SUM OF A PROGRESSION. 261 We see that as we make 7i equal to 5, 6, 7, etc., indefi- nitely, the sum of the series is less than 2 by ^ig-, -^j -^^ etc., indefinitely. Since, by continually halving a quantity, the parts can be made as small as we please, the limit of the sum of the series is 2. Ex. 2. a = 1; r = i. We now have 1 - r and For 7^ = 2, 1 + i :== I - ^; For 7i = 3, 1 + 1 + 1 ^ I _ ^3_.. For 7. = 4, 1 + i + i + ^V = 3 _ ^^. etc. etc. etc. We see how as 7i increases the sum approaches | as its limit. Ex. 3. a = l, r = - I. We then have l-r ^' and For 7i = 2, 1-4 = 1- 3^; For 7^ = 3, 1 - i + J = f + ^\', For ^ = 4, 1 _ 4 + -1 - ^ = I - ^2_. For 7^ = 5, 1 - i + i - i + tV - S + A; etc. etc. etc. We see that as n increases the sums are alternately greater and less than the limit f , but that they continue to apjiroach it as before. EXERCISES. Find the limits of the sums of the following progressions* 1. T + 7^ + Fi'H" ^^^-y ^^ infinitum. 2. -; rTT + 777 — etc., ad infinitum. 4 16 ' 64 -^ 262 PB00BE88I0NS. ^' 's^i~i ~ (s - ly + (s- ly ~ ^*^'- ^^ i^^P^^i^^^^' 3 3^ 3^ 5. T + Ti 4~ "7 3 + 6tc., ad mfinitum. 4 4' 4' 6. — ^2 + "^i ~ 6tc. , ad infinitum, 7. 3 — , h 71 — ^ \i. + 71 — i Ta + etc., «a i7inmtum. 1 + m ' (1 + m)' ' (1 + my ' ' -' 8. - — I 7- — ■ rr + 7- — ; r^ — etc. , ttd ififitiitum. 1 + m (1 + 7ny ' (1 + my -^ 9. From a cistern of water half is drawn into tub A, half of what is left into tub B, half of what is then left into tub 4, and so on alternately. If this were continued without limit, what proportion of the water would go into the respec- tive tubs? 10. A man starts from a point A towards a point B one mile distant. But he stops at a point b two thirds of the way from ^ to ^; then walks to a point c two thirds of the way from b to A; then to a point d two thirds of the way from c to i, and so on alternately, going in each direction two thirds of the way back to the point from which he last set out. To what point would he continually approach, and what is the limit of the distance he could ever walk? Begin by drawing a line to represent the distance from A to B, and mark the points b, c, d, etc., upon it. CHAPTER II. VARIATION. 297. Bef. When the value of one quantity de- pends upon that of another, the one quantity is called a function of the other. Example 1. The time required for a train to perform a journey depends upon the speed of the train. Hence in this case the time is a function of the spewd. Ex. 2. The value of a chest of tea depends upon its weight. Hence the value is a function of the weight. Ex. 3. The weight which a man can carry is a function of his strength. Let the student, as an exercise, name other cases in which one quantity is a function of another. 298. When one quantity -2^ is a function of a second quantity x, then for every value of x there will be a corresponding value of u. Example 1. If tea is worth 50 cents a pound, then to the quantity 1 pound will correspond the value 50 cents; 2 pounds " '' " " $1.00; 3 " " " " " $1.50; etc. etc. etc. Ex. 2. If a train has to make a journey of 60 miles at a uniform speed, then to the speed 60 miles an hour will correspond the time 60 minutes; jr\ i( a a a a a a oa a Ar\ a a a a a ic a qq (( 30 '* ^* ** ** '^ ** ** 120 ** etc. etc. etc. 264 VARIATION. 299 Direct Variation. Def. When two quan- tities are so related that any two values of the one hate the same ratio as the corresponding values of the other, the one is said to vary directly as the other. When one quantity varies directly as another, then by doubling the one we double the other, by trebling the one we treble the other, by halving the one we halve the other, etc. Example. The weight of an article varies directly as the quantity. When we halve, double or treble the quantity, the weight will also be halved, doubled or trebled. Note. Iu speaking of direct variation we may omit the word direct. 300. Expression of the fact that one quantity y varies directly as another quantity x. Let us put a = any one value of y; c = the corresponding value of x; then, by definition, we must have, for all values of x- and y, X : a = y : c. This proportion gives ay = ex, whence y = -x. ^ a . Here we may regard — as a factor by which we multiply any a value of X to obtain the corresponding value of y. Hence The fact that one quantity varies as another is expressed by equating the one to the product of the other into a constant factor. Example. Given that y varies as x, and that when X = 3, then y = 6; it is required to express y in terms of x. Solution. By definition we have, for all values of x and y; y '. 6 = X : 3; .'. 3y = 6x; .-. y = ix, which is the required expression. INVERSE VARIATION. 265 EXERCISES, 1. Given that p varies as r, and that when r = 3, then jf? =: 2; it is required to express j[; in terms of r, and to ii»d the vahies of p corresponding to r == 1, 2, 5, 9, 12, 16 and 33 respectively. 2. If the value of a quantity a of iron is represented by b, what will represent the value of a quantity x of iron? 3. Given that y varies as x, and that when X — a, til en y ~ ^\ it is required to express the vahies of y when (a) X =: 3a; (^) X = 3a -\- ni; (c) X = 3a -{- 27n; [d) X = 3a -\- 3 1)1. 4. Prove that if y varies as x, and that if we suppose x to take a series of values in arithmetical progression, such as X := a, X = a -{- d, X = a -}- 2d, X = a + 3d, etc. etc., then the corresponding values of y will also form an arithmet- ical progression. 5. Prove that if y varies as x, the difference between any two values of y will have the same ratio to the difference between the corresponding values of x which any value of y has to the corresponding value of x. 301. Inverse Variation. One quantity is said to vary inversely as another when the ratio of any two values of the one is the inverse ratio of the corre- sponding values of the other. If ?/, corresponds to x^ and y, '' '' .r„ then,' when y varies inversely as x, y, '-!/. = •'^. : ^- Therefore y^x^ = y^x^. 266 OF VAUIATION. Hence, becftuse x^ and y^ as well as x^ and y^ may be any pair of values of the quantities, In inverse variation tlie product of two corresponding values of the quantities is always the same. Hence, also, in inverse variation One of the quantities can always he found hy dividing some dividend hy the other quantity. This dividend is the product of any two corresponding values of the quantities. Example. If ?/ = 4 when x — 2, and y varies inversely as X, then To a; = 2 corresponds ?/ = 4; " iz; = 4 '' y = 2; *^ ic = 8 '' y = U '' 2; = 16 " y = i; " ic = 32 '' y = h etc. etc. It will be seen that each product of a value of x into the corresponding value of y is 8. EXERCISES. 1. If y varies inversely as x, and when ^ = 3, ?/ = 8, it is required to make a little table showing the values of y for X = 1, 2, 3, etc., to 12. 2. If y varies inversely as x, and when x = a, then y = h, it is required to express the values of y for a; = 1, :?; = 2 and X = ^. 3. Given that u varies inversely as z, and that when 5; = 3 the corresponding value of u is greater by 5 than it is when 2; = 4; it is required to express the relation between u and z by an equation. Note, We may solve this problem with most elegance by taking for the unknown quantity the product of any value of u by the corre- sponding value of z. Let p represent this product, and let «2 represent the value of u when s = 4. We then have the two equations Eliminating u^, we find p = 60. Hence the required equation is _ 7)^ _60 ~ z ~ z ' uz = 60. INVERSE VARIATION. 267 4. Given that u varies inversely as z, and that when z = I- tlie vahie of u is greater by 1 than it is when z = 1; it is required to express the relation between u and z. 5. Given that u varies inversely as z, and that the sum of the values of u tor z = 1 and 2; = 2 is 5; it is required to ex- press the relation between tc and z by an algebraic equation. 6. Show that the quantity of goods which can be pur- chased with a given sum of money varies inversely as the price. 7. Show that the time required to perform a journey varies inversely as the speed. 302. Def. One quantity is said to vary as the square of another when any two values of the one have the same ratio as the squares of the correspond- ing values of the other. If ic and z and w^ and z^ are pairs of corresponding values, then when tc varies as the square of z, we must always have u : u^ = z^ : z^. In the same way as in direct variation (§ 300) may be shown: When one quantity varies as the square of another it is equal to that square multiplied by some constant factor. 303. Def. One quantity is said to vary inversely as the square of another when the ratio of any two values of the one is the inverse ratio of the squares of the two corresponding values of the other. If u varies inversely as the square of z, and if n^ is the value of u when z = z^, then we always have U \ U^ — Z^ \ Z^y which gives z' We therefore conclude: Wlien one quantity varies inversely as the square of another it is equal to some constant quantity divided hy the square of that nf,7ip,t\ 268 vauiatiok EXAMPLES AND EXERCISES. 1. It is found that the attraction between two bodies varies inversely as the square of their distance apart* If when the distance is 1 foot the attraction is represented by 2, it is required to express the attraction at the distances 2, 3, 4, 5 and 10 feet. 2. Supposing the moon to be 60 radii of the earth (that is, 60 times as far from the earth's centre as any point on the earth's surface), what would a ton of coal weigh at the distance of the moon? Note. Because the weight of the coal is equal to the attraction of the earth, it varies inversely as the square of its distance from the earth's centre. 3. The area of a circle varies as the square of its diameter. If the diameter of one circle is 2|^ times that of another, what is the ratio of their areas? 4. The apparent brilliancy of a candle, that is, the amount of light which it will cast upon a small surface held perpen- dicular to the line from the candle to the surface, varies in- versely as the square of the distance. If we represent by 1 the brilliancy of the candle at a distance of 8 feet, what num- bers will represent its brilliancy at the distances of 1 inch, 1 foot, 4 feet and 16 feet respectively? 5. The volume of a sphere varies as the cube of its diam- eter. If one sphere has 3 times the diameter of another, what is the ratio of their volumes? 6. If the diameters of two spheres are to each other as 5:2, what is the ratio of their volumes? 304, Combined Variation. The value of one quantity may depend upon the values of several other quantities. The first quantity is then called the function, and the others the independent variables. We may then suppose all the independent variables but one to be constant, and the remaining one to vary, and ex- press the law of variation of the function. Having thus supposed each of the independent variables in succession to vary, the law of variation of the function is COMBINED VARIATION 269 expressed by stating liow it varies with respect to each inde- pendent variable. Example 1. The tme required to perform a journey is a function of the speed and of the dista7ice. If we suppose the distance to remain constant and the speed to vary, the time will vary inversely as the speed. If we suppose the speed to remain constant and the distance to vary, the ti7ne will vary directly as the distance. These two relations are expressed by saying that the time varies directly as the distance and inversely as the speed. Ex. 2. Suppose a candle at ^ j C to illuminate a surface at /. If we suppose the candle to become 2, 3, 4 or ?^ times as bright, the distance CI remaining constant, the illumination of / will become 2, 3, 4 or ^ times as great. If we suppose that while the brightness of the candle remains constant the distance CI becomes 2, 3 or n times as ffreat, the illumination will be reduced to — , — or — ^ of its first amounto These two facts are expressed by saying that the illumina- tion varies directly as the brightness of the candle and in- versely as the square of the distance. 305. Expression of Comlined Variation. If a function u varies directly as the quantities j3, q, r, etc., then, by §300, its expression must contain each of these quantities as a sim- ple factor. If it varies inversely as the quantities u, v, w, etc., then, by § 301, its expression must contain each of these quantities as a divisor. Therefore the fact that u varies directly as^, q, r . . . and inversely as u^ Vj w , . , is expressed by an equation of the form Avar . . . u = -^^ , uvw . . . in which A represents some constant quantity, the value of which must be chosen so as to fulfil the conditions of each special problem. 270 VARIATION. EXERCISES. 1. Express that the function ^^ varies directly as the square of m and inversely as the cube of x. Ans. u = — ^. X 2. Express that the function s varies directly as m and n and inversely as the square of r, 3. Express that the function q varies directly as m and as the square of x, and inversely as the cube of h and as the fourth power of z. 4. If we represent by unity the brilliancy with which a candle of brightness unity illuminates a page 4 feet away, what number will represent the illumination of a page 8 feet away by a candle of brightness 3 ? 5. If an electric light gives as much light as 2500 candles, at what distance will it illuminate the page of a book as brightly as a candle 5 feet away will ? 6. At C is a candle, and at E is E Y G X an electric light equal to 1600 can- dles, distant 500 feet from C. It is required to find the distances from the candle to the two points X and Y from which the candle and the electric light appear equally bril- liant. 7. If of two lights a feet apart the brighter gives r times as much light as the fainter, it is required to find the positions of the two points on the straight line joining them from which the two lights appear equally bright. 8. An electric light 80 feet distant was found to throw as much light on a book as a candle 2 feet distant. If another candle twice as bright is used, how far must the electric light be placed to give as much light as this brighter candle does at 2 feet distance? CHAPTER III. LOGARITH MS. 306. To every number corresponds a certain "otiier number called its common logarithm. Def. The common logarithm of a number is the exponent with which 10 must be affected in order to produce the number. The term common logarithm is used because there are other loga- rithms than the common ones. Since we are at present only concerned with the latter, we shall drop the adjective common. To express the logarithm of a number n we write log n, so that log n = the logarithm of n. Examples. 10° = 1; .-. log 1 = 0. 10* = 10; .'. log 10 = 1. 10' = 100; .-. log 100 = 2. 10- ■' = T^o; ••• logT*o = - ^. etc. etc. Thus we have Theoeem I. The logarithm of 1 is zero. The logarithm of 10 is 1. etc. etc. (a) 307. If we call any number x, and put y = logx, we have, by definition, 10^ = X. (b) If we suppose y to increase from to 1, a; will increase from 1 to 10. Hence 27^ LOOARTTHMS. Theorem IL llie logarithm of a number hetiven 1 and 10 is a positive fraction between and + 1. 308. If we multiply the equation {h) by 10, we have 10^+1 =: 10^;; that is, log 10a; = ?/ -|- 1. Hence Theorem III. Every time we multij^ly a number by 10 we increase its logarithm by unity. 309. By dividing (b) by 10 we obtain Theorem IV. Every time tve divide a number by 10 we diminish its logarithm by miity. 310. In order that 10*' may be less than unity, y must be negative. Since we have 10-' = ,1, 10 -'=.01, 10-'= .001, lO-'^^.OOOl, etc. etc., we see that as we increase the exponent negatively the num- ber diminishes without limit. Hence Theorem Y. The logarithm of a proper fraction is nega- tive. Theorem VI. The logarithm of is Jiegative infinity. 311. The use of logarithms is founded on the four fol- lowing theorems. Theorem VII. The logarithm of a product is equal to the sum of the logarithns of its factors. Proof. Let p and q be two factors, and suppose h = log p, h = log q. Then lO'^ = p, 10^ = q. Multiplying, lO'^lO'^ = 10^^ + ^^ = pq. Whence, by definition, h ^ h = log {pq), or log^ + log g = log ( pq). This proof may be extended to any number of factors. TBEOUEMS. ^73 Theorem VIII. The loyarithrn of a quotient is found by subtracting the logaritlim of the divisor from that of the divi- dend. Proof. Dividing instead of multiplying the equations in the last theorem, we have lU*^ q Hence, by definition, k — k = log -, or log^ — log q = log -. Theorem IX. The logarithm of any poiver of a number is equal to the logarithm of the number multiplied by the expo- nent of the power. Proof. Let h = logp, and let n be the exponent. Then 10'^ = p. Eaising both sides to the ^^th power, Whence nh = logjt?^ or nlog p = logp"^. Theorem X. The logarithm of a root of a number is equal to the logarithm of the number divided by the index of the root. Proof Let s be the number, and let p be its nth. root, so that p = Ys and s = pl^. Hence log s — logjy" = n \ogp. (Th. IX.) Therefore loff » = — ^, ^ ^ n iogV7 = 5f-^ , 374 LOGARITHMS. EXERCISES. Express the following logarithms in terms of logp, log q, log X, log y, log (2; - y) and log (.t + y) : 1. \ogpy. Ans. logjt; -{- log ^. 2. log qx 3. log^(/?/. 4. \og 2j(x -{- y), 5. log a;^ + xy. Ans. log ^ -|- log {x -\- y). 6. log lOjo. Ans. logjy + l. 7. log lOxy. 8. logjo^ic. Ans. 2 log ;j -f log a;. 9. logp^x. 10. log^o;. 11. log lOj^V. 12. log 10;^"^^^ 1 13. log 10^ 14. log & 10" 15. log-. 16. log^,. q q 17. log 1'. 18. log |.]. 19. log ^? 30. logi5i!I. 21. log-l^. 22. log~^. * lOji?^' ^ 100j^5' 23. logl/J. 24. \ogV{x^ij). 25. log^*?/i 26. logVlO. 27. log Via 28. log Vpq. xy 29. log i/lO(2: -\- y), 30. log , ^^. 31. log (a:'' - y'). 32. log (:?:' - xy). 33. log (2:* - x\f). 34. log jt?(:z:' - xhf'). 35. log (a:' - yy. 36. logjf?^^*(a;^ - :ry)* TABLE OF L0OABITHM8. 276 Table of Logarithms.* 313. The logarithm of a number consists of an integer and a decimal fraction. The decimal fraction is called the mantissa of the loga- rithm. The integer is called the characteristic of the logarithm. A table of logarithms gives only the mantissse. 313. To find the mantissa of the logarithn of a give^i number. Case I. When the number has three or fewer figures. Rule. Find in the left-hand column that line which con" tains the first two figures of the number, and select that column ivhich has the third figure at its top. TJie four figures i7i this = 24; .62 X 24 = 15 mant. log 17762 = .2495 * It is very desirable that the pupil should learn to use logarithms in multiplication and division at the earliest period possible in his studies. The precepts for using this table are therefore arranged so that ^hey may be used before understanding the entire theory of logarithms. 276 LOGARITHMS. 314, Characteristics, The characteristic of a logarithm is less by unity than the number of figures preceding the decimal point iu the number to which it corresponds. Examples. The mantissa last found gives the following logarithms: log 17762 = 4.2495: log 1776.2 ^ 3.2495: log 177.62 = 2.2495: log 1.7762 = 0.2495. Every time we move the decimal point one place toward the left we divide the number by 10 and diminish the loga- rithm by unity, which leaves the mantissa unchanged (§309). If the number is a decimal fraction, the cliaracteristic is — 1 when the figure following the decimal point is dif- ferent from zero; — 2 when one ^ero follows the decimal point; — 3 when two zeros follow it, and so on. In these cases the minus sign is written above the char- acteristic to show that it belongs to the characteristic alone and not to the mantissa. Examples. log 0.17762 = 1.2495; log 0.017762 =2.2495; log 0.0017762 = 3.2495: etc. etc. EXERCISES. Find the logarithms of the following numbers: 1. 701. 6. 7032. 11. 1.0246. 2. 700. 7. 70.32. 12. 2.0324. 3. 70. 8. 0.7032. 13. 51.623. 4. 7. 9. 0.007032. 14. 11.111. 5. 0.7. 10. 2956. 15. 3.243. 315. To find the number corresponding to a given loga- rithm. Rule. 1. Find in the table, if possible, the mantissa of the given logarithm. The corresponding figures in the left- hand column and at the top of the column are the fig^ires of the required number^ TABLE OF LOGARITHMS. 277 2. If the mantissa is not found in the tahle, find the next smaller mantissa. The three correspo?iding figures are the first three figures of the required number. 3. Take the excess of the given mantissa over the next smaller one in the table and divide it by the difference between the tivo consecutive mantissce in the table. 4. The result is a decimal fraction to be written after the three figures already found, the decimal 'point being dropped. 5. Having thus found the figures of the number, insert the decimal point correspondi7ig to the give7i characteristic accord- ing to the rule of § 314. Example 1. Find the number whose logarithm is 4.7267. We find in the tables that to the mantissa .7267 corre- spond the figures 533. The characteristic being 4, there must be five figures before tlie decimal point, so that we have Number = 53300. Ex. 2. Find the number whose logarithm is 1.2491. We find from the tables Next smaller mantissa, 2480; N = 178. _ Given mantissa, 2491 Excess of given mantissa, 11 Difference of mantissas in table = 2504 — 2480 = 24. Figures to be added to iV^ = ^ J = . 46 Therefore the figures of the number are 17846. The characteristic being 1, there are two figures before the deci- mal point, so that Number = 17.846. EXERCISES. Find the numbers corresponding to the following loga- rithms : 1. 1.2032. 6. 4.0343. 2. 0.7916. 7. 3.1922. 3. 0.0212. 8. 2.8282. 4. r.0212. 9. 1.0602. 5. 2.0212. 10. 0.9293. 278 TABLE OF FOUR- PL ACE LOGARITHMS. No. 1 1 2 3 4 5 6 7 8 9 10 oooo 0043 0086 0128 0170 0212 0253 0294 0334 0374 11 12 13 14 15 16 17 18 19 0414 0792 II39 146 1 1761 2041 2304 2553 2788 0453 0828 1173 1492 1790 2068 2330 2577 2810 0492 0864 1206 1523 1818 2095 2355 2601 2833 0531 0899 1239 1553 1847 2122 2380 2625 2856 0569 0934 1271 1584 1875 2148 2405 2648 2878 0607 0969 1303 1614 1903 2175 2430 2672 2900 0645 1004 1335 1644 1931 2201 2455 2695 2923 0682 1038 1367 1673 1959 2227 2480 2718 2945 0719 1072 1399 1703 1987 2253 2504 2742 2967 0755 1 106 1430 1732 2014 2279 2529 2765 2989 20 3010 ! 3032 3054 3075 3096 I 3118 1 3324 1 3522 3711 1 3892 I 4065 ! 4232 4393 4548 ! 4698 3139 3160 318I 3201 21 22 23 24 25 26 27 28 29 3222 3424 3617 3802 3979 4150 4314 4472 4624 3243 3444 3636 3820 3997 4166 4330 4487 4639 3263 3464 3655 3838 4014 4183 4346 4502 4654 3284 3483 3674 3856 4031 4200 4362 4518 4669 3304 3502 3692 3874 4048 4216 4378 4533 4683 3345 3541 3729 3909 4082 4249 4409 4564 4713 3365 3560 3747 3927 4099 4265 4425 4579 4728 3385 3579 3766 3945 4116 4281 4440 4594 4742 3404 3598 3784 3962 4133 4298 4456 4609 4757 30 4771 4786 4800 4814 4829 ' 4843 4857 4871 4886 4900 31 32 33 34 35 36 37 38 39 4914 5051 5185 5315 5441 5563 5682 5798 5911 4928 5065 5198 5328 5453 5575 5694 5809 5922 4942 5079 5211 5340 5465 5587 5705 5821 5933 4955 5092 5224 5353 5478 5599 5717 5832 5944 4969 5105 5237 5366 5490 5611 5729 5843 5955 i 4983 5119 5250 1 5378 5502 5623 5740 5855 5966 4997 5132 5263 5391 5514 5635 5752 5866 5977 501 1 5145 5276 5403 5527 5647 5763 5877 5988 5024 5159 5289 5416 5539 5658 5775 5888 5999 5038 5172 5302 5428 5551 5670 5786 5899 6010 40 6021 6031 6042 6053 6064 i 6075 6085 6096 6107 6117 41 42 43 44 45 46 47 48 49 6128 6232 6335 6435 6532 6628 6721 6812 6902 6138 6243 6345 6444 6542 6637 6730 6821 691 1 6149 6253 6355 6454 6551 6646 6739 6830 6920 6160 6263 6365 6464 6561 6656 6749 6839 6928 6170 6274 6375 6474 6571 6665 6758 6848 6937 1 6180 6284 6385 6484 6580 ! 6675 1 6767 1 6857 ! 6946 6191 6294 6395 6493 6590 6684 6776 6866 6955 6201 6304 6405 6503 6599 6693 6785 6875 6964 6212 6314 6415 6513 6609 6702 6794 6884 6972 6222 6325 6425 6522 6618 6712 6803 6893 6981 50 6990 6998 7007 7016 7024 i 7033 7042 7050 7059 7067 51 52 53 54 7076 7160 7243 7324 7084 7168 7251 7332 7093 7177 7259 7340 7101 7185 7267 7348 7110 7193 7275 7356 1 7118 7202 1 7284 ' 7364 7126 7210 7292 7372 7135 7218 7300 7380 7143 7226 7308 7388 7152 7235 7316 7396 TABLE OF FOUR-PLACE LOGARITHMS 279 No. 1 2 3 4 5 6 7 8 9 55 50 7404 7482 7412 7490 7419 7497 7427 7505 7435 7513 7443 7520 7451 7528 7459 7536 7466 7543 7474 7551 57 58 59 7559 7634 7709 7566 7642 7716 7574 7649 7723 7582 7657 7731 7589 7664 7738 7597 7672 7745 7604 7679 7752 7612 7686 7760 7619 7694 7767 7627 7701 7774 GO 7782 7789 7796 7803 7810 7818 7825 7832 7S39 7846 01 62 63 7853 7924 7993 7860 7931 8000 7868 7938 8007 7875 7945 8014 7882 7952 8021 7889 7959 8028 7896 7966 8035 7903 7973 8041 7910 79S0 8048 7917 7987 8055 64 65 66 8062 8129 8195 8069 8136 8202 8075 8142 8209 8082 8149 8215 8089 8156 8222 8096 8162 8228 8102 8169 8235 8109 8176 8241 8116 8182 8248 8122 8189 8254 67 68 69 8261 8325 8388 8267 8331 8395 8274 8338 8401 8280 8344 8407 8287 8351 8414 8293 8357 8420 8299 8363 8426 8306 8370 8432 8312 8376 8439 8319 8382 8445 70 8451 8457 8463 8470 8476 8482 8543 8603 8663 8488 8494 8500 8506 71 72 73 8513 8573 8633 8519 8579 8639 8525 8585 8645 S531 S591 8651 8537 8597 8657 8549 8609 8669 8555 8615 8675 8561 8621 8681 8567 8627 8686 74 75 76 8692 8751 8808 8698 8756 8814 8704 8762 8820 8710 8768 8825 8716 8774 8831 8722 8779 8837 8727 8785 8842 8733 8791 8848 8739 8797. 8854 8745 8802 8859 77 78 79 8865 8921 8976 8871 8927 8982 8876 8932 8987 8882 8938 8993 8887 8943 8998 8893 8949 9004 8899 8954 9009 8904 8960 9015 8910 8965 9020 8915 8971 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 82 83 9085 9138 9191 9090 9143 9196 9096 9149 9201 9101 9154 9206 9106 9159 9212 9112 9165 9217 9117 9170 9222 9122 9175 9227 9128 9180 9232 9133 9186 9238 84 85 86 9243 9294 9345 9248 9299 9350 9253 9304 9355 9258 9309 9360 9263 9315 9365 9269 9320 9370 9274 9325 9375 9279 9330 9380 9284 9335 9385 9289 9340 9390 87 88 89 9395 9445 9494 9400 9450 9499 9405 9455 9504 9410 9460 9509 9415 9465 9513 9420 9469 9518 9425 9474 9523 9430 9479 9528 9435 9484 9533 9440 9489 9538 90 9542 9547 9552 9557 9562 1 9566 9571 9576 9581 9586 91 92 93 9590 9638 9685 9595 9643 9689 9600 9647 9694 9605 9652 9699 9609 ! 9657 i 9703 9614 9661 9708 9619 9666 9713 9624 9671 9717 9628 9675 9722 9633 9680 9727 94 95 96 9731 9777 9823 9736 9782 9827 9741 9786 9832 9745 9791 9836 9750 9795 i 9841 ; 9754 9800 9845 9759 9805 9850 9763 9809 9854 9768 9814 9859 9773 9818 9863 97 98 99 9868 9912 9956 9872 9917 9961 9877 9921 9965 9881 9926 9969 9886 9930 9974 9890 9934 9978 9894 9939 9983 9899 9943 9987 9903 9948 9991 9908 9952 9996 APPENDIX. Supplementary Exercises. 135. Factoring. 13. 2 - 3a:' + x\ 14. 81 + ISy' + y\ 15. x^ + {m -\- n)x + m7i, 16. x" — {m -\- n)x + wm. 17. ay' - Say + 2a. 18. ???a;' - 13??za: + 40m. 19. m'x' - Smx + 15. 20. b'x' - 7b'x + 10Z>. 21. m'y + 2my - Sy, 22. nY - StiY - 40y. 23. X* + (2a - 3b)x' - 6abx\ 24. x"^^ - 4:^:" - 12. 26. «V + 8«V + 15«V. 28. 3y + ISy + 24. 32. -, + --S. m m, 36.^^:-^. c c {a - by (a + by (a + by (a - by c c c 42. 16a» + a'2 + §^. 25. %' + 6%^ + Sby. 27. S+»S+f- 29. i.+^+«- 31. 2a;' + 10^ + 8. 33. c c 35. a' ¥ ¥ a"' m'r* n's* 37. n' m'* 39. a' 4«^ 4«' x''^ x' '^ X*' 41. c- + c« + i 282 APPENDIX. 135^, Factors of Quadrinomials, Sometimes an expres- sion containing four terms may be expressed as the product of two binomials. Examples. ax — ay -\- bx — by = a(x — y) -f l){x — y) = {a-^b){x- y); a^ — ar — at -\- rt = a{a — r) — t(a — r) = {a — t) (a — r). EXERCISES. Factor: 1. ab -}- ac -{- mb -\- mc. 2. a^ + «^^^ + «^^ + ^^* 3. ex — cy — x^ -\- xy. 4. mV — m^t — joV + p^t, 5. a^ — a'q — ap ~\-pq. 6. ab — bf' + a(^ — V^ff* 7. ax^ + axy + bxy + by\ 8. (a^ + ^'):z^.y - ab{x^ + y'). 9. (a^ -b'')pq-ab{f -q%V), mn{x^-y^)^xy{:)i^-m^xy), 11. a' + jt?' - op — «y. 12. A:ab + 2^?/ - %bx — xy, 13. 6*' + ab{x — y) — y^xy, 14. a^ — a^my -\- a'mx — am^xy. 15. 1 + m H h 1. 16. 2 H 1 \-mn-^l. 17. m^j + 1 + 1 + — . 18. 2 + am H . ' ' mn am 1 Yfl 19. 2w + m'7^ + -. 20. m^ -\ [- mn + 1. 7/1/? mq np nq mp mq np nq 23. f + ^-^-*. 34. 4 + i? + f + 4. m ,n m n ab a b ab MISCEI-l-ANEOUS EXERCISES. Factor: .49 o ^'_^' n' V a' ^' 4-|+^- ^' 8«' + 24^^ + 18:.'. ^- 8""6-+i8- ^- ?"~^"^F* 7. a"* ~ 4a2"* + 4a^'". 8. 3;? + 18;^' + 27;^'. FACTOBINa. 283 9 ifl±il^ + 2^Ji-+lVlO. i'-^. mn ^wf^ n^J 9 z^ 8 "^27* i6 256* \x al \x al ao a' 17. 16f;_£. 18. ^'-9!^'. 19. 9a'^'* + 30«&'c + 26c\ 20. 2?/'' - 4.yz + 2;2». 21. 4 + 8a;'' + 4:x\ 22. 3z/^ + 12?/2 + 12z\ 23. ._L^-.-J_^. 24.^;-^-^ 3^^ 4r 25. (x+yY (x-yf 3 30 75 /VX. 26. x^ a'' 27. 28. 14 5 n^ dn 9* 29. 31. {a -by {a^hy w' + 1. 30. 32. n' - 1. 33. n'' - 1. 34. n^"" - 1. 35. 1 - 7l'\ 36. m^^ - n'\ 37. 39. «» - 2«J + Z^'^ + a - Z>. 38. 40. 1 + ?^^ 41. -^ + 2 + 1 + - + ^. 42. m' — 2?;^?^ + Ti'* - 43. ^^i 2 + ^ + -- m m 44. 4 . ^ , -h. By what binomial factors must we multiply the following expressions that the products may be binomials? (a) X — a, (h) x^ + ax + «^ \c) re -\-n-\-\. (d) m' + m' + m + 1. (e) x^ — ax-\- a'. (/) x^ — ax^ -{- a*:c — a*. 284 APPENDIX. 189. Problems in Ratio and Proportion. 38. Find the ratio of two numbers whose sum is to their difference in the ratio m : n. 39. The speed of the steamship Servia is to that of the Bothnia as 13 to 10; and the first steams 5 miles farther in 8 hours than the second does in 10 hours. What is the speed of each? 40. The speed of two pedestrians was as 4 : 3, and the slower was 5 hours longer in going 36 miles than the slower was in going 24. What was the speed of each? 41. A chemist had two vessels, A, containing acid, and B, an equal quantity of water. He poured one third the acid into the water, and then poured one third of this mixture back into the acid. What was then the ratio of acid to water in A? 43. If 24 grains of gold and 400 grains of silver are each worth one dollar, what will be the weight of a coin containing equal parts of gold and silver and worth a dollar? 43. A chemist has two mixtures of alcohol and water, the one containing 90 per cent, of alcohol, the other 50 per cent. How much of the first must he add to 1 litre of the second to make a mixture containing 80 j)er cent, of alcohol. 44. It is a law of mechanics that the distances through which heavy bodies will fall in a vacuum in different times are proportional to the squares of the times. If a body fall 48 feet farther in 2 seconds than in 1 second, how far will it fall in 1 second? How far in t seconds? 45. On a line are two points whose distance is a. The first point divides the line into parts whose ratio is 2 : 3; the second into parts whose ratio is 5 : 7. What is the length of the line in terms of «? 46. If a line is divided into two parts whose ratio is m : n^ what is the ratio of the length of the whole line to the distance of the point of division from the middle point? 47. A line is divided into three segments proportional to the numbers m, p and q. What is the ratio of the parts into which the middle point of the whole line divides the middle segment? RATIO AND PROPORTION. 285 48. A sailing-ship leaves port,and 12 hours later is fol- lowed by a steamship. The ratio of the speeds being 3 : 8, how long will it take the steamer to overtake the ship? 49. A courier started from his post, going 7 miles in 3 hours. Two hours later another followed, going 7 miles in 2 hours. How long will the second be overtaking the first? 50. The areas of the openings of two water-faucets are in the ratio 3:5; the speeds of flow of the water through the openings are in the ratio 3:4. At the end of an hour 1221 gallons more have flowed through the second than through the first. What was the flew from each ? 51. The speeds of two trains, A and B, are as m to n, and the journeys they have to make as j^ : q. It took train B t hours longer to make its journey than it did train A. What was the length of each journey and the speed of each train? 52. A street-railway runs along a regular incline, in conse- quence of which the speeds of the cars going in the two direc- tions are as 2:3. The cajs leave each terminus at regular intervals of 5 minutes. x\t what intervals of time will a car going up hill meet the successive cars coming down, and vice versa 9 53. The same thing being supposed, two cars starting out simultaneously from the two ends of the route meet in 30 min- utes. How long in time is the journey for each car? 54. Give the algebraic answers to the two preceding ques- tions when the ratio of the speeds is di : n. 55. Find two numbers which are to each other as 4:3, and whose difference is J of the less. 56. li X : y'.'.Q :% and ^x — 3?/ = 7, what is the value of X and ?/? 57. A yearns profits were divided among two partners in the proportion of 3 : 4. If the second should give $425 to the first, their shares would be equal. What was the amount divided? 58. In a first yearns partnership A had 3 shares, and B 4. In the second A had 1, and B 2. In the second year A had $30e less than he had the first, and B had $200 more. What were the profits? 50. In a farm-yard there aic 4 slu^ep to every 3 cattle, and 286 APPENDIX. 5 cattle to 6 Hogs. How many hogs are there to every 2i> sheep ? 60. A drover started to market with a herd of 7 horses to every 5 mules. He sold 27 horses and bought 3 mules, and then had 3 horses to every 4 mules. How many of each had he at first? 61. Find two quantities whose sum, difference and product are proportional to 5, 1 and 12 respectively. 62. What number is that to which if 2, 6 and 10 be sever- ally added, the first sum shall be to the second as the second is to the third ? 63. What two numbers are to each other as 3 to 4, to each of which if 4 be added, the sums will be as 4 to 5? 64. What quantity must be taken from each term of the ratio m : n that it may equal the ratio c : d^ 05. If a : b\)Q the square of the ratio of a-{- c\ b -\- c, show that c is a mean proportional between a and b. 66. li a : b :: c : d, show that a{a -\- b -{■ c -{- d) = {a-^b) {a-\-c). 67. A line is divided by one point into two parts in the ratio of 3 : 5, and by another point into two parts in the ratio of 1 : 3. The distance between the points of division is 1 inch. What is the length of the line? 68. One ingot contains two parts of gold and one of silver, and another two parts of gold and three of silver. If equal parts are taken from each ingot, what will be the proportion of the gold to the silver in the alloy? 69. If two ounces be taken from the first of the above ingots and three from the second, what will be the ratio of the gold to the silver? 70. A cask contains 4 gallons of water and 18 gallons of alcohol. How many gallons of a mixture containing 2 parts ■water to 5 parts alcohol must be put in the cask so that there may be 2 parts water to 7 alcohol? 71. Which is the greater ratio, 1 -^ a :1 — a or l-fa': 1 — a^, a being positive and less than 1? 72. Which is the greater ratio, a^ — ab -]- b"" : a' -\- ab -{- b"* or a^ — a'b'^ + ^* ' f^'' + f^^^"^ + ^S ^ and b having like signs? 73. What number must be taken from the second term of the ratio 2 : 34 and added to the first that it may equal 5 : 6? RATIO AND PROPORTION. 287 74. It iri a theorem of mechanics that, in order that tv»o masses, V and W, at the ends of a lever, AB, may be in equi- librium, the distances of their points of suspension, A and B, from the fulcrum, F, must be inversely proportional to their weights; that is, we must have FA. Weight V A weight W = FB Now, if the length AB of the lever is /, and the weights of \ and W are respectively ni and n, express the lengths AF and FB of the arms of the lever. 75. The weights at the ends of a lever are 8 and 13 kilo- grammes, and the fulcrum is 3 inches from the middle of the iever. What is the length of the lever? 76. The sum of the two weights is 25 pounds, and the ratio of the distance of the fulcrum from the middle point to the length of the lever is ;:i : 9. What are the weights? 77. The weights are m and n (m > n), and one arm of the lever is // longer than the other. Express the length of the lever. 78. A lever was balanced with weights of 7 and 9 kilo- grammes at its ends. One kilogramme being taken from the lesser and added to the greater (making the weights 6 and 10 kilogrammes), the fulcrum had to be moved 2 inches. What was the length of the lever? 79. What number must be taken from each term of the ratio 19 : 30 that it may equal the ratio 1:2? 80. If a : b::( : (I, show that aihw Va' -f c-^ : Vb' + d^ Find the ratio :r : // from tlio following proportions: 81. rr-f y: ./• - // — /// : H. 82. X -f 2y : X — 2y = ?fi : 2n. 83. x-^2y : 2x -|- // = ni : a. 84. 7nx -i- 7iy : my -\- nx — p : q. ' S5. mx 4- nv : mx — ny = n : q. 288 APPENDIX. 272. Quadratic Equations witli Several Un- known Quantities. 14. ^' + !/' = «; x' -y' = l). 16. x^-f = «; 18. xy = h, x''-]-y^ = a; 20. xy = b. x' - y' = m; X* — y* = n. 22. x-\- y — !-^ = mx y 15. ax" + by' = c; a'x" + b'y" = c\ 17. x'-i-y' = 152; xy = 15. 19. af + f = a; X -\-y =b. 21. x^^f = M X — y — k. 23. X -\- Vl^x' = a' y +Vl^y' ___ xy = 71. {x + Vl + x') (y-^ Vl 4- y") = h\ 24. a;' + ?/' + a: + 2/ = ^^^; ^5- ^'' + ?/' + ^^• - y = « + ^. ^'^ — ^^ + :?; — ?/ = ;i. (^j'' + y^) (^ — /y) — ^'■^^ 26. a;^ — xy = a^y; 27. (a:^"^ — y^) {'X — y) — (i\ xy — y^ = b'^x; {x^ + y'') {x -\~ y) — b. 28. x' -j-y' -{-z' = 84; 29. x +y +z = 12; X -\- y -\- z = 14:; ^y -\- yz -{- zx = 47; xy = 8. ^^'^ 4- i/=^ - / = 0. SO. x-\-y= 9; ?i + ^=9; 31. .-^ + ^^ = 13; xy = 35; x'-\-u'= 52; ?/'+v'=341„ /y + i; = 9; nv = 18. 32. The panel in a door is 12 by 18 inches, and it is to be surrounded by a margin of uniform width and equal surface to the panel. How wide must the margin be? 33. The fore wheel of a coach makes 6 more revolutions than the hind wheel in going 160 yards; but if the circumfer- ence of each wheel be increased by 4 feet, the fore wheel will make only 4 more revolutions in 160 yards. What is the cir- cumference of each wheel? 34. The sum of three numbers is 15; the difference between the first and third is 3 more than the difference between the second and third, and the sum of their squares is 93. What are the numbers? 35. A principal of ^6000 amounts with simple interest to $7800 after a certain number of years. Had the rate been 1 per cent, higher and the time 1 year longer, it would have amounted to 1720 more. What was the time and rate? QUADRATIC EQUATIONS. 289 36. A courier left a town riding at a uniform rate. Three hours afterwards another followed, going 1 mile an hour faster. Two hours after the second another started, goiiig 6 miles an hour. They arrive at their destination at the same time. What was the distance and rate of riding? Ans. Dist, = 60 or 6. Speeds, 4, 5 and 6 or 1, 2 and C. 37. In a right-angled triangle the hypothenuse is 5 and the area 6. What are the sides? 38. Find two numbers whose product is 180, and if the greater be diminished by 5 and the less increased by 3 the product of the sum and difference will be 150. 39. Find two numbers whose sum is 100 and the sum of their square roots 14. 40. Find two numbers whose sum is 35 and the sum of their cube roots 5. 41. By selling a horse for $130 I gain as much per cent, as the horse cost me. What did I pay for him ? 42. What is the price of apples a dozen when four less in 20 cents^ worth raises the price 5 cents per dozen? 43. The sum of the squares of three consecutive numbers is 149. What are the numbers? 44. If twice the product of two consecutive numbers be divided by three times their sum the quotient will be f. What are the numbers? 45. A woman bought a number of oranges for 36 cents. If she had bought 2 more for the same money she would have paid \ of a cent less for each orange. How many did she buy? 46. In mowing 60 acres of grass, 5 days less would have been sufficient if 2 acres more a day had been mown. How many acres were mown per day? 47. A broker bought a certain number of shares (par value $100 each) at a discount for $6400. When they were at the same per cent, premium, he sold all but 20 for $7200. How many shares did he buy, and at what price ? 48. If the length and breadth of a rectangle were each increased by 2, the area would be 238; if both were each di- minished by 2, the area would be 130. Find the length and breadth. 49. Twice the product of two digits is equal to the number itself; and 7 times the sum of the digits is equal to the number 290 APPENDIX. 50. The sum of two numbers is | of the greater, and the difference of their squares is 45. What are the numbers? 51. The numerator and denominator of two fractions are each greater by 2 than those of another; and the sum of the two fractions is 2f ; if the denominators were interchanged, the sum of the two fractions would be 3. What are the frac- tions? 52. A man starts from A to go to B. During the first half of the journey he drives ^ mile an hour slower than the other half, and arrives in 5| hours. On his return he travels a mile slower during the first half than when he went in go- ing over tlie same portion, and returned in 6f hours. What was the distance and rate of driving? 53. A person who has $8800 invests a part of it in one enterprise and the rest in another; the dividends differ in rate, but are equal in amount. If the sums invested had ex- changed rates of dividends, the first would have yielded $200 and the other $288. What were the rates? 54. Divide 50 into two such parts that their product may be to the sum of their squares as 6 to 13. 55. A company at a hotel had $12 to pay, but before set- tling 2 left, when those remaining had 30 cents apiece more to pay than before. How many were there? 56. A and B set out from two towns which are 144 miles apart, and travelled until they met. A went 8 miles an hour, and the number of hours they travelled was three times greater than the number of miles B travelled an hour. At what time did they meet, and what was B's speed? 57. In a purse containing 28 pieces of silver and nickel, each silver coin is worth as many cents as there are uickel coins, and each nickel is worth as many cents as there are silver coins, and the whole are worth $1.50. How many are there of each? 58. Find two such numbers that the product of their sum and difference may be 7, and the product of the sum and dif- ference of their squares may be 144. 59. A grocer sold 50 pounds of pepper and 80 pounds of ginger for $26; but he sold 25 pounds more of pepper for $10 than he did of ginger for $4. What was tlie price ]>ei" pound of each? ARITHMETICAL PR00BE8SI0N. 291 2S5, Aritliiiietical Progression. 23. Find the ni\i term of the series 1, 3, 6, 7, etc. 24. Find the sum of n terms of the series 2, 4, 6, 8, etc. 25. Find the sum of n terms of the series 1, 3, 5, 7, etc. 2t). Insert six arithmetical means between 'I and 2o. 27. If a body fall 16 feet during the first second, and 32 feet each second thereafter more than in the immediately preceding second, how far will it fall during the tenth second, and how far in ten seconds? 28. Find the sum of n terms of the series 5, 12, 19, 26, 33, etc. 13 29. Find the sum of 12 terms of the series 7, -x-, 6, etc. 30. What is the expression for the sum of n terms of a 3 series whose first term is — and the difference between the 2 third and seventh terms 3? 31. The difference between the first and tenth term of an increasing A. P. is 18, and the sum of the ten terms is 100. What is the A. P. ? 32. a = 3, and the fourth term is 4, What is the sum of eight terms? 33. There are two arithmetical series which have the same, common difference; the first terms are 2 and 3 respectively, and the sum of five terms of one is to the sum of five terms of the other as 8 : 9. What are the series? 34. There are four numbers in A. P. whose sum is to the sum of their squares as 2 : 12, and the sum of the first three is 12. What is the A. P. ? 35. A number consisting of three digits, which are in arithmetical progression, if divided by the sum of its digits gives 15 for a quotient, and if 396 be added to it the digits will be reversed. What is the number? 36. Find three numbers in arithmetical progression the isumof whose squares shall be 261, and the square of the mean greater than the product of the extremes by 9. 37. Find four numbers in arithmetical progression whose fium is 16 and continued product 105. •292 APPENDIX. 38. A starts from a certain place going 2 miles the first day, tl the second, G the third, and so on; three days after B sets out and travels 15 miles a day. How many days be- fore B overtakes A? 39. A traveller started from a certain place going 2 miles tlie first day, 5 the second, and so on. After two days another followed and went 6 miles the first day, 10 the second, and so on. After how many days will they be together? 40. In a series of five terms the sum of the first three is 24, and the last three 42. What is the series? 41. Find three numbers in arithmetical progression whose sum shall be 36, and the sum of the first and second shall be 4 of the sum of the second and third. 42. How many means must be inserted between 5 and 17 in order that the sum of the first two shall be to the sum of the second two as 4 : 7? 43. In an arithmetical progression of three terms whose common difference is 4, the product of the second and third is greater by 12 than the product of the first and second. What is the series? 44. The gum of the squares of the extremes of an arithmet- ical progression of four terms is 153, and the sum of the squares of the means 117. What is the series? 45. The sum of five terms of an arithmetical progression is 50, and the product of the first and fifth is to the product of the second and fourth as 7 : 8. What is the series? 46. The sum of nine terms of an arithmetical progression is 1)0. What is the middle term, and the sum of the extremes? 47. A and B start together from the same place; A goes 20 miles a day and B 15 miles the first day and | mile more on each succeeding day than on the preceding day. How far apart will they be at the end of 10 days, and which will be in advance? 48. A and B have a distance of 27 miles to walk; A starts at 2^ miles an hour with an hourly increase of i mile; B starts at 5 miles an hour, but falls off i mile every hour. Which will finish first, and by how much? 49. Find a progression of four terms in which the sum of the extremes is 21 and the product of the means 104. GEOMETIUCAL PROGUESSIOJS^. ^9;J 39^. Geometrical Progression. 21- Find three numbers in A. P. which being increased by 1, 3, and 10 respectively the sum will form a geometrical progression whose common ratio is 2. 22. Find three numbers in geometrical progression wliose 19 sum is 19 and the sum of whose reciprocals --. oo 23. Express the sum of n terms of the G. P. whose first >ierm is a and common ratio W, 24. Express the sum of n terms when the second term is 2 and the third is — 6. 25. Show that the sum of ^n terms when divided by the «um of n terms gives the quotient r'^ -\-l. 26. Find five terms in a geometrical progression the sum *f whose three means is 156 and the sum of the two extremes 328. 27. In a geometrical progression of three terms, the sum of the first and second exceeds the third by 2, and the sum of \he first and third exceeds the second by 14, What is the series? 28. The sum of two numbers is 30, and the arithmetical mean exceeds the geometrical by 3. What are the numbers? 29. Having a progression of an odd number of terms, show that if from the sura of the even terms (the second, fourth, etc.) we subtract the sum of all the odd terms except the last (the first, third, etc.), the remainder is equal to the difference between the first and last terms divided by r -f- 1- 30. In a geometrical progression of four terms the first is less than the third by 24, and the sum of the extremes is to the sum of the means as 7 : 3. What is the series? 31. The sum of three numbers in geometrical progression is 38, and the product of the mean by the sum of extremes is 312. What is the series? 32. Find a geometrical progression of three terms whose product is 1728 and the difference of extremes 45. 33. The sum of four terms of a geometrical progression is 80, and the last term divided by the sum of the two means is 2 J. What is the series? 294 APPKMJIX. 34. If 1, 3 and 7 be added to three consecutive terms of an arithmetical progression whose 0. R. is 2, the sums will form a geometrical progression. Find the series? 35. Find four numbers in a geometrical progression such that the fourth shall be 144 more than the second, and the sum of the means be to the sum of the extremes as 3 : 7. 36. Find a geometrical progression in which the sum of the second and third terms is 60, and of the first and third 50. 37. The sum of the first and second of four terms of a geometrical progression is 16, and the sum of the third and fourth terms 144. What is the series? 38. Show that the difference of any two terms of a Gr. P. is divisible by r - 1. (Of. § 136.) 39. Find a geometrical progression of three terms whose product is 729, and the sum of the extremes divided by the means is 3^. 40. The difference of two numbers is 48, and the arith- metic mean exceeds the geometric by 18. What are the numbers? 41. The fifth term of a G: P. exceeds the first by 16, and the fourth exceeds the second by 4 V^. Find the first term and common ratio. 42. In a G. P. the sum of n terms is S, and the sum of 27^ terms in Q8. Express the common ratio and first term. 43. In a G. P. oi'Zn -\-l terms, whose first term is 5, the Bum of the first and last terms is 125 greater than twice the middle term. Find the common ratio. 44. The first term of a G. P. is 2, and the continued product of the first five terms is 128. What is the common ratio? 45. Find that G. P. of which the product of the first and second terms is 3, and that of the third and fourth terms 48. 46. A person who each year gained half as much again as he did the year before, gained $2059 in 7 years. What was his gain the first year? UNIVERSITY OF ^K^- 'Sf^..f ^S^' ^-vs;^. ^4 THIS BOOK IS DUE ^tsT tHB^^A.^^ DAjieBs STAMPEi) BIIJ.OW ^^ AN INITIAL FINE OF 25 GENETS WILL BE ASSESSED FOR FAILURE TO RETURN THIS BOOK ON THE DATE DUE. THE PENALTY WILL INCREASE TO 50 CENTS ON THE FOURTH DAY AND TO $1.00 ON THE SEVENTH DAY OVERDUE. Jtu 2V .#-*iJ JCT 2 1W5 74iaii»61Pte RFeo U5 JAWcT SS ^ t^>6<^ fi4^^ RECD UD ^e?^ '64-2PM IPTWA^^ir SEMTONILL AU6 2 8 1995 U.CfiEBKeuEY LD 21-100w-8.'34 ) O* *i^. v^.. . ^ ,<.- 'A^'V^-' .^•^^^fCj ^\' ■51 r^.;t.^ * i^^^^-c^.,.'. •^^e--^-..^^^.:?^ 'Hi ., -N ■"■ .' ■'.(' '.■ST' ■:*^ ... ': ■ .'■-' '■ 'J .f *;'/;,<;■- >■:: } X"'- ■ '