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MATHEMATICAL TEXT-BOOKS 
 
 BY 
 
 GEORGE A. WENTWORTH 
 
 Elementary Arithmetic 
 
 Practical Arithmetic 
 
 Mental Arithmetic 
 
 Primary Arithmetic (Wentworth and Reed) 
 
 Grammar School Arithmetic 
 
 Advanced Arithmetic 
 
 Exercises in Arithmetic (Wentworth and Hill) 
 
 First Steps in Algebra 
 
 New School Algebra 
 
 Elementary Algebra 
 
 Elements of Algebra 
 
 Shorter Course in Algebra 
 
 College Algebra (Revised Edition) 
 
 Advanced Algebra 
 
 Complete Algebra 
 
 Higher Algebra 
 
 Exercises in Algebra (Wentworth and Hill) 
 
 First Steps in Geometry (Wentworth and Hill) 
 
 Plane and Solid Geometry (Revised) 
 
 Plane Geometry (Revised) 
 
 Solid Geometry (Revised) 
 
 Plane and Solid Geometry and Plane Trigonometry 
 
 (Second Revised Edition) 
 Analytic Geometry 
 Logarithms, Metric Measures, etc. 
 Geometrical Exercises 
 Syllabus of Geometry 
 
 Examination Manual in Geometry (Wentworth and Hill) 
 Exercise Manual in Geometry (Wentworth and Hill) 
 Plane Trigonometry (Second Revised Edition) 
 Plane Trigonometry and Tables (Second Revised 
 
 Edition) 
 Plane and Spherical Trigonometry (Second Revised 
 
 Edition) 
 Plane and Spherical Trigonometry and Tables 
 
 (Second Revised Edition) 
 Plane Trigonometry, Surveying, and Tables (Second 
 
 Revised Edition) 
 Surveying and Tables (Second Revised Edition) 
 Plane and Spherical Trigonometry, Surveying, and 
 
 Tables (Second Revised Edition) 
 Plane and Spherical Trigonometry, Surveying, and 
 
 Navigation (Second Revised Edition) 
 Logarithmic and Trigonometric Tables 
 Seven Tables, Complete (Wentworth and mil) 
 
NEW 
 
 ' SCHOOL ALQEBBA 
 
 BY 
 
 G. A. WENTWOKTH 
 
 AlTTHOB OF A SERIES OF TEXT-BOOKS IN MATHEMATICS 
 
 GINN AND COMPANY 
 
 BOSTON • NEW YORK • CHICAGO • LONDON 
 ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO 
 

 Copyright, 1898, by 
 GEORGE A. WENTWORTH 
 
 ALL BIGHTS RESERVED 
 A 822.5 
 
 TEfte satftenatum jgrcgg 
 
 GINN AND COMPANY- PRO- 
 PRIETORS • BOSTON • U.S.A. 
 
PEEFAOE. 
 
 The first chapter of this book prepares the way for quite 
 a full treatment of simple integral equations with one 
 unknown number. In the first two chapters only positive 
 numbers are involved, and the beginner is led to see the 
 practical advantages of Algebra before he encounters the 
 difficulties of negative numbers. 
 
 The definitions and explanations contained in these 
 chapters should be carefully read at first ; after the learner 
 has become familiar with algebraic operations, special atten- 
 tion should be given to the principal definitions. 
 
 The third chapter contains a simple explanation of nega- 
 tive numbers. The recognition of the fact that the real 
 nature of subtraction is counting backwards, and that the 
 real nature of multiplication is forming the product from 
 the multiplicand precisely as the multiplier is formed from 
 unity, makes an easy road to the laws of addition and sub- 
 traction of algebraic numbers, and to the law of signs in 
 multiplication and division. All the principles and rules 
 of this chapter are illustrated and enforced by numerous 
 examples involving simple algebraic expressions only. 
 
 The ordinary processes with compound expressions, in- 
 cluding cases of resolution into factors, and the treatment 
 of fractions, naturally follow the third chapter. The im- 
 mediate succession of topics that require similar work is of 
 the highest importance to the beginner, and it is hoped that 
 the chapters on compound expressions will prove interest- 
 ing, and give sufficient readiness in the use of symbols. 
 
 53905 
 
iv PREFACE. 
 
 The chapter on Factors has been made as complete as 
 possible for an elementary text-book, with a view to shorten 
 subsequent work. The easy method of resolving quadratic 
 trinomials into factors, whether the coefficient of the square 
 of the letter involved is unity or greater than unity, and the 
 Factor Theorem, explained on page 102, will be found of very 
 great service in abridging algebraic processes. Examples 
 of short methods for finding the highest common factor of 
 compound expressions are given on page 118 ; and examples 
 of short methods for solving quadratic equations by resolv- 
 ing them into factors are given on pages 272 and 273. 
 
 A five-place table of logarithms is placed at the end of 
 the book instead of a four-place table. Five-place loga- 
 rithms are in common use for practical calculations, and 
 are required by most colleges and science schools for the 
 solution of problems set in entrance examination papers. 
 
 The exercises throughout the book are carefully graded. 
 They are sufficiently varied and interesting, and are not so 
 difficult as to discourage the learner, or so easy as to deprive 
 him of the satisfaction of well-earned success. 
 
 The author has spared no pains to make this a model 
 text-book in subject-matter and mechanical execution. The 
 remarkable favor with which his other Algebras have been 
 received is shown by the fact that nearly a million copies 
 have already been sold, and the sale continues to increase 
 from year to year. The author trusts that this new candi- 
 date for favor will have the same generous reception, and be 
 found to meet fully the requirements of the recent advance 
 in the science and method of teaching Elementary Algebra. 
 
 The author is under obligations to many teachers for 
 
 valuable suggestions, and he will be thankful for corrections 
 
 or criticisms. 
 
 G. A. WENTWOETH. 
 
 Exeter, N. H., June, 1898. 
 
CONTENTS. 
 
 CHAPTER PAGE 
 
 I. Definitions and Notation 1 
 
 II. Simple Equations 15 
 
 III. Positive and Negative Numbers .... 33 
 
 IV. Addition and Subtraction 49 
 
 V. Multiplication and Division . . . . .69 
 
 VI. Special Rules 76 
 
 VII. Factors 85 
 
 VIII. Common Factors and Multiples .... 108 
 
 IX. Fractions 123 
 
 X. Fractional Equations 148 
 
 XL Simultaneous Simple Equations . . . .174 
 
 XII. Problems with Two or More Unknown Numbers 190 
 
 XIII. Simple Indeterminate Equations .... 205 
 
 XIV. Inequalities 208 
 
 XV. Involution and Evolution *. . . 212 
 
 XVI. Theory of Exponents ........... 230 
 
 XVII. Radical Expressions 238 
 
 XVIII. Imaginary Expressions . , 256 
 
 XIX. Quadratic Equations . , . . . . ... . . . . 262 
 
 XX. Simultaneous Quadratics 293 
 
 XXI. Ratio, Proportion, and Variation .... 303 
 
 XXII. Progressions 322 
 
 XXIII. Variables and Limits 338 
 
 XXIV. Properties of Series 345 
 
 XXV. Binomial Theorem 352 
 
 XXVI. Logarithms 372 
 
 XXVII. Graphs 409 
 
NOTICE TO TEACHERS. 
 
 Pamphlets containing the answers will be furnished 
 without charge to teachers for their classes, on appli- 
 cation to G-INN £ COMPANY, Publishers. 
 
NEW SCHOOL ALGEBEA. 
 
 CHAPTER I. 
 DEFINITIONS AND NOTATION. 
 
 Numbers and Number-Symbols. 
 
 1. Algebra. Algebra, like Arithmetic, treats of numbers. 
 
 2. Units. In counting separate objects or in measuring 
 magnitudes, the standards by which we count or measure 
 are called units. 
 
 Thus, in counting the boys in a school, the unit is a boy ; in selling 
 eggs by the dozen, the unit is a dozen eggs ; in selling bricks by the 
 thousand, the unit is a thousand bricks ; in expressing the measure of 
 short distances, the unit is an inch, a foot, or a yard ; in expressing 
 the measure of long distances, the unit is a rod, or a mile. 
 
 3. Numbers. Repetitions of the unit are expressed by 
 numbers. 
 
 A single unit and groups of units formed by successive additions of 
 a unit may be represented as follows : 
 
 / // /// //// mi * mi i 
 mil mi in iw mi iw mi 
 
 These representative groups are named one, two, three, four, five, 
 six, seven, eight, nine, ten; and are known collectively under the 
 general name of numbers. It is obvious that these representative 
 groups will have the same meaning, whatever the units may be that 
 are counted. 
 
2 DEFINITIONS AND NOTATION. 
 
 4. Quantities. A number of specified units of any kind 
 is called a quantity ; as 4 pounds, 5 oranges. 
 
 Note. Quantities are often called concrete numbers, the adjective 
 concrete being transferred from the units counted to the numbers 
 that count them; but a number signifies the times a unit is taken, 
 whether the unit is expressed or understood, and is always abstract. 
 
 Thus, 4 barrels of flour means 4 times 1 barrel of flour ; and 10 
 cords of wood means 10 times 1 cord of wood. 
 
 5. Number-Symbols in Arithmetic. Instead of groups of 
 straight marks, we use in Arithmetic the arbitrary symbols 
 1, 2, 3, 4, 5, 6, 7, 8, 9, called Arabic numerals, for the num- 
 bers one, two, three, four, five, six, seven, eight, nine. 
 
 The next number, ten, is indicated by writing the figure 
 1 in a different position, so that it shall signify not one, but 
 ten. This change of position is effected by introducing a 
 new symbol, ; called nought or zero, and signifying none. 
 
 All succeeding numbers up to the number consisting of 10 tens are 
 expressed by writing the figure for the number of tens they contain 
 in the second place from the right, and the figure for the number oi 
 units besides in the first place. The hundreds of a number are written 
 in the third place from the right. The thousands are written in the 
 fourth place from the right ; and so on. 
 
 6. Number-Symbols in Algebra. Algebra employs the 
 letters of the alphabet in addition to the figures of Arith- 
 metic to represent numbers. The letters of the alphabet 
 are used as general symbols of numbers to which any par- 
 ticular values may be assigned. In any problem, however, 
 a letter is understood to have the same value throughout 
 the problem. 
 
 7. Terms Common to Arithmetic and Algebra. Terms com- 
 mon to Arithmetic and Algebra, as addition, sum, subtrac- 
 tion, minuend, subtrahend, difference, etc., have the same 
 meaning in both ; or an extended meaning in Algebra con- 
 sistent with the sense attached to them in Arithmetic. 
 
DEFINITIONS AND NOTATION. 3 
 
 The Principal Signs of Operations. 
 
 The principal signs of operations are the same in Algebra 
 as in Arithmetic. 
 
 8. The Sign of Addition, +. The sign + is read plus. 
 
 Thus, 4 + 3 is read 4 plus 3, and indicates that the number 3 is to 
 be added to the number 4 ; a -f- b is read a plus 6, and indicates that 
 the number 6 is to be added to the number a. 
 
 9. The Sign of Subtraction, — . The sign — is read minus. 
 
 Thus, 4 — 3 is read 4 minus 3, and indicates that the number 3 is 
 to be subtracted from the number 4 ; a — b is read a minus 6, and 
 indicates that the number b is to be subtracted from the number a. 
 
 10. The Sign of Multiplication, X. The sign X is read 
 times, or multiplied by. 
 
 Thus, 4 X 8 is read 4 times 3, and indicates that the number 3 is 
 to be multiplied by 4 ; a X 6 is read a times b, and indicates that the 
 number 6 is to be multiplied by the number a. 
 
 A dot is sometimes used for the sign of multiplication. 
 Thus, 2 • 3 • 4 • 5 means the same as 2 X 3 X 4 X 5. Either 
 sign is read multiplied by when followed by the multiplier. 
 $a X b, or $a • b, is read a dollars multiplied by b. 
 
 11. The Sign of Division, -J-. The sign ■+- is read divided by. 
 
 Thus, 4 -r 2 is read 4 divided by 2, and indicates that the number 
 4 is to be divided by 2 ; a -r 6 is read a divided by 6, and indicates 
 that the number a is to be divided by the number b. 
 
 Division is also indicated by writing the dividend above 
 the divisor with a horizontal line between them; or by 
 separating the dividend from the divisor by an oblique 
 line, called the solidus. 
 
 Thus, -i or a fb y means the same as a •+- b. 
 
 Note. The operation of adding b to a, of subtracting 6 from a, of 
 multiplying a by 6, or of dividing a by 6 is algebraically complete 
 when the two letters are connected by the proper sign. 
 
4 DEFINITIONS AND NOTATION 
 
 12. The Radical Sign, ->/• Tne si g n V is called the 
 radical sign, and denotes that a root of the number before 
 which it is placed is to be found. 
 
 Other Signs Used in Algebra. 
 
 13. The Sign of Equality, =. The sign = is read is 
 equal to, and when placed between two numbers indicates 
 that these two numbers are equal. 
 
 Thus, 8 + 4 = 12 means that the sum of 8 and 4 is equal to 12 ; 
 x + y = 20 means that the sum of x and y is equal to 20 ; and x = a + b 
 means that x is equal to the sum of a and b. 
 
 14. The Sign of Deduction, .•. The sign .'.is read hence 
 or therefore. 
 
 15. The Sign of Continuation, The sign is read 
 
 and so on. * 
 
 Thus, 1, 2, 3, 4, is read one, two, three, four, and so on. oi, 
 
 Oa, as, a n is read, a sub one, a sub two, a sub three, and so on to 
 
 a sub n. a', a", a'", is read a prime, a second, a third, and so on. 
 
 16. The Signs of Aggregation. The signs of aggregation 
 are the parenthesis ( ), the bracket [ ], the brace \ \, the 
 vinculum — , and the bar | . 
 
 These signs mean that the indicated operations in the 
 expressions affected by them are to be performed first, and 
 the result treated as a single number. 
 
 Thus, (a + b) X c means that the sum of a and b is to be multiplied 
 by c ; (a — b) X c means that the difference of a and b is to be multi- 
 plied by c. 
 
 The vinculum is written over the expression that is to be 
 treated as a single number. 
 
 Thus, a — b-rc means the same as a — (b + c), and signifi es that 
 the sum of b and c is to be subtracted from a ; and Va — b means 
 the same as V(# ~~ &)> an d signifies that b is to be subtracted from a, 
 and the square root of the remainder found. 
 
DEFINITIONS AND NOTATION. 5 
 
 Factors, Powers, Roots. 
 
 17. Factors. When a number is the product of two 01 
 more numbers, each of these numbers, or the product of 
 two or more of them, is called a factor of the given number. 
 
 Thus, 2, a, b, 2 a, 2 6, ab are factors of 2 ab. 
 
 18. Factors expressed by letters are called literal factors j 
 factors expressed by figures are called numerical factors. 
 
 19. The sign X is omitted between factors, if the factors 
 are letters, or a numerical factor and a literal factor. 
 
 Thus, we write 63 ab for 63 x a x b ; we write abc for a X b X c. 
 
 20. The expression abc must not be confounded with 
 a + b -f- c. afo is a product ; a + 6 +- c is a sum. 
 
 If a = 2, b = 3, c = 4, 
 
 then afo = 2 X 3 X 4 = 24 ; 
 
 but a + & + c = 2 + 3 + 4 = 9. 
 
 Note. When a sign of operation is omitted in the potation of 
 Arithmetic, it is always the sign of addition; but when a sign of 
 operation is omitted in the notation of Algebra, it is always the sign 
 of multiplication. Thus, 456 means 400 + 50 + 6, but 4 ab means 
 4 X a X 6. 
 
 21. If one factor of a product is equal to 0, the product 
 is equal to 0, whatever the values of the other factors. 
 Such a factor is called a zero factor. 
 
 Thus, abed = 0, if a, 6, c, or d = 0. 
 
 22. Coefficients. Any factor of a product may be con- 
 sidered as the coefficient of the remaining factors ; that is, 
 the co-factor of the remaining factors. Coefficients expressed 
 by letters are called literal coefficients ; expressed by Arabic 
 numerals, numerical coefficients. 
 
 Thus, in 7 x, 7 is the numerical coefficient of x ; in ax, a is the 
 literal coefficient of x. 
 
 If no numerical coefficient is written, 1 is understood. 
 
6 DEFINITIONS AND NOTATION. 
 
 23. Powers and Roots. When a number is taken a num- 
 ber of times as a factor, the result is called a power of the 
 factor. When a number is the product of equal factors, 
 one of the equal factors is called a root of the number. 
 
 24. Indices or Exponents of Powers. An index or exponent 
 of a power is a number-symbol written at the right of and 
 a little above the number. 
 
 If the exponent is a whole number, it shows the number 
 of times the given number is taken as a factor. 
 
 Thus, a 1 , or simply a, denotes that a is taken once as a factor ; a 2 
 denotes that a is taken twice as a factor ; a 8 denotes that a is taken 
 three times as a factor ; and so on. These are read : the first power 
 of a ; the second power of a ; the third power of a ; and so on. We 
 write a 3 for aaa, a 4 for aaaa, a n for aaaa to n factors. 
 
 Note. The second power of a number is often called the square of 
 that number ; thus, a 2 is called the square of a, because if a denotes 
 the number of units of length in the side of a square, a 2 denotes the 
 number of units of surface in the square. The third power of a 
 number is often called the cube of that number ; thus, a 3 is called the 
 cube of a, because if a denotes the number of units of length in the 
 edge of a cube, a 8 denotes the number of units of volume in the cube. 
 
 25. The meaning of coefficient and exponent must be 
 carefully distinguished. Thus, 
 
 4a = a + a + a + a; 
 a* = aXaX.aXa. 
 Ifa = 3, 4a = 3 + 3 + 3 + 3 = 12. 
 
 a* = 3 X 3 X 3 x 3 = 81. 
 
 26. Indices of Roots. An index of a root is a number- 
 symbol written above the radical sign to indicate the re- 
 quired root. 
 
 Thus, Va, or simply Va, means one of the two equal factors of a, 
 
 that is, the square root of a ; Va means one of the three equal factors 
 of a, that is, the cube root of a ; and so on. 
 
DEFINITIONS AND NOTATION. 7 
 
 Algebraic Expressions. 
 
 27. An Algebraic Expression. An algebraic expression is 
 a number written with algebraic symbols. An algebraic 
 expression may consist of one symbol, or of several symbols 
 connected by signs. 
 
 Thus, a, 3 a&c, 5 a + 2 & — 3 c, are algebraic expressions. 
 
 28. Terms. A term is an algebraic expression of one sym- 
 bol, or of several symbols not separated by the sign + or — . 
 
 Thus, a, 5 xy, 2 ab X 4 cd, j—z are algebraic expressions of one term 
 each. A term may be separated into parts by the sign Xor-r. 
 
 29. Similar Terms. If terms have the same letters, and 
 each letter has the same exponent in all the terms, they are 
 called similar terms or like terms. 
 
 Thus, 3x 2 y 3 , 5x 2 y 3 , and 7 x 2 y* are similar terms. 
 
 30. Simple Expressions. An algebraic expression of one 
 term is called a simple expression or a monomial. 
 
 Thus, 5 xy, 7 a X 2 6, 7 a -r 2 6, are simple expressions. 
 
 31. Compound Expressions. An algebraic expression of 
 two or more terms is called a compound expression or a poly- 
 nomial. 
 
 Thus, 5 xy + 7 a, 2 x — y — 3 z, are compound expressions. 
 
 32. A polynomial of two terms is called a binomial ; of 
 three terms, a trinomial. % 
 
 Thus, 3 a — b is a binomial ; and 3a — 6 + cisa trinomial. 
 
 33. Plus and Minus Terms. A term preceded by the 
 sign -f is called a plus term ; and a term preceded by the 
 sign — is called a minus term. The sign + before a single 
 term, and before the first of a series of terms is omitted. 
 
8 DEFINITIONS AND NOTATION. 
 
 34. A plus term and a minus term cancel each other when 
 combined, if both terms stand for the same number. 
 
 35. The Numerical Value of an Expression. The result 
 obtained by putting particular values for the letters of an 
 expression and performing the indicated operations is called 
 the numerical value of the expression. 
 
 Numerical Values of Simple Expressions. 
 
 1. If a = 3, find the numerical values of 4 a and a*. 
 
 4a = 4Xa = 4x3 = 12; 
 and a 4 = aXaXaXa = 3X3x3x3 = 81. 
 
 2. If a = 5, b = 6, c = 7, find the numerical value of the 
 expression J^ abc. 
 
 T 9 ¥ a6c = ^X5x6x7 = 135. 
 
 3. If x = 2, y = 3, find the numerical value of 5 x V. 
 
 5 xV = 5X2 2 X3 8 = 5X4X27 = 540. 
 
 4. If x = 4, y = 6, find the numerical value of f x*y. 
 
 \&y = * X 42 X 6 = f X 16 X 6 = 64. 
 
 5. If x = 2, y = 3, z = 4, find the numerical value of 
 $x?y + 3z*. 
 
 8z*y _ 8 X2X2X3 _1 
 3 z» 3X4X4X4~2* 
 
 6. If x = 3, find the numerical values of ^/Ax 2 ; Via?; 
 and 2 V( 9 a; 2 )- 
 
 V4x 2 = V4 x 3 2 = 2 x 9 = 18. 
 
 Viz 2 = V4 X 3 2 = V§6 =6. 
 
 2 V(9x 2 ) = 2 V(9 X 3 2 ) = 2 X 9 = 18. 
 
 Note. When no vinculum or parenthesis is used, a radical sign 
 affects only the symbol immediately following it. 
 
1. 
 
 15*. 
 
 11. 
 
 f&V. 
 
 2. 
 
 3ab. 
 
 12. 
 
 *«y. 
 
 3. 
 
 7 by. 
 
 13. 
 
 w. 
 
 4. 
 
 5bd. 
 
 14. 
 
 §V- 
 
 5. 
 
 9y 2 . 
 
 15. 
 
 f x V. 
 
 6. 
 
 3b 2 c. 
 
 16. 
 
 tv<w. 
 
 7. 
 
 ±c 2 x 2 . 
 
 17. 
 
 A«*V. 
 
 8. 
 
 2b 6 x. 
 
 18. 
 
 f a 4 z 2 . 
 
 9. 
 
 b 2 cy 2 . 
 
 19. 
 
 *&y. 
 
 10. 
 
 a 4 b 3 c 2 . 
 
 20. 
 
 i«w 
 
 3ft¥ 
 
 26. 
 
 V*2/. 
 
 8 6c 
 
 27. 
 
 ■y/dx. 
 
 U 2 y 
 
 28. 
 
 !/Se. 
 
 2dc 
 
 29. 
 
 ^/bdx 8 . 
 
 x 2 y 2 
 
 30. 
 
 V(%). 
 
 5b 2 c 2 
 
 31. 
 
 ^Jab 2 x 2 . 
 
 x 2 y 2 z 2 
 
 32. 
 
 ■^Jxhfz 3 . 
 
 3c 3 d? 
 
 33. 
 
 Zj3be 2 d. 
 
 10 a 5 b 5 
 
 34. 
 
 2V^W. 
 
 oV>b a 
 
 35. 
 
 c^/dx 2 . 
 
 
 DEFINITIONS AND NOTATION. 
 
 Exercise 1. 
 
 If a = 1, b = 2, c = 3, d = 4:, x = 5, y = 6, z = 0, find 
 the numerical value of : 
 
 21 
 
 22. 
 
 23. 
 
 24. 
 
 25. 
 
 Numerical Values of Compound Expressions. 
 
 36. Each term should be written in the algebraic form 
 by omitting the sign X between a numerical factor and a 
 literal factor or between two literal factors. The opera- 
 tions indicated in a term must be performed before the 
 operation indicated by the sign prefixed to the term. 
 
 37. The parts of a term are combined in the order of the 
 signs X and ■** from left to right. 
 
 The terms of an expression are combined in the order of 
 the signs + and — from left to right. 
 
 Thus, 60 - 40 -i- 5 X 3 - 20 = 60 - ^ X 3 - 20 = 16. 
 
 38. The terms may be arranged in any order before 
 combining them. This is called the commutative law for 
 addition and subtraction. 
 
10 DEFINITIONS AND NOTATION. 
 
 Numerical Values of Compound Expressions. 
 
 1. If b = 10, c = 2, y = 5, find the numerical value of 
 66- (8y + 2c)c-2cy. 
 
 66 — (8y -r 2 c) c — 2 cy = 6 X 10 — - 4 -0 X 2 — 2 X 2 X 5 
 = 60 — 20 — 20 = 20. 
 
 2. If x = 7, y = 5, find the numerical value of • 
 
 x + y 
 
 (x + y)(x — y) + 
 
 « — y 
 
 C^y)(*-y) + ~| = (7 + 6)(7--6) + |i| 
 
 = 12 X 2 + V- = 30. 
 
 Exercise 2. 
 If a = 1, 6 = 2, c = 3, find the value of : 
 
 1. 7 a — be. b.2a — b + c. 9. V 4 a ^ + 2 c. 
 
 2. ac + 6. 6. a6 + 6c — ac. 10. V6abc — 2b. 
 
 3. 4a&-c. 7. 6 2 + a 2 + c 2 . 11. c 3 - 6 3 . 
 
 4. 6a6-6-c. 8. 5 6c 2 -2a6. 12. -^c 2 - a\ 
 
 13. 26 + 3c -a. 16. 6b - lObc -*- 12a 4- 2c. 
 
 14 <> + 6) 2 -h2(c-a) 3 . 17. 5 c h- (6- a) -6 -a. 
 15. V66c" - 6 - c. 18. -vW - VbW. 
 
 If a = 1, b = 2, c = 3, d = 0, find the value of : 
 
 19. la -be + 6d. 25. ^4abcd + 2b*. 
 
 20. ac + 6 — d. 26. -\/Tabcd + 6 C . 
 
 21. ±ab-cd-d. 27. 6 3 - c + d c . 
 
 22. 2a-H^. 28. 3c*-26 2 + a. 
 
 23. ab + bc-ad. 29. 26 +(5c - 3)-h(2c - a). 
 
 24. 2a6-56c£ 30. 3c 2 - 2a 9 - - 26 b . 
 
DEFINITIONS AND NOTATION. H 
 
 Parentheses. 
 
 39. A parenthesis preceded by the sign -[-. If a man has 
 10 dollars and afterwards collects 3 dollars and then 2 
 dollars, it makes no difference whether he puts the 3 
 dollars and the 2 dollars together and adds their sum to 
 his 10 dollars, or adds the 3 dollars to his 10 dollars, and 
 then the 2 dollars. 
 
 The first process is represented byl0-f(34-2). 
 The second process is represented by 10 + 3 4- 2. 
 
 Hence, 10 + (3 + 2) = 10 + 3 + 2. (1) 
 
 If a man has 10 dollars and afterwards collects 3 dollars 
 and then pays a bill of 2 dollars, it makes no difference 
 whether he pays the 2 dollars from the 3 dollars collected 
 and adds the remainder to his 10 dollars, or adds the 3 
 dollars collected to his 10 dollars and pays from this sum 
 his bill of 2 dollars. 
 
 The first process is represented by 10 + (3 — 2). 
 The second process is represented by 10 4- 3 — 2. 
 
 Hence, 10 + (3 - 2) = 10 4- 3 - 2. (2) 
 
 If we use general symbols in (1) and (2), we have, 
 
 « + (J + c) = a + J + c, 
 and a 4- (b — c) — a + b — c. Hence, 
 
 We have the general rule for a parenthesis preceded by 4~ * 
 
 If an expression within a parenthesis is preceded by the 
 sign +, the parenthesis may be removed without making any 
 change in the signs of the terms of the expression. 
 
 Instead of a parenthesis, any other sign of aggregation 
 may be used and the same rule will apply. 
 
12 DEFINITIONS AND NOTATION 
 
 40. A parenthesis preceded by the sign — . If a man with 
 10 dollars has to pay two bills, one of 3 dollars and one of 
 2 dollars, it makes no difference whether he takes 3 dollars 
 and 2 dollars at one time, or takes 3 dollars and 2 dollars 
 in succession, from his 10 dollars. 
 
 The first process is represented by 10 — (3 + 2). 
 The second process is represented by 10 — 3 — 2. 
 
 Hence, 10 - (3 + 2) = 10 - 3 - 2. (1) 
 
 If a man has 10 dollars consisting of two 5-dollar bills, 
 and has a debt of 3 dollars to pay, he can pay his debt by 
 giving a 5-dollar bill and receiving 2 dollars. 
 
 This process is represented by 10 — 5 + 2. 
 
 Since the debt paid is 3 dollars, that is, (5 — 2) dollars, 
 the number of dollars he has left can be expressed by 
 
 10 -(5 -2). 
 Hence, 10 - (5 - 2) = 10 - 5 + 2. (2) 
 
 If we use general symbols in (1) and (2), we have, 
 a — (b -f- c) = a — b — c, 
 and a — (b — c) — a — b + c. Hence, 
 
 We have the general rule for a parenthesis preceded by — : 
 
 If an expression within a parenthesis is preceded by the 
 sign — , the parenthesis may be removed, provided the sign 
 before each term within the parenthesis is changed, the 
 sign -\-to — and the sign — to +. 
 
 Note. If the vinculum is used, the sign prefixed to the first term 
 under the vinculum must he understood as the sign before the 
 vinculum. 
 
 Thus, a + b — c has the same meaning as a + (b — c), 
 and a — b — c has the same meaning as a — (b — c). 
 
DEFINITIONS AND NOTATION. 13 
 
 ♦ 
 Exercise 3. 
 
 Eemove the parentheses and combine : 
 
 1. 9 + (3 + 2). 5. 9 - (8 - 6). 9. (3 - 2) - (2 - 1). 
 
 2. 9 + (3 - 2). 6. 8 - (7 - 5). 10. (7 - 3) - (3 - 2). 
 
 3. 7 + (5 + 1). 7. 9 - (6 + 1). ll. (8 - 2) - (5 - 3). 
 
 4. 7 + (5 - 1). 8. 8 - (3 + 2). 12. 15 - (10 - 3 - 2). 
 
 If a = 10, b = 5, c = 4, d = 2, find the value of: 
 
 13. (a + 6) + (c + d). 15. (a-b)-(c-d). 
 
 14. (a + &)-(<?-<*). 16. (a -$) + (« — d). 
 
 Product of a Compound by a Simple Factor. 
 
 41. In finding the product of 4 (5 + 3), it makes no dif- 
 ference in the result whether we multiply the sum of 5 and 
 3 by 4, or multiply 5 by 4 and 3 by 4 and add the products. 
 
 By the first process, 4(5 + 3) = 4x8 = 32. 
 
 By the second process, 4 (5 + 3) = (4 X 5 + 4 X 3) = 32. 
 
 In like manner, 4 (5 — 3) = 4 X 2 = 8, 
 
 and 4(5-3) = (4x 5-4x3)= 8, 
 
 In general symbols, a (b + c) = ab -f ac, 
 and a(b — c) = ab — ac. 
 
 This is called the distributive law for multiplication. 
 
 42. The order of the factors is immaterial. 
 Thus, 4(5 + 3)^4x5 + 4x3 = 32, ' 
 
 and (5 + 3) 4 = 5 X 4 + 3 X 4 = 32. 
 
 In general symbols, ab = ba. 
 This is called the commutative law for multiplication. 
 Perform the indicated operations : 
 
 J. x + 3(a-b). 2. x-S(a-b). 
 
 1. x + 3 (a — b) = x + (3a - 3 b) = x + 3 a - 36. 
 
 2. x — 3(a-&) = z-(3a-36) = x — 3a + 36. 
 
14 DEFINITIONS AND NOTATION. 
 
 Exercise 4. 
 
 Perform the indicated operations, and find the numerical 
 value of each expression, if a = 5, b = 4, c = 3 : 
 
 1. 3(ab + c). 7. bc + a(b — c). 13. (a-c)b-a. 
 
 2. 4:(ac + b). 8. b + 2(a-c). 14. (a -&)<* + 2c. 
 
 3. 2(aft — c). 9. ac-2(6 + c). 15. (6 + c)c-2&. 
 
 4. 5(ac-&). 10. 5ac-2(b 2 + b). 16. (a-c)7-3e. 
 
 5. 7(Jc-a). 11. 2b±3(a-c). 17. <>-c)c 2 -2c 2 . 
 
 6. ac(b-c). 12. a6 + &(£-c). 18. (a 2 — & 2 )c-a 2 . 
 
 Quotient of a Compound by a Simple Expression. 
 
 43. In finding the quotient of (8 + 4)^-2 it makes no 
 difference in the result whether we divide the sum of 8 and 
 4 by 2, or divide 8 by 2 and 4 by 2, and add the quotients. 
 
 By the first process, (8 + 4) -*- 2 = 12 -h 2 = 6. 
 
 By the second process, (8 + 4) -s- 2 = (8-^2 + 4-^ 2) = 6. 
 
 In general symbols, = --}--, 
 
 , a — b a b 
 
 and = 
 
 c c c 
 
 This is called the distributive law for division. 
 
 Perform the indicated operations : 
 
 1. x + (3a + 3 b) -J- 3. 2. x-(3a + 3b)+3. 
 
 1. x + (3a + 36) -f 3 = » + (a + 6) = x + a + 6. 
 
 2. * — (3 a + 3 fe\ -fr 3 = x — (a + b) = x — a — b. 
 
 Exercise 5. 
 
 Perform the indicated operations, and find the numerical 
 value of each expression, if a = 8, b = 4, c = 2 : 
 
 1. (5 + c)-s-c. 5. O^ + c) 
 
 (a + c) + b. 6. (ac + &) + £. 10. (6 2 - c 2 ) -5- J. 
 
 3. (a-b) + b. 7. (ac - &) 
 
 c. 9. (£ 2 + c 2 ) 
 
 &. 11. (a 2 -c 2 )H-c 2 . 
 
 4. (6-c)-5-c. 8. (oi-o)-HO. 12. (a 1 - &■)■+■ P. 
 
CHAPTER IL 
 SIMPLE EQUATIONS. 
 
 44. Equations. An equation is a statement in symbols 
 that two expressions stand for the same number. 
 
 Thus, the equation Sx + 2 = 8 states that Sx + 2 and 8 stand for 
 the same number. 
 
 45. That part of the equation which precedes the sign 
 of equality is called the first member, or left side, and that 
 part of the equation which follows the sign of equality is 
 called the second member, or right side. 
 
 46. An equation containing letters, if true for all values 
 of the letters involved, is called an identical equation ; but 
 if it is true only for certain particular values of the letters 
 involved, it is called an equation of condition. 
 
 Thus, a + b = b + a, which is true for all values of a and b, is an 
 identical equation ; and 3 x + 2 = 8, which is true only when x stands 
 for 2, is an equation of condition. 
 
 For brevity, an identical equation is called an identity, 
 and an equation of condition is called simply an equation. 
 
 47. We often employ an equation to discover an unknown 
 number from its relation to known numbers. We usually 
 represent the unknown number by one of the last letters of 
 the alphabet, as x, y, z ; and the known numbers by the 
 first letters, a, b, c, and by the Arabic numerals. 
 
 48. Simple Equations. Equations which, when reduced 
 to their simplest form, contain only the first power 
 
16 SIMPLE EQUATIONS. 
 
 of the unknown numbers are called simple equations, or 
 equations of the first degree. 
 
 Thus, 7x + 5 = 4x + 14, and ax + b = c are simple equations in x. 
 
 49. Two or more like terms may be combined to form a 
 single like term by uniting their numerical coefficients. 
 
 Thus, 3 ax + ax = 4 ax ; and box — Sax = 2 ax. 
 
 50. To Solve an Equation with One Unknown Number is to 
 find the unknown number; that is, to find the number 
 which, when substituted for its symbol in the given equa- 
 tion, renders the equation an identity. 
 
 This number is said to satisfy the equation, and is called 
 the root of the equation. 
 
 51. Axioms. In solving an equation, we make use of 
 the following self-evident truths, called axioms : 
 
 Ax. 1. If equal numbers are added to equal numbers, the 
 sums are equal. 
 
 Ax. 2. If equal numbers are subtracted from equal num- 
 bers, the remainders are equal. 
 
 Ax. 3. If equal numbers are multiplied by equal num- 
 bers, the products are equal. 
 
 Ax. 4. If equal numbers are divided by equal numbers, 
 the quotients are equal. 
 
 Ax. 5. If two numbers are equal to the same number, 
 they are equal to each other. 
 
 52. Transposition of Terms. It becomes necessary in 
 solving simple equations to bring all the terms that contain 
 the symbols for the unknown numbers to one side of the 
 equation, and all the other terms to the other side. This 
 process is called transposing the terms. 
 
SIMPLE EQUATIONS. 17 
 
 1. Find the number for which x stands when 
 
 x — b = a. 
 Add b to each side, x — b + b = a + b. (Ax. 1) 
 
 Cancel -b + 5, x = a + b. (§34) 
 
 The result is the same as if we had transposed — b from 
 the left side to the right side and changed its sign. 
 
 2. Find the number for which x stands when 
 
 x + b = a. 
 Subtract b from each side, x + b — b = a — b. (Ax. 2) 
 Cancel + b~b, x = a - b. (§34) 
 
 In this case, we have transposed b from the left side to 
 the right side and changed its sign. 
 
 We can proceed in like manner in any other case. 
 
 Hence, the general rule : 
 
 53. Any term may be transposed from one side of an 
 equation to the other, provided its sign is changed. 
 
 It follows from axioms 1 and 2 that : 
 
 54. Any term that occurs with the same sign on both sides 
 of an equation may be cancelled. 
 
 If we transpose each term of the equation, 
 
 c — x = a — b, (1) 
 
 we have b — a = x — c. 
 
 That is, x — c = b — a. (2) 
 
 Equation (2) is the same as (1) with the sign before 
 each term changed. Hence : 
 
 55. The sign of every term of an equation may be changed 
 without destroying the equality. 
 
18 SIMPLE EQUATIONS. 
 
 56. Numerical Equations. An equation in which all the 
 known numbers are expressed by Arabic numerals is called 
 a numerical equation. 
 
 Solution of Simple Numerical Equations in X 
 
 1. Solve Sx-7 = 14:-Ax. 
 
 Transpose — 4 x to the left side and — 7 to the right side.. 
 
 
 3z + 4x = 14 + 7. 
 
 (§58) 
 
 Combine, 
 
 7x = 21. 
 
 (§ 49) 
 
 Divide by 7, 
 
 x = S. 
 
 (Ax. 4) 
 
 2. Solve the equation 
 
 1 - 4 (x - 2) = Ix - 3 (3 x - 1). 
 Multiply the compound factor by the simple factor in each side, 
 
 1 — (4 x — 8) = 7 x — (9 x — 3). 
 Remove the parenthesis in each side, 
 
 l-4z + 8 = 7a;-9x + 3. (§40) 
 
 Transpose, 9x — 4ce — 7jc = 3 — 1 — 8. 
 Change the signs of all the terms, 
 
 4z + 7x — 9z = l + 8-3. ^§55) 
 
 Combine, 2 x = 6. (§ 49) 
 
 Divide by 2, x = 3. (Ax. 4) 
 
 57. To Solve a Simple Numerical Equation in ^r, therefore : 
 Transpose all the terms that contain x to the left side, and 
 
 all the other terms to the right side. Combine similar terms, 
 and divide both sides by the coefficient of x. 
 
 58. Verification. If the value found for x is substituted 
 for x in the original equation, and the equation reduces to 
 an identity, the value of x, that is, the root of the equation, 
 is said to be verified. 
 
 Note. In verifying a solution, as in solving an equation, it is im- 
 portant to notice that the signs of all the terms may be changed. 
 
SIMPLE EQUATIONS. 
 
 Show that x stands for 3 in the equation 
 
 3a;-7 = 14-4a;. 
 Put 3 for x in this equation, and we have 
 
 3x3-7 = 14 -4X3, 
 or, 9 - 7 = 14 - 12, 
 
 that is, 2 = 2. 
 
 Exercise 6. 
 Find the value of x, and verify the answer : 
 
 1. 5a; -4 = 16. 10. 14a; -79 = 8a: -25. 
 
 2. 3a; + 4 = 25. 11. 5a; -4 = 12- 3a\ 
 
 3. 24a; -7a; = 34. 12. Ix + 4 = 3x + 24. 
 
 4. 16a; = 7a; + 81. 13. 12a - 16 = 8 + 6x. 
 
 5. 3a; = 55 -2a;. 14. 4a; -10 = 14 + 2a;. 
 
 6. 5 a; = 3 a; + 6. 15. 2 x — 5 = 7 — x. 
 
 7. 7 a; = 6a; + 4. 16. 4a; -14 = a; -2. 
 
 8. 5x = 28-2x. 17. 4a; - 11 = 2a; - 5. 
 
 9. 2a; = 11+ a-. 18. 4a; - 10 = 3a; - 5. 
 
 19. 5 (x + 1) + 6 (x + 2) = 7 (a 4- 3). 
 
 20. 4 (a; + 7) - 36 = 13 (x - 2). 
 
 21. 6 (3 aj - 1) -8a; = 140 + 2 (x - 1). 
 
 22. 3 (3x - 2) - 6 (4 - x) = 24a; -4 (7a; -2). 
 
 23. 3 (aj + 13) - 15 = 4 (x - 2) - 9. 
 
 24. 10a; - (x -10) = 3a; + 52. 
 
 25. 3a; - (x + 5) - (x - 3) = 10 - a;. 
 
 26. x 2 + 8a; - (a; 2 - aj) = 5 (x + 3) + 5. 
 
 27. 5 - x + 4 (a; - 1) - (a; - 2) = 15. 
 
 28. 3 (a? + 10) + 4 (x + 20) + 5x = 185 - 3a;. 
 
 29. 2 (a? - 2) + 3 (a? - 3) + 4 (a? - 4) = 3a; + 7. 
 
 30. (5a; + 3) - 2 (x - 1) + (1 - x) = 4 (9 - a;). 
 
 31. 7 - 21 (x + 3) = 13 - 15 (2x - 5). 
 
 32. 5 (a? - 3) - 7 (6 - x) + 29 = 50 - 3 (8 - x). 
 
 & 
 
20 SIMPLE EQUATIONS. 
 
 Statement and Solution of Problems. 
 
 59. To express in algebraic language the conditions of a 
 problem that are stated in common language is generally 
 very difficult for the beginner. We will therefore give an 
 exercise on translating common language into algebraic 
 language before proceeding to the solutions of problems. 
 
 Exercise 7. 
 
 1. Write in symbols : a diminished by b ; a increased 
 by b \ a multiplied by b ; a divided by b ; the square of a ; 
 the square root of a ; the cube root of a ; the square of a 
 multiplied by the fourth power of b. 
 
 2. If a man walks x miles an hour, how many miles 
 will he walk in 4 hours ? in a hours ? \ A 
 
 3. If a man walks 3 miles an hour, how many hours 
 will it take him to walk 12 miles ? x miles ? f 
 
 4. If a man walks x miles an hour, how many hours 
 
 will it take him to walk 20 miles ? y miles ? i> Jk 
 
 n \ 
 
 5. What is the divisor, if the dividend is 20 and the 
 
 quotient 5 ? if the dividend is a and the quotient b ? *\ -*< 
 
 6. What is the dividend, if the divisor is 4, the quotient 
 3, and the remainder 2 ? if the divisor is d, the quotient 
 q, and the remainder r ? \ ^ o*-fy *t ^ 
 
 7. What is the quotient, if the dividend is 22, the divi- 
 sor 4, and the remainder 2 ? if the dividend is p } the divi- 
 sor d, and the remainder r? $~ ^C^ ^ 
 
 8. What is the divisor, if the dividend is 22, the quo- 
 tient 4, and the remainder 2 ? if the dividend is p, the 
 quotient q, and the remainder r ? 5" j_ 
 
 9. If one part of 25 is 10, what is the other part ? 
 
SIMPLE EQUATIONS. 21 
 
 10. If one part of 30 is x, what is the other part ? 
 
 11. If one part of x is c, what is the other part ? '%^'(^ , 
 
 12. If the sum of two numbers is 40, and one of them is 
 25, what is the other ? ' T 
 
 13. If the sum of two numbers is x, and one of them is 
 5, what is the other ? -fi ^3" - >S 
 
 14. If the sum of two numbers is s, and one of them is 
 a, what is the other ? f ^ £. 
 
 15. If the difference of two numbers is 7, and the smaller 
 number is 13, what is the greater number ? *-0 
 
 16. If the difference of two numbers is a, and the smaller 
 number is x, what is the greater number ? °U t- pf 
 
 17. If the difference of two numbers is c, and the greater 
 number is x, what is the smaller number ? ^ - O 
 
 18. Henry is a years old to-day. How old was he 4 
 years ago ? How old will he be in 4 years ? a ~ ^ 
 
 19. John is x years old to-day. How old was he b years 
 ago ? How old will he be a years hence ? "%*~ at n ^ ^ 
 
 20. By how much does 5 x exceed 3 x ? "*- ~fi 
 
 21. By how much does x exceed a ? V s ^ 
 
 22. How much does a lack of being x ? V ~ 
 
 23. Write the excess of 2x -f 3 over a? + 1. 
 
 Note. If the number to be subtracted is a compound expression 
 it must be enclosed by a parenthesis. Thus, the excess of 2 x + 3 
 over x + 1 is 2 x + 3 — (x + 1). 
 
 24. Write the excess of 3a; over 4 (18 — x). 
 
 25. Write the excess of x — 50 over 80 — x. 
 
 26. What is the excess of 2 x — 24 over 80 — 5 a; ? 
 
 27. What is the excess of 5 a; -f- 24 over 60 — x? 
 
 28. Express in cents a half-dollars and b quarters. 
 
22 SIMPLE EQUATIONS. 
 
 29. Express in cents a dollars b dimes and c cents. 
 
 30. A man has a dollars. If lie spends b half-dollars 
 and c dimes, how many cents has he left ? JS"T> '{>-+- j&c) * "k 
 
 31. A man has x dollars y dimes and z cents. If he 
 spends a half-dollars and b quarters, how many cents has 
 he left? /^ +- /O^Z - (i«^ f ^U] -^^ 
 
 32. A man makes a journey of 236 miles. He travels a 
 miles by train, c miles by boat, and the remainder on foot. 
 How far does he go on foot ? %~ % $ (f — ^ ~ v 
 
 33. A train is running at the rate of a miles an hour. 
 How many miles will it travel in m hours ? c ^^^ 
 
 34. The floor of a square room measures a feet each 
 way. How many square yards q| oilcloth will b& required 
 to cover it? b oj- ^ " J> i %^" •<"* 
 
 35. The floor of a rectangular room measures a feet by 
 b feet. How many square yards of. oilcloth will be re- 
 quired to cover it ? S^l^ _2^- ■ J^ ' -~^T 
 
 36. A rectangular floor is a feet long and b feet wide. 
 In the middle of the floor there is a square carpet c feet on 
 a side. How many square yards of the floor are bare ? 
 
 60. In stating problems, x must not be put for money, 
 length, time, weight, etc., but for the required number of 
 specified units of money, length, time, weight, etc. 
 
 Each statement must be made in algebraic symbols, and 
 the meaning of each algebraic statement should be written 
 out in full, in common language. 
 
 After the algebraic statements are written, it is necessary 
 and sufficient, in problems involving only one unknown 
 number, to select two expressions that stand for the same 
 number, and to make them the members of the required 
 equation. (Ax. 5.) 
 
SIMPLE EQUATIONS. 23 
 
 Problems Stated and Solved. 
 
 1. Three times a certain number is equal to the number 
 increased by 20. Find the number. 
 
 Let x = the number. 
 
 Then 3 x = 3 times the number ; 
 
 and x + 20 = the number increased by 20. 
 
 But the last two expressions are equal. 
 Therefore, 3 x — x + 20. 
 
 Transposing, 3 x — x = 20. 
 
 Combining, . 2 x = 20. 
 
 Dividing by 2, x — 10. 
 
 2. John has three times as many oranges as James, and 
 they together have 32. How many has each ? 
 
 Let x stand for the number of oranges James has. 
 
 Then 3 x is the number of oranges John has ; 
 and x + 3 x is the number of oranges they together have. 
 But 32 is the number of oranges they together have. 
 Therefore, <c + 3x = 32. 
 
 Combining, 4ze = 32. 
 
 Dividing by 4, x = 8. 
 
 Multiplying by 3, 3 x = 24. 
 
 Therefore, James has 8 oranges, and John has 24 oranges. 
 
 Note. Beginners in stating the preceding problem generally write: 
 Let x = what James had. 
 
 Now, we know what James had. He had oranges, and we are to 
 discover simply the number of oranges he had. 
 
 3. James and John together have $24, and James has 
 $8 more than John. How many dollars has each ? 
 
 Let x stand for the number of dollars John has. 
 
 Then x + 8 is the number of dollars James has ; 
 
 and x + (x + 8) is the number of dollars they together have. 
 
 But 24 is the number of dollars they together have. 
 
24 SIMPLE EQUATIONS. 
 
 Therefore, x + (x + 8) = 24. 
 
 Kemoving the parenthesis, x + x + 8 = 24. 
 Combining, 2 x = 16. 
 
 Dividing by 2, x — 8. 
 
 Adding 8 to each side, x + 8 = 16. 
 
 Therefore, John has $8, and James has $16. 
 
 Note. The beginner must avoid the mistake of writing 
 Let x = John's money. 
 
 We are required to find the number of dollars John has, and there- 
 fore x must represent this required number. 
 
 4. The sum of two numbers is 18, and three times the 
 greater number exceeds four times the smaller by 5. Find 
 the numbers. 
 
 Let x = the greater number. 
 
 Then, since 18 is the sum, and x is one of the numbers, the other 
 number must be the sum minus x. Hence, 
 
 18 — x = the smaller number. 
 Now, three times the greater number is 3 x, and four times the less 
 number is 4 (18 — x); and 3x — 4 (18 — x) is equal to the excess of 
 three times the greater number over four times the smaller number. 
 But 5 = this excess. 
 
 .-. 3x-4(18-x) = 6. 
 .-. 3x — (72 — 4x) = 5, 
 or . 3x — 72 + 4x = 5. 
 
 .-. 7x = 77, 
 and x = 11. 
 
 Therefore, the numbers are 11 and 7. 
 
 5. A had three times as much money as B. He gave 
 B $10, and then had only twice as much as B. How much 
 had each at first ? 
 
 Let x = the number of dollars B had at first 
 Then 3 x = the number of dollars A had at first. 
 Now, 3 x — 10 is the number of dollars A had after giving $10 to 
 B. and x + 10 is the number of dollars B then had 
 
SIMPLE EQUATIONS. 25 
 
 Since A's money is twice B's, 
 
 3x — 10 = 2(x + 10) 
 3x — 10 = 2x + 20 
 3x-2x = 20 + 10 
 x = 30 
 3x = 90. 
 
 Therefore, B had $30 and A had $90. 
 
 6. A has $7 in half-dollars and quarters. If he has 24 
 coins in all, how many are halves and how many quarters ? 
 
 Let x = the number of halves. 
 
 Then 24 — x = the number of quarters. 
 
 50 x = the value in cents of the halves. 
 25 (24 — x) = the value in cetits of the quarters. 
 Therefore, 50 x + 25 (24 — x) = the value of his money in cents. 
 But 700 = the value of his money in cents. 
 
 Hence, 50 x + 25 (24 — x) = 700 
 50x + 600-25x = 700 
 25x = 100 
 x = 4 
 and 24 — x = 20. 
 
 Therefore, he has 4 half-dollars and 20 quarters. 
 
 7. A man is now twice as old as his son ; 15 years ago 
 he was three times as old. Find the age of each. 
 
 Let x = the number of years in the son's age. 
 
 Then 2 x = the number of years in the father's age. 
 
 x — 15 = the number of years in the son's age 15 years ago. 
 2 x — 15 = the number of years in the father's age 15 years ago. 
 But 15 years ago 3 times the son's age was equal to the father's age. 
 
 Therefore, 3 (x - 15) = 2 x - 16 
 
 3x-45 = 2x-15 
 x = 30 
 2x = 60. 
 
 Therefore, the son is 30 years old and the father 60 
 years oldo 
 
26 SIMPLE EQUATIONS. 
 
 Exercise 8. 
 
 1. If a number is multiplied by 9, the product is 810. 
 Find the number. 
 
 2. If the sum of the ages of a father and son is 56 years, 
 and the father is 7 times as old as the son, what is the age 
 of each ? 
 
 3. The sum of two numbers is 161, and the greater is 6 
 times the less. Find the numbers. 
 
 4. A tree 100 feet high was broken so that the part 
 broken off was 9 times the length of the part left standing. 
 Find the length of each part. 
 
 5. The difference of two numbers is 7, and their sum is 
 63. Find the numbers. 
 
 6. The difference of two numbers is 13, and their sum is 
 59. Find the numbers. 
 
 7. Divide 36 into two parts so that one part shall be 
 greater by 6 than the other part. 
 
 8. Three times a given number is equal to the number 
 increased by 36. Find the number. 
 
 9. Three times a given number diminished by 20 is 
 equal to the given number. Find the number. 
 
 10. One number is 4 times another, and their difference 
 is 24. Find the numbers. 
 
 11. The sum of two numbers is 48, and one of them 
 exceeds the other by 6. Find the numbers. 
 
 12. The sum of two numbers is 42, and 5 times the 
 smaller number is equal to the larger number. Find the 
 numbers. 
 
 13. Find three consecutive numbers, x, x + 1, and x + 2, 
 whose sum is 108. 
 
SIMPLE EQUATIONS. 27 
 
 14. Find five consecutive numbers whose sum is 70. 
 
 15. A man walks 4 miles an hour for x hours, and 
 another man walks 3 miles an hour for a; + 4 hours. If 
 they each walk the same distance, how many miles does 
 each walk? 
 
 16. A farmer employed two men to build 112 rods of 
 wall. One of them built on the average 4 rods a day, and 
 the other 3 rods a day. How many days did they work ? 
 
 17. Two men start from the same place and travel in 
 opposite directions, one 30 miles a day, and the other 20 
 miles a day. In how many days will they be 350 miles apart ? 
 
 18. Two men start from the same place and travel in the 
 same direction, one 30 miles a day, and the other 20 miles 
 a day. In how many days will they be 350 miles apart ? 
 
 19. A man bought 3 equal lots of hay for $255. For 
 the first lot he gave $17 a ton, for the second $16, for the 
 third $18. How many tons of each kind did he buy ? 
 
 20. A farmer sold a quantity of wood for $84, one half 
 of it at $3 a cord, and the other half at $4 a cord. How 
 many cords of each kind did he sell ? . 
 
 21. If 2x — S stands for 37, for what number will 
 7 + x stand ? 
 
 22. At an election two opposing candidates received 
 together 2000 votes, and one received 100 more votes than 
 the other. How many votes did each candidate receive ? 
 
 23. If a number is multiplied by 17, the product is 136. 
 Find the number. 
 
 24. The sum of two numbers is 54, and the greater is 
 seventeen times the smaller number. Find the numbers. 
 
 25. A tank holding 1500 gallons has three pipes. The 
 first lets in 8 gallons a minute, the second 10 gallons, and 
 
28 SIMPLE EQUATIONS. 
 
 the third 12 gallons a minute. In how many minutes will 
 the tank be filled ? 
 
 26. The fore and hind wheels of a carriage are 10 feet 
 and 12 feet respectively in circumference. How many 
 feet will the carriage have passed over when the fore wheel 
 has made 100 revolutions more than the hind wheel ? 
 
 27. Divide a yard of tape into two parts so that one part 
 shall be 6 inches longer than the other part. 
 
 28. Divide 23 into two parts such that the sum of twice 
 the greater and three times the smaller is 57. 
 
 29. Four times the smaller of two numbers is three times 
 the greater, and their sum is 63. Find the numbers. 
 
 30. A farmer sold a sheep, a cow, and a horse for $216. 
 He sold the cow for seven times as much as the sheep, and 
 the horse for four times as much as the cow. How much 
 did he get for each ? 
 
 31. Distribute $15 among Thomas, Richard, and Henry 
 so that Thomas and Richard shall each have twice as much 
 as Henry. 
 
 32. Three men, A, B, and C, pay $1000 taxes. B pays 
 four times as much as A, and C pays as much as A and B 
 together. How much does each pay ? 
 
 33. John's age is three times the age of James, and their 
 ages together are 16 years. What is the age of each ? 
 
 34. Twice a certain number increased by 8 is 40. Find 
 the number. 
 
 35. Three times a certain number is 46 more than the 
 number itself. Find the number. 
 
 36. One number is four times as large as another. If I 
 take the smaller from 12 and the greater from 21 the 
 remainders are equal. What are the numbers ? 
 
SIMPLE EQUATIONS. 29 
 
 37. Thirty yards of cloth and 20 yards of silk together 
 cost $70 ; and the silk costs twice as much per yard as 
 the cloth. How much does each cost per yard ? 
 
 38. In a company of 180 persons, composed of men, 
 women, and children, there are twice as many men as 
 women, and three times as many women as children. 
 How many are there of each ? 
 
 39. Two trains traveling, one at 25 and the other at 30 
 miles an hour, start at the same time from two places 330 
 miles apart, and move toward each other. In how many 
 hours will the trains meet ? 
 
 40. Twelve persons subscribed 'for a new boat, but two 
 being unable to pay, each of the others had to pay $4 more 
 than his share. Find the cost of the boat. 
 
 41. A tree 84 feet high was broken so that the part 
 broken off was five times the length of the part left stand- 
 ing. Required the length of each part. 
 
 42. At an election there were two candidates, and 2800 
 votes were cast. The successful candidate had a majority 
 of 160. How many votes were cast for each ? 
 
 43. Divide 20 into two parts such that four times the 
 greater exceeds three times the smaller by 17. 
 
 44. The sum of two numbers is 50, and seven times the 
 smaller number exceeds three times the greater number by 
 10. Find the numbers. 
 
 45. Divide 19 into two such parts that twice the smaller 
 part exceeds the greater by two. 
 
 46. Three times the excess of a certain number over 6 is 
 equal to the number plus 18. Find the number. 
 
 47. Thirty-one times a number exceeds 80 by as much as 
 nine times the number is less than 80. Find the number. 
 
30 SIMPLE EQUATIONS. 
 
 48. Find the number whose double diminished by 3 ex- 
 ceeds 80 by as much as the number itself is less than 100. 
 
 49. Divide 19 into two parts such that the greater part 
 exceeds twice the smaller part by 1 less than twice the 
 smaller part. , 
 
 50. A man is now twice as old as his son ; 20 years ago 
 he was four times as old as his son. Find the age of each. 
 
 51. A man is four times as old as his son ; in 20 years 
 he will be only twice as old. Find the age of each. 
 
 52. A man was four times as old as his son 7 years ago, 
 and will be only twice .as old as his son 7 years hence. 
 Find the age of each. 
 
 63. A man has 8 hours for an excursion. How far can 
 he ride into the country in a carriage that goes at the rate 
 of 9 miles an hour so as to return in time, walking back at 
 the rate of 3 miles an hour ? 
 
 54. A man was hired for 26 days. Every day he worked 
 he was to receive $3, and every day he was idle he was to 
 pay $1 for his board. At the end of the time he received 
 $62. How many days did he work ? 
 
 55. A, walking 4 miles an hour, starts two hours after B, 
 who walks 3 miles an hour. How many miles must A walk 
 to overtake B ? 
 
 56. A river runs 1 mile an hour. A man swims a certain 
 distance up the river in 3 hours, and the same distance down 
 in 1 hour. Find his rate of swimming in still water. 
 
 57. A man bought 12 yards of velvet. If he had bought 
 1 yard iess for the same money, each yard would have cost 
 $1 more. What did the velvet cost a yard ? 
 
 58. A and B have together $8; A and C, $10; B and 
 C, $12. How much has each ? 
 
SIMPLE EQUATIONS. 31 
 
 59. I have in mind a certain number. If this number is 
 diminished by 8 and the remainder multiplied by 8, the 
 result is the same as if the number were diminished by 6 
 and the remainder multiplied by 6. What is the number ? 
 
 60. A man having only ten-cent pieces and five-cent 
 pieces wished to give some children 15 cents each, but 
 found that he had not money enough by 25 cents; he, 
 therefore, gave them 10 cents each and had 30 cents left. 
 How many children were there ? 
 
 61.* A sum of money was divided among A, B, and C in 
 such a way that A received three times as much as B, and 
 B twice as much as C. If A received $6 more than C, how 
 much did each receive ? 
 
 62. The sum of the ages of a man and his son is 80 
 years ; and the father's age is 2 years more than twice the 
 age of his son. What is the age of each ? 
 
 63. Two casks contain equal quantities of vinegar. From 
 one cask 37 gallons are drawn, and from the other 7 gallons 
 are drawn. The quantity now remaining in one cask is 7 
 times that remaining in the other. How much did each 
 cask contain at first ? 
 
 64. A merchant has two kinds of tea; one worth 50 
 cents a pound, and the other 75 cents a pound. He makes 
 a mixture from these of 100 pounds, worth 60 cenfs a 
 pound. How many pounds of each kind does he take ? 
 
 65. A had $7 and B had $5. B gave A a certain sum ; 
 then A had 3 times as much as B. How many dollars did 
 B give A ? 
 
 66. A boy bought 9 dozen oranges for $2.50. For a part 
 he paid 25 cents a dozen, and for the remainder 30 cents a 
 dozen. How many dozen of each kind did he buy ? 
 
32 SIMPLE EQUATIONS. 
 
 Note. In the following examples express in cents all money values. 
 
 67. How can $2.25 be paid in quarters and ten-cent 
 pieces so as to pay twice as many ten-cent pieces as 
 quarters ? , 
 
 68. I have $1.80 in ten-cent pieces and five-cent pieces, 
 and have four times as many five-cent pieces as ten-cent 
 pieces. How many have I of each ? 
 
 69. I have $6 in silver half-dollars and quarters, and I 
 have 20 coins in all. How many have I of each ? 
 
 70. I have five times as many half-dollars as quarters, 
 and the half-dollars and quarters amount to $11. How 
 many have I of each ? 
 
 71. A man has $65 in ten-dollar bills and one-dollar 
 bills. He has three times as many one-dollar bills as ten- 
 dollar bills. How many bills has he of each kind ? 
 
 72. A sum of money is divided among three persons, 
 A, B, and C, in such a way that A and B together have 
 $6, A and C $6.50, and B and C $7.50. How much has 
 each ? 
 
 73. A purse contains 27 coins which amount to $11.25. 
 There is a certain number of silver dollars, and three times 
 as many half-dollars as dollars; the remaining coins are 
 quarters. Find the number of each. 
 
 74. A man bought 10 yards of calico and 20 yards of 
 cloth for $30.60. The cloth cost as many quarters per 
 yard as the calico cost cents per yard. Find the price of 
 each per yard. 
 
 75. A man has a certain number of dollars, half-dollars, 
 and quarters. The number of quarters is twice the number 
 of half-dollars and four times the number of dollars. If 
 he has $15, how many coins of each kind has he ? 
 
CHAPTER IH. 
 POSITIVE AND NEGATIVE NUMBERS. 
 
 61. Positive and Negative Quantities. If a person is en- 
 gaged in trade, his capital will be increased by his gains, 
 and diminished by his losses. 
 
 Increase in temperature is measured by the number of 
 degrees the mercury rises in a thermometer, and decrease 
 in temperature by the number of degrees the mercury falls. 
 
 In considering any quantity whatever, a quantity that 
 increases the quantity under consideration is called a posi- 
 tive quantity ; and a quantity that decreases the quantity 
 under consideration is called a negative quantity. 
 
 62. The Natural Series of Numbers. If from a given point, 
 marked 0, we draw a straight line to the right, and begin- 
 ning from the zero point lay off units of length on this line, 
 the successive repetitions of the unit will be expressed by 
 the natural series of numbers, 1, 2, 3, 4, etc. Thus : 
 
 1 
 
 L_ 
 
 2 
 
 1 
 
 3 
 
 1 
 
 4 
 
 1 
 
 5 
 
 1 
 
 6 
 
 1 
 
 7 
 1 
 
 8 
 
 1 
 
 9 10 11 
 
 1 1 1 
 
 In this series if we wish to add 2 to 5, we begin at 5, 
 count 2 units forwards, and arrive at 7. If we wish to 
 subtract 2 from 5, we begin at 5, count 2 units backwards, 
 and arrive at 3. If we wish to subtract 5 from 5, we count 
 5 units backwards from 5, and arrive at 0. If we wish to 
 subtract 5 from 2, we cannot do it, because when we have 
 counted backwards from 2 as far as 0, the natural series of 
 numbers comes to an end. 
 
34 POSITIVE AND NEGATIVE NUMBERS. 
 
 63. Positive and Negative Numbers. In order to subtract 
 a greater number from a smaller it is necessary to assume 
 a new series of numbers, beginning at zero and. extending 
 to the left of zero. The series to the left of zero must 
 proceed from zero by the repetitions of the unit, precisely 
 like the natural series to the right of zero ; and the oppo- 
 sition between the right-hand series and the left-hand series 
 must be clearly marked. This opposition is indicated by 
 calling every number in the right-hand series a positive 
 number, and prefixing to it, when written, the sign -f- ; and 
 by calling every number in the left-hand series a negative 
 number, and prefixing to it the sign — . The two series of 
 numbers may be called the algebraic series of numbers, and 
 written thus : 
 
 -4 -3-2-1 *0 +1 + 2 -f3 +4 
 
 I I I I I _J I J I 
 
 If, in this double series of numbers, we wish to subtract 
 4 from 2, we begin at 2 in the positive series, count 4 units 
 in the negative direction (to the left), and arrive at — 2 in 
 the negative series ; that is, 2 — 4 = — 2. 
 
 The result obtained by subtracting a greater number from 
 a less, when both are positive, is always a negative number. 
 
 In general, if a and b represent any two numbers of the 
 positive series, the expression a — b will be a positive num- 
 ber when a is greater than b ; will be zero when a is equal 
 to b ; will be a negative number when a is less than b. 
 
 In counting from left to right in the algebraic series, 
 numbers increase in magnitude ; in counting from right to 
 left, numbers decrease in magnitude. Thus, —3,-1, 0, 
 -f 2, + 4, are arranged in ascending order of magnitude. 
 
 64. The Absolute Value of a Number. The absolute value 
 of a number is its value independent of its sign. 
 
POSITIVE AND NEGATIVE NUMBERS. 35 
 
 65. Every algebraic number, as + 4 or — 4, consists of 
 a sign + or — and the absolute value of the number. The 
 sign shows whether the number belongs to the positive or 
 negative series of numbers ; the absolute value shows the 
 place the number has in the positive or negative series. 
 
 When no sign stands before a number, the sign + is 
 always understood. But the sign — is never omitted. 
 
 66. Two algebraic numbers that have, one the sign -f-, 
 and the other the sign — , are said to have unlike signs.' 
 
 Two algebraic numbers that have the same absolute 
 values, but unlike signs, cancel each other when combined. 
 Thus, + 4-4 = 0; + a - a = 0. 
 
 67. Double Meanings of the Signs + and — . The use of 
 
 the signs + and — to indicate addition and subtraction 
 must be carefully distinguished from the use of the signs 
 + and — to indicate in which series, the positive or the 
 negative, a given number belongs. In the first sense they 
 are signs of operations, and are common to Arithmetic and 
 Algebra ; in the second sense they are signs of opposition, 
 and are employed in Algebra alone. 
 
 Note. In Arithmetic, if the things counted are whole units, the 
 numbers that count them are called whole numbers, integral numbers, 
 or integers, the adjective being transferred from the things counted 
 to the numbers that count them. But if the things counted are only 
 parts of units, the numbers that count them are called fractional num- 
 bers, or simply fractions, the adjective being transferred from the 
 things counted to the numbers that count them. 
 
 Likewise in Algebra, if the units counted are negative, the numbers 
 that count them are called negative numbers, the adjective that defines 
 the nature of the units counted being transferred to the numbers that 
 count them. 
 
 68. Addition of Algebraic Numbers. An algebraic number 
 is often enclosed in a parenthesis, in order that the signs + 
 
36 POSITIVE AND NEGATIVE NUMBERS. 
 
 and — , which are used to distinguish positive and negative 
 numbers, may not be confounded with the + and — signs 
 that denote the operations of addition and subtraction. 
 
 Thus, + 4 + (— 3) expresses the sum, and + 4 — (— 3) expresses 
 the difference, of the numbers + 4 and — 3. 
 
 69. In order to add two algebraic numbers, we begin at 
 the place in the series which the first number occupies, and 
 count, in the direction indicated by the sign of the second 
 number, as many units as there are in the absolute value of 
 the second number. 
 
 -4 -3 -2 -1 +1 +2 +3 +4 
 
 I I I I I I I t [ 
 
 Thus, the sum of+2 + (+3)is found by counting from 
 + 2 three units in the positive direction; that is, to the right 
 and is, therefore, -j- 5. 
 
 The sum of + 2 + ( — 3) is found by counting from + 2 
 three units in the negative direction ; that is, to the left, and 
 is, therefore, — 1. 
 
 The sum of — 2 -f- (+ 3) is found by counting from — 2 
 three units in the positive direction, and is, therefore, + 1. 
 
 The sum of — 2 + (— 3) is found by counting from — 2 
 three units in the negative direction, and is, therefore, — 5. 
 
 70. If a and b represent any two numbers, we have 
 + a + (+ b) = a + b. -a + (+b) = -a + b. 
 + a + (-b) = a-b. -a + (-b) = -a-b. 
 
 Therefore, from these four cases, we have the following 
 Rule for Adding Two Algebraic Numbers: 
 
 1. If the numbers have like signs, find the sum of their 
 absolute values, and prefix the common sign to the result. 
 
 2. If the numbers have unlike signs, find the difference of 
 their absolute values, and prefix the sign of the greater num- 
 ber to the result. 
 
POSITIVE AND NEGATIVE NUMBERS. 37 
 
 71. The result is called the algebraic sum in distinction 
 from the arithmetical sum ; that is, the sum of the absolute 
 values of the numbers. 
 
 Note. If there are more than two numbers to be added, add two 
 of the numbers, and then this sum to a third number, and so on ; or 
 find the sum of the positive numbers and the sum of the negative 
 numbers, then the difference between the absolute values of these two 
 sums, and prefix the sign of the greater sum to the result. 
 
 Exercise 9. 
 Perform mentally the indicated additions : 
 1. 2. 3. 4. 5. 6. 
 
 + 8 
 
 -8 
 
 -7 
 
 -11 
 
 + 11 
 
 -11 
 
 -7 
 
 + 7 
 
 -7 
 
 + 4 
 
 - 4 
 
 - 4 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 
 12. 
 
 + 5 
 
 -8 
 
 -9 
 
 -20 
 
 + 87 
 
 -37 
 
 -7 
 
 -7 
 
 -7 
 
 + 24 
 
 -36 
 
 +40 
 
 -9 
 
 + 9 
 
 -6 
 
 + 36 
 
 -42 
 
 -20 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 17. 
 
 18. 
 
 + 15 
 
 -15 
 
 -21 
 
 -20 
 
 -18 
 
 + 17 
 
 + 12 
 
 -12 
 
 + 12 
 
 -30 
 
 + 32 
 
 -27 
 
 -20 
 
 + 30 
 
 -13 
 
 + 40 
 
 -12 
 
 -19 
 
 19. 
 
 20. 
 
 21. 
 
 22. 
 
 23. 
 
 24. 
 
 31 
 
 -31 
 
 81 
 
 19 
 
 -15 
 
 90 
 
 -17 
 
 -17 
 
 -71 
 
 - 7 
 
 -70 
 
 -30 
 
 -15 
 
 -15 
 
 31 
 
 13 
 
 -40 
 
 -20 
 
 21 
 
 21 
 
 -40 
 
 - 28 
 
 -90 
 
 10 
 
38 POSITIVE AND NEGATIVE NUMBERS. 
 
 Addition of Similar Monomials. 
 
 1. Find the sum of 3 a, 2 a, a, 5 a, 7 a. 
 
 The sum of the coefficients is 3 + 2 + 1+5 + 7 = 1 8. 
 Hence, the sum of the monomials is 18 a. 
 
 2. Find the sum of — 5 c, — c, — 3 c, — 4 c, —2 c. 
 
 The sum of the coefficients is — 5 — 1— 3 — 4 — 2 = — 15. 
 Hence, the sum of the monomials is — 15 c. 
 
 3. Find the sum of Sx, — 9x, — x, 3x, &x, — 12 x, x. 
 
 The sum of the positive coefficients is8 + 3 + 4 + l = 16. 
 The sum of the negative coefficients is — 9 — 1 — 12 = — 22. 
 The difference between 16 and 22 is 6, and the sign of the greater 
 is negative. Hence, the sum is — 6 x. Therefore, 
 
 72. To Find the Sum of Similar Monomials, 
 
 Find the algebraic sum of the coefficients, and annex to 
 this sum the letters common to the terms. 
 
 Exercise 10. 
 Perform mentally the indicated additions : 
 
 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 6. 
 
 la 
 
 7xy 
 
 3yz 
 
 2ab 
 
 13 c 
 
 3xz 
 
 2a 
 
 2xy 
 
 -9yz 
 
 -9ab 
 
 12 c 
 
 7xz 
 
 — 5a 
 
 -3xij 
 
 -2yz 
 
 -7ab 
 
 -24 c 
 
 — 9xz 
 
 — 2a 
 
 ±xy 
 
 Tyz 
 
 Sab 
 
 15 c 
 
 — xz 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 
 12. 
 
 12 « 4 
 
 4 m 2 
 
 11 abc 
 
 27 yV 
 
 17 x 9 
 
 18/ 
 
 - 7* 4 
 
 -9m 2 
 
 3abc 
 
 - 41 y*z* 
 
 -31a; 8 
 
 27 y* 
 
 - 8z* 
 
 -7m 2 
 
 — 7 abc 
 
 - 2yV 
 
 -47* 8 
 
 -43/ 
 
 - z* 
 
 5 m 2 
 
 — 9 abc 
 
 - y 4 * 4 
 
 61 z 8 
 
 - 21 ?/ 
 
POSITIVE AND NEGATIVE NUMBERS. 39 
 
 Exercise 11. 
 Eind the algebraic sum of : 
 
 1. 7 a, 2 a, — 3 a, — 5 a. 9. 3 yz, — 9 yz, 20 yz, 7 ys. 
 
 2. 7 ay, 2 ccy, — 4 #y, — 5 xy. 10. 2 a£, — 10 ab, — lab,3 ah. 
 
 3. 4 a 2 £, - 3 a 2 6, - 5 a 2 b. 11. 14 z 2 , 9 x 2 , - 8 x 2 , - 11 x\ 
 
 4. 3xy,4xy,7xy,-3xy. 12. 17 y s , - 8 y 8 , 5 y\ - f. 
 
 5. 16&, -11&, -25, 3&. 13. 12* 4 , -7z 4 , -Sz 4 , -9s 4 . 
 
 6. 13c, 12c, -24c, 2c. 14. 2\y 2 ,-\7y 2 ,-3f,-^f. 
 
 7. — 3fc*,7a», — 2a», — a*. 15. c 2 , - 2 c 2 , - 15 c 2 , - 18 c fl . 
 
 8. 2 ac, 5 ac, — 9 ac, 3 ac. 16. 4m 2 , — 11 m 2 , — 7m 2 , 5m 2 . 
 
 Express in one term each of the following : 
 
 17. 9z 2 -7a: 2 + 4a: 2 -3a 2 + 3a; 2 -5a; 2 . 
 
 18. 3 a 2 - 18 a 2 + a 2 - 5 a 2 + 6 a 2 - 10 a 2 . 
 
 19. 5 A + 7 a 8 z - 9 a*x - 29 a*u + 4 a 8 x. 
 
 20. - 5 a 2 6 2 + 7 a 2 b 2 + 11 a 2 * 2 - 4 a 2 * 2 - 9 a 2 £-. 
 
 21. - 21 ax s + 20 ar 5 - 6 aa 8 4- 5 az 8 - 13 ax 3 . 
 
 22. — 11 abcx + 3 a£c# — 7 a£ca; + 29 a6c# + obex. 
 
 23. - 3 y 4 z 4 - 27 yV - 2 yV + 41 #V 4 y 4 z 4 . 
 
 24. - 4 xY + 18 afy 5 4- 27 x 6 y* - 43 aV - x 6 if. 
 
 25. - 31 a&2 8 4- 17 abz* - 47 abz 3 4 61 a^ 8 + abz\ 
 
 26. ia + |a- T ^ + |a-fa + a. 
 
 27. |a 2 -5a 2 -f |a 2 + 7a 2 -f a 2 4-£a 2 . 
 
 28. xy-\xy + \xy-\xy + \xy — \\xy. 
 
 29. }«_|s_f«4-3 + }.|s_ js. 
 
 30. ^ ayz -\- ty- ays 4- ty ayz + *■£■ ayz — 7 ays. 
 
 31. 14 a - la - 3a- 15a + 12a. 
 
 32. 4 a 2 c - 10 a 2 c + 6 a 2 c - 9 a 2 c 4- a 2 c. 
 S3. 3aty-4afy + 2afy- a; V + 5a; y 
 
40 POSITIVE AND NEGATIVE NUMBERS. 
 
 Subtraction of Algebraic Numbers. 
 
 73. In order to subtract one algebraic number from 
 another, we begin at the place in the series which the 
 minuend occupies, and count, in the direction opposite to 
 that indicated by the sign of the subtrahend, as many units 
 as there are in the absolute value of the subtrahend. 
 
 -4 -3 -2 -1 +1 + 2 +3 +4 
 
 I I I ! ! I I I I 
 
 Thus, the result of subtracting + 1 from + 3 is found 
 by counting from + 3 one unit in the negative direction ; 
 that is, in the direction opposite to that indicated by the sign 
 + before 1, and is, therefore, + 2. 
 
 The result of subtracting — 1 from + 3 is found by 
 counting from + 3 one unit in the positive direction, and 
 is, therefore, -f 4. 
 
 The result of subtracting 4- 1 from — 3 is found by 
 counting from — 3 one unit in the negative direction, and 
 is, therefore, — 4. 
 
 The result of subtracting — 1 from — 3 is found by 
 counting from — 3 one unit in the positive direction, and 
 is, therefore, — 2. 
 
 If a and b represent any two numbers, we have 
 + a — (+b) = a-b. - a - (+b) = - a -b. 
 
 + a-\-b) = a + b. -a-\-b) = -a + b. 
 
 74. From these four eases we see that subtracting a posi- 
 tive number is equivalent to adding an equal negative num- 
 ber ; and that subtracting a negative number is equivalent 
 to adding an equal positive number. Therefore, 
 
 75. To Subtract One Algebraic Number from Another, 
 Change the sign of the subtrahend, and add the result td 
 
 the minuend. 
 
POSITIVE AND NEGATIVE NUMBERS. 41 
 
 Exercise 12. 
 Perform mentally the indicated subtractions : 
 
 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 6. 
 
 11 
 
 -11 
 
 -11 
 
 11 
 
 3 
 
 - 3 
 
 3 
 
 - 3 
 
 3 
 
 - 3 
 
 11 
 
 -11 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 
 12. 
 
 3 
 
 3 
 
 -3 
 
 3 
 
 -3 
 
 3 
 
 8 
 
 -8 
 
 -8 
 
 -9 
 
 -9 
 
 9 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 17. 
 
 18. 
 
 -8 
 
 -7 
 
 8 
 
 7 
 
 7 
 
 6 
 
 7 
 
 8 
 
 -7 
 
 8 
 
 -3 
 
 -4 
 
 Subtraction of Similar Monomials. 
 
 1. From 15m 2 x 2 take - 7 m 2 x 2 . 
 
 15 mW - (- 7 m 2 ^ 2 ) = 15 m 2 x 2 + 7 m^x 2 
 = 22 m 2 x 2 . Hence, 
 
 76. To Subtract a Monomial from a Similar Monomial, 
 
 Change the sign of the coefficient of the subtrahend ; then 
 add the coefficients, and annex the common letters to the result. 
 
 , 
 
 Exercise 13. 
 
 
 
 Perform mentally the indicated subtractions : 
 
 
 1. 2. 
 
 3. 
 
 4. 
 
 5. 
 
 165-^vwvw, 3yz 
 -115<^,-9y2 
 
 - 10 ab 
 
 - 6 a 2 
 
 -7m 2 
 
 - lab 
 
 -10 a 2 
 
 3 m 2 
 
 6. 7. 
 
 8. 
 
 9. 
 
 10. 
 
 -29 c -f41c 
 
 -29c 
 
 Sabe 
 
 — ac 2 x 
 
 — 41c 29c 
 
 -41c 
 
 — abc 
 
 Sac 2 x 
 
42 POSITIVE AND NEGATIVE NUMBERS. 
 
 If a = 4, 6 = — 2, c = — 3, find the value of: 
 
 11. a + (- b) + c. 14. - (- a ) + (- b) - (- c). 
 
 12. -a-(-b) + c. 15. + (-a)-(-&)-(-c). 
 
 13. a -b + (-c). 16. -(-«)-(-£) + (-/). 
 
 Multiplication of Algebraic Numbers. 
 
 77. Multiplication is generally defined in Arithmetic as 
 the process of finding the result when one number (the 
 multiplicand) is taken as many times as there are units in 
 another number (the multiplier). This definition fails 
 when the multiplier is a fraction. In multiplying by a 
 fraction, we divide the multiplicand into as many equal 
 parts as there are units in the denominator, and take as 
 many of these parts as there are units in the numerator. 
 
 If, for example, we multiply 6 by § , we divide 6 into 
 three equal parts and take two of these parts, obtaining 4 
 for the product. The multiplier, f , is § of 1, and the prod- 
 uct, 4, is £ of 6 ; that is, the product is obtained from the 
 multiplicand precisely as the multiplier is obtained from 1. 
 
 78. Multiplication may be defined, therefore, 
 
 As the process of obtaining the product from the multipli- 
 cand as the multiplier is obtained from unity. 
 
 79. Every extension of the meaning of a term must 
 be consistent with the sense previously attached to the 
 term, and with the general laws of numbers. 
 
 This extension of the meaning of multiplication is con- 
 sistent with the sense attached to multiplication when the 
 multiplier is a positive whole number. 
 
 Thus, 5X7 = 35, 
 
 the multiplier, 5, =1 + 1 + 1 + 1-1-1, 
 
 and the product, 35, =7 + 7 + 7 + 7 + 7. 
 
POSITIVE AND NEGATIVE NUMBERS. 43 
 
 Law of Signs in Multiplication. 
 By the definition of multiplication (§ 78), 
 since + 3 = + l + l + l, 
 
 3 x (+ 8) = + 8 + 8 + 8 = 4- 24, 
 and 3 X (- 8) = - 8 + (- 8) + (- 8) = - 24. 
 
 Again, since — 3 = — 1 — 1 — 1, 
 
 (_ 3) X 8 = - 8 - 8 - 8 = - 24, 
 and (- 3) X (- 8) = - (- 8) - (- 8) - (- 8) 
 = + 8 + 8 + 8= + 24. 
 
 The minus sign before the multiplier, 3, signifies that the 
 repetitions of the multiplicand are to be subtracted. 
 If a and b stand for any two numbers, we have 
 (+ a) X (+ b) = + ab, 
 (H- a) X (— b) = — ab, 
 (— a) X (+ b) = — ab, 
 , (- a) X (- b) = + ab. 
 That is, if two numbers have like signs, the product has the 
 plus sign ; if unlike signs, the product has the minus sign. 
 
 80. The Law of Signs in Multiplication is, therefore, 
 Like signs give +, and unlike signs give —. 
 
 The Index Law in Multiplication. 
 
 Since a 2 = aa, and a % = aaa, 
 
 a 2 X a z = aa X aaa = a b = a 2+i ; 
 a 4 X a = aaaa X a = a h = a 4+1 . 
 
 If then m and n are positive integers, 
 a m X a n = a m + n . 
 
 In like manner, a m X a M X a p = a w + ra+ * > . 
 
 81. The Index Law in Multiplication is, therefore, 
 
 The exponent of a letter in the product is equal to the sum 
 of the exponents of the letter in the factors of the product. 
 
44 POSITIVE AND NEGATIVE NUMBERS. 
 
 Multiplication of Monomials. 
 
 1. Find the product of 6a 2 b* and 7 ab 2 c*. 
 
 Since the order of the factors is immaterial, (§ 42) 
 
 6 a 2 6 8 x 7 abW = 6x7x a 2 xaxbzxWxc* 
 = 42 aWc*. 
 
 2. Find the product of — 3 ab and 7 ab 8 . 
 
 — Sab X7 ab z = — 3x7 XaXaXbxb* ** 
 
 = - 21 a 2 6*. 
 
 3. Find the product of # n and x 8 , and of cc n and x n . 
 
 x» x x 8 — x n + 3 . 
 x n X x n = x n + n = x 2n . Therefore, 
 
 82. To Find the Product of Two Monomials, 
 
 Find the product of the numerical coefficients ; and to 
 this product annex the letters, giving to each letter an ex- 
 ponent equal to the sum of its exponents in the factors. 
 
 83. A product of three or more factors is called the con- 
 tinued product of the factors. 
 
 1. Find the continued product of (— a)x(— b)X(— c). 
 
 By the law of signs, § 80, we have 
 
 (- a) x (- b) = ab, 
 and (ab) X (— c) = — abc. 
 
 C 
 
 2. Find the continued product of 
 
 (-a) X(-b)X (-c) X (-«*)• 
 By the law of signs, (—a) x (— 6) = ab, 
 
 (ab) X (— c) = — abc, 
 (— abc) x (— d) = abed. 
 
 84. From Examples 1 and 2 (§ 83), we see that an odd 
 number of negative factors gives a negative product; and an 
 even number of negative factors gives a positive product. 
 
POSITIVE AND NEGATIVE NUMBERS. 45 
 
 Exercise 14. 
 
 Note. The beginner should first write the sign of the product; 
 then the product of the numerical coefficients after the sign ; and, 
 lastly, the letters in alphabetical order, giving to each letter the 
 proper exponent. 
 
 Find mentally the product of : 
 
 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 6. 
 
 3a 
 
 -3a 
 
 3a 
 
 ~3a 
 
 9 a 2 
 
 -9a 2 
 
 2a 
 
 -2a 
 
 -2a 
 
 2a 
 
 6a 
 
 6 a 2 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 
 12. 
 
 5x 2 
 
 -7 a* 
 
 8 a 8 
 
 -7m 2 
 
 x 2 y 2 
 
 - a W 
 
 lx h 
 
 — a 
 
 -7 a 8 
 
 8 m 4 
 
 — 5 x 2 y 2 
 
 a 2 bc - 
 
 
 
 dr^j-W** 
 
 
 13. 
 
 14. 
 
 15. 
 
 16. 
 
 17. 
 
 18. 
 
 -3a 2 x 
 
 5<zc 3 
 
 7x 5 
 
 -9b 2 
 
 -a*b 
 
 9a m 
 
 -3 ax 2 
 
 6a 2 c 
 
 -3z 7 
 
 -6a 2 
 2 &V* 
 
 -ah* 
 
 6a» 
 
 19. 
 
 20. 
 
 21. 
 
 22. 
 
 23. 
 
 24. 
 
 a m+1 
 
 — x n 
 
 z n+s 
 
 a" 
 
 2/" 
 
 
 <x m_1 
 
 2x m ~ 
 
 z n-2 
 
 x 2 y, xhf, 
 
 a y n+2 
 and — 15 abxy. ^ 3.0 It 
 
 -q^'T* 3 ' 
 
 - z-x 
 
 25. -2 
 
 
 
 26. -8 
 
 a 2 , -2 b 2 , 
 
 — 3 ab, and a 2 b 2 . 
 
 
 27. ab, — ac, — be, and — 3 abe. 
 
 28. — xyz 2 , x 2 yz, — xy 2 z, and 3 xyz. 
 
 29. 4 a 3 , -10a 2 6, 25 a& 2 , and - ab\ 
 
 30. 6 2 , 6a& 2 , -±a 2 b, and -2a& 2 . 
 
 31. a 8n , a 2 "" 1 , a 2 * +1 , a 1 " 2 ", and a. 
 
 32. x m+1 , x m -' z , x m , x m+n , and x m ~ n . 
 
46 POSITIVE AND NEGATIVE NUMBERS. 
 
 Division of Algebraic Numbers. 
 
 85. Division is the operation of finding one of two 
 factors, when their product and the other factor are given. 
 
 86. With reference to this operation the product is called 
 the dividend, the given factor the divisor, and the required 
 factor the quotient. 
 
 Law of Signs in Division. 
 Since (+ a) X (+ b) = + ab, .*. + ab -r- (+ a) = + b. 
 Since (+ a) X (— b) = — ab, .'. — ab -i- (+ a) = — b. 
 Since (— a) X (-f- b) = — ab, .*. — ab •*- (— a) — + b. 
 Since (— a) X (— b) = + ab, .'. + ab -5- (— a) = — b. 
 
 That is, if the dividend and divisor have like signs, the 
 quotient has the plus sign ; and if they have unlike signs, 
 the quotient has the minus sign. 
 
 87. The Law of Signs in Division is, therefore, 
 Like signs give +, and unlike signs give — . 
 
 The Index Law in Division. 
 
 The quotient contains the factors of the dividend that 
 are not found in the divisor. 
 
 1. Divide a 5 by a 2 . 2. Divide a A by a. 
 
 • a 6 aaaaa ft . „ 
 
 1. — = = aaa = a 9 = a b ~ . 
 
 a* aa 
 
 a* aaaa 
 
 J- 
 
 2. — = = aaa 
 
 a a 
 
 If m and n are positive integers, and m is greater than n, 
 a m -T- a n = a m ~ n . 
 
POSITIVE AND NEGATIVE NUMBEBS. 47 
 
 88. The Index Law in Division is, therefore, 
 
 The exponent of a letter in the quotient is equal to the 
 
 exponent of the letter in the dividend minus the exponent of 
 
 the letter in the divisor. 
 
 Division of Monomials. 
 
 1. Divide 24 a 7 by 8 a 6 . 
 
 8 a 6 
 We obtain the factor 3 of the quotient by dividing 24 by 8 ; and the 
 factor a 2 of the quotient, by writing a with an exponent equal to the 
 exponent of a in the dividend minus the exponent of a in the divisor. 
 
 2. Divide 20 a 5 b 6 by - 4 ab\ 
 
 20a»P _ , ... , ... 
 
 — 7—7; = — 5a 5 ~ 1 6 6 - 6 = — 5a 4 &. 
 
 3. Divide -30aW by ~20abc\ 
 
 -30aWe* 
 
 f aWc. 
 
 — 20 abc* 
 
 4. Divide - 57 a*" 1 by — 19 a*- 3 . 
 
 — -Lz — - = 3^-1-^-3) = 3a *-i-*+3 = 3a a # 
 — 19 a x ~ 3 
 
 5! Divide 77 a? m b n <f- by 11 a m £ B c 8 . 
 
 77 a 2 " 1 ^* 
 
 11 fl^C 3 
 
 = 7 a 2m - m &»— n c*- 3 = 7 o^c*— 8 . 
 
 6 n l> n 
 
 Note. Since by division — = 1 ; and by the index law — = &°, it 
 
 follows that 6° as 1. Hence, any letter which by the rule would appear 
 in the quotient with zero for an exponent, may be omitted without affect- 
 ing the quotient. 
 
 89. To Find the Quotient of Two Monomials, therefore, 
 Divide the numerical coefficient of the dividend by the 
 numerical coefficient of the divisor ; and to the result annex 
 the letters, giving to each letter an exponent equal to its 
 exponent in the dividend minus its exponent in the divisor. 
 
48 POSITIVE AND NEGATIVE NUMBEBS. 
 
 Exercise 15. 
 Perform mentally the indicated divisions : 
 
 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 12 ab* x i* 
 6ab 
 
 16 aV . 
 8ae 
 
 20 ay ^.x 
 oxy u 
 
 14 a?y £ 8 
 7 ay« 8 
 
 54 a*ic 8 c 4 
 6 a 8 ic 2 c 
 
 6. 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 - 20 cy 
 
 - 27 a 4 z 2 
 
 — 56 a A x A 
 -8a*x 
 
 63 ^z 8 
 
 -7^ 2 
 
 72 x*yz 2 
 
 5c 2 y 
 
 -9 a 3 
 
 12 xy« 
 
 11. 
 
 12. 
 
 13. 
 
 14. 
 
 15. 
 
 - 9 a 2 b V 
 
 - 27 ab 2 x s 
 Sabx 2 
 
 - 3 xY z K 
 
 -2cy 
 
 -18x n 
 
 — ab 2 c 
 
 x 2 y 2 z z 
 
 -4a;"- s 
 
 16. 
 
 17. 
 
 18. 
 
 19. 
 
 20. 
 
 56 a 2 xY 
 -lx*tf 
 
 - 3 ab V 
 - a6V 
 
 - 2 a 2 6 2 z' 
 ate 2 
 
 1 4aV 
 
 -2«y 
 
 6aV 
 -2aar* 
 
 21. 
 
 22. 
 
 23. 
 
 24. 
 
 25. 
 
 -a 5 x 9 
 -a*x* 
 
 26. 
 
 -51aV 
 17 ay 
 
 27. 
 
 28 c»d 4 
 -Itfd 2 
 
 28. 
 
 16 afc 2 * 3 
 — 4 a# 2 a: 
 
 29. 
 
 12 icVV 
 - 3'xy 2 z* 
 
 30. 
 
 7 a 2 fo n 
 7 afo"- 1 
 
 3a n + 1 : 
 
 L2 £c n i/ n 54 
 6x 2 y 
 
 a"" 2 ?/"- 1 
 
 -9«y 
 
 xY 
 
 x n ~ 2 y n ~ l 
 
 31. 
 
 32. 
 
 33. 
 
 34. 
 
 35. 
 
 -6cy 
 
 36. 
 
 4 a A b A 
 2a 2 b 2 
 
 37. 
 
 -3a 4 & 
 38. 
 
 2y*z 
 39. 
 
 — i/ r ^ 8 
 40. 
 
 -8 C y 
 
 10 aV 
 5aV 
 
 2- 
 
 12 a l2 lf 
 1* 
 
 3ay° 
 — xY 
 
 - x'V 
 
 -4cy 
 
 t/i 1 
 
CHAPTER IV. 
 ADDITION AND SUBTRACTION. 
 
 Integral Compound Expressions. 
 
 90. If an algebraic expression contains no letter in the 
 denominator of any of its terms, it is called an integral 
 expression. Thus, x z -f- 7 ex 2 — c 8 — 5 c 2 x, -J- ax — £ bey, are 
 integral expressions. 
 
 An integral expression may have for some values of the letters a 
 fractional value, and a fractional expression an integral value. If, 
 for instance, a stands for £ and 6 for £, the integral expression 
 
 2 a — 5 b stands f or £ — f = \ ; and the fractional expression — stands 
 
 for ^ -r £ = 5. Integral and fractional expressions, therefore, are so 
 named on account of the form of the expressions, and with no refer- 
 ence whatever to the numerical value of the expressions when definite 
 numbers are put in place of the letters. 
 
 Addition of Integral Compound Expressions. 
 
 91. The addition of two compound algebraic expressions 
 can be represented by connecting the second expression 
 with the first by the sign +. If there are no like terms in 
 the two expressions, the operation is algebraically complete 
 when the two expressions are thus connected (§ 11, Note). 
 
 If, for example, it is required to add m -+- n — p to 
 a + b + c, the result will be a -f- b + c ■+- (m + n — p) ; or, 
 removing the parenthesis (§ 39), a + b-\-c + m + n— p. 
 
50 ADDITION AND SUBTRACTION. 
 
 92. If, however, there are like terms in the expressions, 
 every set of like terms can be replaced by a single term 
 with a coefficient equal to the algebraic sum of the coeffi- 
 cients of the like terms. 
 
 1. Add 5 a 2 + ±a + 3 to 2 a 2 -3a- ±. 
 
 2a 2 -3a-4-f(5a 2 + 4a + 3) 
 
 = 2a 2 -3a-4: + 5a 2 + ±a + 3 (§39) 
 
 = 2a 2 + 5a 2 -3a + 4a-4 + 3 (§38) 
 
 = 7 a 2 + a — 1. 
 
 This process is more conveniently represented by arrang- 
 ing the terms in columns, so that like terms shall stand in 
 the same column, as follows : 
 
 2 a 2 - 3 a - 4 
 5 a 2 + 4 a -f 3 
 7 a 2 + a - 1 
 
 The coefficient of a 2 in the result will be 5 + 2, or 7 ; the coeffi- 
 cient of a will be — 3 + 4, or 1 ; the last term will be — 4 + 3, or — 1. 
 
 2. Ad&2a 3 -3a 2 b + ±ab 2 + b 3 ) a 3 + ±a 2 b - lab 2 - 2b 3 ; 
 3a? + a 2 b-3 ab 2 - 4 b 3 ; and 2 a s + 2 a 2 b -f- 6 a£ 2 - 3 b 3 . 
 
 2a s -3a 2 b + 4,ab 2 + b 3 
 a* + ±a 2 b-l ab 2 -2 b 3 
 3a 3 + a 2 b-3ab 2 -4b* 
 2a 3 + 2a 2 b + 6ab 2 -3b 3 
 8a 3 + 4:a 2 b -8 b 9 
 
 The coefficient of a 8 in the result will be 2 + 1 + 3 + 2, or + 8 ; the 
 coefficient of a 2 6 will be — 3 + 4 + 1 + 2, or +4; the coefficient of 
 ab 2 will be 4 — 7 — 3 + 6, or 0, and, therefore, the term ab 2 will not 
 appear in the result (§ 21) ; and the coefficient of 6 s will be 1 — 2 — 4 
 — 3, or — 8. 
 
ADDITION AND SUBTRACTION. 51 
 
 Exercise 16. 
 
 1. a -+- b ; a — b. 
 
 Add: 
 
 3. 5z 2 + 6z-2; 3z 2 -7a;4-2. #* -^ 
 
 4. 3a; 2 -2^?/ + 2/ 2 ; a; a - 2 a^ 4- 3 y*. ^Y^- ^ X ^ *" "^ 
 
 5. ax 2 + ^-4; 3az 2 -2kc44; - 4az 2 - 2fo 4- 5. -Z'h'T 
 
 6. 5x + 3y + z-, 3x + 2y + 3z; x-3y-5z. 
 
 7. - 3 a£ - 2 az 2 + 3 A + x 3 ; - 4 ab - 6 A + 5 ax 2 ; 
 
 # 3 — ab -f- a 2 # — ace 2 ; ax 2 + 8ab — 5 a 2 #. 
 
 8. a 4 - 2 a 8 + 3 a 2 - a + 7; 2 a 4 - 3 a 8 + 2 a 2 - a + 6; 
 
 a 4 -2a 3 + 2a s -5. 
 
 9. 3a 2 -a0 + ac-3& 2 + 4fo-s 2 ; - Abe + 5c 2 + 2 ao; 
 
 5 a 2 - a& - ac + 5 be ; - 4 a 2 + M - 5 be 4 - 2 c 2 . 
 
 10. x 4 -3z 3 + 2z 2 -4;c + 7; 3x* + 2x 3 4- a 2 - 5a - 6 ; 
 
 4;z 4 + 3z 8 -3z 2 + 9z-2; 2z 4 - X s + a 2 - a + 1. 
 
 11. 72/ 3 -3V-4^V + ^ 3 ; -Sar'-llz?/ 2 -^^ 2 -?/ 8 ; 
 
 x*y — xz 2 — y 2 — Bxy 2 ; — &xz 2 + y 2 —z 3 -\-6xy 2 + 10 a; 3 . 
 
 12. a 4 - 2 a 3 + 3 a 2 - 3 a - 2 ; a 3 + a 8 + a - 3 a* + 3 ; 
 
 4 a 4 + 5 a 8 ; 2a 2 +3a-2; -a 2 - 2 a - 3. 
 
 13. x 3 + 2xy 2 - x 2 y - y 3 ; 2x 3 - 3x 2 y - 4^ - 7^; 
 
 x 3 - 8 V - 7 s/ 8 . 
 
 14. c 4 - 3 c 3 + 2 c 2 - 4 c + 7 ; 2 c 4 + 3 c 3 4- 2 c 2 + 5 c 4- 6 ; 
 
 c 8 -4c 4 -4c 2 -5. 
 
 15. 3 x 2 — xy + #2 — 3 y 2 — £ 2 ; — 5 x 2 — ay — xz + 5 2/2 ; 
 
 tf + 3yz 4-3* 2 ; 6x 2 - 6y - 6 z + ±xz-, ±yz-5xz. 
 
 16. m* — 3 m 4 n — 6 ?^ 8 ti 2 ; m 8 w 2 + raV* — 5 w 4 w - 3» 5 ; 
 
 2mH4m¥-3mw 4 -w 5 ; -2mV-3mw 4 + n 4 ; 
 - m* 4- 2 m?i 4 4- 2 rc 5 + 3 m*n. 
 
 17. 6y-l-2a: 2 y; 5 + 2xy* - ±x 2 y; -afy-5 + 6a;y*i 
 
 2 + x^-^j x 2 i/-2V-5/ + l. 
 
52 ADDITION AND SUBTRACTION. 
 
 Subtraction of Compound Expressions. 
 
 93. The subtraction of one expression from another, if 
 none of the terms are alike, can be represented only by 
 connecting the subtrahend with the minuend by means of 
 the sign — . 
 
 If, for example, it is required to subtract a -\-b + c from 
 m, + n — p, the result will be represented by 
 
 m + n— p — (a + b -\- c); 
 
 or, removing the parenthesis (§ 40), 
 
 m -\- n — p — a — b — c. 
 
 If, however, some of the terms in the two expressions 
 are alike, we can replace like terms by a single term : 
 
 1. Subtract a 8 - 2a 2 -\- 2a -1 from 3 a* - 2 a 2 + a - 2 ; 
 the result may be expressed as follows : 
 
 3 a 8 - 2 a 2 + a - 2 - (a 8 .- 2 a 2 + 2 a - 1) ; 
 or, removing the parenthesis (§ 40), 
 
 3a 8 -2a 2 + a-2-a 8 + 2a 2 -2a + l 
 
 = 3a 8 -a 3 -2a* + 2a 2 + a-2a-2 + l (§38) 
 = 2**-a-l. 
 
 This process is more easily performed by writing the 
 subtrahend below the minuend, mentally changing the sign 
 of each term in the subtrahend, and adding the two expres- 
 sions. Thus, the above example may be written 
 3a 8 -2a 2 + a -2~-^> 
 
 a s -2a 2 + 2a-l ^ % ' 
 2 a 8 - a - 1 
 
 The coefficient of a 8 will be 3 — 1, or 2 ; the coefficient of a 2 will 
 be — 2 + 2, or 0, and therefore the term containing a 2 will not appear 
 in the result ; the coefficient of a will be 1 — 2, or — 1 ; the last term 
 will be — 2 + 1, or — 1. 
 
ADDITION AND SUBTRACTION. 53 
 
 2. Subtract a 5 4 4 a 8 a 2 - 3 A 8 - 4 ax 4 
 
 from a 8 # 2 + 2 a 2 x 3 — 4 owe 4 . 
 Here terms that are alike can be written in columns : 
 A 2 + 2 a 2 x 3 - 4 aa; 4 
 a 5 + 4 A 2 - 3 cV - 4 as 4 
 - a 6 - 3 A 2 + 5 a V 
 
 There is no term a 5 in the minuend, hence the coefficient of a 6 in 
 the result will be — 1, or — 1 ; the coefficient of a s x 2 will be 1 — 4, or 
 — 3 ; the coefficient of a?x z will be 2 + 3, or + 5 ; the coefficient of 
 ax 4 will be — 4 + 4, or 0, and ax 4 will not appear in the result. 
 
 Exercise 17. 
 
 1. From 8a — 45 — 2c take 2a — 3b — 3c. 
 
 2. From 3a-4b + 3c take 2a-Sb-c-d. 
 
 3. From 7 a 2 - 9 x - 1 take 5 a 2 - 6 x - 3. 
 
 4. From 2 a? 2 — 2 ace + a 2 take x 2 — ax — a 2 . 
 
 5. From 4 a — 3 b — 3c take 2a — 3b + ±c. 
 
 6. From5z 2 4-7;z + 4 take 3z 2 - 7x 4 2. 
 
 7. From 2ax + 3by + 5 take 3 a« — 3 by — 5. 
 
 8. From 4 a 2 - 6 a& + 2 6 2 take 3a 2 -\-ab-\- b\ 
 
 9. From 4 a 2 6 + 7 a& 2 + 9 take 8-3 aft 2 . 
 
 10. From 5 a 2 c + 6 a 2 b -8 a 8 take b 3 4 6 a 2 & - 5 a 2 c. 
 
 11. From a 2 - b 2 take S 2 . 13. From b 2 take a 2 - b\ 
 
 12. From a 2 - b 2 take a 2 . 14. From a 2 take a 2 - b 2 . 
 15. From a 4 + 3 ax 3 - 2 foe 2 + 3 ex - 4 d 
 
 take 3 cc 4 4- ax 3 — 4: bx 2 + 6 ex + d. 
 
 IfA = 3a 2 -2ab + 5b\ C = 7 a 2 - 8 ab 4 5 ft*, 
 
 ^ = 9 a 2 - 5 oft 4- 3 b 2 , D = 11 a 2 - 3 ab- 4b 2 , 
 find the expression for : 
 
 1Q. A+C + B-D. 19. A+C- B- D. 
 
 11. A- C-B + D. 20. A- C + B + D. 
 
 18. C- A- B + D. 21. ^ 4 C - B 4- 2>. 
 
54 ADDITION AND SUBTRACTION 
 
 Insertion of Parentheses. 
 We hare the following equivalent expressions : 
 
 a -f- (b + c) = a + b + o, .'.a + Hc = fl + (H«)l 
 
 a + (b — c) = a + ^ — c, .*. a + & — c — a + (£ — c) ; 
 
 a — (b -f c) = a — b — c, .'. a — b — c = a — (b + c); 
 
 a — (b — c) = a — b -f- c, .'.a — b -{- c = a — (b — c). 
 
 94. Hence, a parenthesis preceded by the sign + may 
 not only be removed without changing the sign of any term, 
 but may also be inserted, enclosing any number of terms, 
 without changing the sign of any term. 
 
 And a parenthesis preceded by the sign — may not only 
 be removed, provided the sign of every term within the 
 parenthesis is changed, namely, + to — and — to -+-, but 
 may also be inserted, enclosing any number of terms, pro- 
 vided the sign of every term enclosed is changed. 
 
 95. Expressions may occur having parentheses within 
 parentheses. In such cases signs of aggregation of dif- 
 ferent shapes are used, and the beginner, when he meets 
 with one branch of a parenthesis (, or bracket [, or brace 
 \ , must look carefully for the other branch, whatever may 
 intervene; and all that is included between the twc 
 branches must be treated as the + or — sign before the 
 sign of aggregation directs. It is best to remove each paren- 
 thesis in succession, beginning with the innermost. Thus, 
 
 1. a - \b - (c - d) + e] 
 = a-[J-c + rf-fe] 
 = a — b + c — d — e. 
 
 = a-\b-[c-d + e+f]\ 
 = a—\b — c + d — e—f\ 
 = a — b + c — d + e +/. 
 
ADDITION AND SUBTRACTION. 
 
 55 
 
 Exercise 18. 
 
 Simplify the following by removing the parentheses and 
 combining like terms : 
 
 1. a-b-\_a-(b-c)-c]. OR-^/^ ' ^ 
 
 2. m — \n — (p — w)]. 
 
 3. 2x-jy + [4*-0 + 2z)]g. 
 
 4. 3a-{2d-[5c-(3a + 5)]j. 
 
 5. a—\b + [c — (d-b) + (i]-2b\. 
 
 6. 3aB-[9-(2oj + 7) + 3a;]. 
 
 7. 2x-[ij-(x-2 1/ )l 
 
 8. a -[2o 4- (3c -2 5) 4- a]. 
 
 9. (a — x + 2/) — (b — a? — y) + (a + 6 — 2 y). 
 
 10. 3 a - [- 4 & f (4 a - b) - (2 a - 5 tjj 
 
 11. 4 - [a - (2 ft - 3 c) + c] + [a - (2 & - 5 c - a)]. 
 
 12. « + (y- *) - [(3* -2y) + *]+[>-<? - 2«)]. 
 
 13. a-[2a+(a-2a)4 2<7,]-5a-J6a-[(a+2a)-a]5 
 
 14. 2 a - (3 y + s) - {ft - (c - b) 4- e - [a - (e - ft)jjf. 
 
 15. a — lb -\-c 
 
 (a + J)-c] + (2a-Hc). 
 
 Note. Remember that the sign — which is written in the last 
 problem before the first t erm 6 under the vinculum is really the sign 
 of the vinculum, — b 4 c meaning the same as — (b 4 c). 
 
 lx — (x — 5 — *)]}< 
 
 16. 10-ic- j- 
 
 17. 2x-\2x + (y — «)— 3* + [2oj— (y — «— 2y)— 3«]4-4yJ. 
 
 18. a - {ft - [- c + a - (a - J) - »]} 4- [2 a - (J - a)]. 
 
 19. a — j & — [a — (c — b) + c — a - 
 
 20. 5a-j-3a-[3a-(2a-a 
 
 21. 20 - a- \7 a - [8 a - (9 a- 3 -6a)]J. 
 
 22. a- j5y-[>-(3y-2«) + «-(a;-2y-*)]} 
 
 (a — 6 — c) — a] 4- ftf. 
 !)-;«] 4r#}. 
 
56 ADDITION AND SUBTRACTION. 
 
 Exercise 19. 
 
 In each of the following expressions enclose the last three 
 terms i*. a parenthesis preceded by the sign — , remembering 
 that the sign of each term enclosed must be changed : 
 
 1. 2a-b + 3c-k + 3e + 5f) 
 
 2. x — a — y —lb -\- z -f c J 
 
 3. a + b-(c^±afbTl) 
 
 4. ax ,-f- by -f cz + ibx —\cy -f- cz\ 
 
 5. 3a + 2b + 2c-[&d\- 3e + 4/>) 
 
 6. x — y -f z — [5 xy -\- 4 xz 4^3 ?/«.) 
 
 Considering all the factors that precede x, y, and z, respectively, 
 as the coefficients of these letters, we may collect in parentheses the 
 coefficients of se, y, and z in the following expression : 
 
 ax — by + ay — az — cz + bx = ax + bx + ay — by — az — cz 
 
 = (a + b)x + (a — 6)y — (a + c) z. 
 
 In like manner, collect the coefficients of x, y, and z in 
 the following expressions : 
 
 7. ax -\r by -\- cz -\- bx — cy -f- az. 
 
 8. ax + 2 ay + 4 az — bx -f 3 y — 3 bz — 2 z. 
 
 9. ax — 2 by — 5 cz — 4 £x + 3 cy — 7 az. 
 
 10. ax + 3 ay + 2 by — bz — 11 ex + 2 cy — cz. 
 
 11. 4 £y — 3 ax — 6 cz + 2 foe — 7 ox — 5 cy — x — y — z. 
 
 12. 6 az — 5 by + 3 cz — 2 5z — 3 ay + z — a + y. 
 
 13. z — 6y + 3 az — 3 cy -f 2 ax — 2 HUB — 5 &z. 
 
 14. x + ay — az — acx -f- &cz — mny — y — z. 
 
 15. 2 ax — 6 ay + 46z — 4 foe — 2 ex — 3cy. 
 
 16. ax-foc + 2ay + 3y + 4az-3£z-2z. 
 
 17. ax — 2 £y -f 5 cz — 4 foe — 3 cy + az — 2 ex — ay + 4 £z. 
 
 18. 12^x + 12 j/y + 4*y - 12 5z - l£^cx 4- 6<jy + 3cz. 
 
ADDITION AND SUBTRACTION. 57 
 
 Exercise 20. — Eeview. 
 
 1. Add 4a 8 - 5a 2 - 5a* 2 + 6a 2 *; 6 a 3 + 3* 8 -f 4az 2 ; 
 19ax 2 — llx 2 — 15a 2 x; 10 * 3 + 7 a 2 * + 5 a 8 — 18 ax*. 
 
 2. Add 3a6 + 3a + 66 — 46 2 ; ab + 2a + 45 + 9b 2 ; 
 7a& — 4a — 86 + 13a 2 ; 6a + 12£ — 2 a£ — 11 a 2 . 
 
 Note. Similar compound expressions are added in precisely the 
 same way as simple expressions, by finding the sum of their coeffi- 
 cients. Thus, 3 (x — y) + 5 (x — y) — 2 (x — y) = 6 (x — y). 
 
 3. Add4(5-*); 6(5 -«)j 3 (5 - x) ; -12(5 -a); 
 ,2(5-*); -9(5-*). (/(J^ ' 
 
 4. Add (a -f 5)* 2 + (b + c)y 2 + (a + c)z 2 -, (b + c)* 2 + 
 (a + c)y 2 +(a + £)z 2 ; (a + c)x 2 +(a + &)y 2 -f-(& + c )* 2 . 
 
 5. Add (a + b)x -f- (b + c) y -f (c + a)s; (5 + c)z + 
 (c + a)x —(a + b)y ) (a + c) y + (a + b) z — (b + c)x. 
 
 6. From a 3 — x 2 take a 3 + 2 a* -f- **• 
 
 7. From 3 a 2 + 2 ax -f- * 2 take a 2 — a* — a; 3 . 
 
 8. From8* 2 -3a* + 5 take 5 * 2 + 2 a* -f 5. 
 
 9. From a 3 -f- 3 b 2 o + ab 2 — abc take a£ 2 — abc + 6 s . 
 
 10. From (a -{- b) x -\- (a -\- c) y take (a — h)x — (a — c) y. 
 
 11. Simplify 7a- J3a- [4a- (5a- 2a)]J. - S^£ 
 
 12. Simplify 3a-^a + 5-[a + 5 + c-(a + 5 + c + cT)]5. 
 
 Bracket the coefficients, and arrange according to the 
 descending powers of * : . 
 
 13. x 2 — ax — c 2 x 2 — bx -f- bx 3 — ex 2 -f- a 2 * 8 •'"— a 8 — ex. 
 
 14. 2a*-3&* 2 -7cz 3 -25x + 2c* 2 + 8a* 3 -2e*-a* 2 — foe 8 
 
 If a = 1, b = 3, c = 5, and d = 7, find the value of : 
 
 15. a 2 - (£ 2 - c 2 ) - [J 2 - (c 2 - a 2 )] + [c 2 - (& 2 - a 2 )]. 
 
 16. a - 25 - \S c - d— [3 a - (5 5 - e - 8 <*)] - 2 & j. 
 
58 ADDITION AND SUBTRACTION. 
 
 17. From 2 d + 11 a + 10 b — 5 c' take 2c + 5a-3£; 
 
 and show that the result is numerically correct when 
 a = 1, b = 3, c = 5, d = 7. 
 
 18. If a — 1, b = — 3, c = — 5, d = 0, find the value of 
 
 a 2 + 2b 2 + 3c 2 + ±d 2 . 
 
 If a — 3, J = 4, c = 9, and 2s=a + ^ + c, find the 
 value of: 
 
 19. s (s — a) (s — b) (s — c). 
 
 20. s 2 + - a) 2 + (s - J) 2 + (s- c)\ 
 
 21. s 2 - (5 — a) (s - b) - (s - b) (s - c) - (s - c) (s - a). 
 
 22. Iix = a + 2b-3c, y = b + 2c-3a, z = c + 2a — 3b, 
 show that x + y -f- z = 0. 
 
 23. Ifx = a-2£ + 3c, y = 6-2c + 3a, s = c-2a + 36, 
 
 show that £c-f-2/ + 2 = 2a-j-2&-f-2c. 
 
 24. What must be added to x 2 + 5 y 2 -f 3 z 2 in order that 
 
 the sum may be 2 y 2 — 2 2 ? 
 
 25. What must be added to 5 a 3 — 7 a 2 b + 3 ab 2 in order 
 
 that the sum may be a 3 — 2 a 2 6 — 2 ab 2 + # 3 ? 
 
 If E = 5a 3 + 3a 2 b -2b 3 , F=3a 3 - 7 o?b - b 3 , 
 G = 2a 2 b-a 3 -b 3 , H= a 2 b -2a 3 -3b 3 , 
 
 find the simplest expression for : 
 
 26. E+\F-G-H\. 30. E+G-(F-H). 
 
 27. E-\F+-G-H\. 31. F-H-(E+G). 
 
 28. J£+^-G + J3"g. 32. H-E-(F-G). 
 
 29. ^-^-(^-Jr^. 33. F-G-(E-H). 
 
 34. From 2 z 2 - 2 y 2 - » a take 3 y 2 + 2 x 2 - z 2 , and from 
 
 the remainder take 3 z 2 — 2 y 2 — x\ 
 36. Take the sum of a?c — 2 a* + 2 ac 2 and a 8 — ac* — a*c 
 
 from a 8 — 2 a?c + 3 ac\ 
 
 
CHAPTER V. 
 MULTIPLICATION AND DIVISION. 
 
 
 Multiplication of Compound Expressions. 
 
 96. Degree of a Term. A term that has one letter is said 
 to be of the first degree ; a term that is the product of two 
 letters is said to be of the second degree ; and so on. 
 
 97. Degree of a Compound Expression. The degree of a 
 compound expression is the degree of that term of the 
 expression which is of the highest degree. 
 
 Thus, a 2 x 2 + bx + c is of the fourth degree, since a 2 x 2 is of the 
 fourth degree. 
 
 98. When all the terms of a compound expression are 
 of the same degree, the expression is said to be homogeneous. 
 
 Thus, x s + 3 x 2 y + 3 xy 2 + y* is a homogeneous expression, every 
 term being of the third degree. 
 
 99. Dominant Letter. If there is one letter in an expres- 
 sion of more importance than the rest, it is called the 
 dominant letter ; and the degree of the expression is called 
 by the degree of the dominant letter. 
 
 Thus, a?x 2 + bx + c is of the second degree in x. 
 
 100. Arrangement of a Compound Expression. A compound 
 expression is said to be arranged according to the powers 
 of some letter when the exponents of that letter descend or 
 ascend, from left to right, in the order of magnitude. 
 
 Thus, 3 ax 3 — 4 bx 2 — 6 ax + 8 b is arranged according to the de- 
 scending powers of x ; and 86 — 6 ax — 4 bx 2 + 3 ax 3 is arranged 
 according to the ascending powers of x. 
 
60 MULTIPLICATION AND DIVISION. 
 
 Multiplication of Polynomials by Monomials. 
 
 We have a (b + c) = ab + ac ; (§ 41) 
 
 and a(b — c) = ab — ac. Hence, 
 
 101. To Multiply a Polynomial by a Monomial, 
 
 Multiply each term of the polynomial by the monomial, 
 and connect the partial products with their proper signs. 
 
 Find the product of ab + ac — be and abc. 
 
 ab -\- ac — be 
 abc 
 
 a 2 b 2 c + a 2 bc 2 - ab 2 c* 
 
 Note. We multiply ab, the first term of the multiplicand, by abc, 
 and work to the right. 
 
 Exercise 21. 
 
 i 
 
 Multiply : 
 
 1. 5a + 3b by 2a 2 . 7. a 2 + b 2 - c 2 by a 3 bc 2 . 
 
 2. ab - be by 5 a 3 bc. 8. 5 a 2 - 3 b 2 + 2 c 2 by Aab s c 2 . 
 
 3. aft — ac — be by a&c. 9. abc — 3 a 3 6c 2 by — 2 ab 2 c. 
 
 4. 6 a 5 & - 7 a 2 6 2 c by a 2 6 2 c. 10. xyz 2 + ScV* b y - *V- 
 
 5. cc — ?/ — z by — 3x B y 7 z 9 . 11. 3x - 2 ?/ — 4 by 5a 2 . 
 
 6. x 2 + 2if-zby -3x 2 . 12. 3x 2 - 4 y 2 + 5s 2 by 2arfy. 
 
 13. a 3 x — 5 a 2 # 2 -f #£ 3 + 2 a 4 by ax 2 y. 
 
 14. - 9 a 5 + 3 a 3 6 2 - 4 a 2 b s - b 5 by - 3 a& 4 . 
 
 15. 3z 3 -2z 2 ?/-7£c?/ 2 + ?/ 3 by -5x 2 y. 
 
 16. -4V + 5x 2 // + 8a; 3 by -3 a 2 ?/. 
 
 17. - 3 .+ 2 ab + a 2 ^ 2 by - a 4 . 
 
 18. - * - 2 a* 2 + 5 afyt; 2 by - 3 x*yz. 
 
MULTIPLICATION AND DIVISION. 61 
 
 Multiplication of Polynomials by Polynomials. 
 
 If we have m + n -f p to be multiplied by a -f- J 4- c, we 
 may substitute Jf for the multiplicand m + K+f Then 
 
 (a + b + c) Jf = aM + bM+ cM. (§ 41) 
 
 If now we substitute for M its value m -±-n+p,we have 
 
 aM + &Jf-f cilf 
 
 = a (m -f- w +p) + b(m + n -\-p) + c(m + n +i?) 
 = awi -f- aw + ap + #w + #ra + bp -f- cm -f- en + cjp. 
 
 102. To Find the Product of Two Polynomials, therefore, 
 Multiply every term of the multiplicand by each term of 
 the multiplier, and add the partial products. « 
 
 In multiplying polynomials, it is a convenient arrange- 
 ment to write the multiplier under the multiplicand, and 
 place like terms of the partial products in columns. 
 
 1. Multiply 5a- 6b by 3a- 4& 
 5a — 6b 
 Sa - 4fr 
 15 a 2 - 18 ab 
 
 -20ab + 2±b 2 
 15 a 2 - 38 ab + 24 b* 
 
 We multiply 5 a, the first term of the multiplicand, by 3 a, the first 
 term of the multiplier, and obtain 15a 2 ; then we multiply — 66, the 
 second term of the multiplicand, by 3 a, the first term of the multi- 
 plier, and obtain — 18 ab. The first line of partial products is 15 a 2 
 — 18 ab. In multiplying by — 4 6, we obtain for a second line of 
 partial products — 20 ab + 24 6 2 , and this is put one place to the right, 
 so that the like terms — 18 ab and — 20 ab may stand in the same 
 column. We then add the coefficients of the like terms, and obtain 
 the complete product in its simplest form. 
 
62 MULTIPLICATION AND DIVISION. 
 
 2. Multiply 4 a; + 3 + 5 a; 2 - 6 a; 3 by 4 - 6x 2 - 5x. 
 
 Arrange both multiplicand and multiplier according to 
 the ascending powers of x. 
 
 3+ 4a; + 5x 2 - 6a; 8 
 4- 5x- 6x 2 
 
 12 + 16a; + 20a; 2 -24a; 8 
 
 - 15a; - 20a; 2 - 25a; 8 -f 30a; 4 
 
 - 18 a; 2 - 24 a; 8 - 30 a; 4 + 36 a;* 
 
 12+ x -18a; 2 -73a; 3 + 36a^ 
 
 3. Multiply 1 + 2a; + a; 4 - 3x 2 by x 3 - 2 - 2x. 
 
 Arrange according to the descending powers of x. 
 
 x*-3x 2 + 2x +1 
 x 8 -2x -2 
 
 -2x 5 + 6a; 8 -4a; 2 -2a; 
 
 -2a; 4 +6a; 2 -4a;-2 
 
 -5a* +7a; 3 + 2a; 2 -6a;-2 
 
 4. Multiply a 2 + b 2 + c 2 — ab — be — ac by a -+- b -f- c. 
 
 Arrange according to the descending powers of a. 
 
 a 2 — ab — ac -f- 6 2 — 5c + c 2 
 a + # + £ 
 
 a 8 — a 2 £ — a 2 c -f- a& 2 — afo + ac 2 
 
 + a 2 b -ab 2 - abc +b 8 -b 2 c + bc 2 
 
 + a 2 c — abc — ac 2 + b 2 c — be 2 + c z 
 
 'a 1 -3 abc +b* +? 
 
 Note. The student should observe that, with a view to bringing 
 like terms of the partial products in columns, the terms of the multi- 
 plicand and multiplier are arranged in the same order. 
 
MULTIPLICATION AND DIVISION. 63 
 
 Exercise 22. 
 
 Multiply : 
 
 1. x + 10 by x + 6. 12. 2* -3 by a; + 3. 
 
 2. x — 2 by x - 3. 13. a; - 7 by 2 a; - 1. 
 
 3. se — 3 by x + 5. 14. m — n by 2 m + 1. 
 
 4. a; + 3 by a; — 3. 15. m — a by m + a. 
 
 5. a; - 11 by a; - 1. 16. 3a; + 7 by 2a; - 3. 
 
 6. -a; + 2 by -a;- 3. 17. 5x - 2y by 5x + 2y. 
 
 7. —x- 2 by x — 2. 18. 3a; - 4y by 2a; + 3y. 
 
 8. — a; + 4 by x — 4. 19. a; 2 + y 2 by a; 3 — y 3 . 
 
 9. - a; + 7 by a; + 7. 20. 2 x 2 + 3 y 2 by x 2 + y 2 . 
 10. a; — 7 by a; + 7. 21. x-\-y-\-zbyx — y-\-z. 
 LI. a;-3by2a; + 3. 22. a; + 2y-s by a; — y + 2s, 
 
 23. x 2 — xy + y 2 by a; 2 + a;y + y 2 . 
 
 24. m 2 — mn + w 2 by m + w. 
 
 25. m 2 + mn + n 2 by m — n. 
 
 26. a 2 - 3 ab + b 2 by a 2 - 3 a£ - b 2 . 
 
 27. a 2 -7a + 2 by a 2 - 2 a + 3. 
 
 28. 2x 2 -3xy-{-4,y 2 bj 3x 2 + 4:xy-5y 2 . 
 
 29. a; 2 + xy + y 2 by x 2 — xz — z 2 . 
 
 30. a; 2 4- y 2 + 2 2 — a;y — xz — yz by a; + y + * c 
 
 31. ±a 2 -10ab + 25a 2 b 2 by 5b + 2a. 
 
 32. a; 2 + 4 y by y 2 + 4 x. 
 
 33. a; 2 + 2a;y + 8 by ?/ 2 + 2a;y-8. 
 
 34. a 2 + J 2 + l-a5-a-&bya + l+S. 
 
 35. 3a; 2 -2y 2 + 5s 2 by 8a; 2 + 2i/ 2 -3« 2 . 
 
 36. a 2 + b 2 -{- c 2 — ab — ae — bo by a -{-b + c. 
 
 37. x 2 — xy + yt + x + y + lbyx + y — l. 
 
64 MULTIPLICATION AND DIVISION. 
 
 38. 5x + 4:X 2 + x 8 -2±byx 2 + ll-4:X. 
 
 39. x 8 + llx-4;x 2 -24:by x 2 + 5 + 4x. 
 
 40. x 4 + x 2 -4 : x-ll+2x s by x 2 -2x-\-3. 
 
 41. -5x 4 ~x 2 -x-\-x s -\-13x 3 by x 2 -2-2x. 
 
 42. 3x + x 8 -2x 2 -±by 2x + 4x 3 + 3x 2 + 1. 
 
 43. 5a 4 4-2a 2 ^ 2 4-^ 8 -3a 3 ^ by 5a 8 b - 2 ab 3 + 3 a 2 b 2 + b*. 
 
 44. 4a 7 y-32a?/ 4 -8ay + 16ay bjay + 4ay + 4«y. 
 
 45. 6a 5 b + 3 a 2 b 4 - 2 ab 5 + 6 1 by 4 a 4 - 2 a& 8 - 3 J 4 . 
 
 46. ic 2 + 2/ 2 + 2 ay - 2 a - 2 ?/ - 1 by x + y — 1. 
 
 47. a m + 2 a" 1 ' 1 — 3 a m ~ 2 - 1 by a + 1. 
 
 48. a n - 4 a"" 1 + 5 a n ~ 2 + a n ~ 3 by a - 1. 
 
 49. a 4n+1 — 4 a 3n + 2 a 2 *" 1 — a n ~ 2 by 2 a 3 - a 2 + a. 
 
 50. aJ* — y n+1 by x n -f y n+1 . 
 
 51. a 2 " + 2 ay 1 + y 2 " by x 2n — 2 x n y* + y 2n . 
 
 52. a n - a n+1 - a n+2 + 1 by a n + 1. 
 
 53. a 8n — a 2n + a n — l by a n — 1. 
 
 54. a n — a"" 1 + a n ~ 2 + a n ~ 3 by a 2 - 1. 
 
 Exercise 23. — Review. 
 Simplify : 
 
 1. (x 2 -x -19) (x 2 + 2x-3). 
 
 2. (l + 2a-f a 2 )(l-z s 4-2z 2 -3a;). 
 
 3. (2z 2 4-2-f 3x)(2a-3a 2 4-2ar J ). 
 
 4. (3a 2 + 5-4ic)(8 + 6a; 2 -7ic). 
 
 5. (x 8 4- # — y) — (ic 2 — y 2 4- ay) (a — 1). 
 
 6. (3 4- 7X 2 - 5x) (8 x 2 - 6x -10sc 3 -f 4). 
 
 7. (b 8 + 6 ab 2 - 4 a 2 £) (2 a 2 £ - ab 2 - 8 a 8 ). 
 
 8. (x 2 + ax-b)-(x 2 + 5x-±)(x-3). 
 
 9. (a^ — wa; 2 + nx + r) (x 2 -\- ex + d). 
 10. Jx 2 -(a4-5)a;4-^K a; - c )- 
 
MULTIPLICATION AND DIVISION. 
 
 11. (x s + x*y + xy 2 + y*)(x-y). 
 
 12. (Ax 2 + 9y 2 - 6x7/) (4z 2 + §tf + §xy). 
 
 13. (z 4 -3a; 2 + 5)(a 2 + 4)(a;-2). ? 
 
 14. (z 4 - *%» + y*) (x* + x 2 f + y 4 ). 
 
 15. (2a 5 -3z 3 + 4z 2 -5)(a; 2 -8)(a;-3). ' 
 
 16. (a 2m — a m y m + y 2 ™) (a m + y m )(a + y)> 
 
 17. (a Sx - a 2x + a x - 1) (a x + 1) (a x - 1). 
 
 18. (a 2 + b 2 + c?-h2ab-ac-bc)(a + b + c). 
 
 19. (a-2b)(b-2a)-(a-3b)(±b-a) + 2ab. — 
 
 20. (a + & + c) (a + b - c) - (2 ab - c 2 ). 
 
 21. (m + ri)m — [(w — ri) 2 — n(n — m)~]. — 
 
 22. [ac-(a-ft)(ft4-c)]-ft[ft-(a-c)]. 
 
 23. (a - 1) (a* - 2) - 3x(x + 3) + 2 [(a; + 2) (i + 1)- 3]. 
 
 24. 4 (a - 3 5) (a + 3 J) - 2 (a - 6 ft) 2 - 2 (a 2 + 6 ft 2 ). 
 
 25. (x-\-y + z) 2 —x(y + z — x) — y(x+z — y)—z(x-}-y--z). 
 
 26. 5 j (a — J) a; — c?/ \ — 2 \ a (x — y) — bx \ — (3 ax — 5 cy). 
 
 27. (a - b) x - (ft - c) a - j (ft - x) (ft - c) - (ft - c) (ft + c) \. 
 
 28. (a + b) (b + c) — (c + d) (a + d) - (a + c) (ft — d). 
 
 29. a 2 (b - c) - b 2 (a- c) + <? (a-b)-(a- b)(a -c)(b- c). 
 
 30. (2 a - b) 2 + 2 b (a + b) - 3 a 2 - (a - J) 2 + (a + ft) (a - ft). 
 If a = 0, J = 1, c = — 1, find the value of: 
 
 31. (b-c) 2 +(c-a) 2 + (a-b)\ ' 
 
 32. (b + cY-(o-a) 2 -(a-b). 
 
 33. (a-b)(a-c) + c(3a-b-c) + 2ac-(a-c) + 2b. 
 
 34. (ft - c) 8 + (a - ft) 3 + (c - a)*- 3 (ft - c) (a - ft) (c - a). 
 
 35. If a = 12, and ft = 5, find the values of 
 V^~ 2 + Vft*: and Va 2 + ft 2 . 
 
66 MULTIPLICATION AND DIVISION. 
 
 Division of Compound Expressions. 
 Division of a Polynomial by a Monomial. 
 By the distributive law, (§ 43), 
 
 9a 4 b 2 x- 12 a 3 bx 2 -3a 2 x 9 a 4 b 2 x 12 a s bx 2 3a 2 x 
 
 1. 
 
 3 a 2 x 3 a 2 x 3 a 2 x 3 a 2 # 
 
 = 3 a 2 b 2 -±abx- 1. 
 
 2* 2 *- 1 ""2 a; 2 - 1 2^- 1_ ^ ^* ' 
 
 Note. Here we have 4 n + 1 — (2n — l) = 4n+l— 2n+l = 2n + 2, 
 and 3n— (2n — l) = 3n — 2n + l = n + l, as indices of x in the first 
 and last terms of the quotient respectively. 
 
 103. To Divide a Polynomial by a Monomial, therefore, 
 Divide each term of the dividend by the divisor, and con- 
 nect the partial quotients with their proper signs. 
 
 Exercise 24. 
 Divide : 
 
 1. x 2 + 2xyhyx. 3. 4 x G - 8 x 3 by 2 x s . 
 
 2. a 2 -2abbja. 4. - 6x 3 - 2x by - 2x. 
 
 5. - 8 a 5 - 16 a 10 by - 8 a 3 . 
 
 6. 27 a 3 - 36 a 2 by 9 a 2 . 
 
 7. - 30 a 7 + 20 a 9 by - 10 a 6 . 
 
 8. -12:ry-4xy by -4<bY- 
 9.-3 arV - 6 #V by - 3 af«*. 
 
 10. 3 aW -- 9 aW by 3 aW. 
 
 11. x 2 — xy — xz by — x. 
 
 12. 3a 8 -6a 2 6-9a5 2 by -3 a. 
 
 13. # 5 ?/ 2 — ay — x 2 y 2 by a 2 ?/ 2 . 
 
 14. a% 2 c - a 2 b s c - a 2 bc* by abc. 
 
 15. 8a 8 --4a 2 £-6a& 2 by -2a.- 
 
 16. 5 m z n — 10 m 2 n 2 — 15 wwi 8 by 5 m». 
 
MULTIPLICATION AND DIVISION. 67 
 
 Division of One Polynomial by Another. 
 If the divisor (one factor) = a -f- b -f- c, 
 
 and the quotient (other factor) = n-\-p + g, 
 
 ( + an + bn + en 
 then the dividend (product) — < + ap -\- bp -\- cp 
 
 [ + aq -f hq + c^. 
 
 The first term of the dividend is an ; that is, the product 
 of a, the first term of the divisor, by n, the first term of the 
 quotient. The first term n of the quotient is therefore 
 found by dividing an, the first term of the dividend, by a, 
 the first term of the divisor. 
 
 If the partial product formed by multiplying the entire 
 divisor by n is subtracted from the dividend, the first term 
 of the remainder ap is the product of a, the first term of the 
 divisor, by p, the second term of the quotient ; that is, the 
 second term of the quotient is obtained by dividing the 
 first term of the remainder by the first term of the divisor. 
 In like manner, the third term of the quotient is obtained 
 by dividing the first term of the new remainder by the 
 first term of the divisor ; and so on. Therefore, 
 
 104. We have the following rule for division : 
 
 Arrange both the dividend and divisor in ascending or 
 descending powers of some common letter. 
 
 Divide the first term of the dividend by the first term of 
 the divisor. 
 
 Write the result as the first term of the quotient. 
 
 Multiply all the terms of the divisor by the first term of 
 the quotient. 
 
 Subtract the product from the dividend. 
 
 If there is a remainder, consider it as a new dividend and 
 proceed as before. 
 
68 
 
 MULTIPLICATION AND DIVISION. 
 
 105. It is of fundamental importance to arrange the divi- 
 dend and divisor in the same order with respect to a com- 
 mon letter, and to keep this order throughout the operation. 
 
 The beginner should study carefully the processes in the 
 following examples : 
 
 1. Divide x 2 + 18 x + 77 by x + 7. 
 
 x 2 + 18 x + 77 
 a?+ 7x 
 
 x + 1 
 
 x + 11 
 
 11 a + 77 
 11* + 77 
 
 Note. The student will notice that by this process we have in 
 
 effect separated the dividend into two parts, x 2 + 7x and 11 x + 77, 
 
 and divided each part by x + 7, and that the complete quotient is the 
 
 sum of the partial quotients x and 11. Thus, 
 
 x2 + 18 x + 77 = a* + 7x + Its + 77 = (x 2 + 7x) + (llx + 77); 
 
 x2 + 18x + 77 x 2 + 7x J llx + 77 , in 
 -I rir~ = x + 11. 
 
 x + 7 
 
 x + 7 
 
 x + 7 
 
 2. Divide a 2 - 2 ab + b 2 by a- b. 
 
 a 2 -2ab + b 2 
 
 a — b 
 
 a 2 — ab 
 
 a — b 
 
 - ab + b 2 
 
 - ab + b 2 
 
 
 3. Divide 4 a A x 2 - 4 a 2 x A + x 6 - a« by x 2 - a\ 
 
 Arrange according to descending powers of x. 
 
 x e - 4 a 2 x* + 4 aV - 
 
 <c 4 - 3 a V + a 4 
 
 -3aV + 4aV 
 -3«¥ + 3a¥ 
 
 a¥ 
 
MULTIPLICATION AND DIVISION. 
 
 69 
 
 4, Divide 22 aV 4 15 b* 4 3 a 4 - 10 a 3 J 
 
 Arrange according to descending powers 
 3 a 4 - 10 tf* + 22 a 2 b 2 - 22 aft 3 4 15 b 4 
 
 -22ab 8 
 
 ' a 2 + 3b 2 -2ak 
 of a. 
 a 2 -2ah + 3b' k 
 
 3 a 4 - 6a 3 &4 9^ 2 
 
 3a 2 -±ab + hb' k 
 
 - 4»% + 13 a 2 6 2 - 22 a& 3 4 15 ft 4 
 
 - 4a 8 H 8a 2 6 2 -12a& 8 
 
 
 5a 2 6 2 -10a& 3 + 15& 4 
 5 a 2 6 2 - 10 «& 3 + 15 b* 
 
 5. Divide 5x 8 - x + 1 - Sx 4 hj 1 + 3x 2 - 2x. 
 Arrange according to ascending powers of x. 
 
 1- x + 5x 3 -3x 4 
 l-2x-\-3x 2 
 
 1 - 2 x + 3 a 2 
 
 14- ic — sc a 
 
 z-3a 2 + 5cc 3 
 
 -3z 4 
 
 » - 2 z 2 4 3 x s 
 
 - « 2 + 2a; 8 -3x 4 
 
 - z 2 + 2tf 8 -3a; 4 
 
 6. Divide x 3 -{- y 3 + z 3 — 3 xyz hy x + y + z. 
 Arrange according to descending powers of x. 
 
 x z — 3 xyz 4 y 8 + z 3 
 x 3 -f- x 2 y 4- # 2 ^ 
 
 a; + y 4 « 
 
 — x"y — ar« 
 
 q; 2 y — xy 2 — xyz 
 
 1 — xy — xz + y 2 — yz + z* 
 3xyz + y 3 + z 3 
 
 — x' 2 z 4 xy 2 
 
 — x 2 z 
 
 -2xyz + f + z 8 
 — xyz — xz 2 
 
 xy 2 
 xy 2 
 
 — xyz 4 xz 2 + y 8 + z 8 
 + f + f* 
 
 
 — #?/£ 4 xz 2 — y 2 ^ 4 z 3 
 
 — xyz — y 2 z — yz 2 
 
 xz 2 4 y^ 2 4 £ 8 
 «* 2 4 ys 2 4 z* 
 
70 MULTIPLICATION AND DIVISION. 
 
 7. Divide 4 a x+1 - 30 a* -h 19 a*' 1 -f 5 a*' 2 + 9 a x " 4 
 
 by a*" 3 — 7 a x ~ 4 + 2 a x ~ 5 — 3 a*- 6 . 
 
 4a x+1 -30a a; 4-19a iE - 1 + 5a x - 2 +9a x - 4 
 4a x+1 -28a x -f 8a*- 1 -12a x - 2 
 
 a a> ~ s ~7a x - 4 -\-2a x -' i -3a x -^ 
 
 4a 4 -2a 3 -3a 2 
 
 — 2a x +lla x - 1 +17a x - 2 H-9a x - 4 
 
 - 2a x +Ua x - 1 - 4a x ~ 2 +6a x - 3 
 
 - 3a x - 1 4-21a x - 2 -6a x "- 3 +9a x - 4 
 
 - 3^ x - 1 +21a x - 2 -6a x - 8 +9^ x - 4 
 
 Note. We find the index of a in the first term of the quotient by 
 subtracting the index of a in the first term of the divisor from the 
 index of a in the first term of the dividend. Now, (x + 1) — (x — 3) 
 = x + 1 — £ + 3 = 4. Hence, 4 is the index of a in the first term of 
 the quotient. In the same way the other indices are found. 
 
 Exercise 25. 
 Divide : 
 
 1. a 2 + la + 12 by a + 4. 6. 4cc 2 + 12 x + 9 by 2 x + 3. 
 
 2. a 2 - 5 a + 6 by a - 3. 7. 6 a 2 - 11 a; + 4 by 3 a - 4. 
 
 3. a 2 4- 2 a 7/ + ?/ 2 by aj + y. 8. 8 z 2 -10 aa-3 a 2 by 4 cc+a, 
 
 4. x 2 — 2 xy -\- y 2 by x — y. 9. 3 a 2 — 4 a — 4 by 2 — a. 
 B. x 2 — y 2 by x — y. 10. a 3 — 8 a — 3 by 3 — a. 
 
 11. a 4 + 11 a 2 - 12 a - 5 a 3 + 6 by 3 + a 2 - 3 a. 
 
 12. y 4 - 9f + y* -lGy -4 by ^ + 4 + 43/. 
 
 13. 36 + m 4 - 13 m 2 by 6 -f m 2 + 5 m. 
 
 14. l-s-3s 2 -s 6 byl+2s + s 2 . 
 
 15. £ 6 - 2 5 8 -f- 1 by £ 2 - 2 J + 1. 
 
 16. ar* + 2a-Y + 9 y *by x 2 - 2 ary + 3 ^. 
 
 17. a 5 + £ 5 by a 4 - a*b + « 2 & 2 - ab s + **. 
 
 18. 1 + 5ar* - 6z 4 by 1- x + 3a* 
 
MULTIPLICATION AND DIVISION. 71 
 
 19. 8zY + 92/ 4 -|-16x 4 by 4x 2 + 3y 2 - 4=xy. 
 
 20. x z + tf + z 3 + 3x 2 y + 3xy 2 by x + y + z. ' 
 
 21. a s + b s + c*-3abcby a + b + c. 
 
 22. a 3 + 8y 8 + « 3 -6xys by x 2 + ±tf+ z 2 -xz — 2xy-2yz. 
 
 23. 2a 2 — 3y 2 + a;?/ — jb* — 4=yz — z 2 by 2x + 3y + z. 
 
 24. x 2 — y 2 — 2 yz — z 2 by x-\-y-\- z. 
 
 25. a; 4 + jc 2 ^ 2 + y 4 by a; 2 + xy + 2Z 2 . 
 
 26. a: 4 -9a: 2 + 12a;-4by x 2 + 3x-2. 
 
 27. ^-22/ 4 -6^ + 42/ 2 + 13y + 6by 2/ 3 + 3^ + 3y + l. 
 
 28. 2/ 4 -5?/ 2 « 2 + 4^by y 2 -yz-2z 2 . 
 
 29. a: 2 -4?/ 2 -9s 2 + 12yzby a; + 2?/--3z. 
 
 30. a: 5 -41a; -120 by x 2 + ±x f 5. 
 
 31. cc 4 -3-f 5z + z 3 -4z 2 by 3-2a:-a; 2 . 
 
 32. 6-2z 4 -f 10a 3 -lla: 2 + a;by 4Z-3-2X 2 . 
 
 33. l-6x 5 + 5x 6 by l-2x-^x 2 . 
 
 34. a: 4 + 81 + 9a; 2 by 3a; -a 2 -9. 
 
 35. x z — y 3 by x 2 + xy + 2Z 2 - 
 
 36. a; 6 + / by x 2 + y 2 . 
 
 37. a; 4 + a 2 x 2 + a 4 by x 2 - ax + a 2 . 
 
 38. a 2 - 2 b 2 + a& - 3 c 2 + 1 be + 2ac by 3 c + a - &. 
 
 39. ab + 2a 2 - 3 b 2 - 46c - ac - c 2 by c + 2a + 3b. 
 
 40. 15 a 4 + 10a 3 x + ±a 2 x 2 + 6ax s - 3x 4 by 3a 2 + 2ax- x 2 
 
 41. a s -8b 3 -l-6abby a-2b-l. 
 
 42. 0^ — 3 a; 2n y n + 3 x n y 2n — y 8 " by x n — y n . 
 
 43. a m+n b n — 4 a m+n - 1 & 2n — 27 a m + n - 2 b 3n + 42 a »»+»-3&*» 
 
 by a m + 3 a" 1 " 1 ^ — 6 a 7 "- 2 ^". 
 
72 MULTIPLICATION AND DIVISION. 
 
 106. Integral expressions may have fractional coefficients, 
 since an algebraic expression is integral if it has no letter 
 in the denominator. The processes with fractional coeffi- 
 cients are precisely the same as with integral coefficients, 
 as will be seen by the following examples worked out : 
 
 1. Add £a 2 - £ ab + \ b 2 and | a 2 + § ab - §#. 
 
 %a 2 -%ab + \b 2 
 f a 2 + f «& - f 6 2 
 %a 2 + \ab-\b 2 
 
 2. From \a 2 -\ab + \b 2 take \a 2 -\ab + %b 2 . 
 
 \a 2 -\ab + |P 
 ja'-frq6 + g& 2 
 £a 2 + £«6- T V 2 
 
 3. Multiply \a 2 -\ab + J& 2 by \a-%h 
 
 \a 2 -%ab + \b 2 
 
 \a -%b 
 
 \a z -\a 2 b + \ab 2 
 
 -ja 2 b+ ^ab 2 -^ 
 ±a s -$a 2 b + $§ab 2 -%b 3 
 
 4. Divide f b* + ^-bd 2 - $$%b 2 d - ^d* by }ft - §d. 
 
 3* ~$ d 
 
 g& 3 - tfPd 
 
 tV-$bd+\ffl 
 
 ftPd + *+beP-ih(P 
 A 6 2 dT 4- fa* 2 
 
MULTIPLICATION AND DIVISION. 73 
 
 Exercise 26. 
 
 1 . Add | a 2 b + J W + A and - ^ a 2 & -}^-J. 
 
 2. From f x 2 + 3 ax — f a 2 take 2 a 2 — § ax — £ a 2 . 
 
 3. From \y — %a — \x + \b take J y + *£ a — § x. 
 
 4. Multiply Jc 2 -ic-^by J^-Jc + J. 
 
 6. Multiply \x-\x 2 + \x 8 by ^x + ^tf + ^x*. 
 
 6. Multiply 0.5 m 4 - 0.4 m 8 rc + 1.2 m 2 w 2 + 0.8 mn z - 1.4 rc 4 
 
 by 0.4 m 2 - 0.6 rarc - 0.8 n 2 . 
 
 7. Divide -ft a 4 - { a 8 & + Jf a 2 £ 2 + J a& 8 by f a + £&. 
 
 8. Divide-^ 6 + ^ 2 -|J^ 8 + |^ 4 by - je? + 2d 
 
 Exercise 27. — Review. 
 
 1. Find the value of x 8 + y 8 + * 8 — 3 ai/2, if a; = 1, y = 2, 
 
 and 2 = — 3. 
 
 2. Find the value of V2 6c — a, and of V2 6c — a, if 
 
 6 = 8, c = 9, and a = 23. 
 
 3. Add a 2 b - ab 2 + £ 8 and a 8 - £ a 2 6 + ab 2 - f & 8 . 
 
 4. Multiply a 2m — a" 1 ^ + b 2m by a m + b m . 
 
 5. Multiply 4 a 2m+4 + 6 a" 1 * 8 + 9 a 2 by 2 a m+4 - 3 a 9 . 
 
 6. Divide a 8 + 8 y 8 - 125 z s + 30xyzby x + 2y - 5z. 
 
 7 . Simplify (aj - a) 2 - (a* - b) 2 - (a - b) (a + b - 3 x). 
 
 8. Find the coefficient of x in the expression 
 
 x + a- 2\2 a -b(c- a;)]. 
 
 9. Multiply 4 a^+2"- 1 - 7 iC 2m - 3 »+ 2 + 5 x 2 ** 8 ™- 2 by 5 a? 2 -" 1 - 9 ". 
 
 10. Divide c 2 ^ 2 * - c 4 ^ 4 - 2 * - c 4 ^ 8 -* by c^d 2 - 2 *. 
 
 11. Divide raV — m 1+x 2/ 1+w + m 5 - a y*-* by m s - x y n ^\ 
 
74 MULTIPLICATION AND DIVISION. 
 
 12. Divide a l+3 v - a? -f a 2+5 » by a 2 ~ x+ K 
 
 13. Divide x — a; 1 *-*"" 4 -f a; 2m by a? 2 -™- 2 ". 
 
 14. Divide ^ — %f-*> n + y 4p+1 by y^-** 1 . 
 
 15. Divide 2 a 3n - 6 a; 2 ^" + 6x n y 2n — 2y Sn byx n -y n . 
 
 16. Divide x 8 - 2 asc 2 + a 2 ic - a&a; - 6 2 a; + a 2 b + a& 2 
 
 by x 2 — ax -\-bx — ab. 
 
 17. Divide z 4wi + x 2 " 1 + lby x 2 ™ - x m + 1. 
 
 18. Divide 3 cc m+7 — 4 a TO+6 — 12 x m+5 — 9 x m+4 
 
 by x m+ * — 3 # m+3 . 
 
 19- Divide 6 sc 4 "* 5 - 13 x Sm + s + 13 a 2OT+5 - 13 x m+s - 5 x 6 
 by 2 sc 2 ™ - 3 x m - 1. 
 
 20. Divide 12 a 5 "- 3 - a 4 "" 2 - 20 a 3 "" 1 -f 19 a 2n - 10 a n+1 
 
 by 4 a 2 " - 3 a n+1 + 2 a 2 . 
 
 21. Arrange according to descending powers of x the 
 
 following expression, and enclose the coefficient of 
 each power in a parenthesis with a minus sign 
 before each parenthesis except the first : 
 x s — 2 bx — a 2 x* — ax — ax 2 — ex — a 2 x z — bcx. 
 
 22. Divide 1.2 a*x - 5.494 a s x 2 + 4.8 a 2 x s + 0.9 ax 4 - X s 
 
 hy0.6ax-2x 2 . 
 
 23. Multiply \a?-\ab±\b 2 \)y %a + \b. 
 
 24. Multiply %a 2 + ab + f& 2 by %a-\b. 
 
 25. Divide * a 8 + T ^ a& 2 + T V & 8 by £ a + £ b. 
 
 26. Subtract J a 2 + £ a?y + £ y 2 from £ « 2 - £ xy +\ y 2 . 
 
 27. Subtract o^ -\-\xy — \y 2 from 2 a- 2 — \ xy + y*. 
 
 28. If a = 8, 5 = 6, c = — 4, find the value of 
 
 Va 2 + 2 6c + V6 2 + ac -f Vc 2 + ok 
 
CHAPTER VL 
 SPECIAL RULES. 
 
 Multiplication. 
 
 107. Square of the Sum of Two Numbers. 
 
 (a + by = (a + b)(a+. b) 
 
 = a(a + b) + b(a + b) 
 
 = a 2 + ab + ab + b 2 
 
 = a 2 + 2ab + b 2 . Hence, 
 
 Rule 1. The square of the sum of two numbers is the 
 sum of their squares plus twice their product. 
 
 108. Square of the Difference of Two Numbers. 
 
 (a - b) 2 =(a-b){a-b) 
 
 = a (a — b) — b (a — b) 
 
 = a 2 — ab — ab -\- b 2 
 
 — a 2 — 2ab -f- b 2 . Hence, 
 
 Rule 2. The square of the difference of two numbers is 
 the sum of their squares minus twice their product. 
 
 109. Product of the Sum and Difference of Two Numbers. 
 
 (a + b) (a - b) = a (a - b) + b (a - b) 
 — a 2 — ab -\- ab — b 2 
 = a 2 - b 2 . Hence, 
 
 Rule 3. The product of the sum and difference of two 
 numbers is the difference of their squares. 
 
76 SPECIAL RULES OF MULTIPLICATION. 
 
 110. The following rule for raising a monomial to any- 
 required power will be useful in solving examples in multi- 
 plication : 
 
 Raise the numerical coefficient to the required power, 
 and multiply the exponent of each letter by the exponent of 
 the required power. 
 
 Thus the square of 7 a 2 6 6 is 49 a 4 6 12 . 
 
 Exercise 28. 
 Write the product of : 
 
 1. (x + y)\ 7. (x + y)(x- y). 
 
 2. (x-a)\ 8. (4* -3) (4z + 3). 
 
 3. (x + 2b) 2 . 9. (3a 2 + 4tb 2 ) (3a 2 -±b*). 
 
 4. (Sx -2c) 2 . 10. (Sa-c)(Sa-c). 
 
 5. (4y-5) 2 . 11. (x + lb^ix + lb 2 ). 
 
 6. (3 a 2 + 4 a 2 ) 2 . 12. (ax + 2by)(ax-2by). 
 
 111. If we are required to multiply a + b+cby a + b — c> 
 we may abridge the ordinary process as follows : 
 
 (a + b + c) (a + b - c) = \(a + b) + c\\(a + b) - c\ 
 By Kule 3, = (a + b) 2 - c 2 
 
 By Rule 1, = a 2 + 2ab + b 2 - c 2 . 
 
 If we are required to multiply a + b — chj a — b + c, we 
 may put the expressions in the following forms, and per- 
 form the operation : 
 
 (a + b - c) (a - b + c) = \a + (b - c) \\a - (b - c)\ 
 By Rule 3, = a 2 - (b - c) 2 
 
 By Rule 2, = a 2 - (b 2 -2bc + c*) 
 
 By § 40, = a 2 - b 2 -f 2bc - c*. 
 
SPECIAL RULES OF MULTIPLICATION. 77 
 
 Exercise 29. 
 Find the product of : 
 
 1. x + y 4- z and x — y — z, 
 
 2. x — y 4- z and x — y — z. 
 
 3. owe + by -f 1 and ax + by — 1. 
 
 4. 1 + a; — y and 1 — a; -f y. 
 
 5. a + 2b- 3c and a- 2b + 3c. 
 
 6. a 2 - ab + 6 2 and a 2 + ab + 6 2 . 
 
 7. ra 2 4- mw 4- rc 2 and m 2 — m» 4- n 2 . 
 
 8. 2 + cc + x 2 and 2 - x - x 2 . 
 
 9. a 2 + a 4- 1 and a 2 — a + 1. 
 
 10. 3# + 2y-2 and 3#-2y + s. 
 
 11. 1 + # 4- y and 1 4- x — y. 
 
 12. a 2 -2acc4-4a; 2 anda 2 + 2aa;4-4aj 2 . 
 
 13. x 2 -2xy 4-y 2 andic 2 4-2icy4-y 2 . 
 
 14. x — y 4- 13 3 4 and » — y — 13 z 4 . 
 
 15. x 2 -5y 2 - 7z 9 2Uidx 2 -5y 2 + 7z z . 
 
 112. Square of any Polynomial. If we put x for a, and 
 
 y 4- « f or b, in the identity 
 
 ( a 4- 5) 2 = a 2 + 2 a& 4- #*, 
 we have 
 
 ja + (y 4- *)? 2 = x 2 4- 2a (y 4- *) + (y 4- *) 2 , 
 or (x + y + z) 2 =x 2 + 2xy + 2xz + y 2 + 2yz + z 2 
 = x 2 + y 2 + z 2 + 2xy + 2xz + 2yz. 
 
 The complete product consists of the sum of the squares 
 of the terms of the given expression and twice the product 
 of each term into all the terms that follow it 
 
 
78 SPECIAL RULES OF MULTIPLICATION. 
 
 Again, if we put a — b for a, and c — d for b, in the same 
 identity, we have 
 
 = (a - b) 2 + 2 (a - &) (0 - d) + (c - d)* 
 = (a 2 -2ab + b 2 ) + 2a(c-d)-2b(c-d) + (c 2 -2cd + d 2 ) 
 ±=a 2 -2ab + b 2 + 2ac-2ad-2bc + 2bd'' r c 2 - 2cd + d 2 
 = a 2 + b 2 + c 2 + d 2 -2ab + 2ac-2ad-2bc + 2bd- 2cd. 
 
 Here the same law holds as before, the sign of each 
 double product being + or — , according as the factors com- 
 posing it have like or unlike signs. The same is true for 
 any polynomial. Hence we have the following rule : 
 
 Rule 4. The square of a polynomial is the sum of the 
 squares of the several terms and twice the product obtained 
 by multiplying each term into all the terms that follow it. 
 
 Exercise 30. 
 Write the square of : 
 
 1. 2x — 3y. 12. a -2b -3c. 
 
 2. a + b + c. 13. 3a- b + 2c. 
 
 3. x + y — z. 14. x + 2y — 3z, 
 
 4. x — y + z. 15. x 2 -y 2 -\-z\ 
 
 5. x + y + 5. 16. x-2y-3z. 
 
 6. x + 2y + 3. 17. 2z-y + x. 
 
 7. a-b + c. 18. s-hy + s + l. 
 
 8. 3x -2y + 4. 19. x - y + z - 1. 
 
 9. 2x-3y + 4:Z. 20. 4x + y + z-2. 
 
 10. x 2 + y 2 + z\ 21. 2x-y-z-3. 
 
 11. 2x — y-z. 22. 05-2^-3*4-4. 
 
SPECIAL RULES OF MULTIPLICATION. 79 
 
 113. Product of Two Binomials. The product of two bino- 
 mials which have the form x + a, x + b, should be carefully 
 noticed and remembered. 
 
 1. (x + 5) (x + 3) = x (x + 3) + 5 (x + 3) 
 
 = z 2 + 3;c + 5:z + 15 
 = x 2 + 8 a + 15. 
 
 2. (x - 5) (x - 3) = x (x - 3) - 5 (x - 3) 
 
 = « 2 -3£c- 5z + 15 
 = x 2 -8x + 15. 
 
 3. (a + 5) (cc - 3) = x (x - 3) + 5 (x - 3) 
 
 = ar 2 -3z + 5a;-15 
 = ic 2 + 2^-15. 
 
 4. (x - 5) (a; + 3) = x (x + 3) - 5 (a + 3) 
 
 = a 2 + 3a- 5* -15 
 = a 2 -2x -15. 
 
 Each of these results has three terms. 
 
 The first term of each result is the product of the first 
 terms of the binomials. 
 
 The last term of each result is the product of the second 
 terms of the binomials. 
 
 The middle term of each result has for the coefficient of x 
 the algebraic sum of the second terms of the binomials. 
 
 The intermediate step given above may be omitted, and 
 the products written at once by inspection. Thus, 
 
 1. Multiply x + 8 by x + 7. 
 
 8 + 7 = 15 ; 8 X 7 = 56. 
 .'.(x + 8) (x + 7) - x 2 + 15z + 56. 
 
80 SPECIAL RULES OF MULTIPLICATION. 
 
 2. Multiply x — 8 by x — 7. 
 
 (-8) + (-7) = -15; (-8) (-7) = + 56. 
 .-. (x - 8) (x - 7) = x 2 - 15a + 56. 
 
 3. Multiply x — 7ybyx + 6y. 
 
 -7y + 6y=-y; (- 7 y) (6 y) = - 42 y\ 
 .'. (x - ly) (x + 6y) = x 2 - xy - 42y 2 . 
 
 4. Multiply x 2 + 6 (a + b) by a 2 - 5 (a + J). 
 
 - 5 (a . + J) + 6 (a + b) = (a + b) ; 
 
 - 5 (a + J) X 6 (a + 6) = - 30 (a + J) 2 . 
 .•.^ 2 +6(a + J)^x 2 -5(a + J)5 = a; 4 +(a+^ 2 -30(a+J) J 
 
 
 Exercise 31. 
 
 Find by inspection the 
 
 product 
 
 ;of: 
 
 1. (x + 8) (x + 3). 
 
 15. 
 
 (x + 6y)(x-5y). 
 
 2. (a + 8)(«- 3). 
 
 16. 
 
 (x 2 - 9) (x 2 + 8). 
 
 3. (a-7>(« + 10). 
 
 17. 
 
 (a 2 + 2y 2 )(^-3y 2 ). 
 
 4. (x -9)(x- 5). 
 
 18. 
 
 (x 2 +82/ 2 )(aj 2 -4y 2 ). 
 
 5. (a - 10) (x + 9). 
 
 19. 
 
 (ab - 8) (a& + 5). 
 
 6. (a -10) (a -5). 
 
 20. 
 
 (a£ — 7 icy) (ab 4- 3 ay). 
 
 7. (a - 3a) (a + 2a). 
 
 21. 
 
 (z-32/)(z-3y). 
 
 8. (a + 2 J) (a-4&). 
 
 22. 
 
 (a + 6) (a + 6). 
 
 9. (a -12) (a -3). 
 
 23. 
 
 (a-3£)(a-3£). 
 
 10. (a + 26) (a + 46). 
 
 24. 
 
 (a; — c) (a — d). 
 
 11. (a-3o)(a + 76). 
 
 25. 
 
 (a + a) (x — 5). 
 
 12. (a + 25) (a -9 J). 
 
 26. 
 
 (a: — a) (a + b). 
 
 13. (s-3a) (s-4a). 
 
 27. 
 
 \(a + b) + 2\\(a + b)-± 
 
 14. (s + 4*)(«-2«). 28. \(x+y)-2\\(x+y)+±\. 
 
SPECIAL BULES OF MULTIPLICATION. 81 
 
 114. In like manner the product of any two binomials 
 may be written. 
 
 1. Multiply 2a-bby Sa + Ab. 
 
 (2a-b)(3a + 4:b) = 6a 2 + 8ab-3ab-4tb* 
 = 6a 2 + 5ab-U 2 . 
 
 2. Multiply 2x + 3y by 3x-2y. 
 
 The middle term is 
 
 2x X ( — 2y) + 3y X 3x = 5xy. 
 .-. (2 a; + 3y) (3x -2y) = 6x 2 + 5xy- 6y*. 
 
 Exercise 32. 
 Find the product of : 
 
 1. 3x — y and 2x -\-y. 6. 10 a; — 3y and 10 a; — 1 y. 
 
 2. ±x-3y tm&3x-2y. 7. 3a 2 - 2b 2 and 2a? + 3b\ 
 
 3. 5 x — 4 y and 3 x — 4 y. 8. a 2 -\-b 2 and a — b. - 
 
 4. a; -7?/ and 2a; -5y. 9. 3a 2 - 26 2 and 2a + 3 J. 
 
 5. 11 x — 2 y and 7 x -\- y. 10. a 2 — b 2 and a -{- b. 
 
 11. 4a;-fl and 3a — 2. 
 
 12. 3 a — 5 and x + 1. 
 
 13. 3 a 2 + a; 2 and 4 a 2 -a; 2 . 
 
 14. 2x + y and x + 2y. 
 
 15. 3 J + a and 2o-3«. 
 
 16. 2 a -{-5 b and 4 a — 36. 
 
 17. 4a; + 3^ and 2a; - 7y. 
 
 18. 2y + 3z emd 3y — z. 
 
 19. 2 a; + 7 y and 3 a; — y. 
 
 20. 3 a — 2 c and 2a — 5 c, 
 
82 SPECIAL RULES OF DIVISION. 
 
 Division. 
 
 115. The following rule for finding any required root of 
 a monomial will be found useful in solving examples in 
 division : 
 
 Find the required root of the numerical coefficient, and 
 divide the exponent of each letter by the index of the re- 
 quired root. 
 
 Thus, the square root of 25 » 2 y 4 is 5 xy\ 
 
 116. Difference of Two Squares. 
 
 a 2 — b 2 a 2 — b 2 
 
 — - = a — b: — = a -\-b. Hence, 
 
 a + b a — b 
 
 Kule 1. The difference of the squares of two numbers is 
 divisible by the sum of the numbers, and the quotient is the 
 difference of the numbers. 
 
 The difference of the squares of two numbers is divisible 
 by the difference of the numbers, and the quotient is the 
 sum of the numbers. 
 
 Exercise 33. 
 Write by inspection the quotient of : 
 
 a 2 - A 
 
 7. 
 
 2. — 8. 
 
 a — 
 
 2 
 
 9-; 
 
 X 2 
 
 3 + 
 
 X 
 
 16- 
 
 -a 2 
 
 4 + 
 
 a 
 
 x 2 - 
 
 25 
 
 x — 
 
 5 
 
 36- 
 
 X 2 
 
 6 + 
 
 X 
 
 9a 2 
 
 -b 2 
 
 4. — •• 10. 
 
 x — o 
 
 6 — — — — — • 12 
 
 3a -b ' a-(b-\-c) 
 
 9a 4 - 
 
 -252/* 
 
 3 a 2 
 
 + 5y 2 
 
 4z 16 
 
 -9y« 
 
 2 x s 
 
 -3/ 
 
 4x 10 
 
 -a 8 
 
 2x b 
 
 -a 4 
 
 flW 
 
 t-x™ 
 
 ab*& 
 
 l + z 6 
 
 x 4 a» 
 
 ~b w 
 
 x 2 a* 
 
 -b* 
 
 a 2 - 
 
 (b + c) 2 
 
SPECIAL RULES OF DIVISION. 83 
 
 117. Sum and Difference of Two Cubes. By performing 
 the division, we find that 
 
 a 8 + b 8 a 8 — b 8 
 
 — — = a 2 — ab + b 2 ; — = a 2 -t- ab + b 2 . Hence, 
 
 a -f- b a — b 
 
 Rule 2. The sum of the cubes of two numbers is divisible 
 by the sum of the numbers, and the quotient is the sum of 
 the squares of the numbers minus their product. 
 
 Rule 3. The difference of the cubes of two numbers is 
 divisible by the difference of the numbers, and the quotient 
 is the sum of the squares of the numbers plus their product. 
 
 Exercise 34. 
 
 Write by inspection the quotient of : 
 
 « l-8z 8 a 8 b 8 -c* 
 1. z £ 9. — -7 
 
 ab — c 
 
 a 8 b 8 + c* 
 
 ab + c 
 64-f-y 8 
 
 4 + y ' 
 343 -8 a 8 
 
 7-2a 
 8 a 8 + b« 
 
 2 a +b 2 
 x« + 729y* 
 
 x 2 + 9y 
 a 9 -27 b 8 
 
 a 2 -3b' 
 
 8a 8 -64V* 
 l fi - • 
 
 l + 3« 2x-±y 2 
 
 l-2x 
 
 l + 8z 8 
 
 l + 2x 
 
 27 a 8 - b 8 
 
 3a-b 
 
 27 a 8 + b 8 
 
 3a + b 
 
 64 x 8 + 27 y 8 
 
 4x + 3y 
 
 64 x 8 - 27 y 8 
 
 4:X — 3y 
 
 1-27 z 8 
 
 l-3z 
 
 1 + 27 z 8 
 
 2 - i+jt 10 - 
 
 6a — b 
 
 «• *££ 
 
 „ 64x 8 + 27v 8 
 
 64 a 8 - 27 y 8 
 6. — ; ^-- 14. 
 
 15. 
 
a 
 
 -b 
 
 a 4 
 
 -b 4 
 
 a 
 
 + b 
 
 a 5 
 
 -b 5 
 
 a 
 
 -b 
 
 a" 
 
 + & 5 
 
 84 SPECIAL RULES OF DIVISION. 
 
 118. Sum and Difference of any Two Like Powers. By 
 
 performing the division, we find that 
 
 = a*-a 2 b + ab 2 - £ 8 ; 
 
 = a 4 + a s b -h a*b 2 + ab* + b 4 \ 
 
 = a 4 - a 8 & + a 2 & 2 - a& 8 + £ 4 . 
 <z + b 
 
 We find by trial that 
 
 a 2 + J 2 , a 4 4- b 4 , a 6 + & 6 , and so on 
 are wotf divisible by a -{- b ov by a — b. Hence, 
 
 When n is a positive integer, it is proved in chap, vii, 
 
 1. a n + b n is divisible by a + b if n is odd, and by neither 
 a + b nor a — b if n is even. 
 
 2. a n — b n is divisible by a — b if n is odd, and by both 
 a + b and a — b if n is even. 
 
 Note. It is important to notice in the above examples that the 
 terms of the quotient are all positive when the divisor is a — b, and 
 alternately positive and negative when the divisor is a + b ; also, that 
 the quotient is homogeneous, the exponent of a decreasing and of b 
 increasing by 1 for each successive term. 
 
 Exercise 35. 
 Find the quotient of : 
 
 x«-y* x 4 -l m a; 5 + 32 
 
 4. 
 
 x — y 
 
 X 
 
 -hi 
 
 X 4 
 
 -16 
 
 X 
 
 -2 
 
 a* 
 
 -32 
 
 ^ se 6 — V 5 
 2. y • 
 
 x + y 
 
 3 - x-1 6 ' x-2 9 ' 
 
 X 
 
 1 
 
 + 2 
 — m 4 
 
 1 — m 
 
 1 + m 6 
 
CHAPTER VIL 
 FACTORS. 
 
 119. Rational Expressions. An expression is rational 
 if none of its terms contain square or other roots. 
 
 120. Factors of Rational and Integral Expressions. By fac- 
 tors of a given integral number in Arithmetic we mean 
 integral numbers that will exactly divide the given number. 
 
 Likewise, by factors of a rational and integral expression 
 in Algebra we mean rational and integral expressions that 
 will exactly divide the given expression. 
 
 121. Factors of Monomials. The factors of a monomial 
 may be found by inspection. Thus, the factors of 14 a 2 b 
 are 7, 2, a, a, and b. 
 
 122. Factors of Polynomials. The form of a polynomial 
 that can be resolved into factors often suggests the process 
 of finding the factors. 
 
 123. When the terms have a common monomial factor. 
 Kesolve into factors 2x 2 + 6 xy. 
 
 Since 2 and x are factors of each term, we have 
 
 2x 2 + 6xy 2x2 , 6xy 
 
 2x =Jx- + ^ = X + Sy ' 
 .-. 2x 2 + Qxy = 2x(x + Sy). 
 
 Hence, the required factors are 2 x and x + 3 y. 
 
86 FACTORS. 
 
 Exercise 36. 
 
 Resolve into factors : 
 
 •1. 3a 2 -6a 8 . 13. 8 a 2 x 2 -4a 2 b + 12 ahj\ 
 
 2. 2 a 2 -4 a. 14. 8 a 8 b 2 c 2 - 4 aW + 2 aW. 
 
 3. 5ab-5a 2 b*. 15. 15 a 8 * - 10 aty + 5 a 8 *. 
 
 4. 3a 2 £-4a& 2 . 16. a 8 cy 8 + 2 aVy 2 - a 2 c2/ 4 . 
 
 5. SxY + 4,xy. 17. 3& 8 c 8 + 2& 2 c 2 -6fo 8 . 
 
 6. 3a 8 -a 2 + a. 18. 6a 2 b - Sab - 12 ab 2 . 
 
 7. x* + x 2 y-xy\ 19. 5a¥ + 3a 2 c + 4aV. 
 
 8. a*-a*b + a 2 b 2 . 20. 6 a; 8 / + 3 x 2 y 2 - 15 xy\ 
 
 9. 3 a 4 -9 a 2 -6 a 8 . 21. 7 aW - 14 a£c + 7 ab 2 c\ 
 
 10. a£ 2 -6c 2 + fa;. 22. 8x 2 y 2 + 16xyz-24:X 2 y 2 »*. 
 
 11. 8 a 2 b - 6 a 8 -f 4 a&. 23. a£ 2 c 8 - 2 a 2 fo + 3 a 8 6 8 c 2 . 
 
 12. 4 x 2 y — 8 a??/ 2 — 4 icy. 24. x 2 y 2 z 2 — x 8 y 2 z z -f ic 2 i/ 8 «. 
 
 124. When the terms can be grouped so as to show a common 
 compound factor. 
 
 1. Resolve into factors ac + ad + be -f #<£ 
 
 ac + ad + 6c + bd = (ac + ad) + (be + M) (1) 
 
 = a(c + d) + b(c + d) (2) 
 
 = (a + 6) (c + d). (3) 
 
 Since one factor is seen in (2) to be c .+ d, dividing by c + d we 
 obtain the other factor, a + b. 
 
 2. Resolve into factors 3x 2 + 6ax + bx + 2ab. 
 
 3& + 6ax + bx + 2ab = (Sx* + 6ax) + (bx + 2db) 
 = 3x (x + 2 a) + b (x + 2a) 
 = (3x + 6)(x + 2a). 
 
FACTORS. 87 
 
 3. Find the factors of ac + ad — be — bd. 
 
 ac + ad — bc — bd = (ac + ad) — (bc + bd) 
 = a{c + d)-b(c + d) 
 = (a-b)(c + d). 
 
 Note. Here the last two terms, — be — bd, being put within a 
 parenthesis preceded by the sign — , have their signs changed to +. 
 
 4. Eesolve into factors 3 # 8 — 5 x 2 — 6 x + 10. 
 
 3x« - 5x 2 - 6x + 10 = (3x 8 - 5x 2 ) - (6x - 10) 
 = x 2 (3x-5)-2(3x-5) 
 = (x 2 -2)(3x-5). 
 
 5. Eesolve into factors 5 x z — 15 ax 2, — x + 3 a. 
 
 6x 8 - 15 ox 2 - x + 3 a = (5x 3 - 15 ax 2 ) - (x — 3 a) 
 = 5x 2 (x — 3 a) — 1 (x — 3 a) 
 = (5x 2 -l)(x-3a). 
 
 6. Eesolve into factors 6 y — 27 x 2 y — 10 x + 45 a; 8 . 
 
 6 y - 27 xfy — lOx + 45 x 8 = 6y — lOx — 27 x 2 y + 45 x 8 
 
 = (6y - lOx) - (27 x 2 y - 45 x 8 ) 
 = 2 (3y — 5x) — 9x 2 (3y — 5x) 
 = (2-9x 2 )(3y-5x). 
 
 Exercise 37. 
 
 Eesolve into factors : 
 
 1. ax — focVf ay — by. 
 
 2. ax — bx — ay r by. 
 
 3. ax — cy — ay -\- ex. 
 
 4. se 2 + aa? — &» — ab. 
 
 5. cc 2 + jcy — ax — ay. 
 
 6. x 2 — xy — 6x + 6y. 
 
 7. 2a£-3ac-2&y 4-3cy. 
 
 8. 2 x 2 — 3 ccy + 4 aa? — 6 ay. 
 
 9. ab-3bc-2ae+6e 2 . 
 
 10. x z -\- x — x 2 z — «. 
 
 11. a: 8 4- 4a: 2 4- 3a 4- 12. 
 
 12. 3ae — 3ax — c + x. 
 
88 FACTORS. 
 
 13. a% — abx — ac +- ex. 23. a%x +- # 2 c# — a 2 cy — fofy 
 
 14. 2a 8 -3a 2 -4a;+-6. 24. 3z 2 - 5y 2 - 6x B + 10xy 2 
 
 15. aa 4 + bx* - aa - b. 25. 8 ace - 10 foe - 12 a + 156. 
 
 16. ax 2 + a 2 x + a + x. 26. 6 x 4 + 8 x* - 9 ic 2 - 12 x. 
 
 17. (x-yy + 2y(x-y). 27. 3c# 4 - 2aV - 9cx 2 +- 6dx. 
 
 18. H-15cc 4 -5cc-3ic 8 . 28. (a + b)(c + d)-3c(a + b). 
 
 19. a 2 - « 8 + 1 — x. 29. b s x + bc 2 x — b 2 cy - c z y. 
 
 20. £c 8 -5» 2 -f 2a;- 10. 30. 2ac-bc + 4:a 2 -2ab. 
 
 21. a 8 +-7x 2 +-3a+-21. 31. A + ac 2 -«w-ck 
 
 22. 2 4 -« 8 -s + l. 32. 1+- c - c*xy - c s xy. 
 
 125. When a trinomial is a perfect square. A trinomial is 
 a perfect square if its first and last terms are perfect 
 squares and positive, and its middle term is twice the 
 product of the square roots of the first and last terms. 
 
 Thus, 16 a 2 — 24 ab + 9 b 2 is a perfect square. 
 
 The rule for extracting the square root of a perfect 
 trinomial square is as follows : 
 
 Extract the square roots of the first and last terms, and 
 connect these square roots by the sign of the middle term. 
 
 Thus, if we wish to find the square root of 
 
 16a 2 -24o6 + 962, 
 
 we take the square roots of 16 a 2 and 9 b 2 , which are 4 a and 36, 
 respectively, and connect these square roots by the sign of the middle 
 term. The square root is therefore 
 
 4a-S6. 
 In like manner, the square root of 
 
 16a 2 + 24 06 + 962 is 
 4a + 36. 
 
FACTORS. 89 
 
 1. Resolve into factors x 2 + 2 xy + y 2 . 
 
 The factors of x 2 + 2 xy + y 2 are 
 
 (x + y) (x + y). 
 
 2. Resolve into factors a; 4 — 2 x 2 y + y 2 . 
 
 The factors of x 4 — 2 x 2 y + y 2 are 
 
 (x 2 -y) (x 2 - y). 
 
 Exercise 38. 
 Resolve into factors : 
 
 1. a 2 -6ab + 9b 2 . 18. 121 a 2 + 198 ay + 81 */ 2 . 
 
 2. 4a 2 + 4a& + £ 2 . 19. aW - 2 a&W 5 + a; 16 . 
 
 3. a 2 -4a& + 46 2 . 20. 49 - 140 k* + 100 A; 4 . 
 
 4. x 2 + 6xy + 9y 2 . 21. 49 a 2 + 42 ac 2 + 9 c 4 . 
 
 5. 4a 2 -12az + 9a 2 . 22. 81c 2 -90c + 25. 
 
 6. a 2 - 10 ab + 25b 2 . 23. 121 + 110 a; + 25 a 2 . 
 
 7. 4a a -4a + l. 24. 144 + 168 3 + 49 z 2 . 
 
 8. 49 2/ 2 — 14 yz + z 2 . 25. 36 a 2 - 60 ccy + 25 y 2 . 
 
 9. x 2 -16 a; + 64. 26. y 2 - 50 yz + 625 « 2 . 
 
 10. 9 a: 2 + 24 a:?/ + 16 y 2 . 27. a; 6 - 34 x* + 289. 
 
 11. 16 a 2 + 8 ax + x 2 . 28. 49 a; 2 - 112 xy + 64 y\ 
 
 12. 25 + 80 a; + 64 a; 2 . 29. 49 a 2 b 2 - 28 abc + 4 c 2 
 
 13. 49 a; 2 -28^2/ + 4 2/ 2 . 30. 121 a; 2 - 286 xy + 169 y\ 
 
 14. 1-20 6 + 100& 2 . 31. 4a 4 + 20a 2 a; 2 + 25a; 4 . 
 
 15. 81a 2 + 126a& + 49 6 2 . 32. (x + y) 2 -4z(x + y) + 4:Z 2 . 
 
 16. m 2 ^ 2 - IQmna 2 + 64a 4 . 33. (a - b) 2 - 6 (a - b) + 9. 
 
 17. 4a 2 -20aaj + 25a; 2 . 34. (a + c) 2 + 10(a + c) +25. 
 
90 FACTORS. 
 
 126. When a binomial is the difference of two squares. 
 
 The difference of two squares is the product of two 
 factors, which may be found as follows : 
 
 Take the square root of the first term and the square root 
 of the second term. 
 
 The sum of these roots will form the first factor. 
 
 The difference of these roots will form the second factor. 
 
 Resolve into two factors 16 x 2 — 9 y*. 
 
 The square root of 16 x 2 is 4 x. 
 The square root of 9 y 6 is 3 2/ 3 . 
 The sum of these roots is 4 x + 3 y 8 . 
 The difference of these roots is 4 x — 3 y 8 . 
 Therefore, 16 x 2 — 9 y 6 = (4 x + 3 y s ) (4 z — 3 y 8 ). 
 
 Exercise 39. 
 Eesolve into factors : 
 
 1. a 2 -4. ' 13. 25-16a 2 - 25. 8lx 2 -±y\ 
 
 2. 1-x 2 . 14. 16-25y 2 . 26. 64 a 4 - b\ 
 
 3. x 2 -9y 2 . 15. a 2 b 2 -l. 27. m 2 n 2 - 36. 
 
 4. 4a 2 -496 2 . 16. x 2 - 100. 28. z 4 - 144. 
 
 5. a 2 -4 ?/ 2 . 17. 121 a 2 - 36 6 2 . 29. z 2 -25. 
 
 6. 49- 100 y 2 . 18. 49 a 14 -?/ 12 . 30. 25-64y 2 . 
 
 7. l-49ic 6 . 19. 64, a 2 -9 b 6 . 31. 16x 17 -9xy*. 
 
 8. 4-121?/ 8 . 20. 81a 4 6 4 -c 4 . 32. 25x 10 -16a*x\ 
 
 9. 1-169 a 6 . 21. 4a 2 c-9c 5 . 33. 36 a 2 x 2 - 49 a 4 . 
 
 10. a 2 6 2 - 4 c 6 . 22. 20 a 3 b* - 5 ab. 34. cc 2 - 16 y\ 
 
 11. 9x 8 -a 6 . 23. 3a 5 -12aV. 35. 1-400 a 4 . 
 
 12. 4a 16 -2/ 20 . 24. 9 a 2 -Sib 2 . 36. 4a 2 c-9c 8 . 
 
FACTORS. 91 
 
 127. If the squares are compound expressions, the same 
 method may be employed. 
 
 1. Eesolve into factors (x + 3y) 2 — 16 a 2 . 
 
 The square root of the first term is x + 3 y. 
 
 The square root of the second term is 4 a. 
 
 The sum of these roots is x + 3 y + 4 a. 
 
 The difference of these roots is x + 3 y — 4 a. 
 
 Therefore, (x + Sy) 2 — 16a 2 = (x + 3y + 4a) (x + Sy — 4a). 
 
 2. Eesolve into factors a? — (3b — 5c) 2 . 
 
 The square roots of the terms are a and (3 6 — 5 c). 
 The sum of these roots is a + (3 6 — 5 c), or a + 3 b — 5 c. 
 The difference of these roots is a — (3 6 — 5 c), or a — 3 b + 5 c. 
 Therefore, a 2 - (36 - 5c) 2 = (a + 36 - 5c) (a - 36 + 5c). 
 
 If the factors have like terms, these terms should be 
 llected so as to give the results in the simplest form. 
 
 3. Eesolve into factors (3 a + 5 b) 2 - (2 a - 3 b) 2 . 
 
 The square roots of the terms are 3 a + 5 6 and 2 a — 3 6. 
 The sum of these roots is (3 a + 5 6) + (2 a — 3 6), 
 
 or3a + 56 + 2a — 36 = 5a + 2 6. 
 The difference of these roots is (3 a + 5 6) — (2 a — 3 6), 
 
 or3a + 56 — 2a + 36 = a + 8 6. 
 Therefore, (3a + 5 6) 2 - (2 a - 3 6) 2 = (5 a + 2 6) (a + 8 6). 
 
 Exercise 40. 
 
 Eesolve into factors : 
 (x + y) 2 -z 2 . 5. (a - b) 2 - (c - d) 2 . 
 
 (x-y) 2 -z 2 . 6. (2 a + b) 2 - 25 c 2 . 
 
 . (x-2y) 2 -4z 2 . 7. (x + 2y) 2 -(2x-y) 2 . 
 
 (a + 3b) 2 - 16c 2 . 8. (x + 3) 2 -(3x- 4) 2 . 
 
92 FACTORS. 
 
 9. x 2 -(y-z) 2 . 15. (a + b-cy-(a-b-c)\ 
 
 10. a 2 -(3b -2c) 2 . 16. (a-3a) 2 -(3a-2;c) 2 . 
 
 11. b 2 -(2a + 3c) 2 . 17. (2a- l) 2 - (3a + 1) 2 . 
 
 12. l-(a + 5&) 2 . ' 18. (x-5) 2 -(x + y-5) 2 . 
 
 13. 9a 2 - (x -3c) 2 . 19. (2a + £-c) 2 -(a-2&+c) 2 , 
 
 14. 16a 2 -(2y-3z) 2 . 20. (a + 2b-3cy-(a + 5cy. 
 
 128. By properly grouping the terms, compound expres- 
 sions may often be written as the difference of two squares, 
 and the factors readily found. 
 
 1. Kesolve into factors a 2 — 2ab -\- b 2 — 9 c\ 
 
 a 2 — 2ab + 6 2 — 9c 2 = (a 2 - 2ab + 6 2 ) — 9c 2 
 = (a - b) 2 - 9 c 2 
 = (a-b + 3c)(a-b — Sc). 
 
 2. Kesolve into factors 12 ab + 9 x 2 - 4 a 2 - 9 £ 2 . 
 
 Here 12 ab shows that it is the middle term of the expression which 
 • has in its first and last terms a 2 and 6 2 , and the minus sign before 4 a 2 
 and 9 b 2 shows that these terms must be put in a parenthesis with the 
 minus sign before it, in order that they may be made positive. 
 
 Therefore, the arrangement will be 
 9s2-(4a 2 -12a& + 9& 2 ) = 9x 2 - (2a - 36) 2 
 
 = (3s + 2a — 3&)(3» — 2a + 36). 
 
 3. Kesolve into factors - a 2 + b 2 - c 2 + d? + 2ac + 2bd. 
 
 Here 2 bd must be grouped with 6 2 and d 2 , and 2 ac with — a 4 and 
 — c 2 } and this last group put in a parenthesis preceded by — . 
 
 - a 2 + b 2 - c 2 + d 2 + 2 ac + 2 bd 
 
 = (b 2 + 2 6d + d 2 ) - (a 2 - 2 ac + c 2 ) 
 
 = (6 + d) 2 - (a - c) 2 
 
 sb (& + d + « — c) (b + 4*- a + c). 
 
FACTORS. 93 
 
 Exercise 41. 
 Resolve into factors : 
 
 1. a 2 + 2ab + b 2 -4:C 2 . 5. a 2 - x 2 - y 2 - 2 xy. 
 
 2. x 2 -2xy + y 2 -9a 2 . 6. l-a 2 -2ab-b 2 . 
 
 3. b 2 -x 2 + ±ax-±a 2 . 7. a 2 + b 2 + 2ab - l§a 2 b\ 
 
 4. 4a 2 + 4a& + & 2 -# 2 . 8. Ax 2 - 9 a 2 + 6a - 1. 
 
 9. a 2 + & 2 -c 2 -d 2 -2a&-2ctf\ 
 
 10. x 2 + y 2 - 2xy - 2ab - a 2 - b 2 . 
 
 11. 9a 2 -6a; + l-a 2 -4a&-46 2 
 
 12. a 2 + 2ab-x 2 -6xy-9y 2 + b 2 . 
 
 13. a 2 - 2# + l-& 2 + 2fy -y 2 . 
 
 14. 9-6a-ha 2 -a 2 -8a&-16& 2 . 
 
 15. 4-4« + a 2 - ±ab-b 2 -±a\ 
 
 16. a 4 - a 2 - 9 + V + 6 a - 2 a 2 6 2 . 
 
 17. 4a 2 + 9c 2 -12ac + 12^-9& 2 -4d 2 
 
 18. 4cc 2 -42/ 2 -4a; + l-« 2 + 42/«. 
 19.. ±xy-x 2 + l -±y 2 . 
 
 20. c 2 -2ac-l + 2£-}-a 2 -& 2 . 
 
 21. 4ac-l-6a;-9a; 2 + a 2 + 4c 2 . 
 
 22. 4-9ic 2 -42/ 2 + 12icy. 
 
 23. 4a 4 -12a 2 -9z 2 + 12 2/z-4?/ 2 + 9. 
 
 24. a 4 - b 2 -4 : x i -6a 2 c + 4;bx 2 + 9c 2 . 
 
 25. 25a 2 -l-10a&-9a 2 a 2 + & 2 +6«sc. 
 
 26. 16£c 4 + 30cc 8 + 8c 2 x 2 -25x 6 -9 + c 4 . 
 
 27. 4 a 4 + 9 b« - 12 a 2 6 8 - 81 a 2 ^ 6 . 
 
94 FACTORS. 
 
 129. A trinomial in the form a 4 + a 2 b 2 + 6 4 can be 
 
 written as the difference of two squares. 
 
 Since a trinomial is a perfect square when the middle 
 term is twice the product of the square roots of the first 
 and last terms, it is obvious that we must add a% 2 to the 
 middle term of a 4 + a 2 b 2 + & 4 to make it a perfect square. 
 We must also subtract a 2 b 2 to keep the value of the 
 expression unchanged. 
 
 1. Eesolve into factors a 4 4- a 2 b 2 -f- b\ 
 
 a 4 + a 2 6 2 + & 4 = a 4 + 2 a 2 6 2 + 6 4 - a 2 6 2 
 = (a 2 + 6 2 ) 2 - a 2 6 2 
 = (a 2 + 6 2 + ab) (a 2 + 6 2 - ab) 
 = (a 2 + a6 + 6 2 ) (a 2 - ab + 5 2 ). 
 
 2. Resolve into factors 4 a 4 - 37 »y -f- 9 y 4 . 
 
 Twice the product of the square roots of 4 x 4 and 9 y* is 12 x*y*., 
 We may separate the term — 'Sixty 2 into two terms, — 12x 2 y 2 and 
 — 25 x 2 y 2 , and write the expression 
 
 (4 x 4 - 12 x 2 ^ 2 + 9 y 4 ) - 25 x 2 y 2 
 
 = (2x 2 — 32/ 2 ) 2 -25xV 
 
 = (2x 2 — Sy 2 + 6xy) (2x 2 — Sy 2 — 5xy) 
 
 = (2x 2 + hxy — Sy 2 ) (2x 2 — hxy —■ Sy*). 
 
 Exercise 42. 
 Resolve into factors 
 
 1. x* + xy + y*. 6. 9a* + 26a 2 b* + 25b\ 
 
 2. » 4 + a 2 + 1. 7. 4x* - 21x 2 f + 9y\ 
 
 3. 9a 4 -15a 2 + l. 8. 4a 4 -29a 2 c 2 + 25c 4 . 
 
 4. 16a 4 - 17a 2 + 1. 9. 4 a* + 16 a 2 c 2 + 25 c 4 . 
 
 5. 4a 4 -13a 2 + l. 10. 25 a 4 + 31 *y + 16 y* 
 
FACTOBS. 95 
 
 130. When a trinomial has the form x 2 + ax + 6. 
 
 Where a is the algebraic sum of two numbers, and is 
 either positive or negative ; and b is the product of these 
 two numbers, and is either positive or negative. 
 
 Since (x + 5) (x + 3) = x 2 + Sx + 15, 
 
 the factors of x 2 -f 8 a; + 15 are a; + 5 and sc + 3. 
 
 Since (x + 5) (a - 3) = x 2 + 2x - 15, 
 
 the factors of x 2 + 2 a; — 15 are cc ■+■ 5 and a; — 3. 
 
 Hence, if a trinomial of the form x 2 -\- ax + b is such an 
 expression that it can be resolved into two binomial factors, 
 the first term of each factor will be x ; and the second 
 terms of the factors will be two numbers whose product is b, 
 the last term of the trinomial, and whose algebraic sum is a, 
 the coefficient of x in the middle term of the trinomial. 
 
 1. Eesolve into factors x 2 + 11 x -J- 30. 
 
 We are required to find two numbers whose product is 30 and 
 whose sum is 11. 
 
 Two numbers whose product is 30 are 1 and 30, 2 and 15, 3 and 10, 
 6 and 6 ; and the sum of the last two numbers is 11. Hence.; 
 
 x 2 + 11 x + 30 = (x + 5>(a + 6). 
 
 2. Eesolve into factors x 2 — 7 x + 12. 
 
 We are required to find two numbers whose product is 12 and 
 whose algebraic sum is — 7. 
 
 Since the product is + 12, the two numbers are both positive or both 
 negative ; and since their sum is — 7, they must both be negative. 
 
 Two negative numbers whose product is 12 are — 12 and — 1,-6 
 and — 2,-4 and — 3 ; and the sum of the last two numbers is — 7. 
 Hence, 
 
 x 2 - Ix 4- 12 = (x- 4) (x - 3). 
 
96 FACTORS. 
 
 3. Resolve into factors x 2 + 2 x ~ 24* 
 
 We are required to find two numbers whose product is — 24 and 
 whose algebraic sum is 2. 
 
 Since the product is — 24, one of the numbers is positive and the 
 other negative ; and since their sum is +2, the larger number is 
 positive. 
 
 Two numbers whose product is — 24, and the larger number posi- 
 tive, are 24 and — 1, 12 and — 2, 8 and — 3, 6 and — 4 ; and the sum 
 of the last two numbers is + 2. Hence, 
 
 x 2 -f- 2x - 24: = (x + 6) (x - 4). 
 
 4. Resolve into factors x 2 — 3 x — 18. 
 
 Since the product is — 18, one of the numbers is positive and the 
 other negative ; and since their sum is — 3, the larger number is 
 negative. 
 
 Two numbers whose product is — 18, and the larger number nega- 
 tive, are — 18 and 1, — 9 and 2,-6 and 3 ; and the sum of the last 
 two numbers is — 3. Hence, 
 
 x 2 - Sx - 18 m (x - 6) (x + 3). 
 
 5. Resolve into factors x 2 — 10 xy + 9 y 2 . 
 
 We are required to find two numbers whose product is Oy 2 and 
 whose algebraic sum is — 10 y. 
 
 Since the product is + 9y 2 , and the sum — 10 y, the last two terms 
 must both be negative. 
 
 Two negative numbers whose product is Qy 2 are — Qy and — y, 
 — Sy and — Sy ; and the sum of the first two numbers is — 10 y. 
 Hence, 
 
 x 2 - lOxy + 9y 2 ^(x- 9y) (x - y). 
 
 131. From these examples it will be seen that the 
 following statements are true : 
 
 1. If the third term of a given trinomial is negative, the 
 second terms of its binomial factors have unlike signs. 
 
 2. If the third term of a given trinomial is positive, the 
 second terms of its binomial factors have the same sign, and 
 this sign is the sign of the middle term. 
 
 
FACTORS. 97 
 
 Exercise 43. 
 Resolve into factors : 
 
 1. x 2 + 8x + 15. 24. x 2 + Tx + 10. 
 
 2. x 2 -8x + 15. 25. ic 2 -7# + 10. 
 
 3. a; 2 + 2a; -15. 26. z 2 + 3a;-10. 
 
 4. x 2 -3ic-10. 27. a 2 + ax -6a 2 . 
 6. a; 2 + 5 ax + 6 a 2 . 28. x 2 — ax — 6 a 2 . 
 
 6. x 2 - 5 ax + 6 a 2 . 29. a 2 + 5 scy -f 4 y 2 
 
 7. £c 2 -2x-15. 30. x 2 - 3x7/ -Ay 2 . 
 
 8. x 2 + 5«+6. 31. a 2 -5ajy + 4y 2 . 
 
 9. x 2 -5x + 6. 32. x 2 + 3xy-±y\ 
 
 10. £C 2 + x-6. 33. cc 2 + 3a;y + 2y 2 . 
 
 11. x 2 - x — 6. 34. a 2 -7ab + 10b 2 . 
 
 12. a; 2 + 6a; + 5. 35. a 2 x 2 - 3 aa; - 54. 
 
 13. a; 2 -6a; + 5. 36. a; 2 -7a; -44. 
 
 14. a; 2 + 4a;-5. 37. x 2 + x-132. 
 
 15. a; 2 -4a; -5. 38. x 2 — 15a; + 50. 
 
 16. a; 2 + 9a; + 18. 39. a 2 -23 a + 120. 
 
 17. a; 2 -9a; + 18. 40. a 2 + 17 a -390. 
 
 18. a; 2 + 3a; -18. 41. c 2 + 25c-150. 
 
 19. a; 2 -3a; -18. 42. g 2 - 58c + 57. 
 
 20. x 2 + 9x + 8. 43. a 4 -lla 2 £ 8 + 30&». 
 
 21. x 2 -9x + 8. 44. « 2 + 9«?/ + 202/ 2 . 
 
 22. x 2 + Tx-8. 45. £cV + 19a;^+48«* 
 
 23. a; 2 -7 a; -8. 46. a 2 b 2 — 13 abc + 22 c\ 
 
98 FACTORS. 
 
 132. When a trinomial has the form ax 2 + bx + c. 
 
 1. Find the factors of 8 x 2 - 22 x - 21. 
 
 Multiply by 8, the coefficient of x 2 , and write the result in the 
 following form: 
 
 (8x) 2 -22 X 8x — 168. 
 Put z for 8 x, and we have 
 
 z 2 - 22 z - 168. 
 Resolve this expression into two binomial factors (§ 130) 
 (2 - 28) (s + 6). 
 
 Since we have multiplied by 8, and put z for 8 x, we must reverse 
 this process. Hence, put 8 x for z and divide by 8, and we have 
 (8s — 28) (8s + 6) , 
 8 
 
 As 4 is a factor of (8x — 28), and 2 is a factor of (8x + 6), we 
 divide by 8 by dividing the first factor by 4 and the second factor by 
 2, thus 
 
 (8 x — 28) (8 s + 6) 
 
 4X2 
 = (2x-7)(4x + 3). 
 
 2. Find the factors of 24 a; 2 - 70 xy - 75 y\ 
 
 Multiply by 24, (24 x) 2 — 70 y X 24 x — 1800 y 2 . 
 
 Put z for 24 x, z 2 — 7Qyz — 1800 y 2 . 
 
 Resolve into factors, (z — 90 y) (z + 20 y). (§ 130) 
 
 Put 24 x for z, (24 x — 90 y) (24 x + 20 y). 
 
 Divide by 6X4, (4x — 16y) (6x + 6y). 
 
 3. Find, the factors of 12 cc 2 - 23 a^ -f- 10 y 2 . 
 
 Multiply by 12, (12 x) 2 - 23 y X 12 x + 120 y 2 . 
 
 Put z for 12 x, z 2 — 2Syz + 120 y 2 . 
 
 Resolve into factors, (z — 15 y) (z — 8 y). (§ 130) 
 
 Put 12 x for z, (12 x — 15y)(12x — $y). 
 
 Divide by 3 X 4, (4* — 5 y) (3 x — 2 y). 
 
FACTORS. 99 
 
 Exercise 44. 
 Eesolve into factors : 
 
 1. 2x 2 + 5x + 3. 25. 2x 2 + 5xy + 2y 2 . 
 
 2. 3x 2 -x~2. 26. 6x 2 -7bx-3b 2 . 
 
 3. 5x 2 -Sx + S. 27. 8a 2 -f 14a&-15& 2 . 
 
 4. 6x 2 + 7x + 2. 28. 6a 2 -19ac + 10c 2 . 
 
 5. 6x 2 -x-2. 29. 8a 2 + 34z?/ + 21y 2 . 
 
 6. 15z 2 + 14z-8. 30. Sx 2 -22xy-21y 2 . 
 
 7. 8x 2 -10a + 3. 31. 6x 2 + 19 xy -7 y 2 . 
 
 8. 18« 2 + 9cc-2. 32. 11 a 2 - 2Sab + 2£ 2 . 
 
 9. 12cc 2 -5x-2. 33. 2c 2 - IScd + 6d 2 . 
 
 10. 12z 2 -7cc + l. 34. 6y 2 + 7yz-3z 2 . 
 
 11. 12x 2 -x-l. 35. 15a: 2 -26a;y + 8y 2 
 
 12. 3x 2 -2x-5. 36. 9av 2 + 6a:y-8?/ 2 . 
 
 13. 3x 2 + ±x-4:. 37. 6z 2 -a^-35y 2 . 
 
 14. 6a 2 + 5;c-4. 38. 10 a; 2 - 21^-10^. 
 
 15. 4z 2 -fl3z + 3. 39. Ux 2 -55xy + 21y 2 . 
 
 16. 4ic 2 + ll£c-3. 40. 6x 2 -23xy + 20y*. 
 
 17. 4a; 2 -4a; -3. 41. 6 x 2 + 35 ay - 6 y 2 . 
 
 18. 4x 2 + 8£C + 3. 42. 2±x 2 -14:xy- 5y 2 . 
 
 19. 6a 2 x 2 + az-l. 43. 24 a 2 -38 ay + 15 y 2 . 
 
 20. 6a 2 + 17 a + 12. 44. 24a 2 -2ay-15y 2 . 
 
 21. 12 x 2 - 13 a -14. 45. 36x 2 -19xy-6y 2 . 
 
 22. 10a 2 -23a -5. 46. 15x 2 + 19xy + 6y 2 . 
 
 23. 8c 2 -|- 53c -21. 47. 12 x 2 + 31 xy - 15 y 2 . 
 
 24. 8« 2 -37s-15. 48. 5y 2 + 13y-6. 
 
100 FACTORS. 
 
 133. When a binomial is the sum or difference of two cubes. 
 
 Since 
 
 — -J-j. = «2 _ ab + h 2 
 
 (§117) 
 
 and 
 
 o? — h z , , ^ , M 
 
 r = a 2 + «& 4- 6 2 ; 
 
 a — 6 
 
 (§117) 
 
 
 .-. a 8 + £ 8 = ( a + & ) ( a 2 _ ah + ja) 
 
 (1) 
 
 and 
 
 a 8 _ & 8 = ( a _ j) ( a * + a& + £2). 
 
 (2) 
 
 Therefore, the sum of two perfect cubes is divisible by 
 the sum of their cube roots, and the difference of two per- 
 fect cubes is divisible by the difference of their cube roots. 
 
 1. Eesolve into factors 8 a 8 + 27 b\ 
 
 The cube root of 8 a 8 is 2 a, and of 27 6 s is 3 6 s . 
 By putting 2 a for a and 3 6 s for 6 in (1), we have 
 (2a)« + (362)3 = (2a + 362) (4a 2 -6a62 + 96*). 
 
 2. Eesolve into factors 125 sc 8 — 1. 
 
 The cube root of 125 x 8 is 5x, and of 1 is 1. 
 By putting 5x for a and 1 for 6 in (2), we have 
 126 x 8 - 1 = (5x - 1) (25 x 2 + 5x + 1). 
 
 3. Eesolve into factors x* -f y 9 . 
 
 The cube root of x 6 is x 2 , and of y 9 is y 8 . 
 
 By putting x 2 for a and y s for 6 in (1), we have 
 
 X« + y9 = (x* + y2) ( X 4 — Xty 8 + y«). 
 
 4. Eesolve into factors (x — y) z + « 8 . 
 
 The cube root of (x — y)* is x — y, and of z 8 is z. 
 
 By putting x ; - - y for a and z for 6 in (1), we have 
 
 (x - yf + z» = [(x - y) + z] [(x - y) 2 - (x - y) z + z 2 ] 
 
 = (x — y + z) (x 2 — 2 xy + y 2 — xz + yz + a: 2 ). 
 
factors. [ t , ,\\ ; , , > ; ' ', , • • \10l 
 
 Exercise 45. 
 Kesolve into factors : 
 
 1. a 8 + 8b 8 . 5. 27x 8 y 8 -l. 9. 216 a 6 -J 8 . 
 
 2. a 8 -27 a 6 . 6. a* + 27 b 8 . 10. 64 a 8 - 27 b\ 
 
 3. a 8 + 64. 7. a 8 y 8 -64. 11. 343 -a 8 . 
 
 4. 125 a 8 + 1. 8. 64 a 6 + 125 b 8 . 12. a 8 b 8 + 343. 
 
 13. 8 a 8 - b\ 33. 8 x 8 - (x - y)\ 
 
 14. 216m 8 + ?i 6 . 34. 8 (x + y) 8 + z\ 
 
 15. xY - 512 z 3 . 35. (a -f 6) 8 - (a - &)•. 
 
 16. 729 a 6 + 216 c 6 . 36. (a 2 - 3) 8 - z 8 
 
 17. 729 2/ 8 - 64 s 8 . 37. 6 8 + (a - e) 8 . 
 
 18. 512 a 8 -1. 38. (z-l) 8 -(2-l) 8 . 
 
 19. (a + 6) 8 -l. 39. (a-Sby-c 8 . 
 
 20. (a-6) 8 + l. 40. (2-3a) 8 + 6 8 . 
 
 21. l-(a-by. 41. (3 + 5) 8 -a 8 . 
 
 22. 27 a 6 + 125 y 8 . 42. (y - z) 8 -f (y + z) 8 . 
 
 23. 216 + 343 y\ 43. (2x + y) 8 - (x- y)\ 
 
 24. a 15 -125 a 6 . 44. l-(7a-56) 8 . 
 
 25. 8ic 18 + 273/ 12 . 45. (3a + y) 8 -8« 8 . 
 
 26. a 16 -2166 9 . 46. (2a-b) 8 -c 8 . 
 
 27. c 8 d 8 - 343 2 6 . 47. x 8 - (y - z)\ 
 
 28. ^¥ + 8. 48. 125- (x — 2 y) 8 . 
 
 29. 8 ay* 9 - 1. 49. (2x + y) 8 - 27 z 8 . 
 
 30. 27a;y^ 6 + l. 50. (a — 3) 8 - z 8 . 
 
 31. 64a: 8 -125?/ 9 . 51. (7a-6) 8 + c 8 . 
 
 32. 125 a 6 -}- 27 y 15 . 52. (3c-2y) 8 -8z 8 . 
 
1,02 • FACTORS. 
 
 Theory of Divisors. 
 
 134. Theorem. The expression x — y is an exact divisor 
 of x n — y n when n is any positive integer. 
 
 Since — a"- 1 y + x n - 1 y = 0, (§34) 
 
 x n — y n = x n — x n ~ 1 y + x n ~ 1 y — y™. 
 
 Taking out x n ~ 1 from the first two terms of the right side, and y from 
 the last two terms, we have 
 
 X 7l — yn = j£»-l ( X _ y) -{- y (rffl- 1 _ y»-l). 
 
 Now a; — y is an exact divisor of the right side, if it is an exact divisor 
 of x n - 1 — y n - 1 ; and if x — y is an exact divisor of the right side, it 
 is an exact divisor of the left side ; that is, x — y is an exact divisor 
 of x n — y n if it is an exact divisor of x n - J — y n - K 
 
 But x — y is an exact divisor of x z — y* (§ 117), therefore it is an 
 exact divisor of sc 4 — y* ; and since it is an exact divisor of ce 4 — y 4 , it 
 is an exact divisor of x 5 — y 5 ; and so on, indefinitely. 
 
 The method employed in proving this Theorem is called 
 Proof by Mathematical Induction. 
 
 135. The Factor Theorem. If a rational and integral 
 expression in x vanishes, that is, becomes equal to 0, when r 
 is put for x, then x — r is an exact divisor of the expression, 
 
 Given ax n + bx n - 2 + + hx -f k (1) 
 
 By supposition, ar" + br n - i + + hr + k = 0. (2) 
 
 By subtracting (2) from (1), the given expression assumes the form 
 
 a (x n — r«) + b (x"- 1 — r"- 1 ) + + h (x — r). 
 
 But x — r is an exact divisor of x n — r», x n-1 — r» - J , and so on. 
 Therefore, x — r is an exact divisor of the given expression. 
 
 Note. If x — r is an exact divisor of the given expression, r is an 
 exact divisor of k ; for fc, the last term of the dividend, is equal to r, 
 the last term of the divisor, multiplied by the last term of the quotient, 
 Therefore, in searching for numerical values of x that will make the 
 given expression vanish, only exact divisors of the last term of the 
 expression need be tried. 
 
FACTORS. ; , , /,\ >,,;;' *' , '' i '\i08 
 
 1. Kesolve into factors x s -f- 3 x 2 — 13 x — 15. 
 
 The exact divisors of 15 are 1, — 1, 3, — 3, 5, — 5, 15, — 15. 
 If we put 1 for x in x 8 + 3x 2 — 13 x — 15, the expression does not 
 vanish. If we put — 1 for x, the expression vanishes. 
 Therefore, x — (— 1), that is, x + 1, is a factor. 
 Divide the expression by x + 1, and we have 
 
 x 3 + 3x 2 - 13x - 15 = (x + 1) (x 2 + 2x - 15) 
 = (x + 1) (x - 3) (x + 5). 
 
 Note. An expression can sometimes be resolved into three or 
 more factors. 
 
 2. Kesolve into factors x s — 26 x — 5. 
 
 By trial we find that the only exact divisor of — 5 that makes the 
 expression vanish is — 5. 
 
 Therefore, divide by x + 5, and we have 
 
 x 3 - 26x - 5 = (x + 5) (x 2 - 5 x - 1). 
 
 As neither + 1 nor — 1, the exact divisors of — 1, will make 
 x 2 — 5 x — 1 vanish, this expression cannot be resolved into factors. 
 
 Exercise 46. 
 Eesolve into factors : 
 
 1. a 8 -10a -3. 10. x B -3x 2 + 4:X-2. 
 
 2. x»-26x + 5. 11. « 8 + 9ic 2 + 16ic + 4. 
 
 3. a 8 -15z-4. 12. x s + 2x 2 -S4,x-5. 
 
 4. z 8 -8;z + 3. 13. 4:x*-12x 2 + 9x-l. 
 
 5. z 8 + 2x 2 + 9. 14. x*-2x 2 -23x + 60. 
 
 6. x»-3x + 2. 15. 6ic 8 -23ic 2 + 16cc-3. 
 
 7. z 8 - 12 x + 16. 16. a; 8 -10a; 2 H-33£c-36. 
 
 8. x 8 + 4,x 2 -5. 17. x* + 7 ar 2 + 12 a + 4. 
 
 9. 4a 8 -7a + 3. 18. x 8 + 5 a; 2 + 7 a -f 2. 
 
104 : <<! , FACTORS. 
 
 136. A compound expression involving x and y is divis- 
 ible by x — y if the expression vanishes when + y is put 
 for cc ; and is divisible by x -f- y if the expression van- 
 ishes when — y is put for x. 
 
 If n is a positive integer, prove by the Factor Theorem : 
 
 1. x n + y n is never divisible by x — y. 
 
 Put y for x in ic n + y n ; then x n + y n = y n + y 7l = 2y n . 
 As 2 y" is not zero, x n + y w is not divisible by a; — y. 
 
 2. a;" — y n is divisible by a; + y, if iz is even. 
 
 Put — y for x in cc n — y ra , then x n — y 71 = (— y) n — y 71 . 
 If n is even, (— y) re = y n , and (— y)" — yn — yn — yn m 
 As y n — y n = 0, x n — y n is divisible by x + y, if n is even. 
 
 3. x n + y" is divisible by a; + y, if 22 is odd. 
 
 Put — y for x in « ra + y 71 , then x n + ?/ n = (— y) n + y 71 . 
 If n is odd, (— y) n = — y n , and (— y)» + y n = — y n + y*. 
 As — y" + y 71 = 0, z n + y 71 is divisible by cc + y, if /7 is odd. 
 
 From § 134 and these three cases, we have 
 
 1. For all positive integral values of n y 
 
 x n_ y n = ( x _ y) ^n-1 + x n-2 y + X n-Z y * 4. 4. yn-iy 
 
 2. For all positive even integral values of n, 
 
 x n — y n = (x -f y) (a; n_1 — x n ~ 2 y + x n ~ z y 2 — — #* _1 ). 
 
 3. For all positive odd integral values of n, 
 
 x» + y n = (x + y) (x n ~ l — x n ~ 2 y + a; B -y - + y"" 1 )- 
 
 4. x n + 2/ n is never divisible by x — y ; and is not divisible 
 by x -^ y, if n is even. 
 
 Note. In applying the preceding rules for resolving an expression 
 into factors, if the terms have a common monomial factor, this factor 
 should be removed first. 
 
 When an expression can be expressed as the difference of two per- 
 fect squares, the method of § 126 should be employed in preference to 
 any other. 
 
FACTORS. IO16 
 
 Exercise 47. — Review. 
 Resolve into factors : 
 
 1. a 3 -9 a. 23. x 2 y 2 - 4 xy A - 3 x 2 y*. 
 
 2. ±x A -x 2 . 24. x* + x 2 + x + l. 
 
 25. Sx 2 -\-x-2. 
 
 26. a 2 -24 a; + 95. 
 
 27. 9 a 2 + 12 a + 4. 
 
 28. a 2 -£ 2 -c 2 + 2fo. 
 
 29. x 2 — 2y + 2x-xy. 
 
 30. 3a 8 + 2a; 2 -9a;-6. 
 
 31. (x-y) 2 -b 2 . 
 
 32. m 2 -2raw + 7& 2 - 1. 
 
 33. a 2 + 2aa; + 3&z + 6a&, 
 
 34. z 2 + ra 2 -w 2 -2ma;. 
 
 35. 9x A + 21x 2 y 2 + 25y\ 
 
 36. x 2 — 4t + y 2 + 2xy. 
 
 37. 6z 2 -a;-77. 
 
 38. x 2 -8xy-65y\ 
 
 39. jc 4 - 7a: 2 + l. 
 
 40. 1 - a 2 - b 2 - 2 ah. 
 
 41. 3a; 4 - 6x* + 9x 2 . 
 
 42. a: 8 -5a: 2 -2;c + 10. 
 
 43. x 2 -\- ax — bx — ab. 
 
 44. 2 a; 2 — 3 xy + 4 aa; — 6 ay. 
 
 3. 
 
 x 12 + y 12 . 
 
 4. 
 
 x 4 — y\ 
 
 5. 
 
 x 9 + y\ 
 
 6. 
 
 x 6 — y\ 
 
 7. 
 
 36 a 2 - 49 y 2 . 
 
 8. 
 
 x 6 + y*. 
 
 9. 
 
 a* -14 a -f- 49. 
 
 10. 
 
 x 2 - (a - 6) 2 . 
 
 11. 
 
 a 2 — (m + ?i) 2 . 
 
 12. 
 
 a 2 -lis + 18. 
 
 13. 
 
 » 2 + 4 a; — 45. 
 
 14. 
 
 a 2 + 13 a; + 36. 
 
 15. 
 
 a: 2 -13 a: -48. 
 
 16. 
 
 a: 2 + 9a; -36. 
 
 17. 
 
 x 2 -\-x- 110. 
 
 18. 
 
 2a; 2 + 3z2/-22/ 2 . 
 
 19. 
 
 s 2 - 6* -40. 
 
 20. 
 
 x 4 + x 2 y 2 + y 4 . 
 
 21. 
 
 a; 2 -7a; -60. 
 
 22. 
 
 2a 2 -la + 6. 
 
•106 FACTORS. 
 
 45. a 5 + y 5 . 63. ax* + £a 8 - as - J. 
 
 46. 32 a 5 -c 5 . 64. (a-y) 2 -2y(a-y). 
 
 47. a 6 + 64/. 65. 1 - 10 ay + 25 a 2 y 2 . 
 
 48. 729 -a 6 . 66. a 2 - b 2 + 2bc - c 2 . 
 
 49. a 12 -y 12 . 67. a 2 + 4 y 2 - z 2 - 4 ay. 
 
 50. (a + 6) 4 -l. 68. a 2 - 4 5 2 - 9 c 2 + 12 be. 
 
 51. 16a 4 -81. 69. 4a 2 + 9y 2 -z 2 -12ay. 
 
 52. a 4 + a 2 + l. 70. (a + 5) 2 - (c - d) 2 . 
 
 53. 27a 8 -64a 8 . 71. a 2 + a + Sb-9b 2 . 
 
 54. a 9 + ?/ 9 . 72. 5 c 4 -15 c 8 -90 c 2 . 
 
 55. a 9 — y 9 . 73. a 2 a — c 2 a + a 2 y — c 2 y. 
 
 56. a 9 -256. 74. a 4 + 16 a 2 x 2 + 256 a\ 
 
 57. l-(a-y) 8 . 75. (x + y y + (2x-yy. 
 
 58. a 6 -216. 76. a 2 + b 2 - c 2 + 2ab. 
 
 59. a 2 -4a -77. 77. y 2 - a 2 - c 2 - 2ac. 
 
 60. a* + b B + a + b. 78. (a + 5 a) 2 - 25 a 2 . 
 
 61. a 8 -£ 8 + a-6. 79. 2 ay - a 2 - y 2 + z 2 . 
 
 62. a 2 -6 2 + a-£. 80. 4a 4 - 9a 2 + 6a - 1. 
 
 81. a 2 -2a& + £ 2 + 12ay-4a 2 -9y 2 
 
 82. 2a 2 -4ay + 2y 2 + 2aa-2ay. 
 
 83. (a + b) 2 - 1 - ab (a + b + 1). 
 
 84. a 8 -a 2 + 3a + 5. 
 
 85. a 2 — y 2 — z 2 — 2 yz + a + y + «. 
 
 86. a 2 + y 2 + * 2 -2ay-2a2 + 2y». 
 
 87. 4a 2 # J -(a 2 + & 2 -c 2 ) 2 . 
 
FACTORS. 107 
 
 88. x 2 + 2x-3. 97. x 2 + 7xy + 12y\ 
 
 89. a 2 + 3a -40. 98. » 2 + «2/-2y 2 . 
 
 90. a 2 -9z-10. 99. x 2 + 3a^-4?, 2 . 
 
 91. x 4 + 8x 2 -9. 100. a 4 -(?/ + z) 4 . 
 
 92. a 4 -2z 8 -24z 2 . 101. x 8 - y 8 - 3 xy (x - y). 
 
 93. z 4 -14z 2 - 51. 102. sc 8 -2a: 2 + 2-x. 
 
 94. x*-10ax + 16a 2 . 103. a; 8 - 8 - 6x + 3z 2 . 
 
 95. ic 4 - 15 x V + 9 2/ 4 . 104. 3 a 2 ?/ 2 + 9 «y 8 - 12 y 4 . 
 
 96. l-9a>-10a: 2 . 105. a 4 - a 8 £ + a& 8 - 6 4 . 
 
 106. a 2 -4c 2 + a-2c. 
 
 107. 4ta 2 + 9b 2 -c 2 + 12ab. 
 
 108. 15x 2 -5ax + 3bx — ab. 
 
 109. 3 a* + 15 a»b- 24 a 2 b 2 . 
 
 110. 6a 8 -30a 2 6 + 36a& 2 . 
 
 111. 25a 2 -4x 2 + 4x-10a. 
 
 112. £c 4 y — sr^ 8 — sc 8 y 2 + jc?/ 4 . 
 
 113. 9a 2 -46 2 + 3a + 2&. 
 
 114. x 8 -y 8 - (x 2 -y 2 )-(x- y)\ 
 
 115. (a._y)«_l_2(a>-y-l). 
 
 116. a 8 -2a 2 c + a 2 -4a + 8c-4. 
 
 117. a 2 -& 2 -c 2 + 26c + a. + &-c. 
 
 118. x 8 z 2 - 8 y 8 z 2 - 4 x 8 n 2 + 32 y 8 7i 2 
 
 119. 5ac + 3bc + c + 5ab + 3b 2 + b. 
 
 120. 2 a& - 2 5c - ace + ex + 2 £ 2 - bx. 
 
 121. s 4 - 2 afa 2 - a 4 - a 2 £ 2 - b\ 
 
CHAPTER Vm. 
 COMMON FACTORS AND MULTIPLES. 
 
 Highest Common Factor. 
 
 137. Common Factors. A common factor of two or more 
 integral numbers in Arithmetic is an integral number that 
 divides each of them without a remainder. 
 
 138. A common factor of two or more integral and rational 
 expressions in Algebra is an integral and rational expression 
 that divides each of them without a remainder. 
 
 Thus, 6 a is a common factor of 20 a and 25 a. 
 
 139. Two numbers in Arithmetic are said to be prime to 
 each other when they have no common factor except 1. 
 
 140. Two expressions in Algebra are said to be prime to 
 each other when they have no common factor except 1. 
 
 141. The greatest common factor of two or more integral 
 numbers in Arithmetic is the greatest number that will 
 divide each of them without a remainder. 
 
 142. The highest common factor of two or more integral 
 mnd rational expressions in Algebra is an integral and 
 rational expression of highest degree that will divide each 
 of them without a remainder. 
 
 Thus, 3 et? is the highest common factor of 3 « 2 , 6 a 8 , and 12 a 4 ,- 
 &x*y 2 is the highest common factor of l@»*y 2 and 15x 2 y 2 . 
 
COMMON FACTORS AND MULTIPLES. 109 
 
 For brevity, we use H. C. F. for highest common factor. 
 
 1. Find the H. C. F. of 42 a% 2 and 30 a 2 b\ 
 
 ±2a*b 2 = 2 X 3 X 7 X aaa X bb; 
 
 30a 2 b* = 2x3x5XaaX bbbb. 
 
 .'. the H.C.F. = 2x3XaaXbb = 6a 2 b 2 . 
 
 2. Find the H.C.F. of x 2 - 9y 2 and x 2 + 6xy + 9y\ 
 
 x 2 -9y 2 = (x + 3y)(x-3y); 
 
 x 2 + 6*2/ + % 2 = (a + 3y) (x + 3y). 
 .-.the H.C.F. =x + 3y. 
 
 3. Find the H.C.F. of 
 
 ' 4a; 2 -4a;-80; 2a: 2 -18a; + 40; 2 a; 2 - 24 x + 70. 
 
 4a; 2 - 4* - 80 = 4 (a 2 - a; - 20) 
 
 = 4(a-5)(a;-h4); 
 2x 2 - 18a; + 40 = 2 (a; 2 - 9x + 20) 
 
 = 2 (a; - 5) (a; - 4) ; 
 2 a; 2 - 24 a; + 70 = 2 (x 2 - 12 a; + 35) 
 = 2 (a; - 5) (x - 7). 
 .-. the H.C.F. = 2 (a; - 5). Therefore, 
 
 143. To Find the H.C.F. of Two or More Expressions, 
 
 Resolve each expression into its prime factors. 
 
 The product of all the common factors, each factor being 
 taken the least number of times it occurs in any of the given 
 expressions, is the highest common factor required. 
 
 Note. The highest common factor in Algebra corresponds to the 
 greatest common measure, or greatest common divisor, in Arithmetic. 
 We cannot apply the terms greatest and least to algebraic expres- 
 sions in which particular values have not been given to the letters 
 contained in the expressions. Thus a is greater than a 2 , if a stands 
 for \ j but a is of lower degree than a 2 . 
 
110 COMMON FACTORS AND MULTIPLES. 
 
 Exercise 48. 
 Find the H.C.F. of: 
 
 1. 120 a 2 and 168 a 8 . 4. 36 a*x 2 and 28 x *y. 
 
 2. 36 x B and 27 x\ 5. 48 a 2 b*c and 60 a*c\ 
 
 3. 42 a 2 x* and 60 a*x 2 . 6. 8 (a + &) 2 and 6 (a + ft) 8 . 
 
 7. 12a(a;-f y) 2 and4ft(a; + *,) 8 . 
 
 8. (a? - l) 2 (x + 2) 2 and (a; - 3) (x + 2) 8 . 
 
 9. 24 a 2 ft 8 (a + ft) and 42 a 8 ft (a + ft) 2 . 
 
 10. sc 2 (ic-3) 2 andcc 2 -3ic. 12. x 2 - 4a; and a; 2 - 6a; + 8. 
 
 11. a; 2 -16 and a; 2 + 4 a;. 13. a; 2 - 7x + 12 and x 2 - 16. 
 
 14. 9» 2 -4y 2 and27a; 8 -83/ 8 . 
 
 15. x 2 - 7x -8 and x 2 + 5x + 4. 
 
 16. £c 2 + 3jcy-10y 2 anda; 2 -2«y~35y 2 . 
 
 17. a; 4 -2a; 8 -24a; 2 and 6a; 5 - 6a 4 - 180a; 8 . 
 
 18. a; 8 -3a; 2 y and a; 8 - 27 y 8 . 
 
 19. l + 64a; 8 andl-4a; + 16a; 2 . 
 
 20. a; 4 - 81 and a; 4 + Sx 2 - 9. 
 
 21. a; 2 + 2x - 3 and x 2 + 7x + 12. 
 
 22. a; 2 - 6a; + 5 and x 2 -f Sx - 40. 
 
 23. 3 a 4 + 15 a% - 12 a% 2 and 6 a 8 - 30 a 2 b + 36 aft 2 . 
 
 24. Qx 2 y - 12xy 2 + 6y* and 3ary -f 9a;y 8 - 12 y\ 
 
 25. l-16c 4 andl + c 2 -12c 4 . 
 
 26. 9aa; 8 -aa;; .9a; 2 - 6a; + 1; 27 a; 8 -1. 
 
 27. x 2 - 3a; -54 j a; 2 -a; -42; a; 2 -2a; -48. 
 
 28. 8a; 8 + 27y 8 ; 4a; 2 + 12a;y + 9y 2 ; 4a; 2 - 9y 2 . 
 
 29. a; 8 — x 2 y — xy 2 + y 8 j x 2 — y 2 \ x 2 -\- 2xy -\- y\ 
 
COMMON FACTORS AND MULTIPLES. Ill 
 
 Lowest Common Multiple. 
 
 144. Common Multiples. A common multiple of two 01 
 more integral numbers in Arithmetic is a number that is 
 exactly divisible by each of the numbers. 
 
 A common multiple of two or more integral and rational 
 expressions in Algebra is an integral and rational expression 
 that is exactly divisible by each of the expressions. 
 
 145. The least common multiple of two or more integral 
 numbers in Arithmetic is the least integral number that is 
 exactly divisible by each of the given numbers. 
 
 The lowest common multiple of two or more integral and 
 rational expressions in Algebra is an integral and rational 
 expression of lowest degree that is exactly divisible by each 
 of the given expressions. 
 
 We use L. C. M. for lowest common multiple. 
 
 1. Find the L. C. M. of 42 a*b 2 ; 30 aW ; 66 ab\ 
 
 42a 8 b 2 = 2 X 3 X 7 X a 8 X J 2 ; 
 30aW = 2 X3X 5Xa 2 X6 4 ; 
 66 ab* = 2 X 3 X 11 X a X b\ 
 
 The L. C. M. must evidently contain each factor the greatest num- 
 ber of times that it occurs in any expression. 
 
 .'.the L.C.M. = 2x3x7x5xllXa 8 X& 4 
 
 = 2310 a%\ 
 
 2. Find the L.C.M. of 
 
 4z 2 - 4z - 80 and 2x 2 - 18a; + 40. 
 4z 2 - 4aj - 80 = 4 (x 2 - x - 20) = 4 (x - 5) (x + 4) ; 
 2x* - 18a + 40 = 2 (x* - 9x + 20) = 2 (x - 5) (x - 4). 
 .-. the L. CM. = 4 (x - 5) (x + 4) (x - 4). Hence, 
 
112 COMMON FACTORS AND MULTIPLES. 
 
 146. To Find the L. C. M. of Two or More Expressions, 
 
 Resolve each expression into its prime factors. 
 
 The product of all the different factors, each factor being 
 taken the greatest number of times it occurs in any of the 
 given expressions, is the lowest common multiple required. 
 
 Exercise 49. 
 Find the L. CM. of: 
 
 1. x % y 2 and x 2 y z . 7. ab z , a 2 b 2 c 2 , and abc*. 
 
 2. 5 abc 2 and a 2 b 2 c\ 8. x 2 y, xy 2 z 2 , and x 2 y*z. 
 
 3. 4 x z y and 12 xy s . 9. x 2 and x 2 + x. 
 
 4. 5 a*b z and 10 a 2 b B . 10. x 2 - 1 and x 2 -f x. 
 
 5. 21 xy* and 28 x*y\ 11. a 2 + ab and aft + 6 2 . 
 
 6. 10 x 2 z* and 15 x 2 z 2 . 12. a 2 +2a and (a + 2) 2 . 
 
 13. a 2 + 4a + 4 and a 2 + 5a -f 6. 
 
 14. a 2 + a;-20anda 2 -sc-30. 
 
 15. y 2 -f y - 42 and */ 2 - 11 y + 30. 
 
 16. s 2 - 10* + 24 and 2 2 + z - 20. 
 
 17. (a + b) 2 ; (a-b) 2 ', a 2 - b 2 . 
 
 18. (a + 2c) 2 ; (a -2c) 2 ; a 2 -4c 2 . 
 
 19. 4ay(a; + 2/) 2 ; 2 x 2 (cc 2 - y 2 ) ; a 8 (a + y). 
 
 20. a 2 + 7a + 12; x 2 + 6x + 8; x 2 + 5x + 6. 
 
 21. 1-2/ 2 ; 1-2/ 8 ; 1 + y. 
 
 22. a 2 + 2icy + 2/ 2 ; x 2 -y 2 ; x 2 -2xy-\-y 2 . 
 
 23. a 8 -27; a 2 + 5z; <c 2 + 2a;-15. 
 
 24. y»-l; 2/ 8 + 2/ 2 + y + l; y 8 - y 2 + y - 1. . 
 
 25. (a-f-y) 2 -« 2 ; (x + y + z) 2 -, x + y - z. 
 
 26. a; 2 -(ft + J)a; + ai; cc 2 — (a + c) a -f- ac. 
 
 27. aj 2 + 3ay + 2?/ 2 ; a 2 +-5a^ + 4y 2 ; x 2 -6xy-7y*. 
 
 28. jc 2 -7^ + 123/ 2 ; s 2 - 6a# + 8y 2 ; x 2 -5^ + 6y 3 . 
 
COMMON FACTOBS AND MULTIPLES. 113 
 
 147. The chief difficulty in finding the H.C.F. and the 
 L.C.M. of two or more algebraic expressions consists in 
 resolving the expressions into factors. 
 
 When two numbers in Arithmetic cannot readily be re- 
 solved into their prime factors, we divide the greater num- 
 ber by the smaller; then the divisor by the remainder; 
 and so on until there is no remainder. The last divisor is 
 the greatest common factor. 
 
 Likewise in Algebra when two given expressions cannot 
 readily be resolved into their factors, we arrange the two 
 given expressions in descending powers of a common 
 letter, and divide the expression which is of higher degree 
 in the common letter by the other expression. After the 
 first division, we take the remainder for a new divisor and 
 the divisor for a new dividend, and so proceed until there 
 is no remainder. The last divisor is the highest common 
 factor. 
 
 Note. If the two expressions are of the same degree in the com- 
 mon letter, either expression may be taken for the divisor. 
 
 Find the H.C.E. of 2x 2 + x - 3 and 4a 8 + Sx 2 - x - 6. 
 
 2x* + x — 3)4x* + 8x 2 — x-6(2x + 3 
 4* 8 + 2a; 2 — 6x 
 
 6x 2 + bx — 6 
 6z 2 + 3cc — 9 
 
 2z + 3)2x2+ z-3(s-l 
 2x* + Sx 
 
 — 2x — 3 
 
 — 2x — 3 
 
 .'.the H.C.F. = 2z+3. 
 
 Note. Each division is continued until the first term of the re- 
 mainder is of lower degree than the first term of the divisor. 
 
114 COMMON FACTORS AND MULTIPLES. 
 
 148. This method is of use only to obtain the compound 
 factor of the H. C. F. Monomial factors of the given expres- 
 sions must first be separated from them, and the H. C. F. of 
 these monomial factors must be reserved to be multiplied 
 into the compound factor obtained. Also, at any stage of 
 the operation a monomial factor of either expression may 
 be removed without affecting the compound factor sought. 
 
 1. Find the H.C.F. of 
 
 12a 4 + 30a 8 - 72a 2 and 32a 8 + 84a 2 - 176a. 
 
 12 aj* + 30 «s - 72 z2 = 6x 2 (2x 2 + 5x - 12). 
 32x3 + 8 4 X 2 - I76x = 4x (8x 2 + 21x - 44). 
 6 x 2 and 4 x have 2 x common. 
 
 2x 2 + 6x - 12) 8x 2 + 21 « - 44 (4 
 8x 2 + 20x-48 
 
 x+ 4)2x 2 + 5x-12(2x-3 
 2x 2 + 8x 
 
 -3x-12 
 -3x-12 
 
 .'. the H.C.F. m 2x (a + 4). 
 
 2. Find the H. C. F. of 4 a 2 - 8 a - 5 and 12a 2 - 4a -65. 
 
 4x 2 -8x-5)12x 2 - 4x-65(3 
 12x 2 -24x-15 
 20x-50 
 
 The first division ends here, for 20 x is of lower degree than 4 x 2 . 
 
 We take out the simple factor 10 from 20 x — 50, for 10 is not a 
 factor of the given expressions, and its rejection can in no way affect 
 the compound factor sought, and proceed with 2 x — 6 for a divisor. 
 
 2x-5)4x 2 - 8x-5(2x+l 
 4x 2 — lOx 
 
 2x — 5 
 2x-5 
 
 .\theH.C.F. =2a-5. 
 
COMMON FACTOBS AND MULTIPLES. 115 
 
 3. Find the H.C.F. of 
 
 21a 8 - 4a; 2 - 15a - 2 and 21a 8 - 32 a 2 - 54a - 7. 
 
 21x 8 -4x 2 -15x-2)21x 3 -32x 2 -54x-7(l 
 21 x 3 — 4x 2 — 15x — 2 
 
 
 28x 2 -39x-5 
 
 The difficulty here cannot be obviated by removing a simple factor 
 from tlie remainder, for — 28 x 2 — 39 x — 5 has no simple factor. In 
 this case, to avoid the inconvenience of fractions we multiply the 
 expression 21 x 3 — 4x 2 — 15x — 2 by the simple factor 4 to make its 
 first term exactly divisible by — 28 x 2 . 
 
 The introduction of such a factor can in no way affect the H.C.F. 
 sought, for 4 is not a factor of either of the given expressions ; and if 
 we multiply only one of the expressions by 4 we do not introduce a 
 common factor. 
 
 The signs of all the terms of the remainder may be changed ; for 
 if an expression A is divisible by — F, it is divisible by + F. 
 
 The process then is continued by changing the signs of all the terms 
 of the remainder and multiplying the divisor by 4. 
 
 28x 2 + 39x + 5)84x 3 — 16x 2 — 60x— 8(3x 
 84x 3 + 117x 2 + 16x 
 
 -133x 2 - 75x- 8 
 Multiply by — 4, — 4 
 
 532x2 4-300x4- 32(19 
 532 x 2 4- 741x4- 95 
 Divide by - 63, — 63 ) - 441 x - 63 
 
 7x4- 1 
 
 7x 4- 1)28x2 4-39x4-5(4x4- 5 
 28x 2 + 4x 
 
 35x + 5 
 35x + 5 
 
 .\theH.C.F. = 7a 4-1. 
 
116 
 
 COMMON FACTORS AND MULTIPLES. 
 
 
 
 4. Find the H.C.F. of 
 
 8x 2 + 2x - 3 and 6x* + 5x 2 - 2. 
 
 Multiply by 4, 
 8x 2 + 2x — 
 
 Multiply by 4, 
 
 6x3 + 
 4 
 
 3)24x3 + 
 24x3 + 
 
 5x 2 - 2 
 
 20x2- 8 (3x + 7 
 6x 2 — 9x 
 
 14x2+ 9x_ 8 
 4 
 
 
 56x2 + 36x-32 
 56x 2 + 14x-21 
 
 Divide by 11, 
 
 
 ll)22x-ll 
 
 2a- l)8x 2 + 2x-3 
 8x 2 — 4x 
 
 
 6x-3 
 6x-3 
 
 .the H.C.F. = 2 x 
 
 The following arrangement of the work will be found 
 jnost convenient : 
 
 3x 
 
 8x2 + 2x-3 
 
 6x 3 + 5x 2 — 2 
 
 8x 2 — 4x 
 
 4 
 
 6x-3 
 
 24x 3 + 20x 2 — 8 
 
 6x-3 
 
 24x 8 + 6x 2 — 9x 
 
 
 14x 2 + Q X — g 
 
 
 4 
 
 
 56x 2 + 36x-32 
 
 
 56x 2 + 14x— 21 
 
 
 ll)22x — 11 
 
 
 2x— 1. 
 
 + 7 
 
 4x + 3 
 
 Note. From the nature of division, the successive remainders are 
 expressions of lower and lower degree. Hence, unless at some step 
 the division leaves no remainder, we shall at last have a remainder 
 that does not contain the common letter. In this case the given 
 expressions have no H.C.F. that contains the common letter. 
 
COMMON FACTORS AND MULTIPLES. 117 
 
 149. In the examples worked out we have assumed that 
 the divisor which is contained in the corresponding divi- 
 dend without a remainder is the H.C.F. required. 
 
 The proof may be given as follows : 
 
 Let A and B stand for two expressions which have no 
 monomial factors, and which are arranged according to the 
 descending powers of a common letter, the degree of B 
 being not higher than that of A in the common letter. 
 
 Let A be divided by B, and let Q stand for the quotient, 
 and B for the remainder. Then, since the dividend is 
 equal to the product of the divisor and quotient plus the 
 remainder, we have 
 
 A = BQ + B. (1) 
 
 Since the remainder is equal to the dividend minus the 
 product of the divisor and quotient, we have 
 
 B = A-BQ. (2) 
 
 Now, a factor of each of the terms of an expression is a 
 factor of the expression. Hence, any common factor of B 
 and B is a factor of BQ -f- B, and by (1) a factor of A. 
 That is, a common factor of B and B is also a common 
 factor of A and B. 
 
 Also, any common factor of A and B is a factor of A — BQ, 
 and by (2) a factor of B. That is, a common factor of A 
 and B is also a common factor of B and B. 
 
 Therefore, the common factors of A and B are the same 
 as the common factors of B and B ; and consequently the 
 H.C.F. of A and B is the same as the H.C.F. of B and B. 
 
 The proof for each succeeding step in the process is 
 precisely the same ; so that the H. C. F. of any divisor and 
 the corresponding dividend is the H. C. F. required. 
 
 If at any step there is no remainder, the divisor is a fac- 
 tor of the corresponding dividend, and is therefore the 
 H.C.F. of itself and the corresponding dividend. Heaee, 
 this divisor is the H. C. F. required. 
 
118 
 
 COMMON FACTORS AND MULTIPLES. 
 
 150. The methods of resolving expressions into factors, 
 given in the last chapter, often enable us to shorten the 
 work of finding the H. C. F. required. 
 
 1. Find the H.C.F. of 
 
 a 4 + 3a 8 + 12a; -16; 
 
 a; 8 -13 a; + 12. 
 
 Both of these expressions vanish when 1 is put for x. Therefore, 
 both are divisible by x — 1, § 135. 
 
 The first quotient is x 8 + 4 x 2 + 4 x + 16 = (x 2 + 4) (x + 4). 
 The second quotient is x 2 + x — 12 = (x — 3) (x + 4). 
 
 Therefore, the H.C.F. is (x - 1) (x + 4). 
 
 2. Find the H.C.F. of 
 
 2a; 4 + 9a; 8 + 14a; +-3; 3a; 4 + 14a; 8 + 9x + 2. 
 
 2x4 + 9xs + 14x + 3 
 
 3x 4 + 14x 8 + 9x + 2 
 2 
 
 6x* + 28x 8 + 18x + 4 
 6x 4 + 27x 3 + 42x + 9 
 
 x 8 — 24x — 5 
 The remainder, x 8 — 24 x — 5, vanishes when 5 is put for x. 
 The quotient of x 8 — 24 x — 5 divided by x — 5 is x 2 + 5 x + 1. 
 Since 5 is not an exact divisor of 3, x — 5 is not a factor of 2 x* 
 
 + 9 x 8 + 14 x + 3 ; but x 2 + 5 x + 1 is found by trial to 'be a factor, 
 
 and is, therefore, the H. C. F. required. 
 
 3. Find the H.C.F. of 
 
 28a; 2 + 39a + 5; 84 x* - 16 jc 2 - 60 x - 8. 
 By § 132, the factors of 28 x 2 + 39x + 5 are 7x + 1 and 4x + 5. 
 The factor 7 x + 1 is the H. C. F. required. 
 
 4. Find the H.C.F. of 
 
 2z 4 -6a 8 -a 2 + 15a;-10; 4z 4 + 6x*- 4z 2 - 15a;- 15. 
 
 2x*-6x 8 -x 2 + 16x-10 
 
 4x*+ 6x 8 -4x 2 -15x-15 
 4x* - 12 x 8 - 2x 2 + 30x -20 
 
 18x 8 -2x 2 
 The remaindor = 2 x 2 (9 x — 1) — 6 (9 x — 1) 
 
 = (2x 2 -5)(9x-l). 
 The factor 2 x 2 — 5 is the H. C. F. required. 
 
 45x+ 6 
 
COMMON FACTORS AND MULTIPLES. 
 
 119 
 
 5. Find the H. C. F. and the L. C. M. of : 
 
 6 a; 8 - llx 2 y + 2y* and 9 a 8 - 22 xy 2 - 8y» 
 
 6a 8 — llx*y + 2y 8 
 
 6x 8 — 8x 2 y — 4xy 2 
 
 — 3x 2 y + 4x£/ 2 + 2?/ 8 
 
 - 3x 2 y + 4sy 2 + 2y 8 
 
 9x 3 — 22xi/ 2 — 8y 8 
 2 
 
 
 18 x 8 -44x2/ 2 - 16 1/ 8 
 18x 8 -33x 2 y + 6y 8 
 
 
 lly)33x 2 ?/ — 44 xj/ 2 — 
 
 22 y 3 
 
 3x 2 - 4xy — 
 
 2y 2 
 
 .'. the H. C.F. = 3 a; 2 - 4xy -2y 2 . 
 
 2x-y 
 
 To find the L. C. M., divide each of the expressions by the 
 H.C.F. 
 
 (6x s - llx 2 y -{- 2 y*) + (3 x 2 - ±xy - 2 y 2 ) = 2 x - y' 
 (9 x s - 22 xy 2 - 8 f) -s- (3 x 2 - 4 xy - 2 y 2 ) = 3 x + 4 y. 
 . ' . the L. C . M. = (2 x - y) (3 a; + 4 y) (3 a; 2 - 4 ay - 2 y 2 ). 
 
 Exercise 50. 
 Find the H. C. F. and the L. C. M. of : 
 
 1. 4a; 2 + 3a;-10; 4a; 8 + 7a; 2 -3a; -15. 
 
 2. 2x s -6x 2 + 5x-2; 8a; 8 - 23 x 2 -f 17 a; - 6. 
 
 3. 6a; 8 -7aa; 2 -20a 2 a;; 3 x 2 + aa; - 4 a 2 . 
 
 4. 3a; 8 - 13a; 2 + 23a; -21; 6 a; 8 + a; 2 - 44 x + 21, 
 
 5. c*-2c* + c, 2c 4 -2c 8 -2c -2. 
 
 6. a 8 -6a 2 a; + 12aa; 2 -8a; 8 ; 2 a 2 - 8 ax + 8 a; 2 , 
 
 7. 7a; 8 -2a; 2 -5; 7 a; 8 + 12 a; 2 + 10a; + 5. 
 
 8. x* -13 a; 2 + 36; a; 4 
 
 7a; 2 +.a; + 6. 
 
 9. 2a; 8 + 3a; 2 -7a;-10; 4a; 8 - 4a; 2 - 9a; + 5. 
 
 10. 12a; 8 - x 2 -30a; -16; 6 a; 8 - 2a; 2 - 13 a; - 6. 
 
 11. 6a; 8 + x 2 -5x -2; 6a; 8 + 5a; 2 - 3a; - 2. 
 
 12. a; 8 -9a; 2 + 26a;-24; a; 8 - 12 a; 2 + 47 x - 60. 
 
120 COMMON FACTORS AND MULTIPLES. 
 
 13. 4a; 8 -2a; 2 -16a; -91; 12 x 8 - 28 x 2 - 37 x - 42. 
 
 14. a; 4 -4a: 8 + 10a; 2 -12a;4-9; a; 4 + 2a; 2 -b9. 
 
 15. 2 a; 8 -3a; 2 -16a; + 24; 4a; 5 + 2x 4 -28x*-16x 2 -32x. 
 
 16. 12a; 8 + 4a; 2 + 17 a; -3; 24a; 8 - 52a; 2 + 14a; - 1. 
 
 17. 2a; 8 + 7aa; 2 + 4a 2 a;-3a 8 ; 4 a; 8 + 9 aa; 2 - 2 a 2 a; - a 8 . 
 
 18. 2a; 8 -9aa; 2 + 9a 2 a;-7a 8 ; 4a; 8 - 20 aa; 2 + 20 a 2 a; - 16 a 8 
 
 19. 2a; 4 + 9a; 8 + 14a; + 3; 3a; 4 + 14a; 8 -f 9a; + 2. 
 
 20. 20a; 8 + 2x 2 -18 x + 48 ; 20 a; 4 - 17 a; 2 + 48 a; - 3. 
 
 21. 2a; 8 + a; 2 -12a; + 9; 2a; 8 - 7a; 2 + 12a; - 9. 
 2«. a; 8 -8a; + 3; a; 6 - 3a; 5 + 21a; - 8. 
 
 23. 3x*-3x 2 y + xy* — y s ; 4 a; 8 - a; 2 y - 3 xy\ 
 
 24. 8a; 4 - 6a; 8 - x 2 + 15a; -25; 4a; 8 + 7a; 2 - 3a; - 15. 
 
 25. 4a; 8 -4a; 2 -5a; + 3; 10 a; 2 - 19 x + 6. 
 
 26. 6a; 4 - 13a; 8 + 3a; 2 + 2a;; 6a; 4 - 10a; 8 + 4a; 2 - 6a; + 4 
 
 27. 2a; 4 - 3a; 8 + 2a; 2 -2x -3; 4a; 4 + 3a; 2 + 4a; - 3. 
 
 28. 3a; 4 -a; 8 -2a; 2 + 2a;-8; 6a; 8 + 13 a; 2 + 3 a; + 20. 
 
 29. 3a; 6 + 2a; 4 + a; 2 ; 3a; 4 + 2a; 8 - 3a; 2 + 2a; - 1. 
 
 30. 3 - 2a; + 5a; 2 + 2a; 8 ; 12 - 17a; + 2a; 2 + 3a; 8 . 
 
 31. 10a; -6x 2 - 11a; 8 + 9a; 4 - 6a; 6 ; 
 
 60a; -f 4a; 2 + 10 a; 8 + 10 a; 4 + 4 x\ 
 
 32. a; 4 -a; 8 -14a; 2 + a; + l; a; 5 - 4 a; 4 - x z - 2x* + 8 x + 2. 
 
 33. 2a 4 -2a 8 -3 a 2 -2a; 3a 4 - a 8 - 2a 2 - 16a. 
 
 34. 6a; 8 -14aa; 2 + ^a 2 a;-4a 8 ; a; 4 -ax 8 -a 2 a; 2 -a 8 a;- 2a 4 . 
 
 35. 4-2a;-8a; 2 + 7a; 8 -9a; 6 ; 2+ 5a; - 10a; 2 - 7a; 8 + 6a; 4 . 
 
 36. 2a 4 + 3a 8 a;-9a 2 a; 2 ; 6 a 4 x - 3 ax* - 17 a 8 a; 2 + 14 aV. 
 
 37. 2a 6 - 4a 4 + 8a 8 - 12a 2 + 6a; 
 
 3a«-3a 6 -6a 4 + 9a 8 -3a*. 
 
COMMON FACTORS AND MULTIPLES. 121 
 
 151. The product of the H. C.F. and the L. CM. of two 
 
 expressions is equal to the product of the given expressions. 
 
 Let A and B stand for any two expressions ; and let F 
 stand for their H.C.F. and M for their L.C.M. 
 
 Let a and b be the quotients when A and B respectively 
 are divided by F. Then 
 
 A = aF 
 and B = IF. 
 
 Therefore, AB = F X abF. (1) 
 
 Since F stands for the H. C. F. of A and B, F contains 
 all the common factors of A and B. Therefore, a and b 
 have no common factor, and abF is the L. C. M. of A and B. 
 
 Put M for its equal, abF, in equation (1), and we have 
 
 AB = FM. 
 
 152. Since FM=AB } (§151) 
 
 j t f == ^ = 4x^ = ^X A. That is, 
 F F F 
 
 The lowest common multiple of two expressions may be 
 found by dividing their product by their highest common 
 factor, or by dividing either of them by their highest com- 
 mon factor and multiplying the quotient by the other. 
 
 153. The H. C. F. of three or more expressions is obtained 
 by finding the H. C. F. of two of them ; then the H. C. F. of 
 this result and of the third expression j and so on. 
 
 For, if A, B, and C stand for three expressions, 
 and D for the highest common factor of A and B, 
 and E for the highest common factor of D and (7, 
 then D contains every factor common to A and B f 
 and E contains every factor common to D and C} 
 that is, every factor common to A, B, and C. 
 
122 COMMON FACTORS AND MULTIPLES. 
 
 154. The L. C. M. of three or more expressions is obtained 
 by finding the L. C. M. of two of them ; then the L. C. M. of 
 this result and of the third expression ; and so on. 
 
 For, if A, B, and C stand for three expressions, 
 
 and L for the lowest common multiple of A and B, 
 and M for the lowest common multiple of L and C, 
 then L is the expression of lowest degree that is 
 exactly divisible by A and B, 
 
 and M is the expression of lowest degree that is 
 exactly divisible by L and C. That is, M is the expression 
 of lowest degree that is exactly divisible by A, B, and C. 
 
 Exercise 51. 
 Find the H. C. F. and the L. C. M. of : 
 
 1. 6x 2 + x-2', 2x 2 + 7x-4:' J 2x 2 -7x + 3. 
 
 2. a 2 + 2ab + b 2 ; a 2 -b 2 ; a 8 + 2a 2 b + 2ab 2 + b\ 
 
 3. £c 2 -5aic+4a 2 ; x 2 -3ax + 2a 2 ; 3x* - Wax + 1 a\ 
 
 4. x 2 + x-6-, x*-2x 2 -x + 2; x* + 3x* - 6x - 8. 
 
 5. a 8 -6a 2 + lla-6; a 8 -8a 2 + 19a;-12; 
 
 x*-9x 2 + 26x-24, 
 
 6. 6x 2 + 7xy- 3y 2 ; 3x 2 + llxy - Ay 2 ; 
 
 2x 2 + llxy + 12y 2 
 
 7. 8-14a + 6a 2 ; 4a + 4a 2 -3a 8 ; 4a 2 + 2a 8 -6a 4 . 
 
 8. 6cc 8 + 7a; 2 -3a; 3x 2 + Ux-5; 6 x* + 39 x + 45. 
 
 9. 27 a 8 -a 8 ; 6z 2 + aa;-a 2 ; 15 a 2 - 5 ax + 3 bx - ab. 
 
 10. x z -l;2x 2 -x-l\3x 2 -x-2. 
 
 11. 6a 2 -a-2; 21z 2 -17a + 2; 14z 2 + 5x-l. 
 
 12. 12a 2 + 2a;-4; 12x 2 - 42* - 24; 12 x 2 - 28 x - 24. 
 
 13. 2aj 2 + 3a;-5; 3cc 2 -a;-2; 2x 2 -\-x-3. 
 
 14. <e 8 + 7a; 2 -l-5a; -1; a; 2 + 3x - 3a; 8 - 1 ; 
 
 3z 8 + 5sc 2 + a;-l. 
 
CHAPTER IX. 
 FRACTIONS. 
 
 Definitions. 
 
 155. An algebraic fraction is the indicated quotient of 
 two expressions written in the form -• 
 
 156. The dividend a is called the numerator; the divisor 
 b is called the denominator ; the numerator and denominator 
 are called the terms of the fraction. 
 
 Fundamental Principle of Fractions. 
 
 Let l = x. (1) 
 
 Multiply by b, 
 Multiply by c, 
 
 Divide by bc y r~~ x ' (?) 
 
 From equations (1) and (2), -- = ~ 
 
 o be 
 
 Now — is obtained by multiplying both terms of - by c ; 
 and - is obtained by dividing both terms of -7- by e. Hence, 
 
 157. If the numerator and denominator of a fraction are 
 both multiplied by the same number, or both divided by the 
 same number, the value of the fraction is not altered. 
 
 b 
 
 = X. 
 
 a 
 
 = bx. 
 
 ac 
 
 = hex. 
 
 ac 
 be 
 
 = X. 
 
124 FRACTIONS. 
 
 Reduction of Fractions to their Lowest Terms. 
 
 158. A fraction is in its lowest terms when the numera- 
 tor and denominator have no common factor. Hence, 
 
 To Reduce a Fraction to Lowest Terms, 
 
 Resolve the numerator and denominator into their prime 
 factors, and cancel all the common factors ; or, divide the 
 numerator and denominator by their highest common factor. 
 
 Eeduce the following fractions to their lowest terms : 
 38 a 2 b*c* 2xl9aW 2 6V 
 
 57a 8 bc 2 3x 19 a 8 bc 2 ' Sa 
 
 a 
 
 (a — x) (a 2 + ax + x 2 ) _ a 2 -f- ax + x 2 
 
 a 2 — x 2 (a — x) (a + x) a -\- x 
 
 a 2 + la + 10 (a + 5) (a + 2) a + 5 
 a 2 + 5a + 6 ~ (a + 3) (a -f 2) ~ a + 3* 
 1 
 
 • a-3 _ 2a; 2 4.43.- 3 
 
 We find by the Factor Theorem the H.C.R of the 
 numerator and denominator to be x — 1. 
 
 The numerator divided by x — 1 = x 2 — 3 x -f- 1. 
 The denominator divided by » — 1 = x 2 — x + 3. 
 x*-4x 2 + 4x-l z 2 -3a + l 
 
 Therefore, 
 
 :*-2x 2 + 4x-3 x 2 -x + 3 
 
 Exercise 52. 
 Reduce to lowest terms : 
 
 6ab 2 42m 2 & 34axV 
 
 1# 9a 2 b' 4 ' 49 mn 2 ' ' 51 a 2 xy 7 ' 
 
 m 3ab 2 c 30 x y*z* Q 35aW 
 
 3. 
 
 15 aW ' 18 a 2 ?/ 2 * 2 ' 5a¥c 
 
 26 a; V 21 ^ 2 ^ 2 58 aW 
 
 39 a:/' 6 ' 28 m 2 p' ' 87 aW* 
 
FRACTIONS. 125 
 
 4a; 2 + 12 ax + 9 a 2 
 
 10. — -% zr~-' 19. 
 
 12 x i — loxy 
 
 Aa 2 -9c 2 
 
 8a' + 6a , ^ 
 
 a 2 + 4a + 4 
 
 5 2 -55 
 13 ' 6 2 -46-5* ■ 22 ' 
 
 M ' 4(a 2 + ac + c 2 )' 23 ' 
 
 a; 8 + ?/ 8 
 ^ 5i ^r 24. 
 
 a; 2 + 2 a;?/ + y 2 
 
 16 ' a; 2 + 2a; -15* 25 " 
 
 17 . ^-^ + 1 1 . 
 
 ■to . . 07 
 
 2a, 2 -7a; -15 4a 2 c 2 -(a 2 + c 2 -£ 2 ) 2 
 
 Eeduce by the Factor Theorem j or by finding the H. C. E 
 of the numerator and denominator : 
 
 a: 2 
 
 8 a; 3 + 27 a 8 
 — y 2 — 2 yz — z 2 
 
 x> 
 
 X* 
 
 + 2 xy + y 2 — z 2 
 + a; 2 ?/ 2 + y 4 
 
 x* — y 
 
 2 a 2 + 17 
 
 3 
 
 a + 21 
 
 3 a 2 + 26 
 (a + by - 
 
 a + 35 
 -c 2 
 
 (a 
 
 (x 
 
 + 6 + 
 
 c) 2 
 -b 2 
 
 
 -6) 2 - 
 + a) 2 - 
 
 -a 2 
 -b 2 
 
 (a 
 
 + 6) 2 - 
 + b) 2 - 
 
 -a 2 
 
 -(c + d) 2 
 
 (« 
 
 + c) 2 - 
 + 
 
 - (b + d) 2 
 c) 2 - b 2 
 
 28 - 3x*-8x + &' 32 ' 
 
 x 3 + 4ar — 5 
 
 3 x s — x 2 — x — 1 
 3a; 3 -4a; 2 -a; + 2* 
 
 31 . ^-18«H-36 36 . 
 
 3a; 8 + 17 a; 2 + 22 a; + 8 
 
 6a; 8 + 25 
 
 a; 2 + 23 a; + 6 
 
 a; 8 
 
 -3a; 2 
 
 -15. 
 
 z + 25 
 
 X 
 
 > + 7a; 2 
 
 + 5a 
 
 ;-25 
 
 2x* + x 2 
 
 -Sx 
 
 + 3 
 
 3a; 8 + 8a; 
 
 2 + x 
 
 -2 
 
 a; 8 
 
 + 4a; 2 
 
 -Sx 
 
 + 24 
 
 a; 4 -a; 8 + 8a;-8 
 
126 FRACTIONS. 
 
 159. There are three signs to consider in a fraction, the 
 sign before the fraction, the sign of the numerator, and the 
 sign of the denominator. 
 
 s »"*> i=— b = -— =-— b ' (§ 87 ) 
 
 any two of the three signs may be changed without chang- 
 ing the value of the fraction. 
 
 The sign of a compound expression is changed by chang- 
 ing the sign of every term of the expression. Hence, 
 
 1. We may change the sign of every term of the numera- 
 tor and denominator of a fraction without changing the 
 value of the fraction. 
 
 2. We may change the sign before a fraction and the sign 
 of every term of either the numerator or denominator with- 
 out changing the value of the fraction. 
 
 160. From the Law of Signs, therefore, 
 
 1. We may change the signs of an even number of factors 
 of the numerator, or of the denominator, or of both, without 
 changing the sign of the fraction. 
 
 2. We may change the signs of an odd number of factors 
 of the numerator, or of the denominator, or of both counted 
 together, if we change the sign before the fraction. 
 
 Eeduce to its lowest terms -^ £4 — ; — Sr 
 
 (b — a) (c + d) 
 
 Change the sign of the factor (b — a) of the denominator and the 
 
 sign before the fraction, and we have 
 
 (q — b) (c — d) _ (a-b)(c — d) _ c — d 
 
 (b-a)(c + d) {a-b){c + d) c + d 
 
 In the last fraction change the sign of the numerator, the sign of the 
 
 fraction, and the order of the terms of the denominator, and we have 
 
 c — d _ d — c 
 
 c + d~ d + c' 
 
 Note. Factors and terms must not be confounded. 
 
27 y>- 
 
 -x 3 
 
 abx — 
 
 bx 2 
 
 ex 2 — 
 
 acx 
 
 4ft 2 - 
 
 6ab 
 
 9 b 2 - 
 
 4 a 2 
 
 1- 
 
 X 2 
 
 4x (x 
 
 -1) 
 
 <z 2 + 5^-14 
 
 8. 
 
 9. 
 
 FRACTIONS. 127 
 
 Exercise 53. 
 
 Keduce to lowest terms : 
 
 xy-Sy 2 3ab(b 2 -a*) 
 
 ^(a^-ab 2 ) 
 
 c 2 -(a + b) 2 
 a 2 -f- ab — ac 
 
 ac — be — ax + &b 
 
 x' 2 — c 2 
 
 (b + c + d) 2 -a 2 
 (a-b) 2 -(c + d) 2 ' 
 
 (x - a) 2 - V 2 
 5 * 4 -a 2 ' ' (a-b) 2 -x 2 
 
 161. Mixed Expressions. A mixed expression is an inte 
 gral expression and a fraction. 
 
 Thus, x + - and x + y tj— are mixed expressions. 
 
 x x "v y 
 
 Reduction of Fractions to Integral or Mixed 
 Expressions. 
 
 By the distributive law of division, § 43, 
 
 a 2 + c a 2 c , c 
 
 = h - = a-\ 
 
 a a a a 
 
 Therefore, a fraction whose numerator is of a degree equal 
 to, or higher than, the degree of the denominator is reduced 
 to an integral or mixed expression by division. Hence, 
 
 162. To Reduce a Fraction to an Integral or Mixed Expression, 
 
 Divide the numerator by the denominator. 
 
 mu 10s2 + 15x-7 10x 2 , 15 x 7 _ , 7 
 Thus, = — V — = 2 x + 3 — - — 
 
 'ox ox bx ox ox 
 
128 
 
 FRACTIONS. 
 
 Keduce — — — to an integral or mixed expres- 
 
 sion. 
 
 a* + 4 a 2 
 a 8 + a 2 
 
 2a 
 
 a 2 + a — 2 
 
 a + 3 
 
 3 a 2 + 2 a — 5 
 
 3 a 2 + 3 a — 6 
 
 - a + 1 
 
 The remainder —a + 1 is the numerator of a fraction, and the 
 divisor a 2 + a — 2 is its denominator, to be added to the integral 
 quotient a + 3. Thus,the complete expression required is 
 
 a + 3 + 
 
 a + 1 
 
 a 2 + a — 2 
 
 By changing the sign of each term of the numerator and the sign 
 before the fraction, 
 
 — a + 1 , «, a — 1 
 
 a + 3 + 
 
 a + 3 
 
 a 2 + a-2 a 2 + a — 2 
 
 The last form of the expression is the form usually written. 
 
 Exercise 54. 
 Reduce to an integral or a mixed expression : 
 
 1. 
 
 2. 
 
 3. 
 
 4. 
 
 5. 
 
 4x 2 -f 12x^3 
 
 4a; 
 
 
 3x 2 -6x- 
 
 2 
 
 3x 
 
 
 x 8 + y 8 
 
 
 x — y 
 
 
 x s — y 8 
 
 
 z + y 
 
 
 a s + 2z s 
 
 
 a-\-2x 
 
 6. 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 Ax 2 -3x — 54 
 
 
 x — 4 
 
 
 3 x s — x 2 — 16 x - 
 
 -2 
 
 x 2 + x-3 
 
 
 X* + 4:X*-7 
 
 
 x 2 + x-2 
 4 # 2 + 6 ax - 
 
 21a' 
 
 2x-3a 
 5x 8 + 9x* + 5 
 
 5 x 2 •+ 4 ic — 1 
 
FBACTIONS. 129 
 
 Reduction of Mixed Expressions to Fractions. 
 
 163. The value of a number is unaltered if it is both 
 multiplied and divided by the same number. Hence, 
 To Reduce a Mixed Expression to a Fraction, 
 
 Multiply the integral expression by the denominator, to 
 the product add the numerator, and under the result write 
 the denominator. 
 
 a 2 -ab- b 2 
 
 Reduce to a fraction a — b — 
 
 a + b 
 
 _ b _ a 1 -^- o 2 = (a -b)(a + b)- (a 2 -ab-b 2 ) 
 a + b a + b 
 
 a 
 
 2 + ab + b 2 
 
 a + b 
 
 ab 
 
 a + b 
 
 Note. The dividing line between the terms of a fraction has the 
 force of a vinculum affecting the numerator. If, therefore, a minus 
 sign precedes the dividing line, as in the preceding example, and this 
 line is removed, the numerator of the given fraction must be enclosed 
 in a parenthesis preceded by the minus sign, or the sign of every term 
 of the numerator must be changed. 
 
 Exercise 55. 
 Reduce to a fraction : 
 
 1. a + b--^-- 5. a 2 + ax + x 2 — • 
 
 a + b 
 
 _ , a 2 + x 2 n a — Sx , _ 
 
 1, a + x ■ 6. a + 2 x. 
 
 a + x 4 
 
 3. re + 4 — • 7. 3a — 2b 
 
 x — 3 a + b 
 
 x s 
 a + x 
 
 A 2 ,2 x * o o 9 21 -13 a? 
 4. a 2 — ax + x z ; 8. 2x — 7 — 
 
130 FRACTIONS. 
 
 Reduction of Fractions to Equivalent Fractions 
 Having the Lowest Common Denominator. 
 
 1. Reduce - — 5 ; tt~ » tt~z to equivalent fractions having 
 the lowest common denominator. 
 
 The L. C. M. of the denominators is 12 a 8 . (§ 152) 
 
 Divide the L. C. M. by the denominators 4 a 2 ; 3a; 6 a 8 . 
 The respective quotients are 3a; 4 a 2 ; and 2. 
 Multiply both terms of the given fractions taken in order by the 
 respective quotients, 3a; 4 a 2 ; and 2. 
 We have for the required fractions 
 
 9ox 8a 2 y 10 
 
 12 a 8 ' 12 a» ' and 12 a 8 ' 
 
 2 - ReduCG a* + L + 6 '> s' + 4s + 3 t0 e ^ ivalent frac " 
 tions having the lowest common denominator. 
 
 Express the denominators in their prime factors. 
 
 2 3 2 3 
 
 x 2 + 5x + 6 ' a 2 + 4x + 3 (x + 3) (as + 2) ' (x + 3) (x + 1) 
 The lowest common denominator (L. C. D.) is 
 
 (x + 3) (x + 2) (x + I). 
 The respective quotients are x + 1 ; x + 2. 
 The respective products are 2 (x + 1) ; 3 (x + 2). 
 The required fractions are 
 
 2 (x + 1) 3 (x + 2) 
 
 (x + 3) (x + 2) (x + 1) ; (x '+ 3) (x + 2) (x + 1)' 
 
 164. Therefore, we have the following rnle : 
 
 Find the lowest common multiple of the denominators of 
 the given fractions for the common denominator. Divide 
 this common denominator by each of the given denominators ; 
 and multiply the given numerators each by the correspond- 
 ing quotient for the required numerators. 
 
 Note. Every fraction should be in its lowest terms before the 
 common denominator is found. 
 
FRACTIONS. 131 
 
 Exercise 56. 
 Express with lowest common denominator : 
 
 3a;-7 4o3-9 4 a 2 + c 2 2a + e 
 
 r 6 ; 18 4d 2 -c 2 ' 2a-c" 
 
 a-2aj3a; 2 -2a a 2 -f ?/ 2 1 
 
 3a ' 9ax 25o3 2 -4y 2 ' 5o3 + 2y 
 
 2o3-4y 3o;-8y a; + 2 a; - 2 
 
 3 ' 5cc 2 ' 10» 03-2* 03 + 2* 
 
 4 a — 5 c 3a — 2c 4y 2 xy 
 
 4 * 5ac ' 12a 2 c * 3 (a + ft) ' 6(a 2 -b 2 )' 
 
 5 6 8x + 2 2»-l 3o3 + 2 
 
 5. - : z t,' 15. 
 
 l- x > l-x 2 x-2' 3o3-6 ? 5o3-10 
 
 1 2 a — bm . c — bn 
 
 6 * o7+2 5 oT+T 16 ' 4a. ; 1; 303 
 
 i a a * 1 1 
 
 7. ; -= ;• 17. 
 
 .3 
 
 (a-b)(b-c)> (a-b)(a-c) 
 
 1 X X 2 X 8 
 
 18. 
 
 l+2a' l-4a 2 * x-V x + V x 2 -l 
 
 9 4 — x a b c 
 
 9. — s: — 19. 
 
 16 -a 2 ' 4 + 03 * a-b' a + b J a 2 - b 2 
 
 a 2 a 1 x x 
 
 0. — -: 20 
 
 27 -a*' 3- a x + V (z + l) 2 ' (x + l) 2 
 
 111 
 
 21. 
 22. 
 23. 
 
 (x-y)(x-z)' (x-y)(x-c) ) (x - c) (x - z) 
 
 1 1 1 
 
 x 2 -5x + 6' ic 2 -4o3 + 3 ; 03 2 -3oj + 2* 
 
 1 1 1 
 
 o; — 2a' x 2 — 5 ax -\- 6 a 2i x — 3 a 
 
132 FRACTIONS. 
 
 Addition and Subtraction of Fractions. 
 
 L Find the algebraic sum of - H 
 
 a b c = a+b-c 
 
 XXX X 
 
 2. Find the algebraic sum of 
 
 3a — 4# 2a — b -\- c a — kc 
 4 3 + ~~ 12 
 
 The L. C. D. = 12. 
 The multipliers, that is, the quotients obtained by dividing 12 by 
 4, 3, and 12, are 3, 4, and 1, respectively. 
 Hence, the sum of the fractions is 
 
 9a-126 8a-46 + 4c a-4c 
 12 12 12 
 
 _ 0a — 12 6 — 8a + 46 — 4c + a — 4c 
 
 12 
 _ 2a — 86 — 8c _ a — 46 — 4c 
 12 ~ 6 
 
 The preceding work may be arranged as follows : 
 
 The L. C. D. = 12. 
 
 The multipliers are 3, 4, and 1, respectively. 
 
 3 (3 a — 4 b) = 9a — 12b = 1st numerator. 
 
 — 4 (2 a — b + c)= — 8a + 4 6 — 4 c = 2d numerator. 
 
 1 (a — 4 c) = _a — 4c = 3d numerator. 
 
 2a— 86 — 8c 
 or 2 (a — 4 b — 4 c) = the sum of the numerators. 
 
 ., ^ 2(a — 46 — 4c) a — 46 — 4c 
 
 .*. sum of fractions = — i — '- = • 
 
 12 o 
 
 165. To Add Fractions, therefore, 
 
 Reduce the fractions, if they have different denominators, 
 to equivalent fractions having the lowest common denomina- 
 tor ; and write the algebraic sum of the numerators of these 
 fractions over the common denominator. 
 
 Note. The resulting fraction should be expressed in its lowest 
 terms. 
 
 
FRACTIONS. 
 
 133 
 
 Exercise 57. 
 Find the algebraic sum of : 
 
 x — 1 x — 3 a; — 7 a; — 2 
 2 5 10 + 5 
 
 2x — 1 + 7 — 4 a; — 3 
 2 " 3 ~ 6 4 ~~ 2 ' 
 
 70-5 _ 30 + 2 + 1 _ 5 a; -10 
 
 8 3 4 12 
 
 20 + 3 , — 2 50 + 4 20-4 
 
 9 + 6 12 3 
 
 20 + 3 , + 3 180 + 5 x - 3 
 
 3. 
 
 4. 
 
 5. 
 
 2x 
 
 4a; 
 
 80 s 
 
 x 
 
 x 2x -11 a; + 3 a; — 7 a; - 1 
 6 * 2 3 4 + 6 12 ' 
 
 4a_ 2 _ q + 5 4&» a 2 6 + ab 2 - 4 a 4 
 £ 2 «6 a a + a 2 6 2 
 
 5 a; -11 _ 0-1 11 a; -1 _ 12 a; -5 . 
 8 ' 4 10 12 3 
 
 + 1 , 30-4 , 1 60 + 7 
 
 9. 
 
 10. 
 
 11. 
 
 2 ' 5 + 4 8 
 
 2-0-6 80-4 . 56a; -48 
 
 50 
 
 15 
 
 45 
 
 11 0y + 2 5y 2 -3 60 2 -5 
 
 x*y 2 
 
 xif 
 
 x s y 
 
 a — b , b — e c — a , a£> 2 + &c 2 + ca 2 
 cab abc 
 
 1 2x — g y— 60 
 2az 2 y ' 6y 2 « 2 0s 2 4aV ±x 2 yz 
 
 13. JU 1 
 
134 FBACTIONS. 
 
 166. When the denominators are polynomials arranged 
 in the same order, we first express the denominators in 
 their prime factors. 
 
 Find the algebraic sum of : 
 
 a 4- b a — b Aab 
 a — b a 4 b a 2 — b 2 
 a s - &2 = ( a + 6) ( a _ &). 
 The L.C.D. is (a - 6) (a + 6). 
 
 The multipliers are a 4 6, a — 6, and 1, respectively. 
 (a 4 6) (a 4 b) = a 2 4 2 a& + 6 2 = 1st numerator. 
 
 — (a — b) (a — b) = — a 2 4 2 ab — b 2 = 2d numerator. 
 
 — 1 (4 a&) = —4ab = 3d numerator. 
 
 = sum of numerators. 
 
 .*. sum of fractions = 0. 
 
 Exercise 58. 
 Find the algebraic sum of : 
 
 1-f x 
 
 
 4 z- 8. 
 
 x+6 x-5 1+x + x 2 1-x + x 2 
 
 2 __1 L_. 9 a + y g-y ± x v . 
 
 ' l+a; 1 — a? ' x — y x -\- y x 2 — y 2 
 
 3, l+a; 1 - *•' 10 ' a - a + a + SB + a 2 4 z 2 * 
 
 c a + y s* - y 2 n 1 2 | 1 , 
 
 ' x — y (x — y) 2 " x x 4 1 x + 2 
 
 x-y (x-y) 2 6-2a 2 1_ 
 
 a? +1/ (a + y) 2 ■ 9 -a 2 34 a 3- a 
 
 6 i . i . 13 _^__^ «*!_ 
 
 2a 2 4-2a^" 1r 2a 2 -2a6 a4& (a + i) 2 («46) 8 
 
 1 Q-3c) 2 s + y 2a; x 2 (x-y) 
 
 a-3c a*-21c* ' y x+y yi^-y 2 ) 
 
 Hint. Reduce the last fraction to lowest terms. 
 
FRACTIONS. 
 
 135 
 
 15. 
 
 x + 2y x — 2y x 2 — 4 y a 
 1 1 
 
 16. -L-? + 
 
 17. 
 
 18. 
 
 19. 
 
 y — 1 y y -{ y 2 - 
 
 x 
 
 a;' 5 — x x 
 4 a 5 a 2 
 
 x -1 
 
 3 
 
 x — a (x — a) 2 (x — a) 8 
 
 1 a 2 x + a 
 
 x-2a x s -8a* x 2 + 2ax + 4:a* 
 
 1 
 
 .„ « 2 -2a; + 3, x-2 
 20. s-^— : f- 
 
 OJ 8 -hl 
 
 ar — x+1 x -f- 1 
 
 21. 
 
 22. 
 
 1 . £C-1 , a 2 + a - 3 
 
 + 0.0 . r, + — 1 
 
 »-3ic 2 + 3^4-9 x 8 - 27 
 x 2 + 8 x + 15 x - 1 
 
 x 2 + 7x-f-10 a; -2 
 Hint. Reduce the first fraction to lowest terma 
 
 x 2 — 5 ax + 6 a 2 x — 1 a 
 
 23. 
 
 24. 
 
 ic 2 — 8 ace + 15 a 2 x — 5 a 
 11 2 
 
 a -2 x 2 -3x + 2 x 2 -4x + 3 
 
 Hint. Express the denominators of the last two fractions in 
 ime factors. 
 
 25. 
 26. 
 27. 
 
 a 2 -7a + 12 a 2 -4a-f-3 a 2 -5a + 4 
 
 3 4 
 
 10a 2 + a -3 2a 2 + 7a -4* 
 
 3 1 
 
 2-x-6x 2 \-x-2x 2 
 
136 FRACTIONS. 
 
 167. When the denominators are polynomials not ar- 
 ranged in the same order, we first write the fractions, by 
 § 159, so that the denominators shall be arranged in the 
 same order. 
 
 1 tk a <u * 2 3 L 2*-3 
 
 1. Find the sum of - 7 + z : — ;• 
 
 x 2x — 1 1 — 4x 2 
 
 Change the signs before the terms of the denominator of the third 
 fraction and the sign before the fraction. We now have 
 2_ 3 2x-3 
 
 x 2x — 1 4x 2 — 1* 
 Proceeding as in § 166, we find for the required sum 
 
 -2 2 
 
 x(2x— l)(2x + l) x(l-2x)(l+2x)' 
 if we change the sign of 2 and the signs of (2 x — 1), § 160. 
 
 2. Find the sum of 
 
 1 i * 
 
 a (a — b) (a — c) b(b — a) (b — c) e(c — a) (c — b) 
 
 Change the sign of the factor (b — a) in the denominator of the 
 second fraction, and change the sign before the fraction, § 160. 
 
 Change the signs of the two factors (c — a) and (c — b) in the 
 denominator of the third fraction, § 160. We now have 
 
 * ! + 1 
 
 a(a — 6) (a — c) b{a — 6) (6 — c) c(a — c)(b — c) 
 The L.C.D. = abc (a -b)(a- c) (b - c). 
 
 bc(p — c) = 6 2 c — 6c 2 = 1st numerator. 
 
 — ac(a — c) = — a?c + ac* = 2d numerator. 
 
 afr( a — b) = a 2 b — ab 2 = 3d numerator. 
 
 a 2 b — a?c — ab 2 + ac 2 + b 2 c — be 2 = sum of numerators. 
 = a 2 (b-c)-a (b 2 - c 2 ) + bc{b- c). 
 Divide by the common factor (6 — c), and we have 
 a 2 — ab — ac + be \ 
 and this is equal to (a — b) (a — c). 
 
 .-. (a — 6) (a — c) (6 — c) = sum of numerators. 
 
 tt + . (a-b)(a-c)(b-c) _ 1 m 
 
 ,. sum of fraction* " «bc (a -6)<a-c)^-c) " o6c 
 
FRACTIONS. 137 
 
 Exercise 59. 
 Find the algebraic sum of : 
 
 — 1 x + 1 1 — x 
 a Sa 2 ax 
 
 a — x a-{- x x 2 — a 2 
 3 _ 2 !5 
 
 2a-3 3 + 2a 9-4a 2 
 
 c& — b 2a a 3 + a 2 6 
 T~~*"a~^b + b*- a 2 b 
 
 3 5 2z-7 
 
 x l-2x Ax 2 -1 
 
 (x H- a) 2 (a — #) 2 £p2 — a 
 
 7# 1 s-y a:y-2a; a 
 
 ' x — y x 2 -\- xy -\- y 2 y 8 — x* 
 
 8. , * * 4 
 
 (a; - 2) (* - 3) (aj - 1) (3 - as) (x - 1) (as - 2) 
 
 be ac ab 
 
 (c — a) (a — b) (a — b) (b — c) (b — c) (c — a) 
 
 b -f c a + c , a + b 
 
 (a — b) (a — c) (b — c) (b — a) (c — a) (c — b) 
 
 3 4 6 
 
 (a — b)(b — c) (b — a) (c — a) (a — c) (c — b) 
 
 I + l i 
 
 x(x -y)(x- z) y(y - x) (y - z) xyz 
 
 13 f£zj£ b 2 + ae e 2 + ab 
 
 (a — b) (a — c) (b — a) (b 4- e) (e — a) (c + b) 
 
 11. 
 
 12. 
 
138 FRACTIONS. 
 
 Multiplication of Fractions. 
 Find the product of - X -v 
 
 Let i x i = *- W 
 
 Multiply each of these equals by b X d. 
 
 Since the order of the factors is immaterial, (§ 42) 
 
 ( f X M X (^Xdj=bXdXx. 
 Or, aXc — bXdXx. 
 
 Divide by bxd, f£j=«- V) 
 
 From (1) and (2), ^X°- = ;~ 
 
 T ... . a c e a X c e a X c X e 
 
 Likewise, - X - X - = - ; X - .— ; ; - ; 
 
 b d f bXd J bXd Xf 9 
 
 and so on for any number of fractions. Hence, 
 
 168. To Find the Product of Two or More Fractions, 
 
 Find the product of the numerators for the required 
 numerator, and of the denominators for the required de- 
 nominator. 
 
 In applying the rule, reduce every mixed expression to a fraction, 
 and every integral expression to a fraction with 1 for the denominator. 
 
 Cancel every factor common to a numerator and a denominator, as 
 the cancelling of a common factor before the multiplication is equiva- 
 lent to cancelling it after the multiplication. 
 
 4 _. , J , . 2 a% 6 cH 5 ab*c 
 
 1. Find the product of — X — X g-j^ 
 
 2a?b QcH bab^c _ 2 X 6 X 5aWc*d _ ^ , 
 3 cd* 5 aft 8 a W 3X5X8 aWd* 2 d» * 
 
 
FRACTIONS. 139 
 
 2. Find the product of 
 
 x 2 - y 2 xy-2y 2 x 2 - xy 
 
 x 2 -3xy + 2y 2 x 2 + xy (as - y) 2 
 
 Express the numerators and denominators in prime factors. 
 (x - y) (x + y) x y(x-2y) x x(x-y) _ y _ 
 (x — y)(x — 2y) x(x-\-y) (x — y)(x — y) x — y 
 
 The common factors cancelled are (x — y), (x + y), (x — 2 y), x. 
 and (x — y). 
 
 Exercise 60. 
 
 Find the product of : 
 
 12 x 2 Uxy 
 1. -=-? X r-rj- 10. 
 
 7 y 2 9 ars 8 
 
 3a% 2 c* 20 mV 
 2. -j X — . • 11. 
 
 4 aw» 21 arc 6 
 
 7 rascy 3 a s c* 
 
 9mV 4 a 2 ?/ 2 
 
 8 x 8 y* W mn 
 
 21«y 4a 2 b 
 
 6 . J*K x Mtf. 15 . 
 
 12 « 2 35ic« 
 
 „ x 2 — y 2 4 x 
 
 7. 2 , o X — 16. 
 
 cc 2 + y 2 x + y 
 
 o 3cc 2 -a 2a 
 
 8. ■ — *7n — a — 17 - 
 
 a 2 x 2 — 4 x 
 
 3x* 3x-6 
 
 5* -10* 4x 8 
 
 18. 
 
 a 8 - j« A 
 
 ±a 2 
 
 x 2 4- 2/ 2 x 
 a; 2 — y 2 
 
 x + y 
 3x 2 + 3y 2 
 
 ab-b 2 
 
 a (a -\r I) 
 
 a(a 2 -b 2 ) 
 b 2 
 
 a 2 - Ax 2 
 a 2 4- 4 ax 
 
 ax 4- 4 x 2 
 a 2 — 2 ax 
 
 x * ~~ y A x ~ y 
 
 (x — y) 2 *'\ x 2 4- xy 
 
 x* + a* 
 x 2 -9a 2 
 
 w a? 4- 3 a 
 jc 4- a 
 
 a*-b* 
 a* -f- b* X 
 
 a 2 -ab+ b 2 
 a — b 
 
 x 2 -l 
 
 x> -25 
 
 x 2 — 4:X — 
 
 5"x 8 +2a;-3 
 
 (,-£\ 
 
 x f ."* Y 
 
140 FRACTIONS. 
 
 fa. _ x y ~ y 2 \ ( x - x y 2 ~ y* \ x * 2 
 
 \ x + V ) \ x 2 + y* ) x 2 — xy + y 2 
 
 21 
 
 . 8 a 2 b c 2 d 4 ab bed — cd 2 
 22. X 7T-z X — - X 
 
 8a 8 co" ±(b 2 -bd) 
 
 23 y(**-y*) x fr 2 -*/ 2 ) 2 x fr + y) 2 
 
 x(x + y) x 2 + xy + y 2 (x — y) 2 
 
 ■ (a + b) 2 -e 2 x a x (a-b) 2 -e 2 
 
 a 2 + aft" — ac (a -f c) 2 — ft 3 ab — b 2 — be 
 
 G 
 
 (a + b) 2 -e 2 c 2 -(a-b) 2 
 
 a 2 -(b- e) 2 c 2 -(a + b) 2 ae - a 2 + ab 
 
 (x - a) 2 - b 2 x 2 -(b- a) 2 ax + a 2 - ab 
 (x - bf - a 2 X x 2 - (a - b) 2 X bx-ab + b 2 
 
 27. 
 
 a 2 -2ab + b 2 -c 2 a + b 
 
 a 2 + 2ab + b 2 -c 2 a-b + e 
 
 x * + (x + 1 y X 2 + X 2x + l 
 
 X (x + 1) (x + l) 2 - x 2 x 
 
 2 ax s 4- 2 a 8 x x 2 — a 2 x + a 
 
 29. 
 
 (x - a) 2 (x 4- a) 2 2 (x 2 4- a 2 ) ax 
 
 ^ a 2 -b 2 -e 2 -2be (. 2c \ 
 
 a £ — ab — ac \ a + b + ej 
 
 a A + a 2 b 2 + V a + b a 2 - b 2 
 !r *•-*• X a 8 4-ft 8>< a 
 
 a; 2 4 7o;y + 12y 2 a: 3 + ay - 2 y 8 
 
FRACTIONS. 141 
 
 Division of Fractions. 
 
 169. Reciprocals. If the product of two rmmbers is equal 
 to 1, each of the numbers is called the reciprocal of the other. 
 
 The reciprocal of - is - ; for - X - = -7 = 1. 
 a a ao 
 
 The reciprocal of a fraction, therefore, is the fraction 
 inverted. 
 
 Find the quotient of - ■+■ -• 
 
 Let § + | = * (1) 
 
 Since the dividend is the product of the divisor and 
 quotient, 
 
 a c 
 
 Multiply each of these equals by -• 
 
 mi a d c d _„ 
 
 Then - X - = - X - X x = x. (2) 
 
 c a c v ' 
 
 From (1) and (2), \ -*- 4= T * -' Therefore, 
 
 w v * b d b c 
 
 170. To Divide by a Fraction, 
 
 Multiply the dividend by the reciprocal of the divisor. 
 
 Fmd the result of - X -2 : z h « — 7^ — 
 
 x x* — 4 a; — 5 xr — 25 
 
 1 x 2 -l . gg + 2a?-3 = l x 2 -l x 2 -25 
 
 x x 2 — 4x — 5 * x 2 — 25 a; x 2 — 4x — 5 x 2 + 2x — 3 
 
 ,1 (x-l)(x + l) (g-5)(g + 5> == x + 5 
 x (x-5)(x + l) (x + 3)(x-l) x(x + 3) 
 The common factors cancelled are (x — 1), (x + 1), (x — 6). 
 
 i 
 
142 FRACTIONS. 
 
 Exercise 61. 
 
 Find the quotient of: 
 
 15a 2 . 9x»z 5c^ 3b» She 
 
 ' If : 2Sxy' 2ab X 5ac '' 2a 9 ' 
 
 2. 
 
 3x 2 y 2 z 2 m 18 x 2 yz 2 x 2 -a 2 % . x - a 
 
 ±a 2 b 2 c 2 : 9a*b 2 c' x 2 -la 2 '*' x + 2a 
 
 5oW . 20 'nMp* x 2 y 2 + 3xy , xy + 3 
 
 3m*np* ' 2±abc* ' 4c 2 -l : 2c-l 
 
 16 a W 8 aW 9x 2 ~4y 2 3x-2y 
 
 21 m 2 xY : 7m*xy' 9 * 4 - x 2 ~*~ 2+x 
 
 2& ae y a& a 2 — 4 a 2 + 2 a 
 
 fo ! 3£ 2 5c 2 ' * a 2 + 4a~% 2 -16* 
 
 ft 2 -4a + 3 a 2 -10a + 21 a 2 -7a 
 
 11. 
 
 12. 
 
 a 2 -5a + 4,' a 2 - 9 a + 20 a 2 - 5 a 
 
 b 2 -7b + 6 . £ 2 -145 + 48 & 2 + 66 
 6 2 -f 36-4 : 6 2 + 10& + 24 : b*-8b 2 ' 
 
 13 a; 2 -y 2 % 2 +-a;y x 2 - xy 
 
 x 2 -3xy + 2y 2 ' xy-2y 2 (x-y)* 
 
 (a -by 2ab-2b 2 a? + ab 
 a 2 -b 2 ' 3 X ~a~=T" 
 
 15. 
 16. 
 17. 
 18. 
 
 (a + by-c 2 m c 2 -(a + b)* 
 a?-(b- c) 2 ' c 2 -(a- b) 2 ' 
 
 (x-ay-b 2 m x 2 -(a-b) 2 
 (x-b) 2 -a 2 *' x 2 -(b-ay 
 
 (a + by -(c + d) 2 m (a- c y -(d- by 
 
 (a + cy-(b+d) 2 ' (a-b) 2 -(d-cy 
 
 x 2 — 2 xy + y 2 — z 2 x — y -\-z 
 x 2 + 2xy -f y 2 — z 2 ' x + y — z 
 
FRACTIONS. 143 
 
 Complex Fractions. 
 
 171. A complex fraction is a fraction that has one or 
 more fractions in either or both of its terms. 
 
 3x 
 
 Simplify the complex fraction •? — ^-r • 
 
 4 
 
 3« . 4 s — 1 ' 4 12 x _ 
 
 = 3 x — = 3 x X = • Hence, 
 
 4x — 1 4 Ax — 1 4x — 1 ■ 
 
 172. To Simplify a Complex Fraction, 
 
 Divide the numerator by the denominator. 
 
 173. The shortest way to simplify a complex fraction is 
 
 to multiply both terms of the fraction by the L. G. D. of the 
 
 fractions contained in the numerator and denominator, 
 
 §157. 
 
 a a 
 
 ., a . ,.„ a — x a + x 
 1. Simplify 
 
 a + x 
 
 The L. CM. of a — x and a + x is (a — x) (a + x). 
 Multiply both terms by (a — x) (a + x), and we have 
 a (a + x) — a (a — x) _ a 2 + ax — a 2 + ax __ 2ax _ 
 x(a-{- x) + x(a — x)~ ax + x 2 + ax — x 2 ~ 2ax~ 
 
 2. Simplify =■ • 
 
 1 + 
 
 1 + 
 
 Multiply both terms of the last complex fraction by 1 — x. We 
 
 \ x 
 
 e ? and this put in place of the last complex fraction changes 
 
 A X 1 
 
 given fraction to the form 
 
 i+i. 
 
 2 x 
 
 Multiply both terms of this fraction by 2 — x. We have 
 
 3-2 
 
144 FRACTIONS. 
 
 Exercise 62. 
 Simplify : 
 
 1. 
 
 £ + £ 
 m m 
 
 z 
 
 
 m 
 
 2. 
 
 z 
 
 X 
 
 3. 
 
 ^-Sd 
 c 
 
 ab 
 
 m + n 
 1 
 
 c — 
 
 8. 
 
 2m -f- n . 
 m ■+■ n 
 
 1 n 
 
 9. 
 
 m •+- n 
 
 x s + y* 
 x 2 -y 2 
 
 x 2 — xy-\- y 2 
 
 
 x-y 
 
 1 a 
 
 o 
 
 a-b a 2 -b 2 
 
 
 a b 
 
 1+-^ 4+A+ 1 
 
 11. 
 
 ab + b 2 a 2 + ab 
 
 _I + 1 + I 
 
 ab ac bo 
 
 + c) s 
 
 x + 1 a# 
 
 5 . _^i^. i2. 2 » c , 
 
 i 2>_ - + - + - 
 
 x + y b c a 
 
 mn 
 
 m -\ 
 
 m — n 
 
 6. 13. 
 
 mn 
 m 
 
 14. 
 
 o x — 1 
 
 1 
 
 1 
 
 1 
 
 ab 
 
 ac 
 
 be 
 
 a 2 - 
 
 -(«- 
 
 •c) 2 
 
 
 a 
 
 
 
 1 
 
 
 
 x 2 - 
 
 -1 
 
 
 x+- 
 
 1 
 
FRACTIONS. 
 
 145 
 
 15. 
 
 16. 
 
 x + y 
 
 a + 
 
 1 + 
 
 a + 1 
 3-a 
 
 x + y + 
 
 x-y + 
 
 x+y 
 
 x + 
 
 17. 
 
 x + 
 
 r-M 
 
 y (xyz + x 4- *) 
 
 18. 
 
 3<z6c 
 
 #c 4- ac 4- «6 
 
 ft-1 5-1 c-1 
 
 a b c 
 
 a o c 
 
 20. 
 
 21. 
 
 (m + n m 2 4- w 2 \ m /m — w m 2 + w 2 \ 
 m — n m 2 — t& 2 / ' \ra + n m 2 — n % J 
 
 f x-y _ & - y* \ x f n + y + ^_±_f\ . 
 
 \* + 2/ x* + y*J \x-y x 2 - y 2 J 
 
 1 1 
 
 
 a 6 4-c 
 
 22. 
 
 14- « 4 
 
 1 — a; 4- x' 
 
 23. 
 
 x 2 — y 2 — z 2 — 2yz 
 
 x 2 — y 2 — z 2 + 2yz 
 
 x-y-z 
 
 x + y — z 
 
146 FRACTIONS. 
 
 Exercise 63. — Keview. 
 When a = 2, b = — 2, and c = 4 ; find the value of : 
 
 1. VV + b* + c 8 - (a - b - c)\ 
 
 a 2 + b 2 -c 2 + 2ab 
 • a 2_ b 2_ c 2 + 2 bc' 
 
 3. (b - 5) (J - 4) - 3 (b - 2) (b - 1) + 3 (b + 1) (6,+ 2). 
 
 4. Eednce to lowest terms ^ + 13x2 + 17x + 6 ' 
 Simplify : 
 
 l x— 4 « — a 
 
 5 - "3 ^t*t «c — z tt - ; ;t: + 
 
 (x-S)(x-2) (x-l)(x-3) ' (a;-l)(a;-2) 
 a a 2 a 2 4 a 4 
 
 ' a-& a + 6 a 2 + b 2 a 4 + 6 4 
 
 Hint. Add the first two fractions ; then their sum and the third 
 fraction and this result to the fourth fraction. 
 
 7. ~z ^--^-7-tH 
 
 a 1 -2 a 2 + 2 d z + l 
 
 8 1 • {( 1 I 1 V 2 V 
 
 x 2 -2x + \ ( x 2 -l x + 2 \ 
 (x-2) 2 '' \x 2 -± X x + 1) 
 
 10. 
 
 11. 
 
 a 2 -b 2 ab-2b 2 ^_ (a - b) 2 
 
 a 2 -3ab + 2b 2>< a 2 + ab ' a{a-b) 
 
 (a + b) 2 - c* a 2 b 2 c 2 W 
 
 a 2 + ab — ac a 2 -+- ab + ac a#c 
 
 s 2 -f7a:y + 10y 2 s + 1 . 1 
 
 ' x 2 + 6xy + 5y* x 2 + 4a + 4 * x + 2 
 
FRACTIONS. 147 
 
 13 x 2 -\-yz , y 2j txz . z 2 + xy 
 
 (x-y)(x-z) (j/-z)(y-x) (z-x)(z-y) 
 
 14. ( x ^"^ W ^ i — A . 
 \l + a; x J ' \l-\-x X J 
 
 c\x — c x + 2c) x 2 + ex — 2 c 2 
 
 16 . *-* (j* 1 1 £— ) 
 
 x 2 y 2 \x 2 — y 2 x 2 -\- y 2 J 
 
 3(x 2 + x-2) _ 3(x 2 -x-2) _ Sx 
 x 2 -x-2 x 2 + x-2 x 2 -± 
 
 \» + «/ y \2/ x + yj 
 
 (i-^L_+-^Yi+-i 2-Y 
 
 19. 
 
 20 a: 2 - ay + y 2 ?( s 8 - y 8 ; (y - x) 2 
 x 2 -\- xy + y 2 x 8 -\- y 8 ' (x + y) 2 
 
 2a — b — c 2b — c — a 2c — a — b 
 
 (a — b) (a — c) (b — c) (6 — a) (c — a) (c — £) 
 
 -J ? 1 + 1 1 + 1 
 
 22< s + 1 (s + 2)(g + l) ^ g«^> a r J ft 
 
 a; + 2 (a + l)(a + 2) a 2 £ 2 a 
 
 V 1+* 1-x 2 J\2x + lJ 
 
 25> a;8 ~ 8 y 3 x ^ 2 - x y + y 2 x x ( x * - v*) . 
 
 x (x — y) x 2 + 2 xy + 4: y 2 x 8 + y* 
 
 no 2,2,2 b + c-a c + a-b a + b-c 
 
 26. — ttH ; 7 
 
 a b c be ac ab 
 
CHAPTER X. 
 FRACTIONAL EQUATIONS. 
 
 Reduction of Equations Containing Fractions. 
 
 1. Solve - — = x — 9. 
 
 o 11 
 
 Multiply by 33, the L. C. M. of the denominators. 
 Then, llx - 3x + 3 = 33x — 297, 
 
 11 x — 3x-33x = — 297 — 3, 
 -25x = ~300. 
 .-. x = 12. 
 
 Note. Since the minus sign precedes the second fraction, in 
 removing the denominator the sign of every term of the numerator 
 is changed. 
 
 2. Solve 2x + 1 2*- 1 8 
 
 2x-l 2cc + l 4a 2 -l 
 
 The L. C. D. == (2x + 1) (2x - 1). 
 Multiply by the L. C. D., and we have, 
 
 4x 2 + 4x + l-(4x 2 -4x + l) = 8. 
 
 .-. 4x 2 + 4x + 1-4x2 + 4x- 1 = 8. 
 
 Reducing, x = 1. 
 
 174. To Clear an Equation of Fractions, therefore, 
 Multiply each term by the L. C. M. of the denominators. 
 
 If a fraction is preceded by a minus sign, the sign of 
 every term of the numerator must be changed when the 
 denominator is removed. 
 
Solve 
 
 FRACTIONAL EQUATIONS. 149 
 
 x — 4 x — 5 a; — 7 a; — 8 
 a; — 5 a; — 6 a; — 8 a; — 9 
 
 Note. The solution of this and similar problems will be much 
 easier by combining the fractions on the left side and the fractions 
 on the right side than by the rule. 
 
 (x - 4) (x - 6) - (x - 5)2 _ (x - 7) (x - 9) - (x - 8)2 
 (x - 5) (x - 6) (x — 8) (x — 9) 
 
 By simplifying the numerators, we have 
 -1 -1 
 
 (x - 5) (x — 6) (x — 8) (x — 9) 
 
 Since the numerators are equal, the denominators are equal. 
 
 Hence, (x — 5) (x - 6) = (x — 8) (x — 9). 
 
 Solving, we have x = 7. 
 
 Exercise 64. 
 Solve: 
 3«-l 2x + l Ax-5 . 
 
 h — 3 5~ = 4 * 
 
 5x + l , 19cc + 7 3z-l 7z-l 
 
 3. 
 
 4. 4 + 
 
 3 9 2 6 
 
 x Sx-2 \\x + 2 2-7x 
 
 7 2 14 
 
 9s + 5 8s-7 _ 36a3 + l5 41 
 14 + 14 ' 56 + 56* 
 
 7 a; -4 3a; -1 5(x- 1) _ 3 (Sx - 1) x 
 9 6 6 ~ 20 7* 
 
150 FRACTIONAL EQUATIONS. 
 
 4 5 4 3 
 
 - 10 a; + 11 12 a? -13 . 7 -6a 
 
 9 4 — . 
 
 6 3 4 
 
 3 5 9 1 
 
 10 ' y-4 + 2(y-4)" h 2( 2 /-4) 2* 
 
 2 5 8 x 5_ 
 
 x-1 2 (x-1) 3(a?-l) aj-l + 18* 
 
 2 a; -3 fi a; + 5 11 
 2a; -4 3a; -6 2' 
 
 10 -7a; 5a;-4 , 2x + l 8 2a;-l 
 
 13. — = — = — = 17. 
 
 6 -7a; 5x 2x-l 4a; 2 -1 2x + 1 
 
 5 + 8x 45-Sx 5-2x 2-lx 5a; 2 + 4 
 
 14 ' 3 + 2a;~13-2a;' 1 x-1 x + 1 ~ x 2 - 1 ' 
 
 15 5x-l _ 5x-S _6 *_+? . _£f_ = 
 
 2a; + 3 2a;-3 a; + 2 a;-2 a a -4 
 
 a; a; 2 -5x 2 4 , a? + 1 a 2 -3 
 
 16 ' 3~3^T~3' 20 * l+a; + l-a; l-^- U * 
 
 2a; + l _ 7a;-l 2a; 2 -3a;-45 
 3a;-3~~6a; + 6 4a; 2 -4 
 
 rto x 2 -a; + l , a; 2 + a; + 1 
 
 22. H rr-= = 2a;. 
 
 as— 1 a; 4- 1 
 
 9a; + 5 3s 2 - 51a; - 71 15a; - 7 
 23 ' G^-l) 4 " 18(a; 2 -l) 9(a; + l)' 
 
 24 ; a7+^ + a7+~3~a; 2 + 5a; + 6 ==a 
 
 25 _1 1_ = _1 1_. 
 
 x — 1 a; — 2 a; — 3 a; — 4 
 
FRACTIONAL EQUATIONS. 151 
 
 175. If the denominators contain both simple and com- 
 pound expressions, it is best to remove the simple expres- 
 sions first, and then each compound expression in turn. 
 After each multiplication the result should be reduced to 
 the simplest form. 
 
 i. solve - ir - + 5 _ jrg = _ r -. 
 Multiply both sides by 14, 
 
 Then, 8x + 5 + 7( ' 7X ~ -^- ) = 8x + 12. 
 
 ' 3a: 4- 1 
 
 Transpose and combine, — ^ — ;r— p = 7. 
 
 Divide by 7 and multiply by 3x + 1, 
 
 7x — 3 = 3z + l. 
 i 
 .-. x = 1- 
 
 Q 4a; 7x 
 
 9 Q1 3 ~T l T" 3 
 
 2. Solve _j_ = z jjp- 
 
 Simplify the complex fractions by multiplying both terms of each 
 fraction by 9. 
 
 Solve : 
 
 
 27 -4x 
 
 1 
 
 7x-27 
 
 
 36 
 
 4 
 
 90 
 
 Multiply both sides by 180, 
 
 the L. C. D., 
 
 
 135- 20 x 
 
 = 45 
 
 - 14 x + 54, 
 
 
 — 6x 
 
 = - 
 
 36. 
 
 
 .*. X 
 
 = 6. 
 
 
 
 Exercise 65 
 
 
 e : 
 
 1. 
 
 4a; + 3 2x- 
 
 5 
 
 2a;-l 
 
 10 5x- 
 
 1 
 
 5 
 
 2* 
 
 9a; + 20 4a;- 
 36 5x 
 
 -12 
 -4 
 
 *i; 
 
152 FRACTIONAL EQUATIONS. 
 
 10a; + 17 12a; + 2 5a; -4 
 3 ' 18 13 oj -16" 9 
 
 6a; + 7 7a;-13 2a; + 4 
 
 9 + 6z + 3 " 3 
 6a; + 7 2a;-22a; + l 
 
 15 7a;-6~~ 5 
 
 6a; + l 2a;-4 2x-l 
 
 15 7a;-16~ 5 
 
 11 a; -13 22 a; -75 13 x + 7 
 
 5. 
 
 7. 
 
 9, 
 
 14 28 2(3a + 7) 
 
 2a; + 8^ 13a;-2 x Ix x + 16 
 9 17 x - 32 + 3 7 12 36 
 
 5a; 7-2a; 2 l + 3a; 2 
 
 ./• 
 
 ^+4, 
 
 15 14 (a; -1) 21 6 ^105 
 
 10 2 *- 5 , *~ 3 4 *~ 3 iy 
 10 * 5 + 2a;-15~ 10 1A " 
 
 Literal Equations. 
 
 176. Literal equations are equations in which some or 
 all of the given numbers are represented by letters ; known 
 numbers are represented by the first letters of the alphabet- 
 Solve {a - x) (a + x) = 2 a 2 + 2aa; - x\ 
 
 Then, a 2 — x 2 = 2 a 2 + 2 ax — x 2 , 
 
 — 2 ax = a 2 , 
 a 
 
 Exercise 66. 
 Solve : 
 
 1. ax + 2 J = 3 bx + 4 a. 3. (a + a;) ( b + x) — x (x — c). 
 
 2. x 2 + & 2 = (a-a;) 2 . 4. (x- a)(x + b)=(x-b)(x- c). 
 
ERA C TIONAL EQUA TIONS. 
 
 153 
 
 8. 
 9 
 .0. 
 
 Ll. 
 
 12. 
 13. 
 
 14. 
 
 a + ax c -\- ex 
 
 C + d Til — x 
 
 ab -f bx an ■+- nx 
 
 x -\- 2 _m -{- n 
 x — 2 m — n 
 
 m -\- n m — n 
 
 2-x 
 e -\- dx 
 
 2 + x 
 
 a -\- bx _ 
 a -\- b e -\- d 
 
 6x + a _3x — 
 ±x + b 
 
 X X X 
 
 a b c 
 
 ax — b 
 
 2x — a 
 = d. 
 bx + o 
 
 e 
 
 5 ax 
 
 a — b 
 
 x 
 
 = abc. 
 
 3a = 8#. 
 
 5 a 2bx 
 
 a — 
 I 15. - + 
 
 b a + b a 2 
 n x + n 
 
 16. 
 
 x-\-n 
 
 x 
 
 nx 
 a 
 
 a b — a b + a 
 
 x + a 
 
 17. 
 
 18. 
 
 19. 
 
 20. 
 
 21. 
 
 22. 
 
 23. 
 
 24. 
 
 ic — a 
 
 x — b 
 
 \2a;- 6 
 
 x — a 
 
 2 
 
 2x — a 
 
 m -\- n 
 
 
 X 
 
 a 
 
 1 
 
 b 
 
 ax 
 
 b ax + b 
 
 ex + d 
 
 ex 
 1 
 
 a ■+• x 
 
 a — x ■+- 3 
 3 
 
 a — b 
 bx -+- c 
 
 2rf 
 a 
 
 + 
 
 a# — c 
 
 0. 
 
 25. 
 
 x = b 
 
 -b _ 2(a-\-b) 
 x — a x-\-b x 
 
 ) . c — bx 
 
 20 a — bx , 9 e — asc , 
 
 26. - H s 1- 
 
 6 
 6d — c# 
 
 5a 
 
 27 . -- &-s , a(6-s) 
 
 ax 
 T 
 
 2c 
 
 Se 
 
 b_ 
 Sd 
 
 2d 
 
 = 10. 
 
154 FRACTIONAL EQUATIONS. 
 
 Problems Involving Fractional Equations. 
 Exercise 67. 
 
 1. The sum of the third and fourth parts of a certain 
 number exceeds 3 times the difference of the fifth and sixth 
 parts by 29. Find the number. 
 
 Let x s= the number. 
 
 Then, - + - = the sum of its third and fourth parts, 
 
 x x 
 
 t — - = the difference of its fifth and sixth parts, 
 
 3^- — -J=3 times the difference of its fifth and sixth parts, 
 
 x x /x x\ . 
 
 o + T — of- — -J = the given excess. 
 
 But 29 = the given excess. 
 
 •••i+i-(f-i)=* 
 
 Multiply by 60, the L. C. D. of the fractions. 
 
 20s + 15x — 36x + SOx = 60 X 29. 
 Combining, 29 x = 60 X 29. 
 
 Dividing by 29, x = 60. 
 
 The required number, therefore, is 60. 
 
 2. The difference between the fifth and ninth parts of 
 a certain number is 4. Find the number. 
 
 3. One half of a certain number exceeds the sum of its 
 fifth and ninth parts by 17. Find the number. 
 
 4. The sum of the third and sixth parts of a certain 
 number exceeds the difference of its sixth and seventh parts 
 by 20. Find the number. 
 
 5. There are two consecutive numbers, x and x + 1, such 
 that one half of the larger exceeds one third of the smaller 
 number by 9. Find the numbers. 
 
 
FRACTIONAL EQUATIONS. 155 
 
 6. The sum of two numbers is 63, and if the greater 
 is divided by the smaller number, the quotient is 2 and the 
 remainder 3. Find the numbers. 
 
 Let x = the greater number. 
 
 Then 63 — x = the smaller number. 
 
 Dividend — Remainder 
 
 Since the quotient = 
 
 Divisor 
 
 and since, in this problem, the dividend is x, the remainder is 3, and 
 the divisor is 63 — x, we have 
 
 63 -x 
 Solving, x = 43. 
 
 The two numbers, therefore, are 43 and 20. 
 
 7. The sum of two numbers is 120, and if the greater is 
 divided by the smaller number, the quotient is 2 and the 
 remainder 15. Find the numbers. 
 
 8. The sum of two numbers is 126, and if the greater is 
 divided by the smaller number, the quotient is 3 and the 
 remainder 10. Find the numbers. 
 
 9. The difference of two numbers is 51, and if the greater 
 is divided by the smaller, the quotient is 4 and the remain- 
 der 6. Find the numbers. 
 
 10. The difference of two numbers is 87, and if the 
 greater is divided by the smaller, the quotient is 8 and the 
 remainder 10. Find the numbers. 
 
 11. Divide 450 into two parts such that the smaller part 
 is contained in the larger part 9 times, with a remainder 
 of 20. 
 
 12. The difference of two numbers is 25, and if the 
 greater is divided by the smaller, the quotient is 4 and 
 the remainder 4. Find the numbers. 
 
156 FRACTIONAL EQUATIONS. 
 
 13. Eight years ago a boy was one fourth as old as he 
 will be one year hence. How old is he now ? 
 
 Let x = the number of years old he is now. 
 
 Then x — 8 = the number of years old he was eight years ago, 
 and x + 1 = the number of years old he will be one year hence. 
 .\x — S = i(x + 1). 
 Solving, x — 11. 
 
 Therefore, the boy is 11 years old. 
 
 14. A son is one third as old as his father. In 10 years 
 he will be one half as old. Find the age of the son. 
 
 15. B's age is one fifth of A's age. In 12 years B's age 
 will be one third of A's age. Find their ages. 
 
 16. The sum of the ages of A and B is 60 years, and 10 
 years hence B's age will be one third of A's. Find their 
 
 17. A father is 40 years old, and his son is one fourth of 
 that age. In how many years will the son be half as old 
 as his father ? 
 
 18. A is 30 years old, and B's age is two thirds of A's. 
 How many years ago was B's age one third of A's ? 
 
 19. A son is one fourth as old as his father. Four years 
 ago he was only one fifth as old as his father. What is 
 the age of each ? 
 
 20. A is 40 years old, and B is half as old as A. In how 
 many years will B be two thirds as old as A ? 
 
 21. B is one third as old as A. Ten years ago he was 
 one fourth as old as A. What are their present ages ? 
 
 22. The sum of the ages of a father and his son is 75 
 years. The son's age increased by 5 years is one fourth of 
 the father's age. Find their ages. 
 
FRACTIONAL EQUATIONS. 157 
 
 23. A rectangle has its length 6 feet more and its width 
 5 feet less than the side of the equivalent square. Find the 
 dimensions of the rectangle. 
 
 Let x = the number of feet in a side of the square. 
 
 Then x + 6 = the number of feet in the length of the rectangle, 
 and x — 5 = the number of feet in the width of the rectangle. 
 
 Since the area of a rectangle is equal to the product of the number 
 of units of length in the length and width of the rectangle, 
 
 (x + 6) (x — 5) = the area of the rectangle in square feet, 
 and x x x = the area of the square in square feet. 
 
 But these areas are equal. 
 
 .-. (x + 6) (x - 5) = x 2 . 
 Solving, x = 30. 
 
 Therefore, the dimensions of the rectangle are 36 feet 
 and 25 feet. 
 
 24. A rectangle has its length and breadth, respectively, 
 12 feet longer and 8 feet shorter than the side of the 
 equivalent square. Find its area. 
 
 25. The length of a floor exceeds the breadth by 6 feet. 
 If each dimension were 1 foot more, the area of the floor 
 would be 41 sq. ft. more. Find its dimensions. 
 
 26. A rectangle whose length is 8 feet more than its 
 breadth would have its area 35 sq. ft. more, if each dimen- 
 sion were 1 foot more. Find its dimensions. 
 
 27. The length of a rectangle exceeds its width by 4 feet. 
 If the length were diminished by 2 feet and the width by 
 2 feet, the area would be diminished by 40 sq. ft. Find its 
 dimensions. 
 
 28. The length of a floor exceeds its width by 8 feet. 
 If each dimension were 2 feet more, the area would be 124 
 sq. ft. more. Find its dimensions. 
 
158 FRACTIONAL EQUATIONS. 
 
 29. A can do a piece of work in 2 days, and B can do 
 it in 3 days. How many days will it take both together to 
 do the work ? 
 
 Let x ~ the number of days it will take both together. 
 
 Then - = the part both together can do in one day, 
 
 x 
 
 ■£• = the part A can do in one day, 
 
 £ s= the part B can do in one day, 
 
 and i + i = the part both together can do in one day. 
 
 ,.1+1=1. 
 
 2 3 x 
 
 Solving, x = 1£. 
 
 Therefore, they together can do the work in 1^ days. 
 
 30. A can do a piece of work in 3 days, B in 4 days, and 
 C in 5 days. How many days will it take them to do it 
 working together ? 
 
 31. A can do a piece of work in 6 days, B in 5 days, and 
 C in 4 days. How many days will it take them together 
 to do the work ? 
 
 32. A can do a piece of work in 2\ days, B in 3^ days, 
 and C in 3| days. How many days will it take them 
 together to do the work ? 
 
 33. A can do a piece of work in 8 days, B in 10 days ; 
 A and B together, with the help of C, can do the work in 
 3 days. How many days will it take C alone to do the 
 work? 
 
 .34. A and B together can mow a field in 8 honrs, A and 
 C in 10 hours, and A alone in 15 hours. In how many 
 hours can B and C together mow the field ? . 
 
 ! 35. A and B together can build a wall in 12 days, A and 
 C in 15 days, B and C in 20 days. In how many days can 
 they build the wall if they all work together ? 
 
 Hint. By working 2 days each they build y 1 ^ + ^ 4- ^ of it. 
 
 Hence, m one day they can build ±(fa + T V + **&)• 
 
FRA C TIONAL EQUA TIONS. 159 
 
 36. A cistern can be filled by two pipes in 15 and 20 
 hours, respectively ; and can be emptied by a waste pipe 
 in 30 hours. In how many hours will it be filled if all the 
 pipes together are open ? 
 
 Let x = the number of hours it all the pipes are open. 
 
 Then - = the part filled in one hour if all the pipes are open. 
 
 tV + -fa ~~ ^5 = the part all together can fill in one hour. 
 
 *'*15 + 20 30 x 
 
 Solving, x =3 12. 
 
 Therefore, if all the pipes are open it will be filled in 12 
 hour?. 
 
 37. A cistern can be filled by three pipes in 8, 12, and 
 16 hours, respectively. In how many hours will it be filled 
 by all the pipes together ? 
 
 38. A cistern can be filled by two pipes in 4 hours and 5 
 hours, respectively, and can be emptied by a third pipe in 
 6 hours. In how many hours will the cistern be filled if the 
 pipes are all running together ? 
 
 39. A tank can be filled by three pipes in 1 hour and 40 
 minutes, 3 hours and 20 minutes, and 5 hours, respectively. 
 In what time will the tank be filled if all three pipes are 
 running together ? 
 
 40. A cistern can be filled by three pipes in 2\ hours, Z\ 
 hours, and 4§ hours, respectively. In how many hours will 
 the cistern be filled if all the pipes are running together ? 
 
 41. A cistern has three pipes. The first pipe will fill the 
 cistern in 6 hours, the second in 10 hours, and all three 
 pipes together will fill it in 3 hours. How long will it take 
 the third pipe alone to fill it ? 
 
160 FRACTIONAL EQUATIONS. 
 
 42. A courier who travels 6 miles an hour is followed, 
 after 2 hours, by a second courier who travels 7£ miles an 
 hour. In how many hours will the second courier overtake 
 the first ? 
 
 Let x = the number of hours the first travels. 
 
 Then x — 2 = the number of hours the second travels, 
 
 6 x = the number of miles the first travels, 
 and (x — 2) 7£ = the number of miles the second travels. 
 
 They both travel the same distance. 
 
 .:6x= (x — 2)7*, 
 or 12x=15x-30. 
 
 .-. x = 10. 
 
 Therefore, the second courier will overtake the first in 
 10 — 2, or 8 hours. 
 
 43. A sets out from Boston and walks towards Portland 
 at the rate of 3 miles an hour. Three hours afterward B 
 sets out from the same place and walks in the same direc- 
 tion at the rate of 3^ miles an hour. How far from Boston 
 will B overtake A ? 
 
 44. A courier who goes at the rate of 4 \ miles an hour is 
 followed, after 4 hours, by another who goes at the rate of 
 5* miles an hour. In how many hours will the second 
 overtake the first ? 
 
 45. A person walks to the top of a mountain at the rate 
 of 1^ miles an hour, and down the same way at the rate of 
 4* miles an hour. If he is out 6 hours, how far is it to the 
 top of the mountain ? 
 
 46. In going a certain distance, a train traveling at the 
 rate of 50 miles an hour takes 2 hours less than a train 
 traveling 40 miles an hour. Find the distance. 
 
FRACTIONAL EQUATIONS. 161 
 
 47. Find the time between 2 and 3 o'clock when the 
 hands of a clock are together. 
 
 At 2 o'clock the hour-hand is 10 minute-spaces ahead of the 
 minute-hand. 
 
 Let x = the number of spaces the minute-hand moves over. 
 
 Then x — 10 = the number of spaces the hour-hand moves over. 
 Now, as the minute-hand moves 12 times as fast as the hour-hand, 
 12 (x — 10) = the number of spaces the minute-hand moves over. 
 
 .-. 12 (x - 10) = x, 
 and 11 x = 120. 
 
 .-. x = lOf-f 
 
 Therefore, the time is 10}£ minutes past 2 o'clock. 
 
 48. Find the time between 4 and 5 o'clock when the 
 hands of a clock are together. 
 
 49. Find the time between 3 and 4 o'clock when the 
 hands of a clock are at right angles to each other. 
 
 Hint. In this case the minute-hand is 16 minutes ahead of the 
 hour-hand. 
 
 50. Find the time between 2 and 3 o'clock when the 
 hands of a clock point in opposite directions. 
 
 Hint. In this case the minute-hand is 30 minutes ahead of the 
 hour-hand, or 30 minutes behind it. 
 
 51. Find the times between 4 and 5 o'clock when the 
 hands of a clock are at right angles to each other. 
 
 52. Find the time between 1 and 2 o'clock when the 
 hands of a clock point in opposite directions. 
 
 53. At what time between 6 and 7 o'clock are the hands 
 of a watch together ? 
 
162 FRACTIONAL EQUATIONS. 
 
 54. A hare takes 4 leaps to a greyhound's 3; but 2 of 
 the greyhound's leaps are equivalent to 3 of the hare's. 
 The hare has a start of 50 of her own leaps. How many 
 leaps must the greyhound take to catch the hare ? 
 
 Let 3 x = the number of leaps taken by the greyhound. 
 
 Then 4 x = the number of leaps of the hare in the same time. 
 Also, let a = the number of feet in one leap of the hare. 
 
 Then — = the number of feet in one leap of the hound. 
 
 Therefore, 3 x X — - or — — = the whole distance. 
 
 As the hare has a start of 50 leaps, and takes 4 x leaps more before 
 
 she is caught, and as each leap is a feet, 
 
 (50 + 4 x) a = the whole distance. 
 
 9 ax 
 .-. -7r-= (50 + 4x)a. 
 
 Multiply by 2, 9 ax = (100 + 8 x) a. 
 
 Divide by a, 9x "= 100 + 8x, 
 
 x == 100, 
 .'.Sx = 300. 
 
 Therefore, the greyhound must take 300 leaps. 
 
 55. A hound takes 3 leaps while a rabbit takes 5 j but 
 1 of the hound's leaps is equivalent to 2 of the rabbit's. 
 The rabbit has a start of 120 of her own leaps. How 
 many leaps will the rabbit take before she is caught ? 
 
 56. A rabbit takes 6 leaps to a dog's 5, and 7 of the dog's 
 leaps are equivalent to 9 of the rabbit's. The rabbit has a 
 start of 60 of her own leaps. How many leaps must the 
 dog take to catch the rabbit ? 
 
 57. A dog takes 4 leaps while a rabbit takes 5 j but 3 
 of the dog's leaps are equivalent to 4 of the rabbit's. The 
 rabbit has a start of 90 of the dog's leaps. How many 
 leaps will each take before the rabbit is caught ? 
 
FRACTIONAL EQUATIONS. 163 
 
 58. A merchant adds yearly to his capital one third of 
 it, but takes from it, at the end of each year, $ 5000 for 
 expenses. At the end of the third year, after deducting 
 the last $5000, he has twice his original capital. How 
 much had he at first? 
 
 Let x = number of dollars he had at first. 
 
 4x ^™ 4 a; -15000 
 Then -g- — 5000, or > 
 
 will stand for the number of dollars at the end of first year, 
 4 /4« - 15000 \ ennn 16 x - 105000 
 
 "^ s( 8 ) ~ 5 ° 00 ' ° r 9 ' 
 
 will stand for the number of dollars at the end of second year, 
 4 /16s- 1 05000 \ CAAA 64^-555000 
 
 ftnd 8 ( ) "" 500 °' ° r _ "27 ' 
 
 will stand for the number of dollars at the end of third year. 
 
 But 2 x stands for the number of dollars at the end of third year. 
 64 a; -555000 . 
 
 '•• 27 = 2X ' 
 
 Whence x = 55,500. 
 
 Therefore, the merchant had $55,500 at first. 
 
 59. A trader adds yearly to his capital one fourth of it, 
 but takes from it, at the end of each year, $800 for ex- 
 penses. At the end of the third year, after deducting the 
 last $800, he has 1$ of his original capital. How much 
 had he at first ? 
 
 60. A trader adds yearly to his capital one fifth of it, 
 but takes from it, at the end of each year, $2500 for ex- 
 penses. At the end of the third year, after deducting 
 the last $2500, he has 1 T 7 ^ of his original capital. Find 
 his original capital. 
 
 61. A trader maintained himself for three years at an ex- 
 pense of $500 a year; and each year increased that part of 
 his stock which was not so expended by one third of it. 
 At the end of the third year his original stock was doubled. 
 What was his original stock ? 
 
164 FRACTIONAL EQUATIONS. 
 
 62. The sum of the third, fourth, and fifth parts of a 
 number exceeds the half of the number by 17. Find the 
 number. 
 
 63. There are two consecutive numbers, x and a? -f 1, 
 such that one fourth of the smaller exceeds one ninth of the 
 larger by 11. Find the numbers. 
 
 64. Find three consecutive numbers such that if they are 
 divided by 7, 10, and 17, respectively, the sum of the quo- 
 tients will be 15. 
 
 65. In a mixture of alcohol and water the alcohol was 
 24 gallons more than half the mixture, and the water was 
 4 gallons less than one fourth the mixture. How many 
 gallons were there of each ? 
 
 66. The width of a room is three fourths of its length. 
 If the width was 4 feet more and the length 4 feet less, the 
 room would be square. Find its dimensions. 
 
 67. The difference of two numbers is 40, and if the greater 
 is divided by the less the quotient is 4, and the remainder 
 4. Find the numbers. 
 
 68. Divide the number 240 into two parts such that the 
 smaller part is contained in the larger part 5 times, with a 
 remainder of 6. 
 
 69. A can do a piece of work in 3 days, B in 4 days, and 
 C in 6 days. How many days will it take them to do it 
 working together ? 
 
 70. At what time between 2 and 3 o'clock are the hands 
 of a watch at right angles ? 
 
 71. Find a number such that the sum of its sixth and 
 ninth parts shall exceed the difference of its ninth and 
 twelfth parts by 9. 
 
FRACTIONAL EQUATIONS. 165 
 
 72. The sum of two numbers is 91, and if the greater is 
 divided by the less the quotient is 2, and the remainder 
 
 I is 1. Find the numbers. 
 
 73. A is 60 years old, and B is three fourths as old. 
 How many years since B was just half as old as A ? 
 
 74. A can do a piece of work in 2£ days, B in 3£ days, 
 and C in 4§ days. How long will it take them to do it 
 working together ? 
 
 75. A and B can separately do a piece of work in 12 
 days and 20 days, and with the help of C they can do it in 
 6 days. How long would it take C alone to do the work ? 
 
 76. A rectangle has its length 4 feet longer and its width 
 3 feet shorter than the side of the equivalent square. Find 
 its area. 
 
 77. A person has 6 hours at his disposal. How far may 
 he ride at 6 miles an hour so as to return in time, walking 
 
 j 'back at the rate of 3 miles an hour ? 
 
 78. A boy starts from Exeter and walks towards Ando- 
 ver at the rate of 3 miles an hour, and 2 hours later another 
 boy starts from Andover and walks towards Exeter at the 
 
 * rate of 2\ miles an hour. The distance from Exeter to 
 
 t A ndover is 28 miles. How far from Exeter will they meet ? 
 79. A dog makes 4 leaps while a hare makes 5, but 3 of 
 e dog's leaps are equal to 4 of the hare's. The hare has a 
 start of 60 of the dog's leaps. How many leaps will each 
 make before the hare is caught ? 
 
 80. At what time between 3 and 4 o'clock are the hands 
 f a watch pointing in opposite directions ? 
 
 81. In going from Boston to Portland, a passenger train, 
 | 36 miles an hour, occupies 1 hour less time than a freight 
 
 in at 27 miles an hour. Find the distance from Boston 
 Portland. 
 
166 FRACTIONAL EQUATIONS. 
 
 82. A cistern can be filled by three pipes in 15, 20, and 
 30 minutes, respectively. In what time will it be filled if 
 the pipes are all running together ? 
 
 83. A cistern can be filled by two pipes in 25 minutes 
 and 30 minutes, respectively, and emptied by a third in 20 
 minutes. In what time will it be filled if all three pipes 
 are running together ? 
 
 84. A hare takes 7 leaps while a dog takes 5, and 5 of 
 the dog's leaps are equal to 8 of the hare's. The hare has 
 a start of 50 of her own leaps. How many leaps will the 
 hare take before she is caught ? 
 
 85. The width of a rectangle is an inch more than half 
 its length, and if a strip 5 inches wide is taken off from the 
 four sides, the area of the strip is 510 square inches. Find 
 the dimensions of the rectangle. 
 
 86. A and B together can do a piece of work in 10 days, 
 A and C in 12 days, and A by himself in 18 days. How 
 many days will it take B and C together to do the work ? 
 How many days will it take A, B, and C together ? 
 
 87. A and B can do a piece of work in 10 days, A and 
 G in 12 days, B and C in 15 days. How long will it take 
 them to do the work if they all work together ? 
 
 88. A sets out and travels at the rate of 9 miles in 2 
 hours. Seven hours afterwards B sets out from the same 
 place and travels in the same direction at the rate of 5 
 miles an hour. In how many hours will B overtake A ? 
 
 89. A man walks to the top of a mountain at the rate of 
 2\ miles an hour, and down the same way at the rate of 4 
 miles an hour, and is out 5 hours. How far is it to the top 
 of the mountain ? 
 
FRACTIONAL EQUATIONS. 167 
 
 90. A tank can be filled by three pipes in 1 hour 20 
 minutes, 2 hours 20 minutes, and 3 hours 20 minutes, 
 respectively. In how many minutes can it be filled by 
 all three together? 
 
 91. A's age now is two fifths of B's. Eight years ago 
 A's age was two ninths of B's. Find their ages. 
 
 92. A had five times as much money as B. He gave B 
 5 dollars, and then had only twice as much as B. How 
 much had each at first ? 
 
 93. At what time between 12 and 1 o'clock are the hour 
 and minute-hands pointing in opposite directions ? 
 
 94. Eleven sixteenths of a certain principal was at in- 
 terest at 5 per cent, and the balance at 4 per cent. The 
 entire income was $1500. Find the principal. 
 
 95. A train that travels 36 miles an hour is f of an 
 hour in advance of a second train that travels 42 miles 
 an hour. In how long a time will the last train overtake 
 the first? 
 
 96. An express train that travels 40 miles an hour starts 
 from a certain place 50 minutes after a freight train, and 
 overtakes the freight train in 2 hours 5 minutes. Find 
 the rate per hour of the freight train. 
 
 97. If 1 pound of tin loses / T of a pound, and 1 pound 
 of lead loses -fa of a pound, when weighed in water, how 
 many pounds of tin and of lead in a mass of 60 pounds that 
 loses 7 pounds when weighed in water ? 
 
 98. If 19 ounces of gold lose 1 ounce, and 10 ounces of 
 silver lose 1 ounce, when weighed in water, how many 
 ounces of gold and of silver in a mass of gold and silver 
 that weighs 530 ounces in air and 495 ounces in water ? 
 
168 FRACTIONAL EQUATIONS. 
 
 99. A messenger starts to carry a despatch, and 5 hours 
 later a second messenger sets out to overtake the first in 8 
 hours. In order to do this, he is obliged to travel 2\ miles 
 an hour more than the first. How many miles an hour- 
 does the first travel ? 
 
 100. The fore and hind wheels of a carriage are respec- 
 tively 9^ feet and llf feet in circumference. What distance 
 will the carriage have made when one of the fore wheels 
 has made 160 revolutions more than one of the hind 
 wheels ? 
 
 101. When a certain brigade of troops is formed in a 
 solid square there is found to be 100 men over ; but when 
 formed in column with 5 men more in front and 3 men less 
 in depth than before, the column needs 5 men to complete 
 it. Find the number of troops. 
 
 102. An officer can form his men in a hollow square 14. 
 deep. The whole number of men is 3136. Find the num- 
 ber of men in the front of the hollow square. 
 
 103. A trader increases his capital each year by one 
 fourth of it, and at the end of each year takes out $2400 
 for expenses. At the end of 3 years, after deducting the 
 last $2400, he finds his capital to be $10,000. Find his 
 original capital. 
 
 104. A and B together can do a piece of work in 1^ days, 
 A and C together in If days, and B and C together in If 
 days. How many days will it take each alone to do the 
 work? 
 
 105. A fox pursued by a hound has a start of 100 of her 
 leaps. The fox makes 3 leaps while the hound makes 2; 
 but 3 leaps of the hound are equivalent to 5 of the fox 
 How many leaps will each take before the hound catches 
 the fox? 
 
FRACTIONAL EQUATIONS. 169 
 
 Formulas and Rules. 
 
 177. When the given numbers of a problem are repre- 
 sented by letters, the result obtained from solving the prob- 
 lem is a general expression which includes all problems of 
 that class. Such an expression is called a formula, and the 
 translation of this formula into words is called a rule. 
 
 We will illustrate by examples: 
 
 1. The sum of two numbers is s, and their difference d) 
 
 find the 
 
 numbers. 
 
 
 Let 
 
 
 x = the smaller number ; 
 
 then, 
 
 
 x + d = the larger number. 
 
 Hence, 
 
 
 x + x + d = a, 
 
 or 
 
 
 2x — s — d. 
 s — d 
 
 2 
 s + d 
 
 Therefore, the numbers are — - — and — - — 
 
 As these formulas hold true whatever numbers s and d 
 stand for, we have the general rule for finding two numbers 
 when their sum and difference are given : 
 
 Add the difference to the sum and take half the result for 
 the greater number. 
 
 Subtract the difference from the sum and take half the 
 result for the smaller number. 
 
 2. If A can do a piece of work in a days, and B can do 
 the same work in b days, in how many days can botb 
 together do it? 
 
170 FRACTIONAL EQUATIONS. 
 
 Let x = the required number of days. 
 
 Then, - = the part both together can do in one day. 
 
 x 
 
 Now, - = the part A can do in one day, 
 
 and - as the part B can do in one day ; 
 
 therefore, - + - = the part both together can do in one day. 
 
 ,.! + i = i. 
 
 a b x 
 
 Whence, 
 
 ab 
 
 a + b 
 
 The translation of this formula gives the following rule 
 for finding the time required by two agents together to 
 produce a given result, when the time required by each 
 agent separately is known : 
 
 Divide the product of the numbers that express the units 
 of time required by each to do the work by the sum of these 
 numbers ; the quotient is the time required by both together. 
 
 Exercise 68. 
 
 1. A person has a hours at his disposal. How far may 
 he ride in a coach that travels b miles an hour, so as to 
 return home in time, walking back at the rate of c miles aa 
 hour ? 
 
 2. A courier who travels a miles a day is followed after 
 c days by another who travels b miles a day. In how many 
 days will the second overtake the first ? 
 
 3. A has a dollars and B has b dollars. B gives A a cer- 
 tain number of dollars, and then has c times as much as A 
 How many dollars does A receive from B ? 
 
FRACTIONAL EQUATIONS. 171 
 
 4. The fore wheel of a carriage is a feet in circumfer- 
 ence, and the hind wheel is b feet. Find the distance 
 traveled when the fore wheel has made c revolutions more 
 than the hind wheel. 
 
 5. Two towns, P and Q, are a miles apart. One person 
 sets out from P and travels towards Q at the rate of b miles 
 an hour, and at the same time another person sets out from 
 Q and travels towards P at the rate of c miles an hour. 
 How many miles from P will they meet ? 
 
 6. A person was employed a days on these conditions : 
 for each day he worked he was to receive b cents, and for 
 each day he was idle he was to forfeit c cents. At the end 
 of a days he received d cents. How many days did he work ? 
 
 7. A banker has two kinds of coins : it takes a pieces of 
 the first to make a dollar, and b pieces of the second to make 
 a dollar. A person wishes to obtain c pieces for a dollar. 
 How many pieces of each kind must the banker give him ? 
 
 Interest Formulas. 
 
 178. The elements involved in computation of interest 
 are the principal, rate, time, interest, and amount. 
 
 Let p = the principal, 
 
 r = the interest of $1 for 1 year, at the given rate, 
 y^f = the time expressed in years, 
 
 i = the interest for the given time and rate, 
 Sa = the amount (sum of principal and interest). 
 
 179. Given the Principal, Rate, and Time ; to Find the Interest. 
 
 Since r is the interest of $ 1 for 1 year, pr is the interest 
 of $p for 1 year, and prt is the interest of %p for t years. 
 
 .\ i = prt. (Formula 1) 
 
 Rule. Find the product of the principal, rate, and time. 
 
172 FRACTIONAL EQUATIONS. 
 
 180. Given the Interest, Rate, and Time ; to Find the Principal, 
 
 By formula 1, prt = i. 
 
 Divide by rt, p = — • (Formula 2) 
 
 181. Given the Amount, Rate, and Time ; to Find the Principal. 
 
 From formula 1, p -\-prt = a, 
 or p (1 -f- rt) = a. 
 
 a 
 
 Divide by 1 + rt, p = • (Formula 3) 
 
 182. Given the Amount, Principal, and Rate ; to Find the Time. 
 
 From formula 1, p +prt = a. 
 Transpose p } prt = a —p. 
 
 Divide by pr, t = — —' (Formula 4) 
 
 183. Given the Amount, Principal, and Time ; to Find the Rate. 
 
 From formula 1, p -\-prt = a. 
 Transpose p, prt = a —p. 
 
 Divide by pt, r = — £*• (Formula 5) 
 
 Exercise 69. 
 Solve by the preceding formulas : 
 
 1. The sum of two numbers is 40, and their difference is 
 10. Find the numbers. 
 
 2. The sum of two angles is 100°, and their difference is 
 21° 30'. Find the angles. 
 
 3. The sum of two angles is 116° 24' 30", and their 
 difference is 56° 2V 44". Find the angles. 
 
FRACTIONAL EQUATIONS. 173 
 
 4. A can do a piece of work in 6 days, and B in 5 days. 
 How long will it take both together to do it ? 
 
 5. Find the interest of $2750 for 3 years at 4£ per 
 cent. 
 
 6. Find the interest of $950 for 2 years 6 months at 5 
 per cent. 
 
 7. Find the amount of $2000 for 7 years 4 months at 6 
 per cent. 
 
 8. Find the rate if the interest on $680 for 7 months is 
 $35.70. 
 
 9. Find the rate if the amount of $750 for 4 years is 
 $900. 
 
 10. Find the rate if a sum of money doubles in 16 years 
 8 months. 
 
 11. Find the time required for the interest on $2130 to 
 be $436.65 at 6 per cent. 
 
 12. Find the time required for the interest at 5 per cent 
 on a sum of money to be equal to the principal. 
 
 13. Find the principal that will produce $161.25 interest 
 in 3 years 9 months at 8 per cent. 
 
 14. Find the principal that will amount to $1500 in 3 
 years 4 months at 6 per cent. 
 
 15. How much money is required to yield $2000 interest 
 annually if the money is invested at 5 per cent ? 
 
 16. Find the time in which $640 will amount to $1000 
 at 6 per cent. 
 
 17. Find the principal that will produce $100 per month, 
 at 6 per cent. 
 
 18. Find the rate if the interest on $700 for 10 months 
 is $25. 
 
CHAPTER XL 
 SIMULTANEOUS SIMPLE EQUATIONS. 
 
 184. If we have two unknown numbers and but one rela- 
 tion between them, we can find an unlimited number of 
 pairs of values for which the given relation will hold true. 
 
 Thus, if x and y are unknown, and we have given only the one 
 relation x + y = 10, we can assume any value for x, and then from 
 the relation x + y = 10 find the corresponding value of y. For from 
 x + y = 10 we find y = 10 — x. If x stands for 1, y stands for 9 ; if 
 x stands for 2, y stands for 8 ; if x stands for — 2, y stands for 12 ; 
 and so on without end. 
 
 185. We may, however, have two equations that express 
 different relations between the two unknowns. • Such equa- 
 tions are called independent equations. 
 
 Thus, x + y = 10 and x — y = 2 are independent equations, for 
 they evidently express different relations between x and y. 
 
 186. Independent equations involving the sameunknowns 
 are called simultaneous equations. 
 
 If we have two unknowns, and have given two independ- 
 ent equations involving them, there is but one pair of values 
 which will hold true for both equations. 
 
 Thus, if in § 184, besides the relation x + y = 10, we have also the 
 relation x — y = 2, the only pair of values for which both equations 
 will hold true is the pair x = 6, y = 4. 
 
 Observe that in this problem x stands for the same number in both 
 equations ; so also does y. 
 
SIMULTANEOUS SIMPLE EQUATIONS. 175 
 
 187. Simultaneous equations are solved by combining 
 the equations so as to obtain a single equation with one 
 unknown number; this process is called elimination. 
 
 There are three methods of elimination in general use : 
 
 I. By Addition or Subtraction. 
 II. By Substitution. 
 III. By Comparison. 
 
 188. Elimination by Addition or Subtraction. 
 
 1. Solve: 5x-3y = 20\ (1) 
 2x + 5y = 39) (2) 
 
 Multiply (1) by 5, and (2) by 3, 
 
 25* -15?/ = 100 (3) 
 
 6* + 15y = 117 (4) 
 
 Add (3) and (4), 31* =217 
 
 .-. * = 7. 
 
 Substitute the value of * in (2), 
 
 14 + 5y = 39. 
 
 ••• y = 5. 
 
 In this solution y is eliminated by addition. 
 
 2. Solve: 6x + 35y = 177\ (1) 
 Sx-21y = 33) (2) 
 
 Multiply (1) by 4, and (2) by 3, 
 
 24x + 140y = 708 (3) 
 
 24*- 63 y= 99 (4) 
 
 Subtract, 203 y = 609 
 
 .-. y = 3. 
 
 Substitute the value of y in (2), 
 
 8* — 63 = 33. 
 
 .-. * = 12. 
 
 In this solution x is eliminated by subtraction. 
 
176 SIMULTANEOUS SIMPLE EQUATIONS, 
 
 189. To Eliminate by Addition or Subtraction, therefore, 
 
 Multiply the equations by such numbers as will make the 
 coefficients of one of the unknown numbers equal in the 
 resulting equations. 
 
 Add the resulting equations, or subtract one from the other, 
 according as these equal coefficients have unlike or like signs. 
 
 Note. It is generally best to select the letter to be eliminated 
 that requires the smallest multipliers to make its coefficients equal ; 
 and the smallest multiplier for each equation is found by dividing 
 the L. C. M. of the coefficients of this letter by the given coefficient in 
 that equation. Thus, in example 2, the L. C. M. of 6 and 8 (the co- 
 efficients of x) is 24, and hence the smallest multipliers of the two 
 equations are 4 and 3, respectively. 
 
 Sometimes the solution is simplified "by first adding the 
 given equations or by subtracting one from the other. 
 
 Solve: ' x + 49 y = 51) (1) 
 
 49a; + y = 99 ) (2) 
 
 Add (1) and (2), 50 X + 50 y = 150. (3) 
 
 Divide (3) by 50, X + y = 3. (4) 
 
 Subtract (4) from (1), 48 y = 48. 
 
 .-. V = 1. 
 Subtract (4) from (2), 48 x = 96. 
 
 .-. x = 2. 
 
 Exercise 70. 
 Solve by addition or subtraction: 
 
 1. 5x + 2y = 39\ 4. 4a-5y = 26) 
 2 X - y= 3> Sx-6y = 15i 
 
 2. x + 3y = 22\ 5. x + 2y = 35\ 
 2x-4y= 4) 3x-2y = 17$ 
 
 3. 7x-2y = ll\ 6. x + 4y = 35\ 
 
 x + 5y = 28 ) 2x - 3y = 26 ) 
 
 
 
SIMULTANEOUS SIMPLE EQUATIONS. 177 
 
 7. 
 
 3x + 5y = 50\ 
 
 11. 
 
 « + 2y = 
 
 9 1 
 
 90) 
 
 
 
 
 3x- 3y = 
 
 
 8. 
 
 5x + 2y = 36\ 
 2x + 3y = 43) 
 
 12. 
 
 4cc — 3y = 
 
 39 1 
 17) 
 
 
 
 
 3x — 4y = 
 
 
 9. 
 
 3x + 7y = 50\ 
 5x-2y = 15)> 
 
 13. 
 
 7x- 2y = 
 
 69 "> 
 
 39) 
 
 
 
 
 x — 10 y = 
 
 
 10. 
 
 2x+ y = 3\ 
 lx + 5y = 2l) 
 
 14. 
 
 3a+ 7y = 
 
 16 \ 
 13) 
 
 
 
 
 2z + 5y = 
 
 
 190. 
 
 Elimination by Substitution. 
 
 
 
 
 Solv 
 
 e: 5sc + 4 
 
 y = 32 
 
 } 
 
 
 (i) 
 
 
 4z + 3 
 
 y = 25 
 
 
 (2) 
 
 
 Transpose 4 y in (1), 
 
 6x- 
 
 = 32-4y. 
 
 
 (3) 
 
 
 Divide by coefficient of 
 
 X, X : 
 
 _32-4y_ 
 5 
 
 
 <*) 
 
 
 Substitute the value of : 
 
 c in (2), 
 
 
 - 
 
 
 
 <(^) + *- 
 
 = 25, 
 
 
 
 
 128 - 16 
 6 
 
 V + 3y-- 
 
 = 25, 
 
 
 
 
 128-165 
 
 — y- 
 
 = 125, 
 = -3. 
 = 3. 
 
 
 
 
 Substitute the value of ? 
 
 /in (2), 
 
 4» + 9: 
 
 = 25. 
 = 4. 
 
 
 
 To Eliminate by Substitution, therefore, 
 
 From one of the equations obtain the value of one of the 
 cnknown numbers in terms of the other. 
 
 Substitute for this unknown number its value in the othei 
 equation, and reduce the resulting equation. 
 
178 SIMULTANEOUS SIMPLE EQUATIONS. 
 
 Exercise 71. 
 
 Solve by substitution : 
 
 1. 2x- ly= 0} 8. 3x- 2y = 2S\ 
 
 3x- 5y = ll) 2x + 5y = 63) 
 
 
 Ax- 5y = 4| 9. 2x- 3y = 23~> 
 
 Sx- 2y = 10) 5x+ 2y = 2$i 
 
 2x- 3y = 1} 10. 6x- 7y = ll\ 
 
 3a>-.2y = 29) 5x- 6y = 8) 
 
 4. x+ y = 19\ 11. 7x+ 6y = 20\ 
 
 2x+ 7y = 88) 2x+ 5y = 32) 
 
 5. 2x- y= 5) 12. «+ 5y = 37) 
 
 x+ 2y = 25) 3x + 2y = 46) 
 
 6. 19a;-15y = 23"> 13. 3cc- 7?/= 40) 
 = 21) 4»- 3y= 9) 
 
 13a;- 5^ = 21) 4a;- 3y 
 
 jc + 10y = 73) 14. 5a> + 9y 
 
 7x- 2y= 7 S 3a;-f-lly 
 
 191. Elimination by Comparison. 
 
 Solve: 2x-5y = 66\ (1) 
 
 3x + 2y = 23) (2) 
 
 Transpose 6 y in (1) and 2 y in (2), 
 
 2s = 66 + 5y, (8) 
 
 8s = 23-2y. (4) 
 
 Divide (3) by 2, x = 2S±*E (5 ) 
 
 Divide (4) by 8, x = 23 ~ 2 ? . (6) 
 
 t * 66 + 5y 23 — 2y _ 
 
 Equate the values of x, = - — » (7) 
 
SIMULTANEOUS SIMPLE EQUATIONS. 179 
 
 Keduce (7), 198 + 15 y = 46 - 4 y, 
 
 19 y = - 152. 
 
 .-.y = - 8. 
 Substitute the value of y in (1), 
 
 2 x + 40 = 06. 
 
 .-. x = 13. 
 
 192. To Eliminate by Comparison, therefore, 
 
 From each equation obtain the value of one of the unknown 
 numbers in terms of the other. 
 
 Form an equation from these equal values and reduce the 
 equation. 
 
 Exercise 72. 
 Solve by comparison : 
 1. 
 
 2. 
 
 4. 
 
 6. 
 
 7. 11 
 
 X + y = 30 \ 
 
 Sx- 2y = 25$ 
 
 9. 
 
 2a;- 3y= lj 
 5x+ 2y = 126i> 
 
 ? x + Sy = 70} 
 5x — 4=y = 7 ) 
 
 10. 
 
 50a;- 9y = 1} 
 7x- y= 3) 
 
 9x + 4?/ = 54| 
 4a; + 9y = S9i 
 
 11. 
 
 x + 21y= 2 } 
 27 ?/+ 2 a; = 19) 
 
 7x + 2y = 6S} 
 Sx- y= 3) 
 
 12. 
 
 10a; + 3?/ = 174| 
 3a; + 10?/ = 125) 
 
 2 x -33 ?/ = 29 > 
 
 3 z _ 47 y = 46 ) 
 
 13. 
 
 6a;-13y = 2) 
 5x-12y = 4) 
 
 2x- y= 9} 
 5 X - 3yiUy 
 
 14. 
 
 2a; + 2/ = 108 ]. 
 10a; + 2y= 60) 
 
 la;— 7y= 6") 
 9a;- 5y = 10) 
 
 15. 
 
 3a?- 52/= 5\ 
 7x+ y = 2Q>5) 
 
 8. 5a; + 92/ = 188) 16. 12a; + 7y=176| 
 
 13a;-22/= 57) 3y-19a;= 3) 
 
180 
 
 SIMULTANEOUS SIMPLE EQUATIONS. 
 
 193. Each equation must be simplified, if necessary, 
 before the elimination. 
 
 Solve: ia>-i(y + l) = l> 
 i(s + l) + *(y-l) = 9j 
 
 (1) 
 
 (2) 
 
 Multiply (1) by 4, and (2) by 12, 
 
 
 Sx — 2y — 2 = 4, 
 
 (3) 
 
 4x + 4 + 9y — 9 = 108. 
 
 (4) 
 
 From (3), 3x — 2y = 6. 
 
 (5) 
 
 From (4), 4x + 9y = 113. 
 
 (6) 
 
 Multiply (5) by 4, and (6) by 3, 
 
 
 12x- 8y= 24 
 
 (7) 
 
 12x + 27y = 339 
 
 (8) 
 
 Subtract (7) from (8), 35 y = 315 
 
 
 .-. y = 9. 
 
 
 Substitute value of y in (1), x = 8. 
 
 
 Exercise 73. 
 
 Solve : 
 
 3^2 3 
 
 - + £ = ! 
 
 2~ r 3 6 
 
 *_1 £-2 
 
 8^5 
 
 2a; -f 
 
 2jr-5_ 
 
 = 21 
 
 2 - 3 + 2 
 
 2^ 9 
 
 4a; + 5y 
 
 =^+2„=i 
 
 3z-5y 2s + y 
 6 - 2 Xo ~ 5 
 
 a:-2y a; , y 
 J 8 4 ~2 + 3 J 
 
 6. *£$**.*] 
 x — y 
 
 7* -13 
 
 3^-5 " J 
 
 - 
 
7. 
 
 SIMULTANEOUS SIMPLE EQUATIONS. 
 + 1 y + 2 , 2(a?-y) 
 
 181 
 
 3 4 5 
 
 4 3 2 
 
 g + 2y + l 
 °* 2x-y + l 
 
 x-y + 3 
 
 x-2 10 - x y - 10 
 5 3 ~ 4 
 
 2y + 4 4a + y + 26 
 3 8 
 
 10. 
 
 2a-y + 3 _ a;-2y + 19 
 
 3 " 4 
 3a:-4y + 3 _ 2y-4a; + 21 
 
 4 3 
 
 11. 
 
 7 + sc 
 
 ^ = 3y-5 
 
 12. 
 
 5y — 7 , 4x — 3 . Q 1 
 2" + _ 6- 
 
 26 + 5a:-6y 
 
 13 
 
 4y — 3sc 
 
 5a-6y 3(* + 2y) 
 1J+ 6 ~ 4 
 
 t o * + 8 3y-a . 
 
 13 - ^r - 2 e 
 
 2a; + y 9a? — 7 7y-4a; + 36 
 2 8 16 
 
 14. 
 
 a; + 2y + 3 = 5y-4a; — 6 
 13 " 3 
 
 6a?-5y + 4 3a? + 2y-H 
 3 19 
 
182 
 
 SIMULTANEOUS SIMPLE EQUATIONS. 
 
 15. 
 
 x -f y 5 
 y -x~ 3 
 
 16. 
 
 y — 4 _ # 4- 1 
 
 „. »£^l2 + 2«.3(j,-l) 
 
 5a; + 6y 4a; — 3y _ 
 
 10 3 y 
 
 18. 
 
 4a;-3y-7 9a;-4y-25 
 
 5 
 
 30 
 
 y-1 10a;-3y--20 ^ 3a;-f 2y + 3 
 3 + 20 ~ 30 
 
 Note. In solving the following problems proceed as in § 175. 
 
 6y + 5 4a;-5y + 3 ^ 9y-4 
 8 4a; -2y . 12 
 
 8a; + 3 a;-3y 6a;-l 
 4 + 7-x ~ 3 
 
 «** 2y — x 
 
 20 -*-23^ = 
 -3 
 
 y + 
 
 a -18 
 
 a;-9£ 
 
 4 a; -|- 7 5a;-4y = 17 + 8s 
 
 3 + 2a; + 8 ~~ 6 
 g a; - 12 4 a; - 6 y - 13 _ 10 x - 53 
 
 4 2a;-3y ~ 8 
 
 7 + 8a; 3(a;-2y) ^ ll4-4a; 
 
 10 2 (a; -4) 5 
 
 8(2y + 3) ^ 6y + 21 3y-h5a; 
 4 4 2(2y-3)J 
 
SIMULTANEOUS SIMPLE EQUATIONS. 183 
 
 194. Literal Simultaneous Equations. 
 
 Solve : ax + by = c > 
 
 a'a; + 6'?/ == c' ) 
 
 Note. The letters a', b' are read a prime, b prime. In like man- 
 ner, a", a"' are read a second, a third, and oi, a 2 , a 3 are read a sub 
 one, a sub two, a sub three. It is sometimes convenient to represent 
 different numbers that have a common property by the same letter 
 marked by accents or suffixes. Here a and a' have a common prop- 
 erty as coefficients of x. 
 
 ax +by = c. (1) 
 
 a'x + b'y = c'. (2) 
 
 To find the value ofy, multiply (1) by a' and (2) by a. 
 aa f x + a'by — a'c 
 aa'x + ab'y = ac' 
 
 Subtract, 
 
 Divide by a'b — 
 
 a'by — ab'y 
 ab', y 
 
 — a'c - 
 a'c - 
 a'b - 
 
 - ac' 
 
 - ac' 
 -ab' 
 
 
 To find the value 
 
 of x, multiply (1) by 
 ab'x + bb'y — b'c 
 a'bx -f bb'y = be' 
 
 b', and 
 
 (2) by b. 
 
 Subtract, 
 
 ab'x — a'bx 
 
 = b'c- 
 
 -be' 
 
 
 Divide by ab' — a'b, 
 
 ab' — a'b 
 Exercise 74. 
 
 Solve 
 
 1. x 4- y = s > 5. bx -b ay = abc ") 
 
 x — y = d ) x = dy ) 
 
 2. race + ny = r > 6. foe -f- a?/ = 1 > 
 ra'cc -f- w f ?/ = r' ) b'x — a'y = 1 ) 
 
 3. ax — by = c ") 7. Sbx -\- 2 ay = 3ab) 
 a'x + #V = £ J 4:bx — 3ay = %ab ) 
 
 4. a> — y = raw ") 8. 2 cc — 3y = a — b\ 
 % cx 4- aoy = ras) 3 cc — 2y = a-r-b ) 
 
184 
 
 SIMULTANEOUS SIMPLE EQUATIONS, 
 bx cy 1 
 
 9. 
 
 a?-b 2 
 
 + 
 
 a + b 
 
 10. 
 
 bx + cy = a + b 
 
 a 14. ax + by = c 
 
 y — n b 
 bx + ay = c 
 
 a o 
 a b~ 3 
 
 12. 
 
 + 
 
 V 
 
 a + b ' a — b 
 x -\- y = 2a 
 
 = 2 
 
 x — y = a — b 
 
 bx + ay 
 
 :: 
 
 15. 3a 2 + ax = b 2 + by 
 ax + 2by = d 
 
 16. 
 
 // 
 
 17. 
 
 a -\-b a — b a + b \ 
 
 a-6j 
 2/ 
 
 a + b a — b 
 
 + — 
 
 m — a m 
 
 + 
 
 = 1 
 
 1 
 
 n — a n 
 
 195. Fractional simultaneous equations, of which the 
 denominators are simple expressions and contain the un- 
 known numbers, may be solved as follows: 
 
 >olve : - + - = m 
 x y 
 
 
 0) 
 
 c , d 
 
 - + - = n 
 
 x y 
 
 
 (2) 
 
 To find the value of y. 
 
 
 Multiply (1) by c, 1 = cm. 
 
 x y 
 
 (3) 
 
 Multiply (2) by a, 1 — aw. 
 
 x y 
 
 (*) 
 
 Subtract (4) from (3), bc ~ °^ 
 
 = cm — an. 
 
 • 
 
SIMULTANEOUS SIMPLE EQUATIONS. 
 
 185 
 
 Multiply both sides by y, and we have 
 
 be — ad = (cm — an) y. 
 
 ■~. . , ■. be — ad 
 
 Divide by cm — em, 
 
 To find the value ofx 
 Multiply (1) by a\ 
 
 Multiply (2) by 6, 
 
 Subtract (6) from (5), 
 
 Multiply both sides by x, and we have 
 
 ad — be = (dm — bn) x. 
 ad — be 
 
 y _ - ■ 
 
 cm — an 
 
 
 ad . bd , 
 
 1 = dm. 
 
 x y 
 
 (5) 
 
 be , bd 
 
 — 1 = bn. 
 
 x y 
 
 (6) 
 
 ad — be . , 
 = dm — bn. 
 
 
 Divide by dm — bn, 
 2. Solve: 
 
 dm — bn 
 
 Sx hy 
 
 "7 1 
 
 = 3 
 
 (1) 
 
 (2) 
 
 6x 10 y 
 
 Multiply (1) by 15, the L. C. M. of 3 and 6 ; and (2) by 30, the 
 C. M. of 6 and 10, 
 
 ^ + « = 105. 
 
 x y 
 
 ^-§ = 90. 
 
 x y 
 
 (3) 
 (4) 
 
 Multiply (4) by 2, and add the result to (3), 
 
 55 = 285. 
 
 X 
 
 Substitute the value of x in (1), and we obtain 
 
 1 
 
186 
 
 SIMULTANEOUS SIMPLE EQUATIONS. 
 
 Solve 
 
 Exercise 75. 
 
 1. 
 
 M=2 
 
 x y 
 
 
 x y ¥ 
 
 2. 
 
 « + ^ = 49 
 
 x y 
 
 
 1 + ^ = 23 
 x y 
 
 3. 
 
 2 + « = 3 1 
 
 15_4 = 
 
 1 
 
 a; y 
 
 *-2 = 7 
 
 a y j 
 
 - 
 
 5. 
 
 A+ifh 5 " 
 
 
 _5_ _4_ _. 11 
 
 
 4x 5y 20 
 
 6. 
 
 2^ + 3i; =3 
 
 
 4# oy 
 
 3.9 
 
 7. — I — = m 
 x y 
 
 8. ^ + ± = b 
 x y 
 
 n 1 
 -4- - = c 
 
 
 m 
 
 ■ w 
 
 
 9. 
 
 
 + - = 
 
 = a 
 
 
 x 
 
 y 
 
 
 
 r 
 
 . 5 
 
 
 
 
 -h- = 
 
 b 
 
 
 X 
 
 2/ 
 
 
 
 a 
 
 6 
 
 ac 
 
 0. 
 
 — 
 
 — — — 
 
 : 
 
 
 X 
 
 y 
 
 b 
 
 
 b 
 
 a 
 
 be 
 
 
 X 
 
 y~ 
 
 a 
 
 11. 
 
 ax 
 
 2^ 
 
 by 
 
 ax by 
 
 12. ± + JL = a + h 
 
 bx ay 
 
 b.a 2 
 - + - = a 2 
 
 * y 
 
SIMULTANEOUS SIMPLE EQUATIONS. 
 
 187 
 
 196. If three simultaneous equations are given, involving 
 three unknown numbers, one of the unknowns must be 
 eliminated between two pairs of the equations; then a 
 second unknown between the two resulting equations. 
 
 Likewise, if four or more equations are given, involving 
 four or more unknown numbers, one of the unknowns must 
 be eliminated between three or more pairs of the equations ; 
 then a second between the pairs that can be formed of the 
 resulting equations ; and so on. 
 
 Note. The pairs chosen to eliminate from must be independent 
 pairs, so that each of the given equations shall be used in the process 
 of the eliminations. 
 
 Solve: 2x -3y + 4«= 4" 
 
 5x — y — &z = 5. 
 
 Eliminate z between the equations (1) and (3). 
 
 Multiply (1) by 2, 4x - Qy + 82 = 8 
 
 (3) is 5x — y — %z= 5 
 
 Add, 9x-7y =13 
 
 Eliminate z between the equations (1) and (2). 
 
 Multiply (1) by 7, 14 x - 21 y + 28 z = 28 
 
 Multiply (2) by 4, 
 Add, 
 
 12s + 20y-28z = 48 
 
 26 x 
 
 y 
 
 = 76 
 
 (1) 
 (2) 
 (3) 
 
 (4) 
 (5) 
 
 (6) 
 
 We now have two equations (5) and (6) involving two un- 
 knowns, x and y. 
 
 Multiply (6) by 7, 182 x - 7 y = 532 (7) 
 
 (5) is 9x-7y =, 13 
 
 Subtract (5) from (7), 173 x 
 
 Substitute the value of x in (6), 
 
 = 519 
 .-. x = 3. 
 78 - y = 76. 
 
 .-. y = 2. 
 
 Substitute the values of x and y in (1), 
 
 6-6 + 42 = 4. 
 .% * m % 
 
188 SIMULTANEOUS SIMPLE EQUATIONS. 
 
 Solve: 
 
 Exercise 76. 
 
 1. x + y- 8 = 0^1 
 
 10. 5x + 2y-20z = 20 
 
 y + * - 28 = 
 
 3x-6y + 7z = 51 
 
 <c-fs-14 = oJ 
 
 4x + 8y- 9* = 53 
 
 6. 
 
 
 2. 4a + 32/ + 2« = 25^j 11. x + 2y + 10z = 44"i 
 3a-2y + 5s = 20 J> 3a-f-3y + 7z = 384f- 
 
 10x-5y + 3* = 17J 2a + y + s = 256J 
 
 3. 5o;-2y- 2^ = 12^1 12. 10a; = y + ±z + 56^ 
 
 # + y + * = 8 f 3y = 2z + 3s-98 J> 
 
 7x + 3y + 4z = 42J 2* = ic-3y-18J 
 
 4. x- y+ «=Hi 13. 3«-5y-2« = 14^ 
 3z + 32/-2z = 60 f- 5^-8?/- z = 12 I 
 
 10x-5y-3«= OJ x-3y-3z = lJ 
 
 5. 10 a- 2/ + 3« = 42^ 14. 2x + 3y+ z = Sl^i 
 
 7^ + 2*/ + s = 51f a;- y + 3* = 13l 
 
 3a + 3y- z = 24 J lOy + 5cc - 2* = 48J 
 
 5x + 2y- 3z = 160^ 15. 2cc + 3y- 4z = l 
 3a + 9# + 8s = 115 >- 10x- 6y + 12z = 6 
 
 2x-3y-5z= 45J x + 12y+ 2z = 5. 
 
 7. 6x-2y + 5z = 53^ 16. 3a + 6y + 2z = 3" 
 5x + 3y + 7z = 33> 12y+ 4s- 6a = 2 
 
 cc 4" 2/ + 2 = 5J 9#-|-18y — 4s = 4. 
 
 8. 3x- 3y'+ 42 = 20^ 17 * 2x + V + 2z==S 
 6x + 2y-7z = 5> 5y-4:X-4:Z = l 
 2x- y + $z = 4:5J 3x + 9y+ z = 9 
 
 9. 2x + 7y + 10z = 25^ 18. 3x + 2y+ z = 20^ 
 
 a+ y _ «= 9^ 2x- y+ 3* = 26£ V 
 
 7s-7y-ll* = 73J a+ y + 10* = 55 J 
 
SIMULTANEOUS SIMPLE EQUATIONS. 
 
 189 
 
 19. ---+ 4 = 
 x y 
 
 1-1 + 1 = 
 
 y z 
 
 Z X 
 
 24. 1 + 1-1-0 
 
 x y a 
 
 i + l-i=o 
 
 X Z 
 
 1+1-1=0 
 
 . 1 + i + i =36 
 x y z 
 
 M-i =28 
 
 x y z 
 
 l + f + f = 20 
 
 x 3y 2z 
 
 2 , 1 + | + |=62 
 
 .!+?-«= i 
 
 x y z 
 
 5 + 1 + ^ = 24 
 
 x y z 
 
 26. 5 + 1-2 = 
 
 a; y * 
 
 2-5-2=0 
 
 * 2/ 
 
 1+1-^=0 
 as z 3 
 
 «. 1+1-1=. 
 
 a y z 
 
 x y z 
 
 1 ^ 1 1 
 -H = c 
 
 y z x 
 
 „. «_4 + 5 = 38 
 
 x y z 
 
 5 + 2 + 15= 61 
 
 x y z 
 
 *-« + *2 = 161 
 
 a; y z 
 
 53. 
 
 4 3 
 
 1 
 
 
 
 x y 
 
 20 
 
 2 3 
 
 1 
 
 — — — ^^ 
 
 — 
 
 £ £C 
 
 15 
 
 4 5_ 
 
 1 
 
 « 2/ 
 
 12 
 
 j 
 
 28 . 5_§ + 4 = 29 
 
 a: y * 
 
 5_6_r = _ 104 
 
 a; y * 
 
 9 + 10_8 = U9 
 
 y z x 
 
CHAPTER XH. 
 
 PROBLEMS INVOLVING TWO OR MORE UNBLOWN 
 NUMBERS. 
 
 197. It is often necessary in the solution of problems to 
 employ two or more letters to represent the numbers to be 
 found. In all cases the conditions must be sufficient to 
 give just as many equations as there are unknown numbers 
 employed. If there are more equations than unknown 
 numbers, some of them are superfluous or inconsistent ; if 
 there are fewer equations than unknown numbers, the 
 problem is indeterminate. 
 
 Exercise 77. 
 
 1. If A gave B $10, B would have three times as much 
 money as A. If B gave A $10, A would have twice as much 
 money as B. How much has each ? 
 
 Let x = the number of dollars A has, 
 
 and y = the number of dollars B has. 
 
 Then, if A gave B $10, 
 
 x — 10 = the number of dollars A would have, 
 y + 10 = the number of dollars B would have. 
 .-. y + 10 = 3(x-10). (1) 
 
 If B gave A $10, 
 
 x + 10 = the number of dollars A would have, 
 y — 10 = the number of dollars B would have. 
 .-.x + 10 = 2(y-10). (2) 
 
 From the solution of equations (1) and (2), * = 22, and y = 26. 
 
 Therefore, A has $22, and B has $26. 
 

 PROBLEMS. 191 
 
 2. If the smaller of two numbers is divided by the 
 greater, the quotient is 0.21, and the remainder 0.0057; 
 but if the greater is divided by the smaller, the quotient is 
 4 and the remainder 0.742. Find the numbers. 
 
 Let x = the greater number, 
 
 and y = the smaller number. 
 
 Then , Inmost =021) (1) 
 
 X 
 
 x — 0.742 
 and = 4. (2) 
 
 .-. y - 0.21 x = 0.0067, (3) 
 
 x-4y = 0.742. (4) 
 
 Multiply (3) by 4, 4 y — 0.84 x = 0.0228 (6) 
 
 (4) is — 4y+ x = 0.742 
 
 Add, 0.16 x = 0.7648 
 
 .-. x = 4.78. 
 Put the value of x in (2), 4 y = 4.038, 
 
 .-. y = 1.0095. 
 
 Therefore, the numbers are 4.78 and 1.0095. 
 
 3. If A gave B $ 100, A would then have half as much 
 money as B ; but if B gave A $100, B would have one third 
 as much as A. How much has each ? 
 
 4. If the greater of two numbers is divided by the 
 smaller, the quotient is 7 and the remainder 4; but if three 
 times the greater number is divided by twice the smaller, 
 the quotient is 11 and the remainder 4. Find the numbers. 
 
 5. If the greater of two numbers is divided by the 
 smaller, the quotient is 4 and the remainder 0.37; but if 
 the smaller is divided by the greater, the quotient is 0.23 
 and the remainder 0.0149. Find the numbers. 
 
 6. If A gave B $5, he would have $ 6 less than B ; but if 
 he received $5 from B, three times his money would be $20 
 more than four times B's. How much has each ? 
 
192 PROBLEMS. 
 
 7. If the numerator of a fraction is doubled and its 
 denominator diminished by 1, its value will be \. If its 
 denominator is doubled and its numerator increased by 1, 
 its value will be \. Find the fraction. 
 
 Let x = the numerator, 
 
 and y = the denominator. 
 
 Then, j-^L = ,, (1) 
 
 and 2_L1 = |. (2) 
 
 The solution of equations (1) and (2) gives 5 for x and 21 for y. 
 Therefore, the required fraction is / T . 
 
 8. A certain fraction becomes equal to \ if 3 is added 
 to its numerator and 1 to its denominator, and equal to \ 
 if 3 is subtracted from its numerator and from its denomi- 
 nator. Find the fraction. 
 
 9. A certain fraction becomes equal to T 9 T if 1 is added 
 to double its numerator, and equal to £ if 3 is subtracted 
 from its numerator and from its denominator. Find the 
 fraction. 
 
 10. There are two fractions with numerators 11 and 5, 
 respectively, whose sum is If ; but if their denominators 
 are interchanged their sum is 2J. Find the fractions. 
 
 11. A certain fraction becomes equal to \ when 7 is added 
 to its denominator, and equal to 2 when 13 is added to its 
 numerator. Find the fraction. 
 
 12. A certain fraction becomes equal to £ when the 
 denominator is increased by 4, and equal to f £ when the 
 numerator is diminished by 15. Find the fraction. 
 
 13. A certain fraction becomes equal to § if 7 is added 
 to the numerator, and equal to § if 7 is subtracted from 
 the denominator. Find the fraction. 
 
PROBLEMS. 193 
 
 14. A certain number is expressed by three digits. The 
 sum of the digits is 21. The sum of the first and last digits 
 is twice the middle digit. If the hundreds' and tens' digits 
 are interchanged, the number is diminished by 90. Find 
 the number. 
 
 Let x = the hundreds' digit, 
 
 y = the tens' digit, 
 z = the units' digit. 
 Then, 100 x + 10 y + z = the number. 
 
 By the conditions, x + y + z = 21, (1) 
 
 x + z = 2y, (2) 
 
 and 1002/ + 10z + z = 100x+ 10y + z-90. (3) 
 
 Solving these equations, x = 8, y = 7, z = 6. 
 
 Therefore, the number is 876. 
 
 15. The sum of the two digits of a number is 9, and if 
 27 is subtracted from the number, the digits will be reversed. 
 Find the number. 
 
 16. The sum of the two digits of a number is 9, and if 
 the number is divided by the sum of the digits, the quotient 
 is 5. Find the number. 
 
 17. A certain number is expressed by two digits. The 
 sum of the digits is 11. If the digits are reversed, the new 
 number exceeds the given number by 27. Find the number. 
 
 18. A certain number is expressed by three digits, the 
 units' digit being zero. If the hundreds' and tens' digits 
 are interchanged, the number is diminished by 180. If the 
 hundreds' digit is halved, and the tens' and units' digits are 
 interchanged, the number is diminished by 336. Find the 
 number. 
 
 19. A number is expressed by three digits. If the digits 
 are reversed, the new number exceeds the given number by 
 99. If the number is divided by nine times the sum of its 
 digits, the quotient is 3. The sum of the hundreds' and 
 units' digits exceeds the tens' digit by 1. Find the number 
 
194 PROBLEMS. 
 
 20. A boatman rows 20 miles down a river and back in 
 8 hours. He finds that he can row 5 miles down the river 
 in the same time that he rows 3 miles up the river. Find 
 the time he was rowing down and up respectively. 
 
 Let x = the boatman's rate per hour in still water, 
 
 and y = the rate per hour of the current. 
 
 20 
 Then, ■ = the number of hours he was rowing down, 
 
 20 
 and _ = the number of hours he was rowing up. 
 
 Therefore, -~- + -^- = 8, (1) 
 
 x + y x—y v ' 
 
 5 3 
 
 and — I — = (2) 
 
 x + y x — y v ' 
 
 Solving these equations, x = 5i, y = 1£. 
 
 20 20 
 
 Therefore, — ; — = 3, = 5. 
 
 x+y x—y 
 
 It takes him 3 hours to row down and 5 hours to row 
 back. 
 
 21. A boat's crew which can pull down a river at the 
 rate of 10 miles an hour finds that it takes twice as long to 
 row a mile up the river as to row a mile down. Find the 
 rate of their rowing in still water and the rate of the 
 stream. 
 
 22. A boatman rows down a stream, which runs at the 
 rate of 2\ miles an hour, for a certain distance in 1 hour 
 30 minutes; it takes him 4 hours 30 minutes to return. 
 Find the distance he pulled down the stream and his rate 
 of rowing in still water. 
 
 23. A person rows down a stream a distance of 20 miles 
 and back again in 10 hours. He finds he can row 2 miles 
 against the stream in the same time he can row 3 miles with 
 it. Find the time of his rowing down and of his rowing 
 up the stream ; and also the rate of the stream. 
 
 
PROBLEMS. 195 
 
 24. A and B can do a piece of work together in 3 days, 
 A and C in 4 days, B and C in 4£ days. How long will it 
 take each alone to do the work ? 
 
 Let x, y, z = the number of days in which A, B, C can do the 
 work, respectively. 
 
 Then, -*-»- = the parts A, B, C can do in 1 day, respectively. 
 
 And - + - = the part A and B together can do in one day. 
 x y 
 
 But £ = the part A and B together can do in 1 day. 
 
 Therefore, 1 + 1 = 1, (1) 
 
 x y 3 v ' 
 
 Likewise, - + - = -, (2) 
 
 x z 4 
 
 and 1 + 1=4 (li = l) < 3 > 
 
 y z 9 \4£ 9/ * ' 
 
 Add, and divide by 2, - + - + - = f£ (4) 
 
 x y z 7 2 
 
 Subtract (1), (2), and (3), separately from (4), and we have 
 
 !=-JL i_ii !-.!§. 
 
 z 72' y 72' x 72* 
 Therefore, z = 14f , y - 6 T 6 T , a; = 5 T \. 
 
 Therefore, A can do the work in 5 fa days, B in 6/j 
 days, and C in 14f days. 
 
 25. A cistern has three pipes, A, B, and C. A and B 
 will fill the cistern in 1 hour 10 minntes, A and C in 1 
 hour 24 minutes, B and C in 2 hours 20 minutes. How long 
 will it take each pipe alone to fill it ? 
 
 26. A and B can do a piece of work in 2\ days, A and 
 C in 3^ days, B and C in 4 days. How long will it take 
 each alone to do the work ? 
 
 27. A and B can do a piece of work in a days, A and C 
 in b days, B and C in c days. How long will it take each 
 alone to do the work ? 
 
196 PROBLEMS. 
 
 28. A sum of money, at simple interest, amounted to 
 $2480 in 4 years, and to $2600 in 5 years. Find the sum 
 and the rate of interest. 
 
 Let x = the number of dollars in the principal, 
 
 and y = the rate of interest. 
 
 The interest for one year is r^r of the principal, = -r^r of x ; for 4 
 100 100 
 
 years, = — of x ; and for 5 years, = — ^ of x. 
 The amount is principal + interest. 
 Therefore, x + |g| = 2480. 
 
 * + |H = 2 aoo. 
 
 Hence, 100* + 4«y = 248,000. (1) 
 
 100 x + 5 xy = 260,000. (2) 
 
 Multiply (1) by 5 and (2) by 4, and we have 
 
 500x + 20xy = 1,240,000 
 
 400x + 20xy = 1,040,000 
 Subtract, 100 x = 200,000. 
 
 Therefore, x = 2000. 
 
 Substitute the value of x in (1), y = 6. 
 Therefore, the sum is $2000, and the rate 6%. 
 
 29. A sum of money, at simple interest, amounted in 4 
 years to $29,000, and in 5 years to $30,000. Find the sum 
 and the rate of interest. 
 
 30. A sum of money, at simple interest, amounted in 10 
 months to $2100, and in 18 months to $2180. Find the 
 sum and the rate of interest. 
 
 31. A man has $10,000 invested. For a part of this 
 sum he receives 5 per cent interest, and for the rest 6 per 
 cent; the income from his 5 per cent investment is $60 
 more than from his 6 per cent. How much has he in each 
 investment ? 
 
 
PROBLEMS. 197 
 
 32. In a mile race A gives B a start of 20 yards and 
 beats him by 30 seconds. At the second trial A gives B a 
 start of 32 seconds and beats him by 9^ T yards. Find the 
 number of yards each runs a second. 
 
 Let x = the number of yards A runs a second, 
 
 and y = the number of yards B runs a second. 
 
 Since there are 1760 yards in a mile, 
 
 = the number of seconds it takes A to run a mile. 
 
 x 
 
 Since B has a start of 20 yards, he runs 1740 yards the first trial ; 
 
 and as he was 30 seconds longer than A, 
 
 h 30 = the number of seconds B was running. 
 
 x 
 
 But = the number of seconds B was running. 
 
 ,!™ = «29 + 80 . (1) 
 
 y x v ' 
 
 In the second trial B runs (1760 — 9 T 5 T ) yards = 1750 T 6 T yards. 
 
 ,.l™&=l™ + 32. (2) 
 
 y x w 
 
 From the solution of equations (1) and (2), x = 5}f , and y = 5 T 3 T . 
 
 Therefore, A runs 5|f yards a second, and B runs 5^ 
 r ards a second. 
 
 33. Two men, A and B, run a mile, and A wins by 2 
 seconds. In the second trial B has a start of 18J yards, 
 and wins by 1 second. Find the number of yards each 
 runs a second, and the number of miles each would run in 
 an hour. 
 
 34. In a mile race A gives B a start of 3 seconds, and is 
 beaten by 12^ yards. In the second trial A gives B a start 
 of 10 yards, and the race is a tie. Find the number of 
 yards each runs a second. At this rate, how many miles 
 could each run in an hour ? 
 
198 PROBLEMS. 
 
 35. A train, after traveling an hour from A towards B, 
 meets with an accident which detains it half an hour ; after 
 which it proceeds at four fifths of its usual rate, and arrives 
 an hour and a quarter late. If the accident had happened 
 30 miles farther on, the train would have been only an hour 
 late. Find the usual rate of the train. 
 
 Let y = the number of miles from A to B, 
 
 and 5 x = the number of miles the train travels per hour. 
 
 Then, 4 x = the rate of the train after the accident. 
 
 Then, y — 5 x = the number of miles the train has to go after 
 the accident. 
 
 Hence, *—z — the number of hours required usually, 
 
 5 x 
 
 and *— = the number of hours actually required. 
 
 y — bx y — 5x ' ' . ., , 
 
 .*. — 2 — = the loss in hours of running time. 
 
 4a; bx 
 
 But since the train was detained £ an hour and arrived 1£ hours 
 late, the running time was £ of an hour more than usual. That is, 
 £ = loss in hours of running time. 
 
 y — bx y — bx _S m 
 
 *'• 4x bx 4' K} 
 
 If the accident had happened 30 miles farther on, the remainder of 
 the journey would have been y — (5 x + 30) miles, and the loss in 
 running time would have been f an hour. 
 
 y - (5s + 30) y - (5x + 30) _ 1 , 
 
 4x bx 2 K ' 
 
 From the solution of equations (1) and (2), x = 6, and bx = 30. 
 
 Therefore, the usual rate of the train is 30 miles an hour. 
 
 36. An express train, after traveling an hour from A 
 towards B, meets with an accident which delays it 15 min- 
 utes. It afterwards proceeds at two thirds its usual rate, 
 and arrives 24 minutes late. If the accident had happened 
 5 miles farther on, the train would have been only 21 
 minutes late. Find the usual rate of the train. 
 
PROBLEMS. 199 
 
 37. If 3 yards of velvet and 12 yards of silk cost $60, 
 and 4 yards of velvet and 5 yards of silk cost $58, what is 
 the price of a yard of velvet and of a yard of silk ? 
 
 &8. If 5 bushels of wheat, 4 of rye, and 3 of oats are sold 
 for $9 ; 3 bushels of wheat, 5 of rye, and 6 of oats for $8.75 ; 
 and 2 bushels of wheat, 3 of rye, and 9 of oats for $7.25 ; 
 what is the price per bushel of each kind of grain ? 
 
 39. A train proceeded a certain distance at a uniform 
 rate. If the speed had been 6 miles an hour more, the time 
 occupied would have been 5 hours less ; but if the speed had 
 been 6 miles an hour less, the time occupied would have 
 been 1\ hours more. Find the distance. 
 
 Hint. If x = the number of hours the train travels, and y the 
 number of miles per hour, then xy = the distance. 
 
 40. A certain number of persons paid a bill. If there had 
 been 10 persons more, each would have paid $2 less ; but if 
 there had been 5 persons less, each would have paid $2.50 
 more. Find the number of persons and the amount of the bill. 
 
 41. A man bought 10 cows and 50 sheep for $750. He 
 sold the cows at a profit of 10 per cent, and the sheep at a 
 profit of 30 per cent, and received in all $875. Find the 
 average cost of a cow and of a sheep. 
 
 42. It is 40 miles from Dover to Portland. A sets out 
 from Dover, and B from Portland, at 7 o'clock a.m., to meet 
 each other. A walks at the rate of 3^ miles an hour, but 
 stops 1 hour on the way ; B walks at the rate of 2£ miles 
 an hour. At what time of day and how far from Portland 
 will they meet ? 
 
 43. A number is expressed by three digits. The sum of 
 the digits is 21 ; the sum of the first and second exceeds the 
 third by 3 ; and if 198 is added to the number, the digits in 
 the units' and hundreds' places will be interchanged. 
 Find the number. 
 
200 PROBLEMS. 
 
 44. If the length of a rectangular field is increased by 5 
 yards and its breadth by 10 yards, its area is increased by 
 450 square yards ; but if its length is increased by 5 yards 
 ana its. breadth diminished by 10 yards, its area is dimin- 
 ished by 350 square yards. Find its dimensions. 
 
 45. If the floor of a certain hall had been 2 feet longer 
 and 4 feet wider, it would have contained 528 square feet 
 more ; but if the length and width were each 2 feet less, it 
 would contain 316 square feet less. Find its dimensions. 
 
 46. If the length of a rectangle was 4 feet less and the 
 width 3 feet more, the figure would be a square of the same 
 area as the given rectangle. Find the dimensions of the 
 rectangle. 
 
 47. If a certain number is divided by the sum of its two 
 digits diminished by 2, the quotient is 5 and the remainder 
 1 ; if the digits are interchanged, and the resulting number 
 is divided by the sum of the digits increased by 2, the 
 quotient is 5 and the remainder 8. Find the number. 
 
 48. A person has a certain capital invested at a certain 
 rate per cent. Another person has $2000 more capital 
 invested at one per cent better than the first, and receives 
 $150 more income. A third person has $3000 more capital 
 invested at two per cent better than the first, and receives 
 $280 more income. Find the capital of each, and the rate 
 at which it is invested. 
 
 49. A man makes an investment at 4 per cent, and a 
 second investment at 4^ per cent. His income from the 
 two investments is $715. If the first investment had been 
 made at 4£ per cent and the second at 4 per cent, his income 
 would have been $15 greater. Find the amount of each 
 investment 
 
PROBLEMS. 201 
 
 50. A number is expressed by two digits, the units' digit 
 being the larger. If the number is divided by the sum of 
 its digits, the quotient is 4. If the digits are reversed and 
 the resulting number is divided by 2 more than the differ- 
 ence of the digits, the quotient is 14. Find the number. 
 
 51. An income of $335 a year is obtained from two in- 
 vestments, one in 4^ per cent stock and the other in 5 per 
 cent stock. If the 4£ per cent stock should be sold at 110, 
 and the 5 per cent at 125, the sum realized from both stocks 
 together would be $8300. How much of each stock is there ? 
 
 52. A sum of money, at simple interest, amounted in m 
 years to c dollars, and in n years to d dollars. Find the 
 sum and the rate of interest. 
 
 53. A sum of money, at simple interest, amounted in m 
 months to a dollars, and in n months to b dollars. Find 
 the sum and the rate of interest. 
 
 54. A person has $18,375 to invest. He can buy 3 per 
 cent bonds at 75, and 5 per cent bonds at 120. How much 
 of his money must he invest in each kind of bonds in order 
 to have the same income from each investment ? 
 
 Hint. Notice that the 3 per cent bonds at 75 pay 4 per cent on the 
 loney invested, and 5 per cent bonds at 120 pay 4| per cent. 
 
 55. In a mile race A gives B a start of 44 yards, and is 
 beaten by 1 second. In a second trial A gives B a start of 
 6 seconds, and beats him by 9£ yards. Find the number 
 of yards each runs a second. 
 
 56. A train, after running 2 hours from A towards B, 
 meets with an accident which delays it 20 minutes. It 
 afterwards proceeds at four fifths its usual rate, and arrives 
 1 hour 40 minutes late. If the accident had happened 
 40 miles nearer A, the train would have been 2 hours late. 
 Find the usual rate of the train. 
 
202 PROBLEMS. 
 
 57. A boy bought some apples at 3 for 5 cents, and 
 some at 4 for 5 cents, paying $1 for the whole. He sold 
 them at 2 cents apiece, and cleared 40 cents. How many 
 of each kind did he buy ? 
 
 58. Find the area of a rectangular floor, such that if 3 
 feet were taken from the length and 3 feet added to the 
 breadth, its area would be increased by 6 square feet, but 
 if 5 feet were taken from the breadth and 3 feet added to 
 the length, its area would be diminished by 90 square feet. 
 
 59. A courier was sent from A to B, a distance of 147 
 miles. After 7 hours, a second courier was sent from A, 
 who overtook the first just as he was entering B. The time 
 required by the first to travel 17 miles added to the time 
 required by the second to travel 76 miles is 9 hours 
 40 minutes. How many miles did each travel per hour ? 
 
 60. A box contains a mixture of 6 quarts of oats and 9 
 of corn, and another box contains a mixture of 6 quarts of 
 oats and 2 of corn. How many quarts must be taken from 
 each box in order to have a mixture of 7 quarts, half oats 
 and half corn ? 
 
 61. A train traveling 30 miles an hour takes 21 minutes 
 longer to go from A to B than a train which travels 36 
 miles an hour. Find the distance from A to B. 
 
 62. A man buys 570 oranges, some at 16 for 25 cents, 
 and the rest at 18 for 25 cents. He sells them all at the 
 rate of 15 for 25 cents, and gains 75 cents. How many of 
 each kind does he buy ? 
 
 63. A and B run a mile race. In the first heat B 
 receives 12 seconds start, and is beaten by 44 yards. In 
 the second heat B receives 165 yards start, and arrives at 
 the winning post 10 seconds before A. Find the time in 
 which each can run a mile. 
 
PROBLEMS. 203 
 
 198. The discussion of a problem consists in making 
 various suppositions as to the relative values of the given 
 numbers, and explaining the results. We will illustrate by 
 the following example : 
 
 Two couriers are traveling along the same road, in the 
 same direction. A travels m miles an hour, and B travels^ 
 n miles an hour. At 12 o'clock B is d miles in advance of A. 
 When will the couriers be together ? 
 
 Suppose they will be together x hours after 12. Then A has trav- 
 eled mx miles, and B has traveled nx miles, and as A has traveled 
 d miles more than B 
 
 mx — nx = d. .-. x 
 
 Discussion. 1. If m is greater than n, the value of x is positive, 
 id A will overtake B after 12 o'clock. 
 
 2. If m is less than n, the value of x is negative. In this case B 
 travels faster than A, and as he is d miles ahead of A at 12 o'clock, 
 A cannot overtake B after 12 o'clock, but B passed A before 12 
 o'clock. The supposition, therefore, that the couriers are together 
 after 12 o'clock is incorrect, and the negative value of x points to an 
 error in the supposition. 
 
 3. If m equals n, then the value of x assumes the form - • Now, if 
 
 the couriers are d miles apart at 12 o'clock, and if they travel at the 
 same rates, it is obvious that they never will be together, so that the 
 
 symbol - may be regarded as the symbol of impossibility. 
 
 4. If m equals n and d is 0, then becomes - • Now, if the 
 
 * ' m — n 
 
 couriers are together* at 12 o'clock, and if they travel at the same 
 rates, it is obvious that they will be together all the time, so that x 
 
 may have an indefinite number of values. Hence, the symbol - may 
 
 be regarded as the symbol of indetermination. 
 
204 PROBLEMS. 
 
 Exercise 78. 
 
 1. A train traveling b miles per hour is m hours in 
 advance of a second train that travels a miles per hour. 
 In how many hours will the second train overtake the first ? 
 
 . bm 
 % Ans. r* 
 
 a — o 
 
 Discuss the problem (1) when a is greater than b ; (2) when a is equal 
 to b ; (3) when a is less than b. 
 
 2. A man setting out on a journey drove at the rate of 
 a miles an hour to the nearest railway station, distant b 
 miles from his house. On arriving at the station he found 
 that the train left c hours before. At what rate per hour 
 should he have driven in order to reach the station just 
 in time for the train ? 
 
 ab 
 
 Ans. 
 
 b 
 
 Discuss the problem (1) when c = ; (2) when c = - ; (3) when 
 
 c = In case (2), how many hours did the man have to drive 
 
 from his house to the station ? In case (3), what is the meaning of 
 the negative value of c ? 
 
 3. A wine merchant has two kinds of wine which he sells, 
 one at a dollars, and the other at b dollars per gallon. He 
 wishes to make a mixture of I gallons, that shall cost him 
 on the average m dollars a gallon. How many gallons 
 must he take of each ? 
 
 Ans. * ?- of the first ; ^ —/- of the second. 
 
 a — b a — b 
 
 Discuss the problem (1) when a = b; (2) when a or b = m; (3) 
 when a = 6 = m ; (4) when a is greater than b and less than m ; (6) 
 when a is greater than b and 6 is greater than m. 
 
CHAPTER XIII. 
 SIMPLE INDETERMINATE EQUATIONS. 
 
 199. If a single equation is given with two unknown 
 numbers, and no other condition is imposed, the number of 
 its solutions is unlimited ; for, if any value is assigned to 
 one of the unknown numbers a corresponding value may be 
 found for the other. An equation that has an indefinite 
 number of solutions is said to be indeterminate. 
 
 200. The values of the unknown numbers in an inde- 
 terminate equation are dependent upon each other ; so that 
 they are confined to a particular range. 
 
 This range may be still further limited by requiring 
 these values to satisfy some given condition; as, for 
 instance, that they shall be positive integers. 
 
 1. Solve Sx + Ay = 22, in positive integers. 
 
 Transpose, 3 x = 22 — 4y. 
 
 Divide by 3, x = 7 — y + 1 
 
 Transpose, x + y — 7 = 
 
 3 
 
 l-y 
 
 1 —y 
 Since x and y are integers, — r- 2 - is an integer, 
 o 
 
 Let — sr* = to, an integer. 
 
 o 
 
 Then, y = 1 — 3 to. (1) 
 
 Put this value for y in the given equation. 
 Then, x = 6 + 4 m. (2) 
 
 In equations (1) and (2), 
 
 If to = 0, then y = 1 and x = 6. 
 
 If to = — 1, then y = 4 and x = 2. ' 
 
 No other value of m gives positive integers for both x and y. 
 
206 SIMPLE INDETERMINATE EQUATIONS. 
 
 2. Solve 5 x — 14 y = 11, in positive integers. 
 Transpose, 5x = 11 + 14 y. 
 
 Divide by 5, x = 2 + 2 y + * * ** ■ 
 
 o 
 
 1 + 4 w 
 Transpose, x — 2 y — 2 = — ■*- — • 
 
 5 
 
 1+42/ 
 
 Then, — - — must be integral. 
 
 T 1 + 4y ._ 5m — 1 . . . . 
 Let — - — = m, then y = — > a fraction in form. 
 
 To avoid this difficulty, it is necessary to make the coefficient of y 
 
 equal to unity. Since — — - is integral, any multiple of — ■—- is 
 o 5 
 
 integral. Multiply the numerator of the fraction, then, by a number 
 
 that will make the division of the coefficient of y give a remainder of 1. 
 
 In this case, multiply by 4. 
 
 We have i±J2* = g„ + i±l. 
 
 5 5 
 
 4 + y 
 Let — - — = ra, an integer, 
 
 o 
 
 Then, y = 5 ra — 4. (1) 
 
 Since x = I (11 + 14?/), from the original equation, 
 
 x = 14 ra — 9. (2) 
 
 Here it is obvious that ra may have any positive value. 
 
 If ra = 1, x = 5, y = 1 
 
 If ra = 2, x = 19, y = 6 
 
 If m = 3, a; = 33, y = 11 
 
 and so on. 
 
 3. Solve 5x + 6 y = 30, so that sc may be a multiple of y f 
 and both positive. 
 
 Let x = ray. 
 
 Put this value of x in the given equation. 
 Then, (5 ra + 6) y = 30. 
 
 30 
 
 
 .-.y = 
 and x = 
 
 5ra + 6' 
 30 to 
 
 5to + 6 
 
 Ifra = 2, x = 3f, y = l|; 
 
 If ra = 3, x = 4f T y = If 
 
SIMPLE INDETERMINATE EQUATIONS. 207 
 
 Exercise 79. 
 Solve in positive integers : 
 
 1. 2x + 11 y = 49. 5. 3x + 8y = 61. 
 
 2. 7cc + 3?/ = 40. 6. $x + 5y = 97. 
 
 3. 5x + 7y = 53. 7. 16aj + 7y = 110. 
 
 4. x + 10y = 29. 8. 7z + 10^ = 206. 
 
 Solve in least positive integers : 
 
 9. 12a - ly = 1. 12. 23ic-9y = 929. 
 
 10. 5x-17y = 23. 13. 23x-33y = ±3. 
 
 11. 23y-13a; = 3. 14. 555 x - 22 y = 73. 
 
 15. A man spent $114 in buying calves at $5 apiece, 
 and pigs at $3 apiece. How many did lie buy of each ? 
 
 16. In bow many ways can a man pay a debt of $87 
 with, five-dollar bills and two-dollar bills ? 
 
 17. Find the smallest number that, when divided by 5 
 
 or when divided by 7, gives 4 for a remainder. 
 
 n — 4 7t ~~ 4 
 
 Let n = the number, then — - — = x, and — = — = y. 
 o 7 . 
 
 18. A farmer sold 15 calves, 14 lambs, and 13 pigs for 
 $200. Some days after, at the same price for each kind, 
 he sold 7 calves, 11 lambs, and 16 pigs, and received $141. 
 What was the price of each ? 
 
 First eliminate one of the unknowns from the two equations. 
 
 19. A number is expressed by three digits. The sum of 
 the digits is 20. If 16 is subtracted from the number and 
 the remainder divided by 2, the digits will be reversed. 
 Find the number. 
 
 20. In how many ways may 100 be divided into two 
 parts, one of which shall be a multiple of 7 and the other 
 a multiple of 9 ? 
 
CHAPTER XIV. 
 ^EQUALITIES. 
 
 201. If a — b is positive, a is said to be greater than b ; 
 if a — b is negative, a is said tp be less than b. 
 
 Note. Letters in this chapter are understood to stand for positive 
 numbers, unless the contrary is expressly stated. 
 
 202. An Inequality is a statement in symbols that one of 
 two numbers is greater than or less than the other. 
 
 203. The Sign of an Inequality is >, which always points 
 toward the smaller number. Thus, a > b is read a is 
 greater than b ; c < d is read c is less than d. 
 
 204. The expressions that precede and follow the sign of 
 an inequality are called, respectively, the first and second 
 members of the inequality. 
 
 205. Two inequalities are said to subsist in the same sense 
 if the signs of the inequalities point in the same direction ; 
 and two inequalities are said to be the reverse of each other 
 if the signs point in opposite directions. 
 
 Thus, a > b and c > d subsist in the same sense, but a > b and c < d 
 are the reverse of each other. 
 
 206. If the signs of all the terms of an inequality are 
 changed, the inequality is reversed. Thus, if a>^, then 
 — a < — b. 
 
 207. If the members of an inequality are interchanged, 
 the inequality is reversed. Thus, if a > b, then b < a. 
 
INEQUALITIES. 
 
 209 
 
 208. An inequality will continue to subsist in the same 
 sense if each member is increased, diminished, multiplied, or 
 divided, by the same positive number. 
 
 Thus, if a > 6, then a + c > 6 + c ; a — c > 6 — c ; oc>6c; 
 
 aTC>6-rc. Therefore, 
 
 209. A term can be transposed from one member of an 
 inequality to the other, provided the sign of the term is 
 changed. 
 
 Thus, if a — c>b, 
 
 by adding c to both members, a > b + c. 
 
 (§ 208) 
 
 210. -4w inequality will be reversed if its members are 
 subtracted from equal numbers; or if its members are 
 multiplied or divided by the same negative number. 
 
 Thus, iix = y and a > b, then x — a<y — b; — ac< — 6c; and 
 a-f(-c)<&-f-(-c). 
 
 211. The sum or product of the corresponding members 
 of two inequalities that subsist in the same sense is an 
 inequality in the same sense. 
 
 Thus, if a > b and c > d, then a + c > 6 + eZ, and ac > bd. 
 
 212. The difference or quotient of the corresponding 
 embers of two inequalities that subsist in the same sense 
 
 may be an inequality in the same sense, or the reverse, or 
 
 may be an equality. 
 
 Thus, 
 
 7>4 
 
 5>4 
 
 By subtraction, -^ By subtraction, -^ 
 
 4>2 J '2 = 2 
 
 7>4 5>4 
 
 By division, 3>2 By division, -^^ 
 
 2£>2 lf<2 
 
210 INEQUALITIES. 
 
 1. Find one limit for x, if 
 
 
 4sc 
 
 3> 2 5 
 
 Multiply by 10, 
 Transpose, 
 Divide by 25, 
 
 
 40x- 
 
 -30>15x- 
 25 x > 24. 
 
 Find the limits of 
 
 X, 
 
 
 
 x - 
 
 -4> 
 
 2-3x, 
 
 
 Sx- 
 
 -2< 
 
 a + 3. 
 
 Transpose in (1), 
 Divide by 4, 
 Transpose in (2), 
 Divide by 2, 
 
 
 
 4x>6. 
 
 x>H. 
 2x<5. 
 
 x<2|. 
 
 6. 
 
 given a; — 4 > 2 — 3 x, (1) 
 
 and 3a-2<a + 3. (2) 
 
 Therefore, the value of x lies between 1^ and 2\. 
 
 Exercise 80. 
 Find one limit for x, given : 
 
 1. (x + iy<x 2 + 3x-5. 3. x + 2b>7x. 
 
 4tX — 2 S — 5x _ _ x , _, 
 
 2. — 3— > 7 4 - 3x-2<- + 7£. 
 
 5. Find the limiting values of x, 
 given 4 x — 6 < 2 # + 4, 
 and 2z + 4>16-2a;. 
 
 6. If a < ft, find the limiting values of #, 
 
 ax . 7 7 « 2 
 
 given — + ox — a& > — > 
 
 , to b* 
 
 and y — a# + ab < y 
 
 7. Find the integral value of x, 
 
 given J (a + 2) + £z <•£- (x - 4) -f 3, 
 
 and i (x + 2) + %x > £ (x + 1) + i- 
 
INEQUALITIES. 211 
 
 8. Twice a certain integral number increased by 7 is not 
 greater than 19 ; and three times the number diminished 
 by 5 is not less than 13. Find the number. 
 
 9. Twice the number of pupils in a certain class is less 
 than 3 times the number minus 39 ; and 4 times the num- 
 ber plus 20 is greater than 5 times the number minus 21. 
 Find the number of pupils in the class. 
 
 213. Theorem. If a and b are unequal, a 2 + b 2 >2 ab. 
 
 For (a — b) 2 must be positive, whatever the values of a aud b. 
 That is, (a — b) 2 > 0. 
 
 Squaring, a 2 - 2 ab + b 2 > 0. 
 
 Transposing — 2 ab, a 2 + b 2 > 2 ab. 
 
 If a and b are positive and unequal, show that 
 a* + b*>a 2 b + ab 2 . 
 
 Now, a 3 + 6 3 > a 2 b + ab 2 , 
 
 if (dividing each side by a + b) 
 
 a 2 — ab + b 2 > ab, 
 if (transposing — ab) a 2 + b 2 > 2 a&. 
 
 But a 2 + b 2 > 2 a&. (Theorem) 
 
 Therefore, a 3 4- 6 s > a 2 b 4- ao 2 . 
 
 Exercise 81. 
 If the letters are unequal and positive, show that : 
 
 1. a 2 + 3b' 2 >2b (a + b). 
 
 2. (a 2 + b 2 ) (a 4 + 6 4 ) > (a 8 + J 8 ) 2 . 
 
 3. a 2 b + a 2 c + a& 2 + & 2 c + ac 2 + be 2 > 6 afo. 
 
 4. The sum of any fraction and its reciprocal > 2. 
 
 5. ab + ac +bc < (a + ft - c) 2 + (a + c-b) 2 + (b + c-a) 2 
 
 6. (a 2 + b 2 ) (e 2 + d 2 ) > (ac 4- 6d) 2 . 
 
 a + & 2ab a b 1,1 
 
CHAPTER XV. 
 INVOLUTION AND EVOLUTION. 
 
 Involution. 
 
 214. The operation of raising an expression to any re- 
 quired power is called Involution. 
 
 215. Index Law for Involution. If m is a positive integer, 
 
 a m = a X a X a to m factors. 
 
 Consequently, if m and n are both positive integers, 
 (a n ) m = a n X a n X a n tow factors 
 
 — qTi + n + n to m terms 
 
 = a 7 ™. Hence, 
 
 Any required power of a given power of a number is found 
 by multiplying the exponent of the given power by the expo- 
 nent of the required power. 
 
 216. To find (ab) n . 
 (ab) n = ab X cub to n factors 
 
 = (aX a to n factors) (b X b to n factors) 
 
 = a n b n . 
 
 In like manner, (abc) n = a n b n e n \ and so on. Hence, 
 Any required power of a product is found by taking the 
 product of its factors each raised to the required power. 
 
 217. In the same way it may be shown that 
 Any required power of a fraction is found by taking the 
 
 required power of the numerator and of the denominator. 
 
INVOLUTION AND EVOLUTION. 213 
 
 218. From the Law of Signs in multiplication it is evi- 
 dent that all even powers of a number are positive ; all odd 
 powers of a number have the same sign as the number itself. 
 
 Hence, no even power of any number can be negative ; 
 and the even powers of two compound expressions that 
 have the same terms with opposite signs are identical. 
 
 Thus, (b - af = (a- b)\ 
 
 219. Binomials. By actual multiplication we obtain, 
 
 (a + iy^at + Zab + b 2 ) 
 
 (a + b) 9 = a 8 + 3 a% + 3 ab 2 + b* ; 
 
 (a + by = a 4 + ±a*b + 6a 2 b 2 + ±ab* + #. 
 
 In these results it will be observed that : 
 
 1. The number of terms is greater by one than the ex« 
 ment of the binomial. 
 
 2. In the first term the exponent of a is the same as 
 the exponent of the binomial, and the exponent of a 
 decreases by one in each succeeding term. 
 
 3. b appears in the second term with 1 for an exponent, 
 id its exponent increases by 1 in each succeeding term. 
 
 4. The coefficient of the first term is 1. 
 
 5. The coefficient of the second term is the same as the 
 exponent of the binomial. 
 
 6. The coefficient of each succeeding term is found from 
 Le next preceding term by multiplying the coefficient of 
 Lat term by the exponent of a and dividing the product 
 
 )y a number greater by one than the exponent of b. 
 
 220. If b is negative, the terms in which the odd powers 
 >f b occur are negative. Thus, 
 
214 INVOLUTION AND EVOLUTION. 
 
 1. (a - b) s = a 8 - 3 a% -{-Sab 2 - b*. 
 
 2. (a -by = a 4 -4: a 8 b + 6a 2 b 2 _4aS 8 + b\ 
 
 By the above rules any power of a binomial of the form 
 a ± b may be written at once. 
 
 Note. The double sign ± is read plus or minus ; and a ± b means 
 a + b or a — b. 
 
 221. The same method may be employed when the terms 
 of a binomial have coefficients or exponents. 
 
 1. Find the third power of 5 a; 2 — 2y*. 
 Since (a - b)» = a* - 3 a 2 b + 3ab 2 - b*, 
 
 putting 5 x 2 for a, and 2 y* for J, we have 
 (5x 2 -2y*y 
 = (5a 2 ) 8 - 3 (5x*) 2 (2y*) + 3 (5a 2 ) (2y 8 ) 2 - (2y*) s 
 = 125 x« - 150 a?y + 60 x 2 y« - 8 f. 
 
 2. Find the fourth power of x 2 — % y. 
 
 Since (a - by = a 4 - 4 a 8 6 + 6 a 2 6 2 -4ab* + b% 
 putting x 2 for a, and | y ? or &» we nave 
 
 = (a 2 ) 4 - 4 (a 2 ) 8 (Jy) + 6 (a5«) 9 (iy)» - 4a 2 (Jy) 8 + (Jy) 4 
 = x 9 - 2x*y + f «y - }sY + ^y 4 . 
 
 222. In like manner, & polynomial of three or more terms 
 may be raised to any power by enclosing its terms in paren- 
 
 1 theses, so as to give the expression the form of a binomial. 
 
 1. (a + b + cy = [a + (b + c)y 
 
 = a* + 3a 2 (b+c)+3a(b + cy+(b+cy 
 
 = a 9 + 3 a 2 b + 3 a*c + 3 ab 2 + 6 afo 
 
 + 3 ac 2 + b* + 3 b 2 c + 3 6c 2 + c«„ 
 
 
INVOLUTION AND EVOLUTION. 
 
 215 
 
 2. (x s - 2x 2 -\- 3x + 4y 
 
 = [(x 3 -2a; 2 ) + (3a; + 4)] 2 
 
 = (x s - 2 a: 2 ) 2 + 2 (x* -2a: 2 ) (3x + 4) + (3x -f- 4) 2 
 = a: 6 - 4x 5 + 4z 4 + 6z 4 - 4a; 8 - 16x 2 + 9x 2 + 24a + 16 
 = a: 6 - 4a; 5 + 10a; 4 -4a; 8 -- 7 x 2 + 24 a; + 16. 
 
 Exercise 82. 
 Eaise to the required power : 
 
 1. (ay. 
 
 2. (a 2 b*) b . 
 
 3 ' \3ab*J 
 
 4. (-5a6 2 c 8 ) 4 . 
 
 5. {-lx 2 yz*)\ 
 /_3aWY 
 
 7. (-2cc 2 2/ 4 ) 6 . 
 
 8. (- 3 a 2 6 8 a; 4 ) 5 . 
 
 »■ (-sw- 
 
 10. (cc + 2) 6 . 
 
 11. (a; 2 -2) 4 . 
 
 12. (x + 3)\ 
 
 13. (2a; + l) 6 . 
 
 14. (2m 2 -l) 5 . 
 
 15. (2x + 3yy. 
 
 16. (2 a;-?,) 8 . 
 
 17. (xy-2)\ 
 
 18. (1 - a; + a; 2 ) 2 . 
 
 19. (l-2a; + 3a; 2 ) 3 . 
 
 20. (1-a-fa 2 ) 8 . 
 
 21. (3-4a; + 5a; 2 ) 2 . 
 
 Evolution. 
 
 223. The operation of finding any required root of an 
 expression is called Evolution. A root of an imperfect 
 power cannot be found exactly. 
 
 Thus, the exact value of the square root of 2 can be written only 
 as V2, and the exact value of the cube root of 4 can be written only 
 
 as v4. Approximate values of these expressions, however, can be 
 found by annexing ciphers and extracting the root. 
 
 1 
 
216 INVOLUTION AND EVOLUTION. 
 
 224. Index Law for Evolution. If m and n are positive 
 integers, we have 
 
 
 (a m ) n = a mn . 
 
 (§215) 
 
 Converselj, 
 Also, 
 
 Converselj, 
 and 
 
 n, ■* 
 
 ■y/a mn = a ■ = a m . 
 (ab) n = a n 6 w . 
 y~tfW= ^fa n X V^" = 
 Va6 = V^ X J/b. 
 
 (§ 216) 
 
 : ab, 
 
 
 Thus, the c^&e n>o£ of a 6 is a? = a 2 ; the fourth root of 
 81 a 12 is found bj taking the fourth root of 81 and of a 12 ; 
 and is 3 a 8 . Hence, 
 
 225. To Find the Root of a Simple Expression, 
 
 Take the required root of the numerical coefficient, and 
 divide the exponent of each letter by the index of the required 
 root. 
 
 226. From the Law of Signs it is evident that : 
 
 1. Anj even root of a positive number will have the double 
 sign, ±. 
 
 2. There can be no even root of a negative number. 
 
 For V— x 2 is neither + x nor — x ; since the square of + x = + x 2 , 
 and the square of — x = + x 2 . 
 
 The indicated even root of a negative number is call 
 an imaginary number. 
 
 3. Anj odd root of a number will have the same sign as 
 the number. 
 
 © n a n 
 — ~Tn • (§ 217) 
 
 n \ n n ~\l tjft n. 
 
 Converselj, a/— = -— — -r ' That is, 
 
 Any required root of a fraction is found by taking the 
 required root of the numerator and of the denominator. 
 
INVOLUTION AND EVOLUTION. 
 
 217 
 
 „ ./16 a; 2 4 a; 3/ 
 
 Thus ' ^Slf = ± % ; V - 27mV = - 3 
 
 mn 2 ; 
 
 *! l6xy 2 2xhf 
 
 \ 81 a 16 ~ ± 3 a 4 ' 
 
 227. If the root of a number expressed in figures is not 
 readily detected, it may be found by resolving the number 
 into its prime factors. Thus, to find the square root of 
 3,415,104: 
 
 2 8 
 
 3415104 
 
 2 8 
 
 426888 
 
 3 2 
 
 53361 
 
 7 
 
 5929 
 
 7 
 
 847 
 
 11 
 
 121 
 
 11 
 
 3,415,104 = 2 6 X 3 2 X 7 2 X ll 2 . 
 V3,415,104 = 2 8 x 3 X7 X 11 = 1848. 
 
 Simplify : 
 
 i. V4^y. 
 
 2. V64z 9 . 
 
 3. Vl6xy 2 
 
 4. V-32a 10 . 
 
 5. V-27ic a 
 
 6. V25a 4 . 
 
 7. v'- 8 a 8 6 6 . 
 
 8. V64* 12 . 
 
 Exercise 83. 
 
 9. -v^-216 
 
 10. V729* 18 . 
 
 6, 
 
 11. V243^ 10 . 
 
 12. V-1728^ 8 . 
 
 13. ■v'- 343 a\ 
 
 14. •^81^ 24 . 
 
 15. -\/512a 12 6 16 . 
 
 16. V* 8B i/ 12m . 
 
 17, V 
 
 9a 2 ^ 
 
 18 
 
 16ic 4 2/ 2 
 
 »< 8a?y 
 \ 27 * 9 
 
 20. 
 
 21 
 
 4 
 ■4 
 
 S2a 1Q 
 243 x™ 
 
 16 x 4 
 
 81a«b 12 ' 
 
 216 a 24 
 
218 INVOLUTION AND EVOLUTION 
 
 Square Roots of Compound Expressions. 
 
 . Since the square of a -f- b is a 2 + 2 ab + b 2 , the 
 square root of a*\+ 2 ab + J 2 is a 4- b. 
 
 It is required to find a method for extracting the square 
 root, a + 5, when a 2 -\- 2 ab -\- b 2 is given : 
 
 The first term, a, of the root is obviously the square root of the 
 
 first term, a 2 , of the expression. 
 
 a 2 + 2 ab + b 2 [a + 6 If the a 2 is subtracted from the given 
 
 a 2 expression, the remainder is 2ab + b 2 . 
 
 2a + b 20& + 6 2 Therefore, the second term, &, of the root 
 
 2ab + b 2 is obtained when the first term of this 
 
 remainder is divided by 2 a, that is, by 
 
 double the part of the root already found. Also, since 
 
 2ab + b 2 = (2a + b)b, 
 
 the divisor is completed by adding to the trial divisor the new term of 
 the root. 
 
 Find the square root of 25 x 2 — 20 x*y + 4 x*y*. 
 
 25 x 2 — 20 x*y + 4 x i y 2 \5x — 2x 2 y ' 
 25 x 2 
 
 — 20x*y + 4x*y* 
 
 — 20s 8 y + 4s 4 y 2 
 
 10a; — 2cc 2 y 
 
 The expression is arranged according to the ascending powers of x. 
 
 The square root of the first term is 6 », and 5 x is placed at the 
 right of the given expression, for the first term of the root. 
 
 The second term of the root, — 2x 2 y, is obtained by dividing 
 — 20 x s y by 10 x, and this new term of the root is also annexed to the 
 divisor, 10 x, to complete the divisor. 
 
 229. The same method will apply to longer expressions, 
 if care is taken to obtain the trial divisor at each stage of 
 the process, by doubling the part of the root already found, 
 and to obtain the complete divisor by annexing the new term 
 of the root to the trial divisor. 
 
INVOLUTION AND EVOLUTION 
 
 219 
 
 Find the square root of 
 
 1 + 10a 2 + 25x* + 16a 6 - 2±x 5 - 20 a 8 -4a. 
 
 Arrange the expression in descending powers of x. 
 
 16x6-24x 6 + 25x*-20x 3 + 10x 2 -4x+11 4x 3 -3x 2 +2x-l 
 16 x« 
 
 $x*-~3x? 
 
 -24x6 + 25^ 
 -24x* + 9x< 
 
 — 20 x 3 + 10x 2 
 -12x3+ 4x 2 
 
 8x 3 - 
 
 6x 2 + 2x 
 
 16x^ 
 16x4 
 
 8x 8 — 6x 2 + 4x 
 
 — 1 
 
 — 8x 3 + 6x 2 -4x + l 
 
 - 8x 3 + 6x 2 -4x + l 
 
 It will be noticed that each successive trial divisor may be obtained 
 by taking the preceding complete divisor with its last term doubled. 
 
 Exercise 84. 
 Find the square root of : 
 
 1. a 4 -8a 8 + 18a 2 -8a + l. 
 
 2. 9 
 
 6 a s + 13 a 2 - 4 a + 4. 
 
 3. 4:X*-12x s y + 29xy-30xtf + 25y\ 
 
 4. l + 4a + 10a 2 + 12a 8 + 9a 4 . 
 
 6. 16-96a + 216a 2 -216a 8 + 81a 4 
 
 6. a 4 - 22 x s + 95 x 2 + 286a + 169. 
 
 7. 4a 4 - 11a 2 + 25- 12a 8 + 30a. 
 
 8. 9a 4 + 49 -12a 8 - 28a + 46a 2 . 
 
 9. 49 a 4 + 126 a 8 + 121 -73 a 2 -198 a. 
 
 10. 16a 4 -30a-31a 2 + 24a 8 + 25. 
 
 11. a 4 - 2 ax* + 3 a 2 a 2 - 2 a 8 x + a\ 
 
 12. 9a 4 - 18a + 1+87 a 2 - 54a 8 . 
 
220 
 
 INVOLUTION AND EVOLUTION. 
 
 230. If an expression contains powers and reciprocals 
 of powers of the same letter, the order of arrangement 
 in descending powers of the letter is as follows : 
 
 •Ju • •£>• J_« 
 
 111 
 
 Find the square root of 
 
 9c 2 4as 2 
 a 2 + 9c 2 
 
 4x 101 
 
 15 o 25 
 
 Arrange in descending powers of x. 
 
 4x 2 
 9 c 2 
 i& 
 9c 2 
 
 4x 101 6 c 9c 2 
 15 c 25 5x x 2 
 
 2x 
 3c 
 
 6c 
 5x 
 
 1 3c 
 5 + x 
 
 4x 1 
 
 4x 
 
 101 
 25 
 
 1 
 25 
 
 
 3c 5 
 
 15c ' 
 15 c 
 
 
 4x 
 3c' 
 
 2 3c 
 6 x 
 
 4 
 
 4 
 
 6c 9c* 
 5x x 2 
 6 c 9c 2 
 5x + x 2 
 
 231. An approximate value of an imperfect square can 
 be found to any required number of terms as follows ; 
 Find to three terms the square root of x 2 +px. 
 
 x 2 +px 
 
 x + 
 
 8x 
 
 •t? 
 
 px 
 
 px 
 
 *? 
 
 
 2x4 
 
 ■p 
 
 4 
 
 
 
 
 *> 2 
 4 
 
 8x 
 
INVOLUTION AND EVOLUTION. 221 
 
 Exercise 85. 
 Find the square root of : 
 
 1. ±X*+-4:X*-}sX + ~ 
 
 4a 2 , 46 2 
 
 2. -77- + 8+—- 
 
 b 2 a 2 
 
 n „ , 3 a 2 a , 1 
 
 4. ^ + -3- +—- + J + -. 
 
 tf 3j 41 3, y* 
 y* y lo 4a; 4r 
 
 4 2 4 
 
 7. 16x* + iJ-x 2 y + 8x 2 + $ y a + ty + L 
 
 9x* Sx 8 43<k 2 7x 49 
 4 . 2 + 4 2 + 4 ' 
 
 9. 4a 2 + -^---ll + 4a. 
 or a 
 
 Find to three terms the square root of : 
 
 10. a 2 + b. 13. 1 + a. 16. 4 a 2 + 3. 
 
 11. x 2 + \y. 14. l-2a. 17. 4 -3a. 
 
 12. l + 2a. 15. 4a 2 + 26. 18. 4a 2 - 1. 
 
 232. Arithmetical Square Roots. In extracting the square 
 root of an arithmetical number, the first step is to arrange 
 the figures in groups. 
 
 Since 1 = l 2 , 100 = 10 2 , 10,000 = 100 2 , and so on, the 
 square root of a number between 1 and 100 lies between 1 
 and 10 ; of a number between 100 and 10,000 lies between 
 
 1 
 
222 INVOLUTION AND EVOLUTION. 
 
 10 and 100. In other words, the square root of a number 
 expressed by one or two figures is a number of one figure .; 
 of a number expressed by three or four figures is a num- 
 ber of two figures ; and so on. 
 
 If, therefore, an integral square number is divided into 
 groups of two figures each, from the right to the left, the 
 number of figures in the root will be equal to the number 
 of groups of figures. The last group to the left may have 
 one figure or two figures. 
 
 Find the square root of 3249. 
 
 32 49 (57 In this case, a in the typical form a 2 + 2 ab + fc 2 
 
 25 represents 5 tens, that is, 50, and b represents 7. The 
 
 107) 7 49 25 subtracted is really 2500, that is, a 2 , and the cora- 
 
 7 49 plete divisor 2a + 6is2X50 + 7 = 107. 
 
 233. The same method will apply to numbers of more 
 than two groups of figures by considering a in the typical 
 form to represent at each step the part of the root already 
 found. 
 
 It must be observed that a represents so many tens with 
 respect to the next figure of the root. 
 
 Find the square root of 5,322,249. 
 
 5 32 22 49(2307 
 
 4 
 43)132 
 
 129 
 4607)3 22 49 
 3 22 49 
 
 234. If the square root of a number has decimal places, 
 the number itself will have twice as many. Thus, if 0.21 is 
 the square root of some number, this number will be (0.21) 2 
 = 0.21 X 0.21 = 0.0441 ; and if 0.111 is the root, the num- 
 ber will be (0.111 ) 2 = 0.111 X 0.111 = 0.012321. 
 
INVOLUTION AND EVOLUTION. 
 
 223 
 
 Therefore, the number of decimal places in every square 
 decimal will be even, and the number of decimal places in 
 the root will be half as many as in the given number itself. 
 
 Hence, if a given number contains a decimal, we divide the 
 number into groups of two figures each, by beginning at the 
 decimal point and marking toward the left for the integral 
 number, and toward the right for the decimal. We must 
 be careful to have the last group on the right of the deci- 
 mal point contain two figures, annexing a cipher when 
 necessary. 
 
 Find the square root of 41.2164 ; of 965.9664. 
 
 41.21 64 (6.42 9 65.96 64 (31.08 
 
 36 9 
 
 124)6 21 61)65 
 
 4 96 61 
 
 1282)25 64 
 25 64 
 
 6208)4 96 64 
 4 96 64 
 
 If a number contains an odd number of decimal 
 places, or if any number gives a remainder when as many 
 figures in the root have been obtained as the given number 
 has groups, then its exact square root cannot be found. We 
 may, however, approximate to its exact root as near as we 
 please by annexing ciphers and continuing the operation. 
 
 The square root of a common fraction whose denominator 
 
 is not a perfect square can be found approximately by 
 
 jducing the fraction to a decimal and then extracting the 
 
 )t ; or by reducing the fraction to an equivalent fraction 
 
 rhose denominator is a perfect square, and extracting the 
 
 square root of both terms of the fraction. Thus, 
 
 /(X625 = 0.79057 ; 
 
 4 
 
 VlO _^ VlO _ 3.16227 
 Vl6~ 4 4 
 
 = 0.79057. 
 
224 
 
 INVOLUTION AND EVOLUTION. 
 
 Find the square root of 3 ; of 357.357. 
 
 3.(1.732. 
 
 1 
 27)2 00 
 
 189 
 343) 11 00 
 10 29 
 
 3462) 71 00 
 69 24 
 
 3 57.35 70 (18.903. 
 
 1 
 28)2 57 
 
 2 24 
 369)33 35 
 33 21 
 
 37803) 14 7000 
 11 3409 
 
 Exercise 86 
 Find the square root of : 
 
 1. 289. 
 
 2. 1225. 
 
 3. 12,544. 
 
 4. 253,009. 
 
 5. 529,984. 
 
 6. 150.0625. 
 
 7. 118.1569. 
 
 8. 172.3969. 
 
 11. 640.343025. 
 
 12. 100.240144. 
 
 13. 316.021729. 
 
 9. 5200.140544. 14. 454.585041. 
 10. 1303.282201. 15. 5127.276025. 
 
 Find to four decimal places the square root of : 
 
 16. 10. 19. 0.5. 22. 0.607. 25. f. 28. f 
 
 17. 3. 20. 0.7. 23. 0.521. 26. f. 29. f. 
 
 18. 5. 21. 0.9. 24. 0.687. 27. £. 30. f T . 
 
 Cube Roots of Compound. Expressions. 
 
 Since the cube of a + b is a 9 4- 3 a 2 b 4- 3 ab 2 + b Q , 
 the cube root of 
 
 a* + 3 a 2 b -f 3 ab 2 + b* is a + b. 
 

 INVOLUTION AND EVOLUTION 225 
 
 It is required to devise a method for extracting the cube 
 root, a + b. when a 8 4- 3 a 2 b 4- 3 a& 2 4- & 8 is given : 
 
 The first term, a, of the root is obviously the cube root of the first 
 term, a 8 , of the given expression. 
 
 a 8 + 3a 2 6 + 3 ad 2 + W \a + b 
 8 a 2 a 8 
 
 4 3 a& + 62 
 
 3a 2 + 3a& + 6 2 
 
 3a26 + 3a&2 + 68 
 3 a 2 6 + 3 a& 2 -f &3 
 
 If a 8 is subtracted, the remainder is 3 a?b + 3 a& 2 -f &s . therefore, 
 the second term, 6, of the root is obtained by dividing the first term of 
 this remainder by three times the square of a. 
 
 Also, since 3 a% + 3 a& 2 + &3 = (3 a 2 + 3 a & + 52) & ? t he complete 
 divisor is obtained by adding 3 ab 4- 6 2 to the £ria£ divisor 3 a 2 . 
 
 Find the cube root of 8 x* 4- 36 cc 2 y 4- 54 a^ 2 4- 27 y 8 . 
 
 8x 8 + 36x2y + 54*y 2 + 27y 8 j2x + 3^ 
 12x 2 8x 8 
 
 (6x + 3y)3y = +18xy + 9y 2 
 
 12x 2 + 18x2/ + 9 2/ 2 
 
 36 x 2 2/4- 64xy 2 + 27^ 
 36 x 2 y + 54x2/2 + 27 2/ 8 
 
 The cube root of the first term is 2 x, and 2 x is therefore the first 
 term of the root. 8 x 8 , the cube of 2 x, is subtracted. 
 
 The second term of the root, 3 2/, is obtained by dividing 36 x 2 y by 
 8 (2 x) 2 = 12 x 2 , vhich corresponds to 3 a 2 in the typical form, and the 
 divisor is completed by annexing to 12 x 2 the expression 
 
 {3(2x) + 32/}32/ = 18x2/ + 92/ 2 . 
 
 237. The same method may be applied to longer expres- 
 jions by considering a in the typical form 3 a 2 4- 3 ab 4- b 2 
 to represent at each stage of the process the part of the root 
 already found. Thus, if the part of the root already found 
 is x 4- y, then 3 a 2 of the typical form will be represented 
 by S(x 4- yf\ and if the third term of the root is 4- z, the 
 3 ab 4- b 2 will be represented by 3 (as 4- y) z 4- z 2 . So that 
 the complete divisor, 3 a 2 4- 3 ab 4- b 2 9 will be represented 
 by 3(x + y) 2 4- S(x 4- y) z 4- z\ 
 
226 
 
 INVOLUTION AND EVOLUTION. 
 
 Find the cube root of x e — 3 x 5 + 5 x z — 3 x — 1. 
 
 jx 2 -x-l 
 
 3x* 
 
 (3x2 _ X )(_ b) = -3x8+ x s 
 
 3x 4 -3x 3 + x 2 
 
 x6-3x 6 + 5x3-3x~l 
 x 6 
 
 — 3x 6 + 5x 8 
 
 — 3x 5 + 3x 4 — x 3 
 
 3 (x 2 - x) 2 = 3x 4 - 6x 8 + 3x 2 
 (3x 2 -3x- !)(-!) = -3x 2 + 3x + l 
 
 3x 4 -6x 8 
 
 + 3x+l 
 
 -3x 4 + 6x 8 -3x-l 
 -3x 4 + 6x 8 — 3x-l 
 
 The root is placed above the given expression for convenience of 
 arrangement on the page. 
 
 The first term of the root, x 2 , is obtained by taking the cube root 
 of the first term of the given expression ; and the first trial divisor, 
 3 x 4 , is obtained by taking three times the square of this term. 
 
 The first complete divisor is found by annexing to the trial divisor 
 (3x 2 — x) (— x), which expression corresponds to (3 a + b) b in the 
 typical form. 
 
 The part of the root already found (a) is now represented by x 2 — x; 
 therefore, 3 a 2 is represented by 3 (x 2 — x) 2 = 3 x 4 — 6 x 8 + 3 x 2 , the 
 second trial divisor ; and (3 a + b) b by (3 x 2 — 3 x — 1) (— 1) ; there- 
 fore, in the second complete divisor, 3 a 2 + (3 a + b) b is represented by 
 (3x 4 - 6x 8 + 3x 2 ) + (3x 2 - 3x - 1) (- 1) = 3x 4 — 6x 8 + 3x + 1. 
 
 Exercise 87. 
 Find the cube root of : 
 
 1. a* + 3a 2 x + 3ax 2 + x\ 
 
 2. S + 12x + 6x 2 + x\ 
 
 3. x 6 - 3 ax 5 + 5 a*x 3 — 3a 5 x- a\ 
 
 4. 1 - 6x + 21x 2 - Ux* + 63a 4 - 54a B + 27a* 
 6. l-3x + 6x 2 -7x* + 6x i -3x* + x\ 
 
 6. » 8 -r-l-6a;-6a; 6 4-15a; 2 -f-15ic 4 -20a; 8 . 
 
INVOLUTION AND EVOLUTION. 227 
 
 7. 64a; 6 - 144a: 5 + 8 - 36a; + 102 a; 2 - 171 a; 3 + 204a; 4 . 
 
 8. 27 a G - 27 a 5 - 18 a 4 + 17 a 3 + 6 a 2 - 3 a - 1. 
 
 9. 8 a; 6 - 36a; 5 + 66a; 4 - 63a; 3 + 33a; 2 - 9sc + 1. 
 10. 27 + 108a; + 90a; 2 - 80a; 3 - 60a; 4 + 48a; 5 -8x\ 
 
 a; 9 3a; 8 5a; 6 3a; 4 x s 
 
 y\h yU y\1 ^10 ^9 
 
 238. Arithmetical Cube Roots. In extracting the cube root 
 of an arithmetical number, the first step is to arrange the 
 figures in groups. 
 
 Since 1 = l 3 , 1000 = 10 3 , 1,000,000 = 100 8 , and so on, it 
 follows that the cube root of any number between 1 and 
 1000, that is, of any number which has' one, two, or three 
 figures, is a number of one figure ; and that the cube root 
 of any number between 1000 and 1,000,000, that is, of any 
 number which has fowr, five, or six figures, is a number of 
 two figures ; and so on. 
 
 If, therefore, an integral cube number is divided into 
 groups of three figures each, from right to left, the number 
 of figures in the root will be equal to the number of groups. 
 The last group to the left may have one, two, or three 
 figures. 
 
 239. If the cube root of a number has decimal places, 
 the number itself will have three times as many. Thus, 
 if 0.11 is the cube root of a number, the number is 
 0.11 X 0.11 X 0.11 = 0.001331. Hence, if a given number 
 contains a decimal, we divide the number into groups 
 of three figures each, by beginning at the decimal point 
 and marking toward the left for the integral number, and 
 
228 
 
 INVOLUTION AND EVOLUTION. 
 
 toward the right for the decimal. We must be careful to 
 have the last group on the right of the decimal point con- 
 tain three figures, annexing ciphers when necessary. 
 
 240. Notice that if a denotes the first term, and b the 
 second term of the root, the first complete divisor is 
 
 3a* + 3ab + b% 
 
 and the second trial divisor is 3 (a -+- by, that is, 
 
 3a 2 + 6ab + 3b% 
 
 which may be obtained by adding to the preceding complete 
 divisor its second term and twice its third term. 
 
 Extract the cube root of 5 to five places of decimals. 
 
 
 
 5.000(1.70997 
 1 
 
 3 X 10 2 = 300 
 
 4000 
 
 3(iOX 7) = 210 
 
 
 7 2 = 49^) 
 559 > 
 
 3913 
 
 259 J 
 
 
 87 000 000 
 
 3 X 1700 2 = 8670000 
 
 
 3(1700X9)= 45900 
 
 
 9 2 = 81 1 
 
 
 8715981 >. 
 
 78 443 829 
 
 45981 J- 
 
 8 556 1710 
 
 3 X 1709 2 = 8762043 
 
 7 885 8387 
 
 
 670 33230 
 
 
 
 613 34301 
 
 After the first two figures of the root are found, the next trial 
 divisor is obtained by bringing down the sum of the 210 and 49 
 obtained in completing the preceding divisor ; then adding the three 
 numbers connected by the brace, and annexing two ciphers to the result. 
 
 The last two figures of the root are found by division. The rule in 
 such cases is that two less than the number of figures already obtained 
 may be found by division without error, the divisor being three times 
 the square of the part of the root already found. 
 
INVOLUTION AND EVOLUTION. 
 
 229 
 
 Exercise 88. 
 Find the cube root of : 
 . 4913. 3. 1,404,928. 
 
 .5. 385,828.352. 
 6. 1838.265625. 
 
 2. 42,875. 4. 127,263,527. 
 
 Find to four decimal places the cube root of : 
 87. 9. 3.02. 11. 0.05. 13. §. 16. f T . 
 
 . 10. 10. 2.05. 12. 0.677. 14. f. 16. ^. 
 
 241. Since the fourth power is the square of the square, 
 and the sixth power the square of the cube, the fourth root 
 Is the square root of the square root, and the sixth root is 
 the cube root of the square root. In like manner, the 
 eighth, ninth, twelfth, root may be found. 
 
 Exercise 89. 
 Find the fourth root of : 
 81a 4 + 108 a 8 + 54 a; 2 -f 12 a 4- 1. 
 16x*-32ax* + 24ta 2 x 2 -8a 8 x + a\ 
 . 14- 4a 4- a 8 4- 4a 7 4- 10 a 6 4- 16a 8 4- 10 a 2 4- 19a 4 4- 16a 5 . 
 
 Find the sixth root of : 
 4. l + 6rf + d 6 + 6d 5 + 15^ + 20<2 8 -f 15^. 
 ►. 729 - 1458 x + 1215 a 2 - 540 a 8 -f 135 a 4 - 18 a 5 4- a 6 . 
 1 - I82/ + 135?/ 2 - 540 y* 4- 1215 y 4 - 1458 y 5 + 729y 6 . 
 
 Find the eighth root of: 
 '. 1 - Sy 4- 28i/ 2 - 56y* 4- 70y 4 - 56y* 4- 28y«- 8y 7 4-y 8 . 
 
CHAPTER XVI. 
 THEORY OF EXPONENTS. 
 
 242. If n is a positive integer, we have defined a n to 
 mean the product obtained by taking a as a factor n times, 
 and called a n the nt\i power of a; we have also defined -^/a 
 as a number which taken n times as a factor gives the prod- 
 uct a, and called y/a the nth. root of a. 
 
 243. By this definition of a n the exponent n denotes 
 simply repetitions of a as a factor; and such expressions 
 
 as a , a~ 8 have no meaning. It is found convenient, how- 
 ever, to extend the meaning of a n to include fractional and 
 negative values of n. 
 
 244. If we do not define the meaning of a n when n is 
 fractional or negative, but require that the meaning of a n 
 must in all cases be such that the fundamental index law 
 shall always hold true, namely, 
 
 a m X a n = a m+ » 
 
 we shall find that this condition alone will be sufficient to 
 define the meaning of a n for all cases. 
 
 245. Meaning of a Zero Exponent. By the index law, 
 
 
 a X a n = a + n = a n . 
 
 Cb 
 
 Divide by a n , a° = — = 1. 
 
 Therefore, the zero power of any number is equal to 
 unity. 
 
THEORY OF EXPONENTS. 
 
 231 
 
 246. Meaning of a Fractional Exponent. By the index law, 
 
 *_„*+*_„§ _ 
 
 a* X cr = a' a = a 3 = a ; 
 
 a § X a 1 X a* = a l+i+4 = a § = 
 
 a f X a* X a 1 X a 1 
 
 a f + f + f+l = a ¥ = a8 
 
 - - i+- ton terms 2 
 
 a n X a n to n factors = a n n = a n = a 
 
 a n X a n to n factors = a n n 
 
 provided ra and n are positive integers. 
 
 That is, a* = s/a ; a* = -^a ; 
 
 = a ■ = a* 
 
 .?_ i 
 
 Va 8 ; « n= V^ 
 
 The meaning, therefore, of a", where ra and w are posi- 
 tive integers, is the nth root of the rath power of a. Hence, 
 The numerator of a fractional exponent indicates a power 
 md the denominator a root. 
 
 247. Meaning of a Negative Exponent. By the index law, 
 f n is a positive integer, 
 
 But 
 
 a n X a- n = a n - n = a°. 
 a° = l. 
 
 .'. a n X a~ n = 1. 
 
 (§ 245) 
 
 That is, a n and a~ n are reciprocals of each other (§ 169), 
 that a~ u = — , and a n = 
 
 248. Hence, we can change any factor from the numerator 
 )f a fraction to the denominator, or from the denominator 
 to the numerator, provided we change the sign of its exponent. 
 
 ab 2 ' 1 
 
 Thus ' cW may be written a62c " Sd ~ 3 ' or a-i6-2c»d8 ■ 
 
232 THEORY OF EXPONENTS. 
 
 249. We have now assigned definite meanings to frac- 
 tional exponents and negative exponents, by assuming that 
 the index law for multiplication, a m X a n = a m + n , is true for 
 all values of the exponents m and n. 
 
 It remains to show that the index laws established for 
 division, involution, and evolution apply to fractional and 
 negative exponents. 
 
 250. Index Law of Division for all Values of m and n. To 
 divide by a number is to multiply the dividend by the 
 reciprocal of the divisor. 
 
 Therefore, for all values of m and n, 
 
 Qtn 1 
 
 - = a»X- = «»Xa- (§217) 
 
 = a m ~ n . 
 
 251. Index Law of Involution and Evolution for all Values 
 of 222 and 22. 
 
 To prove (a m ) n = a mn for all values of m and n. 
 
 Case 1. Let m have any value, and let n be a positive 
 integer. 
 
 Then, (a m ) n = a m X a m X a m to n factors 
 
 = a* 
 
 i + m + m to n terms 
 
 P 
 
 Case 2. Let m have any value, and n = -,p and q being 
 positive integers. 
 
 Then, (a m )« = y/(a m y = vS 
 
 mp 
 
 = aT. 
 
 Case 3. Let m have any value, and n = — r, r being a 
 positive integer or a positive fraction. 
 
THEORY OF EXPONENTS. 
 
 (a m )~ r = 
 
 Then ' (a^ * 
 
 Therefore, (a m ) n = a mn for all values of m and n. 
 
 233 
 (§ 247) 
 
 252. To prove (ab) n = a n b n for any value of n. 
 Case 1. Let wbea positive integer. 
 Then, (ab) n = ab X ab X ab to n factors 
 
 = (a X a ton factors) (b X b to n factors) 
 
 = a n b n . 
 
 P 
 Case 2. Let n--,p and q being positive integers. 
 
 Then, by Case 1, § 251, since q is a positive integer, the qth 
 p p p 
 power of (aby = (aby X (ab)* to q factors 
 
 z t\- + - too term* 
 
 = (ab)i « 
 
 = (aby = a*>b*>. (By Case 1) 
 
 Also by Case 1, § 251, since q is a positive integer, the ^th 
 power of 
 
 p p p p p p 
 a<ib* = (a« X a« to q factors) (6« X 6« to q factors) 
 
 = a*>b p . 
 
 p 
 But the qth. power of (ab)* = (aby = a?b*. 
 
 That is, [(«*)*]* = \_a^f. 
 
 Extracting the ^th root of each member, we have 
 
 p p p 
 (ab)* = am. 
 
 Case 3. Let n = — r,r being a positive integer or fraction, 
 1 1 
 
 Then, 
 
 (ab)- 
 
 (ab) r a r b r 
 
 — a~ r b' 
 
234 
 
 THEORY OF EXPONENTS. 
 
 EXAMPLES. 
 
 1. 27-* = -^ 
 
 27* -^27 3 
 
 2. 16 f =(v / 16) 8 = (±2) 8 = d=8. 
 
 3. a % X aT l = a 1 ' 1 = a** = %. 
 
 4. a 1 X a* X a _i = a 8 **"* = a = 1. 
 
 5. (O 
 
 6. 
 
 i\6_ „-ix6 
 
 •*-(-«) = a - 3 + s 
 
 7. </a*b- 3 c~ 4 d = a%~ kT W. 
 
 8 f4a-V § = 1 = J = 1 =- 
 
 k } (4a"»)* 4« a -» x * 8«" x 8 
 
 / lBa-A -*. / giy.V 27 6 f 27 a 8 ^ 1 
 \ 816 8 / \16a~V 
 
 10. (3V 3 )- 
 
 8a- f 
 
 a' a' 
 
 (3 s a- s ) § 3 §xf a~ 2 3* ^ 
 
 Exercise 90. 
 Express with fractional exponents : 
 
 1. ■&?. 3. ■$€?. 5. -v^ 5 . 7. V^ + ■% 4- ■v / 16« 
 
 2. Vol 4. ^—8. 6. "V / o i , 8. VaV + Va 8 ^. 
 
THEORY OF EXPONENTS. 235 
 
 Express with radical signs : 
 9. a*. 11. ah*. 13. x*y~*. 15 - a* - xh*. 
 
 10. c* 12. a*£*. 14. 3aj*y~*. 16. a* + »V. 
 
 Express with positive exponents : 
 
 2a - V 
 17. a~ 8 . 19. 3x~Y- 21. 4z-y- 2 . 23. 
 
 3-2^2^-6 
 
 18. a~*. 20. 4cc?/- 7 . 22. 3 a" 4 ft* 
 
 Write without denominators : 
 
 24 - -6 • 26 - — ££=•? 
 
 as -6 a -J oc 2 a 
 
 25. — : — r 27. 
 
 
 28. 
 
 X 
 
 a~ 
 
 -2£-f 
 
 
 29. 
 
 a" 
 
 a" 
 
 
 40. 
 
 dW" 
 
 1 x ( A )* 
 
 -2^-2 c -4 
 
 Find the value of : 
 
 30. 8*. 35. (-27)1 
 
 31. 16"*. 36. (- 27)* X 25*. 41. flf"M X ah*. 
 
 32. 27*. 37. 81 _l X 16*. 42. (a~k*)~K 
 
 33. (-8)"*. 38. 8 } X4 4 . 43. (cTV 1 )- 2 . 
 
 34. 36 f . 39. (^V)^Xl6~l 44. (»*y"***)H 
 
 If a = &, b = 2, c = 1, find the numerical value of : 
 
 45. aV 1 . 47. a~^ 2 . 49. 3 (abf. 51. (a£ 2 c)*. 
 
 46. a£- 2 - 48. a~V~* 50. 2 (a&)~*. 52. (ab 2 c)~*. 
 
236 THEORY OF EXPONENTS. 
 
 Multiplication. 
 Multiply a?* + as*y* + y* by «* - » V 4- y* 
 «* + &M + y* 
 
 x + x*yi + »*y* 
 
 + a?*y* + g*y* + y 
 
 a; + x V + y 
 
 Exercise 91. 
 Multiply : 
 
 1. a* + J* by a* - &* 3. a* - ft* by a* - b*. 
 
 2. a* + 6* by a* + 6*. 4. a; 1 + 2* by x 1 - 2 a. 
 
 5. x~ 2 + £C _1 y" * 4- y -2 by a; -2 — ar^y 1 + y~\ 
 
 6. a;* — aj*y* + y* by a;* -f y* 
 
 7. «* + ojV + y* by »* - y* 
 
 8. l + a-i + a-ityi^a-i+fr-t 
 
 9. aV* + 2a*-3ft*by 2*"*- 4«T*- 6«rV. 
 Division. 
 
 Divide Vaj 1 + Va; - 12 by v£ - 3. 
 x i + x*-12|x*-3 
 
 «* — 3** x* + 4 
 
 + 4x*-12 
 + 4x*-12 
 
 r 
 
 Hereaj* + ^ = **-**=«*; 4a; 8 + a; 8 = 4 a:'" * = 4a*° 
 
THEORY OF EXPONENTS. 
 
 237 
 
 Exercise 92. 
 
 3. a — b by a* — b . 
 
 4. a + Jbyc^ + 6*. 
 
 Divide : 
 
 1. a-b by a* — fr* 
 
 2. a + # by a* + **. 
 
 5. aj-3aj* + 3a;*-l by »* - 1 
 
 6. a; -f a? V + ^ b y x * y + y • 
 
 7. aj* - 4sc* 4- 1 4- 6aT* by a£ - 2. 
 
 8. 9x-12x* -2 + ±x~* + x- 1 by 3 a;* - 2 - a;""*. 
 
 9. 2ar- 2 + 6ar-y i -16ar 5 2r 4 by 2 a; 4- 2 a: 2 ?/- x 4- 4 a; V* 
 
 ill i i l 
 
 10. x + y 4- s — 3 x J y J z J by as* 4- y + * . 
 
 Square Boot. 
 Find the square root of 
 
 9 a;- 4 - 18 aj~«y* + 15 «T> ~ 6 x ~ V + 2/ 2 - 
 9a ._4 — I8x- «y* 4- 15x- 2 y — 6 a- ty* +y 2 |3x-2— 3^-iyi+y 
 
 9x-* 
 
 —lSx-tyt + lbx-tp 
 -18x-^4- 9x- 2 y 
 
 6 s- 2 — 6a- ^ +2/ 
 
 6x-2 y _ 6x-!^4-y 2 
 6x~ 2 y— 6a-V+y 2 
 
 Exercise 93. 
 Find the square root of : 
 
 1. a?*4-2aj*4-i 
 
 4**4-4. 
 
 2. 4a*-4a*J*4-J* 
 
 3. a ; 
 
 4. 4 a- 2 4- 4 a' 1 4-1. 
 
 5. 9a -12a* 4- 10 -4a"* 4- a" 1 . 
 
 6. m 2 4-2m — l-2m- 1 4-m- 2 . 
 
 7. 49^-28x4-18^-4^4-1. 
 
CHAPTER XVII. 
 RADICAL EXPRESSIONS. 
 
 253. A radical expression is an expression affected by 
 the radical sign ; as Va, V5, w, Va~+~b, V32. 
 
 254. An indicated root that cannot be exactly obtained 
 is called a surd. An indicated root that can be exactly 
 obtained is said to have the form of a surd. 
 
 The required root shows the order of a surd ; and surds 
 are named quadratic, cubic, biquadratic, according as the 
 second, third, or fourth roots are required. 
 
 The product of a rational factor and a surd factor is 
 called a mixed surd; as 3V2, b^fit. The rational factor 
 of a mixed surd is called the coefficient of the surd. 
 
 When there is no rational factor outside of the radical, 
 sign, that is, when the coefficient is 1, the surd is said to be 
 entire; as V2, Va. 
 
 255. A surd is in its simplest form when the expression 
 under the radical sign is integral and as small as possible. 
 
 Surds are said to be similar if they have the same surd 
 factor when reduced to the simplest form. 
 
 Note. In operations with surds, arithmetical numbers contained 
 in the surds should be expressed in their prime factors. 
 
 Reduction of Radicals. 
 
 256. To reduce a radical is to change its form without 
 changing its value. 
 
RADICALS. 239 
 
 Case 1. 
 
 257. When the radical is a perfect power and has for an 
 exponent a factor of the index of the root. 
 
 1. ■fa~* = J = a* = Va. 
 
 2. -y/MtfP = ^(6 aby = (6 ah) 1 = (6 abf = V6~a~b. 
 
 3. ~Z/25 JW = ^(6a a *c 4 ) 2 = (5 a^c 4 ) 1 = (5 a 2 ^ 4 )* 
 
 = -v / 5^ 2 ^ 4 . 
 We have, therefore, the following rule : 
 Divide the exponent of the power by the index of the root. 
 
 
 Exercise 94. 
 
 Simplify : 
 
 
 l. ^25. 
 
 6. -yfiM*. 
 
 2. ^16. 
 
 7. -v^ 5 . 
 
 3. ^27. 
 
 8. -\/M\ 
 
 • 4. y/t&. 
 
 9. -f/27 a 8 b 6 . 
 
 5. -^64. 
 
 10. "^16a 4 6 4 . 
 
 11. 
 
 e / 25 a 2 
 ^646 2 ' 
 
 Af(as 
 
 12. 16 ^ 
 
 3) 5 
 
 Case 2. 
 
 ^ 
 
 13. ' ^ 
 
 8scy 
 
 258. When the radical is the product of two factors, one of 
 which is a perfect power of the same degree as the radical. 
 
 Since -\/aFb = Va n X\ / J = (fv / J(§ 224), we have 
 
 1. Va% = Va*xVb==aVb] 
 
 2. -^108 = ^27 X 4 = -^27 X -y/i = 3^4 ; 
 
240 RADICALS. 
 
 3. 4 V72 a 2 b* = 4 V36 a% 2 X2b = 4 V36 a% 2 X V2b 
 
 = 4X 6oiV2J = 24aiV26; 
 
 4. 2^54 a 4 6 = 2^27 a*x2ab = 2^/2T~a~ 8 X V^S ' 
 
 = 2 X Sa-^/2ao = 6a^/2aJ. 
 We have, therefore, the following rule : 
 
 Resolve the radical into two factors, one of which is the 
 greatest perfect power of the same degree as the radical. 
 
 Remove this factor from under the radical sign, extract 
 the required root, and multiply the coefficient of the surd by 
 the root obtained. 
 
 Exercise 95. 
 
 Simplify : 
 
 1. V28. 13. 7V144. 25 it Ux y - 
 
 r- . V27m 8 n* 
 
 2. V72. 14. 8VwA. 
 
 3. </72. 15. 3</bW. 26. ^p; 
 
 4. -v^OO. 16. 2^V. . 
 
 8j - 27 »/ mg 
 
 5. V432. 17. 11 -y/a 12 b 16 . ' \'216^S' 
 
 3/ 
 
 6. Vl92. 18. 7V8^. ^ 2 «/^ 
 
 7. -\/l28. 19. 6^27 mV. ^ 243 
 
 8. "^243. 20. 4-V^Sy. 29 *EZ.. 
 
 , 8/ VlMfi 
 
 9. V176. 21. V1029. 
 
 10. V405. 22. ^2187. 30. jj&l 
 
 11. 2^112. 23. "^1250. 
 
 12. 3^/S6l. 24. 4^648. 
 
 1296 
 "612 
 
 3 ah 
 31. 
 
 Sab I 20c 2 
 2c \9a 2 b 2 ' 
 
RADICALS. 
 
 241 
 
 Case 3. 
 
 259. When the radical expression is a fraction, the denomi- 
 nator of which is not a perfect power of the same degree as the 
 radical. 
 
 V| = V| = Viox^ = iVIo. 
 
 2 VA-V^-^-v^-*v* 
 
 3 - <§=^^^r 4 =4 
 
 5X3X4 
 
 27 X 8 
 1 3 
 
 60 X 
 
 27 X 8 
 
 3x2 
 
 60 = J V60. 
 
 We have, therefore, the following rule : 
 
 Multiply both terms of the fraction by a number that 
 will make the denominator a perfect power of the same 
 degree as the radical ; and then proceed as in Case 2. 
 
 
 Exercise 96. 
 
 
 
 
 Simplify : 
 
 
 
 
 
 i. 2Vf 
 
 4. 7VJ: 7. -v^f. 
 
 
 10. 
 
 2^|. 
 
 2. fVf. 
 
 4/ 3/- 
 
 5. Vff 8. Vf. 
 
 
 11. 
 
 3^A- 
 
 3- iVj. 
 
 6. 3V^. 9. %/fc 
 
 
 12. 
 
 2^£ 
 
 i3 -Vf 
 
 15. \S£t . 17. 
 
 4 
 
 > 2 cy 2 
 6d 2 
 
 
 14. &. 
 
 16. J/JS-. 18 . 
 
 21 
 
 /2 a 8 6 2 c 
 
 
 'ys e 
 
242 RADICALS. 
 
 Case 4. 
 
 260. To reduce a mixed surd to an entire surd. 
 
 Since a^/b = Va" X Vd = "VaFb, we have 
 
 1. 3VS = V32~x~5 = V9~X~5 = V45; 
 
 2. a% Vbc" = V(a 2 6) 2 X be = s/aW X be = VaWc ; 
 
 3. 2a-v / cciy=:^(2a;) 8 X ^ = "V / 8cc 8 X xy = J/S x*y ; 
 
 4. 3 y 2 ^ = *t/(3 y 2 Y X x 8 = ^81 y 8 ^ 8 . 
 We have, therefore, the following rule : 
 
 liaise the coefficient to a power of the same degree as the 
 radical, multiply this power by the given surd factory and. 
 indicate the required root of the product. f 
 
 Exercise 97. 
 Express as an entire surd : 
 
 1. 5VE. 5. 2^3. 9. -2Vy. 13. iVa. 
 
 2. 3Vll. 6. 3^2. 10. -3VsA 14. -JVa 2 . 
 
 3. 3^3. 7. 2-Z/2. 11. -m-v^O. 15. |V»T 8 . 
 
 4. 2^4. 8. 2^4. 12. -2-s/x. 16. - i^/m\ 
 
 Case 5. 
 
 261. To reduce radicals to a common index. 
 
 Keduce V2 and V3 to a common index. 
 
 V2 = 2* = 2 8 = v^» = V8. 
 
 V3 = 3* = 38 = V32 = V9. 
 We have, therefore, the following rule : 
 
RADICALS. 243 
 
 Write the radicals with fractional exponents, and change 
 these fractional exponents to equivalent exponents having the 
 least common denominator. Raise each radical to the power 
 denoted by the numerator, and indicate the root denoted by 
 the common denominator. 
 
 Exercise 98. 
 Eeduce to surds of the same order : 
 
 I. 
 
 ■^3 and y/B. 
 
 7. 
 
 V2, ^3, and yfc . 
 
 2. 
 
 vTIand V6. 
 
 8. 
 
 Va 2 , Vft, and Vc. 
 
 3. 
 
 V2 and y/Z. 
 
 9. 
 
 Vo*, Vc®, and Vcc 8 . 
 
 4. 
 
 Va- and V#~ 2 . 
 
 10. 
 
 y/x 2 y, Vak, and "v2». 
 
 5. 
 
 V5 and VfE. 
 
 11. 
 
 6/ 4/ 
 
 Vsc — y and Vsc + y. 
 
 3. 
 
 2* 2 f , and 2* 
 
 12. 
 
 3/ I 
 
 Va + b and V a — 6. 
 
 Note. Surds of different orders may be reduced to surds of the 
 same order and then compared in respect to magnitude. 
 
 Arrange in order of magnitude : 
 
 13. -v^lS and V6. 15. V^O, V^, and V& 
 
 14. yfl and V§. 16. V3, y%, and yft. 
 
 Addition and Subtraction of Radicals. 
 
 262. In the addition of surds, each surd must be reduced 
 to its simplest form ; and, if the resulting surds are similar, 
 
 Find the algebraic sum of the coefficients, and to this sum 
 annex the common surd factor. 
 
 If the resulting surds are not similar, 
 
 Connect them with their proper signs. 
 
244 RADICALS. 
 
 1. Simplify V27 + V48 + Vl47. 
 
 V27 = (32 x 3)* = 3 X 3* = 3 V3; 
 
 V48 = (2* X 3)* = 2 2 X 3* = 4 X 3* = 4 V3; 
 
 Vl47 = (7 2 X 3)* = 7 X 3* = 7 V3. 
 
 V27 + V48 + Vl47 = (3 + 4 + 7) V3 = 14 V3. 
 
 2. Simplify 2^320 -3 -^40. 
 
 2V320 = 2(2° X 5)* = 2 X 2 2 X 5* = sVS; 
 3V4O = 3(2 8 X 5)* = 3 X 2 X 5* = 6 VS. 
 .-. 2 V320 — 3 ViO = (8 - 6) V6 = 2 VS 
 
 3. Simplify 2V£- 3 Vf + V^. 
 
 2 Vf = 2 V^ = 2V15 X I = f Vl5; 
 3V| = 3 V|| = 3 V15 X £ = I Vl5; . 
 
 .•.2V|-3V|+VS=(^-3 +T 2 5) Vl6 = |Vl6. 
 
 Exercise 99. 
 Simplify : 
 
 1. 4VlT + 3VTT-5VlT. 
 
 2. 2V3 -5V3 + 9V3. 
 
 3. 5"V / 4 + 2^32 --5^108. 7. V27 + V48 + V75. 
 
 4. 3^2 + 4-^2— "^64. 8. 4Vl47-f 3V75 -f VT92. 
 
 5. ^-5 / 5 + 2^-v / 5 + iV / 40. 9. V^ + |V^-ffV^. 
 
 6. 3^5 — 5^+ -^243. 10. -V^ + i^-S-J/^T^. 
 
RADICALS. 
 
 24:5 
 
 11. Va* + bVa - 3Va. 
 
 12. \ r 25b + 2V9b-3VIb. 
 
 13. 2VI75-3V63 + 5V28. 
 
 14. V2 + 3V32 + ^Vl28-6Vl8. 
 
 15. V75 + Vi8 - Vl47 + V300. 
 
 16. 20V245- V5 + Vl25-2^Vl80. 
 
 17. 2V20 + ^Vl2-2V27 4-5V45-9Vl2. 
 
 18. 7V25 + 4V45- V9-2V80 + V20-4VSI. 
 
 19. -#54 + V£- V^-fVf. 
 
 20. 2V| + V60- Vl5 + Vf + V^. 
 
 21. ■v / 27V - -#8^ + -^125^. 
 
 22. Vo^-V^+^32J. 
 
 23. Va 4 ^ + V^ - V4a%. 
 
 24. V4 ic 2 2/ 2 ^ + Vy 4 ^ 4- Va^s. 
 
 25. VaWc - aVIc + bVa?c. 
 
 26. -#81 a 5 - ^16^ + ^256 a 5 . 
 
 27. "#27 m 4 - -#125 m + "#216 m. 
 
 28. V8^ - V50 a 8 - 3 VlSU. 
 
 29. 6 a V63 a& 8 - 3 V112 a% z + 2ab V343 a*. 
 
 30. 3 V125 mV + rc V20 m 8 - V500 m 8 ^ 2 . 
 
 31. V32 a 4 6 5 -f 6 Vm + 3 V128 a?b\ 
 
 32. 2-J^ - 3a 2 ^64l -f5ftvS + 2a 2 ^1255. 
 
246 RADICALS. 
 
 Multiplication of Radicals. 
 263. Since Va X Vb = Vab, we have 
 
 1. 3V8 X 5V2 = 3 X5 X V8 X V2 = 15 Vl6 = 60 ; 
 
 2. 3V2 X 4-^ = 3 vT 8 X 4^=12^72. 
 
 We have, therefore, the following rule : 
 
 Express the radicals with a common index. Find the 
 product of the coefficients for the required coefficient, and the 
 product of the surd factors for the required surd factor. 
 
 Reduce the result to its simplest form. 
 
 Exercise 100. 
 Find the product of : 
 
 1. V3 X V27. 7. vlxv^. 13. -^54 X V& 
 
 2. V5xV20. 8. ^27X^9. 14. 2 V8 X VS. 
 
 3. VS X Vl8. 9. VS X Vl2. 15. a/8 X V^4. 
 
 4. V3 X -vft 10. V3 X VS. 16. ^7 X V-49. 
 
 5. ^2x^32. 11. V^X-v^. 17. ^/SlX^Id. 
 
 6. ^27X^3. 12. ^6x^8. 18. §^18X1^3. 
 
 19. (V18 + 2V72-3V8) X VS. 
 
 20. (V^2-^V864 + 3^4) X $2. 
 
 21. (J V27 - J V2187 + } V432) X V3. 
 
 22. VH X y/t. 25. V3 X -\/72. 28. V81 X V& 
 
 23. VT6 X V250. 26. Vf X -v^. 29. V^ X Vf . 
 
 24. V"64 X ^lfr 27. V^ X V^. 30. Vf X Vj. 
 
RADICALS. 247 
 
 264. Compound radicals are multiplied as follows : 
 
 Multiply 2 V3 + 3 Vx by 3 V3 - 4 Vac. 
 
 2V3 + 3VX 
 3V3-4Vx 
 
 18 + 9V3x 
 
 — 8Vsx — 12* 
 
 18+ V3x-12x 
 Exercise 101. 
 
 Multiply 
 
 1. V 5 + ^* b y V 5 _ ^ 4 - 8 + 3V2by 2- V2. 
 
 2. V 9 - ^ h Y V 9 + ' v/ ^- 5 - 5 + 2V3by3-5V3. 
 
 3. 3 + 2V5 by 2 - VS. 6. 3 - V6 by 6 - 3 Vo\ 
 
 7. 2V6-3V5 by V3 + 2V2. 
 
 8. 7 - V3 by V2 + V5. 
 
 9. V / 9-2v / 4by 4-^3+^2. 
 
 LO. 2V30-3V5 + 5V3 by V8 + V3- VB. 
 
 .1. 3V5-2V3 + 4V7 by 3V7-4VS-5V3. 
 
 L2. 4 V8 + i Vl2 - i V32 by 8 V32 - 4V50 - 2 V2. 
 
 L3. ^6-^3 + ^16 by ^36 + ^9-^4. 
 
 L4. 2Vf-8Vf + 3V| by 3Vf- Vl2- V6. 
 
 L5. 2V|-4V|-7Vf by 3V|-5V30-2VY. 
 
 L6. 2Vl2 + 3V3 + 6Vi by 2Vl2 + 3V3 + 6V£. 
 
248 RADICALS 
 
 Division of Radicals. 
 
 -Vab Va X -Vb nr- 
 265. Since — — = = Wb, we nave 
 
 2V2 
 2 4^3 = 4^3* = 4^3 2 x 2 3 _ t^ 
 
 2V2 2W 2-v^ 
 
 We have, therefore, the following rule : 
 
 Express the radicals with a common index. Find the 
 
 quotient of the coefficients for the required coefficient, and the 
 
 quotient of the surd factors for the required surd factor. 
 
 Reduce the result to its simplest form. 
 
 Exercise 102. 
 Divide : 
 
 1. V243 by VS. 4. V£ by Vf 7. V}$ by Vffj 
 
 2. -v^T by VS. 5. V# by V§. 8. Vjf by V^U 
 
 3. VFa 7 by Va 5 . 6. V^ by Vf 9. Vjf by V$£, 
 
 10. 3V6 + 45V2 by 3V3. 
 
 11. 42V5-30V3 by 2Vl5. 
 
 12. 84V15 + 168V6 by 3V21. 
 
 13. 30-5^4 — 36^10 + 30^90 by 3V^0. 
 
 14. 50VT8 + 184 / 20-48^ / 5 by 2^30. 
 
 15. V54 by "^36. 17. VT2 by Vo*. 19. Vf by ^5|. 
 
 16. -V^49 by sfl. 18. V^ by "V^. 20. V^a- by Vx*. 
 21. VU064 by VlO. 22. V» a - y 3 by x + y r 
 
RADICALS. 249 
 
 The quotient of one surd divided by another may 
 be found by rationalizing the divisor ; that is, by multiply- 
 ing the dividend and divisor by a factor that will free the 
 divisor from surds. This method is of great utility when 
 we wish to find the approximate numerical value of the 
 quotient of two simple surds, and is the method required 
 when the divisor is a compound surd. 
 
 1. Divide 3 V8 by V5. 
 
 3V8 6V2_ 6V2 X V6_ 6Vl2 r - _ 
 
 2. Divide 3VS - 4 V2 by 2V5 + 3V2. 
 
 Multiply the dividend and the divisor by 2V5 — 3 V2, 
 
 (3V5 - 4 V2) (2 V5 - 3V2) _ 54 - 17 VlO 
 (2V6 + 3V2)(2V6-3V2) * 20-18 
 
 64-17V10 n „ 17 rrz „ 
 = - = 27 — V- v 10. Hence, 
 
 267. When the divisor is a binomial containing surds of the 
 second order only, 
 
 Multiply the dividend and the divisor by the divisor, with 
 he sign between the terms changed. 
 
 Exercise 103. 
 Divide : 
 
 Va + V& by Vab. 7. 3 + 5 Vf by 3 - 5V7. 
 
 a. Vl25by5V65. 8. 21 V3 by 4a/3 - 3V2. 
 
 3. 3 by 11 + 3 V7. 9. 75VII by 8V2 + 2Vt. 
 
 4. 3V2-lby3V2 + l. 10. V5 - Vs by VE + V3. 
 6. 17 by 3 V7 + 2 V§. 11. V8 + Vf by V7 - V2. 
 6, lby\/2 + V3. 12. 7-3Vl0by 5 + 4V5. 
 
250 RADICALS. 
 
 Given V2 = 1.41421, V§ = 1.73205, V5 = 2.23607; 
 
 find to four places of decimals the value of : 
 
 13 . i° 16 . i . 19 . i . 22 . 7-3VB . 
 
 V2 V500 3V2 5 + 4V5 
 
 l4 . 4. 17 ; 1 . 20 . 1 . 23 . iws. 
 
 V3 V243 V125 V5-2 
 
 - c 12 ,0 l 1 « 3V2-1 
 
 15. -t=« 18. =• 21. =*• 24. = 
 
 V5 2V3 4V5 3V2 + 1 
 
 Involution and Evolution of Radicals. 
 
 268. Any power or root of a radical is easily found by 
 using fractional exponents. 
 
 1. Find the square of 2 Va. 
 
 (2 Va) 2 = (2 a*) 2 = 2 2 a s = 4 a* = 4 Va 2 . 
 
 2. Find the cube of 2 Va. 
 
 (2 Va) 8 = (2 a*) 8 = 2 3 a* = 8 a* = 8 a Va. 
 
 3. Find the square root of 4 a? ~Va 8 b s . 
 
 (4 a Vo 3 ^)* = (4 xa*&$)* = 4* xM&* = 4* xMfc* = 2 V^x 2 . 
 
 4. Find the cube root of 4ccVa 8 & 8 . 
 
 (4x Va 8 ^)* = (4 xo*6*)* = 4Ma*6* = **a&W = Vl6 a^x 2 . 
 
 Exercise 104. 
 Perform the operations indicated : 
 1. (^T 2 ) 8 . 3. ("Vx*) 15 . 5. VV(x-y) 8 . 
 
 2. (Vm 5 ) 6 . 4. (Vy*) 12 . 6. V-V 7 ^-^) 80 . 
 
RADICALS. 251 
 
 7. (V2^b)\ 10. V^. 13. V^(3a-2^. 
 
 8 . (^2 _ y y, 11# V-^729. 14. V^32a 46 . 
 
 9. (Vz) 4 . 12. VV125. 15. Vl28v'243a^. 
 
 Properties of Quadratic Surds. 
 
 269. A quadratic surd is the indicated root of an imperfect 
 
 square, as V2. 
 
 270. Theorem 1. The product or quotient of two dis- 
 similar quadratic surds will be a quadratic surd. 
 
 Thus, Vab X -Vatic = ab ypj 
 
 ^s/abc -*- -\fa~b = Vc. 
 
 Two dissimilar quadratic surds cannot have all the fac- 
 tors under the radical sign alike. Hence, their product or 
 quotient will contain the first power only of at least one 
 factor, and will therefore be a surd. 
 
 271. Theorem 2. The sum or difference of two dis- 
 similar quadratic surds cannot be a rational number, nor 
 can the sum or difference be expressed as a single surd. 
 
 For, if Va ± V# could equal a rational number c, we 
 should have, by squaring, 
 
 a±2Vab + b = c*; 
 that is, ± 2 ^fab = c 2 — a — b. 
 
 Now, as the right side of this equation, is rational, the 
 left side would be rational ; but, by § 270, ~va~b cannot be 
 rational. Therefore, Va ± V# cannot be rational. 
 
 In like manner it may be shown that Va ± V# cannot 
 be expressed as a single surd Vc. 
 
252 RADICALS. 
 
 272. Theorem 3. A quadratic surd cannot equal the 
 sum of a rational number and a surd. 
 
 For, if Va could equal c -f V#, we should have, by 
 squaring, 
 
 a = c* + 2cVb + b, 
 and, by transposing, 
 
 2cVb = a-b-c 2 . 
 
 That is, a surd equal to a rational number, which is 
 impossible. 
 
 273. Theorem 4. If a + Vb = x + Vy, then a will 
 equal x, and b will equal y. 
 
 For, by transposing, V& — Vy = x — a ; and if b were 
 not equal to y, the difference of two unequal surds would 
 be rational, which by § 271 is impossible. 
 
 .-. b = y, and a = x. 
 
 In like manner, if a — "Vo = x — Vy, a will equal x, 
 and b will equal y. 
 
 274. Theorem 5. ijf \a + V£ = vS 4- Vy, **«» 
 
 Y« — V# = Va! — Vy. 
 
 Square both sides of the .given equation, 
 
 a + Vb = x + 2 Vxy + y. 
 
 Therefore, by § 273, a = x + y, (1) 
 
 and V6 = 2VSy. (2) 
 
 Subtract (2) from (1), 
 
 a — Vft = x — 2 Vxy + y, 
 
 Extract the square root of both sides, 
 
 V a — V& e= Vz — Vy. 
 
RADICALS. 253 
 
 275. To Extract the Square Root of a Binomial Surd. 
 
 1. Extract the square root of 7 + 4 Vo. 
 
 Let Vx + y/y = V7 + 4V3. (1) 
 
 Then, by § 274, Vi-Vy = V7-4V3. (2) 
 
 Multiply (1) by (2), x - y = V49 - 48. 
 
 .\x — y = l. 
 Square (1), then § 273, x + y = 7. 
 
 .-. x = 4, and y = 3. 
 .-. Vx + Vy = Vi -£ V3. 
 .-. V7 + 4V3 = 2 + V3. 
 
 A root may be found by inspection, when the given expression can 
 be written in the form a + 2 V6, by finding two numbers that have 
 their sum equal to a and their product equal to 6. 
 
 2. Find by inspection the square root of 75 — 12 V21. 
 
 It is necessary that the coefficient of the surd b e 2 ; therefore, 
 75 — 12 V21 must be put in the form 75 — 2V6 2 X 21 ; 
 
 that is, 75 — 2V756. 
 
 Two numbers whose sum is 75 and product 756 are 63 and 12. 
 
 Then, 75 — 2 V756 = 63 — 2 V63 X 12 + 12 
 
 = (V63— Vl2)2. 
 That is, V63 — Vl2 = the square root of 75 — 12 V2T ; 
 
 or 3 V7 — 2 V3 = the square root of 75 — 12 V2I. 
 
 3. Extract the square root of 11 + 6 V2. 
 
 11 + 6V2 = ll + 2Vl8. 
 Two numbers whose sum is 11 and product 18 are 9 and 2. 
 
 'Then, 11 + 2 Vl8 = 9 + 2 V9 X 2 + 2 
 
 = (V9 + V2) 2 . 
 That is, V9 + V2 = the square root of 11 + 6 V5; 
 or 3 + V2 = the square root of 11 + 6 V2. 
 
254 RADICALS. 
 
 Exercise 105. 
 Find the square root of : 
 
 1. 7-4V3. 7. 16 + 5V7. 13. 94 + 42V5. 
 
 2. 11 + V72. 8. 75 + 12V2I. 14. 11-2V30. 
 
 3. 7 + 2ViO. 9. 19 + 8V3. 15. 47-4V33. 
 
 4. 18 + 8V5. 10. 8V6 + 20. 16. 29 + 6V22. 
 
 5. 8+2VI5. 11. 28-16V3. 17. 83 + 12V35. 
 
 * 
 
 6. 15-4VH 12. 51+36V2. 18. 55- 12 V2l. 
 
 Equations Containing Radicals. 
 
 276. An equation containing a single radical may be 
 solved by arranging the terms so as to have the radical 
 alone on one side, and then raising both sides to a power 
 corresponding to the order of the radical. 
 
 Solve V# 2 -9 + x = 9. 
 
 Transpose x, Vx 2 — 9 = 9 — x. 
 
 Square, x 2 — 9 = 81 — 18 x + aft 
 
 18s = 90. 
 
 .-. x = 5. 
 
 277. If two radicals are involved, two steps may be 
 necessary. 
 
 Solve Va; + 15 + Vx = 15. 
 
 Square, x + 15 + 2 Vx 2 + 15 x + x = 225. 
 
 Transpose, 2 Vx 2 + 15 x = 210 — 2 x. 
 
 Divide by 2, Vx 2 +15x = 105 - x. 
 
 Square, x 2 + 15 x = 11026 — 210 x + aft 
 
 225 x = 11025. 
 \ x = 49. 
 
RADICALS. 
 
 255 
 
 Exercise 106. 
 Solve: 
 
 1. 2^x + 5 = V28. 8. -^33 + 7 = 3. 
 
 2. 3V4as~8 = Vl3a!-a 9. 14 + -^4 a; - 40 = 10. 
 
 3. Vz + 9 = 5Vo; -3. 
 
 4. 4 = 2Va--3. 
 
 5. 5-V3^ = 4. 
 
 6. 7 + 2^3^ = 5. 
 
 7. ^/2 x - 3 = - 3. 
 
 10. Vl0y-4 = V7y + ll. 
 
 11. 2V*-2 = V32(a:-2)«. 
 
 12. V^4 + x = | + V» 
 
 13. V32 + a; = 16 - VS. 
 
 14. VS — Vaj — 5 = V5. 
 15. V» + 20 - -yJx - 1 - 3 = 0. 
 
 16. Vx + 15 - 7 = 7 - Va; - 13. 
 
 17. x = 7 - Va 2 - 7. 
 
 19. 
 
 18. Va-7 = Vaj + 1 
 
 Va- — 3 VS + 1 
 
 Va- + 3 "" Vz - 2 
 
 2i i + q-*)* 
 
 1 - (1 - x)* 
 
 20.,- 
 
 |-:->IR^»*^-s 
 
 3)* 
 
 23. \/a + VS + ya — Va = VS. 
 
 24. Vox — 1=4 + £ Vox — -J. 
 
 25. 3 Vie — 3Va= VS — Va + 2Va. 
 
 26. V9 + 2a;- V2x = —j= 
 
 V9 + 2a 
 
CHAPTER XVm. 
 IMAGINARY EXPRESSIONS. 
 
 278. An imaginary expression is any expression which 
 involves the indicated even root of a negative number. 
 
 It will be shown hereafter that any indicated even root 
 of a negative number may be made to assume a form which 
 involves only an indicated square root of a negative num- 
 ber. In considering imaginary expressions we accordingly 
 need consider only expressions which involve the indicated 
 square roots of negative numbers. 
 
 Imaginary expressions are also called imaginary numbers 
 and complex numbers. In distinction from imaginary num- 
 bers all other numbers are called real numbers. 
 
 279. Imaginary Square Roots. If a and b are both posi- 
 tive, we have 
 
 Vab = Va X Vb. 
 If one of the two numbers a and b is positive and the 
 other negative, it is assumed that the law still holds true ; 
 we have, accordingly : 
 
 V^l = V4 (- 1) = VI X 
 
 V^5 = V5 (- 1) = VE x 
 
 V^a = Va (— 1) = Va X 
 and so on. 
 
 It appears, then, that every imaginary square root can 
 be made to assume the form a V — 1, where a is a real 
 number. 
 
IMAGINARY EXPRESSIONS. 
 
 257 
 
 280. The symbol V— 1 is called the imaginary unit, and 
 may be defined as an expression the square of which is — 1. 
 
 Hence, V^l X V^ = ( V^) 2 = - 1 ; 
 
 V^a X V^~b = Va X V^l X Vb X V^l 
 
 = -\fab X (- 1) 
 = - Vab. 
 
 281. It will be useful to form the successive powers of 
 the imaginary unit. 
 
 (V^l) = + V=l; 
 
 (V^) 2 =-1; 
 
 ( V^i) 8 = ( V^l) 2 V^I = (- 1) V^i = - V^i ; 
 
 (V^l)« = ( V^l) 2 ( V^T) 2 = (- 1) (- 1) = + 1 ; 
 
 (V^i) 5 = (V^T) 4 V^i = (+ i) V=i = + vpij 
 
 and so on. If, therefore, n is zero or a positive integer, 
 
 (V^~l) 4n+1 = + ps/SI'j 
 
 (V^T) 4n + 2 =-l; 
 (V^l) 4n+3 =- V^; 
 
 (V^~i) 4w+4 = + i. 
 
 Every imaginary expression may be made to assume 
 the form a + b V— 1, where a and b are real numbers, and 
 may be integers, fractions, or surds. 
 
 If b = 0, the expression consists of only the real part a, 
 and is therefore real. 
 
 If a = 0, the expression consists of only the imaginary 
 part b V— 1, and is called a pure imaginary. 
 
258 IMAGINARY EXPRESSIONS. 
 
 283. The form a -f b V— 1 is the typical form of imaginary- 
 expressions. 
 
 Reduce to the typical form 6 + V— 8. 
 This may be written 6 J- Vs X V— 1, or 6 + 2 V2 X V^T ; 
 here a = 6, and b = 2 V2. 
 
 284. Two expressions of the form a •+■ b V— 1, a — b V— 1, 
 are called conjugate imaginaries. 
 
 To find the sum and product of two conjugate imagi- 
 naries : 
 
 The sum is 
 
 a + W-l 
 a — &V-1 
 
 2a 
 
 a -b V^l 
 
 a 2 + a&V^l 
 
 The product is a 2 + b 2 
 
 From the above it appears that the sum and product of 
 two conjugate imaginaries are both real. 
 
 285. Theorem 1. An imaginary expression cannot be 
 equal to a real number. 
 
 For, if possible, let 
 
 a ■+- b V— 1 = c. 
 Transpose a, #V— 1 = c — a. 
 
 Square, — b 2 = (c — a)'' 
 
 J 
 
 Since J 2 and (c — a) 2 are both positive, we have a nega- 
 tive number equal to a positive number, which is impossible. 
 
IMAGINARY EXPRESSIONS. 259 
 
 286. Theorem 2. If two imaginary expressions are 
 equal, the real parts are equal and the imaginary parts are 
 equal. 
 
 For, let a + b V^T = c -f dV^T. 
 
 Then, (b - d)V^l = c - a; 
 
 Square, — (b — d) 2 = (c — a) 2 , 
 
 which is impossible unless & = d and a = c. 
 
 287. Theorem 3. Ifxandyarerealandx + yw — l — O^ 
 then x = awcZ y = 0. 
 
 For, y V— 1 = — sc. 
 
 Square, — y 2 = ic 2 . 
 
 Transpose — y 2 , sc 2 + y 2 = 0, 
 
 which is true only when x = and y = 0. 
 
 Operations with Imaginaries. 
 
 1. Add 5 + 7 VS-1 and 8 - 9 V^l. 
 
 The sum is 5 + 8 + 7 V 1 ^! — 9V^1, 
 
 13 — 2V :r T. 
 
 2. Multiply 3 + 2 V^l by 5- 4V =: 1. 
 (3 + 2 V^T) (5 - 4 vpl) 
 = 15 — 12 V^~l + 10 V^T — 8 (— 1) 
 = 23 — 2V^T. 
 
 3. Divide 14 + 5 V^l by 2 - 3 V^l. 
 
 14 + 5 V^T (14 + 5 V^\) (2 + 3 V^T) 
 2-3V^T ~~ (2-3V^T)(2 + 3V^l) 
 
 _13_+52_V^T 
 4-(-9) 
 
 _ 13 + 52V^1 
 13 
 
 = 1 + 4V =: T, 
 
260 
 
 IMAGINARY EXPRESSIONS. 
 
 Exekcise 107. 
 Eeduce to the form b V— 1 : 
 
 i. V^. 
 
 2. V-16. 
 
 9. V-625. 
 10. V-36. 
 
 17. V- x 18 . 
 
 18. V^±. 
 
 3. V-25. 
 
 11. V-64. 
 
 19. V- a 4 b~ 2 . 
 
 4. V- 144. 12. V- 729. 
 
 20. V-9x 4 . 
 
 5. V-169. 13. V- 289. 21. V-(2a?-3 
 
 6. V^ 
 
 7. V-81. 
 
 8. V- 256. 
 Add: 
 
 14. 7-1024. 
 
 15. V— a; 8 . 
 
 16. V-a 9 . 
 
 22. V-(«-2,y) 2 
 
 23. V- (x 2 + y 2 ). 
 
 24. V- (x 2 - y 2 ). 
 
 25. V- 25 + V- 49 - V- 121. 
 
 26. V- 64 + V^ - V- 36. 
 
 27. V— a 4 -f- V-4a 4 -f V— 16 a 4 . 
 
 28. V^a 2 + V- 81 a 2 - V^ 
 
 29. a-bV^ + a + bV^l. 
 
 30. 2 + 3V =: l-2 + 3V^l. 
 
 31. a + bV^l + c-dV^. 
 
 32. SaV^ — (2a- ^V^. 
 Multiply : 
 
 33. V^ by V^. 
 
 34. -V^by V^. 37. V^ 2 by V^^y 5 . 
 
 35. V-16 by V^9. 38. V^ by V-16. 
 
 36. V-ic 2 by V^" 
 
IMAGINARY EXPRESSIONS. 
 
 261 
 
 39. V- 25 by V-64. 41. 3V^3 by 2-^2. 
 
 40. V- (a + b) by V- (a - b). 42. - 5V^2 by 2 VhS. 
 
 43. V^ + V^ by V^ - V^5. 
 
 1 + V-^3 _j 1 - V^3 
 
 44. jt by x g 
 
 45. a V— a + b V— # by a V— a — J V— 6. 
 
 46. 2V^2 + 3V^ by S-^l — 2V^~5. 
 
 47. -y/3 + 2^3 by V3-2V^3. 
 
 48. m — 3 V— 6 by w + 4 V— c. 
 
 Perform the divisions indicated : 
 
 49. 
 
 a 
 
 50. 
 
 b 
 
 V-fc 2 
 
 
 
 51. 
 
 G 
 
 V=4 
 
 52. 
 
 V=9 
 
 V-81 
 
 53. 
 54. 
 
 V=6 
 
 V— ax 
 
 55. 
 
 56. 
 
 57. 
 
 58. 
 
 59. 
 
 60. 
 
 V-a? 
 
 V- 
 
 V-lOx 8 
 
 V— 5cc 
 
 8V-a; 2 
 
 2Vx 
 
 61. 
 
 1 
 
 3- V^2 
 
 62. 
 
 2 + V^2 
 
 i'-V=I 
 
 63. 
 
 a+x V— 1 
 
 a — ccV— 1 
 
 64. 
 
 VB + V^6 
 
 V6- V^~8 
 
 CK 
 
 2a + 36V^l 
 
 66. 
 
 2a-3&V^T 
 4a-£&V^l 
 
CHAPTER XIX. 
 QUADRATIC EQUATIONS. 
 
 288. We have already considered equations of the first 
 degree in one or more unknowns. We pass now to the 
 treatment of equations containing one or more unknowns 
 to a degree not exceeding the second. An equation which 
 contains the square of the unknown, but no higher power, 
 is called a quadratic equation. 
 
 289. A quadratic equation which involves but one un- 
 known number can contain only : 
 
 1. Terms involving the square of the unknown number. 
 
 2. Terms involving the first power of the unknown 
 number. 
 
 3. Terms which do not involve the unknown number. 
 
 Collecting similar terms, every quadratic equation can be 
 made to assume the form 
 
 ax 2 -f- bx -f- c = 0, 
 
 where a, b, and c are known numbers, and x the unknown 
 number. 
 
 If a, b, c are numbers expressed by figures, the equation 
 is a numerical quadratic. If a, b, c are numbers represented i 
 wholly or in part by letters, the equation is a literal quadratic. 
 
 In the equation ax 2 -f- bx -+■ c = 0, a, b, and c are called 
 the coefficients of the equation. The third term c is called 
 the constant term. 
 
 
QUADRATIC EQUATIONS. 263 
 
 290. If the first power of x is wanting, the equation is a 
 pure quadratic ; in this case b = 0. 
 
 If the first power of x is present, the equation is an 
 affected or complete quadratic. 
 
 Pure Quadratic Equations. 
 
 1. Solve the equation 
 
 5x* 
 
 -48 
 
 = 2x*. 
 
 Collect the terms, 
 
 
 3x2 = 
 
 48. 
 
 Divide by 3, 
 
 
 X* = 
 
 16. 
 
 Extract the square root, 
 
 
 X = 
 
 ±4. 
 
 It will be observed that there are two roots, and that these are 
 numerically equal, but of opposite signs. There can be only two 
 roots, since any number has only two square roots. 
 
 It may seem as though we ought to write the sign ± before the x 
 as well as before the 4. If we do this, we have + cc = + 4, — x = — 4, 
 + x = — 4, — x = + 4. 
 
 From the first and second equations, x = 4 ; from the third and 
 fourth, x = — 4 ; these values of x are both given by the equation 
 x = ± 4. Hence it is unnecessary to write the ± sign on both sides of 
 the reduced equation. 
 
 2. Solve the equation 3 x 2 — 15 = 0. 
 
 Transpose, 3 x 2 = 15. 
 
 Divide by 3, a; 2 = 5. 
 
 Extract the square root, x = ± V5. 
 
 The roots cannot be found exactly, since the square root of 5 can- 
 not be found exactly ; it can, however, be determined approximately 
 any required degree of accuracy ; for example, the roots lie between 
 .23606 and 2.23607 ; and between — 2.23606 and — 2.23607. 
 
 3. Solve the equation 3 x 2 + 15 = 0. 
 
 Transpose, 
 
 Divide by 3, 
 
 Extract the square root, 
 
264 QUADRATIC EQUATIONS. 
 
 There is no square root of a negative number, since the square of 
 any number, positive or negative, is necessarily positive. 
 
 The square root of — 5 differs from the square root of + 5 in that 
 the latter can be found as accurately as we please, while the former 
 cannot be found at all. 
 
 291. A root that can be found exactly is called an exact 
 or rational root. Such roots are either whole numbers or 
 fractions. 
 
 A root that is indicated but can be found only approxi- 
 mately is called a surd. Such roots involve the roots of 
 imperfect powers. 
 
 Eational and surd roots are together called real roots. 
 
 A root that is indicated but cannot be found, either 
 exactly or approximately, is called an imaginary root. Such 
 roots involve the even roots of negative numbers. 
 
 Exercise 108. 
 Solve : 
 
 1. 3x 2 -2 = x 2 + 6. n 3 - x 2 , x 2 + 5 
 
 9. — — I 7, — — o. 
 
 2. 5a 2 -r-l0 = 6z 2 + l. 
 
 11 ' 6 
 
 3. 7 x 2 - 50 = Ax 2 + 25. 5x 2 + 3 17 - x 2 
 
 1 11 84 
 
 4. 6x 2 — - = 4ic 2 + tt' 
 
 6 9 3 17 
 
 „ x 2 + l , A U : 4z 2 6z 2- 3* 
 
 5. — - — = 10. 
 
 5 
 
 5 
 
 6. 3g;2 ~ 8 = 4. 12 ' 3z 2 5x 2 15 
 
 10 
 x 2 -9 _ x 2 + l i3. 
 
 1 
 
 7- — 4— = — 5 ' x-1 x + 1 4 
 
 Q 2^ 2 -4 , o: 2 + 4 15'- 7 _ n 
 
 8. — = 8. 14. h 7: r— = Z. 
 
 7 5 8 — x 2 — 3x 
 
QUADRATIC EQUATIONS. 265 
 
 15. 3ft 2 4-llft = 10ft4-84-ft 2 4-a. 
 
 16. (ft + 4) (x + 5) = 3 (x + 1) (x + 2) - 4. 
 
 17. 3 (a - 2) (ft + 3) = (ft + 1) (ft + 2) -h a; 2 + 5. 
 
 18. (2 ft + 1) (Sx - 2) + (1 - x) (3 + 4ft) = 3 ft 2 - 15. 
 ft 2 + 9 2x*-5 , 3x* + 10 ,. 
 
 3 ft 2 -5 , 2a; 2 + 4 ft 2 - 3 M 
 20. —^- + —9 J—* 
 
 10ft 2 + 7 12a; 2 + 2 _ 5a: 2 - 9 
 r 18 lift 2 - 8 ~ 9 
 
 nn x — 1 , ft + 1 5 „ a , ft 9 a 2 — ft 2 
 
 22. — — r + r = o- 26. - + - = 
 
 ft + 1 ft — 12 ft a aa; 
 
 23. aft 2 + & = c. oiy fl + a , ft — a _5 
 
 ft — a ft + a J 
 
 24. aft 2 + ^ = ^ 2 + a. 
 
 2ft 5* + 2ft _ 
 
 25. ft 2 + 2to + c = &(2ft + l). ft-£> 3ft 
 
 29. 2|(ft + a) (ft + 5) + (* - a) (ft - 5)j = a 2 4- 4 J 2 . 
 
 30. 2 ^ (ft - a) (ft 4- &) + (ft 4- a) (x - b) \ = 9 a 2 4- 2 ab + b\ 
 
 Affected Quadratic Equations. 
 
 292. Since (x ± b) 2 = ft 2 =fc 2 Jft 4- J 2 , it is evident that 
 the expression ft 2 ±2 foe lacks only the third term, b 2 , of 
 being a perfect square. 
 
 This third term is the square of half the coefficient of x. 
 
 Every affected quadratic may be made to assume the 
 form a; 2 ± 2 bx = c, by dividing the equation through by the 
 coefficient of ft 2 . 
 
 I 
 
266 QUADRATIC EQUATIONS. 
 
 293. To solve such an equation : 
 
 The first step is to add to both members the square of 
 half the coefficient ofx. This is called completing the square. 
 
 The second step is to extract the square root of each mem- 
 ber of the resulting equation. 
 
 The third step is to reduce the two resulting simple 
 equations. 
 
 1. Solve the equation x 2 — 8 x = 20. 
 
 Complete the square, x 2 — 8 x + 16 — 36. 
 
 . Extract the square root, x — 4 = ± 6. 
 
 Reduce, x = 4 + 6 = 10, 
 
 or x = 4 — 6 = — 2. 
 
 The roots are 10 and — 2. 
 
 Verify by putting these numbers for x in the given equation. 
 
 x = 10, 
 
 10 2 - 8 (10) = 20, 
 
 100 - 80 = 20. 
 
 x = -2, 
 (- 2) 2 - 8 (- 2) = 20, 
 4 + 16 = 20. 
 
 ooi xi- ,.- x -hi 4cc — 3 
 
 2. Solve the equation = — — • 
 
 * x — 1 a; -f- 9 
 
 Free from fractions, (x + 1) (x + 9) = (a; — 1) (4x — 3). 
 Simplify, — 3 x 2 + 17 x = — 6. 
 
 Divide by— 3, x 2 — ±£x = 2. 
 
 Half the coefficient of x is £ of — J^- = — V> an d the square of — *j 
 is ^ 9 -. Add the square of — ^ to both sides, and we have 
 
 X 3 f UJ "^ 36 36 
 
 Extract the root, 
 Transpose — ty 
 
 17 
 6 
 
 ■?.*»■ 
 
 
 X 
 
 17 ±19 
 6 
 
 • 
 
 .'. X 
 
 = 6, or — 
 
 1 
 
QUADRATIC EQUATIONS. 267 
 
 
 EXERCISI 
 
 s 109 
 
 
 Solve : 
 
 
 
 
 
 1. cc 2 + 2ic = 8. 
 
 
 7. 
 
 2a; 2 -fa; = 15. 
 
 
 2. x 2 -6x = 7. 
 
 
 8. 
 
 5a; 2 + 3a; = 2. 
 
 
 3. a; 2 -4a; = 12. 
 
 
 9. 
 
 a; 2 + \x = 40. 
 
 
 4. jc 2 -f-4a; = 5. 
 
 
 10. 
 
 3a; 2 -4a; = 4. 
 
 
 5. a; 2 + 5a; = 14. 
 
 
 11. 
 
 6a; 2 + a; = l. 
 
 
 6. a; 2 -3a; = 28. 
 
 
 12. 
 
 6 a; 2 -x = 2. 
 
 
 13. 12a; 2 -11a;- 
 
 4-2 
 
 = 0. 
 
 
 
 14. 15x 2 -2x- 
 
 1 = 
 
 :0. 
 
 
 
 {x + 1){x + 2) 
 
 (x- 
 
 ■l)(--2) 3 
 
 
 5 
 
 
 
 2 
 
 
 (2 a; -3) a; 
 
 (* + 4) ( 
 
 *-l) . 
 
 
 1C ' 4 
 
 
 6 
 
 
 17. 
 
 3a + 5 2a;-5 
 a; + 4 ' a? -2 
 
 
 22. 
 
 a; + 2 4-a; 7 
 a;-l 2a; 3 
 
 18. 
 
 a; — 6 a; 4- 5 . 
 a; -2 f 2a; + 1 
 
 
 23. 
 
 a; + 3 5a; + 8 
 a; — 2 a; + 4 
 
 19. 
 
 4-3a; l + 2a; 9 
 2 + x 1-x 2 
 
 
 24. 
 
 2a;-l 3 
 3 2a;-l~ 
 
 20. 
 
 x x + 1 13 
 a; + 1 x 6 
 
 
 25. 
 
 x+1 x+2 13 
 a; + 2 a; + l~ 6 
 
 21. 
 
 5 a; 2 — 4 a; = 1. 
 
 
 26. 
 
 7a; 2 -8a; = -l. 
 
268 QUADRATIC EQUATIONS. 
 
 294. If the coefficient of x 2 is 4, 9, 16, or any other 
 perfect square, we may complete the square by adding to 
 each side the square of the quotient obtained from dividing 
 the second term by twice the square root of the first term. 
 
 Solve 4« 2 - 23 x = - 30. 
 
 23 
 The square root of 4a; 2 is 2x, and 23 x divided by twice 2x is — • 
 
 23 
 Add the square of — to both sides. 
 
 Then, ^- 2 3* + (fy = f- 3 = f 
 
 23 7 
 
 Extract the root, 2x — —= ± -' 
 
 4 4 
 
 _ 23 ±7 30 ftr 16 
 
 Transpose, 2 x = — - — = — ' or "T" 
 
 .-. x = 3f, or 2. 
 
 If the coefficient of x 2 is not a perfect square, we may 
 multiply the equation by a number that will make the 
 coefficient of x 2 a perfect square. 
 
 Solve -3x 2 + 5x = -2. 
 
 Since the even root of a negative number is impossible, it is neces- 
 sary to change the sign of each term. The resulting equation is 
 
 3x 2 — bx = 2. 
 Multiply by 3, 9 x 2 — 15 x = 6. 
 
 Complete the square, 9x 2 — 15jc + — = — • 
 
 Extract the square root, 
 
 »H=±l 
 
 Reduce, 
 
 *-*p- 
 
 
 3x = 6, or — 1. 
 
 
 .-. x = 2, or - - • 
 
QUADRATIC EQUATIONS. 269 
 
 Exercise 110. 
 
 Solve : 
 
 1. 3a 2 -2a = 8. 2 ,2x OK 
 
 14. 3 x 2 + — = 25. 
 
 2. 5a 2 -6a = 27. 
 
 3. 2a 2 + 3a = 5. 15 ' *"--f = 3* + l. 
 
 4. 2a 2 -5a = 7. 16 . ^ - | = 2 (^ - 2). 
 
 5. 3a 2 + 7a = 6. 
 
 6. 5a 2 -7a = 24. 
 
 7. 8a 2 -h 3a = 26. 
 
 8. 7 a 2 + 5a = 150. 
 
 9. 6a 2 + 5a = 14. 
 
 10. 7a 2 - 2a = £. 
 
 11. 8a 2 + 7a = 51. 
 
 12. 7a 2 -20a = 75. 
 
 13. 11 a 2 - 10 a = 24. 3 2a; 3 
 
 23. (x + 2) (2 a + 1) + (x - 1) (3a + 2) = 57. 
 
 24. 3a (2a + 5) - (a + 3) (3a - 1) = 1. 
 
 25 (2a + 5)(a-3) a(3a + 4) _ 5> 
 3 5 
 
 26. £ (5a 2 -8a- 6)- i (a 2 - 3) = 2a + 1. 
 
 L •+*,.* 29 . -i-^^-i-. 
 
 a; +3 a: a— 1 as— 2 a;— 4 
 
 28 _J___3 4 g+,2 a +i 3 
 
 17. 
 
 2a 2 , 3a + M 
 
 18. 
 
 3* „ 2 1 
 
 T- 2a= i6 
 
 19. 
 
 |a 2 + |a = y 
 
 20. 
 
 2a-3 = -- 
 
 X 
 
 21. 
 
 7 a 5 20 
 5 3a~ 3 * 
 
270 QUADRATIC EQUATIONS. 
 
 Another Method of Completing the Square. 
 
 295. If a complete quadratic is multiplied by four times 
 the coefficient of x 2 , fractions will be avoided. 
 
 So\ve3x 2 -5x = 2. 
 
 Multiply by 12, 36 x 2 — 60 x = 24. 
 
 Complete the square, 36 x 2 — 60 x + 25 = 49. 
 Extract the square root, 6 x — 5 = ± 7. 
 
 Reduce, 6x = 5±7. 
 
 6 x = 12, or — 2. 
 
 .-. x = 2, or — - • 
 
 The number added to complete the square by this last method is 
 the square of the coefficient of x in the original equation 3 x 2 — 5 x = 2. 
 
 296. If tbe coefficient of x is an even number, we may 
 multiply by the coefficient of x 2 , and add to each member 
 the square of half the coefficient of x in the given equation. 
 
 Solve 3 z 2 + 4 a = 20. 
 
 Multiply by the coefficient of x 2 and add to each side the square oi 
 half the coefficient of x, 
 
 9x 2 + ( ) + 4 = 64. 
 Extract the square root, 3 x + 2 — ± 8. 
 
 Reduce, 3 x = — 2 ± 8. 
 
 3 x = 6, or — 10. 
 .-. x = 2, or — 3f 
 
 Note. If a trinomial is a perfect square, its root is found by taking 
 the square root of the first and third terms and connecting these roots 
 by the sign of the middle term. It is not necessary, therefore, in com- 
 pleting the square, to write the middle term, but its place may be 
 indicated by a parenthesis, as in this example. 
 
 Verify by putting the values of x in the given equation. 
 
 x = 2. x = — 3i. 
 
 3 (2) 2 + 4 (2) = 20. 3 (- 8i) 2 + 4 (- 3*) = 20. 
 
 12 + 8 = 20. 33* - 13* = 20. 
 
QUADRATIC EQUATIONS. 271 
 
 Exercise 111. 
 
 Solve 
 
 2x , 1 „ x + 1 2x 
 
 L ** 3 + 12~°- 4 * ^+4 _ a^ + 6 
 
 a; 2 a; a; a + 3 1 
 
 2. --- = 2(* + 2). 5 . ______ = __, 
 
 3£ _4_ JS 2 3 2 
 
 * 4+33: 6* «-l~.z-2 *-4* 
 
 3a? 5 3a 2 23 
 
 2(a? + l) 8 x 2 -l 4(jc — 1) 
 g 11-303 | 2(7-4g) _ L 
 
 ..^+* 
 
 a3 2 -4 ' a; +2 5(0 — 2) 
 
 10. - H — — = 5. 
 
 2 x — 3 a; -f- 1 
 
 2a; + 3 1-x = 7-3a? 
 
 * 2(20-1) 2(a;^-l)~4-3a;' , 
 
 --fc 2».— 1 3 a;-2 , , 
 
 12 ' — ^73^ = ^8 + 5. 
 
 3a: + 2 7 - ar _ 7s-l 
 2#-l' r 2a;-f-l 4» a -l" 1 " 
 
 x —5 ; 0—8 ^ SO 1 
 
 ' a;+3a5-3 _ a; 2 -92' 
 
 M 2gfl 4a; + 1 45 
 
 7-a: 7 + a; 49 -a 2 "*" 1 ' 
 
272 QUADRATIC EQUATIONS. 
 
 Solution by Resolving into Factors. 
 
 297. A quadratic which has been reduced to its simplest 
 form, and has all its terms written on one side, may often 
 have that side resolved by inspection into factors, and the 
 roots found by putting each factor equal to zero. 
 
 1. Solve x 2 + 7 x- 60 = 0. 
 
 Since x 2 + 7 x — 60 = (x + 12) (x — 5), (§ 130) 
 
 the equation x 2 + 7 x — 60 = 
 
 may be written (x + 12) (x — 5) = 0. 
 
 If either of the factors x + 12 or x — 5 is 0, the product of the two 
 factors is 0, and the equation is satisfied. 
 
 Hence, x + 12 = 0, or x — 5 = 0. 
 
 .-. x = — 12, or x = 6. 
 
 2. Solve x s - x 2 - 6x = 0. 
 
 The equation x 3 — x 2 — 6 x = 
 
 may be written x (x 2 — x — 6) = 0, 
 
 or x(x - 3) (x + 2) = 0, (§ 130) 
 
 and is satisfied if x = 0, 3, or — 2. 
 
 Hence, the equation has three roots, 0, 3, — 2. 
 
 3. Solve x s - 2 x 2 - Vlx + 12 = 0. 
 
 By the Factor Theorem (§ 135), we find that 1 put in place of x 
 satisfies the equation, and is therefore a root of the equation. 
 Divide by x — 1, and resolve the quotient into its factors. 
 We have (x - 1) (x - 4) (x + 3) = 0. 
 
 Hence, the roots of the equation are 1, 4, — 3. 
 
 4. Solves* + 3 x -10 = 0. 
 
 If we add -> the square of half the coefficient of x, to the first two 
 
 9 
 terms, we have a perfect trinomial square. Add and subtract -> 
 
QUADBATIC EQUATIONS. 273 
 
 x 2 + 3x + |-|-10 = 0, 
 3\2 49 
 
 x + -j — — = 0. 
 We now have the difference of two squares, and the factors are 
 
 that is, (x + 5) (x — 2) = 0. 
 
 .-. x = — 5, or 2. 
 5. Solve 3z 2 - 2# -2-0. 
 
 Divide by 3, x 2 — \x — § = 0. 
 
 Add and subtract the square of half the coefficient of x, 
 
 *-!»+(f)'-l-N 
 
 / 1\ 2 7 
 that is, ^x — -J — - = 0. 
 
 7 /l -i 
 
 The square root ofq = A/-X7 = i"^* 
 
 Hence, (x -| + |Vf) (x - |-|V?) =0. 
 
 Therefore, x = - — - V7, or - + r V7. 
 
 6. Solve a 2 - a; + 1 = 0. 
 
 x 2 -x + 7-4 + l=0. 
 4 4 
 
 ference of tw( 
 
 (HH-fy 
 
 In order to make this the difference of two squares, write it 
 1\ 2 / 3" 
 
 3 R 1 
 
 The square root of — - = a/- X (— 3) = ^—3- 
 
 Hence, (x -\ + \V=j) (x -\ -|V=s) =0. 
 
 Therefore, x = | - 1 V=8, or | + 1 V=3. 
 
274 QUADEATIC EQUATIONS. 
 
 Exercise 112. 
 Resolve into factors, and find the values of x : 
 
 1. x 2 - 5x + 4 = 0. 5. x s -f x 2 -6x = 0. 
 
 2. 6x 2 -5x-6 = 0. 6. a 8 -8 = 0. 
 
 3. 2x 2 - x -3 = 0. 7. a 8 + 8 = 0. 
 
 4. 10x 2 + a;-3 = 0. 8. x* - 16 = 0. 
 
 9. (a? - 1) (x - 3) (a 2 + 5x + 6) = 0. 
 
 10. (2x -l)(x- 2) (3ar> - 5x - 2) = 0. 
 
 11. (x 2 + aj - 2) (2 a 2 + 3 x - 5) = 0. 
 
 12. z 8 + ;c 2 -4(z + l) = 0. 
 
 13. 3z 8 + 2x 2 -(3a; + 2) = 0. 
 
 14. x 8 - 27 - 13 (a* - 3) = 0. 
 
 15. a 8 + 8 + 3(;z 2 -4) = 0. 17. 2x*- 2x'- (x*- 1)- 0. 
 
 16. x(x 2 -l)-6(x-l) = 0. 18. * 8 -3a:-2 = 0. 
 
 19. 2cc 8 + 2a; 2 + (x 2 -5a;- 6) = 0. 
 
 298. Any quadratic trinomial of the form ax 2 + bx + o can 
 be resolved into two factors by writing it as the difference 
 of two squares. Thus, 
 
 Sx 2 + 7 x - 6 = 3 (x 2 + % x - 2) 
 
 = 3[(z + £) 2 -W] 
 
 = 3(a + S + V)(* + £- J 5 L > 
 = 3 (x + 3) (a - 3) 
 = (aj -(-3) (3 a; -2). 
 
QUADRATIC EQUATIONS. 275 
 
 Exercise 113. 
 Resolve into factors : 
 
 1. 8x 2 -26x + 21. 5. 5« 2 -15z + ll. 
 
 2. 6x 2 -x-7. 6. 3x 2 + 5x-2. 
 
 3. 2x 2 + 7x + 6. 7. 3aj 2 -cc-l. 
 
 4. 5a 2 + 26a + 24, 8. x 2 -Sx-5. 
 
 Literal Equations. 
 
 1. Solve the equation adx — acx 2 = hex — bd. 
 
 Transpose hex and change the signs, 
 
 acx 2 + bcx — adx — bd. 
 Express the left member in two terms, 
 
 acx 2 + (be — ad)x = bd. 
 Multiply by 4 times the coefficient of x 2 , 
 
 4 a 2 c 2 x 2 + 4 ac (be — ad)x = 4 abed. 
 Complete the square, 
 
 4 a 2 c 2 x 2 + ( ) + (bc-ad) 2 = b 2 c 2 + 2 abed + a 2 <P. 
 Extract the root, 2 acx + (6c — ad) = ± (be + ad). 
 Reduce, 2 acx = — (6c — ad) ± (6c + ad) 
 
 = 2 ad, or — 2 be. 
 
 d 6 
 
 .-. x = -i or 
 
 c a 
 
 2. Solve the equation px 2 — px-\- qx 2 + qx — -*■■ — • 
 
 (p + g)x 2 -(p-g)x=^j^- 
 
 4 (p + g) 2 x 2 — 4 (p 2 — g 2 ) x = 4pg. 
 
 4(p + g) 2 x 2 -( ) + (p-g) 2 =p 2 + 2pg + g2. 
 
 2 (P + g) x — (p - q) = ± (p + g). 
 
 2(p + g)x = (p - q) ± (p + q). 
 
 P Q 
 
 .-. x = , - or *■— • 
 
 p + g p+q 
 
 Note. The left-hand member of the equation when simplified 
 must be expressed in two terms, simple or compound, one term con- 
 taining x 2 and the other term containing x. 
 
276 
 
 QUADRATIC EQUATIONS. 
 
 Exercise 114. 
 
 Solve : 
 
 1. x 2 + 2ax = 3a 2 . 
 
 2. x 2 - 4 ax = 12 a 2 . 
 
 3. x 2 + 8bx = 9b 2 . 
 
 4. x 2 + 3bx = 10b 2 . 
 
 6. x 2 + 5ax = Ua 2 . 
 
 6. 3X 2 + 4:CX = 4:C 2 . 
 
 7. 5ax-2x 2 = 2a 2 . 
 
 8. 6a 2 -a;c-a 2 = 0. 
 
 x 2 -{- ax + a 9 
 
 9. 2 a 2 x 2 + ax - 1 = 0. 
 10. 12 b 2 x 2 - 5bx = 3. 
 
 2x? 
 
 ax 
 
 11. 
 
 — + T = lla ( 
 
 12. 
 
 3x 2 x 3 
 
 4 + 2a~2a 2 
 
 13. 
 
 2 • 3 
 a 4r 
 
 3 ax 2 , 2x 13 
 14 ' -4- + T = 37 
 
 15. 
 
 16. 2cc 2 -a + a 
 x (a — x) . x 
 
 x z 
 
 3 
 
 ■■2 a 2 . 
 
 ax. 
 
 17. 
 
 18. 
 
 + 3 = 5«. 
 
 a 4" X 
 
 x(3x — a) _ a 
 
 19. x 2 -\ 
 20. 
 
 4# + a 
 
 m 2 — ?i s 
 
 6b 2 
 
 mn 
 5x 
 
 12 
 
 cc = l. 
 1 
 
 4 
 
 22. a 2 + (a - b) x = ab. 
 2a + x a-2x 8 
 2a-a; a + 2a;~~3' 
 2#-3a . 3<r4-2a 10 
 7' 
 
 23 
 
 24. 
 
 bab a z 
 
 — a x 4- a 
 
 25. 
 
 26. 
 
 27. 
 
 sc + 4a 
 a + 26 
 36-a 
 
 <7 
 
 4ic — a 
 
 a + 5 
 
 « + 2a 
 
 b -{• x a -\-b 
 ~~ x + b 
 
 a -\- b -\- x 
 
 1 
 a + b ■+■ x 
 
 
 2. 
 
 a 6 sc 
 
 28. 9ic 2 -3(a + 2&)* + 2a6 = 0. 
 
 29. (2 a + l)a; 2 + 3 a 2 x + a* - a 2 = 0. 
 
 30. (1 - a 2 ) x 2 - 2 (1 4- a 2 ) x + 1 — a 2 = 0. 
 
 31. (a + b) 2 x 2 - (a 2 -b 2 )x = ab. 
 
 32. (a 4- &) a 2 - (2 a + £) a + a = 0. 
 
33. 
 
 34. 
 35. 
 36. 
 37. 
 38. 
 
 39. 
 
 47. 
 48. 
 49. 
 50. 
 
 51. 
 52. 
 53. 
 54. 
 55. 
 56. 
 
 x x + b a 
 1 1 
 
 QUADRATIC EQUATIONS. 
 1 
 
 277 
 
 a + b 
 
 _ 3 + x 2 
 a — x a + x a 2 — x 2 
 
 x 2 -\-2ab(a 2 -\-b 2 ) 
 a 2 + b 2 -**- 
 
 (2x-ay _ 
 2x-a + 2b~ ' 
 
 (a-iyx* + 2(3a-l)x 
 4a -1 
 
 x 
 
 x — a 
 
 ex = ax 2 + bx 2 
 
 2c 
 b x — c 
 
 ac 
 
 40. ^ = a -j — • 
 
 x — 3 x 4- 3 
 
 41. mar — 1 = — » L 
 
 mn 
 
 42. (odi - 5) (bx - a) = c 2 . 
 ax -\-b mx + w 
 
 4o. 
 44. 
 45. 
 46. 
 
 bx + 
 
 a 7l£C ■+- m 
 
 
 m 
 
 1 ^ 
 
 
 m + 
 
 cc m — x 
 
 & 
 
 (••- 
 
 - 5 s2 ) (x 2 + 1) 
 
 
 
 a 2 4-5 2 
 
 
 « 2 - 
 
 4mraas 
 
 
 a + 5 
 
 | (a 2 4- a 2 4- a5) = J* (20a + 45). 
 jc 2 — 2 mcc = (n — p -\- m) (n — p — m). 
 a 2 — (m + w) a = I (p + a + m + w) (p + q — m — n). 
 mnx 2 — (m 4- n) (mn 4- 1) x 4- (m 4- w) 2 = 0. 
 x 2 m 2 — 4: a 2 x 
 
 3m — 2a 4a — 6m 2 
 
 R , (a + 5) 2 K/ .v, 25 aft 
 
 6 x 4- * *- = 5 (a — 5) 4- 
 
 x 
 
 2b-x-2a x 45 
 a — 25 — x 
 
 6x 
 x — 4a 
 
 ax — bx ab — 5 2 
 55 — x . 2a — a; — 195 
 
 a 2 — 4 5 2 ace 4- 2 5x 
 
 2bx 
 
 a-25 
 a; 4-25 
 
 cc + 13a4-35 
 5a — 35 ^- sc 
 
 a + 35 
 Sa 2 -12ab 9b 2 -±a 2 (2a + 35) (x - 35) 
 
 35 
 
 a4-35 
 
278 QUADRATIC EQUATIONS. 
 
 Solution by a Formula. 
 
 299. Every affected quadratic can be reduced to the form 
 
 ax 2 + bx + c = 0. 
 
 Solve ax 2 + bx + c = 0. 
 
 Transpose c, ax 2 + bx = — c. 
 
 Multiply the equation by 4 a and add the square of 6, 
 
 4a2a: 2 +( ) + 6 2 = 6 2 -4ac. 
 Extract the root, 2ax + b= ± V6 2 — 4 ac# 
 
 — & ± V&2 — 4 ac 
 •*- X = 2^ ■• 
 
 By this formula, the values of x in an equation of the 
 form ax 2 + bx 4- c = may be written at once. 
 
 Solve Sx 2 - 5a + 2 = 0. 
 
 Here a = 3, 6 = — 5, c = 2. 
 
 Putting these values for the letters in the above formula, we have 
 
 5 + V25 — 24 5— V25 — 24 
 X = j .or - 
 
 = I, or f 
 = 1, or f. 
 
 Exercise 115. 
 Solve by the above formula : 
 
 1. 2aj a + 3aj = 14. 7. 5x 2 -Ix = -2. 
 
 2. Sx 2 -5x = 12. 8. 4x 2 -9x = 28. 
 
 3. <c 2 -7a = 18. 9. 5x 2 +Tx = 12. 
 
 4. 5z 2 -a; = 42. 10. II* 2 - 9x = - -g- 
 
 5. 6z 2 -7a; = 10. 11. 7* 2 -f-5x = 38. 
 
 6. 3x 2 -lla; = -6. 12. 5x 2 -7x = & 
 
 
QUADRATIC EQUATIONS. 279 
 
 Equations Involving Two or More Radicals. 
 
 JOO. Solve Vx+i + V2x + 6 = V7 x -f- 14. 
 
 Square, x + 4 + 2V(x + 4)(2x + 6) + 2 x + = 7 x + 14. 
 Simplify, V(x + 4)(2x + t>) = 2x-f2. 
 
 Square, (x + 4) (2 x + C) = (2 x + 2)2. 
 
 Solve, x = 5, or — 2. 
 
 Of these two values, only 5 will satisfy the original equation. 
 
 Squaring both members of the original equation is equivalent to 
 transposing V7x + 14 to the left member, and then multiplying by 
 the rationalizing factor Vx + 4 +'V2x + G + V7x + 14, so that 
 
 (Vx + 4 + V2x + G — V7x+14)(Vx + 4 + V2x + 6 + V~7x + 14)=0, 
 
 and this reduces to V(x + 4)(2x + 0) — (2 x + 2) = 0. 
 
 Transposing and squaring again is equivalent to multiplying by 
 
 (Vx + 4 — V2 x + 6 + V7x + 14)(Vx + 4 — V2x + C — V7x+ 14). 
 Reducing, x 2 — 3 x — 10 = 0. 
 
 Therefore, the equation x 2 — 3 x — 10 = is really obtained from 
 
 (Vx + 4 -f V2x + 6 - vTx + 14) 
 
 X (Vx + 4 + V2x + 6 + V7 x + 14) 
 
 X (Vx + 4 - V2x + 6 — V7x+ 14) 
 
 X (Vx + 4 — V2x + 6 + V7 x + 14) = 0. 
 
 This last equation is satisfied by any value that will satisfy any one 
 )f the four factors of its left member. The first factor is satisfied by 
 6, and the last factor by — 2, while no values can be found to satisfy 
 le second or third factor. 
 
 As 5 is the only value of x that will satisfy the original equation, 
 
 other values must be rejected. 
 
280 QUADRATIC EQUATIONS. 
 
 Exercise 116. 
 Solve : 
 
 1. V9 x + 40 - 2 Va; + 7 = Val 
 
 2. Va + a; + Va — x = V#. 
 
 o 3x+W4:X-x 2 
 
 3. = — J. 
 3 ic — V4 a; — x 2 
 
 4. Va; - 3 - Va; - 14 = V4a; - 155. 
 
 5. VaT+4 — Va: = VaT+|. 
 
 3Vc-4 15 + 3V s 
 2+Vz 40+Vc 
 
 7. V14 a; + 9 + 2 Va7+1 t V3 a; + 1 = 0. 
 
 8. V5a; + 1 - 2 — Vaj + 1 = 0. 
 
 9. Va; - 2 + Va; + 3 — V4a; + 1 = 0. 
 
 10. V7 - a; + V3a; + 10 + Va; + 3 = 0. 
 
 11. 3Vc 8 + 17 + Va; 8 + 1 + 2V5a; 8 + 41 = 
 
 12. 2x—W2x-l=x + 2. 
 
 "2-V2a^ = 0. 
 1 
 
 12. 
 
 - V<c a -8 
 

 QUADRATIC EQUATIONS. 281 
 
 Equations in the Quadratic Form. 
 
 301. An equation is in the quadratic form if it contains 
 but two powers of the unknown number, and the exponent 
 of one power is exactly twice that of the other power. 
 
 302. Equations in the quadratic form may be solved by 
 the methods for solving quadratics. 
 
 1. Solve 8 a 6 + 63 x* = 8. 
 
 Multiply by 32 and complete the square, 
 
 256 a; 6 + ( ) + 63* = 4225. 
 Extract the square root, 16 x 3 + 63 = ± 65. 
 
 Hence, x 8 = -» or — 8. 
 
 o 
 
 Extracting the cube root, two values of x are i and — 2. There 
 are four other values of x which may be found by § 297. 
 
 2. Solve Vx 1 - 3^/x 8 = 40. 
 Using fractional exponents we have, 
 
 x*-3x* = 40. 
 Complete the square, 4 x* — 12 x* + 9 = 169. 
 
 Extract the root, 
 
 2x^-3 = ±13. 
 
 Transpose — 3, 
 
 2x J = 16, or — 10. 
 
 Divide by 2, 
 
 x* = 8, or — 5. 
 
 Extract the cube root, 
 
 x* = 2, or — 5*. 
 
 Raise to the fourth power, 
 
 x — 16, or 5VH. 
 
 3. Solve (2x-Sy- (2x 
 
 - 3) = 6. 
 
 Put y for 2 x — 3, and therefore y 2 for (2 x — 3) 2 . 
 
 We have 
 
 y 2 -y = G. 
 
 Solving, 
 
 y = 3, or — 2. 
 
 Put2x — 3fory, 
 
 
 2x-3 = 3, 
 
 2x-3 = ~ 
 
 , * = 3. 
 
 x = i 
 
282 QUADRATIC EQUATIONS. 
 
 4. Solve 7a 2 -5a + 8V7a 2 -5a + l = -8. 
 
 By adding 1 to both sides, we have 
 
 7x 2 -5x + l + 8V7x 2 -5x + l = -7. 
 Put y for V7x 2 — 5x + 1, and hence y 2 for 7 x 2 — 5* + 1. 
 Then, y 2 + 8y = -7. 
 
 Solving, y = - l, or — 7. 
 
 Therefore, y 2 = 1, or 49. 
 
 We now have 7x 2 — 5x + 1 = 1, or 7x 2 — 5x + 1 = 49. 
 
 Solving these equations, we find for the values of x, 
 
 0,§;or3,-f 
 
 These values all satisfy the given equation when we take the nega- 
 tive value of the square root of the expression 7 x 2 — 5 x + 1 ; they 
 are in fact the four roots of the biquadratic obtained by clearing the 
 given equation of radicals. 
 
 5. Solve x* - 10a 8 + 35 a 2 - 50a: + 24 = 0. 
 Take the square root of the left side. 
 
 a 4 - 10a 8 + 35 a 2 - 50 a + 24|a 2 _-5a_+5 
 
 2a 2 -5a 
 
 - 10a 3 + 35 a 2 
 
 - 10a 8 + 25 a 2 
 
 2a 2 -10a + 5 
 
 10a 2 -50a + 24 
 10a 2 -50a + 25 
 
 It is now seen that if 1 is added, the square will be complete and 
 the equation will be 
 
 a 4 - 10a 8 + 35a 2 - 50a + 25 = 1. 
 
 Extract the square root, and the result is 
 
 x 2 -5x + 6 = ±l. 
 Solving, x = 4, 1, 3, or 2. 
 
QUADRATIC EQUATIONS. 283 
 
 Exercise 117. . 
 Solve: 
 
 1. a; 4 -5a; 2 + 4 = 0. 6. 10 x* - 21 = x\ 
 
 2. x 4 - 13a; 2 + 36 = 0. 7. -</tf 2 + 3Vz = lf. 
 
 3. a; 4 -21 a; 2 = 100. 8. 3Va-2Va^ = -20. 
 
 4. 4a 6 -3a- 8 = 27. 9. 5 x*» + 3 x n = 6£ . 
 
 5. 2a; 4 + 5a; 2 = 21£. 10. (8a; + 3) 2 +(8a; + 3)=30. 
 
 11. 2(« 2 -£C + l)-ViC 2 -iC + l = l. 
 
 12. a; 6 -9a: 8 + 8 = 0. 14. (x + 1) + Va; + 1 = 6. 
 
 13 
 12. a*4-ic-^ = ~- 15. a; 4 -13 a; 2 =-36. 
 
 O 
 
 16. 2a; 2 + 4a; + 9 + 3V2a: 2 + 4a; + 9 = 40. 
 
 17. 2a; 2 + 3a;-5V2a; 2 + 3a; + 9 = -3. 
 
 18. 3a; 2 + 15a;-2Va; 2 + 5a; + l = 2. 
 
 19. x 2 -§x + 3y/2x 2 -3x + 2 = 7. 
 
 20. 2 a; 2 -Va; 2 - 2a; - 3 = 4a; + 9. 
 
 21. 3a; 2 - 4a; + V3a; 2 - 4a; - 6 = 18. 
 
 22. 3a; 2 -7 + 3V3a; 2 -16a; + 21 = 16x. 
 
 23. a; 4 - 2a; 8 - 13a; 2 + 14a; + 24 = 0. 
 
 24. a; 4 - 4a; 8 - 10a; 2 + 28a; -15 = 0. 
 
 25. 4a; 4 -20a; 8 + 23a; 2 + 5a;-6 = 0. 
 
 26. 4a; 4 - 12a; 8 + 5a; 2 + 6x -15 = 0. 
 
 27. 4a; 4 - 12a; 8 + 17a; 2 -12a; -12 = 0. 
 
 28. 6sc 2 + 6a; + Va;(x + 1) = 7, 
 
284 QUADRATIC EQUATIONS. 
 
 Character of the Roots. 
 
 The two roots of ax 2 + bx + c = are 
 
 b , Wb^-Tac , b ^/b 2 -±ac 
 
 ^- + ^~5 and-- - (§299) 
 
 2a 2a 2a 2a v ' 
 
 303. The character of the two roots depends wholly upon 
 b 2 — 4 ac, the expression under the radical sign. 
 
 1. If b 2 — 4 ac is positive and not zero, the roots are real 
 and unequal. 
 
 The roots are real, since the square root of the positive 
 number, 6 2 — 4 ac, can be found exactly or approximately. 
 
 The roots are unequal, since V# 2 — 4 ac is not zero. 
 
 If b 2 — 4 ac is a perfect square, the roots are rational; if 
 b 2 — 4 ac is not a perfect square, the roots are surds. 
 
 2. If b 2 — 4 ac is zero, the two roots are real and equal, 
 
 since they both become — — • Hence, 
 
 2 a 
 
 The roots of ax 2 -f foe + c = are re<z£, if # 2 = or > 4 ac, 
 
 3. If b 2 — 4 ac is negative, the roots are imaginary, since 
 they involve the square root of a negative number. Hence, 
 
 The roots of ax 2 -f bx -f c = are imaginary, if £ 2 < 4 ac. 
 
 The two imaginary roots of a quadratic cannot be equal, 
 since b 2 — 4 ac is not zero. They have, however, the same 
 real parts, and the same imaginary parts with opposite signs, 
 and are, therefore, conjugate imaginaries, § 284. 
 
 The expression b 2 — 4 ac is called the discriminant of the 
 expression ax 2 + bx -f- c. 
 
 
QUADRATIC EQUATIONS. 285 
 
 304. The above cases may be summarized as follows : 
 
 Case 1. If b 2 — 4 ac > 0, the roots are real and unequal. 
 Case 2. If b 2 — 4 ac = 0, the roots are real and equal. 
 Case 3. If b 2 — 4 ac < 0, the roots are imaginary. 
 
 305. By finding the value of b 2 — 4 ac we can determine 
 at once the character of the roots of a given equation. 
 
 1. x 2 - 5a + 6 = 0. 
 
 Here a = 1, b — — 5, c = 6. 
 
 & 2 - 4 ac = 25 - 24 = 1. 
 The roots are real and unequal, and rational. 
 
 2. 3a 2 + 7a:-l = 0. 
 
 Here a = 3, b = 7, c = — 1. 
 
 62 _ 4 ac = 49 + 12 = 61. 
 
 The roots are real and unequal, and are both surds. 
 
 3. 4a 2 -12z + 9 = 0. 
 
 Here a = 4, b = — 12, c = 9. 
 
 62 - 4 ac = 144 - 144 = 0. 
 The roots are real and equal. 
 
 4. 2x 2 - Sx + 4 = 0. 
 
 Here a = 2, b = - 3, c = 4. 
 
 6 2 — 4 ac = 9 — 32 = - 23. 
 The roots are both imaginary. 
 
 5. Find the values of m for which 
 
 2mx 2 + (5m4-2)« + (4m + l) =0 
 has its two roots equal . 
 
 Here • a = 2m,6 = 5m + 2,c = 4m + l. 
 
 If the roots are to be equal, we must have 
 
 62 — 4ac = 0, or (5m + 2) 2 — 8m(4m + l)=0. 
 
 2 
 Solving, m = 2, or — - • 
 
 s 
 
286 QUADRATIC EQUATIONS. 
 
 For these values of m the equation becomes 
 
 4x* + 12x + 9 = 0, and 4x 2 — 4x + 1 =0, 
 each of which has its roots equal. 
 
 Exercise 118. 
 
 Determine witliout solving the character of the roots of 
 each of the following equations : 
 
 1. jc 2 + 5a; + 6 = 0. 6. 6x 2 -7x-3 = 0. 
 
 2. a 2 + 2a; -15 = 0. 7. 5x 2 - 5x - 3 = 0. 
 
 3. x 2 + 2x + 3 = 0. 8. 2x 2 -x + 5 = 0. 
 
 4. 3 a; 2 + 7 a; + 2 = 0. 9. 6a 2 + a - 77 = 0. 
 
 5. 9a; 2 + 6a; + 1 = 0. 10. 5a; 2 + 8a; + ^ = 0. 
 
 5 
 
 Determine the values of m for which the two roots of 
 each of the following equations are equal : 
 
 11. (m + l)a; 2 + (m-l)a; + ra + l = 0. 
 
 12. (2 m - 3) x 2 + mx + m - 1 = 0. 
 
 13. 2mx 2 + a: 2 + 4a; + 2ma;-|-2m-4 = 0. 
 
 14. 2 mx 2 + 3 ma; — 6 = Sx — 2 m — x 2 . 
 
 15. mx 2 + 9x - 10 = 3mx-2x 2 + 2m. 
 
 Relations of Roots and Coefficients. 
 
 306. If we divide the general equation ax 2 + bx + c — 
 
 b c 
 by a, we have the equation a; 2 + -a; + - = 0; this may be 
 
 be 
 written x 2 + jpx -\- q = 0, where p = -> ^ = - • 
 
 ot At 
 
QUADRATIC EQUATIONS. 287 
 
 307. By solving x 2 -f- px -f- q = 0,"and denoting the first 
 value of x by r lf and the second value by r 2 , we have 
 
 1 2^ 2 
 
 r *~ 2 2 
 
 Add, rj + r 2 = — ^?. 
 
 Multiply, r\Tt = y. 
 
 It appears, then, that if any quadratic equation is made 
 to assume the form x 2 -{-px -f- q = 0, the following relations 
 hold between the coefficients and roots of the equation : 
 
 1. The sum of the two roots is equal to the coefficient 
 of x with its sign changed. 
 
 2. The product of the two roots is equal to the constant 
 term. 
 
 
 308. If r x and r 2 are the roots of the equation x 2 -{-px 
 + q = 0, the equation may be written 
 
 (x - rj) (x - r 2 ) = 0. 
 
 309. Form the equation of which the roots are 3 and — 2. 
 
 The equation is (x — 3) (x + 2) = 0, 
 
 or x 2 — x — 6 = 0. 
 
 Exercise 119. 
 Form the equation of which the roots are : 
 1. 7, 6. 5. ljfc - 1J. 9. 3 + V2, 3 - V2. 
 
 2. 5,-3. 
 
 6. -li,-lf. 
 
 10. 
 
 l + V^i , l - 
 
 -V=I 
 
 3. l*,-2. 
 
 7. 13,-4*. 
 
 n. 
 
 a, a — b. 
 
 
 4. 4, 2J. 
 
 ft 3 2 
 
 8 * IT II 
 
 12. 
 
 a -{- b, a — b. 
 
 
288 QUADRATIC EQUATIONS. 
 
 Problems Involving Quadratics. 
 
 310. Problems that involve quadratic equations appar- 
 ently have two solutions, since a quadratic equation has 
 two roots. If both roots of the quadratic equation are 
 positive integers, they will, generally, both be admissible 
 solutions. 
 
 Fractional and negative roots will in some problems give 
 admissible solutions ; in other problems they will not. 
 
 No difficulty will be found in selecting the result which 
 belongs to the particular problem we are solving. Some- 
 times, by a change in the statement of the problem, we may 
 form a new problem which corresponds to the result that 
 was inapplicable to the original problem. 
 
 Imaginary roots indicate that the problem is impossible. 
 
 Here, as in simple equations, x stands for an unknown 
 number. 
 
 1. The sum of the squares of two consecutive numbers 
 is 481. Find the numbers. 
 
 Let x — one number, 
 
 and x + 1 = the other. 
 
 Then, x 2 + (x + l) 2 = 481, 
 
 or 2x 2 + 2x + l =481. 
 
 The solution of which gives x = 15, or — 16. 
 
 The positive root 15 gives for the numbers, 15 and 16. 
 
 The negative root — 16 is inapplicable to the problem, as consecu- 
 tive numbers are understood to be integers which follow one another 
 in the common scale, 1, 2, 3, 4 
 
 2. A pedler bought a number of knives for $2.40. Had 
 he 'bought 4 more for the same money, he would have paid 
 3 cents less for each. How many knives did he buy, and 
 what did he pay for each ? * 
 
 Let x = number of knives he bought. 
 
 240 
 Then, = number of cents he paid for each 
 
QUADRATIC EQUATIONS. 289 
 
 But if x + 4 = number of knives he bought, 
 
 — — =5 number of cents he paid for each, 
 x + 4 
 
 
 —7 = the difference in price. 
 
 x x + 4 
 
 But 
 
 3 = the difference in price. 
 
 
 240 240 
 x x + 4 
 
 Solving, 
 
 x = 16, or — 20. 
 
 He bought 16 knives, therefore, and paid *$*, or 15 
 cents for each. 
 
 If the problem is changed so as to read : A pedler bought 
 a number of knives for $ 2.40 ; if he had bought 4 less 
 for the same money, he would have paid 3 cents more for 
 each, the equation will be 
 
 240 240 _ _ 
 
 - — — o. 
 
 x — 4 x 
 Solving, x = 20, or — 16. 
 
 This second problem is therefore the one which the nega- 
 tive answer of the first problem suggests. 
 
 3. What is the price of eggs per dozen when 2 more in 
 a shilling's worth lowers the price 1 penny per dozen ? 
 
 Let x = number of eggs for a shilling. 
 
 Then, - = cost of 1 egg in shillings, 
 
 x 
 
 12 
 and — = cost of 1 dozen in shillings. 
 
 But if x + 2 = number of eggs for a shilling, 
 
 12 
 
 = cost of 1 dozen in shillings. 
 
 X T £i 
 
 12 12 1 
 
 •*• — - J-^2 = 12 (1 penny being * ° f a shillin S)' 
 
 The solution of which gives x = 16, or — 18. 
 
 And, if 16 eggs cost a shilling, 1 dozen will cost 9 pence. 
 
 Therefore, the price of the eggs is 9 pence per dozen. 
 
290 QUADRATIC EQUATIONS. 
 
 If the problem is changed so as to read : What is the 
 price of eggs per dozen when two less in a shilling's worth 
 raises the price 1 penny per dozen ? the equation will be 
 
 12 12 _ 1 
 x — 2 x 12 ' 
 The solution of which gives x = 18, or — 16. 
 
 Hence, the number 18, which had a negative sign and was inappli- 
 cable in the original problem, is here the true result. 
 
 Exercise 120. 
 
 1. The sum of two squares of two consecutive integers is 
 761. Find the numbers. 
 
 2. The sum of the squares of two consecutive numbers ex- 
 ceeds the product of the numbers by 13. Find the numbers. 
 
 3. The square of the sum of two consecutive even num- 
 bers exceeds the sum of their squares by 336. Find the 
 numbers. 
 
 4. Twice the product of two consecutive numbers ex^ 
 ceeds the sum of the numbers by 49. Find the numbers. 
 
 5. The sum of the squares of three consecutive numbers 
 is 110. Find the numbers. 
 
 6. The difference of the cubes of two consecutive odd 
 numbers is 602. Find the numbers. 
 
 7. The length of a rectangular field exceeds its breadth 
 by 2 rods. If the length and breadth of the field were 
 each increased by 4 rods, the area would be 80 square rods. 
 Find the dimensions of the field. 
 
 8. The area of a square may be doubled by increasing 
 its length by 10 feet and its breadth by 3 feet. Find the 
 length of its side. 
 
QUADRATIC EQUATIONS. 291 
 
 9. A rectangular grass plot 12 yards long and 9 yards 
 wide has a path around it. The area of the path is § of 
 the area of the plot. Find the width of the path. 
 
 10. The perimeter of a rectangular field is 60 rods. Its 
 area is 200 square rods. Find its dimensions. 
 
 11. The length of a rectangular plot is 10 rods more 
 than twice its width, and the length of a diagonal of the 
 plot is 25 rods. What are the dimensions of the plot ? 
 
 12. The denominator of a certain fraction exceeds the 
 numerator by 3. If both numerator and denominator are 
 increased by 4, the fraction will be increased by J. Find 
 the fraction. 
 
 13. The numerator of a fraction exceeds twice the de- 
 nominator by 1. If the numerator is decreased by 3, and 
 the denominator increased by 3, the resulting fraction will 
 be the reciprocal of the given fraction. Find the fraction. 
 
 14. A farmer sold a number of sheep for $120. If he 
 had sold 5 less for the same money, he would have received 
 $2 more a sheep. How much did he receive a sheep ? 
 
 State the problem to which the negative solution applies. 
 
 15. A merchant sold a certain number of yards of silk 
 for $40.50. If he had sold 9 yards more for the same 
 money, he would have received 75 cents less per yard. 
 How many yards did he sell ? 
 
 16. A man bought a number of geese for $27. He sold 
 all but two for $25, thus gaining 25 cents on each goose 
 sold. How many geese did he buy ? 
 
 17. A man agrees to do a piece of work for $48. It 
 takes him 4 days longer than he expected, and he finds 
 that he has earned $1 less per day than he expected. In 
 how many days did he expect to do the work ? 
 
292 QUADRATIC EQUATIONS. 
 
 18. Find the price of eggs per dozen when 10 more in 
 one dollar's worth lowers the price 4 cents a dozen. 
 
 19. A man sold a horse for $171, and gained as many 
 per cent on the sale as the horse cost dollars. How much 
 did the horse cost ? 
 
 20. A drover bought a certain number of sheep for $160. 
 He kept four, and sold the remainder for $10.60 per head, 
 and made on his investment f as many per cent as he paid 
 dollars for each sheep bought. How many sheep did he buy ? 
 
 21. Two pipes running together can fill a cistern in 5| 
 hours. The larger pipe will fill the cistern in 4 hours less 
 time than the smaller. How long will it take each pipe 
 running alone to fill the cistern ? 
 
 22. A and B can do a piece Of work together in 18 days, 
 and it takes B 15 days longer to do it alone than it does A. 
 In how many days can each do it alone ? 
 
 23. A boat's crew row 4 miles down a river and back 
 again in 1 hour and 30 minutes. Their rate in still water 
 is 2 miles an hour faster than twice the rate of the current. 
 Find the rate of the crew and the rate of the current. 
 
 24. A number is formed by two digits. The units' digit 
 is 2 more than the square of half the tens' digit, and if 18 
 is added to the number, the order of the digits will be- 
 reversed. Find the number. 
 
 25. A circular grass plot is surrounded by a path of a 
 uniform width of 3 feet. The area of the path is $ the area 
 of the plot. Find the radius of the plot. 
 
 26. If a carriage wheel 11 feet round took \ of a second 
 less to revolve, the rate of the carriage would be five miles 
 more per hour. At what rate is the carriage traveling ? 
 
CHAPTER XX. 
 SIMULTANEOUS QUADRATICS. 
 
 311. Quadratic equations involving two unknown num- 
 bers require different methods for their solution, according 
 to the form of the equations. 
 
 Case 1. 
 
 312. When one of the equations is a simple equation. 
 
 Solve Sx 2 - 2x7/ = 5^ (1) 
 
 x-y = 2 J (2) 
 
 Transpose x in (2), y = x — 2. 
 
 In (1) put x — 2 for y, 
 
 3x 2 — 2x(x — 2) = 5. 
 The solution of which gives x = 1, or x = — 5. 
 If x = 1, 
 
 y = l-2 = -l; , 
 and if a = — 5, 
 
 y = -5-2 = -7. 
 We have, therefore, the pairs of values, 
 
 x = l I x = — 5 
 
 1 ( J or » 
 
 y — — 1 ) ' y — — l 
 
 The original equations are both satisfied by either pair of values, 
 But the values x = l,y = — 7, will not satisfy the equations ; nor will 
 the values x = — 5, y = — 1. 
 
 The student must be careful to join to each value of x 
 the corresponding value of y. 
 
294 
 
 SIM UL TANEOUS Q UADRA TICS. 
 
 Case 2. 
 
 313. When the left side of each of the two equations is homo- 
 geneous and of the second degree. 
 
 Solve 
 
 2y 2 - Axy + Sx 2 = 17 1 
 
 y 2 -x 2 = 16 
 
 Let y = vx and substitute vx for y in both equations. 
 From (1), 2 v*x 2 — 4 ux 2 + 3 x 2 = 17. 
 
 r.2 — 
 
 17 
 
 .-. x* 
 
 From (2), 
 
 2t> 2 — 4t> + 3 
 tj2x 2 — x 2 = 16. 
 16 
 
 .-. x 2 = 
 
 v 2 — 1 
 
 17 1ft 
 
 Equate the values of x 2 , — = — r-s as — — -» 
 
 2 v 2 — 4 v + 3 u 2 — 1 
 
 32« 2 -64u + 48 = 17t> 2 -17, 
 
 15 u 2 — 64 u = — 65, 
 
 225u 2 -960u = -976, 
 
 225u 2 -() + 32 2 = 49, 
 
 15v-32 = ±7. 
 
 If 
 
 y =VX = 
 
 5x 
 
 Substitute 
 
 in (2), 
 
 
 25 x 2 
 
 
 
 9 
 
 -x 2 = 
 
 16, 
 
 
 x 2 = 
 
 9, 
 
 
 x = 
 
 ±3, 
 
 V~ 
 
 5x 
 '3 
 
 ±5. 
 
 5 
 
 > = -, or 
 
 13 . 
 5 ' 
 
 
 If 
 
 V 
 
 13 
 
 y 
 
 = VX 
 
 _13x % 
 6 
 
 Substitute in 
 
 (2), 
 
 169x2 
 25 
 
 — X 2 
 
 = 16, 
 
 
 X 2 
 
 25 
 
 
 X 
 
 =4 
 
 y = 
 
 13 x 
 5 
 
 -*¥ 
 
 (1) 
 
 (2) 
 
SIMULTANEOUS QUADRATICS. 
 
 295 
 
 Case 3. 
 
 314. When the two equations are symmetrical with respect 
 to x and y ; that is, when x and y are similarly involved. 
 
 Thus, the expressions 
 
 2x» + 3«V + 2y s , 2xy-3x-3y + l,x*- 3x 2 y -3xy 2 + y 4 
 
 are symmetrical expressions. In this case the general rule 
 is to combine the equations in such a manner as to remove 
 the highest powers of x and y. 
 
 Solve 
 
 x* + ^ = 3371 
 «.+Jf = 7J 
 
 To remove x 4 and y 4 , raise (2) to the fourth power, 
 Add (1), 
 
 x 4 + 4x 3 2/ + 6x 2 2/ 2 + 4x2/ 8 + 2/4 = 2401 
 x* + 2/4= 337 
 
 2x 4 + 4x 3 j/ + 6xV + 4xy 3 + 22/ 4 = 2738 
 Divide hy 2, x 4 + 2 x 3 2/ + 3 xfy 2 + 2 x?/ 8 + 2/ 4 = 1369. 
 Extract the square root, x 2 + xy + y 2 = ± 37. 
 
 Subtract (3) from (2) 2 , xy = 12 or 86. 
 We now have to solve the two pairs of equations, 
 
 From the first, 
 
 From the second, 
 
 a + y = 7) x + y = 7) 
 xy = 12 J ' xy = 86 ) ' 
 
 2/ = 3f ; ° r 2/ = 4 } 
 
 7 ± V- 295 
 
 v = 
 
 7qzV-295 
 
 (1) 
 
 (2) 
 
 (3) 
 
296 SIMULTANEOUS QUADRATICS. 
 
 315. The preceding cases are general methods for the 
 solution of equations that belong to the kinds referred to ; 
 often, however, in the solution of these and other kinds of 
 simultaneous equations involving quadratics, a little inge- 
 nuity will suggest some step by which the roots may be 
 found more easily than by the general method. 
 
 1. Solve x + y = 40 T (1) 
 
 xy = 300 J (2) 
 
 Square (1), x 2 + 2 xy + y* = 1600. (3) 
 
 Multiply (2) by 4, 4xy = 1200. (4) 
 
 Subtract (4) from (3), 
 
 x 2 -2xy + ?/ 2 = 400. (5) 
 
 Extract the square root, x — y = ± 20. (6) 
 
 From (1) and (6), x = 30 ) . 
 
 y = io r 
 
 x = 10 
 % = 30 
 
 I; 
 
 
 I 1 9 
 
 2. Solve £ + £ = ^ 
 jc ?/ 20 
 
 II 41 
 
 ar J ~*V~400 J 
 
 ' 
 
 
 (i) 
 
 (2) 
 
 c ™ * j_ 2 a. * - 81 
 Square(l), _+- + ___. 
 
 
 
 (3) 
 
 2 40 
 
 Subtract (2) from (3), — = ^ ■ 
 
 
 
 W 
 
 Subtract (4) from (2), 
 
 
 
 
 J___2 i_j_. 
 x 2 xy y 2 400 
 
 
 
 
 Extract the square root, - — = ± — • 
 x y £\j 
 
 
 
 (6) 
 
 From (1) and (6), x = 4 i ; or 
 
 y = 5 > ' 
 
 x = 5) t 
 
 y = 4i ' 
 
 
 
SIMULTANEOUS QUADRATICS. 297 
 
 x — y — 41 
 x 2 + y 2 = 40 J 
 
 3. Solve x -y = 4\ (1) 
 
 (2) 
 
 Square (1), x 2 - 2 xy + y 2 = 16. (3) 
 
 Subtract (2) from (3), — 2xy = - 24. (4) 
 
 Subtract (4) from (2), 
 
 x 2 + 2 xy + y 2 = 64. 
 Extract the root, x + y = ± 8. (5) 
 
 From (1) and (5), x = 6) x= ~ 2 ) 
 
 y = 2) ; ° r y=~6J 
 
 4. Solve a? + y* = 91\ (1) 
 
 «: + Jf = 7 J (2) 
 
 Divide (1) by (2), x 2 - xy + y 2 = 13. (3) 
 
 Square (2), x 2 + 2xy + y 2 = 49. (4) 
 
 Subtract (3) from (4), 3xy = 36. 
 
 Divide by — 3, — xy — — 12. (6) 
 
 Add (5) and (3), x 2 — 2 xy + y 2 = 1. 
 
 Extract the root, x — y = db 1. (6) 
 
 From (2) and (6), x = 4 ) a; = 3 ) 
 
 „=3J ; ° r y=4> 
 
 5. Solve « , + y , = 18ajy1 (1) 
 
 a; +y =12 J (2) 
 
 Divide (1) by (2), a 2 - xy + y 2 = $&• • (3) 
 
 Square (2), x 2 + 2xy + y 2 = 144. (4) 
 Subtract (4) from (3), — Sxy = -|* - 144, 
 
 which gives xy = 32 
 
 ry= 12 
 xy = 32 
 x = 8| 
 y = 4 J ' y 
 
 We now have, x + y = 12 ) 
 
 Solving, we find, x = 8^ x = 4 ) 
 
298 
 
 SIM UL TANE OUS Q UADEA TICS. 
 
 
 Exercise 121. 
 
 
 Solve : 
 
 
 
 
 1. aj + y = 7l 3. 
 
 x - 
 
 -y = 6) 
 
 5. x + y = 12 
 
 xy = 10 J 
 
 xy 
 
 = -*} 
 
 X 2 + y 2 = 80 
 
 2. a + y = 12| 4. 
 
 x - 
 
 -y = 10 
 
 1 6. a; 4 2/ = 3 
 
 a;y = 27 J 
 
 xy 
 
 = 11 
 
 J a; 2 + ?/ 2 = 29 
 
 7. x -y = 9l 
 
 
 17. 
 
 a; 2 -^ 2 = 9l 
 
 a; 2 + y 2 = 45j 
 
 
 
 a; -y =lj 
 
 8. a + 2y = 71 
 
 
 18. 
 
 a; 2 + 3?/ + 17 = 0J 
 
 a 2 + 2/ 2 = 10 J 
 
 
 
 3a; — y = 3 
 
 9. 3a; -y =12l 
 
 Z 2 _^2 =16 J 
 
 10. y = 3x + l 1 
 a; 2 + a;y = 33J 
 
 11. 5a; — 4> = 10*1 
 
 
 19. 
 20. 
 
 a; y 
 
 1 1 / 
 
 a; 2 ?/ 2 
 
 a; y 6 
 
 
 3« a -42/ 2 = 8J 
 12. x + 7y = 2S'\ 
 
 
 
 11 13 
 
 a; 2 + y 2 36 
 
 
 xy = 6 J 
 
 
 21. 
 
 3a; + 2y = 2a-?/j 
 
 13. 2« -3y =2 
 
 a; 2 -2a;2/=-7 
 
 14. 2a; -Sy = l] 
 3a; 2 -4a;*/ = 32 J 
 
 } 
 
 22. 
 
 xy = 6 
 
 l+l=ii 
 
 a; 2 y* 
 
 J 
 
 15. x 2 -xy + y 2 = 21 
 
 •1 
 
 23. 
 
 8a; + 6y = 4a;yj 
 
 x + y = 9 
 
 J 
 
 
 a;y = 16 
 
 16. a; 2 -3a;y + 22/ 2 = 
 
 .01 
 
 24. 
 
 a;» + 2/ 8 = 35\ 
 
 2a; + 3y = 7 
 
 J 
 
 
 a; +y = 5 
 
 J 
 
SIMULTANEOUS QUADRATICS. 
 
 299 
 
 Z'l} 
 
 25. x 8 — y 8 
 
 x -y = 
 
 26. x* + y 8 = 65 
 # 4- y = 5 
 
 27. a 2 ?/ + ay 2 = 120 1 
 x + y = 8 
 
 28. aj «~ y » = ^' 
 a -y = i 
 
 29. a 8 -fy 8 = 126 
 
 x 2 — xy + y 2 = 21 
 
 28 
 
 30. 
 
 a? 3 — t/ 3 = 56 
 
 
 
 a; 2 + xy -f- y 2 = 
 
 31. 
 
 ^ ?^_35" 
 y x 3 
 
 « y. 12 
 
 - 
 
 32. 
 
 .a 2 y 2 _ 19 ' 
 y x 2 
 1 11 
 
 y #"18^ 
 
 - 
 
 33. 
 
 z 2 +a;y = 24l 
 
 
 #y + y 2 = 40 J 
 
 3*4. 
 
 x 2 — xy = 8 ^ 
 
 
 ay — y 2 = 7 
 
 
 35. z 2 + 2zy = 24 1 
 2ay + 4y 2 = 120J 
 
 36. ±x 2 +5 
 7*y4-9 
 
 xy = 14 1 
 y 2 =50j 
 
 37. a 2 + a;y + y 2 = 39 
 2x 2 + 3xy + y 2 = 63 
 
 38. 
 
 x* -f 3y 2 = 52 
 
 
 ay + 2 y 2 = 40 
 
 39. 
 
 2 a 2 - y 2 = 46 ' 
 
 
 xy + y 2 = 14 
 
 40. cc 2 + scy + 2y 2 = 
 2cc 2 -3zy + 2 
 
 = 44 1 
 
 y 2 = 16j 
 
 41. 
 
 x* + 3y 2 = 31 
 
 
 ±xy + y 2 = 33 
 
 42. 
 
 3x*+7xy = 8 
 
 
 x 2 4- 5 xy 4- 9 y 
 
 43. 
 
 a 4 4-y 4 = 97J 
 
 
 a; 4-y = 5J 
 
 44. 
 
 * 4 4-y 4 = 17J 
 ^ 4-y = 3 J 
 
 45. z 4 + y 4 = 881*1 
 x -y = 1J 
 
 ;} 
 
 46. a 6 + y 6 = 211 
 
 x 4-y 
 
 47. a 5 -y 5 = 242*1 
 x -y = 2 J 
 
 48. a; 2 4- y 2 = #y 
 a; 4-y = ay 
 
 . a; 8 — y 3 = 7 ay 1 
 x -y =2 J 
 
300 
 
 SIMUL TANEOUS Q UADRA TICS. 
 61 
 
 60. x* + y* = 36xy] 61. x*-y 8 = a 8 \ 
 
 x +y = 24 J x - y = a ] 
 
 61. x 8 + 3xy 2 = 62 
 3x 2 y + y 8 = 63 
 
 52. x 2 + xy +y 2 = 61 1 
 (c 4 + £cV+2/ 4 = 1281J 
 
 53. a; 2 - xy + y 2 = 3 
 a; 4 + x 2 y 2 + y 4 = 21 
 
 54. x +v | ^ - y __ 10 
 
 #•* i- y 2 
 
 2/ « +y 
 
 20 
 
 55, 
 
 x — y x -f- y _ 24 
 x -\- y x — y 5 
 3a + 4y = 36 
 
 56. x 2 + y 2 + x + y = 
 xy + l(S = 
 
 57. a? - y - 3 = 
 2(x 2 -y 2 ) = 3xy 
 
 32 
 
 58. i + i = 7 
 
 x y 
 
 x+1 y + 1 
 
 31 
 
 20 
 
 59. a 4 -f y 4 = 272 
 
 # 2 + y 2 — 3 xy — 4 
 
 62. * + f = l 
 
 a b 
 
 as y 
 
 63. a? 2 = ax 4" 6y 
 
 y 2 = bx + ay 
 
 64. a 2 -f y 2 = 2 (a 2 + 
 xy = a 2 — b 2 
 
 a 4 + b* 
 
 65. x 2 + 2/ 2 = 
 
 aw 
 
 66. x 2 -y 2 = 
 
 a 2 b 2 
 
 a — b 
 
 a + b 
 
 xy = 
 
 ab 
 
 (a + by 
 
 } 
 
 67. x 2 — xy = 
 
 ay - y 2 
 
 = 2a& + 2& 2 ] 
 = 2a&-2& 2 J 
 
 68. x 2 — 
 xy = b 
 
 tf = a* 
 
 69. x 2 -y 2 = 8ab^ 
 xy = a 2 — 4:b 2 J 
 
 60. x 2 + 2/ 2 = £cy + ll 70. ic 8 + y 8 = a 8 + 5 8 l 
 
 «+y=2icy — lj x + y = a +b \ 
 
SIMULTANEOUS QUADRATICS. 301 
 
 Exercise 122. 
 
 1. The area of a rectangle is 60 square feet, and its 
 perimeter is 34 feet. Find the length and breadth of the 
 rectangle. 
 
 2. The area of a rectangle is 108 square feet. If the 
 length and breadth of a rectangle are each increased by 
 3 feet, the area will be 180 square feet. Find the length 
 and breadth of the rectangle. 
 
 3. If the length and breadth of a rectangular plot are 
 each increased by 10 feet, the area will be increased by 400 
 square feet. But "if the length and breadth are each dimin- 
 ished by 5 feet, the area will be 75 square feet. Find the 
 length and breadth of the plot. 
 
 4. The area of a rectangle is 168 square feet, and the 
 length of its diagonal is 25 feet. Find the length and 
 breadth of the rectangle. 
 
 5. The diagonal of a rectangle is 25 inches. If the 
 rectangle were 4 inches shorter and 8 inches wider, the 
 diagonal would still be 25 inches. Find the area of 
 the rectangle. 
 
 6. A rectangular field, containing 180 square rods, is 
 surrounded by a road 1 rod wide. The area of the road is 
 58 square rods. Find the dimensions of the field. 
 
 7. Two square gardens have a total surface of 2137 
 square yards. A rectangular piece of land whose dimen- 
 sions are respectively equal to the sides of the two squares 
 will have 1093 square yards less than the two gardens 
 united. What are the sides of the two squares ? 
 
 8. The sum of two numbers is 22, and the difference of 
 their squares is 44. Find the numbers. 
 
302 SIMULTANEOUS QUADRATICS. 
 
 9. The difference of two numbers is 6, and their product 
 exceeds their sum by 39. Find the numbers. 
 
 10. The sum of two numbers is equal to the difference 
 of their squares, and the product of the numbers exceeds 
 twice their sum by 2. Find the numbers. 
 
 11. The sum of two numbers is 20, and the sum of their 
 cubes is 2060. Find the numbers. 
 
 12. The difference of two numbers is 5, and the differ- 
 ence of their cubes exceeds the difference of their squares 
 by 1290. Find the numbers. 
 
 13. A number is formed of two digits.. The sum of the 
 squares of the digits is 58. If twelve times the units' 
 digit is subtracted from the number, the order of the digits 
 will be reversed. Find the number. 
 
 14. A number is formed of three digits, the third digit 
 being twice the sum of the other two. The first digit plus 
 the product of the other two digits is 25. If 180 is added 
 to the number, the order of the first and second digits will 
 be reversed. Find the number. 
 
 15. There are two numbers formed of the same two 
 digits in reverse order. The sum of the numbers is 33 
 times the difference of the two digits, and the difference of 
 the squares of the numbers is 4752. Find the numbers. 
 
 16. The sum of the numerator and denominator of a cer- 
 tain fraction is 5 ; and if the numerator and denominator 
 are each increased by 3, the value of the fraction will be 
 increased by £. Find the fraction. 
 
 17. The fore wheel of a carriage turns in. a mile 132 
 times more than the hind wheel ; but if the circumferences 
 were each increased by 2 feet, it would turn only 88 times 
 more. Find the circumference of each. 
 
CHAPTER XXL 
 RATIO, PROPORTION, AND VARIATION. 
 
 316. The relative magnitude of two numbers is called 
 their ratio, when expressed by the fraction which the first 
 is of the second. 
 
 Thus, the ratio of 6 to 3 is indicated by the fraction f , which is 
 sometimes written 6 : 3. 
 
 317. The first term of a ratio is called the antecedent, 
 and the second term the consequent. When the antecedent 
 is equal to the consequent, the ratio is called a ratio of 
 equality ; when the antecedent is greater than the conse- 
 quent, the ratio is called a ratio of greater inequality ; when 
 less, a ratio of less inequality. 
 
 318. When the antecedent and consequent are inter- 
 changed, the resulting ratio is called the inverse of the 
 given ratio. 
 
 Thus, the ratio 3 : 6 is the inverse of the ratio 6 : 3. 
 
 319. The ratio of two quantities that can be expressed 
 in integers in terms of a common unit is equal to the ratio 
 of the two numbers by which they are expressed. 
 
 Thus, the ratio of $9 to $11 is equal to the ratio of 9 : 11 ; and the 
 ratio of a line 2| inches long to a line 3| inches long, when both are 
 expressed in terms of a unit T 2 f of an inch long, is equal to the ratio 
 of 32 : 45. 
 
 320. Two quantities different in kind can have no ratio, 
 for then one cannot be a fraction of the other. 
 
304 RATIO. 
 
 321. Two quantities that can be expressed in integers 
 in terms of a common unit are said to be commensurable. 
 The common unit is called a common measure, and each 
 quantity is called a multiple of this common measure. 
 
 Thus, a common measure of 2| feet and 3| feet is \ of a foot, which 
 is contained 15 times in 2£ feet, and 22 times in 8f feet. Hence, 2\ 
 feet and 3| feet are multiples of J of a foot, 2-J- feet being obtained by 
 taking £ of a foot 15 times, and 3| feet by taking i of a foot 22 times. 
 
 322. When two quantities are incommensurable, that is, 
 have no common unit in terms of which both quantities 
 can be expressed in integers, it is impossible to find a frac- 
 tion that will indicate the exact value of the ratio of the 
 given quantities. It is possible, however, by taking the 
 unit sufficiently small, to find a fraction that shall differ 
 from the true value of the ratio by as little as we please. 
 
 Thus, if a and & denote the diagonal and side of a square, 
 
 Now V2 = 1.41421356 , a value greater than 1.414213, but less 
 
 than 1.414214. 
 
 If, then, a millionth part of b is taken as the unit, the value of the 
 
 ratio - lies between } ftfl | f |g and $$$$tBi and therefore differs from 
 
 either of these fractions by less than 100 ^ 000 . 
 
 By carrying the decimal further, a fraction may be found that will 
 differ from the true value of the ratio by less than a billionth, tril- 
 lionth, or any other assigned value whatever. 
 
 323. Expressed generally, when a and b are incommen- 
 surable, and b is divided into any integral number (n) of 
 equal parts, if one of these parts is contained in a more 
 than m times, but less than m + 1 times, then 
 
 a m , m + 1 
 
 T >->but< ; 
 
 b n n 
 
 that is, the value of - lies between — and • 
 
 ' b n n 
 

 RATIO. 305 
 
 The error, therefore, in taking either of these values for 
 
 - is less than the difference between — and ; that is, 
 
 b n n ' 
 
 less than -• But by increasing n indefinitely, - can be 
 
 made to decrease indefinitely, and to become less than any 
 assigned value, however small, though it cannot be made 
 absolutely equal to zero. 
 
 324. The ratio between two incommensurable quantities 
 is called an incommensurable ratio. 
 
 325. Theorem. Two incommensurable ratios are equal 
 if, when the unit of measure is indefinitely diminished, their 
 approximate values constantly remain equal. 
 
 Let a : b and a' : V be two incommensurable ratios whose 
 true values lie between the approximate values — and 
 
 > when the unit of measure is indefinitely dimin- 
 ished. Then they cannot differ by so much as - • 
 
 Now the difference (if any) between the fixed values a : b 
 and a' : b' is a fixed value. Let d denote this fixed value. 
 
 1 
 
 Then, d< 
 
 n 
 
 But if d has any value, however small, — > which by 
 
 hypothesis can be made less than any value, however small, 
 can be made less than d. 
 
 Therefore, d cannot have any value ; that is, d = 0, and 
 ere is no difference 
 therefore, a : b = a' : b\ 
 
 there is no difference between the ratios a:b and a'lb'; 
 
306 RATIO. 
 
 326. A ratio will not be altered if both its terms are 
 multiplied by the same positive number. 
 
 For the ratio a : b is represented by - ■» the ratio ma : mb is repre- 
 sented by — - : and since — - = - ' therefore, ma : mb = a : b. 
 J mb' mb b ' 
 
 327. A ratio will be altered if its terms are multiplied 
 by different positive numbers ; and will be increased or 
 diminished according as the multiplier of the antecedent is 
 greater than or less than, that of the consequent. 
 
 For ma : nb > or < a : b 
 
 ,. ma . a / na\ 
 
 according as ri >0T< b{ = nb} 
 
 according as ma > or < na, 
 
 according as m > or < n. 
 
 328. A ratio of greater inequality will be diminished, 
 and a ratio of less inequality increased, by adding the same 
 positive number to both its terms. 
 
 For 
 
 a + 
 
 x : & + x > or <C.a:b. 
 
 according as 
 
 
 a + x ^ a 
 6 + x>° r< 6' 
 
 according as 
 
 
 ab + bx > or < ab + ax, 
 
 according as 
 
 
 bx > or < ax, 
 
 according as 
 
 
 6 > or < a. 
 
 329. A ratio of greater inequality will be increased, and 
 a ratio of less inequality diminished, by subtracting the same 
 positive number from both its terms. 
 
 For a — x:b — x > or < a : &, 
 
 _. a — x^ .a 
 
 according as , _ > or < - » 
 
 according as ab — bx > or < ab — ax, 
 according as ax > or < 6x, 
 
 according as a >• or < 6. 
 
RA TIO. 307 
 
 330. Katios are compounded by taking the product of the 
 fractions that represent them. 
 
 Thus, the ratio compounded of a : b and c : d is found by taking the 
 
 , , a ,c ac 
 
 product of - and - = ?-. • 
 b a bd 
 
 The ratio compounded of a : b and a ibis the duplicate ratio a 2 : 6 2 , 
 
 and the ratio compounded of a : 6, a : b, and a : 6 is the triplicate ratio 
 
 a 3 : & 8 . 
 
 331. Ratios are compared by comparing the fractions that 
 represent them. 
 
 Thus, a : b > or < c : d 
 
 a. e 
 
 according as - > or < - » 
 
 ad ^bc 
 
 M> 0I< bd 
 according as ad > or < be. 
 
 according as — > or < ■ 
 
 Exercise 123. 
 
 1. Write the ratio compounded of 3 : 5 and 8 : 7. Which 
 of these ratios is increased, and which is diminished by the 
 composition ? 
 
 2. Compound the duplicate ratio of 4 : 15 with the tripli- 
 cate of 5 : 2. 
 
 3. Show that a duplicate ratio is greater or less than its 
 simple ratio according as it is a ratio of greater or less 
 inequality. 
 
 4. Arrange in order of magnitude the ratios 3:4; 23 : 25 ; 
 10 : 11 ; and 15 : 16. 
 
 5. Arrange in order of magnitude 
 
 a -f b : a — b and a 2 -f b 2 : a 2 — b 2 , if a > b. 
 Find the ratio compounded of: 
 
 6. 3:5; 10:21; 14:15. 7. 7:9; 102:105; 15:17. 
 
BATIO. 
 
 a 2 + ax -\- x 2 ^ a 2 — ax -\- x* 
 
 8. -= - — ■ r and ■ 
 
 a 9 — a z x + aar — x 6 a + x 
 
 n ic 2 -9cc + 20 , x 2 -13x + 42 
 
 9. = - and = = 
 
 ar — o x ar — ox 
 
 10. a + bia-b; a 2 + b 2 : (a + b) 2 -, (a 2 - b 2 ) 2 : a 4 - £ 4 . 
 
 11. Two numbers are in the ratio 2 : 3, and if 9 is added 
 to each, they are in the ratio 3 : 4. Find the numbers. 
 
 Let 2 x and Sx represent the numbers. 
 
 12. Show that the ratio a : b is the duplicate of the ratio 
 a + c : b 4- c, if c 2 = ab. 
 
 ] 3. Two numbers are in the ratio 3 : 4. Their sum is to 
 the sum of their squares as 7 : 50. Find the numbers. 
 
 14. If five gold and four silver coins are worth as much 
 as three gold and twelve silver coins, find the ratio of the 
 value of a gold coin to that of a. silver coin. 
 
 15. If eight gold and nine silver coins are worth as much 
 as six gold and nineteen silver coins, find the ratio of the 
 value of a silver coin to that of a gold coin. 
 
 16. There are two roads from A to B, one of them 14 
 miles longer than the other ; and two roads from B to C, 
 one of them 8 miles longer than the other. The distance 
 from A to B is to the distance from B to C, by the shorter 
 roads, as 1 to 2 ; by the longer roads, as 2 to 3. Find the 
 distances. 
 
 17. What must be added to each of the terms of the ratio 
 m : n, that it may become equal to the ratio p : q? 
 
 18. A rectangular field contains 5270 acres, and its length 
 is to its breadth in the ratio of 31 : 17. Find its dimensions. 
 
PROPORTION. 309 
 
 Proportion. 
 
 332. An equation consisting of two equal ratios is called 
 a proportion ; and the terms of the ratios are called propor- 
 tionals. 
 
 333. The algebraic test of a proportion is that the two 
 fractions which represent the ratios of the quantities com- 
 pared shall be equal. 
 
 Thus, the ratio a : b is equal to the ratio c : d if the fraction that 
 represents the ratio a : b is equal to the fraction that represents the 
 ratio c : d. Then the four quantities, a, 6, c, d, are called proportionals, 
 or are said to be in proportion. 
 
 334. If the ratios a : b and c : d form a proportion, the 
 proportion is written 
 
 a : b = c : d 
 (read the ratio of a to b is equal to the ratio of c to d), 
 or a : b : : c : d 
 
 (read a is to b in the same ratio as c is to d). 
 
 The first and last terms, a and d, are called the extremes. 
 
 The two middle terms, b and c, are called the means. 
 
 335. In the proportion a : b = c : d ; d is called a fourth 
 proportional to a, b, and c. 
 
 In the proportion a : b = b : c ; c is called a £AiVd propor- 
 tional to a and J. 
 
 In the proportion a : b = b : c ; 5 is called a meaw propor- 
 tional between a and c. 
 
 336. A continued proportion is a series of equal ratios in 
 which each consequent is the same as the next antecedent. 
 
 Thus, a:& = 6:c = c:d = d:e = e:/isa continued proportion. 
 
310 PROPORTION. 
 
 337. When four quantities are in proportion, the product 
 of the extremes is equal to the product of the means. 
 
 For, if a : b = c : d, 
 
 ^ a - c . 
 
 then, l -- 
 
 Multiply by bd, ad = be. 
 
 The equation ad = be gives a = — > b = — ; so that an 
 
 a c 
 
 extreme may be found by dividing the product of the means 
 by the other extreme ; and a mean may be found by divid- 
 ing the product of the extremes by the other mean. 
 
 Note. By the product of two quantities we mean the product 
 of the two numbers that represent them when the quantities are 
 expressed in a common unit. 
 
 338. If the product of two quantities is equal to the prod* 
 net of two others, either two may be made the extremes of 
 a proportion and the other two the means* 
 
 For, if 
 then, divide by bd, 
 
 or 
 
 339. Transformations of a Proportion. If four quantities, 
 a, b, c, d, are in proportion, they will be in proportion by : 
 
 I. Inversion ; that is, b will be to a as d is to c. 
 For, if a:b = c:d, 
 
 ±v a c 
 
 ad — 
 
 be, 
 
 ad 
 bd~ 
 
 be 
 bd! 
 
 a. 
 b~ 
 
 c 
 3" 
 
 a\b — 
 
 ■ c:d. 
 
PROPORTION. 311 
 
 and the reciprocals of these fractions are equal ; 
 
 that is, - = -• 
 
 a c 
 
 .'. b:a = d : c. 
 II. Composition ; that is, a + b will be to b as c + d is to d. 
 For, if a:b = c:d f 
 
 then, 
 
 a _c 
 b~ d 
 
 and 2 + 1 = ^ + 1* 
 
 or 
 
 then, 
 
 a + b _ £_+d 
 6 ~~ _ d 
 .'. a + b:b = c + did. 
 
 III. Division ; that is, <x — & will be to b as c — d is to d. 
 
 For, if a:b = c:d, 
 
 a _c 
 b~ d' 
 
 and F" 1 ^- 1 ' 
 
 a — b c ~ d 
 
 or 
 
 '.a — b :b = c — did. 
 
 IY. Composition and Division ; that is, a + b will be to 
 a — b as c + d is to c — d. 
 
 For, from II, a + b C + d 
 
 and from III, 
 
 Divide, 
 
 a — b c — d 
 
 o\ a + b :a — b = c + d: c — d. 
 
 
 b 
 
 d 
 
 a 
 
 -b 
 
 c — d 
 
 
 b 
 
 d 
 
 a 
 
 + b 
 
 c + d 
 
SV2 
 
 PROPORTION. 
 
 V. Alternation; 
 
 that is, a will be to c as b is to 
 
 For, if 
 
 a : b = e : d, 
 
 then, 
 
 a e 
 b~d' 
 
 Multiply by -> 
 
 ab be 
 be cd 
 
 or 
 
 a b 
 
 c d 
 
 
 . \ a: e = b :d. 
 
 Note. In order for four quantities, a, 6, c, d, to be in proportion, 
 a and b must be of the same kind and c and d of the same kind ; but 
 c and d need not necessarily be of the same kind as a and 6. In 
 applying alternation, however, all four quantities must be of the same 
 kind. 
 
 340. In a Series of Equal Ratios, the sum of the antecedents 
 is to the sum of the consequents as any antecedent is to its 
 
 -n •* a e e g 
 
 r may be put for each of these ratios. 
 
 mi a c e a 
 
 Then, - = r , 5 = r ,- = r ,| = n 
 
 .*. a — br, c = dr, e =fr, g = hr. 
 .'. a + c + e + gr = (b + d +f+ h)n 
 
 a + c-\- e + g _ _ a 
 '''b + d+f + h~ r ~V 
 .'. a + c+e + gib + d +/+ A = a : &. 
 
 In like manner it may be shown that 
 
 ma + nc +j>e-h qg :mb + nd + pf + qhs=a:b. 
 
PROPORTION. 313 
 
 341. A mean proportional between two quantities is equal 
 to the square root of their product. 
 
 For, if a:b = b:c, 
 
 xv a b 
 
 then, T = ~' 
 
 ' b o 
 
 Clear of fractions, b 2 = ac. 
 
 Extract the square root, b = Vac. 
 
 342. The products of the corresponding terms of two or 
 more proportions are in proportion. 
 
 For, if a:b = c.d i 
 
 e:f=g:h, 
 k:l = m:n, 
 a c e g k m 
 
 Take the product of the left members, and also of the 
 right members of these equations, 
 aek _ cgm 
 bfl dhn 
 .*. aek : bfl = cgm : dhn. 
 
 343. Like powers, or like roots, of the terms of a propor- 
 tion are in proportion. 
 
 For, if a : b = c : d, 
 
 then, 
 
 a 
 b~ 
 
 e 
 *3 r 
 
 Raise both sides to the nth. 
 
 power, 
 
 
 a n 
 b n ~~ 
 
 c n 
 ~ d» 
 
 ,» a n : b n = c n : d» 
 
314 PROPORTION. 
 
 Extract the nth root, -j = ■— • 
 
 1111 
 .*. a n : b n = c n : d n . 
 
 344. The laws that have been established for ratios 
 should be remembered when ratios are expressed in frac- 
 tional form. 
 
 , . x 2 + x + 1 x 2 - x -f 2 
 1. Solve -j r = -j— -• 
 
 By composition and division, 
 
 2 x 2 2 x 2 
 
 2 (x + 1) - 2 (x - 2) 
 
 This equation is satisfied when x = 0. For any other value of x, 
 we may divide by x 2 . 
 
 We then have — j-r = » 
 
 x + 1 2 — a; 
 
 and therefore, x = ^. 
 
 2. If a : b = c : <£, show that 
 
 a 2 + ab:b 2 -ab = c 2 + cd:d 2 - cd. 
 
 If 
 
 o_ c 
 b~d* 
 
 
 then, 
 
 a + b c + d 
 a — b c—d 
 
 (§ 339, IV) 
 
 and 
 
 a c 
 -b -d 
 
 
 
 " — 6 a — b — d c — d* 
 
 (§342) 
 
 that is, 
 
 a 2 + a&_c 2 + cd^ 
 V> — ab d* — cd 
 
 
 or 
 
 a 2 + ab i ft 2 - a& = c 2 + cd : d 2 - cd„ 
 
 
PROPORTION. 315 
 
 Exercise 124. 
 
 1. Find a third proportional to 21 and 28. 
 
 2. Find a mean proportional between 6 and 24 
 
 3. Find a fourth proportional to 3, 5, and 42. 
 
 4. Find a; if 5 + x : 11 - x = 3 : 5. 
 
 If a : b = c : <7, show that : 
 
 5. ac:bd = c 2 :d 2 . 7. a 2 - b 2 : c 2 - d 2 = a 2 : c*. 
 
 6. ab:cd = a 2 :c 2 . 8. 2a + b :2c + d = b ;d. 
 9. 5a — b : 5c — d = a: c. 
 
 10. a-3£:a + 36 = c-3d:c + 3d. 
 
 11. a 2 + a& + 6 2 :a 2 -aM-& 2 = c 2 + cd + d 2 :c 2 -cd + d a . 
 
 Find x in the proportion : 
 
 12. 45:68 = 90:z. 14. x : 1\ = If : If . 
 
 13. 6:3 = a: 7. 15. 3 : ic = 7 : 42. 
 
 16. Find two numbers in the ratio 2 : 3, the sum of whose 
 squares is 325. 
 
 17. Find two numbers in the ratio 5 : 3, thd difference 
 of whose squares is 400. 
 
 18. Find three numbers which are to each other as 
 2:3:5, such that half the sum of the greatest and least 
 exceeds the other by 25. 
 
 19. A and B trade with different sums. A gains $200 
 and B loses $50 and now A's stock : B's : : 2 : £. But, if A 
 had gained $100 and B lost $85, their stocks would have 
 been as 15 : 3£, Find the original stock of each. 
 
316 VARIA TION. 
 
 20. Find a; if 6x - a : ±x - b = 3x + b :2x + a. 
 
 21. Find x and y from the proportions 
 
 x :y = a; -f y > 42 ; x :y = x — y :6. 
 
 22. Find <r and y from the proportions 
 
 2x + y:y = 3y:2y — x', 
 2x + l:2x + 6 = y:y + 2. 
 
 a+b+c+d a—b+c—d a e 
 
 23. If — — ■ - = 1 —> show that - = -=• 
 
 a + b — c — d a — b — c + d b d 
 
 Variation. 
 
 345. One quantity is said to vary as another, when tho 
 two quantities are so related that the ratio of any two values 
 of the one is equal to the ratio of the corresponding values 
 of the other. 
 
 Thus, if it is said that the weight of water varies as its volume, the 
 meaning is, that one gallon of water is to any number of gallons as the 
 weight of one gallon is to the weight of the given number of gallons. 
 
 346. Two quantities may be so related that when a value 
 of one is given, the corresponding value of the other can be 
 found. In this case one quantity is said to be a function 
 of the other ; that is, one quantity depends upon the other 
 for its value. Thus, if the rate at which a man walks is 
 known, the distance he walks can be found when the time 
 is given ; the distance is in this case a funetion of the time. 
 
 347. There is an unlimited number of ways in which 
 tw© quantities may be related. We shall consider in this 
 chapter only a few of these ways. 
 
 348. When x and y are so related that their ratio is 
 constant } y is said to vary as x \ this is abbreviated thus : 
 
VARIATION. 317 
 
 y oo x. The sign oo, called the sign of variation, is read 
 varies as. 
 
 Thus, the area of a triangle with a given base varies as its altitude ; 
 for, if the altitude is changed in any ratio, the area will be changed in 
 the same ratio. 
 
 In this case, if we represent the constant ratio by m, 
 
 V 
 y : x = m, or - = m ; .*. y == mx. 
 x 
 
 Again, if y\ x' and y", x" are two sets of corresponding 
 values of y and x, 
 
 then, y 1 : x' = y" : x", 
 
 or y':y" = x':x". (§339, V) 
 
 349. When x and y are so related that the ratio of y to - 
 
 * x 
 
 is constant, y is said to vary inversely as x ; this is written 
 
 1 
 
 y go -• 
 7 x 
 
 Thus, the time required to do a certain amount of work varies 
 inversely as the number of workmen employed ; for, if the number of 
 workmen is doubled, halved, or changed in any other ratio, the time 
 required will be halved, doubled, or changed in the inverse ratio. 
 
 In this case, y :— = m \ .*. y = — > and xy = m; that is, 
 x x 
 
 the product xy is constant. 
 
 1 1 
 
 As before, y : - = 2/ " :: _, 
 
 Jj JO 
 
 x'y' = x"y", 
 or y' : y" = x" : x'. (§ 338) 
 
 350. If the ratio of y : xz is constant, then y is said to 
 vary jointly as x and z. 
 
 In this case, y = mxz, 
 
 and y' :y" = x'z' : x"z". 
 
318 VARIATION. 
 
 x 
 
 351. If the ratio y : - is constant, then y varies directly 
 z 
 
 as x and inversely as z. 
 
 TYbX 
 
 In this case, y = — > 
 
 and ?/ : y" = — r : — r- = — : — • 
 
 J 9 z' z" z' z" 
 
 352. Theorem 1. Ifyaox, and xao z, then y oo z. 
 
 For y = mx and x = nz. 
 
 .'. y = mnz\ 
 since mw is constant, y varies as z. 
 
 353. Theorem 2. If y go x, and zaox, then (y ± 2) oo a?. 
 For 1/ = m# and z = wsc. 
 
 .'. y ± z = (m ± w) a; ; 
 since m + nis constant, y ±z varies as x. 
 
 354. Theorem 3. If y 00 x when z is constant, and y 00 z 
 when x is constant, then y 00 xz when x and z are both 
 variable. 
 
 Let x', y', z', and x", y", z" be corresponding values of the 
 variables. 
 
 Let x change from x' to x", z remaining constant, and let 
 the corresponding value of y be Y. 
 
 Then, y':Y=x':x". (1) 
 
 Now let z change from z' to z", x remaining constant. 
 
 Then, Y:y" = *':z". (2) 
 
 Multiply (1) and (2), 
 
 y'Y:y"Y=x'z , :x"z", (§342) 
 
 or y' : y" = x'z' : x"z", 
 
 or y' : x'z' = y" : x"z". (§ 339, V) 
 
 .*. the ratio y : xz is constant, and y varies as xz. 
 
VARIATION. 319 
 
 In like manner it may be shown that if y varies as each 
 of any number of quantities x, z, u, etc., when the rest are 
 unchanged, then when they all change, y oo xzu, etc. 
 
 Thus, the volume of a rectangular solid varies as Jhe length when 
 the width and thickness remain constant; as the width when the 
 length and thickness remain constant; as the thickness when the 
 length and width remain constant; but as the product of length, 
 breadth, and thickness when all three vary. 
 
 1. If y varies inversely as x, and when y = 2 the cor- 
 responding value of x is 36, find the corresponding value 
 
 of x when y = 9. 
 
 m 
 Here y = — » or m — xy. 
 
 .-. m = 2 X 36 = 72. 
 
 If 9 and 72 are substituted for y and m, respectively, 
 
 72 
 the result is 9 = — i or 9 x = 72. 
 
 x 
 
 .-. x = S. 
 
 2. The weight of a sphere of given material varies as 
 its volume, and its volume varies as the cube of its diam- 
 eter. If a sphere 4 inches in diameter weighs 20 pounds, 
 find the weight of a sphere 5 inches in diameter. 
 
 Let W represent the weight, 
 
 V represent the volume, 
 and D represent the diameter. 
 
 Then, W oo V and V oo D 8 . 
 
 .-. TTooD 8 . (§352) 
 
 Put W = mD 3 ; 
 
 then, since 20 and 4 are corresponding values of W and D, 
 20 = m x 64. 
 20 _ 5 
 ••• m -64-16* 
 .*. IF = £ 2* 
 .-. when D = 5, W ** A °* 125 = 39 xV 
 
320 VARIATION. 
 
 Exercise 125. 
 
 1. If x oo y, and if y — 3 when x = 5, find x when y is 5. 
 
 2. If JF varies inversely as P, and W is 4 when P is 
 15, find JFvhenPis 12. 
 
 3. If x oo y and y oo z, show that xz oo y 2 . 
 
 4. If x oo - and y oo -? show that a; oo 2. 
 
 5. If x varies inversely as y 2 — 1, and is equal to 24 
 when y — 10, find x when y = 5. 
 
 6. If sc varies as - -+- -> and is equal to 3 when y = 1 
 
 y ■;-* 
 
 and « = 2, show that xyz — 2(g + z). 
 
 7. If x — y varies inversely as z + -> and x + y varies 
 
 z 
 
 inversely as z > find the relation between x and z if 
 
 x = l } y = 3, when £ = - • 
 
 8. The area of a circle varies as the square of its radius, 
 and the area of a circle whose radius is 1 foot is 3.1416 
 square feet. Find the area of a circle whose radius is 20 
 feet. 
 
 9. The volume of a sphere varies as the cube of its 
 radius, and the volume of a sphere whose radius is 1 foot is 
 4.1888 cubic feet. Find the volume of a sphere whose radius 
 is 2 feet. 
 
 10. If a sphere of given material 3 inches in diameter 
 weighs 24 pounds, how much will a sphere of the same 
 material weigh if its diameter is 5 inches ? 
 
VARIA TION. 321 
 
 11. The velocity of a falling body varies as the time 
 during which it has fallen from rest. If the velocity of a 
 falling body at the end of 2 seconds is 64 feet, what is its 
 velocity at the end of 8 seconds ? 
 
 12. The distance a body falls from rest varies as the 
 square of the time it is falling. If a body falls through 
 144 feet in 3 seconds, how far will it fall in 5 seconds ? 
 
 The volume of a right circular cone varies jointly as its 
 height and the square of the radius of its base. 
 
 13. Compare the volume of two cones, one of which is 
 twice as high as the other, but with one half its diameter. 
 
 If the volume of a cone 7 feet high with a base whose 
 radius is 3 feet is 66 cubic feet : 
 
 14. Find the volume of a cone 9 feet high with, a base 
 whose radius is 3 feet. 
 
 15. Find the volume of a cone 7 feet high with a base 
 whose radius is 4 feet. 
 
 16. Find the volume of a cone 9 feet high with a base 
 whose radius is 4 feet. 
 
 17. The volume of a sphere varies as the cube of its 
 radius. If the volume is 179§ cubic feet when the radius 
 is 3£ feet, find the volume when the radius is 1 foot 6 
 inches. 
 
 18. Find the radius of a sphere whose volume is the sum 
 of the volumes of two spheres with radii 3£ feet and 6 feet, 
 respectively. 
 
 19. The distance of the offing at sea varies as the square 
 root of the height of the eye above the sea level, and the 
 distance is 3 miles when the height is 6 feet. Find the 
 distance when the height is 24 feet. 
 
CHAPTER XXTL 
 PROGRESSIONS. 
 
 355. A succession of numbers that proceed according to 
 some fixed law is called a series ; the successive numbers are 
 called the terms of the series. 
 
 A series that ends at some particular term is a finite series ; 
 a series that continues without end is an infinite series. 
 
 Arithmetical Progression. 
 
 356. A series is called an arithmetical series or an arith- 
 metical progression when each term after the first is obtained 
 by adding to the preceding term a constant difference. 
 
 The general representative of such a series is 
 
 1st 2d 3d 4th 
 
 a, a + d , a-{-2d, a + 3 d , 
 
 in which a is the first term and d the common difference ; 
 the series is increasing or decreasing according as d is 
 positive or negative, 
 
 357. The nth Term. Since each succeeding term of the 
 series is obtained by adding d to the preceding term, the 
 coefficient of d is always one less than the number of 
 the term, so that the nth term is a -f- (n — 1) d. 
 
 If the nth. term is represented by I, we have 
 I = a -f* (n — l)d. 
 
ARITHMETICAL PROGRESSION. 323 
 
 358. Sum of the Series. If I denotes the nth. term, a the 
 first term, n the number of terms, d the common difference, 
 and s the sum of n terms, it is evident that 
 
 s= a + (a + d) + (a + 2d) + + (l-d) + I, or 
 
 s = I 4. (Z -.^-f-(7 _2<Z) + + (a + rf)+ a, 
 
 S.2s=(a + l) + (a + l)-\-(a + l) + + (a+Q+(a + 
 
 = n(a + l). 
 
 359. From the equations 
 
 I, l = a + (n-l)d, 
 
 any two of the five numbers a, d, I, n, s may be found when 
 the other three are given. 
 
 1. Find the thirteenth term of an arithmetical progres- 
 sion, if the first term is 3 and the common difference 5. 
 
 Here a = 3, d = 5, n = 13. 
 
 From I, I = 3 + (13 — 1) 5 = 63. 
 
 2. Find the arithmetical series, if the tenth term is 31 
 and the twentieth term 61. 
 
 From I, a + 19 d = 61 
 
 and a+ 9d = 31 
 
 Subtract 10 d = 30 
 
 Whence d = 3. 
 
 Therefore, a = 4. 
 
 Therefore, the series is 4, 7, 10 
 
324 ARITHMETICAL PROGRESSION. 
 
 3. Find the sum of ten terms of the series 2, 5, 8, 11 
 
 Here a = 2, d = 3, n = 10. 
 
 From I, I = 2 + 27 = 29. 
 
 Substitute in n, s = ~ (2 + 29) = 155. 
 
 So 
 
 The sum of ten terms is 155. 
 
 4. The first term of an arithmetical series is 3, the last 
 term 31, and the sum of the series 136. Find the series. 
 
 From I, 
 
 
 
 81 = 3 + (n - 1 
 
 From II, 
 
 
 
 136 = | (3 + 31) 
 
 From (2), 
 
 
 
 n = 8. 
 
 Substitute 
 
 in 
 
 (1), 
 
 d = 4. 
 
 The series is 3, 7, 11, 15, 19, 23, 27, 31. 
 
 5. How many terms of the series 5, 9, 13 must 1 
 
 taken in order that their sum may be 275 ? 
 
 From I, J = 5+(n-l)4. 
 
 .-. Z = 4n + 1. (1) 
 
 FromH, 275 = | (5 + 0- (2) 
 
 Substitute in (2) the value of I found in (1), 
 275 = |(4n + 6), 
 or 2 n 2 + 3 n = 275. 
 
 Complete the square, 
 
 16n 2 + ( ) + 9 = 2209. 
 Extract the root, 4 n + 3 = ± 47. 
 
 Therefore, n = 11, or — 12£. 
 
 We use only the positive result. 
 Therefore, 11 terms must be taken. 
 
ARITHMETICAL PROGRESSION. 
 
 325 
 
 6. Find n when d, l f s are given. 
 From I, 
 From II, 
 
 a = I — (n — l)<L 
 2 s — In 
 
 Therefore, 
 
 l-(n — l)d 
 
 n 
 2 s — In 
 
 In — dn 2 + dn — 2s — ln. 
 dn 2 — (2l + d)n- —2s. 
 Complete the square, 
 
 4d 2 n 2 - ( ) + (2 1 + eZ) 2 = (21 + d)* - Sds. 
 Extract the root, 
 
 2dn— {21 + d) = ±V(2J + d) 2 - 8ds. 
 
 .'. n 
 
 21 + d ± V(2 I + d) 2 - sTs 
 
 2d 
 
 360. The arithmetical mean between two numbers is the 
 number which stands between them, and makes with them 
 an arithmetical series. 
 
 If a and b represent two numbers, and A their arithmet- 
 ical mean, then a, A, b are in arithmetical progression. 
 
 .*. A — a = d f and b — A = d. 
 
 .'. A — a = b — A. 
 
 • - A = — r 
 
 361. Sometimes it is required to insert several afithmet- 
 teal means between two numbers. 
 
 Insert six arithmetical means between 3 and 17. 
 Here the whole number of terms is eight ; therefore, by I, 
 17 = 3 + 7tf. 
 .\ d = 2. 
 
 The series is 3, [5, 7, 9, 11, 13, 15,] 17, the terms in 
 brackets being the means required. 
 
 Note. The student should work out all the problems on the follow- 
 page, using the formulas I and H. 
 
326 
 
 ARITHME TICAL PROGRESSION. 
 
 No. 
 
 Given. 
 
 Required. 
 
 Formulas. 
 
 1 
 2 
 3 
 4 
 
 adn 
 ads 
 an s 
 d n s 
 
 1 
 
 l = a+(n-l)d. 
 
 l = —±d± V2ds-f (a — id)*. 
 
 I = a. 
 
 8 + .(n-l)d. 
 n 2 
 
 5 
 6 
 
 7 
 8 
 
 adn 
 a d I 
 a n I 
 d n I 
 
 s 
 
 s = in[2a + (n — l)d]. 
 
 l + a . Z 2 -a 2 
 S= 2 + 2d ' 
 
 s=(l + a)^ 
 
 s = in[2l—(n — l)d]. 
 
 9 
 10 
 11 
 12 
 
 d n I 
 dn s 
 d Is 
 n I s 
 
 a 
 
 a = I — (n — 1) d. 
 s (n — 1) d 
 a= n 2 * 
 
 a = id±V(l + id)2-2ds. 
 
 2s 
 
 a = 1. 
 
 n 
 
 13 
 14 
 15 
 16 
 
 a n I 
 an s 
 a I s 
 n Is 
 
 d 
 
 , I — a 
 d= • 
 n — 1 
 
 d _2(s-an) i 
 n(n — l) 
 P - a 2 
 28 — I — a 
 
 M 2 (In- s) 
 n (n — 1) 
 
 17 
 18 
 19 
 20 
 
 a d I 
 ads 
 a Is 
 dls 
 
 n 
 
 n = — -=— + 1. 
 a 
 
 d - 2 a ± V(2 a — d) 2 + 8 ds 
 n ~ 2d 
 
 2s 
 
 n = — — • 
 l + a 
 
 2Z + d±V(2Z + d) 2 -8ds 
 n = 2d 
 
ARITHMETICAL PROGRESSION. 327 
 
 Exercise 126. 
 
 1. Find the tenth term of 9, 13, 17 
 
 2. Find the thirteenth term of — 3, — 1, -f- 1 
 
 3. Find the ninth term of — 5, — 8, — 11 
 
 4. Find the eighth term of a, a + 3 b, a -f 6 b 
 
 6. Find the fifteenth term of 1, f , f 
 
 6. Find the fourteenth term of - 44, - 40, — 36 
 
 7. The first term of an arithmetical series is 3, the thir- 
 teenth term is 55. Find the common difference. 
 
 8. Find the arithmetical mean between — 5 and 17 ; 
 between a 2 + ab + b 2 and a 2 - ab + b 2 . 
 
 9. Insert three arithmetical means between 1 and 19; 
 and four between — 4 and 17. 
 
 10. The first term of a series is 2, and the common differ- 
 ence £. What term will be 12 ? 
 
 1 1 . The seventh term of a series, whose common difference 
 is 3, is 11. Find the first term. 
 
 Find the sum of : 
 
 12. 5 + 8 + 11 + to ten terms. 
 
 13. — 4 — 1 -f 2 + to seven terms. 
 
 14. a + 4«-f 7a+ to n terms. 
 
 15. § + ^3- + ^ + to twenty-one terms. 
 
 16. 1 -f 2§ + 4£ + to twenty terms. 
 
 17. The sum of six terms of an arithmetical series is 27, 
 and the first term is 1. Find the series. 
 
 18. How many terms of the series — 5, — 2, + 1 
 
 must be taken that their sum may be 63 ? 
 
 19. The first term of an arithmetical series is 12, and the 
 sum of ten terms is 10. Find the tenth term. 
 
328 ARITHMETICAL PROGRESSION. 
 
 20. When a train arrives at the top of a long slope, the 
 last car is detached and begins to descend, passing over 3 
 feet in the first second, three times 3 feet in the second 
 second, five times 3 feet in the third second, etc. At the 
 end of 30 seconds it reaches the bottom of the slope. Find 
 its velocity the last second. 
 
 21. Insert eleven arithmetical means between 1 and 12. 
 
 22. The first term of an arithmetical series is 3, and the 
 sum of six terms is 28. What term will be 9 ? 
 
 23. The arithmetical mean between two numbers is 10, 
 and the mean between the double of the first and the 
 triple of the second is 27. Find the numbers. 
 
 24. The first term of an arithmetical progression is 3, 
 and the third term is 11. Find the sum of seven terms. 
 
 25. A common clock strikes the hours from 1 to 12. 
 How many times does it strike every 24 hours ? 
 
 26. The Greenwich clock strikes the hours from 1 to 24. 
 How many times does it strike in 24 hours ? 
 
 27. Find three numbers in arithmetical progression of 
 which the sum is 21, and the sum of the first and second 
 j of the sum of the second and third. 
 
 Let x — y, x, and x + y stand for the numbers. 
 
 28. The sum of three numbers in arithmetical progres- 
 sion is 33, and the sum of their squares is 461. Find the 
 numbers. 
 
 29. The sum of four numbers in arithmetical progres- 
 sion is 12, and the sum of their squares is 116. What are 
 the numbers ? 
 
 Let x — 3 y, x — y, z + y, and x + 3 y stand for the numbers. 
 
GEOMETRICAL PROGRESSION. 329 
 
 Geometrical Progression. 
 
 362. A series is called a geometrical series or a geometrical 
 progression when each succeeding term is obtained by mul 
 tiplying the preceding term by a constant multiplier. 
 
 The general representative of such a series is 
 
 a } ar, ar 2 , ar z , ar A , 
 
 in which a is the first term and r the constant multiplier, or 
 ratio. 
 
 The terms increase or decrease in numerical magnitude 
 according as r is numerically greater than or numerically 
 less than unity. 
 
 363. The nth Term. Since the exponent of r increases 
 by one for each succeeding term after the first, the exponent 
 is always one less than the number of the term, so that 
 the nth term is ar n ~ x . 
 
 If the nth term is represented by I, we have 
 I, 1 = ar"" 1 . 
 
 364. Sum of the Series. If I represents the nth. term, a the 
 first term, n the number of terms, r the common ratio, and 
 s the sum of n terms, then, 
 
 s = a + ar + ar 2 + + ar"" 1 . (1) 
 
 Multiply by r, 
 
 rs = ar + ar 2 + ar 8 + •— • + ar"" 1 + ar", (2) 
 Subtract (1) from (2), 
 rs — s — ar 11 — a, 
 or (r — 1) s = a (r* — 1). 
 
 n, .-..- 'fr- 1 * - 
 
 ■ r — 1 
 
330 GEOMETRICAL PROGRESSION. 
 
 365. When r < 1, formula II may be more conveniently 
 written 
 
 a (1 - r") 
 1 — r 
 
 366. I = ar n ~\ (§ 363) 
 Multiply by r, rl = ar n . 
 
 By putting rl for ar n in formula II, we have 
 
 ttt rl — a 
 
 III, s = r - 
 
 r — 1 
 
 When t* < 1, formula III is more conveniently written 
 
 a — ri 
 
 367. From the two formulas I and II, or the two for- 
 mulas I and III, any two of the five numbers a, r, I, n T s 
 may be found when the other three are given. 
 
 1. Find the tenth term of a geometrical series if the 
 first term is 3 and the ratio 2. 
 
 Here a = 3, r = 2, n = 10. 
 
 From I, I = 3 X 2 9 = 3 X 512 = 1536. 
 
 2. Find the geometrical series if the third term is 20 and 
 the sixth term 160. 
 
 Let a = the first term, and r = the ratio. 
 
 Then, ar 2 = the third term, and ar 5 = the sixth term. 
 
 _ , ar* 160 
 Therefore, - = — - 
 
 r« = 8. 
 . . r ~ 2. 
 Since ar 2 = 20, a = 20 + 4 = 6. 
 
 The series is 5, 10, 20, 40 
 
 3. Find the sum of six terms of the series 3, 6, 12 
 Here a = 3, r — 2, n = 6. 
 
 1 r— 1 r— 1 
 
192 
 
 = 3r»-i. 
 
 
 381 
 
 192 r — 
 
 3 
 
 r-1 
 
 
 r 
 
 = 2. 
 
 
 2»-i 
 
 = 64. 
 
 
 .-. n 
 
 = 7. 
 
 
 GEOMETRICAL PROGRESSION. 331 
 
 4. The first term of a geometrical series is 3, the last 
 
 term 192, and the sum of the series 381. Find the number 
 
 of terms and the ratio. 
 
 From I, 192 = 3r»-i. (1) 
 
 i qo r o 
 
 From III, 381 = m _ , . (2) 
 
 From (2), 
 Substitute in (1), 
 
 The series is 3, 6, 12, 24, 48, 96, 192. 
 
 368. The geometrical mean between two numbers is the 
 number which stands between them, and makes with them 
 a geometrical series. 
 
 If a and b denote two numbers, and G their geometrical 
 mean, then a, G, b are in geometrical progression, and by 
 the definition of a geometrical series (§ 362), 
 
 G , b 
 
 — = r, and — = r» 
 
 a G 
 
 • ~ = A 
 "a G 
 
 .'. G=^fab. 
 
 369. Sometimes it is required to insert several geomet- 
 rical means between two numbers. 
 
 Insert three geometrical means between 3 and 48. 
 
 Here the whole number of terms is five ; 3 is the first term and 48 
 the fifth. 
 
 By I, 48 = 3r*. 
 
 r* = 16. 
 .-.r=±2. 
 
 The series is 3, [ 6, 12, 24], 48; 
 
 or 3, [-6, 12, -24], 48. 
 
 The terms in brackets are the means required 
 
 In working out the following results, the student shoald make use 
 of formulas I, II, and ILL 
 
332 
 
 GEOMETRICAL PROGRESSION. 
 
 No. 
 
 Given. 
 
 Required. 
 
 Formulas. 
 
 1 
 
 2 
 3 
 4 
 
 am 
 ar s 
 an s 
 r n s 
 
 1 
 
 1 = ar 1 - 1 . 
 l _a + (r-l)s_ 
 r 
 
 I (s — O"" 1 — a (s — a) n ~ l = 0. 
 _ (r - 1) sr*- 1 . 
 r n — 1 
 
 5 
 6 
 
 7 
 
 8 
 
 ar n 
 a r I 
 
 a n I 
 
 ml 
 
 8 
 
 . _ a (r" - 1) 
 r — 1. 
 rl — a 
 S= r-l ' 
 
 n — 1, — n — 1. — 
 
 S — : : 
 
 n — 1 . n— 1 ,— 
 
 Vi - v^ 
 
 lr*-l 
 
 8 = r * 
 
 yn — pa — 1 
 
 9 
 10 
 11 
 12 
 
 r n I 
 r n s 
 r I s 
 n I s 
 
 a 
 
 _ I 
 f-n — 1 
 
 (r-l) s 
 r n — 1 
 
 a = rl — (r — 1) s. 
 
 a(s — a)"- 1 - J (s - l)»- J = 0. 
 
 13 
 14 
 15 
 16 
 
 a n I 
 an s 
 a I s 
 n I 8 
 
 r 
 
 s . s — a rt 
 
 r n r -J = 0. 
 
 a a 
 s — a 
 
 r" . r"- 1 H . = 0. 
 
 s — l s — l 
 
GEOMETRICAL PROGRESSION. 333 
 
 Exercise 127. 
 
 X. Find the seventh term of 2, 6, 18 
 
 2. Find the sixth term of 3, 6, 12 
 
 3. Find the ninth term of 6, 3, lj- 
 
 4. Find the eighth term of 1, — 2, 4 
 
 5. Find the fifth term of 4 a, — 6 ma 2 , 9 m 2 a 8 
 
 6. Find the geometrical mean between 18 x*y and 3<)xy z z. 
 
 7. Find the ratio when the first and third terms are 5 
 and 80, respectively. 
 
 8. Insert two geometrical means between 8 and 125; 
 and three between 14 and 224. 
 
 9. If a = 2 and r = 3, which term will be equal to 162 ? 
 
 10. The fifth term of a geometrical series is 48, and the 
 ratio 2. Find the first and seventh terms. 
 
 Find the sum of : 
 
 11. 3 + 6 + 12 + to eight terms. 
 
 12. 1 — 34-9 — to seven terms. 
 
 13. 8 + 4 + 2 -f to ten terms. 
 
 14. 0.1 + 0.5 4- 2.5 4- to seven terms. 
 
 15. m — 7 4- rrz — to five terms. 
 
 4 16 
 
 16. The population of a city increased in four years from 
 10,000 to 14,641. What was the annual rate of increase ? 
 
 17. The sum of four numbers in geometrical progression 
 is 200, and the first term is 5. Find the ratio. 
 
 18. Find the sum of eight terms of a geometrical series 
 whose last term is 1, and fifth term £. 
 
334 GEOMETRICAL PROGRESSION. 
 
 19. In an odd number of terms, show that the product of 
 the first and last is equal to the square of the middle term. 
 
 20. The product of four terms of a geometrical series is 
 4, and the fourth term is 4. Find the series. 
 
 21. If from a line one third is cut off, then one third of 
 the remainder, and so on, what fraction of the whole will 
 remain when this has been done five times ? 
 
 22. Of three numbers in geometrical progression, the 
 sum of the first and second exceeds the third by 3, and the 
 sum of the first and third exceeds the second by 21. What 
 are the numbers ? 
 
 23. Find two numbers whose sum is 3£ and geometrical 
 mean l£. 
 
 24. The sum of the squares of two numbers exceeds twice 
 their product by 576 ; the arithmetical mean of the two 
 numbers exceeds the geometrical mean by 6. Find the 
 numbers. 
 
 25. There are four numbers such that the sum of the 
 first and the last is 11, and the sum of the other two is 10. 
 The first three of these four numbers are in arithmetical 
 progression, and the last three are in geometrical progres- 
 sion. Find the numbers. 
 
 26. Find three numbers in geometrical progression such 
 that their sum is 13 and the sum of their squares 91. 
 
 27. The difference between two numbers is 48, and the 
 arithmetical mean exceeds the geometrical mean by 18. 
 Find the numbers. 
 
 28. There are four numbers in geometrical progression, 
 the second of which is less than the fourth by 24, and the 
 sum of the extremes is to the sum of the means as 7 to 3. 
 Find the numbers. 
 
GEOMETRICAL PROGRESSION. 335 
 
 Infinite Geometrical Series. 
 
 370. When r is less than 1, the successive terms of a 
 geometrical series become numerically smaller; by taking n 
 large enough we can make the nth. term, ar n ~ 1 f as small as 
 we please, although we cannot make it absolutely zero. 
 
 The sum of n terms, -r— * — (§ 366), may be written 
 
 : - ; this sum differs from : by - : 
 
 1 — r 1 — r\ 1 — r J 1 — r' 
 
 by taking enough terms we can make I, and consequently 
 j as small as we please. Hence, may be con- 
 sidered the sum of an infinite number of terms of the series. 
 
 1. Find the sum of 1 - j + J - J -f 
 
 Here a = 1, and r = — £. 
 
 _ a _ 1 _ 1 _2 
 
 2. Find the value of 0.2363636 
 
 The terms after the first form a decreasing geometrical series in 
 which a = 0.036, and r = 0.01. 
 
 _ a 0.036 _ 0.036 _ 36 _ 2 
 
 *'• S 1 - r ~ 1 - 0.01 ~ 0.99 ~ 990 ~ 55*' 
 
 2 2 22 + 4 13 
 Therefore, the required value is — + — = rrr = — * 
 
 10 55 110 55 
 
 Exercise 128. 
 Find the sum of the following infinite series : 
 1. 4 + 2 + 1 + 5. 2-li +f- 
 
 2 - £ + £ + ! + 6. 0.1 + 0.01 + 0.001 + 
 
 3 - i - rV + & - 7. 0.868686 
 
 4 - 1 - * + A - 8 - 0.54444 
 
336 HABMONICAL PROGRESSION. 
 
 ♦Harmonica! Progression. 
 
 371. A series is called a harmonical series or a harmonical 
 progression when the reciprocals of its terms form an arith- 
 metical series. 
 
 Hence, the general representative of such a series is 
 
 1 1 1_ 1 
 
 a a-\r d a -{-2d ' a -{- (n — 1) d 
 
 372. Questions relating to harmonical series should be 
 solved by writing the reciprocals of its terms so as to form 
 an arithmetical series. 
 
 373. If a and b denote two numbers, and H their har- 
 monical mean, then, by the definition of a harmonical series. 
 
 H a b H 
 
 . 1. _ 1 4. 1 - a + b - 
 ' ' H a b ab 
 
 2ab 
 
 \H = 
 
 a + b 
 
 374. Sometimes it is required to insert several harmon- 
 ical means between two numbers. 
 
 Let it be required to insert three harmonical means be- 
 tween 3 and 18. 
 
 
 Find the three arithmetical means between £ and T \. 
 
 These are found to be |f, ^f , fy ; therefore, the harmonical means 
 *»**.**.¥; or8H,6fc8. 
 
 * A harmonical series is so called because musical strings of uniform 
 thickness and tension produce harmony when their lengths are repre- 
 sented by the reciprocals of the natural series of numbers ; that is, by 
 the harmonical series 1, i, £, £, §, etc, 
 
HARMONICAL PROGRESSION. 337 
 
 Exercise 129. 
 
 1. Insert four harmonical means between 2 and 12. 
 
 2. Find two numbers whose difference is 8 and har- 
 monical mean If. 
 
 3. Find the seventh term of the harmonical series 3, 
 
 3M 
 
 4. Continue to two terms each way the harmonical 
 series two consecutive terms of which are 15, 16. 
 
 5. The first two terms of a harmonical series are 5 and 
 6. Which term will equal 30 ? 
 
 6. The fifth and ninth terms of a harmonical series are 
 8 and 12. Find the first four terms. 
 
 7. The difference between the arithmetical and har- 
 monical means between two numbers is If, and jone of the 
 numbers is four times the other. Find the numbers. 
 
 8. Find the arithmetical, geometrical, and harmonical 
 means between two numbers, a and b ; and show that the 
 geometrical mean is a mean proportional between the 
 arithmetical and harmonical means. Also, arrange these 
 means in order of magnitude. 
 
 9. The arithmetical mean between two numbers exceeds 
 the geometrical by 13, and the geometrical exceeds the 
 harmonical by 12. What are the numbers ? 
 
 10. The sum of three terms of a harmonical series is 11, 
 and the sum of their squares is 49. Find the numbers. 
 
 11. When a, b, c are in harmonical" progression, show 
 that a : c : : a — b : b — c. 
 
CHAPTER XXIII. 
 VARIABLES AND LIMITS. 
 
 375. Constants and Variables. A number that, under the 
 conditions of the problem into which it enters, may be 
 made to assume any one of an unlimited number of values 
 is called a variable. 
 
 A number that, under the conditions of the problem into 
 which it enters, has a fixed value is called a constant. 
 
 Variables are represented by x, y, z ; constants by a, b, c, 
 and by the Arabic numerals. 
 
 376. Limits. When the value of a variable, measured at 
 a series of definite intervals, can by continuing the series 
 be made to differ from a given constant by less than any 
 assigned quantity, however small, but cannot be made abso- 
 lutely equal to the constant, the constant is called the limit 
 of the variable, and the variable is said to approach indefi- 
 nitely to its limit. 
 
 Consider the repetend 0.33&.-.-, which may be written 
 
 ■$5 + "dhy "+" TtfW + 
 
 The value of each fraction after the first is one tenth of 
 
 the preceding fraction, and by continuing the series we 
 shall reach a fraction less than any assigned value, how- 
 ever small ; that is, the values of the successive fractions 
 approach as a limit. 
 
 The sum of these fractions will always be less than £ ; 
 but the more terms we take, the nearer does the sum 
 approach ^ as a limit. 
 
VARIABLES AND LIMITS. 339 
 
 Suppose a point to move from A toward B, under the con- 
 ditions that the first A mm- m- b 
 second it shall move ~~ ' ' ' 
 one half the distance from A to B, that is, to 31; the next 
 second, one half the remaining distance, that is, to M' ; the 
 next second, one half the remaining distance, that is, to M " ; 
 and so on indefinitely. 
 
 Then it is evident that the moving point may approach 
 as near to B as we please, but will never arrive at B. For, 
 however near it may be to B at any instant, the next second it 
 will pass over one half the interval still remaining ; it must, 
 therefore, approach nearer to B, since half the interval 
 still remaining is some distance, but will not reach B since 
 half the interval still remaining is not the whole distance. 
 
 Hence, the distance from A to the moving point is an 
 increasing variable, which indefinitely approaches the con- 
 stant AB as its limit ; and the distance from the moving 
 point to B is a decreasing variable, which indefinitely 
 approaches the constant zero as its limit. 
 
 If the length of AB is two inches, and the variable is 
 denoted by x, and the difference between the variable and 
 its limit by y : 
 
 after one second, x = 1, y=X 
 
 after two seconds, x = 1 + \, y = \ 
 
 after three seconds, x = \-\-\ + \, V — \ 
 after four seconds, « = l + £ + i+£, y ' = \ 
 and so on indefinitely. 
 
 Now the sum of the series 1 + \ 4- \ + \> etc., is less than 
 2 ; but by taking a great number of terms, the sum can be 
 made to differ from 2 by as little as we please. Hence, 2 is 
 the limit of the sum of the series, when the number of the 
 terms is increased indefinitely; and is the limit of the 
 difference between this variable sum and 2. 
 
340 VARIABLES AND LIMITS. 
 
 377. Test for a Limit. In order to prove that a variable 
 approaches a constant as a limit, it is necessary and suffi- 
 cient to prove that the difference between the variable and 
 the constant can be made as near to as we please, but 
 cannot be made absolutely equal to 0. 
 
 A variable may approach a constant without approaching it as a 
 limit. Thus, in the last example x approaches 3, but not as a limit ; 
 for 3 — x cannot be made as near to as we please, since it cannot 
 be made less than 1. 
 
 378. Infinites. As a variable changes its value, it may 
 constantly increase in numerical value ; if the variable can 
 become numerically greater than any assigned value, how- 
 ever great this assigned value may be, the variable is said 
 to increase without limit, or to increase indefinitely. 
 
 When a variable is conceived to have a value greater 
 than any assigned value, however great, the variable is said 
 to become infinite; such a variable is called an infinite 
 number, or simply an infinite, and is denoted by oo. 
 
 379. Infinitesimals. As a variable changes its value, it 
 may constantly decrease in numerical value; if the vari- 
 able can become numerically less than any assigned value, 
 however small this assigned value may be, the variable is 
 said to decrease without limit, or to decrease indefinitely. 
 
 In this case the variable approaches as a limit. 
 
 When a variable which approaches as a limit is con- 
 ceived to have a value less than any assigned value, how- 
 ever small this assigned value may be, the variable is said 
 to become infinitesimal ; such a variable is called an infini- 
 tesimal number, or simply an infinitesimal. 
 
 380. Infinites and infinitesimals are variables, not con- 
 stants. There is no idea of fixed value implied in either 
 an infinite or an infinitesimal. 
 
VARIABLES AND LIMITS. 341 
 
 381. An infinitesimal is not 0. An infinitesimal is a 
 variable arising from the division of a quantity into a con- 
 stantly increasing number of parts ; is a constant arising 
 from' taking the difference of two equal quantities. 
 
 382. Finite Numbers. A number which cannot become 
 infinite is said to be finite. 
 
 Theorems of Infinites and Infinitesimals. 
 
 383. Theorem 1. If x is infinitesimal, and a is finite 
 and not 0, then ax is infinitesimal. 
 
 For ax can be made as small as we please since x can be 
 made as small as we please. 
 
 384. Theorem 2. If X is infinite, and a is finite and 
 not 0, then aX is infinite. 
 
 For aX can be made as large as we please since Xcan be 
 made as large as we please. 
 
 385. Theorem 3. Ifx is infinitesimal, and a is finite and 
 not 0, then — is infinite. 
 
 For - can be made as large as we please since x can be 
 
 made as small as we please. 
 t 
 
 386. Theorem 4. If X is infinite, and a is finite and 
 
 not 0, then — is infinitesimal. 
 
 For — can be made as small as we please since X can be 
 
 made as large as we please. 
 
 In the above theorems a may be a constant or a variable ; 
 the only restriction on the value of a is that it shall not 
 become either infinite or 0. 
 
342 VARIABLES AND LIMITS. 
 
 387. Abbreviated Notation. The expression - cannot be 
 
 interpreted literally, since we cannot divide by ; neither 
 
 can — = be interpreted literally, since we can find no 
 oo 
 
 number such that the quotient obtained by dividing a by 
 
 that number is 0. 
 
 The expression - = oo is simply an abbreviated way of writing : 
 
 Lf- = X, and x approaches as a limit, X increases without limit, 
 x 
 
 The expression — = is simply an abbreviated way of writing : 
 If -=. = x, and X increases without limit, x approaches Oas a limti. 
 
 388. Approach to a Limit. A variable may approach its 
 limit in one of three ways : 
 
 1. The variable may be always less than its limit. 
 
 2. The variable may be always greater than its limit. 
 
 3. The variable may be sometimes less and sometimes 
 greater than its limit. 
 
 If x represent the sum of n terms of the series 1 + i + i + \ + , 
 
 x is always less than its limit 2. 
 
 If x represent the sum of n terms of the series 3 — ■$- — £■ — \ — , 
 
 x is always greater than its limit 2. 
 
 If x represent the sum of n terms of the series 3 — | + | — f+' , 
 
 we have (§ 364) 
 
 As n is indefinitely increased, x evidently approaches 2 as a limit. 
 If n is even, x is less than 2 ; if n is odd, x is greater than 2. 
 Hence, if n is increased by taking each time one more term, x will 
 be alternately less than and greater than 2. If, for example, 
 
 n= 2, 3, 4, 5, 6, 7, 
 
 x = li, 2i, 1J, 2^, 1|1, W 
 
VARIABLES AND LIMITS. 343 
 
 Indeterminate Forms. —Vanishing Fractions. 
 
 389. When one or more variables are involved in both 
 numerator and denominator of a fraction, it may happen that 
 for certain values of the variables both numerator and denom- 
 inator of the fraction vanish. The fraction then assumes 
 
 the indeterminate form - • If there is but one variable 
 
 involved, we may obtain a definite value as follows : 
 
 Let x be the variable, and a the value of x "for which the 
 
 fraction assumes the form-- Give to a; a value a little 
 
 greater than a, as a + h ; the fraction will now have a defi- 
 nite value. The limit of this last value, as h is indefinitely 
 decreased, is called the limiting value of the fraction. 
 
 The fundamental indeterminate form is -> and all other 
 
 indeterminate forms may be reduced to this. 
 
 rrn on a b a 
 
 ^ i = o"o = o x *=o' 
 
 _ a X a 
 0xco = ox- = — = ~ 
 
 a b X a — X S 
 
 °°- 00 = o"o = o = 0' 
 
 1. Find the limiting value of as x approaches a. 
 
 When x has the value a, the fraction assumes the form - • » 
 Put x = a + h ; the fraction becomes 
 
 (a + h) — a h 
 
 Since h is not 0, we can divide by h and obtain 2 a + h. 
 As A is indefinitely decreased, this approaches 2 a as a limit. 
 
344 VARIABLES AND LIMITS. 
 
 2. Find the limiting value of x X 2 — - when x 
 
 ox -|- Z x — 1 
 
 becomes infinite. 
 
 Divide each term of the numerator and denominator by x 8 . Then, 
 
 2x*-4x + 5 _ -x 2 x« 
 3x 3 + 2x 2 — 1 ~ 2 1 ' 
 
 O T 1 
 
 X X 3 
 
 As x increases indefinitely, each term that contains x of the last 
 
 fraction approaches as a limit (Theorem 4), and the fraction ap- 
 
 2 
 
 proaches - as a limit. 
 
 o 
 
 Exercise 130. 
 Find the limiting value of : 
 
 1. ^ _ a — Q — , . ' when x becomes infinitesimal. 
 
 7 x 8 — 6 x + 4 
 
 fo 2 -5)(s 2 + 7) 
 
 2. ■* 4 \; QK - when a; becomes infinite. 
 
 X -|~ oo 
 
 (a; 4- 2) 8 
 
 3 « o,! when a; becomes infinitesimal. 
 a; 2 + 4 
 
 4- a , 2 _ 7a , + 12 wnen - approaches 3. 
 
 x 2 — 9 
 6 * s 2 + 9 x + 18 When X a PP roaches ~ 3 - 
 
 6 ' tt» + 3s« + 5s + 3 When X a PP roaches " *- 
 
 7> s»+*2* 2 -2a;-l When * a PP roaclies L 
 
 a 4aj + Vaj— 1, .. . „ 
 
 8. . when x approaches 1. 
 
 2x- V«+l 
 
CHAPTER XXIV. 
 PROPERTIES OF SERIES. 
 
 Convergent and Divergent Series. 
 
 390. By performing the indicated division, we obtain 
 
 from - the infinite series 1 + x + x 2 + x z + This 
 
 1 — x 
 
 series, however, is not equal to the fraction for all values of x. 
 
 391. If x is numerically less than 1, the series is equal to 
 the fraction. In this case we can obtain an approximate 
 value for the sum of the series by taking the sum of a 
 number of terms ; the greater the number of terms taken, 
 the nearer will this approximate sum approach the value of 
 the fraction. The approximate sum will never be exactly 
 equal to the fraction, however great the number of terms 
 taken ; but by taking enough terms it can be made to differ 
 from the fraction by as little as we please. 
 
 Thus, if x = -y the value of the fraction is 2, and the 
 
 . . Ill 
 
 series is 1+- + -. + -+ 
 
 The sum of four terms of this series is l£ ; the sum of 
 five terms, ljj; the sum of six terms, 1§£; and so on. 
 The successive approximate sums approach, but never 
 reach, the finite value 2. 
 
 An infinite series is said to be convergent when the 
 sum of the terms, as the number of terms is indefinitely 
 increased, approaches some fixed finite value ; this finite 
 value is called the sum of the series. 
 
346 PROPERTIES OF SERIES, 
 
 393. In the series 1 + x + x 2 + x s + suppose x 
 
 numerically greater than 1. In this case the greater the 
 number of terms taken, the greater will their sum be ; by 
 taking enough terms we can make their sum as large as 
 we please. The fraction, on the other hand, has a definite 
 value. Hence, when x is numerically greater than 1, the 
 series is not equal to the fraction. 
 
 Thus, if x = 2, the value of the fraction is — 1, and the 
 series is 
 
 1 + 2 + 4 + 8 + 
 
 The greater the number of terms taken, the larger the sum. 
 Evidently, the fraction and the series are not equal. 
 
 394. In the same series suppose x = 1. In this case the 
 
 1 1 
 
 fraction is = -, and the series 1+1+1+lH ■ 
 
 The more terms we take, the greater will the sum of the 
 series be, and the sum of the series does not approach a 
 fixed finite value. 
 
 If x, however, is not exactly 1, but is a little less than 1, 
 
 the value of the fraction will be very great, and the 
 
 JL X 
 
 fraction will be equal to the series. 
 
 Suppose x = — 1. In this case the fraction is . . = - , 
 
 1 + 1 Z 
 
 and the series 1 — 1 + 1 — 1 + If we take an even 
 
 number of terms, their sum is ; if an odd number, their 
 
 sum is 1. Hence, the fraction is not equal to the series. 
 
 395. A series is said to be divergent when the sum of the 
 terms, as the number of terms is indefinitely increased, 
 either increases without end, or oscillates in value without 
 approaching any fixed finite value. 
 
PROPERTIES OF SERIES. 347 
 
 No reasoning can be based on a divergent series ; hence, 
 in using an infinite series it is necessary to make such 
 restrictions as will cause the series to be convergent. Thus, 
 
 we can use the infinite series 1 + x + x 2 + x 8 + when, 
 
 and only when, x lies between -f 1 and — 1. 
 
 396. Theorem. If two series, arranged by powers ofx, 
 are equal for all values of x that make both series con- 
 vergent, the corresponding coefficients are equal each to each. 
 
 For, if A + Bx + Cx 2 + = A f + B'x + Cx 2 + • ., 
 
 by transposition, 
 
 A-A f = (B'-B)x + (C - C)x 2 + ..... 
 
 Now, by taking x sufficiently small, the right side of this 
 equation can be made less than any assigned value what- 
 ever, and therefore less than A — A 1 , if A — A r has any 
 value whatever. Hence, A — A' cannot have any value. 
 
 Therefore, A - A' = 0, or A = A\ 
 
 Hence, Bx + Cx 2 + Dx* + = B'x + Cx 2 + D'x* + , 
 
 or (B-B T )x = (C - C) x 2 + (D' -D)x* + ..... 
 Divide by x, 
 
 B-B' = (C'-C)x + (£>' - D)x 2 + ...» 
 By the same proof as for A — A\ 
 
 B-B' = 0,otB = BK 
 In like manner, 
 
 C= C, D = D f ', and so on. 
 Hence, the equation 
 
 A + Bx+ Cx 2 + = A' + B'x + C'x 2 + »..., 
 
 if true for all finite values of x, is an identical equation ; 
 that is, the coefficients of like powers of x are equaL 
 
348 PROPERTIES OF SERIES. 
 
 Indeterminate Coefficients. 
 
 397. Expand — - — — — in ascending powers of x. 
 
 JL ~t~ 35 "i »C 
 
 Assume ■ ^ + ^ x , = A + Bx + Cx* + Dx» + ; 
 
 1 + X + x* 
 
 then, by clearing of fractions, 
 
 2 + Sx = A + Bx + Cx* + Dx* + 
 
 + Ax+ Bx* + Cxs + 
 
 + Ax* + Bx* + 
 
 .:2 + 3x = A + (B + A)x + (C + B + A)x* + (D+C+B)x* + 
 
 By § 396, A =2, B + A = 3, C + B + A- 0, D+C + £ = 0; 
 whence, 2? = 1, C = — 3, D = 2 ; and so on. 
 
 2 + 3* 
 
 ' * 1 + X + *2 
 
 = 2 + x - 3 x 2 + 2 x 3 + 
 
 The series is of course equal to the fraction for only such values of 
 x as make the series convergent. 
 
 Note. In employing the method of Indeterminate Coefficients, 
 the form of the given expression must determine what powers of the 
 variable x must be assumed. It is necessary and sufficient that the 
 assumed equation, when simplified, shall have in the right member 
 all the powers of x that are found in the left member. 
 
 If any powers of x occur in the right member that are not in the 
 left member, the coefficients of these powers in the right member will 
 vanish, so that in this case the method still applies ; but if any powers 
 of x occur in the left member that are not in the right member, then 
 the coefficients of these powers of x must be put equal to in equating 
 the coefficients of like powers of x ; and this leads to absurd results. 
 Thus, if it were assumed that 
 
 2 + 3x 
 
 1+X+X2 
 
 = Ax + Bx* + Ox 3 + 
 
 there would be in the simplified equation no term on the right cor- 
 responding to 2 on the left ; so that, in equating the coefficients of 
 like powers of x, 2, which is 2x°, would have to be put equal to 0x°; 
 that is, 2 = 0, an absurdity. 
 
PROPERTIES OF SERIES. 349 
 
 10. 
 
 11. 
 
 
 Exercise 131. 
 
 
 
 Expand to four terms : 
 
 
 
 1 
 
 1-x 
 
 7. 
 
 3 + x 
 
 * 1 + 2* 
 
 ™ 1+x + x 2 
 
 1 — x — X 2 
 
 1 
 
 5-2x 
 & ' 1+x-x 2 
 
 8. 
 
 1+x 
 
 ' 2-3x 
 
 1+ X + X 2 
 
 1+x 
 
 2—3a 
 
 9. 
 
 l-8z 
 
 ' 2 + 3a 
 
 l-2x + 3x 2 
 
 1 — x-6x 2 
 
 Expand to five terms : 
 
 
 
 4 
 
 12 5-20. 
 
 14. 
 
 3x-2 
 
 ' 2+x 
 
 A ~* l+3a-z 2 
 
 x(x-iy 
 
 2 -x 
 ' 3+x' 
 
 aj(a?-2) 
 
 15. 
 
 x 2 -x + l 
 (*-l)(x 2 + l) 
 
 Partial Fractions. 
 
 398. To resolve a fraction into 'partial fractions is to 
 express it as the sum of a number of fractions of which 
 the respective denominators are the factors of the denom- 
 inator of the given fraction. This is the reverse of the proc- 
 ess of adding fractions that have different denominators. 
 
 Resolution into partial fractions may be easily accom- 
 plished by the use of indeterminate coefficients and the 
 theorem of § 396. 
 
 In decomposing a given fraction into its simplest partial 
 fractions, it is important to determine what form the assumed 
 fractions must have. Since the given fraction is the sum of 
 the required partial fractions, each assumed denominator 
 must be a factor of the given denominator ; moreover, all 
 the factors of the given denominator must be taken as 
 denominators of the assumed fractionso 
 
350 PROPERTIES OF SERIES. 
 
 Since the required partial fractions are to be in their 
 simplest form, incapable of further decomposition, the nu- 
 merator of each required fraction must be assumed with 
 reference to this condition. Thus, if the denominator is 
 x n or (x ± a) n , the assumed fraction must be of the form 
 
 A A ' A Ax + B Ax + B 
 
 — or - — ■ — — : for, if it had the form — or — > 
 
 x n (x±a) ni ' x n (x±.a) n 
 
 it could be decomposed into two fractions, and the partial 
 
 fractions would not be in the simplest form possible. 
 
 When all the monomial factors, and all the binomial 
 
 factors, of the form x ± a, have been removed from the 
 
 denominator of the given expression, there may remain 
 
 quadratic factors that cannot be further resolved; and 
 
 the numerators corresponding to these quadratic factors 
 
 may each contain the first power of x, so that the assumed 
 
 fractions must have either the form -= —7 or the 
 
 ar ± ax + 
 
 . Ax + B 
 form • 
 
 x 2 + b 
 
 3 
 
 1. Resolve into partial fractions. 
 
 X ~\~ JL 
 
 Since x 8 + 1 = (x + 1) (x 3 — x + 1), the denominators will be x + 1 
 and x 2 — x + 1. 
 
 3 A 1 Bx + C 
 
 Assume -ttt = — r - : + 
 
 x 8 + l x + lxs-x + l' 
 then, 3 = A(x* - x + 1) + (Bx + C) (x + 1) 
 
 = (A + B)x* +(B+C-A)x + (A + C); 
 whence, 3 = A + C, B+ C-A = 0, A + B = t 
 
 and -4«=1, B= — 1, (7 = 2. 
 
 Therefore, 
 
 3 _1 x-2 
 
 x 8 + l x + 1 x 3 -x + l 
 
PROPERTIES OF SERIES. 351 
 
 2. Kesolve 2 into partial fractions. 
 
 4 x s -x 2 -Sx -2 . 
 
 a 2 (x 4- 1) 
 
 The denominators may be x, x 2 , x 4- 1, (x + l) 2 
 
 4x 8 -x 2 -3x-2 ^L , B , C 
 Assume tz — r~ 777 = — K + M ■ , + 
 
 x 2 (x + l) 2 x x 2 x + 1 (x + l) 2 
 
 .-. 4x3 - x 2 - 3x - 2 =Ax{z + l) 2 +B{x + l) 2 + Cx 2 (x + 1)+Dx a 
 = (^ + C)x8 + (2^ + B + C + D)x 2 + (^ + 2B)x + JB; 
 whence, A + C = 4, 
 
 A + 2£ = -3, 
 B = -2; 
 and .-. B = - 2, ^i = 1, C = 3, D = - 4. 
 
 Therefore 4x«-x 2 -3x-2 =: l_2 _3 4 ■ 
 
 lnerelore, JC 2 (x+1)2 x x 2 + x + 1 (x + 1)* 
 
 Exercise 132. 
 Kesolve into partial fractions : 
 
 1. 
 
 7z + l 
 
 4. 
 
 5. 
 
 (x + 4) (x - 5) 
 
 7a-l 
 (l-2z)(l-3a:) 
 
 5x-l 
 
 (2x-l)(x-5)' 
 
 x-2 
 (x-5)(x + 2)' 
 
 3 
 
 x 2 - x - 3 
 6. 
 
 7. 
 
 3ce 2 -4 
 
 x 2 (x + 5) 
 
 8. 
 
 7cc 2 -a: 
 
 (x-l) 2 (z-f-2) 
 
 9. 
 
 2z 2 -7a + l 
 
 * 8 -l 
 
 
 7 a;- 1 
 
 1U. 
 
 (6ic-hl)(o;-l) 
 
 11. 
 
 a 2 -3 
 
 (a 2 + 1) (x + 2) 
 
 1 
 
 x 2 -x + l 
 
 x (x 2 - 4) (x 2 + !)(»- I) 5 
 
CHAPTER XXV. 
 BINOMIAL THEOREM. 
 
 399. Binomial Theorem, Positive Integral Exponent. By suc- 
 cessive multiplication we obtain the following identities : 
 
 (a + by = a* + 2ab + b*; 
 
 (a + b) 8 = a* + 3 a% +3ab 2 + b 8 -, 
 
 (a + by = a* + 4a*b + 6a 2 b 2 + 4a* 8 + b\ 
 
 The expressions on the right may be written in a form 
 better adapted to show the law of their formation : 
 
 (a + by = a* + 2ab + |4&V 
 
 / _i_*\« 4.j 8a. 4 ' 3 2?2 , 4-3-2 ,„ 4-3-2-l r , 
 
 Note. The dot between the Arabic figures means the same as the 
 sign X. 
 
 400. Let n represent the exponent of (a -f b) in any one 
 of these identities; then, in the expressions on the right, 
 we observe that the following laws hold true : 
 
 1. The number of terms is n + 1. 
 
 2. The first term is a n , and the exponent of a is one less 
 in each succeeding term. 
 
 3. The first power of b occurs in the second term, the 
 second power in the third term, and the exponent of b is 
 one greater in each succeeding term. 
 
 4. The sum of the exponents of a and b in any term is ft. 
 
BINOMIAL THEOREM. 353 
 
 5. The coefficients of the terms taken in order are 1 ; n ; 
 »» *-■!): n(n-l)(n-2) d 
 
 1-2 ' 1-2-3 ' anasoon - 
 
 401. Consider the coefficient of any term ; the number 
 of factors in the numerator is the same as the number of 
 factors in the denominator, and the number of factors in 
 each is the same as the exponent of b in that term ; this 
 exponent is one less than the number of the term. 
 
 402. Proof of the Theorem. That the laws of § 400 hold 
 true when the exponent is any positive integer is shown 
 as follows : 
 
 We know that the laws hold for the fourth power; 
 suppose, for the moment, that they hold for the &th power. 
 We shall then have 
 
 (a + b) k = a* + ka*~ l b + ^"^ a*-*b* 
 
 + *(*-l)(*-2) at _, & , + (1) 
 
 Multiply both members of (1) by a + b ; the result is 
 (a + b)* + * = a k + l + (k + l) a*b 4- QL+3^ #~ip 
 
 + ( * + i)*(*-i W + (2) 
 
 In (1) put h -b 1 f or h ; this gives 
 (a + b)*' +l *=a*+* + (k + 1) a*b + (fr+'iK* + 1 - *) a k-ip 
 
 + (* + !)(* + ! -1)^ + 1-2) ak _ 2b8 + 
 
 1 * Z ' o 
 
 = a** 1 + (k + 1) aH + ( k + W ak -i h 2 
 
 + (. + i)M.-i) ^, + (3) 
 
 Equation (3) is seen to be the same as equation (2). 
 
354 BINOMIAL THEOREM. 
 
 Hence, (1) holds when we put k -f- 1 for k ; that is, if the 
 laws of § 400 hold for the kth. power, they must hold for 
 the (k + l)th power. 
 
 But the laws hold for the fourth power ; therefore, they 
 must hold for the fifth power. 
 
 Holding for the fifth power, they must hold for the sixth 
 power ; and so on for any positive integral power. 
 
 Therefore, they must hold for the wth power, if n is a 
 positive integer ; and we have 
 
 (a + b) n = a n + na*~ l b + %& ^ a* 
 
 1 • J!i 
 
 + re(ra -l)( r 2 W + . (A) 
 
 Note. The above proof is an example of a proof by mathematical 
 induction. See § 134. 
 
 403. This formula is known as the binomial theorem. 
 
 The expression on the right is known as the expansion of 
 (a -f- b) n ; this expansion is a finite series when n is a positive 
 integer. That the series is finite may be seen as follows : 
 
 In writing out the successive coefficients we shall finally 
 arrive at a coefficient that contains the factor n — n ; and, 
 therefore, this term will vanish. The coefficients of all the 
 succeeding terms likewise contain the factor n — n, and, 
 therefore, all these terms will vanish. 
 
 404. If a and b are interchanged, the identity (A) may 
 be written 
 
 (a + b) n = (b + a) n = b n + nb^a + W ^~ 1 V ~V 
 
 . n (n — 1) (n — 2) . . R , 
 + K 1-2-3 lb "-' a ' + 
 
BINOMIAL THEOREM. 355 
 
 This last expansion is the expansion of (A) written in 
 reverse order. Comparing the two expansions, we see 
 that the coefficient of the last term is the same as the 
 coefficient of the first term; the coefficient of the last term 
 but one is the same as the coefficient of the first term but 
 one; and so on. 
 
 In general, the coefficient of the rth term from the end 
 is the same as the coefficient of the rth term from the 
 beginning. In writing an expansion by the binomial 
 theorem, after arriving at the middle term, we can shorten 
 the work by observing that the lemaining coefficients are 
 those already found, taken in reverse order. 
 
 405. If b is negative, the terms that involve even powers 
 of b will be positive, and the terms that involve odd powers 
 of b will be negative. Hence, 
 
 (a - b) n = a n - na*-*b + n( \~ V) a»-%* 
 
 1 * 2i 
 
 n(n — l)(n — 2) ... rtr . 
 — 1-2-3 + W 
 
 Also, putting 1 for a and x for b in (A) and (B). 
 
 (1 + x) n = 1 + nx 4- n ^ l ~ V) x i 
 
 + ^~i 1} 2 ( .r 2) * 8 + < c ) 
 
 (1 - x)« = 1 - nx + *fe -j& a* 
 
 1-2 
 
 _ n(n-l)Jn-2) xi+ ^ 
 
 1-2-3 
 
356 BINOMIAL THEOREM. 
 
 1. Expand (1 + 2x)\ 
 
 In (C) put 2 x f or x and 5 for n. The result is 
 
 (1 + 2x)« = 1 + 5(2x) + f^f 4*2 + rfr| 8x8 
 
 + l-2-3-4 16iC +1-2-3-4-6 323 * 
 = 1 + 10s + 40*2 + 80*8 + 80x* + 32 gfi. 
 
 2. Expand to three terms 
 
 /l _ 2£>y 
 
 . ... 
 
 1 2x 2 
 
 Put a for - , and 6 for — ; then, by (B), 
 x <-> 
 
 (« - 6)6 = a 6 _ 6 a 5 & + 15 a 4&2 _ 
 
 Replace a and 6 by their values, 
 x 6 x» + 3 
 
 406. Any Required Term. From (A) it is evident (§ 400) 
 that the (r + l)th term in the expansion of (a + b) n is 
 
 7t (n — 1) (n — 2) to r factors n _ r , r 
 
 1x2x3 r a 
 
 Note. In finding the coefficient of the (r + l)th term, write the 
 
 series of factors 1 X 2 X 3 r for the denominator of the coefficient, 
 
 then write over this series the factors n (n — 1) (n — 2), etc., writing 
 just as many factors in the numerator as there are in the denominator. 
 
 The (r -f l)th term in the expansion of (a — b) n is the 
 same as the above if r is even, and the negative of the 
 above if r is odd. 
 
BINOMIAL THEOREM. 357 
 
 Find the eighth term 
 
 °'M)» 
 
 x 2 
 Here a = 4, b = — » n = 10, r = 7. 
 
 3 
 
 ™ * 1 ^- 10-9-8-7-6-5 -4 /J%lt / x 2 \* 
 
 The term required is —- y— ^ (4) 8 ( - -) , 
 
 which reduces to — 60 x 14 . 
 
 407. A trinomial may be expanded by the binomial 
 theorem as follows : 
 Expand (1 + 2 x - a 2 ) 8 . 
 
 Put 2 x — x 2 = z. 
 
 Then, (1 + zf = 1 + 3z + 3* 2 + 2«. 
 
 Replace z by 2 a; — x 2 . 
 
 .% (1 + 2x-x 2 ) 3 = 1 + 3(2x-x 2 ) + 3(2x-x 2 ) 2 + (2X-X 2 ) 8 
 = 1 + 6x + 9x 2 — 4x 3 — 9x 4 + 6x 6 — x°. 
 
 Exercise 133. 
 Expand : 
 
 1. (a + b)\ 6. (a» + by. II. ( m -J + w 2)4. 
 
 2. (x-2) 5 . ?> ( m « + w «)8 # 12 . (ar-» + **)•. 
 
 3. (Sx -2yY. , 
 
 8. (a -by. 13. (2x 2 + ^ 6 . 
 
 / £C "2/ 1 
 
 4 ' \y"*y 9 - (** + * 1 ) 5 - l4 - (»*-cV. 
 
 5. (4 + 3y) 4 . 10. (a-^ + fl- 2 ) 8 . 15. (2 a 2 - £V^) 5 . 
 
 18. (2x 2 y- 1 -y^/y) 4 . 21. (2^- 2 -^' t 
 
358 BINOMIAL THEOREM. 
 
 24. Find the fourth term of (2 a; - Sy)\ 
 
 25. Find the ninety-seventh term of (2 a — b) 100 . 
 
 Note. As the expansion has 101 terms, the ninety-seventh term 
 from the beginning is the fifth term from the end. 
 
 26. Find the eighth term of (3 x — y) l \ 
 
 27. Find the tenth term of (2 a 2 - -J-a) 20 . 
 
 28. Find the fifth term of (a - 2 V&) 25 . 
 
 29. Find the eleventh term of (2 — a) 16 . 
 
 30. Find the fifteenth term of (x + y) 20 - 
 
 31. Find the fourth term of (3 - 2 ic 2 ) 9 . 
 
 32. Find the twelfth term of (a? - a Vic) 17 . 
 
 33. Find the seventh term of (y 2 - l) 88 . 
 
 34. Find the fifth term of Qa-bVb) 21 . 
 
 35. Find the fourth term of (VS - V^ 2 ) 20 . 
 
 36. Find the third term of ( Va — V^~6) 7 . 
 
 37. Find the sixth term of (Va 3 - V^) 9 . 
 
 38. Find the eighth term of (Vfa + Vfa-) 20 . 
 
 39. Find the ninth term of (x^T^T + y V^I) 16 . 
 
 a*b - - > • 
 
 of (t 
 
 vw 
 
 41. Find the seventh term of (x •+• aj _1 ) 2n . 
 
BINOMIAL THEOREM. 359 
 
 408. Binomial Theorem, Any Exponent. We have seen 
 (§ 402) that when n is a positive integer we have the 
 identity 
 
 ( l +x y = l + nx + ^^x> + n{n - 1 l%- 2 h >+ 
 
 We proceed to the case of fractional and negative expo- 
 nents. 
 
 I. Suppose n is a positive fraction, — > in which p and 
 
 q are positive integers. We may assume that 
 
 (i + xy = (A + Bx+ cx 2 + Dx* + y, (1) 
 
 provided x is so taken that the series 
 
 A + Bx -f Cx* + Dx % + 
 
 is convergent, § 392. 
 
 That this assumption is allowable may be seen as follows : 
 Expand both members of (1). We obtain 
 
 1+ ^ + £^, 2 + E(£z^|^ JC 3 + 9 
 
 + 
 
 and A« + qA«~ l Bx+ I fM ^^ Ai-'P+qA*- 1 ^] x 
 
 In the first k coefficients of the second series there enter 
 
 only the first k of the coefficients A, B, C, D, If, then, 
 
 we equate the coefficients of corresponding terms in the 
 two series (§ 396) as far as the &th term, we shall have just 
 k equations to find k unknown numbers A, B, C, D 9 •••% 
 Hence, the assumption made in (1) is allowable. 
 
 Equating the two first terms and the two second terms, 
 we obtain 
 
 A* = 1, .\ A = 1 ; 
 qA q " 1 B=p 1 or qB=p f .*. J5 = ^- 
 
360 BINOMIAL THEOREM. 
 
 Extracting the ^th root of both members of (1), we have 
 
 (1 + x)\ = l +^x + Cx 2 + Dx* + , (2) 
 
 where x is to be so taken that the series on the right is 
 convergent. 
 
 II. Suppose n is a negative number, integral or frac- 
 tional. Let n = — ra, so that m is positive ; then, 
 
 1 
 
 (1 + x) n = (1+ x)~ m 
 
 (i + xy 
 
 From (2), whether m is integral or fractional, we may 
 assume 
 
 1 = 1 
 
 (1+ x) m ~ 1 + mx + cx 2 + efo 8 + * 
 
 By actual division this gives an equation in the form 
 
 (1 + x)~ m = l-mx + Cx 2 + Dx 8 + (3) 
 
 409. It appears from (2) and (3) (§ 408) that whether n is 
 integral or fractional, positive or negative, we may assume 
 
 (1 + x) n = 1 + nx 4- Cx 2 + Dx 8 4- , 
 
 provided the series on the right is convergent. 
 
 Square both members, 
 
 (l + 2x + x*) n = l + 2nx + 2Cx 2 + 2Dx* + ••••• 
 
 + n 2 x 2 +2nCx* + (1) 
 
 Also, since 
 
 (1 + y)» = 1 + ny 4- Cy 2 4- Dy s + , 
 
 we have, putting 2 x 4- x 2 for y> 
 
 (1 4- 2 x 4- xy = 1 4- n (2 x + x 2 ) 4- C (2 x -\- x 2 ) 2 
 
 + D(2x \-x*)* 
 
 = l+2nx + nx 2 +4Ccc 8 4- 
 
 + 4:Cx 2 +$Dx* + (2) 
 
BINOMIAL THEOREM. 361 
 
 Equate corresponding coefficients in (1) and (2), 
 
 n + ±C = 2C + n 2 , 
 ±C + 8D = 2D + 2nC. 
 
 ,'. 2 <7 = »*-*, and C = W ^ 1) ; 
 
 3i> = (,-2)fiandi>^ M^-^-^ . 
 
 and so on. 
 
 Hence, whether n is integral or fractional, positive or 
 negative, we have 
 
 (1 + *)- = 1 + nx + S^V + ^-l)fc-2) a , + , 
 
 provided, always, x is so taken that the series on the right 
 is convergent. 
 
 The series obtained will be an infinite series unless n is a 
 positive integer, § 403. 
 
 410. If x is negative, 
 
 /-. n„ i , n(n — l) 2 n(n — l)(n — 2) 8 
 
 (1 - x) n = 1 - nx + v V v 1-2-3 ~*~ 
 
 Also, if x < a, 
 
 .= an L 1 + "« + 1^2 s + J 
 
 if x > a, 
 
 = a» + fl»r^ + SfergS i^.^ + ; 
 
 r | . • . - • \ 
 
 (a + x)' =(x + a)« = x»(l+^Y 
 
 |_ # 12 x 2 J 
 
 * - s» ■+ nax*- 1 + "^"^ aV- 2 + . 
 
362 BINOMIAL THEOREM. 
 
 1. Expand (1 + x)*. 
 
 The above equation is true only for those values of x that make 
 the series convergent. 
 
 2. Expand — • 
 
 T== = (!-*>-* 
 
 VI — X 
 
 = i-(-^+=ti li ^- ~^ 1 ~ f 3 ~ f » 8 + 
 
 if x is so taken that the series is convergent. 
 
 A root may often be extracted by means of an expansion* 
 
 3. Extract the cube root of 344 to six decimal places. 
 
 844 = 843 ( 1 + 8i3) = 78 ( 1 + 3i3)- 
 
 '[_ 3\343>M 1-2 1,343/ + J 
 
 = 7 (1 + 0.000971817 - 0.000000944 + ) 
 
 ' = 7.006796. 
 
 4. Find the eighth term of f x -= ) . 
 
 \ 4Va?/ 
 
 Here 
 
 a 
 
 = x, 
 
 4Vx 
 
 8. R . _ 
 
 ? 7 
 
 3 
 
 — , n = 
 
 4x 
 
 1 
 
 T 
 
 r=7. 
 
 > 
 
 
 1-2-3 
 1-3-5-7 
 
 4-5-6-7 
 •9-1113 
 
 3 7 
 
 X 
 
 1-3-5-7-9-11-13-3 7 
 2-4-6-8-10-12-14-47-x u 
 
BINOMIAL THEOREM. 363 
 
 Exercise 134. 
 Expand to four terms : 
 
 1. (1+*)* 6. (1+a)* 11. (2a + 3y)* 
 
 2. (1+tf)- 2 . 7. (l+x)~\ 12. (2a* + 3y)~* 
 
 3. (1 +*)"*. 8. (1+z)- 8 . 13. 
 
 Va 2 — cc* 
 
 4. (1-s)* 9. (l + 5aj)- 5 . j 
 
 14 - - 
 
 5. (1 - x)~*' 10. (1 + 5 a) 1 . V(a - xf 
 
 15. Find the fourth term of 
 
 16. Find the fifth term of 
 
 \ 2>/xJ 
 
 ^/(a-2xy 
 
 17. Find the third term of (4 - 7 x)\ 
 
 18. Find the sixth term of (a 2 — 2 axy. 
 
 19. Find the fifth term of (1 - 2 xfK 
 
 20. Find the fifth term of (1 - x)~\ 
 
 21. Find the seventh term of (1 — x) . 
 
 ! 
 
 22. Find the third term of (1 -f x) 2 ». 
 
 23. Find the fourth term of (1 4- x)~K 
 
 24. Find the sixth term of (2 V • 
 
 25. Find the fifth term of (2 x - 3 y)~*, 
 
 26. Find the fourth term of (1 — 5x)~K 
 
364 MISCELLANEOUS PROBLEMS. 
 
 Exercise 135. — General Be view. 
 
 1. Add (a - b) x 2 + (b - c) y 2 + (c - a) z* ; (b - c) x 2 
 
 + (e - a) y 2 + (a - b) z 2 ; (c - a) x 2 + (a - b) y 2 
 + (b-c)z\ 
 
 2. Add(a + b)x + (b + c)y- (c + a)z; (b + c)z+ (c + a)x 
 
 — (a + b)y\ (a + c)y+ (a + b)z— (b + c)x. 
 
 3. From 4a; 8 — §x 2 + %x — 7 take the sum of 
 
 'Sx 8 + 7 - 8a; 2 + 7x and - 9 a; 8 - 8x 2 + 4a; + 4. 
 
 4. Find the product of a p — Sal 1 '- 1 + ±a p ~ 2 — 6a p ~ 8 + 5a p ~ 4 
 
 and 2 a* — a 2 + a. 
 
 5. Divide 1 - 6x 5 + 5a; 6 by 1 - 2x + x 2 . 
 
 6. Divide 4 /** +1 - 30 fc* + 19 A*- 1 + 5 A*- 2 + 9 A*~ 4 
 
 by h*-* - 7 h*-* + 2 7^- 5 - 3 h x ~\ 
 
 7. Simplify 3Ja - 2 (J - c) J - [45 + \2 b - (c - a)}]. 
 
 8. Find the factors of 10 x 2 + 79 a; - 8. 
 
 9. Find the H.C.F. of 2a; 8 + a; 2 + 4a; - 7 
 
 and a; 8 - 2a; 2 + 1. 
 
 10. Keduce to lowest terms ■»,,,>. — si * 
 
 5 a 4 + 9 a 8 — 64 
 
 11. If a = 4, b = £, c = — 2, find the numerical value of 
 
 C £ 2 
 
 12. 
 
 T , a + 5 , ^x/^ - a Y a; — 2a + & A 
 
 If x = — ^r— > show that I r ; ^77 = 0. 
 
 2 \x — by x-\- a — 2 b 
 
 13. If a; = 2 y + 3 z, show that 
 
 x*-8y*-27z*- lSxyz = 0. 
 
 14. Eesolve into factors 6 a; 2 -f 5 a; — 4. 
 
 15. Eesolve into factors a; 4 + 2 x % - 13 x 2 - 38 a; - 24. 
 
 16. Eesolve into factors 12 a; 8 4- 20 x 2 — x — 6. 
 
MISCELLANEOUS PROBLEMS. 365 
 
 17. A boy bought a number of apples at the rate of 5 for 
 2 cents. He sold half of them at the rate of 2 for a cent 
 and the rest at the rate of 3 for a cent, and cleared a cent 
 by the transaction. How many did he buy ? 
 
 Find the H.C.F. and the L.C.M. of: 
 
 18. 3x 6 -5x* + 2, and2a; 6 -5a; 2 + 3.. 
 
 19. 3x» + 10x 2 + 7x-2, and3z 8 -f 13z 2 + 17a; + 6. 
 
 20. Ax 4 - 9x 2 -f 6x — 1, and 6x*- 7 x 2 -f 1. 
 
 21. x 6 + 11 cc - 12, and « 6 + 11 x z + 54. 
 
 22. 2z 8 + 5a 2 y-5iC2/ 2 + 2/ 8 , and 2 a; 8 - 7 x 2 y + 5 a^ 2 - y\ 
 Simplify : 
 
 x + 1 3aj + 2 , 2a;-l 
 
 23. 
 
 x(x — 2) x (x + 1) ^ a; 2 
 
 _ ■ 1+aj . 1-z 1 + x* 1 - x* 
 24. f- 
 
 1-x ' l-\-x 1-a: 2 1+a; 2 
 
 2 s-1 2(x + 2) x + 5 
 
 ' (x + 2)(x + 5) (x + fytx-l)^ (x-l)(x + 2) 
 
 27. 
 
 ax — a 2 ax -\-2a 2 x 2 -\- ax — 2 a 2 
 
 x - 1 a; — 3 
 
 (a + 3) (3-1) (a> + 3) (2 -a) (2-x)(l-x) 
 
 I + i + l t. i-ia-*) 
 
 a o c „ 2 
 
 28 -T-T7T-T- 
 
 «» I 7.0 O "T~ 
 
 « 
 
 2 ' 7,2 ..2 
 
 £ 2 c 2 ' ab 
 
 1 X i + — 
 
 A ^ 1 — a 1 + a . a a£ 8 
 
 29. - 31. 
 
 *-i+l 
 
 1 — a 1 4- a b 
 
366 
 
 MISCELLANEOUS PROBLEMS. 
 
 32. 
 
 33. 
 
 1 + 
 
 1 + 
 
 1 + 
 
 4 — x 
 
 l + a + 
 
 2 a? 
 1+a 
 
 34. Two passengers have together 400 pounds of bag- 
 gage. One pays .$1.20, the other $1.80, for excess above the 
 weight allowed. If all the baggage had belonged to one 
 person, he would have had to pay $4.50. How much bag- 
 gage is allowed free ? 
 
 Solve : 
 
 37. 
 
 oc 6 a; + 13 9z + 15 . 2x + 15 
 
 35. 3~ — j;^ + o — ' 
 
 15 
 
 5 x — 25 
 
 2x + a , Sx 
 36. 7tz r + 
 
 S(x-a) 2(x + a) 
 
 V 
 
 = 4 1 
 
 1+^ = 2* 
 
 y 
 
 6 
 
 = 2J. 
 
 
 OA 2x t 3y kz . 
 
 39. \--f- = 1 
 
 a o c 
 
 a b c 
 
 a b c 
 
 40. 
 
 a 
 
 b 
 
 e 
 
 
 
 - + 
 
 - + 
 
 
 = 
 
 3 
 
 iC 
 
 y 
 
 z 
 
 
 
 o . 
 
 b 
 
 c 
 
 
 
 - + 
 
 
 
 ss 
 
 1 
 
 X 
 
 y 
 
 z 
 
 
 
 2a 
 
 b 
 
 c 
 
 
 
 — 
 
 — _ 
 
 
 
 = 
 
 
 
 X 
 
 y 
 
 z 
 
 
 
 Find the arithmetical value of : 
 
 41. 36*; 27*; 16*; 32*; 4 f ; 8 1 ; 27*; 64*. 
 
 42. 32*; 64*; 81*; (3#j (5 T V)*; (l*) 1 . 
 
MISCELLANEOUS PROBLEMS. 367 
 
 43. (0.25) 1 ; (0.027)*; 49" *; 32~ f ; 81~ f . 
 
 44. 36"*; 27~*; (A)"*» C - 116 )" 1 ; (0.0016) - *. 
 
 Arrange in ascending order of magnitude : 
 45. 6V7; 9V3; 5VlO. 46. 4V6; 3^3; 5^2. 
 
 Simplify : 
 
 /r /= /r 2VlO 7V48 4Vl5 
 
 47. *V3 X 4 VO -t-4. VS. 48. 7 = X = -fc 7==" 
 
 3 * 7 3V27 5Vl4 15V21 
 
 49. = • 50. =• 51. = 52. 
 
 2 + V3 ' 2-V3 ' 2V5 + 3 ! 2V5-3 
 
 1 1 
 
 53. {C*W) f X (a 56 )- 2 p«-2. 54. (a; 18o Xa;- 12 )3«-2. . 
 
 55. 3 (a* + ft*) 2 - 4 (a* + ft*) (a* - ft*) + (a* - 2 ft 4 ) 2 . 
 
 56. {(a")*-m}S7i. 58. [\(a- m )- n \*>y+[\(a m ) n \-P~\-* 
 
 57 f xP + 9 \ p . / £j V~ a £c 2 iJ ( ? -i)_^2</(p-i) 
 
 \ a* / + \a«"V * 59 ' iC p(3 - 1 >+^^- 1 > ' 
 
 60. 5^320 - 2^- 1715 + 3^^135. 
 
 61. 2V18-3V8 + 2V50; -^81 + v"24 - J/192. 
 
 62. |Vj + V80-iV20; 8V^ + 10 Vf - 2Vj. 
 
 Extract the square root of : 
 
 63. 9ar 4 -18a- 8 y* + 15a;-V-6ar- 1 ^ + 2A 
 
 Extract the cube root of : 
 
 64. 8a* 4- 12a 2 - 30a - 35 + 45a- 1 + 27a" 2 - 27a" 8 . 
 
 
368 MISCELLANEOUS PROBLEMS. 
 
 Resolve into prime factors with fractional exponents: 
 65. -^12, ^72, -n/^.V'oT; and find their product. 
 Solve: 
 
 66. 
 
 67. 
 
 68. 
 
 69. 
 
 70. 
 
 5 
 
 x -2 
 
 4 3 
 
 £C 03+6 
 
 71. 
 
 6 c 2 
 
 ace 2 — = c# — 6cc 2 . 
 
 a + b 
 
 jc + 3 
 
 2z-l 
 
 72. 
 
 3xy-5if = ll 
 
 2«-7 
 
 a-3 " a 
 
 5xy + 3x 2 = 22) 
 
 x+4 
 x — 4 
 
 ? ~ 2 - fiJL 
 •cc-3- 6 *' 
 
 73. 
 
 74. 
 
 x 2 + 10xy = ll] 
 5xy-3y 2 = 2} 
 
 x — a 
 
 Va + y = Vy + 2 1 
 
 cc — 6 
 
 x — a ab 
 
 
 x-y=7 , J 
 
 ace 4- b 
 
 ex -\- d 
 
 75. 
 
 a 2 + a;i/ + 2/ 2 = 52l 
 
 a -\-bx 
 
 c + dx 
 
 
 icy — tc 2 = 8 
 
 Form the equation of which the roots are : 
 
 76. a - 5, a + £. 79. 1 + VS, 1 - V5. 
 
 77. a -26, a + 36. 80. - 1 + V3, - 1 - VS. 
 
 78. a + 26, 2a + 6. 81. 1 -f V^, 1 - V^3. 
 
 82. A vessel that has two pipes can be filled in 2 hours 
 less time by one than by the other, and by both together 
 in 1 hour 52 minutes 30 seconds. How long will it take 
 each pipe alone to fill the vessel ? 
 
 83. A number is expressed by two digits, the second of 
 which is the square of the other, and when 54 is added its 
 digits are interchanged. Find the number. 
 
 84. Divide 35 into two parts such that the sum of the 
 two fractions formed by dividing each part by the other 
 may be 2^. 
 
MISCELLANEOUS PROBLEMS. 369 
 
 85. A number consists of three digits in arithmetical 
 progression ; and this number divided* by the sum of its 
 digits is equal to 26 ; but if 198 is added to the number, 
 the digits in the units' and hundreds' places will be inter- 
 changed. Find the number. 
 
 i 86. The sum of the squares of the extremes of four num- 
 bers in arithmetical progression is 200, and the sum of the 
 squares of the means is 136. What are the numbers ? 
 
 87. Show that if any even number of terms of the series 
 
 1, 3, 5 is taken, the sum of the first half is to the sum 
 
 of the second half in the ratio 1 : 3. 
 
 88. If a horse took 1 second more for each rod, he would 
 travel 1| miles less per hour. Find his rate of traveling. 
 
 Solve : 
 
 89. x 8 + y 3 = l$V2] 90. 2x 2 + Sxy = 
 
 z + 2/ = 3V2 
 
 1 90. 2x 2 + 3xy = 8^ 
 
 J y*-2xy = 20 J 
 
 Expand : 
 
 91. (x-x-y. 93. (2a i -a~ i y. 95. QVx' - i^x)\ 
 
 92. (^--£z) 4 . 94. (2x- 2 + x*y. 96. (jV^F 3 + ^ar 1 ) 8 . 
 
 97. If a, b, e, d are in continued proportion, prove that 
 b + c is a mean proportional between a + b and c -+- d. 
 
 98. If a + b : b + c = c + d : d + a, 
 
 prove that a = c, OYa + b-\-c-\-d = 0. 
 
 99. The number of eggs which, can be bought for 1 
 dollar is equal to twice the number of cents which 32 eggs 
 cost. How many eggs can be bought for 1 dollar ? 
 
 100. Find two fractions whose sum is ^ ¥ , and whose 
 difference is equal to their product. 
 
370 MISCELLANEOUS PROBLEMS. 
 
 101. The velocity of a falling body varies as the time 
 during which it has fallen from rest, and the velocity at 
 the end of 2 seconds is 64 ft. Find the velocity at the 
 end of 6 seconds. 
 
 102. The distance through which a body falls from rest 
 varies as the square of the time it falls ; and a body falls « 
 144 ft. in 3 seconds. How far does it fall in 4 seconds ? 
 
 103. The volume of a gas varies directly as the absolute 
 temperature and inversely as the pressure. If the volume 
 of a gas is 1 cubic foot, when the pressure is 15 and the 
 temperature 280, what will be the volume when the pres- 
 sure is 35 and the temperature 320 ? 
 
 104. The difference between the first and second of four 
 numbers in geometrical progression is 96 \ the difference 
 between the third and fourth is 6. Find the numbers. 
 
 105. If « 2 , b 2 , c 2 are in arithmetical progression, prove 
 that b + c, c+a, a + b are in harmonical progression. 
 
 When x = oo, and when x = 0, find the limit of : 
 
 (2x -3) (3 -5x) x*-x + l , 
 
 1Ub * 7s a -6aj + 4 107, (z 2 + l)(*-l) a 
 
 Kesolve into factors and find all the values of x : 
 
 108. a; 4 -5sc 2 + 4 = 0. 114. 3^-5-^ + 2 = 0. 
 
 109. x 6 -9 x s + 8 = 0. 115. 6^-3^-45 = 0. 
 
 110. 9o; 4 -13a: 2 + 4 = 0. 116. 21 Va- 2 - 5V« - 74=0. 
 
 111. 4z 4 -17a; 2 + 4 = 0. 117. 3 Vx 5 + 4^ - 20 = 0. 
 
 112. 2a 4 -5a; 2 + 2 = 0. 118. 2x« - 19a 8 + 24 = 0. 
 
 113. 2^-3x^ + 1=0. 119. a 4 - 1=0. 
 
§ MISCELLANEOUS PROBLEMS. 371 
 
 120. « 6 -l=0. 124. 4af *-3af^-27=0. 
 
 121. x s + 8 as* - 9 = 0. 125. z 2n + 3 x n - 4 = 0. 
 
 122. 16^-17^ + 1 = 0. 126. 3x* -2ax^ -a 2 = 0. 
 
 123. x~* + 5of* -14 = 0. 127. V2a- - V^2» - 2 = 0. 
 Solve : 
 
 128. Va; + 4 + V3£c + l=V9z + 4. 
 
 129. V5 x + 1 + 2 V4z - 3 = lOVic - 2. 
 
 130. 2 VaT+ 2-3V3«-54-V5z + 1 = 0. 
 
 131. Vll -a;+V8-2a;- V21 + 2 a; = 0. 
 Expand: 
 
 132. (J-x*y-, (3a*-26 2 ) 5 ; (a* -J ft 1 ) 7 * 
 (2-$x*)«; (3 -***)'; (x* - x" 2 ) 8 . 
 
 133. (a 2 -i?/ 8 ) 6 ; (2a 2 +V3a) 4 ; (V^ + l + a 8 ) 6 ; 
 (l + 2a-£c 2 -a 8 ) 8 . 
 
 134. Expand to four terms 
 
 (l-3a>)~*; (l-4a; 2 )~ f ; (1 — fac 4 )*; («r- 2*~*>- 
 
 135. Eind the eighty-seventh term of (2x — y) 90 . 
 
 136. Eesolve into partial fractions 
 
 3-2<c S-2x 1 
 
 1-3Z + 2* 2 ' Q.-x)(l-3xy 1-a; 8 
 3-2x 
 
 137. Expand to five terms 
 
 l-3x + 2x- 
 
CHAPTER XXVI. 
 LOGARITHMS. 
 
 411. If numbers are regarded as powers of ten, the expo- 
 nents of the powers are the Common or Briggs Logarithms of 
 the numbers. 
 
 If A and B denote two numbers, a and b their loga- 
 rithms, then 10° = A, 10 6 = B ; or, written in logarithmic 
 form, log A = a, log B=b. 
 
 412. The logarithm of a product is found by adding the 
 logarithms of its factors. 
 
 For AX B = 10° X 10 & = 10« +& . (§ 244) 
 
 Therefore, log (A X B) = a + b = log A + log B. 
 
 413. The logarithm of a quotient is found by subtract- 
 ing the logarithm of the divisor from that of the dividend. 
 
 A 10° 
 
 For !=S =10 °"*- (§250) 
 
 Therefore, log — = a — b — log A — log B. 
 B 
 
 414. The logarithm of a power of a number is found by 
 multiplying the logarithm of the number by the exponent 
 of the power. 
 
 For A* = (10 a ) n = 10™. (§ 251) 
 
 Therefore, log A n = na = n log A. 
 
LOGARITHMS. 373 
 
 415. The logarithm of the root of a number is found by 
 dividing the logarithm of the number by the index of the 
 root. 
 
 For 1/2 = l/lfr = 10». (§ 251) 
 
 Therefore, log^I = ^ = ^A 
 n n 
 
 416. The logarithms of 1, 10, 100, etc., and of 0.1, 0.01, 
 0.001, etc., are integral numbers. The logarithms of all 
 other numbers are fractions. 
 
 Since 10° = 1, 10" 1 (= ^) = 0.1, 
 
 10* m 10, 10- 2 (= T U) =0.01, 
 
 10 2 = 100, io-»(= TXF v*) = o.ooi, 
 
 therefore, log 1 = 0, log 0.1 = — 1, 
 
 log 10 = 1, log 0.01 =-2, 
 
 log 100 = 2, log 0.001 = - 3. 
 
 Hence, the common logarithms of all numbers between 
 
 1 and 10 will be + a fraction, 
 
 10 and 100 will be 1 -f a fraction, 
 
 100 and 1000 will be 2 + a fraction,' 
 
 1 and 0.1 will be — 1 + a fraction, 
 
 0.1 and 0.01 will be — 2 + a fraction, 
 
 0.01 and 0.001 will be - 3 + a fraction. 
 
 417. If the number is less than 1, the logarithm is 
 negative (§ 416), but is written in such a form that the 
 fractional part is always positive. 
 
 418. Every logarithm, therefore, consists of two parts : 
 a positive or negative integral number, which is called the 
 characteristic, and a positive decimal fraction, which is 
 called the mantissa. 
 
37£ LOGARITHMS. 
 
 Thus, in the logarithm 3.52184, the integral number 3 
 is the characteristic, and the fraction .52184, the mantissa. 
 In the logarithm 0.78256 — 2, which is sometimes written 
 2.78256, the integral number — 2 is the characteristic, and 
 the fraction .78256 is the mantissa. 
 
 419. If the logarithm has a negative characteristic, it 
 is customary to change its form by adding 10, or a mul- 
 tiple of 10, to the characteristic, and then indicating the 
 subtraction of the same number from the result. 
 
 Thus, the logarithm 2.78256 is changed to 8.78256 - 10 
 by adding 10 to the characteristic and writing — 10 after 
 the result. The logarithm 13.92732 is changed to 7.92732 
 — 20 by adding 20 to the characteristic and writing — 20 
 after the result. 
 
 420. The following rules are derived from § 416. 
 
 Rule 1. If the number is greater than 1, make the 
 characteristic of the logarithm one unit less than the num- 
 ber of figures on the left of the decimal point. 
 
 Rule 2. If the number is less than 1, make the char- 
 acteristic of the logarithm negative, and one unit more than 
 the number of zeros between the decimal point and the 
 first significant figure of the given number. 
 
 Rule 3. If the characteristic of a given logarithm is 
 positive, make the number of figures in the integral part of 
 the corresponding number one more than the number of 
 units in the characteristic. 
 
 Rule 4. If the characteristic is negative, make the 
 number of zeros between the decimal point and the first 
 significant figure of the corresponding number one less than 
 the number of units in the characteristic. 
 
LOGARITHMS. 875 
 
 Thus, the characteristic of log 7849.27 is 3 ; the char- 
 acteristic of log 0.037 is - 2, or 8.00000 - 10. If the 
 characteristic is 4, the corresponding number has five 
 figures in its integral part. If the characteristic is — 3, 
 that is, 7.00000 — 10, the corresponding fraction has two 
 zeros between the decimal point and the first significant 
 figure. 
 
 421. The mantissa of the common logarithm of any- 
 integral number, or decimal fraction, depends only upon 
 the digits of the number, and is unchanged so long as the 
 sequence of the digits remains the same. ' 
 
 For, changing the position of the decimal point in a 
 number is equivalent to multiplying or dividing the num- 
 ber by a power of 10. Its common logarithm, therefore, 
 will be increased or diminished by the exponent of that 
 power of 10 ; and since this exponent is integral, the man- 
 tissa, or decimal part of the logarithm, will be unaffected. 
 
 Thus, 271,960 = 10 s - 48451 , 2.7196 = 10 - 48451 , 
 
 27,196 = 10 4 - 48451 , 0.27196 = 10 948451 " 10 , 
 
 2719.6 = 10 8 - 48451 , 0.027196 = lO 8 - 484 * 1 - 10 , 
 
 271.96 = 10 2 - 48461 , 0.0027196 = 10 7 - 48451 - 10 , 
 
 27.196 = 10 1 - 48451 , 0.00027196 = 10 6 - 48461 " 10 . 
 
 One advantage of using the number ten as the base of 
 a system of logarithms consists in the fact that the man- 
 tissa depends only on the sequence of digits, and the char-, 
 aeteristic on the position of the decimal point. 
 
 422. The characteristic of the common logarithm of a 
 number can be found by the rules given in § 420 ; but the 
 mantissa of the logarithm is found by means of a Table of 
 Logarithms. 
 
376 LOGARITHMS. 
 
 423. A Five-place Table of Logarithms. In this table 
 (pp. 389-407) the vertical columns headed N contain 
 numbers, the columns headed 0, 1, 2, 3, etc., contain the loga- 
 rithms, and the columns headed D contain the tabular 
 differences of the logarithms. On page 389 both the char- 
 acteristic and the mantissa are printed. On pages 390-407 
 the mantissa only is printed, and the first two figures of 
 the mantissa are printed in the left-hand column only. 
 
 A star prefixed to the last three figures of a logarithm 
 indicates that the first two figures are in the line below. 
 
 The fractional part of a logarithm can be expressed only 
 approximately, and in a five-place table all figures that 
 follow the fifth are rejected. Whenever the sixth figure is 
 6 or more, the fifth figure is increased by 1, 
 
 Thus, if the mantissa of a logarithm written to seven places is 
 6328732, it is written in a five-place table 53287. If the mantissa is 
 5328751, it is written in a five-place table 53288. 
 
 To Find the Logarithm of a Given Number. 
 
 424. If the given number consists of one or two figures, 
 the logarithm is given on page 389. If zeros follow the 
 significant figures, or if the number is a proper decimal 
 fraction, the characteristic is determined by § 420. 
 
 425. If the given number has three significant figures, 
 the number will be found in the column headed N (pp. 
 390-407), and the mantissa of its logarithm in the next 
 column to the right, and on the same line. Thus, 
 
 Page 390. log 145 = 2. 16137, log 14,500 = 4. 16137. 
 
 Page 395. log 364 = 2.56110, log 0.0364 = 8.56110 - 10. 
 
 Page 402. log 716 = 2.85491, log 0.716 =9.85491-10. 
 
 Page 406. log 926 = 2.96661, log 9260 =3.96661. 
 
LOGARITHMS. 377 
 
 426. If the given number has four significant figures, the 
 first three figures will be found in the column headed N, 
 and the fourth figure at the top of the page in the line 
 containing the figures 0, 1, 2, 3, etc. The mantissa will be 
 found in the column headed by the fourth figure, and on 
 the same line with the first three figures. Thus, 
 
 Page 403. log 7682 = 3.88547, log 76.85 = 1.88564. 
 Page 406. log 93,280 = 4.96979, log 0.9468 = 9.97626 - 10. 
 
 427. If the given number has five or more significant 
 figures, a process called interpolation is required. 
 
 Interpolation is based on the assumption that between 
 two consecutive mantissas of the table the change in the 
 mantissa is directly proportional to the change in the 
 number. 
 
 1. Find the logarithm of 34,237. 
 
 The required mantissa is (§ 421) the same as the mantissa for 3423.7; 
 therefore, it will be found by adding to the mantissa for 3423 seven 
 tenths of the difference between the mantissas for 3423 and 3424. 
 
 The mantissa for 3423 is 53441. 
 
 The difference between the mantissas for 3423 and 3424 is 12. 
 
 Hence, the mantissa for 3423.7 is 53441 + (0.7 of 12) = 53449. 
 
 Therefore, the logarithm of 34,237 is 4.53449. 
 
 2. Find the logarithm of 0.001576,4. 
 
 The required mantissa is the same as the mantissa for 1576.4. 
 The mantissa for 1576 is 19756. 
 
 The difference between the mantissas for 1576 and 1577 is 27. 
 Hence, the mantissa for 1576.4 is 19756 + (0.4 of 27) = 19767. 
 Therefore, the logarithm of 0.0015764 is 7.19767 — 10. 
 
 3. Find the logarithm of 32.6708. 
 
 The required mantissa is the same as the mantissa for 3267.08. 
 The mantissa for 3267 is 51415. 
 
 The difference between the mantissas for 3267 and 3268 is 13. 
 Hence, the mantissa for 3267.08 is 51415 + (0.08 of 13) = 51416. 
 Therefore, the logarithm of 32.6708 is 1.51416. 
 
378 LOGARITHMS. 
 
 428. When the fraction of a unit in the part to be added 
 to the mantissa for four figures is less than 0.5, it is to be 
 neglected; when it is 0.5 or more than 0.5, it is to be taken 
 as one unit. 
 
 Thus, in example 1, § 427, the part to be added to the mantissa for 
 3423 is 8.4, and the .4 is rejected. In example 2, the part to be added 
 to the mantissa for 1576 is 10.8, and 11 is added. 
 
 To Find the Antilogarithm ; that is, the Number 
 Corresponding to a Given Logarithm. 
 
 429. If the given mantissa can be found in the table, 
 the first three figures of the required number will be found 
 in the same line with the mantissa in the column headed 
 N, and the fourth figure at the top of the column containing 
 the mantissa. 
 
 The position of the decimal point is determined by the 
 characteristic (§ 420). 
 
 1. Find the number corresponding to the logarithm 
 0.92002. 
 
 Page 404. The number for the mantissa 92002 is 8318. 
 The characteristic is ; hence (§ 420), the number is 8.318. 
 
 2. Find the number corresponding to the logarithm 
 6.09167. 
 
 Page 390. The number for the mantissa 09167 is 1235. 
 
 The characteristic is 6 ; hence (§ 420), the number is 1,235,000. 
 
 3. Find the number corresponding to the logarithm 
 
 2.51055. 
 
 Page 394. The number for the mantissa 51055 is 3240. 
 The characteristic is 2 ; hence (§ 420), the number is 324. 
 
 4. Find the number corresponding to the logarithm 
 7.50325 - 10. 
 
 Page 394. The number for the mantissa 50325 is 3186. 
 
 The characteristic is — 3 ; hence (§ 420), the number is 0.003186. 
 
LOGARITHMS. 379 
 
 430. If the given mantissa cannot be found in the table, 
 find in the table the two adjacent mantissas between which 
 the given mantissa lies, and the four figures corresponding 
 to the smaller of these two mantissas will be the first four 
 significant figures of the required number. If more than 
 four figures are desired, they may be found by interpolation, 
 as in the following examples : 
 
 1. Find the number corresponding to the logarithm 
 1.48762. 
 
 Here the two adjacent mantissas of the table, between which 
 the given mantissa 48762 lies, are found to be (p. 394) 48756 and 
 48770. The corresponding numbers are 3073 and 3074. The 
 smaller of these, 3073, contains the first four significant figures of the 
 required number. 
 
 The difference between the two adjacent mantissas is 14, and the 
 difference between the corresponding numbers is 1. 
 
 The difference between the smaller of the two adjacent mantissas, 
 48756, and the given mantissa, 48762, is 6. Therefore, the number 
 to be annexed to 3073 is T \ of 1 = 0.428, and the fifth significant 
 figure of the required number is 4. 
 
 Hence, the required number is 30.734. 
 
 2. Find the number corresponding to the .logarithm 
 7.82326 - 10. 
 
 Here the two adjacent mantissas of the table, between which the 
 given mantissa 82326 lies, are found to be (p. 401) 82321 and 82328. 
 The corresponding numbers are 6656 and 6657. The smaller of 
 these, 6656, contains the first four significant figures of the required 
 number. 
 
 The difference between the two adjacent mantissas is 7, and the 
 difference between the corresponding numbers is 1. 
 
 The difference between the smaller mantissa, 82321, and the 
 given mantissa, 82326, is 5. Therefore, the number to be annexed 
 to 6656 is f of 1 = 0.7, and the fifth significant figure of the required 
 number is 7. 
 
 Hence, the required number is 0.0066567. 
 
380 LOGARITHMS. 
 
 Cologarithms. 
 
 431. The logarithm of the reciprocal of a number is 
 called the cologarithm of the number. 
 If A denotes any number, then 
 
 colog A = log j = log 1 - log A (§ 413) = - log A (§ 416). 
 
 Hence, the cologarithm of a number is equal to the loga- 
 rithm of the number with the minus sign prefixed, which 
 sign affects the entire logarithm. 
 
 In order to avoid a negative mantissa in the cologarithm, 
 it is customary to substitute for — log A its equivalent 
 (10 - log A) - 10. 
 
 Hence, the cologarithm of a number is found by sub- 
 tracting the logarithm of the number from 10, and then 
 annexing — 10 to the remainder. 
 
 The best way to perform the subtraction is to begin on 
 the left and subtract from 9 each significant figure of log 
 A except the last, and subtract this from 10. 
 
 If log A is greater in absolute value than 10 and less 
 than 20, then in order to avoid a negative mantissa it is 
 necessary to write — log A in the form 
 (20 - log A) - 20. 
 So that, in this case, colog A is found by subtracting log A 
 from 20, and then annexing — 20 to the remainder. 
 
 1. Find the cologarithm of 4007. 
 
 10. - 10 
 
 Page 396. log 4007 = 3.60282 
 
 colog 4007= 6.39718-10 
 
 2. Find the cologarithm of 103,992,000,000. 
 
 20. -20 
 
 Page 390. log 103992000000 = 11.01700 
 
 colog 103992000000 = 8.98300 — 20 
 
LOGARITHMS. 
 
 381 
 
 If the characteristic of log A is negative, then the sub- 
 trahend, — 10 or — 20, will vanish in finding the value of 
 colog A. 
 
 3. Find the cologarithm of 0.004007. 
 
 Page 396. 
 
 log 0.004007 
 
 10. - 10 
 
 7.60282 - 10 
 
 colog 0.004007 = 2.39718 
 
 With practice, the cologarithm of a number can be taken 
 from the table as rapidly as the logarithm itself. 
 
 By using cologarithms the inconvenience of subtracting 
 the logarithm of a divisor is avoided. For dividing by a 
 number is equivalent to multiplying by its reciprocal. 
 Hence, instead of subtracting the logarithm of a divisor its 
 cologarithm may be added. 
 
 Exercise 136. 
 Find the logarithm of : 
 
 1. 6170. 4. 85.76. 7. 0.8694. 
 
 2. 0.617. 5. 296.8. 8. 0.5908. 
 
 3. 2867. 6. 7004. 9. 73,243. 
 
 Find the cologarithm of ■ 
 
 13. 72,433. 16. 869.278. 
 
 14. 802.376. 17. 154,000. 
 
 15. 15.7643. 18. 70.0426. 
 
 Find the antilogarithm of : 
 
 22. 2.47246. 25. 1.26784. 
 
 23. 7.89081. 26. 3.79029. 
 
 24. 2.91221. 27. 5.18752. 
 
 
 10. 67.3208. 
 
 
 11. 18.5283. 
 
 
 12. a0042003 
 
 19. 
 
 0.002403. 
 
 20. 
 
 0.000777. 
 
 21. 
 
 0.051828. 
 
 28. 
 
 9.79029 - 10. 
 
 29. 
 
 7.62328 - 10. 
 
 30. 
 
 6.15465 - 10. 
 
382 LOGARITHMS 
 
 Computation by Logarithms. 
 
 1. Find the value of x, if x = 72,214 x 0.08203. 
 
 Page 402. log 72214 = 4.85862 
 
 Page 404. log 0.08203 = 8.91397 - 10 
 
 By §412. logx =3.77259 
 
 Page 399. x = 5923.6 
 
 In using a five-place table the numbers corresponding to 
 mantissas may be carried to five significant figures, and in 
 the first part of the table to six figures. 
 
 2. Find the product of - 908.4 x 0.05392 X 2.117. 
 
 Page 406. log 908.4 == 2.95828" 
 
 Page 398. log 0.05392 = 8.73175 - 10 
 
 Page 392. log 2.117 = 0.32572 
 
 Page 390. 2.01575 = log 103.69. - 103.69. Ans. 
 
 When any of the factors are negative, find their logarithms, with- 
 out regard to the signs ; write n after the logarithm that corresponds 
 to a negative number. If the number of logarithms so marked is odd, 
 the product is negative ; if even, the product is positive. 
 
 3. Find the value of x, if x = 5250 -f- 23,487. 
 
 )8. log 5250 = 3.72016 
 
 Page 392. colog 23487 = 5.62917 — 10 
 
 Page 392. log x = 9.34933 - 10 = log 0.22353. 
 
 .-. x = 0.22353 
 
 a jk a a i * •* 7.56 X 4667 X 567 
 
 4. Find the value of *, if x m 899>1 x 0<0 0337 X 23435 
 
 Page 403. log 7.56 = 0.87852 
 
 Page 397. log 4667 = 3.66904 
 
 Page 399. log 567 = 2.75358 
 
 Page 405. colog 899.1 = 7.04619 - 10 
 
 Page 394. colog 0.00337 = 2.47237 
 
 Page 392. colog 23435 = 5.63013 - 10 
 
 Page 393. log x =2.44983 = log 281.73. 
 
 .-.x =281.73. 
 
LOGARITHMS. 383 
 
 5. Find the cube of 376. 
 
 Page 395. log 376 = 2.57519 
 
 Multiply by 3 (§ 414), 3 
 
 Page 398. log 376 3 = 7.72557 = log 53159000. 
 
 ... 3763 = 53,159,000. 
 
 6. Find the square of 0.003278. 
 
 Page 394. log 0.003278 = 7.51561 - 10 
 
 Multiply by 2 (§ 414), 2 
 
 Page 390. log 0.003278 2 = 15.03122 - 20 = log 0.000010745. 
 
 .-. 0.003278 2 = 0.000010745. 
 
 7. Find the square root of 8322. 
 
 Page 404. log 8322 = 3.92023 
 
 Divide by 2 (§ 415), 2 ) 3.92023 
 
 Page 406. log V 8322 = 1.96012 = log 91.226. 
 
 .-. V8322 '=91.226. 
 
 If the given number is a proper fraction, its logarithm will have as 
 a subtrahend 10 or a multiple of 10. In this case, before dividing the 
 logarithm by the index of the root, both the subtrahend and the 
 number preceding the mantissa should be increased by such a number 
 as will make the subtrahend, when divided by the index of the root, 
 10 or a multiple of 10. 
 
 8. Find the square root of 0.000043641. 
 
 Page 396. 
 
 Divide by 2 
 Page 401. 
 
 log 0.000043641 
 
 (§ 415), 
 log Vo. 000043641 
 
 = 5.63989- 
 
 10. 
 2)15.63989- 
 = 7.81995- 
 
 -10 
 
 - 10 
 -20 
 
 - 10 
 
 10 = log 0.0066062. 
 ... VO. 000043641 = 0.0066062. 
 
 9. Find the sixth root of 0.076553. 
 
 Page 403. log 0.076553 = 8.88397-10 
 
 50. — 50 
 
 Divide by 6 (§ 415), 6 ) 58.88397 - 60 
 
 Page 400. log Vo.076553 = 9.81400 - 10 = log 0.65163. 
 .-. Vo. 076553 = 0.65163. 
 
384 LOGARITHMS. 
 
 1A F ,, ■ , * 3.1416 X 4771.21 x 2.7183* 
 10. Find the value of V 3 0,103* x 0.4343* x 69.897* ' 
 
 log 3.1416 =0.49715 == 0.49715 
 
 log 4771.21 = 3.67863 = 3.67863 
 
 i log 2.7183 =i (0.43430) = 0.21715 
 
 4 colog 30.103 = 4 (8.52139 - 10) = 4.08556 - 10 
 
 i colog 0.4343 =£ (0.36221) = 0.18111 
 
 4 colog 69.897 = 4 (8.15554 - 10) = 2.62216 - 10 
 
 11.28176 - 20 
 
 
 30. - 30 
 
 
 5)41.28176-50 
 
 
 8.25635 - 10 
 
 
 = log 0.018046. 
 
 
 11. Find the value of x in 81 x = 10. 
 
 
 81* = 10. 
 
 
 .-. log 81* = log 10. 
 
 
 x log 81 = log 10. 
 
 (§ 4H) 
 
 . = '<>« i? =]-™» = 0.52397. 
 
 
 log 81 1.90849 
 
 Exercise 137. 
 Find the value of : 
 
 1. 849.7 X 0.7834. 10. 6078 -^- 8703. 
 
 2. 3.709x0.08673. 11. 8.326 -=- 0.1978. 
 
 3. 83.75 X 0.009376. 12. 0.6539 +- 0.9761. 
 
 4. 8593 X 0.0008974. 13. - 2.567 h- 0.6785. 
 
 5. - 0.007634 X 6457. 14. (39.47 X 5.938) -^ 76.54. 
 
 6. - 0.07843 X 48.66. 15. (5674 x 0.763) h- 0.9803. 
 
 7. - 0.8734 X 0.4378. 16. 357 -f- (7069 X 0.07948). 
 
 8. - 7.384 X (- 5.837). 17. 8.9 h- (17.81 X 0.002831). 
 
 9. 4657 X 3145. 18. 51.98 -h (81.71 X 0.0008002). 
 
LOGARITHMS. 385 
 
 79.32 x 0.005763 x 0.8064 
 0.4273 X 0.8462 X 0.01 
 
 72.56 X 0.0005723 X 8979 
 42.28 X 4.745 X 0.006158 ' 
 
 0.01723 X 34.^9 X 0.5477 
 0.07692 X 37.69 X 0.7733* 
 
 7.126 X 0.7213 X 0.7583 
 0.4684 X 7.385 X 0.9673* 
 
 2.057 X 77.12 X 0.004896 X 4.771 
 7.582 X 97.33 X 0.008697 X 0.4963* 
 
 24. 5.03 8 . 28. 0.6787 8 . 32. 9.068*. 
 
 25. 15.01 5 . 29. 0.9679 5 . 33. 0.0635*. 
 
 26. 76.85 4 . 30. 0.07867 8 . 34. 0.008721* 
 
 27. 8.713 2 . 31. 0.008546 2 , 35. 0.6543*. 
 
 83.25 X 8375 X 0.008576 
 
 19. 
 20. 
 21. 
 22. 
 2?. 
 
 3b. 
 
 37. 
 
 38. 
 
 39. 
 
 40. 
 
 
 0327 X 687.5 X 0.005003 
 163 2 X 17.74 X 0.7183* 
 
 3.013 2 X 34.34 X 0.08137* 
 / 0.7132 X 9.245 X 0.5477 2 
 *V 76.93 X 0.000173* X 0.01 
 
 65.02 x 0.002753 X 97.98 
 7.298 2 X 0.04754 X 8.156 2 
 
 3> 
 
 ,06012 X VU6( 2 X V06( 
 
 \0J 
 
 .5926 x V0.5926 X V0.5926 
 Find x from the equation : 
 
 41. 5* = 10. 43. 7* = 40. 45. (0.4)~* = 3. 
 
 42. 4* = 20. 44. (1.3)* = 4.2. 46. (0.9)~* = 2. 
 
386 LOGARITHMS. 
 
 Compound Interest and Annuities. 
 
 432. The amount of $ P at compound interest at r per 
 cent 
 
 for 1 year is P (1 + r) ; 
 
 for 2 years is P (1 + r) 2 ; 
 
 for n years is P (1 -J- r) n . 
 
 Hence, if th'e amount for n years is represented by A, 
 A = P(1 + r) w . 
 
 Note. If the interest is compounded semi-annually, 
 A =P(1 + £r) 2 «. 
 
 Find the amount of $150 for 6 years at 4% compound 
 interest 
 
 A = P(l + r) n = 150(1.04) 6 . 
 log 150 =2.17609 
 log 1.046 = 0.10218 
 log A = 2.27827 = log 189.79. 
 
 Hence, the required amount is $189.79. 
 
 433. The present worth, P, of $A, payable in n years at 
 r per cent, must just amount in n years to A. 
 
 Hence, F '= ~y,' (§432) 
 
 434. An annuity is a sum of money to be paid at regular 
 intervals of time, as years, half years, quarter years. 
 
 435. To find the present value of an annuity of %A per annum 
 for n years, at r per cent. 
 
 The present value of the 1st payment is ; (§ 433) 
 
LOGARITHMS. 387 
 
 The present value of the 2d payment is . r 2 ; 
 The present value of the rith payment is 
 
 (1 4- r)> 
 Hence, the present value of all the payments is 
 
 A +*+& + - 
 
 . (1 + r) (1 + r)' (1 + rf 
 
 J^\}-\$Tr)\. (§365) 
 
 1 =— 
 
 1 + r 
 Multiply both numerator and denominator by 1 + r, 
 
 r 
 
 i ^r (i+r)"-i "i 
 
 "71 (l+r)« J 
 
 Find the present value of an annuity of $500 for 5 years, 
 if money is worth 4%. 
 
 ir (l+ r)» - 1 -1 _ 500 / 1.045 _x v 
 rL (l + r)» J 0.04 V 1.04* / 
 log 1.045 = 5 X 0.01703 = 0.08515 = log 1.2166. 
 
 500 0.2166 
 *' 0.04 X 1.2166' 
 
 log 500 = 2.69897 
 
 log 0.2166 = 9.33566 -10 
 colog 0.04 = 1.39794 
 colog 1.2166 = 9.91485 - 10 
 
 3.34742 = log 2226.5. 
 
 Therefore, the present value of the annuity is $2225.50. 
 
388 LOGARITHMS. 
 
 Exercise 138. 
 Find the amount at compound interest : 
 
 1. Of $8764 for 9 years at 5%. 
 
 2. Of $16,470 for 17 years at 3£%. 
 
 3. Of $12,000 for 12 years at 4%. 
 
 4. Of $976.45 for 9 years 6 months at 4£%. 
 Find the principal that will : 
 
 5. Amount to $1200 in 7 years at 5% compound interest. 
 
 6. Amount to $18,740 in 12 years at 4% compound 
 
 interest. 
 
 7. Amount to $847.55 in 5 years 3 months at §\°] com- 
 
 pound interest. 
 
 Find the rate of compound interest : 
 
 8. If $1296 amounts to $1576.75 in 5 years. 
 
 9. If $4830 amounts to $6472.70 in 6 years. 
 
 10. If $4625 amounts to $7404.80 in 12 years. 
 
 11. In what time at 3$% will $2225 amount to $3225 at 
 
 compound interest ? 
 
 12. In what time at 5% will $1640 amount to $3000, 
 
 interest being compounded semi-annually ? 
 
 Find the present value of an annuity : 
 
 13. Of $750 for 12 years, if money is worth 4%. 
 
 14. Of $1200 for 10 years, if money is worth 5J%. 
 
 15. Of $1875 for 6 years, if money is worth 4%. 
 
 16. Of $3200 for 14 years, if money is worth 3£%. 
 
 17. Of $2500 for 8 years, if money is worth 3%. 
 
 18. Of $612.50 for 18 years, if money is worth 3£%. 
 
A TABLE 
 
 OOMMON LOGARITHMS 
 
 NUMBERS 
 
 LOGARITHMS OF NUMBERS TO 100. 
 
 I 
 
 O.OOOOO 
 
 26 
 
 1.41497 
 
 5i 
 
 1.70757 
 
 76 
 
 1. 8808 1 
 
 2 
 
 O.3OIO3 
 
 27 
 
 1 -43 1 36 
 
 52 
 
 1. 71600 
 
 7 l 
 
 1.88649 
 
 3 
 
 O.47712 
 
 28 
 
 1.44716 
 
 53 
 
 1.72428 
 
 78 
 
 1.89209 
 
 4 
 
 0.60206 
 
 29 
 
 1.46240 
 
 54 
 
 1.73239 
 
 79 
 
 1.89763 
 
 5 
 
 O.69897 
 
 30 
 
 1.47712 
 
 55 
 
 1.74036 
 
 80 
 
 1.90309 
 
 6 
 
 O.77815 
 
 3i 
 
 1.49136 
 
 56 
 
 1.74819 
 
 81 
 
 1.90849 
 
 7 
 
 O.845 IO 
 
 32 
 
 I-505I5 
 
 57 
 
 I-75587 
 
 82 
 
 1.91381 
 
 8 
 
 O.9O309 
 
 33 
 
 1.51851 
 
 58 
 
 I-76343 
 
 83 
 
 1.91908 
 
 9 
 
 O.95424 
 
 34 
 
 1.53148 
 
 59 
 
 1.77085 
 
 84 
 
 1.92428 
 
 IO 
 
 1. 00000 
 
 35 
 
 1.54407 
 
 60 
 
 I-778I5 
 
 85 
 
 1.92942 
 
 ii 
 
 I.O4I39 
 
 36 
 
 1.55630 
 
 61 
 
 ?-78533 
 
 86 
 
 1-9345° 
 
 12 
 
 i. 079 1 8 
 
 37 
 
 1.56820 
 
 62 
 
 1.79239 
 
 ! 7 
 
 1.93952 
 
 n 
 
 I. "394 
 
 3* 
 
 1.57978 
 
 63 
 
 J-79934 
 
 88 
 
 1.94448 
 
 14 
 
 1.14613 
 
 39 
 
 1.59106 
 
 64 
 
 1.80618 
 
 89 
 
 1.94939 
 
 15 
 
 1. 17609 
 
 40 
 
 1.60206 
 
 65 
 
 1.81291 
 
 90 
 
 1.95424 
 
 16 
 
 1.20412 
 
 4i 
 
 1. 61278 
 
 66 
 
 1.81954 
 
 9i 
 
 1.95904 
 
 17 
 
 1.23045 
 
 42 
 
 1.62325 
 
 67 
 
 1.82607 
 
 92 
 
 1.96379 
 
 18 
 
 I-25527 
 
 43 
 
 1.63347 
 
 68 
 
 1.83251 
 
 93 
 
 1.96848 
 
 19 
 
 1.27875 
 
 44 
 
 1.64345 
 
 69 
 
 1.83885 
 
 94 
 
 I-973I3 
 
 20 
 
 1.30103 
 
 45 
 
 1.65321 
 
 7° 
 
 1.845 10 
 
 95 
 
 1.97772 
 
 21 
 
 1.32222 
 
 46 
 
 1.66276 
 
 7i 
 
 1.85 126 
 
 96 
 
 1.98227 
 
 22 
 
 1.34242 
 
 47 
 
 1.67210 
 
 72 
 
 1.85733 
 
 97 
 
 1.98677 
 
 23 
 
 I-36I73 
 
 48 
 
 1. 68 1 24 
 
 . 73 
 
 1.86332 
 
 98 
 
 1.99 1 23 
 
 24 
 
 1. 38021 
 
 49 
 
 1 .69020 
 
 74 
 
 1.86923 
 
 99 
 
 1.99564 
 
 25 
 
 1-39794 
 
 50 
 
 1.69897 
 
 75 
 
 1.87506 
 
 100 
 
 2.00000 
 
390 
 
 100—150 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 IOO 
 
 00000 
 
 043 
 
 087 
 
 130 
 
 173 
 
 217 
 
 260 
 
 303 
 
 346 
 
 389 
 
 43 
 
 IOI 
 
 432 
 
 475 
 
 518 
 
 * 6 ol 
 
 604 
 
 647 
 
 689 
 
 732 
 
 775 
 
 817 
 
 43 
 
 1 02 
 
 860 
 
 903 
 
 945 
 
 988 
 
 *030 
 
 ♦072 
 
 *n 5 
 
 *i 57 
 
 *i 99 
 
 ♦242 
 
 42 
 
 103 
 
 01284 
 
 326 
 
 368 
 
 410 
 
 45 2 
 
 494 
 
 536 
 
 578 
 
 620 
 
 662 
 
 42 
 
 104 
 
 7°3 
 
 745 
 
 787 
 
 828 
 
 870 
 
 912 
 
 953 
 
 995 
 
 *o 3 6 
 
 ♦078 
 
 42 
 
 io 5 
 
 021 19 
 
 160 
 
 202 
 
 243 
 
 284 
 
 325 
 
 366 
 
 407 
 
 449 
 
 49o 
 
 41 
 
 106 
 
 53i 
 
 572 
 
 612 
 
 653 
 
 694 
 
 735 
 
 776 
 
 816 
 
 857 
 
 898 
 
 4i 
 
 107 
 
 938 
 
 979 
 
 ♦019 
 
 *o6o 
 
 *IOO 
 
 *i 4 i 
 
 *i8i 
 
 *222 
 
 ♦262 
 
 *302 
 
 40 
 
 108 
 
 03342 
 
 3^3 
 
 423 
 
 463 
 
 5°3 
 
 543 
 
 583 
 
 623 
 
 663 
 
 703 
 
 40 
 
 109 
 
 743 
 
 782 
 
 822 
 
 862 
 
 902 
 
 941 
 
 981 
 
 *02I 
 
 *o6o 
 
 *IOO 
 
 40 
 
 no 
 
 04139 
 
 179 
 
 218 
 
 258 
 
 297 
 
 336 
 
 376 
 
 415 
 
 454 
 
 493 
 
 39 
 
 III 
 
 532 
 
 57i 
 
 610 
 
 650 
 
 689 
 
 727 
 
 766 
 
 805 
 
 844 
 
 883 
 
 39 
 
 112 
 
 922 
 
 961 
 
 999 
 
 *o 3 8 
 
 *o 77 
 
 *ii 5 
 
 *i54 
 
 ♦192 
 
 ♦231 
 
 * 2 69 
 
 39 
 
 "3 
 
 05308 
 
 346 
 
 385 
 
 423 
 
 461 
 
 500 
 
 538 
 
 576 
 
 614 
 
 652 
 
 38 
 
 114 
 
 690 
 
 729 
 
 767 
 
 805 
 
 843 
 
 881 
 
 918 
 
 95 6 
 
 994 
 
 ♦032 
 
 3S ; 
 
 "5 
 
 06070 
 
 108 
 
 H5 
 
 183 
 
 221 
 
 258 
 
 296 
 
 333 
 
 311 
 
 408 
 
 38 
 
 116 
 
 446 
 
 483 
 
 893 
 
 558 
 
 595 
 
 £>33 
 
 670 
 
 707 
 
 744 
 
 781 
 
 37 
 
 117 
 
 819 
 
 856 
 
 930 
 
 967 
 
 ♦004 
 
 ♦041 
 
 ♦078 
 
 *n 5 
 
 ♦151 
 
 37 
 
 118 
 
 07188 
 
 225 
 
 262 
 
 298 
 
 335 
 
 372 
 
 408 
 
 445 
 
 482 
 
 5 l8 
 
 37 
 
 119 
 
 555 
 
 59i 
 
 628 
 
 664 
 
 700 
 
 737 
 
 773 
 
 809 
 
 846 
 
 882 
 
 36 
 
 120 
 
 918 
 
 954 
 
 99o 
 
 ♦027 
 
 ♦063 
 
 ♦099 
 
 *i35 
 
 *i 7 i 
 
 ♦207 
 
 ♦243 
 
 36 
 
 121 
 
 08279 
 
 3H 
 
 35o 
 
 386 
 
 422 
 
 458 
 
 493 
 
 529 
 
 (565 
 
 600 
 
 36 
 
 122 
 
 636 
 
 672 
 
 707 
 
 743 
 
 778 
 
 814 
 
 849 
 
 884 
 
 920 
 
 955 
 
 35 
 
 123 
 
 991 
 
 ♦026 
 
 *o6i 
 
 ♦096 
 
 ♦ 132 
 
 ♦167 
 
 *202 
 
 *237 
 
 ♦272 
 
 *307 
 
 35 
 
 124 
 
 09342 
 
 377 
 
 412 
 
 447 
 
 482 
 
 5i7 
 
 552 
 
 587 
 
 621 
 
 656 
 
 35 
 
 125 
 
 691 
 
 726 
 
 760 
 
 795 
 
 830 
 
 864 
 
 899 
 
 934 
 
 968 
 
 ♦003 
 
 35 
 
 126 
 
 10037 
 
 072 
 
 106 
 
 140 
 
 175 
 
 209 
 
 243 
 
 278 
 
 312 
 
 346 
 
 34 
 
 III 
 
 380 
 
 4i5 
 
 449 
 
 483 
 
 5i7 
 
 55i 
 
 585 
 
 619 
 
 653 
 
 687 
 
 34 
 
 721 
 
 755 
 
 789 
 
 823 
 
 857 
 
 890 
 
 924 
 
 958 
 
 992 
 
 ♦025 
 
 34 
 
 129 
 
 1 1059 
 
 093 
 
 126 
 
 160 
 
 193 
 
 227 
 
 26l 
 
 294 
 
 327 
 
 361 
 
 34 
 
 I30 
 
 394 
 
 428 
 
 461 
 
 494 
 
 528 
 
 893 
 
 594 
 
 628 
 
 661 
 
 694 
 
 33 
 
 131 
 
 727 
 
 760 
 
 793 
 
 826 
 
 860 
 
 926 
 
 959 
 
 992 
 
 ♦024 
 
 33 
 
 132 
 
 12057 
 
 090 
 
 123 
 
 156 
 
 189 
 
 222 
 
 254 
 
 287 
 
 320 
 
 352 
 
 33 
 
 133 
 
 385 
 
 418 
 
 45° 
 
 483 
 
 516 
 
 548 
 
 581 
 
 613 
 
 646 
 
 678 
 
 33 
 
 134 
 
 710 
 
 743 
 
 775 
 
 808 
 
 840 
 
 872 
 
 905 
 
 937 
 
 969 
 
 *OOI 
 
 32 
 
 J 35 
 
 13033 
 
 066 
 
 098 
 
 130 
 
 162 
 
 194 
 
 226 
 
 258 
 
 290 
 
 322 
 
 32 ; 
 
 136 
 
 354 
 
 386 
 
 418 
 
 45° 
 
 481 
 
 5i3 
 
 545 
 
 577 
 
 609 
 
 640 
 
 32 
 
 137 
 
 672 
 
 704 
 
 735 
 
 767 
 
 799 
 
 830 
 
 862 
 
 893 
 
 925 
 
 956 
 
 32 i 
 
 138 
 
 988 
 
 ♦019 
 
 ♦051 
 
 *o82 
 
 *»4 
 
 *i45 
 
 ♦176 
 
 ♦208 
 
 ♦239 
 
 ♦270 
 
 3i 
 
 139 
 
 14301 
 
 333 
 
 364 
 
 395 
 
 426 
 
 457 
 
 489 
 
 520 
 
 55i 
 
 582 
 
 3i 
 
 140 
 
 613 
 
 644 
 
 675 
 
 706 
 
 737 
 
 768 
 
 799 
 
 829 
 
 860 
 
 891 
 
 3i 
 
 141 
 
 922 
 
 953 
 
 983 
 
 ♦014 
 
 ♦045 
 
 ♦076 
 
 *io6 
 
 *i3* 
 
 *i68 
 
 *i 9 8 
 
 31 j 
 
 142 
 
 15229 
 
 259 
 
 290 
 
 320 
 
 35i 
 
 381 
 
 412 
 
 442 
 
 473 
 
 503 
 
 31 
 
 H3 
 
 534 
 
 5 6 4 
 
 594 
 
 625 
 
 655 
 
 685 
 
 715 
 
 746 
 
 776 
 
 806 
 
 30 ■ 
 
 144 
 
 836 
 
 866 
 
 897 
 
 927 
 
 957 
 
 987 
 
 ♦017 
 
 *047 
 
 *o77 
 
 ♦107 
 
 3° 
 
 H5 
 
 16137 
 
 167 
 
 197 
 
 227 
 
 256 
 
 286 
 
 3i6 
 
 346 
 
 376 
 
 406 
 
 30 ' 
 
 146 
 
 435 
 
 465 
 
 495 
 
 524 
 
 554 
 
 584 
 
 613 
 
 643 
 
 673 
 
 702 
 
 30 
 
 147 
 
 732 
 
 761 
 
 791 
 
 820 
 
 850 
 
 879 
 
 909 
 
 938 
 
 967 
 
 997 
 
 29 
 
 148 
 
 17026 
 
 056 
 
 085 
 
 114 
 
 143 
 
 173 
 
 202 
 
 231 
 
 260 
 
 289 
 
 29 
 
 149 
 
 319 
 
 348 
 
 377 
 
 406 
 
 435 
 
 464 
 
 493 
 
 522 
 
 55 1 
 
 580 
 
 29 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 7 
 
 8 
 
 9 
 
 D. 
 

 
 
 
 150— 
 
 200 
 
 
 
 
 
 391 
 
 N. 
 
 1 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 150 
 
 17609 
 
 638 
 
 667 
 
 696 
 
 725 
 
 754 
 
 782 
 
 811 
 
 840 
 
 869 
 
 29 
 
 151 
 
 898 
 
 926 
 
 955 
 
 984 
 
 *oi3 
 
 ♦041 
 
 ♦070 
 
 ♦099 
 
 ♦127 
 
 *i 5 6 
 
 29 
 
 I5 2 
 
 18184 
 
 213 
 
 241 
 
 270 
 
 298 
 
 327 
 
 355 
 
 384 
 
 412 
 
 441 
 
 29 
 
 153 
 
 469 
 
 498 
 
 ¥a 
 
 554 
 
 583 
 
 611 
 
 639 
 
 667 
 
 696 
 
 724 
 
 28 
 
 154 
 
 752 
 
 780 
 
 808 
 
 ^37 
 
 865 
 
 893 
 
 921 
 
 949 
 
 977 
 
 *oo5 
 
 28 
 
 155 
 
 19033 
 
 061 
 
 089 
 
 117 
 
 145 
 
 173 
 
 201 
 
 229 
 
 257 
 
 285 
 
 28 
 
 156 
 
 312 
 
 340 
 
 368 
 
 396 
 
 424 
 
 45i 
 
 479 
 
 507 
 
 535 
 
 562 
 
 28 
 
 157 
 
 590 
 
 618 
 
 645 
 
 673 
 
 700 
 
 728 
 
 is*> 
 
 783 
 
 
 838 
 
 28 
 
 158 
 
 866 
 
 893 
 
 921 
 
 948 
 
 976 
 
 *oo3 
 
 ♦030 
 
 ♦058 
 
 "085 
 358 
 
 *II2 
 
 27 
 
 159 
 
 20140 
 
 167 
 
 194 
 
 222 
 
 249 
 
 276 
 
 303 
 
 330 
 
 385 
 
 27 
 
 160 
 
 412 
 
 439 
 
 466 
 
 493 
 
 520 
 
 548 
 817 
 
 575 
 
 602 
 
 629 
 
 656 
 
 27 
 
 161 
 
 683 
 
 710 
 
 737 
 
 7 6 3 
 
 790 
 
 844 
 
 871 
 
 898 
 
 925 
 
 27 
 
 162 
 
 95 2 
 
 978 
 
 *oo5 
 
 ♦032 
 
 *059 
 
 "085 
 
 *II2 
 
 *I39 
 
 *i65 
 
 *I92 
 
 27 
 
 163 
 
 21219 
 
 245 
 
 272 
 
 299 
 
 325 
 
 352 
 
 378 
 
 405 
 
 43i 
 
 458 
 
 27 
 
 164 
 
 484 
 
 5" 
 
 537 
 
 5 6 4 
 
 590 
 
 617 
 
 643 
 
 669 
 
 696 
 
 722 
 
 26 
 
 165 
 
 748 
 
 775 
 
 801 
 
 827 
 
 854 
 
 880 
 
 906 
 
 932 
 
 958 
 
 985 
 
 26 
 
 166 
 
 2201 1 
 
 037 
 
 063 
 
 089 
 
 "5 
 
 141 
 
 167 
 
 194 
 
 220 
 
 246 
 
 26 
 
 167 
 
 272 
 
 298 
 
 324 
 
 35° 
 
 376 
 
 401 
 
 427 
 
 453 
 
 479 
 
 505 
 
 26 
 
 168 
 
 53i 
 
 557 
 
 840 
 
 608 
 
 634 
 
 660 
 
 686 
 
 712 
 
 737 
 
 763 
 
 26 
 
 169 
 
 789 
 
 814 
 
 866 
 
 891 
 
 917 
 
 943 
 
 968 
 
 994 
 
 *OI9 
 
 26 
 
 170 
 
 23045 
 
 070 
 
 096 
 
 121 
 
 147 
 
 172 
 
 198 
 
 223 
 
 249 
 
 274 
 
 25 
 
 171 
 
 3°o 
 
 325 
 
 35° 
 
 376 
 
 401 
 
 426 
 
 452 
 
 477 
 
 502 
 
 528 
 
 25 
 
 172 
 
 805 
 
 578 
 
 603 
 
 629 
 
 654 
 
 679 
 
 7°4 
 
 729 
 
 754 
 
 779 
 
 25 
 
 173 
 
 830 
 
 855 
 
 880 
 
 905 
 
 93o 
 
 955 
 
 980 
 
 ♦005 
 
 ♦030 
 
 25 
 
 174 
 
 24055 
 
 080 
 
 i°5 
 
 130 
 
 155 
 
 180 
 
 204 
 
 229 
 
 254 
 
 279 
 
 25 
 
 175 
 
 304 
 
 329 
 
 353 
 
 378 
 
 403 
 
 428 
 
 452 
 
 477 
 
 502 
 
 527 
 
 25 
 
 176 
 
 55i 
 
 576 
 
 601 
 
 625 
 
 650 
 
 674 
 
 699 
 
 724 
 
 748 
 
 773 
 
 25 
 
 177 
 
 797 
 
 822 
 
 846 
 
 871 
 
 895 
 
 920 
 
 944 
 
 969 
 
 993 
 
 *oi8 
 
 25 
 
 178 
 
 25042 
 
 066 
 
 091 
 
 "5 
 
 139 
 
 164 
 
 188 
 
 212 
 
 237 
 
 261 
 
 24 
 
 179 
 
 285 
 
 310 
 
 334 
 
 358 
 
 382 
 
 406 
 
 431 
 
 455 
 
 479 
 
 503 
 
 24 
 
 180 
 
 527 
 
 55i 
 
 575 
 
 600 
 
 624 
 
 648 
 
 672 
 
 696 
 
 72b. 
 
 744 
 
 24 
 
 181 
 
 768 
 
 792 
 
 816 
 
 840 
 
 864 
 
 888 
 
 912 
 
 935 
 
 959 
 
 983 
 
 24 
 
 182 
 
 26007 
 
 031 
 
 055 
 
 079 
 
 102 
 
 126 
 
 150 
 
 174 
 
 198 
 
 221 
 
 24 
 
 183 
 
 245 
 
 269 
 
 293 
 
 316 
 
 34o 
 
 364 
 
 387 
 
 411 
 
 435 
 
 458 
 
 24 
 
 184 
 
 482 
 
 5°5 
 
 529 
 
 553 
 
 576 
 
 600 
 
 623 
 
 647 
 
 670 
 
 694 
 
 24 
 
 185 
 
 717 
 
 741 
 
 764 
 
 788 
 
 811 
 
 834 
 
 858 
 
 881 
 
 905 
 
 928 
 
 23 
 
 186 
 
 95i 
 
 975 
 
 998 
 
 *02I 
 
 *045 
 
 *o68 
 
 ♦091 
 
 •114 
 
 *i 3 8 
 
 *i6i 
 
 23 
 
 187 
 
 27184 
 
 207 
 
 231 
 
 254 
 
 277 
 
 300 
 
 323 
 
 346 
 
 37o 
 
 393 
 
 23 
 
 188 
 
 416 
 
 439. 
 
 462 
 
 485 
 
 508 
 
 531 
 
 554 
 
 577 
 807 
 
 600 
 
 623 
 
 23 
 
 189 
 
 646 
 
 669 
 
 692 
 
 715 
 
 738 
 
 761 
 
 784 
 
 830 
 
 852 
 
 23 
 
 190 
 
 875 
 
 898 
 
 921 
 
 944 
 
 967 
 
 989 
 
 *OI2 
 
 *o35 
 
 ♦058 
 
 *o8i 
 
 23 
 
 191 
 
 28103 
 
 126 
 
 149 
 
 I7 o 
 
 194 
 
 217 
 
 24O 
 
 262 
 
 285 
 
 307 
 
 23 
 
 192 
 
 330 
 
 353 
 
 375 
 
 398 
 
 421 
 
 443 
 
 466 
 
 488 
 
 5" 
 
 533 
 
 23 
 
 193 
 
 556 
 
 578 
 
 601 
 
 623 
 
 646 
 
 668 
 
 69I 
 
 713 
 
 735 
 
 n s 
 
 22 
 
 194 
 
 780 
 
 803 
 
 825 
 
 847 
 
 870 
 
 892 
 
 914 
 
 937 
 
 959 
 
 981 
 
 22 
 
 195 
 
 29003 
 
 026 
 
 048 
 
 070 
 
 092 
 
 "5 
 
 137 
 
 159 
 
 181 
 
 203 
 
 22 
 
 196 
 
 226 
 
 248 
 
 270 
 
 292 
 
 314 
 
 336 
 
 358 
 
 380 
 
 403 
 
 425 
 
 22 
 
 197 
 
 447 
 
 469 
 
 491 
 
 513 
 
 535 
 
 557 
 
 579 
 
 601 
 
 623 
 
 Si 5 
 
 22 
 
 198 
 
 667 
 
 688 
 
 710 
 
 732 
 
 754 
 
 776 
 
 798 
 
 820 
 
 842 
 
 863 
 
 22 
 
 199 
 
 885 
 
 907 
 
 929 
 
 95 x 
 
 973 
 
 994 
 
 *oi6 
 
 ♦038 
 
 *o6o 
 
 *o8i 
 
 22 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
392 
 
 200—250 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 7 
 
 8 
 
 9 
 
 D. 
 
 200 
 
 30103 
 
 125 
 
 146 
 
 168 
 
 190 
 
 211 
 
 233 
 
 255 
 
 276 
 
 298 
 
 22 
 
 201 
 
 320 
 
 34i 
 
 363 
 
 384 
 
 406 
 
 428 
 
 449 
 
 47i 
 
 492 
 
 5H 
 
 22 
 
 202 
 
 535 
 
 557 
 
 578 
 
 600 
 
 621 
 
 643 
 
 664 
 
 685 
 
 707 
 
 728 
 
 21 
 
 203 
 
 75o 
 
 771 
 
 792 
 
 814 
 
 835 
 *o 4 8 
 
 856 
 
 878 
 
 899 
 
 920 
 
 942 
 
 21 
 
 204 
 
 963 
 
 984 
 
 *oo6 
 
 ♦027 
 
 ♦069 
 
 ♦091 
 
 *II2 
 
 *i33 
 
 *i54 
 
 21 
 
 205 
 
 3U75 
 
 197 
 
 218 
 
 239 
 
 260 
 
 281 
 
 302 
 
 323 
 
 345 
 
 366 
 
 21 
 
 206 
 
 387 
 
 408 
 
 429 
 
 45° 
 
 471 
 
 492 
 
 513 
 
 534 
 
 555 
 
 576 
 
 21 
 
 207 
 
 597 
 806 
 
 618 
 
 639 
 
 660 
 
 681 
 
 702 
 
 723 
 
 744 
 
 765 
 
 785 
 
 21 
 
 208 
 
 827 
 
 848 
 
 869 
 
 890 
 
 911 
 
 931 
 
 952 
 
 973 
 
 994 
 
 21 
 
 209 
 
 32015 
 
 035 
 
 056 
 
 077 
 
 098 
 
 118 
 
 139 
 
 160 
 
 181 
 
 201 
 
 21 
 
 210 
 
 222 
 
 243 
 
 263 
 
 284 
 
 305 
 
 325 
 
 346 
 
 366 
 
 387 
 
 408 
 
 21 
 
 211 
 
 428 
 
 449 
 
 469 
 
 490 
 
 5 J o 
 
 531 
 
 552 
 
 572 
 
 593 
 
 613 
 
 20 
 
 212 
 
 634 
 
 654 
 
 675 
 
 695 
 
 7i5 
 
 736 
 
 756 
 
 777 
 
 797 
 
 818 
 
 20 
 
 213 
 
 838 
 
 858 
 
 879 
 
 899 
 
 919 
 
 940 
 
 960 
 
 980 
 
 *OOI 
 
 *02I 
 
 20 
 
 214 
 
 33041 
 
 062 
 
 082 
 
 102 
 
 122 
 
 143 
 
 163 
 
 183 
 
 203 
 
 224 
 
 20 
 
 215 
 
 244 
 
 264 
 
 284 
 
 304 
 
 325 
 
 345 
 
 365 
 
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 486 
 
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 242 
 
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 282 
 
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 218 
 
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 276 
 
 295 
 
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 334 
 
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 19 
 
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 291 
 
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 328 
 
 346 
 
 365 
 
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 438 
 
 457 
 
 18 
 
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 548 
 
 566 
 
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 238 
 
 256 
 
 274 
 
 292 
 
 310 
 
 328 
 
 346 
 
 364 
 
 18 
 
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 382 
 
 399 
 
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 435 
 
 453 
 
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 489 
 
 507 
 
 525 
 
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 18 
 
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 596 
 
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 668 
 
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 703 
 
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 129 
 
 146 
 
 164 
 
 182 
 
 199 
 
 217 
 
 235 
 
 252 
 
 18 
 
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 270 
 
 287 
 
 305 
 
 322 
 
 340 
 
 358 
 
 375 
 
 393 
 
 410 
 
 428 
 
 18 
 
 248 
 
 445 
 
 463 
 
 480 
 
 498 
 
 515 
 
 533 
 
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 585 
 
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 18 
 
 249 
 
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 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
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 D. 
 
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 393 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
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 8 
 
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 252 
 
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 192 
 
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 226 
 
 243 
 
 261 
 
 278 
 
 295 
 
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 253 
 
 312 
 
 329 
 
 346 
 
 364 
 
 381 
 
 398 
 
 415 
 
 432 
 
 449 
 
 466 
 
 17 
 
 254 
 
 483 
 
 500 
 
 518 
 
 535 
 
 552 
 
 5 6 9 
 
 586 
 
 603 
 
 620 
 
 637 
 
 17 
 
 255 
 
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 705 
 
 722 
 
 739 
 
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 256 
 
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 257 
 
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 17 
 
 258 
 
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 179 
 
 196 
 
 212 
 
 229 
 
 246 
 
 263 
 
 280 
 
 296 
 
 3i3 
 
 17 
 
 259 
 
 33o 
 
 347 
 
 363 
 
 380 
 
 397 
 
 414 
 
 430 
 
 447 
 
 464 
 
 481 
 
 17 
 
 260 
 
 497 
 
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 547 
 
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 780 
 
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 16 
 
 264 
 
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 177 
 
 193 
 
 210 
 
 226 
 
 243 
 
 259 
 
 275 
 
 292 
 
 308 
 
 16 
 
 265 
 
 325 
 
 341 
 
 357 
 
 374 
 
 390 
 
 406 
 
 423 
 
 439 
 
 455 
 
 472 
 
 16 
 
 266 
 
 488 
 
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 521 
 
 537 
 
 553 
 
 570 
 
 586 
 
 602 
 
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 16 
 
 267 
 
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 667 
 
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 700 
 
 716 
 
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 16 
 
 268 
 
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 185 
 
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 217 
 
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 271 
 
 297 
 
 313 
 
 329 
 
 345 
 
 361 
 
 377 
 
 393 
 
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 16 
 
 272 
 
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 473 
 
 489 
 
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 569 
 
 584 
 
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 616 
 
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 807 
 
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 107 
 
 122 
 
 138 
 
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 185 
 
 201 
 
 217 
 
 232 
 
 16 
 
 277 
 
 248 
 
 264 
 
 279 
 
 295 
 
 311 
 
 326 
 
 342 
 
 358 
 
 373 
 
 389 
 
 16 
 
 278 
 
 404 
 
 420 
 
 436 
 
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 467 
 
 483 
 
 498 
 
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 529 
 
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 279 
 
 560 
 
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 592 
 
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 117 
 
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 378 
 
 240 
 
 255 
 
 271 
 
 286 
 
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 332 
 
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 484 
 
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 120 
 
 135 
 
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 165 
 
 180 
 
 195 
 
 210 
 
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 240 
 
 255 
 
 270 
 
 285 
 
 300 
 
 315 
 
 330 
 
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 479 
 
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 202 
 
 217 
 
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 246 
 
 261 
 
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 276 
 
 290 
 
 305 
 
 319 
 
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 349 
 
 363 
 
 378 
 
 392 
 
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 298 
 
 422 
 
 436 
 
 451 
 
 465 
 
 480 
 
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 538 
 
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 299 
 
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 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
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 300—350 
 
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 1 
 
 2 
 
 3 
 
 4 
 
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 301 
 
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 087 
 
 101 
 
 116 
 
 130 
 
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 144 
 
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 173 
 
 187 
 
 202 
 
 216 
 
 230 
 
 244 
 
 259 
 
 273 
 
 14 
 
 304 
 
 287 
 
 302 
 
 316 
 
 330 
 
 344 
 
 359 
 
 373 
 
 387 
 
 401 
 
 416 
 
 14 
 
 305 
 
 430 
 
 444 
 
 458 
 
 473 
 
 487 
 
 501 
 
 515 
 
 530 
 
 544 
 
 558 
 
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 306 
 
 572 
 
 586 
 
 601 
 
 615 
 
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 307 
 
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 164 
 
 178 
 
 192 
 
 206 
 
 220 
 
 234 
 
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 276 
 
 290 
 
 304 
 
 318 
 
 332 
 
 346 
 
 360 
 
 374 
 
 388 
 
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 312 
 
 415 
 
 429 
 
 443 
 
 457 
 
 471 
 
 485 
 
 499 
 
 513 
 
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 317 
 
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 215 
 
 229 
 
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 243 
 
 256 
 
 270 
 
 284 
 
 297 
 
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 338 
 
 352 
 
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 319 
 
 379 
 
 393 
 
 406 
 
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 501 
 
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 705 
 
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 974 
 
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 108 
 
 121 
 
 135 
 
 148 
 
 162 
 
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 325 
 
 188 
 
 202 
 
 348 
 
 228 
 
 242 
 
 255 
 
 268 
 
 282 
 
 295 
 
 308 
 
 13 
 
 326 
 
 322 
 
 335 
 
 362 
 
 37 S 
 
 388 
 
 402 
 
 415 
 
 428 
 
 441 
 
 13 
 
 327 
 328 
 
 455 
 
 468 
 
 481 
 
 495 
 
 508 
 
 521 
 
 534 
 
 548 
 
 561 
 
 574 
 
 13 
 
 587 
 
 601 
 
 614 
 
 627 
 
 640 
 
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 667 
 
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 329 
 
 720 
 
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 330 
 
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 140 
 
 153 
 
 166 
 
 179 
 
 192 
 
 205 
 
 218 
 
 231 
 
 13 
 
 333 
 
 244 
 
 257 
 388 
 
 270 
 
 284 
 
 297 
 
 310 
 
 323 
 
 336 
 
 349 
 
 362 
 
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 334 
 
 375 
 
 401 
 
 414 
 
 427 
 
 440 
 
 453 
 
 466 
 
 479 
 
 492 
 
 13 
 
 335 
 
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 530 
 
 543 
 
 556 
 
 569 
 
 582 
 
 595 
 
 608 
 
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 336 
 
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 337 
 
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 338 
 
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 122 
 
 135 
 
 13 
 
 340 
 
 148 
 
 161 
 
 173 
 
 186 
 
 199 
 
 212 
 
 224 
 
 237 
 
 250 
 
 263 
 
 13 
 
 34i 
 
 275 
 
 288 
 
 301 
 
 3H 
 
 326 
 
 339 
 
 352 
 
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 377 
 
 390 
 
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 342 
 
 403 
 
 4i5 
 
 428 
 
 441 
 
 580 
 
 466 
 
 479 
 
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 517 
 
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 343 
 
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 542 
 
 555 
 
 567 
 
 593 
 
 605 
 
 618 
 
 631 
 
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 344 
 
 656 
 
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 706 
 
 719 
 
 732 
 
 744 
 
 757 
 
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 345 
 
 782 
 
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 832 
 
 845 
 
 857 
 
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 895 
 
 13 
 
 346 
 
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 983 
 
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 347 
 
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 108 
 
 120 
 
 133 
 
 145 
 
 13 
 
 348 
 
 283 
 
 170 
 
 195 
 
 208 
 
 220 
 
 233 
 
 245 
 
 2 l S 
 
 27O 
 
 12 
 
 349 
 
 295 
 
 307 
 
 320 
 
 332 
 
 345 
 
 357 
 
 370 
 
 382 
 
 394 
 
 12 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
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 D. 
 
350—400 
 
 395 
 
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 2 
 
 3 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
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 419 
 
 432 
 
 444 
 
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 469 
 
 481 
 
 494 
 
 506 
 
 518 
 
 12 
 
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 53i 
 
 543 
 
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 568 
 
 593 
 
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 617 
 
 630 
 
 642 
 
 12 
 
 352 
 
 654 
 
 667 
 
 679 
 
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 704 
 
 716 
 
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 765 
 
 12 
 
 353 
 
 777 
 
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 888 
 
 12 
 
 354 
 
 900 
 
 913 
 
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 355 
 
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 108 
 
 121 
 
 133 
 
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 356 
 
 145 
 
 157 
 
 169 
 
 182 
 
 194 
 
 206 
 
 218 
 
 230 
 
 242 
 
 255 
 
 12 
 
 357 
 
 267 
 
 279 
 
 291 
 
 303 
 
 315 
 
 328 
 
 340 
 
 352 
 
 364 
 
 376 
 
 12 
 
 358 
 
 388 
 
 400 
 
 413 
 
 425 
 
 437 
 
 449 
 
 461 
 
 473 
 
 485 
 
 497 
 
 12 
 
 359 
 
 509 
 
 522 
 
 534 
 
 540 
 
 558 
 
 570 
 
 582 
 
 594 
 
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 618 
 
 12 
 
 360 
 
 630 
 
 642 
 
 654 
 
 666 
 
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 703 
 
 l l $ 
 
 727 
 
 739 
 
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 361 
 
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 787 
 
 799 
 
 811 
 
 823 
 
 835 
 
 847 
 
 859 
 
 12 
 
 362 
 
 871 
 
 883 
 
 895 
 
 907 
 
 919 
 
 931 
 
 943 
 
 955 
 
 967 
 
 979 
 
 12 
 
 363 
 
 991 
 
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 *o86 
 
 ♦098 
 
 12 
 
 364 
 
 56110 
 
 122 
 
 134 
 
 146 
 
 158 
 
 170 
 
 182 
 
 194 
 
 205 
 
 217 
 
 12 
 
 365 
 
 229 
 
 241 
 
 253 
 
 265 
 
 277 
 
 289 
 
 301 
 
 312 
 
 324 
 
 336 
 
 12 
 
 366 
 
 348 
 
 360 
 
 372 
 
 384 
 
 396 
 
 407 
 
 419 
 
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 443 
 
 455 
 
 12 
 
 367 
 
 467 
 
 478 
 
 490 
 
 502 
 
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 526 
 
 538 
 
 549 
 
 561 
 
 573 
 
 12 
 
 368 
 
 585 
 
 597 
 
 608 
 
 620 
 
 632 
 
 644 
 
 656 
 
 667 
 
 679 
 
 691 
 
 12 
 
 369 
 
 703 
 
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 726 
 
 738 
 
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 761 
 
 773 
 
 785 
 
 797 
 
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 12 
 
 37° 
 
 820 
 
 832 
 
 844 
 
 855 
 
 867 
 
 879 
 
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 902 
 
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 101 
 
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 136 
 
 148 
 
 159 
 
 12 
 
 373 
 
 171 
 
 183 
 
 194 
 
 206 
 
 217 
 
 229 
 
 241 
 
 252 
 
 264 
 
 276 
 
 12 
 
 374 
 
 287 
 
 299 
 
 310 
 
 322 
 
 334 
 
 345 
 
 357 
 
 368 
 
 380 
 
 392 
 
 12 
 
 375 
 
 403 
 
 415 
 
 426 
 
 438 
 
 449 
 
 461 
 
 473 
 
 484 
 
 496 
 
 5°7 
 
 12 
 
 376 
 
 519 
 
 530 
 
 542 
 
 553 
 
 565 
 
 576 
 
 588 
 
 600 
 
 611 
 
 623 
 
 12 
 
 377 
 
 634 
 
 646 
 
 657 
 
 669 
 
 680 
 
 692 
 
 703 
 
 1 15 
 830 
 
 It 
 
 738 
 
 11 
 
 378 
 
 749 
 
 761 
 
 772 
 
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 795 
 
 807 
 
 818 
 
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 379 
 
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 380 
 
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 138 
 
 149 
 
 161 
 
 172 
 
 184 
 
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 11 
 
 382 
 
 206 
 
 218 
 
 229 
 
 240 
 
 252 
 
 263 
 
 274 
 
 286 
 
 297 
 
 309 
 
 11 
 
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 320 
 
 331 
 
 343 
 
 354 
 
 365 
 
 377 
 
 388 
 
 399 
 
 410 
 
 422 
 
 11 
 
 384 
 
 433 
 
 444 
 
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 467 
 
 478 
 
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 501 
 
 512 
 
 524 
 
 535 
 
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 546 
 
 557 
 
 5 6 9 
 
 580 
 
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 602 
 
 614 
 
 625 
 
 636 
 
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 386 
 
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 681 
 
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 726 
 
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 387 
 
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 388 
 
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 173 
 
 184 
 
 195 
 
 207 
 
 11 
 
 391 
 
 218 
 
 229 
 
 240 
 
 251 
 
 262 
 
 273 
 
 284 
 
 295 
 
 306 
 
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 11 
 
 392 
 
 329 
 
 340 
 
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 362 
 
 373 
 
 384 
 
 395 
 
 406 
 
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 428 
 
 11 
 
 393 
 
 439 
 
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 461 
 
 472 
 
 483 
 
 494 
 
 506 
 
 5*7 
 
 528 
 
 539 
 
 11 
 
 394 
 
 550 
 
 561 
 
 572 
 
 583 
 
 594 
 
 605 
 
 616 
 
 627 
 
 638 
 
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 11 
 
 395 
 
 660 
 
 671 
 
 682 
 
 693 
 
 704 
 
 715 
 
 726 
 
 737 
 
 748 
 
 759 
 
 11 
 
 396 
 
 770 
 879 
 
 780 
 
 791 
 
 802 
 
 813 
 
 824 
 
 835 
 
 846 
 
 857 
 
 868 
 
 11 
 
 397 
 
 890 
 
 901 
 
 912 
 
 923 
 
 934 
 
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 11 
 
 398 
 
 988 
 
 999 
 
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 11 
 
 399 
 
 60097 
 
 108 
 
 119 
 
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 141 
 
 152 
 
 163 
 
 173 
 
 184 
 
 195 
 
 11 
 
 N. 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
396 
 
 400—450 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
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 7 
 
 8 
 
 9 
 
 D. 
 
 400 
 
 60206 
 
 217 
 
 228 
 
 239 
 
 249 
 
 260 
 
 271 
 
 282 
 
 293 
 
 304 
 
 11 
 
 401 
 
 314 
 
 325 
 
 336 
 
 347 
 
 358 
 
 369 
 
 379 
 
 390 
 
 401 
 
 412 
 
 11 
 
 402 
 
 423 
 
 433 
 
 444 
 
 455 
 
 466 
 
 477 
 
 487 
 
 498 
 
 5°9 
 
 520 
 
 11 
 
 403 
 
 53i 
 
 54i 
 
 552 
 
 563 
 
 574 
 
 584 
 
 595 
 
 606 
 
 617 
 
 627 
 
 11 
 
 404 
 
 638 
 
 649 
 
 660 
 
 670 
 
 681 
 
 692 
 
 703 
 
 713 
 
 724 
 
 735 
 
 11 
 
 405 
 
 746 
 
 756 
 
 767 
 
 778 
 
 788 
 
 799 
 
 810 
 
 821 
 
 831 
 
 842 
 
 11 
 
 406 
 
 853 
 
 863 
 
 874 
 
 885 
 
 895 
 
 906 
 
 917 
 
 927 
 
 938 
 
 949 
 
 11 
 
 407 
 
 959 
 
 970 
 
 981 
 
 991 
 
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 11 
 
 408 
 
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 119 
 
 130 
 
 140 
 
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 162 
 
 11 
 
 409 
 
 172 
 
 183 
 
 194 
 
 204 
 
 215 
 
 225 
 
 236 
 
 247 
 
 257 
 
 268 
 
 11 
 
 410 
 
 278 
 
 289 
 
 300 
 
 310 
 
 321 
 
 33i 
 
 342 
 
 352 
 
 363 
 
 374 
 
 11 
 
 411 
 
 384 
 
 395 
 
 405 
 
 416 
 
 426 
 
 437 
 
 448 
 
 458 
 
 469 
 
 479 
 
 11 
 
 412 
 
 490 
 
 500 
 
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 521 
 
 532 
 
 542 
 
 553 
 
 563 
 
 574 
 
 584 
 
 11 
 
 413 
 
 595 
 
 606 
 
 616 
 
 627 
 
 637 
 
 648 
 
 658 
 
 669 
 
 679 
 
 690 
 
 11 
 
 414 
 
 700 
 
 711 
 
 721 
 
 73i 
 
 742 
 
 752 
 
 763 
 
 773 
 
 784 
 
 794 
 
 10 
 
 415 
 
 805 
 
 815 
 
 826 
 
 836 
 
 847 
 
 857 
 
 868 
 
 878 
 
 888 
 
 899 
 
 10 
 
 416 
 
 909 
 
 920 
 
 930 
 
 941 
 
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 962 
 
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 10 
 
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 118 
 
 128 
 
 138 
 
 149 
 
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 170 
 
 180 
 
 190 
 
 201 
 
 211 
 
 10 
 
 419 
 
 221 
 
 232 
 
 242 
 
 252 
 
 263 
 
 273 
 
 284 
 
 294 
 
 3°4 
 
 3i5 
 
 10 
 
 420 
 
 428 
 
 335 
 
 346 
 
 356 
 
 366 
 
 377 
 
 387 
 
 397 
 
 408 
 
 418 
 
 10 
 
 421 
 
 439 
 
 449 
 
 459 
 
 469 
 
 480 
 
 490 
 
 500 
 
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 521 
 
 10 
 
 422 
 
 53i 
 
 542 
 
 552 
 
 562 
 
 572 
 
 583 
 
 593 
 
 603 
 
 613 
 
 624 
 
 10 
 
 423 
 
 634 
 
 644 
 
 655 
 
 665 
 
 675 
 
 685 
 
 696 
 
 706 
 
 716 
 
 726 
 
 10 
 
 424 
 
 737 
 
 747 
 
 757 
 
 767 
 
 778 
 
 788 
 
 798 
 
 808 
 
 818 
 
 829 
 
 10 
 
 425 
 
 839 
 
 849 
 
 859 
 
 870 
 
 880 
 
 890 
 
 900 
 
 910 
 
 921 
 
 931 
 
 10 
 
 426 
 
 941 
 
 95 l 
 
 961 
 
 972 
 
 982 
 
 992 
 
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 427 
 
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 114 
 
 124 
 
 134 
 
 10 
 
 428 
 
 144 
 
 155 
 
 165 
 
 175 
 
 185 
 
 J95 
 
 205 
 
 215 
 
 225 
 
 236 
 
 10 
 
 429 
 
 246 
 
 256 
 
 266 
 
 276 
 
 286 
 
 296 
 
 306 
 
 317 
 
 327 
 
 337 
 
 10 
 
 430 
 
 347 
 
 357 
 
 3^7 
 
 377 
 
 387 
 
 397 
 
 407 
 
 417 
 
 428 
 
 438 
 
 10 
 
 431 
 
 448 
 
 458 
 
 468 
 
 478 
 
 488 
 
 498 
 
 508 
 
 518 
 
 528 
 
 538 
 
 10 
 
 432 
 
 548 
 
 558 
 
 568 
 
 579 
 
 589 
 
 599 
 
 609 
 
 619 
 
 629 
 
 639 
 
 10 
 
 433 
 
 649 
 
 659 
 
 669 
 
 679 
 
 689 
 
 699 
 
 709 
 
 719 
 
 729 
 
 739 
 
 10 
 
 434 
 
 749 
 
 759 
 
 769 
 
 779 
 
 789 
 
 799 
 
 809 
 
 819 
 
 829 
 
 839 
 
 10 
 
 435 
 
 849 
 
 859 
 
 869 
 
 879 
 
 889 
 
 899 
 
 909 
 
 919 
 
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 939 
 
 10 
 
 436 
 
 949 
 
 959 
 
 969 
 
 979 
 
 988 
 
 998 
 
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 10 
 
 437 
 
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 058 
 
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 108 
 
 118 
 
 128 
 
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 10 
 
 438 
 
 H7 
 
 157 
 
 167 
 
 177 
 
 187 
 
 197 
 
 207 
 
 217 
 
 227 
 
 237 
 
 10 
 
 439 
 
 246 
 
 256 
 
 266 
 
 276 
 
 286 
 
 296 
 
 306 
 
 3i6 
 
 326 
 
 335 
 
 10 
 
 440 
 
 345 
 
 355 
 
 365 
 
 375 
 
 3 ? 5 
 
 395 
 
 404 
 
 414 
 
 424 
 
 434 
 
 10 
 
 441 
 
 444 
 
 454 
 
 464 
 
 473 
 
 483 
 
 493 
 
 5°3 
 
 5*3 
 
 523 
 
 532 
 
 10 
 
 442 
 
 542 
 
 552 
 
 562 
 
 572 
 
 582 
 
 59i 
 
 601 
 
 611 
 
 621 
 
 631 
 
 10 
 
 443 
 
 640 
 
 650 
 
 660 
 
 670 
 
 680 
 
 689 
 
 699 
 
 709 
 
 719 
 
 729 
 
 10 
 
 444 
 
 738 
 
 748 
 
 758 
 
 768 
 
 777 
 
 787 
 
 797 
 
 807 
 
 816 
 
 826 
 
 10 
 
 445 
 
 836 
 
 846 
 
 856 
 
 865 
 
 875 
 
 885 
 
 895 
 
 904 
 
 914 
 
 924 
 
 10 
 
 446 
 
 933 
 
 943 
 
 953 
 
 963 
 
 972 
 
 982 
 
 992 
 
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 10 
 
 447 
 
 65031 
 
 040 
 
 050 
 
 060 
 
 070 
 
 079 
 
 089 
 
 O99 
 
 108 
 
 Il8 
 
 10 
 
 443 
 
 128 
 
 137 
 
 147 
 
 157 
 
 167 
 
 176 
 
 186 
 
 I96 
 
 205 
 
 215 
 
 10 
 
 449 
 
 225 
 
 234 
 
 244 
 
 254 
 
 263 
 
 273 
 
 283 
 
 292 
 
 302 
 
 312 
 
 10 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
450—500 
 
 397 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 45° 
 
 65321 
 
 331 
 
 341 
 
 35° 
 
 360 
 
 369 
 
 379 
 
 389 
 
 398 
 
 408 
 
 10 
 
 45i 
 
 418 
 
 427 
 
 437 
 
 447 
 
 45 6 
 
 466 
 
 475 
 
 485 
 
 495 
 
 5°4 
 
 10 
 
 452 
 
 5H 
 
 523 
 
 533 
 
 543 
 
 552 
 
 562 
 
 57i 
 
 581 
 
 59i 
 
 600 
 
 10 
 
 453 
 
 610 
 
 619 
 
 629 
 
 639 
 
 648 
 
 658 
 
 667 
 
 677 
 
 686 
 
 696 
 
 10 
 
 454 
 
 706 
 
 715 
 
 725 
 
 734 
 
 744 
 
 753 
 
 763 
 
 772 
 
 782 
 
 792 
 
 9 
 
 455 
 
 801 
 
 811 
 
 820 
 
 830 
 
 839 
 
 849 
 
 858 
 
 868 
 
 877 
 
 887 
 
 9 
 
 45 6 
 
 896 
 
 906 
 
 916 
 
 925 
 
 935 
 
 944 
 
 954 
 
 963 
 
 973 
 
 982 
 
 9 
 
 457 
 
 992 
 
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 9 
 
 458 
 
 66087 
 
 096 
 
 106 
 
 "5 
 
 124 
 
 134 
 
 143 
 
 '53 
 
 162 
 
 172 
 
 9 
 
 459 
 
 181 
 
 191 
 
 200 
 
 210 
 
 219 
 
 229 
 
 238 
 
 247 
 
 257 
 
 266 
 
 9 
 
 460 
 
 276 
 
 285 
 
 295 
 
 304 
 
 3H 
 
 323 
 
 332 
 
 342 
 
 35i 
 
 361 
 
 9 
 
 461 
 
 370 
 
 380 
 
 389 
 
 398 
 
 408 
 
 417 
 
 427 
 
 436 
 
 445 
 
 455 
 
 9 
 
 462 
 
 464 
 
 474 
 
 483 
 
 492 
 
 502 
 
 5" 
 
 52i 
 
 53o 
 
 539 
 
 549 
 
 9 
 
 463 
 
 558 
 
 567 
 
 577 
 
 586 
 
 596 
 
 605 
 
 614 
 
 624 
 
 (>33 
 
 642 
 
 9 
 
 464 
 
 652 
 
 661 
 
 671 
 
 680 
 
 689 
 
 699 
 
 708 
 
 717 
 
 727 
 
 736 
 
 9 
 
 465 
 
 745 
 839 
 
 755 
 
 764 
 
 773 
 
 783 
 
 ll 2 
 
 801 
 
 811 
 
 820 
 
 829 
 
 9 
 
 466 
 
 848 
 
 857 
 
 867 
 
 876 
 
 885 
 
 894 
 
 904 
 
 913 
 
 922 
 
 9 
 
 467 
 
 932 
 
 941 
 
 95° 
 
 960 
 
 969 
 
 978 
 
 987 
 
 997 
 
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 9 
 
 468 
 
 67025 
 
 034 
 
 043 
 
 052 
 
 062 
 
 071 
 
 080 
 
 089 
 
 099 
 
 108 
 
 9 
 
 469 
 
 117 
 
 127 
 
 136 
 
 145 
 
 154 
 
 164 
 
 173 
 
 182 
 
 191 
 
 201 
 
 9 
 
 470 
 
 210 
 
 219 
 
 228 
 
 237 
 
 247 
 
 256 
 
 265 
 
 274 
 
 284 
 
 293 
 
 9 
 
 47i 
 
 302 
 
 3" 
 
 321 
 
 33o 
 
 339 
 
 348 
 
 357 
 
 367 
 
 376 
 
 385 
 
 9 
 
 472 
 
 394 
 
 403 
 
 413 
 
 422 
 
 43i 
 
 440 
 
 449 
 
 459 
 
 468 
 
 477 
 
 9 
 
 473 
 
 486 
 
 495 
 
 5°4 
 
 5H 
 
 523 
 
 532 
 
 54i 
 
 55o 
 
 560 
 
 5 6 9 
 
 9 
 
 474 
 
 578 
 
 587 
 
 596 
 
 605 
 
 614 
 
 624 
 
 *>33 
 
 642 
 
 651 
 
 660 
 
 9 
 
 475 
 
 669 
 
 679 
 
 688 
 
 697 
 
 706 
 
 806 
 
 724 
 
 733 
 
 742 
 
 752 
 
 9 
 
 476 
 
 761 
 
 770 
 
 779 
 
 788 
 
 797 
 
 815 
 
 825 
 
 834 
 
 843 
 
 9 
 
 477 
 
 852 
 
 861 
 
 870 
 
 879 
 
 888 
 
 897 
 
 906 
 
 916 
 
 925 
 
 934 
 
 9 
 
 478 
 
 943 
 
 952 
 
 961 
 
 970 
 
 979 
 
 988 
 
 "2 
 
 *oo6 
 
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 ♦024 
 
 9 
 
 479 
 
 68034 
 
 043 
 
 052 
 
 061 
 
 070 
 
 079 
 
 088 
 
 097 
 
 106 
 
 "5 
 
 9 
 
 480 
 
 124 
 
 m 
 
 142 
 
 '5* 
 
 160 
 
 169 
 
 178 
 
 187 
 
 196 
 
 205 
 
 9 
 
 481 
 
 215 
 
 224 
 
 233 
 
 242 
 
 251 
 
 260 
 
 269 
 
 278 
 
 287 
 
 296 
 
 9 
 
 482 
 
 305 
 
 3H 
 
 323 
 
 332 
 
 34i 
 
 35° 
 
 359 
 
 368 
 
 377 
 
 386 
 
 9 
 
 483 
 
 395 
 
 404 
 
 413 
 
 422 
 
 43i 
 
 440 
 
 449 
 
 458 
 
 467 
 
 476 
 
 9 
 
 484 
 
 485 
 
 494 
 
 502 
 
 5" 
 
 520 
 
 529 
 
 538 
 
 547 
 
 556 
 
 565 
 
 9 
 
 4$ ai 
 
 574 
 
 583 
 
 592 
 
 601 
 
 610 
 
 619 
 
 628 
 
 637 
 
 646 
 
 655 
 
 9 
 
 486 
 
 664 
 
 673 
 
 681 
 
 690 
 
 699 
 
 708 
 
 717 
 
 726 
 
 735 
 
 744 
 
 9 
 
 487 
 488 
 
 753 
 
 762 
 
 771 
 
 780 
 
 789 
 
 797 
 
 806 
 
 815 
 
 824 
 
 833 
 
 9 
 
 842 
 
 851 
 
 860 
 
 869 
 
 878 
 
 886 
 
 895 
 
 904 
 
 9i3 
 
 922 
 
 9 
 
 489 
 
 93i 
 
 940 
 
 949 
 
 958 
 
 966 
 
 975 
 
 984 
 
 993 
 
 *002 
 
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 9 
 
 490 
 
 69020 
 
 028 
 
 037 
 
 046 
 
 °55 
 
 064 
 
 073 
 
 082 
 
 O9O 
 
 099 
 
 9 
 
 491 
 
 108 
 
 117 
 
 126 
 
 *35 
 
 144 
 
 152 
 
 161 
 
 170 
 
 179 
 
 188 
 
 9 
 
 492 
 
 197 
 
 205 
 
 214 
 
 223 
 
 232 
 
 241 
 
 249 
 
 258 
 
 267 
 
 276 
 
 9 
 
 493 
 
 285 
 
 294 
 
 302 
 
 3" 
 
 320 
 
 329 
 
 338 
 
 346 
 
 355 
 
 364 
 
 9 
 
 494 
 
 373 
 
 381 
 
 39o 
 
 399 
 
 408 
 
 417 
 
 425 
 
 434 
 
 443 
 
 452 
 
 9 
 
 495 
 
 461 
 
 469 
 
 478 
 
 487 
 
 496 
 
 5°4 
 
 5*3 
 
 522 
 
 53i 
 
 539 
 
 9 
 
 496 
 
 548 
 
 557 
 
 566 
 
 574 
 
 583 
 
 592 
 
 601 
 
 609 
 
 618 
 
 627 
 
 9 
 
 497 
 
 636 
 
 644 
 
 653 
 
 662 
 
 671 
 
 679 
 
 688 
 
 697 
 
 705 
 
 is 
 
 9 
 
 498 
 
 723 
 810 
 
 732 
 
 740 
 
 749 
 836 
 
 758 
 
 767 
 
 775 
 
 # 
 
 793 
 880 
 
 9 
 
 499 
 
 819 
 
 827 
 
 845 
 
 854 
 
 862 
 
 888 
 
 9 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
398 
 
 500—550 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 500 
 
 69897 
 
 906 J 914 
 
 923 
 
 932 
 
 940 
 
 949 
 
 958 
 
 966 
 
 975 
 
 9 
 
 5°i 
 
 984 
 
 992 
 
 *OOI 
 
 *OIO 
 
 *oi8 
 
 ♦027 
 
 *o 3 6 
 
 ♦044 
 
 *o 53 
 
 ♦062 
 
 9 
 
 502 
 
 70070 
 
 079 
 
 088 
 
 096 
 
 io 5 
 
 114 
 
 122 
 
 131 
 
 140 
 
 148 
 
 9 
 
 5°3 
 
 157 
 
 165 
 
 174 
 
 183 
 
 191 
 
 200 
 
 209 
 
 217 
 
 226 
 
 234 
 
 9 
 
 5°4 
 
 243 
 
 252 
 
 260 
 
 269 
 
 278 
 
 286 
 
 295 
 
 303 
 
 312 
 
 321 
 
 9 
 
 505 
 
 329 
 
 338 
 
 346 
 
 355 
 
 364 
 
 372 
 
 38i 
 
 389 
 
 398 
 
 406 
 
 9 
 
 506 
 
 415 
 
 424 
 
 432 
 
 441 
 
 449 
 
 458 
 
 467 
 
 475 
 
 484 
 
 492 
 
 9 
 
 507 
 
 508 
 
 5 o5 
 
 509 
 
 518 
 
 526 
 
 535 
 
 544 
 
 552 
 
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 578 
 
 9 
 
 586 
 
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 603 
 
 612 
 
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 629 
 
 638 
 
 646 
 
 655 
 
 663 
 
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 672 
 
 680 
 
 689 
 
 697 
 
 706 
 
 714 
 
 723 
 
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 740 
 
 749 
 
 9 
 
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 757 
 
 766 
 
 774 
 
 783 
 
 791 
 
 800 
 
 808 
 
 817 
 
 825 
 
 834 
 
 9 
 
 5" 
 
 842 
 
 851 
 
 859 
 
 868 
 
 876 
 
 885 
 
 893 
 
 902 
 
 910 
 
 919 
 
 9 
 
 5*2 
 
 927 
 
 935 
 
 944 
 
 952 
 
 961 
 
 969 
 
 978 
 
 986 
 
 995 
 
 ♦003 
 
 9 
 
 5i3 
 
 71012 
 
 020 
 
 029 
 
 037 
 
 046 
 
 054 
 
 063 
 
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 079 
 
 088 
 
 8 
 
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 096 
 
 105 
 
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 122 
 
 130 
 
 139 
 
 147 
 
 155 
 
 164 
 
 172 
 
 8 
 
 5I I 
 
 181 
 
 189 
 
 198 
 
 206 
 
 214 
 
 223 
 
 231 
 
 240 
 
 248 
 
 257 
 
 8 
 
 516 
 
 265 
 
 273 
 
 282 
 
 290 
 
 299 
 
 307 
 
 315 
 
 324 
 
 332 
 
 341 
 
 8 
 
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 349 
 
 357 
 
 366 
 
 374 
 
 3^3 
 
 391 
 
 399 
 
 408 
 
 416 
 
 425 
 
 8 
 
 433 
 
 441 
 
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 458 
 
 466 
 
 475 
 
 483 
 
 492 
 
 500 
 
 508 
 
 8 
 
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 525 
 
 533 
 
 542 
 
 55° 
 
 559 
 
 567 
 
 575 
 
 584 
 
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 8 
 
 520 
 
 600 
 
 609 
 
 617 
 
 625 
 
 634 
 
 642 
 
 650 
 
 659 
 
 667 
 
 675 
 
 8 
 
 521 
 
 684 
 
 692 
 
 700 
 
 709 
 
 717 
 
 725 
 
 734 
 
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 75° 
 
 759 
 
 8 
 
 522 
 
 767 
 
 775 
 
 784 
 
 792 
 
 800 
 
 809 
 
 817 
 
 825 
 
 834 
 
 842 
 
 8 
 
 5 2 3 
 
 850 
 
 858 
 
 867 
 
 875 
 
 883 
 
 892 
 
 900 
 
 908 
 
 917 
 
 925 
 
 8 
 
 524 
 
 933 
 
 941 
 
 95° 
 
 958 
 
 966 
 
 975 
 
 983 
 
 991 
 
 999 
 
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 8 
 
 525 
 
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 024 
 
 032 
 
 041 
 
 049 
 
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 066 
 
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 082 
 
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 8 
 
 526 
 
 099 
 
 107 
 
 "5 
 
 123 
 
 132 
 
 140 
 
 148 
 
 156 
 
 165 
 
 173 
 
 8 
 
 527 
 
 181 
 
 189 
 
 198 
 
 206 
 
 214 
 
 222 
 
 230 
 
 239 
 
 247 
 
 255 
 
 8 
 
 528 
 
 263 
 
 272 
 
 280 
 
 288 
 
 296 
 
 3°4 
 
 313 
 
 321 
 
 329 
 
 337 
 
 8 
 
 529 
 
 346 
 
 354 
 
 362 
 
 370 
 
 378 
 
 3S7 
 
 395 
 
 403 
 
 411 
 
 419 
 
 8 
 
 53o 
 
 428 
 
 436 
 
 444 
 
 452 
 
 460 
 
 469 
 
 477 
 
 485 
 
 493 
 
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 8 
 
 53i 
 
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 518 
 
 526 
 
 534 
 
 542 
 
 55° 
 
 558 
 
 567 
 
 575 
 
 583 
 
 8 
 
 532 
 
 59i 
 
 599 
 
 607 
 
 616 
 
 624 
 
 632 
 
 640 
 
 648 
 
 656 
 
 665 
 
 8 
 
 533 
 
 673 
 
 681 
 
 689 
 
 697 
 
 705 
 
 7i3 
 
 722 
 
 73o 
 
 738 
 
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 8 
 
 534 
 
 754 
 
 762 
 
 770 
 
 779 
 
 787 
 
 795 
 
 803 
 
 811 
 
 819 
 
 827 
 
 8 
 
 535 
 
 835 
 
 843 
 
 852 
 
 860 
 
 868 
 
 876 
 
 884 
 
 892 
 
 900 
 
 908 
 
 8 
 
 536 
 
 916 
 
 925 
 
 933 
 
 941 
 
 949 
 
 957 
 
 965 
 
 973 
 
 981 
 
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 537 
 
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 538 
 
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 in 
 
 119 
 
 127 
 
 135 
 
 143 
 
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 8 
 
 539 
 
 159 
 
 167 
 
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 183 
 
 191 
 
 199 
 
 207 
 
 215 
 
 223 
 
 231 
 
 8 
 
 540 
 
 239 
 
 247 
 
 255 
 
 263 
 
 272 
 
 280 
 
 288 
 
 296 
 
 304 
 
 312 
 
 8 
 
 54i 
 
 320 
 
 328 
 
 336 
 
 344 
 
 352 
 
 360 
 
 368 
 
 376 
 
 384 
 
 392 
 
 8 
 
 542 
 
 400 
 
 408 
 
 416 
 
 424 
 
 432 
 
 440 
 
 448 
 
 45 6 
 
 464 
 
 472 
 
 8 
 
 543 
 
 480 
 
 488 
 
 496 
 
 5°4 
 
 5 12 
 
 520 
 
 528 
 
 536 
 
 544 
 
 552 
 
 8 
 
 544 
 
 560 
 
 568 
 
 576 
 
 584 
 
 592 
 
 600 
 
 608 
 
 616 
 
 624 
 
 632 
 
 8 
 
 545 
 
 640 
 
 648 
 
 656 
 
 664 
 
 672 
 
 679 
 
 687 
 
 695 
 
 703 
 
 711 
 
 8 
 
 546 
 
 719 
 
 727 
 
 735 
 
 743 
 
 75i 
 
 759 
 
 767 
 
 775 
 
 7^3 
 
 791 
 
 8 
 
 547 
 
 799 
 
 807 
 
 815 
 
 823 
 
 830 
 
 838 
 
 846 
 
 854 
 
 862 
 
 870 
 
 8 
 
 548 
 
 878 
 
 886 
 
 894 
 
 902 
 
 910 
 
 918 
 
 926 
 
 933 
 
 941 
 
 949 
 
 8 
 
 549 
 
 957 
 
 965 
 
 973 
 
 981 
 
 989 
 
 997 
 
 ♦005 
 
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 8 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
550 — 600 
 
 399 
 
 1 ■ " 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 55° 
 
 74036 
 
 044 
 
 052 
 
 060 
 
 068 
 
 076 
 
 084 
 
 092 
 
 099 
 
 107 
 
 8 
 
 551 
 
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 123 
 
 131 
 
 139 
 
 147 
 
 155 
 
 162 
 
 170 
 
 178 
 
 186 
 
 8 
 
 55 2 
 
 194 
 
 202 
 
 210 
 
 218 
 
 225 
 
 233 
 
 241 
 
 249 
 
 257 
 
 265 
 
 8 
 
 553 
 
 273 
 
 280 
 
 288 
 
 296 
 
 304 
 
 312 
 
 320 
 
 327 
 
 335 
 
 343 
 
 8 
 
 554 
 
 35i 
 
 359 
 
 367 
 
 374 
 
 382 
 
 390 
 
 398 
 
 406 
 
 414 
 
 421 
 
 8 
 
 55 l 
 
 429 
 
 437 
 
 445 
 
 453 
 
 461 
 
 468 
 
 476 
 
 484 
 
 492 
 
 500 
 
 8 
 
 556 
 
 5°7 
 
 5 J 5 
 
 523 
 
 53i 
 
 539 
 
 547 
 
 554 
 
 562 
 
 57o 
 
 578 
 
 8 
 
 557 
 
 586 
 
 593 
 
 601 
 
 609 
 
 617 
 
 624 
 
 632 
 
 640 
 
 648 
 
 656 
 
 8 
 
 558 
 
 663 
 
 671 
 
 679 
 
 687 
 
 695 
 
 702 
 
 710 
 
 718 
 
 726 
 
 733 
 
 8 
 
 559 
 
 74i 
 
 749 
 
 757 
 
 764 
 
 772 
 
 780 
 
 788 
 
 796 
 
 803 
 
 811 
 
 8 
 
 560 
 
 819 
 
 827 
 
 834 
 
 842 
 
 850 
 
 858 
 
 865 
 
 873 
 
 881 
 
 889 
 
 8 
 
 561 
 
 896 
 
 904 
 
 912 
 
 920 
 
 927 
 
 935 
 
 943 
 
 95° 
 
 958 
 
 966 
 
 8 
 
 562 
 
 974 
 
 981 
 
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 563 
 
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 059 
 
 066 
 
 074 
 
 082 
 
 089 
 
 O97 
 
 i°5 
 
 "3 
 
 120 
 
 8 
 
 564 
 
 128 
 
 136 
 
 H3 
 
 151 
 
 159 
 
 166 
 
 174 
 
 . 182 
 
 189 
 
 197 
 
 8 
 
 s6 A 
 
 205 
 
 213 
 
 220 
 
 228 
 
 236 
 
 243 
 
 25 o 
 
 259 
 
 266 
 
 274 
 
 8 
 
 566 
 
 282 
 
 289 
 
 297 
 
 305 
 
 312 
 
 320 
 
 328 
 
 335 
 
 343 
 
 35i 
 
 8 
 
 567 
 
 358 
 
 366 
 
 374 
 
 38i 
 
 389 
 
 397 
 
 404 
 
 412 
 
 420 
 
 427 
 
 8 
 
 568 
 
 435 
 
 442 
 
 45° 
 
 458 
 
 465 
 
 473 
 
 481 
 
 488 
 
 496 
 
 5°4 
 
 8 
 
 5 6 9 
 
 5" 
 
 519 
 
 526 
 
 534 
 
 542 
 
 549 
 
 557 
 
 565 
 
 572 
 
 580 
 
 8 
 
 57o 
 
 587 
 
 595 
 
 603 
 
 610 
 
 618 
 
 626 
 
 633 
 
 641 
 
 648 
 
 656 
 
 8 
 
 57i 
 
 664 
 
 671 
 
 679 
 
 686 
 
 694 
 
 702 
 
 709 
 
 717 
 
 724 
 
 732 
 
 8 
 
 572 
 
 740 
 
 747 
 
 755 
 
 762 
 
 770 
 
 778 
 
 785 
 
 793 
 
 800 
 
 808 
 
 8 
 
 573 
 
 l 1S 
 
 823 
 
 S3* 
 
 838 
 
 846 
 
 853 
 
 861 
 
 868 
 
 876 
 
 884 
 
 8 
 
 574 
 
 891 
 
 899 
 
 906 
 
 914 
 
 921 
 
 929 
 
 937 
 
 944 
 
 952 
 
 959 
 
 8 
 
 575 
 
 967 
 
 974 
 
 982 
 
 989 
 
 997 
 
 ♦005 
 
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 8 
 
 576 
 
 76042 
 
 050 
 
 o57 
 
 065 
 
 072 
 
 080 
 
 087 
 
 095 
 
 103 
 
 no 
 
 8 
 
 577 
 
 118 
 
 I2 5 
 
 i33 
 
 140 
 
 148 
 
 155 
 
 163 
 
 I70 
 
 178 
 
 185 
 
 8 
 
 578 
 
 193 
 
 200 
 
 208 
 
 215 
 
 223 
 
 230 
 
 238 
 
 245 
 
 253 
 
 260 
 
 8 
 
 579 
 
 268 
 
 275 
 
 283 
 
 290 
 
 298 
 
 305 
 
 313 
 
 320 
 
 328 
 
 335 
 
 8 
 
 580 
 
 343 
 
 35o 
 
 358 
 
 365 
 
 373 
 
 380 
 
 388 
 
 395 
 
 403 
 
 410 
 
 8 
 
 5 ?' 
 
 418 
 
 425 
 
 433 
 
 440 
 
 448 
 
 455 
 
 462 
 
 470 
 
 477 
 
 485 
 
 7 
 
 582 
 
 492 
 
 500 
 
 507 
 
 5 o 5 
 
 522 
 
 53o 
 
 537 
 
 545 
 
 552 
 
 559 
 
 7 
 
 583 
 
 567 
 
 574 
 
 582 
 
 589 
 
 597 
 
 604 
 
 612 
 
 619 
 
 626 
 
 634 
 
 7 
 
 584 
 
 641 
 
 649 
 
 656 
 
 664 
 
 671 
 
 678 
 
 686 
 
 693 
 
 701 
 
 708 
 
 7 
 
 5?5 
 
 716 
 
 723 
 
 73o 
 
 738 
 
 745 
 
 753 
 
 760 
 
 768 
 
 775 
 
 782 
 
 7 
 
 586 
 
 790 
 
 797 
 
 805 
 
 812 
 
 819 
 
 827 
 
 834 
 
 842 
 
 849 
 
 856 
 
 7 
 
 5f7 
 
 864 
 
 871 
 
 879 
 
 886 
 
 893 
 
 901 
 
 908 
 
 916 
 
 923 
 
 930 
 
 7 
 
 588 
 
 938 
 
 945 
 
 953 
 
 960 
 
 967 
 
 048 
 
 982 
 
 989 
 
 997 
 
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 7 
 
 589 
 
 77012 
 
 019 
 
 026 
 
 034 
 
 041 
 
 056 
 
 063 
 
 070 
 
 078 
 
 7 
 
 59o 
 
 085 
 
 093 
 
 100 
 
 107 
 
 *I5 
 
 122 
 
 129 
 
 m 
 
 144 
 
 '5* 
 
 7 
 
 59i 
 
 159 
 
 166 
 
 173 
 
 181 
 
 188 
 
 195 
 
 203 
 
 210 
 
 217 
 
 225 
 
 7 
 
 592 
 
 232 
 
 240 
 
 247 
 
 254 
 
 262 
 
 269 
 
 276 
 
 283 
 
 291 
 
 298 
 
 7 
 
 593 
 
 305 
 
 313 
 
 320 
 
 327 
 
 335 
 
 342 
 
 349 
 
 357 
 
 364 
 
 31i 
 
 7 
 
 594 
 
 379 
 
 386 
 
 393 
 
 401 
 
 408 
 
 4i5 
 
 422 
 
 430 
 
 437 
 
 444 
 
 7 
 
 595 
 
 452 
 
 459 
 
 466 
 
 474 
 
 481 
 
 488 
 
 495 
 
 5°3 
 
 5 i° 
 
 517 
 
 7 
 
 596 
 
 525 
 
 532 
 
 539 
 
 546 
 
 554 
 
 561 
 
 568 
 
 576 
 
 583 
 
 590 
 
 7 
 
 597 
 
 597 
 
 605 
 
 612 
 
 619 
 
 627 
 
 634 
 
 641 
 
 648 
 
 656 
 
 663 
 
 7 
 
 598 
 
 670 
 
 677 
 
 685 
 
 692 
 
 699 
 
 706 
 
 7 ol 
 
 721 
 
 728 
 
 l 3 i 
 
 7 
 
 599 
 
 743 
 
 75° 
 
 757 
 
 764 
 
 772 
 
 779 
 
 786 
 
 793 
 
 801 
 
 808 
 
 7 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
400 
 
 600—650 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 ! 4 
 
 5 ! 6 
 
 7 8 9 
 
 D. 
 
 600 
 
 77 f£ 5 
 
 822 
 
 830 
 
 837 
 
 844 
 
 851 
 
 859 
 
 866 
 
 873 
 
 880 
 
 7 
 
 601 
 
 887 
 
 895 
 
 902 
 
 909 
 
 916 
 
 924 
 
 93i 
 
 938 
 
 945 
 
 952 
 
 7 
 
 602 
 
 960 
 
 967 
 
 974 
 
 981 988 
 
 996 
 
 *oo3 
 
 *OIO 
 
 ♦017 
 
 ♦025 
 
 7 
 
 603 
 
 78032 
 
 039 
 
 046 
 
 053 
 
 061 
 
 068 
 
 o75 
 
 082 
 
 089 
 
 097 
 
 7 
 
 604 
 
 104 
 
 in 
 
 118 
 
 125 
 
 132 
 
 140 
 
 147 
 
 154 
 
 161 
 
 168 
 
 7 
 
 605 
 
 176 
 
 183 
 
 190 
 
 197 
 
 204 
 
 211 
 
 219 
 
 226 
 
 233 
 
 240 
 
 7 
 
 606 
 
 247 
 
 254 
 
 262 
 
 269 
 
 276 
 
 283 
 
 290 
 
 297 
 
 305 
 
 312 
 
 7 
 
 607 
 
 319 
 
 326 
 
 333 
 
 340 
 
 347 
 
 355 
 
 362 
 
 369 
 
 376 
 
 383 
 
 7 
 
 608 
 
 390 
 
 398 
 
 405 
 
 412 419 
 
 426 
 
 433 
 
 440 
 
 447 
 
 455 
 
 7 
 
 609 
 
 462 
 
 469 
 
 476 
 
 483 
 
 490 
 
 497 
 
 5°4 
 
 512 
 
 519 
 
 526 
 
 7 
 
 610 
 
 533 
 
 540 
 
 547 
 
 554 
 
 561 
 
 569 
 
 576 
 
 583 
 
 590 
 
 597 
 
 7 
 
 611 
 
 604 
 
 611 
 
 618 
 
 625 
 
 633 
 
 640 
 
 647 
 
 654 
 
 661 
 
 668 
 
 7 
 
 612 
 
 675 
 
 682 
 
 689 
 
 696 
 
 704 
 
 711 
 
 718 
 
 725 
 
 732 
 
 739 
 
 7 
 
 613 
 
 746 
 
 753 
 
 760 
 
 767 
 
 774 
 
 78i 
 
 789 
 
 796 
 
 803 
 
 810 
 
 7 
 
 614 
 
 817 
 
 824 
 
 831 
 
 838 
 
 845 
 
 852 
 
 859 
 
 866 
 
 ^73 
 
 880 
 
 7 
 
 615 
 
 888 
 
 895 
 
 902 
 
 909 
 
 916 
 
 923 
 
 930 
 
 937 
 
 944 
 
 951 
 
 7 
 
 616 
 
 958 
 
 965 
 
 972 
 
 979 
 
 986 
 
 993 
 
 *ooo 
 
 ♦007 
 
 ♦014 
 
 *02I 
 
 7 
 
 617 
 
 79029 
 
 036 
 
 043 
 
 050 
 
 o57 
 
 064 
 
 071 
 
 078 
 
 085 
 
 O92 
 
 7 
 
 618 
 
 099 
 
 106 
 
 "3 
 
 120 
 
 127 
 
 134 
 
 141 
 
 148 
 
 155 
 
 162 
 
 7 
 
 619 
 
 169 
 
 176 
 
 183 
 
 190 
 
 197 
 
 204 
 
 211 
 
 218 
 
 225 
 
 232 
 
 7 
 
 620 
 
 239 
 
 246 
 
 253 
 
 260 
 
 267 
 
 274 
 
 28l 
 
 288 
 
 295 
 
 302 
 
 7 
 
 621 
 
 309 
 
 316 
 
 323 
 
 330 
 
 337 
 
 344 
 
 351 
 
 358 
 
 365 
 
 372 
 
 7 
 
 622 
 
 379 
 
 386 
 
 393 
 
 400 
 
 407 
 
 414 
 
 421 
 
 428 
 
 435 
 
 442 
 
 7 
 
 623 
 
 449 
 
 45 6 
 
 463 
 
 470 
 
 477 
 
 484 
 
 491 
 
 498 
 
 5°5 
 
 5 11 
 
 7 
 
 624 
 
 518 
 
 525 
 
 532 
 
 539 
 
 546 
 
 553 
 
 560 
 
 567 
 
 574 
 
 581 
 
 7 
 
 625 
 
 588 
 
 595 
 
 602 
 
 609 
 
 616 
 
 623 
 
 63O 
 
 637 
 
 644 
 
 650 
 
 7 
 
 626 
 
 657 
 
 664 
 
 671 
 
 678 
 
 685 
 
 692 
 
 699 
 
 706 
 
 7i3 
 
 720 
 
 7 
 
 627 
 
 727 
 
 734 
 
 74i 
 
 748 
 
 754 
 
 761 
 
 768 
 
 775 
 
 782 
 
 789 
 
 7 
 
 628 
 
 796 
 
 803 
 
 810 | 817 
 
 824 
 
 831 
 
 837 
 
 844 
 
 851 
 
 858 
 
 7 
 
 629 
 
 865 
 
 872 
 
 879 
 
 886 
 
 893 
 
 900 
 
 906 
 
 913 
 
 920 
 
 927 
 
 7 
 
 630 
 
 934 
 
 941 
 
 948 
 
 955 
 
 962 
 
 969 
 
 975 
 
 982 
 
 989 
 
 996 
 
 7 
 
 631 
 
 80003 
 
 010 
 
 017 
 
 024 
 
 030 
 
 o37 
 
 044 
 
 051 
 
 058 
 
 065 
 
 7 
 
 632 
 
 072 
 
 079 
 
 085 
 
 092 
 
 099 
 
 106 
 
 "3 
 
 • 120 
 
 127 
 
 i34 
 
 7 
 
 633 
 
 140 
 
 *47 
 
 154 
 
 161 
 
 168 
 
 «75 
 
 182 
 
 188 
 
 195 
 
 202 
 
 7 
 
 634 
 
 209 
 
 216 
 
 223 
 
 229 
 
 236 
 
 243 
 
 250 
 
 257 
 
 264 
 
 271 
 
 7 
 
 635 
 
 277 
 
 284 
 
 291 
 
 298 
 
 305 
 
 312 
 
 3i8 
 
 325 
 
 332 
 
 339 
 
 7 
 
 636 
 
 346 
 
 353 
 
 359 
 
 366 
 
 373 
 
 380 
 
 387 
 
 393 
 
 400 
 
 407 
 
 7 
 
 637 
 
 414 
 
 421 
 
 428 
 
 434 
 
 441 
 
 448 
 
 455 
 
 462 
 
 468 
 
 475 
 
 7 
 
 638 
 
 482 
 
 489 
 
 496 
 
 502 
 
 5°9 
 
 516 
 
 523 
 
 530 
 
 536 
 
 543 
 
 7 
 
 639 
 
 550 
 
 557 
 
 5 6 4 
 
 570 
 
 577 
 
 584 
 
 59i 
 
 598 
 
 604 
 
 6u 
 
 7 \ 
 
 640 
 
 618 
 
 625 
 
 632 
 
 638 
 
 645 
 
 652 
 
 659 
 
 665 
 
 672 
 
 679 
 
 7 j 
 
 641 
 
 686 
 
 693 
 
 699 
 
 706 
 
 7i3 
 
 720 
 
 726 
 
 733 
 
 740 
 
 747 
 
 7 ! 
 
 642 
 
 821 
 
 760 
 
 767 
 
 774 
 
 781 
 
 787 
 
 794 
 
 801 
 
 808 
 
 814 
 
 7 
 
 643 
 
 828 
 
 835 
 
 841 
 
 848 
 
 855 
 
 862 
 
 868 
 
 875 
 
 882 
 
 7 1 
 
 644 
 
 889 
 
 895 
 
 902 
 
 909 
 
 916 
 
 922 
 
 929 
 
 936 
 
 943 
 
 949 
 
 7 | 
 
 645 
 
 956 
 
 963 
 
 969 
 
 976 
 
 983 
 
 990 
 
 996 
 
 ♦003 
 
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 *oi7 
 
 7 1 
 
 646 
 
 81023 
 
 030 
 
 037 
 
 043 
 
 050 
 
 o57 
 
 064 
 
 070 
 
 077 
 
 084 
 
 7 1 
 
 647 
 
 090 
 
 097 
 
 104 
 
 in 
 
 117 
 
 124 
 
 131 
 
 137 
 
 144 
 
 * S l 
 
 7 j 
 
 648 
 
 158 
 
 164 
 
 171 
 
 178 
 
 184 
 
 191 
 
 198 
 
 204 
 
 211 
 
 218 
 
 7 : 
 
 649 
 
 224 
 
 231 
 
 238 
 
 245 
 
 251 
 
 258 
 
 265 
 
 271 278 
 
 285 
 
 7 ; 
 
 N. 
 
 
 
 1 
 
 2 3 
 
 4 
 
 5 
 
 6 
 
 7 8 
 
 9 
 
 d - ; 
 
650—700 
 
 401 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 650 
 
 81291 
 
 298 
 
 305 
 
 3" 
 
 3i8 
 
 325 
 
 331 
 
 338 
 
 345 
 
 35 J 
 
 
 651 
 
 358 
 
 365 
 
 37i 
 
 378 
 
 385 
 
 391 
 
 398 
 
 405 
 
 411 
 
 418 
 
 
 652 
 
 425 
 
 43i 
 
 438 
 
 445 
 
 45 o 
 
 458 
 
 465 
 
 47i 
 
 478 
 
 485 
 
 
 653 
 
 491 
 
 498 
 
 5°5 
 
 5 11 
 
 518 
 
 525 
 
 53 i 
 
 538 
 
 544 
 
 55 1 
 
 
 654 
 
 558 
 
 564 
 
 57i 
 
 578 
 
 584 
 
 591 
 
 598 
 
 604 
 
 611 
 
 617 
 
 
 655 
 
 624 
 
 631 
 
 637 
 
 644 
 
 6 5I 
 
 657 
 
 664 
 
 671 
 
 677 
 
 684 
 
 
 656 
 
 690 
 
 697 
 
 704 
 
 710 
 
 717 
 
 723 
 
 730 
 
 737 
 
 743 
 
 v>°. 
 
 
 657 
 
 757 
 
 763 
 
 770 
 
 776 
 
 783 
 
 790 
 
 796 
 
 803 
 
 809 
 
 816 
 
 
 658 
 
 823 
 
 829 
 
 836 
 
 842 
 
 849 
 
 856 
 
 862 
 
 869 
 
 875 
 
 882 
 
 
 659 
 
 889 
 
 895 
 
 902 
 
 908 
 
 915 
 
 921 
 
 928 
 
 935 
 
 941 
 
 948 
 
 
 660 
 
 954 
 
 961 
 
 968 
 
 974 
 
 981 
 
 987 
 
 994 
 
 *ooo 
 
 *oo7 
 
 ♦014 
 
 
 661 
 
 82020 
 
 027 
 
 033 
 
 040 
 
 046 
 
 053 
 
 060 
 
 066 
 
 073 
 
 079 
 
 
 662 
 
 086 
 
 092 
 
 099 
 
 105 
 
 112 
 
 119 
 
 125 
 
 132 
 
 138 
 
 145 
 
 
 663 
 
 151 
 
 158 
 
 164 
 
 171 
 
 178 
 
 184 
 
 191 
 
 197 
 
 204 
 
 210 
 
 
 664 
 
 217 
 
 223 
 
 230 
 
 236 
 
 243 
 
 249 
 
 256 
 
 263 
 
 269 
 
 276 
 
 
 665 
 
 282 
 
 289 
 
 295 
 
 302 
 
 308 
 
 315 
 
 321 
 
 328 
 
 334 
 
 341 
 
 
 666 
 
 347 
 
 354 
 
 360 
 
 367 
 
 373 
 
 380 
 
 387 
 
 393 
 
 400 
 
 406 
 
 
 66? 
 
 413 
 
 419 
 
 426 
 
 432 
 
 439 
 
 445 
 
 452 
 
 458 
 
 465 
 
 47i 
 
 
 668 
 
 478 
 
 484 
 
 491 
 
 497 
 
 5°4 
 
 5io 
 
 5 o 7 
 
 523 
 
 530 
 
 536 
 
 
 669 
 
 543 
 
 549 
 
 556 
 
 562 
 
 5 6 9 
 
 575 
 
 582 
 
 588 
 
 595 
 
 601 
 
 
 670 
 
 607 
 
 614 
 
 620 
 
 627 
 
 633 
 
 640 
 
 646 
 
 65 § 
 
 659 
 
 666 
 
 7 
 
 671 
 
 672 
 
 679 
 
 685 
 
 692 
 
 698 
 
 7°5 
 
 711 
 
 7 S 8 
 
 724 
 
 730 
 
 6 
 
 672 
 
 737 
 
 743 
 
 75° 
 
 75 6 
 
 763 
 
 769 
 
 776 
 
 782 
 
 789 
 
 l 9 J 
 
 6 
 
 673 
 
 802 
 
 808 
 
 814 
 
 821 
 
 827 
 
 834 
 
 840 
 
 847 
 
 853 
 
 860 
 
 6 
 
 674 
 
 866 
 
 872 
 
 879 
 
 885 
 
 892 
 
 898 
 
 905 
 
 911 
 
 918 
 
 924 
 
 6 
 
 b j$ 
 
 930 
 
 937 
 
 943 
 
 950 
 
 956 
 
 963 
 
 969 
 
 975 
 
 982 
 
 988 
 
 6 
 
 676 
 
 995 
 
 *OOI 
 
 *oo8 
 
 ♦014 
 
 *020 
 
 ♦027 
 
 *©33 
 
 ♦040 
 
 ♦046 
 
 ♦052 
 
 6 
 
 677 
 
 83059 
 
 065 
 
 072 
 
 078 
 
 085 
 
 091 
 
 097 
 
 104 
 
 no 
 
 117 
 
 6 
 
 678 
 
 123 
 
 129 
 
 136 
 
 142 
 
 149 
 
 155 
 
 161 
 
 168 
 
 174 
 
 181 
 
 6 
 
 679 
 
 187 
 
 193 
 
 200 
 
 206 
 
 213 
 
 219 
 
 225 
 
 232 
 
 238 
 
 245 
 
 6 
 
 680 
 
 251 
 
 257 
 
 264 
 
 270 
 
 276 
 
 283 
 
 289 
 
 296 
 
 302 
 
 308 
 
 6 
 
 681 
 
 315 
 
 321 
 
 327 
 
 334 
 
 340 
 
 347 
 
 353 
 
 359 
 
 366 
 
 372 
 
 6 
 
 682 
 
 378 
 
 385 
 
 39i 
 
 398 
 
 404 
 
 410 
 
 417 
 
 423 
 
 429 
 
 436 
 
 6 
 
 683 
 
 442 
 
 448 
 
 455 
 
 461 
 
 467 
 
 474 
 
 480 
 
 487 
 
 493 
 
 499 
 
 6 
 
 684 
 
 506 
 
 512 
 
 518 
 
 525 
 
 531 
 
 537 
 
 544 
 
 55o 
 
 556 
 
 563 
 
 6 
 
 685 
 
 569 
 
 575 
 
 582 
 
 588 
 
 594 
 
 601 
 
 607 
 
 613 
 
 620 
 
 626 
 
 6 
 
 686 
 
 632 
 
 639 
 
 645 
 
 651 
 
 658 
 
 664 
 
 670 
 
 677 
 
 683 
 
 689 
 
 6 
 
 687 
 
 696 
 
 702 
 
 708 
 
 715 
 
 721 
 
 727 
 
 734 
 
 740 
 
 746 
 
 m 
 
 6 
 
 688 
 
 759 
 
 765 
 
 77i 
 
 778 
 
 784 
 
 790 
 
 797 
 
 803 
 
 809 
 
 816 
 
 6 
 
 689 
 
 822 
 
 828 
 
 835 
 
 841 
 
 847 
 
 853 
 
 860 
 
 866 
 
 872 
 
 879 
 
 6 
 
 690 
 
 885 
 
 891 
 
 897 
 
 904 
 
 910 
 
 916 
 
 923 
 
 929 
 
 935 
 
 942 
 
 6 
 
 691 
 
 948 
 
 954 
 
 960 
 
 967 
 
 973 
 
 979 
 
 985 
 
 992 
 
 998 
 
 ♦004 
 
 6 
 
 692 
 
 840 1 1 
 
 017 
 
 023 
 
 029 
 
 036 
 
 042 
 
 048 
 
 o55 
 
 061 
 
 067 
 
 6 
 
 693 
 
 073 
 
 080 
 
 086 
 
 092 
 
 098 
 
 105 
 
 in 
 
 117 
 
 123 
 
 130 
 
 6 
 
 994 
 
 136 
 
 142 
 
 148 
 
 155 
 
 161 
 
 167 
 
 173 
 
 180 
 
 186 
 
 192 
 
 6 
 
 695 
 
 198 
 
 205 
 
 211 
 
 217 
 
 223 
 
 230 
 
 236 
 
 242 
 
 248 
 
 255 
 
 6 
 
 696 
 
 261 
 
 267 
 
 273 
 
 280 
 
 286 
 
 292 
 
 298 
 
 3o5 
 
 3" 
 
 317 
 
 6 
 
 697 
 
 323 
 
 330 
 
 336 
 
 342 
 
 348 
 
 354 
 
 361 
 
 3^7 
 
 373 
 
 379 
 
 6 
 
 698 
 
 386 
 
 392 
 
 398 
 
 404 
 
 410 
 
 417 
 
 423 
 
 429 
 
 435 
 
 442 
 
 6 
 
 699 
 
 448 
 
 454 
 
 460 
 
 466 
 
 473 
 
 479 
 
 485 
 
 49i 
 
 497 
 
 5°4 
 
 6 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
402 
 
 700—750 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 700 
 
 84510 
 
 5l6 o 
 
 522 
 
 528 
 
 535 
 
 541 
 
 547 
 
 553 
 
 559 
 
 566 
 
 6 
 
 701 
 
 572 
 
 578 
 
 584 
 
 590 
 
 597 
 
 603 
 
 609 
 
 615 
 
 621 
 
 628 
 
 6 
 
 702 
 
 634 
 
 640 
 
 646 
 
 652 
 
 658 
 
 665 
 
 671 
 
 677 
 
 683 
 
 689 
 
 6 
 
 703 
 
 696 
 
 702 
 
 708 
 
 714 
 
 720 
 
 726 
 
 733 
 
 739 
 
 745 
 
 75i 
 
 6 
 
 704 
 
 757 
 
 763 
 
 770 
 
 776 
 
 782 
 
 788 
 
 794 
 
 800 
 
 807 
 
 813 
 
 6 
 
 705 
 
 819 
 
 825 
 
 831 
 
 837 
 
 844 
 
 850 
 
 856 
 
 862 
 
 868 
 
 874 
 
 6 
 
 706 
 
 880 
 
 887 
 
 893 
 
 899 
 
 905 
 
 911 
 
 917 
 
 924 
 
 930 
 
 936 
 
 6 
 
 707 
 
 942 
 
 948 
 
 954 
 
 960 
 
 967 
 
 973 
 
 979 
 
 985 
 
 991 
 
 997 
 
 6 
 
 708 
 
 85003 
 
 009 
 
 016 
 
 022 
 
 028 
 
 034 
 
 040 
 
 046 
 
 052 
 
 058 
 
 6 
 
 709 
 
 065 
 
 071 
 
 077 
 
 083 
 
 089 
 
 095 
 
 IOI 
 
 107 
 
 114 
 
 120 
 
 6 
 
 710 
 
 126 
 
 132 
 
 138 
 
 144 
 
 150 
 
 156 
 
 163 
 
 169 
 
 175 
 
 181 
 
 6 
 
 711 
 
 187 
 
 193 
 
 199 
 
 205 
 
 211 
 
 217 
 
 224 
 
 230 
 
 236 
 
 242 
 
 6 
 
 712 
 
 248 
 
 254 
 
 260 
 
 266 
 
 272 
 
 278 
 
 285 
 
 291 
 
 297 
 
 35 ! 
 
 303 
 
 6 
 
 713 
 
 309 
 
 315 
 
 321 
 
 327 
 
 333 
 
 339 
 
 345 
 
 352 
 
 364 
 
 6 
 
 714 
 
 37o 
 
 37° 
 
 382 
 
 388 
 
 394 
 
 400 
 
 406 
 
 412 
 
 418 
 
 425 
 
 6 
 
 715 
 
 43i 
 
 437 
 
 443 
 
 449 
 
 455 
 
 461 
 
 467 
 
 473 
 
 479 
 
 485 
 
 6 
 
 716 
 
 491 
 
 497 
 
 5°3 
 
 509 
 
 516 
 
 522 
 
 528 
 
 534 
 
 540 
 
 546 
 
 6 
 
 717 
 
 552 
 
 558 
 
 5 6 4 
 
 570 
 
 576 
 
 582 
 
 588 
 
 594 
 
 600 
 
 606 
 
 6 
 
 718 
 
 612 
 
 618 
 
 625 
 
 631 
 
 637 
 
 643 
 
 649 
 
 655 
 
 661 
 
 667 
 
 6 
 
 719 
 
 673 
 
 679 
 
 685 
 
 691 
 
 697 
 
 703 
 
 709 
 
 715 
 
 721 
 
 727 
 
 6 
 
 720 
 
 733 
 
 739 
 
 745 
 
 751 
 
 757 
 
 763 
 824 
 
 769 
 
 775 
 
 781 
 
 788 
 
 6 
 
 721 
 
 794 
 
 800 
 
 806 
 
 812 
 
 818 
 
 830 
 
 836 
 
 842 
 
 848 
 
 6 
 
 722 
 
 854 
 
 860 
 
 866 
 
 872 
 
 878 
 
 884 
 
 890 
 
 896 
 
 902 
 
 908 
 
 6 
 
 7 2 3 
 
 914 
 
 920 
 
 926 
 
 932 
 
 938 
 
 944 
 
 95° 
 
 95 6 
 
 962 
 
 968 
 
 6 
 
 724 
 
 974 
 
 980 
 
 986 
 
 992 
 
 998 
 
 ♦004 
 
 *OIO 
 
 *oi6 
 
 *022 
 
 ♦028 
 
 6 
 
 725 
 
 86034 
 
 040 
 
 046 
 
 052 
 
 058 
 
 064 
 
 070 
 
 076 
 
 082 
 
 088 
 
 6 
 
 726 
 
 094 
 
 100 
 
 106 
 
 112 
 
 118 
 
 124 
 
 130 
 
 136 
 
 HI 
 
 H7 
 
 6 
 
 727 
 
 *53 
 
 159 
 
 165 
 
 171 
 
 177 
 
 183 
 
 189 
 
 195 
 
 20I 
 
 207 
 
 6 
 
 728 
 
 213 
 
 219 
 
 225 
 
 231 
 
 237 
 
 243 
 
 249 
 
 255 
 
 26l 
 
 267 
 
 6 
 
 729 
 
 273 
 
 279 
 
 285 
 
 291 
 
 297 
 
 303 
 
 308 
 
 3H 
 
 320 
 
 326 
 
 6 
 
 730 
 
 332 
 
 338 
 
 344 
 
 350 
 
 356 
 
 362 
 
 368 
 
 374 
 
 380 
 
 386 
 
 6 
 
 731 
 
 392 
 
 398 
 
 404 
 
 410 
 
 4i5 
 
 421 
 
 427 
 
 433 
 
 439 
 
 445 
 
 6 
 
 732 
 
 45i 
 
 457 
 
 463 
 
 469 
 
 475 
 
 481 
 
 487 
 
 493 
 
 499 
 
 5°4 
 
 6 
 
 733 
 
 5io 
 
 516 
 
 522 
 
 5 ? 
 
 534 
 
 540 
 
 546 
 
 552 
 
 558 
 
 564 
 
 6 
 
 734 
 
 57o 
 
 576 
 
 581 
 
 587 
 
 593 
 
 599 
 
 605 
 
 611 
 
 617 
 
 623 
 
 6 
 
 735 
 
 629 
 
 635 
 
 641 
 
 646 
 
 652 
 
 658 
 
 664 
 
 670 
 
 676 
 
 682 
 
 6 
 
 736 
 
 688 
 
 694 
 
 700 
 
 705 
 
 711 
 
 717 
 
 723 
 
 7 o2 
 
 735 
 
 741 
 
 6 
 
 737 
 
 747 
 
 753 
 
 759 
 
 764 
 
 77o 
 
 776 
 835 
 
 782 
 
 788 
 
 794 
 
 800 
 
 6 
 
 738 
 
 806 
 
 812 
 
 817 
 
 823 
 
 829 
 
 841 
 
 847 
 
 853 
 
 859 
 
 6 
 
 739 
 
 864 
 
 870 
 
 876 
 
 882 
 
 888 
 
 894 
 
 900 
 
 906 
 
 911 
 
 917 
 
 6 
 
 740 
 
 923 
 
 929 
 
 935 
 
 941 
 
 947 
 
 953 
 
 958 
 
 964 
 
 970 
 
 976 
 
 6 
 
 74i 
 
 982 
 
 988 
 
 994 
 
 999 
 
 *oo5 
 
 ♦on 
 
 ♦017 
 
 ♦023 
 
 ♦029 
 
 *o35 
 
 6 
 
 742 
 
 87040 
 
 046 
 
 052 
 
 058 
 
 064 
 
 070 
 
 o75 
 
 081 
 
 087 
 
 093 
 
 6 
 
 743 
 
 099 
 
 i°5 
 
 III 
 
 116 
 
 122 
 
 128 
 
 134 
 
 140 
 
 146 
 
 l 5 l 
 
 6 
 
 744 
 
 157 
 
 163 
 
 169 
 
 175 
 
 181 
 
 186 
 
 192 
 
 198 
 
 204 
 
 210 
 
 6 
 
 745 
 
 216 
 
 221 
 
 227 
 
 233 
 
 239 
 
 245 
 
 251 
 
 256 
 
 262 
 
 268 
 
 6 
 
 746 
 
 274 
 
 280 
 
 286 
 
 291 
 
 297 
 
 303 
 
 309 
 
 315 
 
 320 
 
 326 
 
 6 
 
 747 
 
 332 
 
 338 
 
 344 
 
 349 
 
 355 
 
 361 
 
 367 
 
 373 
 
 379 
 
 384 
 
 6 
 
 748 
 
 390 
 
 396 
 
 402 
 
 408 
 
 413 
 
 419 
 
 425 
 
 431 
 
 437 
 
 442 
 
 6 
 
 749 
 
 448 
 
 454 
 
 460 
 
 466 
 
 47i 
 
 477 
 
 483 
 
 489 
 
 495 
 
 500 
 
 6 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
750—800 
 
 403 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 75° 
 
 87506 
 
 512 
 
 518 
 
 523 
 
 529 
 
 535 
 
 54i 
 
 547 
 
 552 
 
 558 
 
 6 
 
 751 
 
 564 
 
 57o 
 
 576 
 
 581 
 
 587 
 
 593 
 
 599 
 
 604 
 
 610 
 
 616 
 
 6 
 
 752 
 
 622 
 
 628 
 
 633 
 
 639 
 
 645 
 
 651 
 
 656 
 
 662 
 
 668 
 
 674 
 
 6 
 
 753 
 
 679 
 
 685 
 
 691 
 
 697 
 
 7°3 
 
 708 
 
 714 
 
 720 
 
 726 
 
 73i 
 
 6 
 
 754 
 
 737 
 
 743 
 
 749 
 
 754 
 
 760 
 
 766 
 
 772 
 
 777 
 
 783 
 
 789 
 
 6 
 
 755 
 
 795 
 
 800 
 
 806 
 
 812 
 
 818 
 
 823 
 
 829 
 
 835 
 
 841 
 
 846 
 
 6 
 
 75 6 
 
 852 
 
 858 
 
 864 
 
 869 
 
 875 
 
 881 
 
 887 
 
 892 
 
 898 
 
 904 
 
 6 
 
 757 
 758 
 
 910 
 
 915 
 
 921 
 
 927 
 
 933 
 
 938 
 
 944 
 
 95° 
 
 955 
 
 961 
 
 6 
 
 967 
 
 973 
 
 978 
 
 984 
 
 990 
 
 996 
 
 *OOI 
 
 ♦007 
 
 ♦013 
 
 *oi8 
 
 6 
 
 759 
 
 88024 
 
 030 
 
 036 
 
 041 
 
 047 
 
 o53 
 
 058 
 
 064 
 
 070 
 
 076 
 
 6 
 
 760 
 
 081 
 
 087 
 
 093 
 
 098 
 
 104 
 
 no 
 
 116 
 
 121 
 
 127 
 
 133 
 
 6 
 
 761 
 
 138 
 
 144 
 
 150 
 
 156 
 
 161 
 
 167 
 
 173 
 
 178 
 
 184 
 
 190 
 
 6 
 
 762 
 
 195 
 
 201 
 
 207 
 
 213 
 
 218 
 
 224 
 
 230 
 
 235 
 
 241 
 
 247 
 
 6 
 
 763 
 
 252 
 
 258 
 
 264 
 
 270 
 
 275 
 
 281 
 
 287 
 
 292 
 
 298 
 
 304 
 
 6 
 
 764 
 
 3°9 
 
 315 
 
 321 
 
 326 
 
 332 
 
 338 
 
 343 
 
 349 
 
 355 
 
 360 
 
 6 
 
 765 
 
 366 
 
 372 
 
 377 
 
 383 
 
 389 
 
 395 
 
 400 
 
 406 
 
 412 
 
 417 
 
 6 
 
 766 
 
 423 
 
 429 
 
 434 
 
 440 
 
 446 
 
 45i 
 
 457 
 
 463 
 
 468 
 
 474 
 
 6 
 
 767 
 
 480 
 
 485 
 
 491 
 
 497 
 
 502 
 
 508 
 
 5*3 
 
 5*9 
 
 5 o 5 
 
 530 
 
 6 
 
 768 
 
 536 
 
 542 
 
 547 
 
 553 
 
 559 
 
 564 
 
 57o 
 
 576 
 
 581 
 
 587 
 
 6 
 
 769 
 
 593 
 
 598 
 
 604 
 
 610 
 
 615 
 
 621 
 
 627 
 
 632 
 
 638 
 
 643 
 
 6 
 
 770 
 
 649 
 
 655 
 
 660 
 
 666 
 
 672 
 
 677 
 
 683 
 
 689 
 
 694 
 
 700 
 
 6 
 
 771 
 
 705 
 
 7" 
 
 717 
 
 722 
 
 728 
 
 734 
 
 739 
 
 745 
 
 75° 
 
 756 
 
 6 
 
 772 
 
 762 
 818 
 
 767 
 
 773 
 
 779 
 
 784 
 
 790 
 
 795 
 
 801 
 
 807 
 
 812 
 
 6 
 
 773 
 
 824 
 
 829 
 
 835 
 
 840 
 
 846 
 
 852 
 
 857 
 
 863 
 
 868 
 
 6 
 
 774 
 
 874 
 
 880 
 
 885 
 
 891 
 
 897 
 
 902 
 
 908 
 
 913 
 
 919 
 
 925 
 
 6 
 
 775 
 
 930 
 
 936 
 
 941 
 
 947 
 
 953 
 
 958 
 
 964 
 
 969 
 
 975 
 
 981 
 
 6 
 
 776 
 
 986 
 
 992 
 
 997 
 
 ♦003 
 
 ♦009 
 
 ♦014 
 
 *020 
 
 *025 
 
 ♦031 
 
 ♦037 
 
 6 
 
 777 
 
 89042 
 
 048 
 
 o53 
 
 o59 
 
 064 
 
 070 
 
 O76 
 
 081 
 
 087 
 
 092 
 
 6 
 
 778 
 
 098 
 
 104 
 
 109 
 
 "5 
 
 120 
 
 126 
 
 131 
 
 137 
 
 143 
 
 148 
 
 6 
 
 779 
 
 154 
 
 159 
 
 165 
 
 170 
 
 176 
 
 182 
 
 187 
 
 193 
 
 198 
 
 204 
 
 6 
 
 780 
 
 209 
 
 215 
 
 221 
 
 226 
 
 232 
 
 237 
 
 243 
 
 248 
 
 254 
 
 260 
 
 6 
 
 7S o l 
 
 265 
 
 271 
 
 276 
 
 282 
 
 287 
 
 293 
 
 298 
 
 3°4 
 
 310 
 
 315 
 
 6 
 
 782 
 
 321 
 
 326 
 
 332 
 
 337 
 
 343 
 
 348 
 
 354 
 
 360 
 
 365 
 
 37 1 
 
 6 
 
 7 P 
 
 376 
 
 382 
 
 387 
 
 393 
 
 398 
 
 404 
 
 409 
 
 415 
 
 421 
 
 426 
 
 6 
 
 784 
 
 432 
 
 437 
 
 443 
 
 448 
 
 454 
 
 459 
 
 465 
 
 470 
 
 476 
 
 481 
 
 6 
 
 785 
 
 487 
 
 492 
 
 498 
 
 504 
 
 5°9 
 
 515 
 
 520 
 
 526 
 
 53i 
 
 537 
 
 . 6 
 
 786 
 
 542 
 
 548 
 
 553 
 
 559 
 
 5 6 4 
 
 57o 
 
 575 
 
 581 
 
 586 
 
 592 
 
 6 
 
 18 
 
 597 
 
 603 
 
 609 
 
 614 
 
 620 
 
 625 
 
 631 
 
 636 
 
 642 
 
 647 
 
 6 
 
 653 
 
 658 
 
 664 
 
 669 
 
 675 
 
 680 
 
 686 
 
 691 
 
 697 
 
 702 
 
 6 
 
 789 
 
 708 
 
 713 
 
 719 
 
 724 
 
 730 
 
 735 
 
 741 
 
 746 
 
 752 
 
 757 
 
 6 
 
 790 
 
 763 
 818 
 
 768 
 
 774 
 
 779 
 
 785 
 
 790 
 
 796 
 
 801 
 
 807 
 
 812 
 
 5 
 
 791 
 
 823 
 
 829 
 
 834 
 
 840 
 
 845 
 
 851 
 
 856 
 
 862 
 
 867 
 
 5 
 
 792 
 
 873 
 
 878 
 
 883 
 
 889 
 
 894 
 
 900 
 
 905 
 
 911 
 
 916 
 
 922 
 
 5 
 
 793 
 
 927 
 
 933 
 
 938 
 
 944 
 
 949 
 
 955 
 
 960 
 
 966 
 
 971 
 
 977 
 
 5 
 
 794 
 
 982 
 
 988 
 
 993 
 
 998 
 
 *oo4 
 
 ♦009 
 
 ♦015 
 
 *020 
 
 ♦026 
 
 *o3i 
 
 5 
 
 795 
 
 90037 
 
 042 
 
 048 
 
 053 
 
 059 
 
 064 
 
 069 
 
 075 
 
 080 
 
 086 
 
 5 
 
 796 
 
 091 
 
 097 
 
 102 
 
 108 
 
 "3 
 
 119 
 
 124 
 
 129 
 
 135 
 
 140 
 
 5 
 
 797 
 
 146 
 
 151 
 
 157 
 
 162 
 
 168 
 
 173 
 
 179 
 
 I84 
 
 189 
 
 195 
 
 5 
 
 798 
 
 200 
 
 206 
 
 211 
 
 217 
 
 222 
 
 227 
 
 233 
 
 238 
 
 244 
 
 249 
 
 5 
 
 799 
 
 255 
 
 260 
 
 266 
 
 271 
 
 276 
 
 282 
 
 287 
 
 293 
 
 298 
 
 304 
 
 5 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
404 
 
 800—850 
 
 N. 
 
 
 
 1 2 | 3 | 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 800 
 
 90309 
 
 314 
 
 320 
 
 325 
 
 33i 
 
 336 
 
 342 
 
 347 
 
 352 
 
 358 
 
 5 
 
 801 
 
 363 
 
 369 
 
 37 i 
 
 380 
 
 385 
 
 390 
 
 396 
 
 401 
 
 407 
 
 412 
 
 5 
 
 802 
 
 417 
 
 423 
 
 428 
 
 434 
 
 439 
 
 445 
 
 45° 
 
 455 
 
 461 
 
 466 
 
 5 
 
 803 
 
 472 
 
 477 
 
 482 
 
 488 
 
 493 
 
 499 
 
 5°4 
 
 5°9 
 
 515 
 
 520 
 
 5 
 
 804 
 
 526 
 
 53i 
 
 536 
 
 542 
 
 547 
 
 553 
 
 558 
 
 563 
 
 5 6 9 
 
 574 
 
 5 
 
 805 
 
 580 
 
 585 
 
 590 
 
 596 
 
 601 
 
 607 
 
 612 
 
 617 
 
 623 
 
 628 
 
 5 
 
 806 
 
 634 
 
 639 
 
 644 
 
 650 
 
 655 
 
 660 
 
 666 
 
 671 
 
 677 
 
 682 
 
 5 
 
 !°Z 
 
 687 
 
 693 
 
 698 
 
 703 
 
 709 
 
 714 
 
 720 
 
 725 
 
 730 
 
 736 
 
 5 
 
 808 
 
 74i 
 
 747 
 
 752 
 
 757 
 
 763 
 
 768 
 
 773 
 
 779 
 
 784 
 
 789 
 
 5 
 
 809 
 
 795 
 
 800 
 
 806 
 
 811 
 
 816 
 
 822 
 
 827 
 
 832 
 
 838 
 
 843 
 
 5 
 
 810 
 
 849 
 
 854 
 
 859 
 
 865 
 
 870 
 
 875 
 
 881 
 
 886 
 
 891 
 
 897 
 
 5 
 
 8n 
 
 902 
 
 907 
 
 913 
 
 918 
 
 924 
 
 929 
 
 934 
 
 940 
 
 945 
 
 950 
 
 5 
 
 812 
 
 956 
 
 961 
 
 966 
 
 972 
 
 977 
 
 982 
 
 988 
 
 993 
 
 998 
 
 ♦004 
 
 5 
 
 813 
 
 91009 
 
 014 
 
 020 
 
 025 
 
 030 
 
 036 
 
 041 
 
 046 
 
 052 
 
 057 
 
 5 
 
 814 
 
 062 
 
 068 
 
 073 
 
 078 
 
 084 
 
 089 
 
 094 
 
 100 
 
 io 5 
 
 no 
 
 5 
 
 815 
 
 116 
 
 121 
 
 126 
 
 132 
 
 *37 
 
 142 
 
 148 
 
 153 
 
 158 
 
 164 
 
 5 i 
 
 816 
 
 169 
 
 174 
 
 180 
 
 185 
 
 190 
 
 196 
 
 201 
 
 206 
 
 212 
 
 217 
 
 5 1 
 
 817 
 
 222 
 
 228 
 
 233 
 
 238 
 
 243 
 
 249 
 
 254 
 
 259 
 
 265 
 
 270 
 
 5 ! 
 
 818 
 
 275 
 
 281 
 
 286 
 
 291 
 
 297 
 
 302 
 
 307 
 
 312 
 
 318 
 
 323 
 
 5 • 
 
 819 
 
 328 
 
 334 
 
 339 
 
 344 
 
 35° 
 
 355 
 
 360 
 
 365 
 
 37i 
 
 376 
 
 5 i 
 
 820 
 
 381 
 
 387 
 
 392 
 
 397 
 
 403 
 
 408 
 
 413 
 
 418 
 
 424 
 
 429 
 
 5 
 
 821 
 
 434 
 
 440 
 
 445 
 
 45° 
 
 455 
 
 461 
 
 466 
 
 47i 
 
 477 
 
 482 
 
 5 
 
 822 
 
 487 
 
 492 
 
 498 
 
 5°3 
 
 508 
 
 SH 
 
 519 
 
 524 
 
 529 
 
 535 
 
 5 
 
 823 
 
 540 
 
 545 
 
 551 
 
 556 
 
 561 
 
 566 
 
 572 
 
 577 
 
 582 
 
 587 
 
 5 
 
 824 
 
 593 
 
 598 
 
 603 
 
 609 
 
 614 
 
 619 
 
 624 
 
 630 
 
 635 
 
 640 
 
 5 j 
 
 825 
 
 645 
 
 651 
 
 656 
 
 661 
 
 666 
 
 672 
 
 677 
 
 682 
 
 687 
 
 693 
 
 5 
 
 826 
 
 698 
 
 703 
 
 709 
 
 7H 
 
 719 
 
 724 
 
 730 
 
 735 
 
 740 
 
 745 
 
 5 
 
 827 
 
 75 1 
 
 756 
 
 761 
 
 766 
 
 772 
 
 777 
 
 782 
 
 787 
 
 793 
 
 798 
 
 5 
 
 828 
 
 803 
 
 808 
 
 814 
 
 819 
 
 824 
 
 829 
 
 834 
 
 840 
 
 845 
 
 850 
 
 5 
 
 829 
 
 855 
 
 861 
 
 866 
 
 871 
 
 876 
 
 882 
 
 887 
 
 892 
 
 897 
 
 903 
 
 5 
 
 830 
 
 908 
 
 913 
 
 918 
 
 924 
 
 929 
 
 934 
 
 939 
 
 944 
 
 95° 
 
 955 
 
 5 
 
 831 
 
 960 
 
 965 
 
 971 
 
 976 
 
 981 
 
 986 
 
 991 
 
 997 
 
 *002 
 
 ♦007 
 
 5 1 
 
 832 
 
 92012 
 
 018 
 
 023 
 
 028 
 
 o33 
 
 038 
 
 044 
 
 049 
 
 054 
 
 059 
 
 5 
 
 833 
 
 065 
 
 070 
 
 075 
 
 080 
 
 085 
 
 091 
 
 096 
 
 101 
 
 IO6 
 
 in 
 
 5 
 
 834 
 
 117 
 
 122 
 
 127 
 
 132 
 
 m 
 
 143 
 
 148 
 
 153 
 
 I58 
 
 163 
 
 5 
 
 835- 
 
 169 
 
 174 
 
 179 
 
 184 
 
 189 
 
 195 
 
 200 
 
 205 
 
 2IO 
 
 215 
 
 5 
 
 836 
 
 221 
 
 226 
 
 231 
 
 236 
 
 241 
 
 247 
 
 252 
 
 257 
 
 262 
 
 267 
 
 5 1 
 
 ! 37 
 
 273 
 
 278 
 
 283 
 
 288 
 
 293 
 
 298 
 
 304 
 
 309 
 
 3H 
 
 3i9 
 
 5 
 
 838 
 
 324 
 
 330 
 
 335 
 
 340 
 
 345 
 
 35° 
 
 355 
 
 361 
 
 366 
 
 37i 
 
 5 
 
 839 
 
 376 
 
 381 
 
 387 
 
 392 
 
 397 
 
 402 
 
 407 
 
 412 
 
 418 
 
 423 
 
 5 
 
 840 
 
 428 
 
 433 
 
 438 
 
 443 
 
 449 
 
 454 
 
 459 
 
 464 
 
 469 
 
 474 
 
 5 
 
 841 
 
 480 
 
 485 
 
 490 
 
 495 
 
 500 
 
 5°5 
 
 5 11 
 
 516 
 
 521 
 
 526 
 
 5 
 
 842 
 
 53i 
 
 536 
 
 542 
 
 547 
 
 552 
 
 557 
 
 562 
 
 567 
 
 572 
 
 578 
 
 5 
 
 843 
 
 583 
 
 588 
 
 593 
 
 598 
 
 603 
 
 609 
 
 614 
 
 619 
 
 624 
 
 629 
 
 5 
 
 844 
 
 634 
 
 639 
 
 645 
 
 650 
 
 655 
 
 660 
 
 665 
 
 670 
 
 675 
 
 681 
 
 5 
 
 845 
 
 686 
 
 691 
 
 696 
 
 701 
 
 706 
 
 711 
 
 716 
 
 722 
 
 727 
 
 732 
 
 5 
 
 846 
 
 737 
 
 742 
 
 747 
 
 752 
 804 
 
 758 
 
 763 
 
 768 
 
 773 
 
 778 
 
 783 
 
 5 
 
 847 
 
 788 
 
 793 
 
 799 
 
 809 
 
 814 
 
 819 
 
 824 
 
 829 
 
 834 
 
 5 
 
 848 
 
 840 
 
 845 
 
 850 
 
 855 
 
 860 
 
 865 
 
 870 
 
 875 
 
 88l 
 
 886 
 
 5 
 
 849 
 
 891 
 
 896 
 
 901 
 
 906 
 
 911 
 
 916 
 
 921 
 
 927 
 
 932 
 
 937 
 
 5 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
850—900 
 
 405 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 850 
 
 92942 
 
 947 
 
 952 
 
 957 
 
 962 
 
 967 
 
 973 
 
 978 
 
 983 
 
 988 
 
 5 
 
 851 
 
 993 
 
 998 
 
 ♦003 
 
 *oo8 
 
 *oi3 
 
 *oi8 
 
 ♦024 
 
 *029 
 
 *034 
 
 *Q39 
 
 5 
 
 852 
 
 93044 
 
 049 
 
 054 
 
 059 
 
 064 
 
 069 
 
 o75 
 
 080 
 
 085 
 
 090 
 
 5 
 
 853 
 
 095 
 
 100 
 
 105 
 
 no 
 
 "5 
 
 120 
 
 125 
 
 131 
 
 136 
 
 141 
 
 5 
 
 854 
 
 146 
 
 "S* 
 
 156 
 
 161 
 
 166 
 
 171 
 
 176 
 
 181 
 
 186 
 
 192 
 
 5 
 
 855 
 
 197 
 
 202 
 
 207 
 
 212 
 
 217 
 
 222 
 
 227 
 
 232 
 
 237 
 
 242 
 
 5 
 
 856 
 
 247 
 
 252 
 
 25 ? 
 
 263 
 
 268 
 
 273 
 
 278 
 
 283 
 
 288 
 
 293 
 
 5 
 
 ! 5 Z 
 
 298 
 
 303 
 
 308 
 
 313 
 
 3rt 
 
 323 
 
 328 
 
 334 
 
 339 
 
 344 
 
 5 
 
 858 
 
 349 
 
 354 
 
 359 
 
 364 
 
 369 
 
 374 
 
 379 
 
 384 
 
 389 
 
 394 
 
 5 
 
 859 
 
 399 
 
 404 
 
 409 
 
 414 
 
 420 
 
 425 
 
 430 
 
 435 
 
 440 
 
 445 
 
 5 
 
 860 
 
 45° 
 
 455 
 
 460 
 
 465 
 
 47° 
 
 475 
 
 480 
 
 485 
 
 490 
 
 495 
 
 5 
 
 861 
 
 500 
 
 505 
 
 5io 
 
 515 
 
 520 
 
 526 
 
 53i 
 
 536 
 
 54i 
 
 546 
 
 5 
 
 862 
 
 55i 
 
 556 
 
 561 
 
 566 
 
 57i 
 
 576 
 
 58i 
 
 586 
 
 59i 
 
 596 
 
 5 
 
 863 
 
 601 
 
 606 
 
 611 
 
 6l6 
 
 621 
 
 626 
 
 631 
 
 636 
 
 641 
 
 646 
 
 5 
 
 864 
 
 651 
 
 656 
 
 661 
 
 666 
 
 671 
 
 676 
 
 682 
 
 687 
 
 692 
 
 697 
 
 5 
 
 865 
 
 702 
 
 707 
 
 712 
 
 717 
 
 722 
 
 727 
 
 732 
 
 737 
 
 742 
 
 747 
 
 5 
 
 866 
 
 752 
 
 757 
 
 762 
 
 767 
 
 772 
 
 777 
 
 782 
 
 787 
 
 792 
 
 797 
 
 5 
 
 867 
 
 802 
 
 807 
 
 812 
 
 817 
 
 822 
 
 827 
 
 832 
 
 837 
 
 842 
 
 847 
 
 5 
 
 868 
 
 852 
 
 857 
 
 862 
 
 867 
 
 872 
 
 877 
 
 882 
 
 887 
 
 892 
 
 897 
 
 5 
 
 869 
 
 902 
 
 907 
 
 912 
 
 917 
 
 922 
 
 927 
 
 932 
 
 937 
 
 942 
 
 947 
 
 5 
 
 870 
 
 952 
 
 957 
 
 962 
 
 967 
 
 972 
 
 977 
 
 982 
 
 987 
 
 992 
 
 997 
 
 5 
 
 871 
 
 94002 
 
 007 
 
 012 
 
 017 
 
 022 
 
 027 
 
 032 
 
 037 
 
 042 
 
 047 
 
 5 
 
 S o 72 
 
 052 
 
 057 
 
 062 
 
 067 
 
 072 
 
 077 
 
 082 
 
 086 
 
 091 
 
 096 
 
 5 
 
 873 
 
 101 
 
 106 
 
 in 
 
 116 
 
 121 
 
 126 
 
 131 
 
 136 
 
 141 
 
 146 
 
 5 
 
 874 
 
 "S* 
 
 156 
 
 161 
 
 166 
 
 171 
 
 176 
 
 181 
 
 186 
 
 191 
 
 196 
 
 5 
 
 I™ 
 
 201 
 
 206 
 
 211 
 
 216 
 
 221 
 
 226 
 
 231 
 
 236 
 
 240 
 
 245 
 
 5 
 
 876 
 
 250 
 
 255 
 
 260 
 
 265 
 
 270 
 
 275 
 
 280 
 
 285 
 
 290 
 
 295 
 
 5 
 
 877 
 
 300 
 
 305 
 
 310 
 
 3i5 
 
 320 
 
 325 
 
 330 
 
 335 
 
 340 
 
 345 
 
 5 
 
 878 
 
 349 
 
 354 
 
 359 
 
 364 
 
 369 
 
 374 
 
 379 
 
 384 
 
 389 
 
 394 
 
 5 
 
 879 
 
 399 
 
 404 
 
 409 
 
 414 
 
 419 
 
 424 
 
 429 
 
 433 
 
 438 
 
 443 
 
 5 
 
 880 
 
 448 
 
 453 
 
 458 
 
 463 
 
 468 
 
 473 
 
 478 
 
 483 
 
 4§8 
 
 493 
 
 5 
 
 881 
 
 498 
 
 5°3 
 
 507 
 
 5!2 
 
 517 
 
 522 
 
 527 
 
 532 
 
 537 
 
 542 
 
 5 
 
 882 
 
 547 
 
 552 
 
 557 
 
 562 
 
 567 
 
 57i 
 
 576 
 
 58i 
 
 586 
 
 59i 
 
 5 
 
 8 Jt 3 
 
 596 
 
 601 
 
 606 
 
 6ll 
 
 616 
 
 621 
 
 626 
 
 630 
 
 635 
 
 640 
 
 5 
 
 884 
 
 645 
 
 650 
 
 655 
 
 660 
 
 665 
 
 670 
 
 675 
 
 680 
 
 685 
 
 689 
 
 5 
 
 885 
 
 694 
 
 699 
 
 704 
 
 709 
 
 714 
 
 719 
 
 724 
 
 729 
 
 734 
 
 n s 
 
 5 
 
 886 
 
 743 
 
 748 
 
 753 
 
 758 
 
 763 
 
 768 
 
 773 
 
 778 
 
 783 
 
 787 
 
 5 
 
 887 
 
 792 
 
 797 
 
 802 
 
 807 
 
 812 
 
 817 
 
 822 
 
 827 
 
 832 
 
 836 
 
 5 
 
 888 
 
 841 
 
 846 
 
 851 
 
 856 
 
 861 
 
 866 
 
 871 
 
 876 
 
 880 
 
 885 
 
 5 
 
 889 
 
 890 
 
 895 
 
 900 
 
 905 
 
 910 
 
 915 
 
 919 
 
 924 
 
 929 
 
 934 
 
 5 
 
 890 
 
 939 
 
 944 
 
 949 
 
 954 
 
 959 
 
 963 
 
 968 
 
 973 
 
 978 
 
 983 
 
 5 
 
 891 
 
 988 
 
 993 
 
 998 
 
 *002 
 
 ♦007 
 
 *OI2 
 
 ♦017 
 
 *022 
 
 *027 
 
 ♦032 
 
 5 
 
 892 
 
 95036 
 
 041 
 
 046 
 
 05I 
 
 056 
 
 06l 
 
 066 
 
 07I 
 
 075 
 
 080 
 
 5 
 
 393 
 
 085 
 
 090 
 
 095 
 
 IOO 
 
 io 5 
 
 IO9 
 
 114 
 
 119 
 
 124 
 
 129 
 
 5 
 
 894 
 
 134 
 
 139 
 
 143 
 
 I48 
 
 153 
 
 I58 
 
 163 
 
 168 
 
 173 
 
 177 
 
 5 
 
 895 
 
 182 
 
 187 
 
 192 
 
 197 
 
 202 
 
 207 
 
 211 
 
 2l6 
 
 221 
 
 226 
 
 5 
 
 896 
 
 231 
 
 236 
 
 240 
 
 245 
 
 250 
 
 255 
 
 260 
 
 265 
 
 270 
 
 274 
 
 5 
 
 897 
 
 279 
 
 284 
 
 289 
 
 294 
 
 299 
 
 303 
 
 308 
 
 313 
 
 318 
 
 323 
 
 5 
 
 898 
 
 328 
 
 332 
 
 337 
 
 342 
 
 347 
 
 352 
 
 357 
 
 361 
 
 366 
 
 371 
 
 5 
 
 899 
 
 376 
 
 38i 
 
 386 
 
 390 
 
 395 
 
 4OO 
 
 405 
 
 4IO 
 
 415 
 
 419 
 
 5 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
406 
 
 900—950 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 900 
 
 95424 
 
 429 
 
 434 
 
 439 
 
 444 
 
 448 
 
 453 
 
 458 
 
 463 
 
 468 
 
 5 
 
 901 
 
 472 
 
 477 
 
 482 
 
 487 
 
 492 
 
 497 
 
 501 
 
 506 
 
 5" 
 
 516 
 
 5 
 
 902 
 
 521 
 
 525 
 
 530 
 
 535 
 
 540 
 
 545 
 
 550 
 
 554 
 
 559 
 
 564 
 
 5 
 
 903 
 
 5^9 
 
 574 
 
 578 
 
 583 
 
 588 
 
 593 
 
 598 
 
 602 
 
 607 
 
 612 
 
 5 
 
 904 
 
 617 
 
 622 
 
 626 
 
 631 
 
 636 
 
 641 
 
 646 
 
 650 
 
 655 
 
 660 
 
 5 
 
 905 
 
 665 
 
 670 
 
 674 
 
 679 
 
 684 
 
 689 
 
 694 
 
 698 
 
 703 
 
 708 
 
 5 
 
 906 
 
 713 
 
 718 
 
 722 
 
 727 
 
 732 
 
 737 
 
 742 
 
 746 
 
 75 1 
 
 756 
 
 5 
 
 907 
 
 761 
 
 766 
 
 770 
 
 775 
 
 780 
 
 785 
 
 789 
 
 794 
 
 799 
 
 804 
 
 5 
 
 908 
 
 809 
 
 813 
 
 818 
 
 823 
 
 828 
 
 832 
 
 837 
 
 842 
 
 847 
 
 852 
 
 5 
 
 909 
 
 856 
 
 861 
 
 866 
 
 871 
 
 875 
 
 880 
 
 885 
 
 890 
 
 895 
 
 899 
 
 5 
 
 910 
 
 904 
 
 909 
 
 914 
 
 918 
 
 923 
 
 928 
 
 933 
 
 938 
 
 942 
 
 947 
 
 5 
 
 911 
 
 952 
 
 957 
 
 961 
 
 966 
 
 971 
 
 976 
 
 980 
 
 985 
 
 990 
 
 995 
 
 5 
 
 912 
 
 999 
 
 ♦004 
 
 ♦009 
 
 ♦014 
 
 ♦019 
 
 ♦023 
 
 *028 
 
 ♦033 
 
 ♦038 
 
 ♦042 
 
 5 
 
 913 
 
 96047 
 
 052 
 
 057 
 
 061 
 
 066 
 
 071 
 
 076 
 
 080 
 
 085 
 
 090 
 
 5 
 
 914 
 
 095 
 
 099 
 
 104 
 
 109 
 
 114 
 
 118 
 
 123 
 
 128 
 
 133 
 
 137 
 
 5 
 
 915 
 
 142 
 
 147 
 
 !52 
 
 156 
 
 161 
 
 166 
 
 I7 o 
 
 175 
 
 180 
 
 185 
 
 5 
 
 916 
 
 190 
 
 194 
 
 199 
 
 204 
 
 209 
 
 213 
 
 218 
 
 223 
 
 227 
 
 232 
 
 5 
 
 917 
 918 
 
 237 
 
 242 
 
 246 
 
 25 i 
 
 256 
 
 261 
 
 265 
 
 270 
 
 275 
 
 280 
 
 5 
 
 284 
 
 289 
 
 294 
 
 298 
 
 303 
 
 308 
 
 313 
 
 317 
 
 322 
 
 327 
 
 5 
 
 919 
 
 332 
 
 336 
 
 341 
 
 346 
 
 350 
 
 355 
 
 360 
 
 365 
 
 369 
 
 374 
 
 5 
 
 920 
 
 379 
 
 384 
 
 388 
 
 393 
 
 398 
 
 402 
 
 407 
 
 412 
 
 417 
 
 421 
 
 5 
 
 921 
 
 426 
 
 43 o 
 
 4 P 
 
 440 
 
 445 
 
 45° 
 
 454 
 
 459 
 
 464 
 
 468 
 
 5 
 
 922 
 
 473 
 
 478 
 
 483 
 
 487 
 
 492 
 
 497 
 
 5 QI 
 
 506 
 
 5I o 
 
 515 
 
 5 
 
 923 
 
 520 
 
 525 
 
 530 
 
 534 
 
 5 £? 
 
 544 
 
 548 
 
 553 
 
 558 
 
 562 
 
 5 
 
 924 
 
 567 
 
 572 
 
 577 
 
 581 
 
 586 
 
 591 
 
 595 
 
 600 
 
 605 
 
 609 
 
 5 
 
 925 
 
 614 
 
 619 
 
 624 
 
 628 
 
 633 
 
 638 
 
 642 
 
 647 
 
 652 
 
 656 
 
 5 
 
 926 
 
 661 
 
 666 
 
 670 
 
 675 
 
 680 
 
 685 
 
 689 
 
 694 
 
 699 
 
 7°3 
 
 5 
 
 927 
 
 708 
 
 713 
 
 717 
 
 722 
 
 727 
 
 73i 
 
 736 
 
 7 il 
 
 745 
 
 75° 
 
 5 
 
 928 
 
 755 
 
 759 
 
 764 
 
 769 
 
 774 
 
 778 
 
 783 
 
 788 
 834 
 
 792 
 
 797 
 
 5 
 
 929 
 
 802 
 
 806 
 
 811 
 
 816 
 
 820 
 
 825 
 
 830 
 
 839 
 
 844 
 
 5 
 
 930 
 
 848 
 
 853 
 
 858 
 
 862 
 
 867 
 
 872 
 
 876 
 
 881 
 
 886 
 
 890 
 
 5 
 
 931 
 
 895 
 
 900 
 
 904 
 
 909 
 
 914 
 
 918 
 
 923 
 
 928 
 
 932 
 
 937 
 
 5 
 
 932 
 
 942 
 
 946 
 
 95i 
 
 95 6 
 
 960 
 
 965 
 
 970 
 
 974 
 
 979 
 
 984 
 
 5 
 
 933 
 
 988 
 
 993 
 
 997 
 
 *002 
 
 ♦007 
 
 *OII 
 
 *oi6 
 
 *02I 
 
 ♦025 
 
 *030 
 
 5 
 
 934 
 
 97°35 
 
 039 
 
 044 
 
 O49 
 
 053 
 
 058 
 
 063 
 
 067 
 
 072 
 
 077 
 
 5 
 
 935 
 
 081 
 
 086 
 
 090 
 
 095 
 
 100 
 
 104 
 
 109 
 
 114 
 
 118 
 
 123 
 
 5 
 
 936 
 
 128 
 
 132 
 
 137 
 
 142 
 
 146 
 
 I5 1 
 
 155 
 
 l6o 
 
 165 
 
 169 
 
 5 
 
 937 
 
 174 
 
 179 
 
 183 
 
 188 
 
 192 
 
 197 
 
 202 
 
 206 
 
 211 
 
 216 
 
 5 
 
 938 
 
 220 
 
 225 
 
 230 
 
 234 
 
 239 
 
 243 
 
 248 
 
 253 
 
 257 
 
 262 
 
 5 
 
 939 
 
 267 
 
 271 
 
 276 
 
 280 
 
 285 
 
 290 
 
 294 
 
 299 
 
 304 
 
 308 
 
 5 
 
 940 
 
 3i3 
 
 317 
 
 322 
 
 327 
 
 33i 
 
 336 
 
 340 
 
 345 
 
 35o 
 
 354 
 
 5 
 
 941 
 
 359 
 
 364 
 
 368 
 
 373 
 
 377 
 
 382 
 
 387 
 
 39i 
 
 396 
 
 400 
 
 5 
 
 942 
 
 405 
 
 410 
 
 414 
 
 419 
 
 424 
 
 428 
 
 433 
 
 437 
 
 442 
 
 447 
 
 5 
 
 943 
 
 45i 
 
 45 6 
 
 460 
 
 465 
 
 470 
 
 474 
 
 479 
 
 483 
 
 488 
 
 493 
 
 5 
 
 944 
 
 497 
 
 502 
 
 506 
 
 5 11 
 
 516 
 
 520 
 
 525 
 
 529 
 
 534 
 
 539 
 
 5 
 
 945 
 
 543 
 
 548 
 
 552 
 
 557 
 
 562 
 
 566 
 
 571 
 
 575 
 
 I 8 ? 
 
 585 
 
 5 
 
 946 
 
 589 
 
 594 
 
 598 
 
 603 
 
 607 
 
 612 
 
 617 
 
 621 
 
 626 
 
 630 
 
 5 
 
 947 
 
 635 
 
 640 
 
 644 
 
 649 
 
 653 
 
 658 
 
 663 
 
 667 
 
 672 
 
 676 
 
 5 
 
 948 
 
 681 
 
 685 
 
 690 
 
 695 
 
 699 
 
 704 
 
 708 
 
 7i3 
 
 717 
 
 722 
 
 5 
 
 949 
 
 727 
 
 731 
 
 736 
 
 740 
 
 745 
 
 749 
 
 754 
 
 759 
 
 763 
 
 768 
 
 5 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
950—1000 
 
 407 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
 95° 
 
 97772 
 818 
 
 777 
 
 782 
 
 786 
 
 791 
 
 795 
 
 800 
 
 804 
 
 809 
 
 813 
 
 5 
 
 95 ' 
 
 823 
 
 827 
 
 832 
 
 836 
 
 841 
 
 845 
 
 850 
 
 855 
 
 859 
 
 5 
 
 95 2 
 
 864 
 
 868 
 
 873 
 
 877 
 
 882 
 
 886 
 
 891 
 
 896 
 
 900 
 
 905 
 
 5 
 
 953 
 
 909 
 
 914 
 
 918 
 
 923 
 
 928 
 
 932 
 
 937 
 
 941 
 
 946 
 
 95° 
 
 5 
 
 954 
 
 955 
 
 959 
 
 964 
 
 968 
 
 973 
 
 978 
 
 982 
 
 987 
 
 991 
 
 996 
 
 5 
 
 955 
 
 98000 
 
 005 
 
 009 
 
 014 
 
 019 
 
 023 
 
 028 
 
 032 
 
 037 
 
 041 
 
 5 
 
 95 6 
 
 046 050 
 
 °55 
 
 o59 
 
 064 
 
 068 
 
 073 
 
 078 
 
 082 
 
 087 
 
 5 
 
 957 
 
 091 
 
 096 
 
 100 
 
 io 5 
 
 109 
 
 114 
 
 118 
 
 123 
 
 127 
 
 132 
 
 5 
 
 958 
 
 137 
 
 141 
 
 146 
 
 150 
 
 155 
 
 159 
 
 164 
 
 168 
 
 173 
 
 177 
 
 5 
 
 959 
 
 182 
 
 186 
 
 191 
 
 !95 
 
 200 
 
 204 
 
 209 
 
 214 
 
 218 
 
 223 
 
 5 
 
 960 
 
 227 
 
 232 
 
 236 
 
 241 
 
 245 
 
 250 
 
 254 
 
 259 
 
 263 
 
 268 
 
 5 
 
 961 
 
 272 
 
 277 
 
 281 
 
 286 
 
 290 
 
 295 
 
 299 
 
 3°4 
 
 308 
 
 3*3 
 
 5 
 
 962 
 
 318 
 
 322 
 
 327 
 
 33* 
 
 336 
 
 340 
 
 345 
 
 349 
 
 354 
 
 358 
 
 5 
 
 963 
 
 363 
 
 367 
 
 372 
 
 376 
 
 38i 
 
 385 
 
 390 
 
 394 
 
 399 
 
 403 
 
 5 
 
 964 
 
 408 
 
 412 
 
 417 
 
 421 
 
 426 
 
 430 
 
 435 
 
 439 
 
 444 
 
 448 
 
 5 
 
 965 
 
 453 
 
 457 
 
 462 
 
 466 
 
 47i 
 
 475 
 
 480 
 
 484 
 
 489 
 
 493 
 
 4 
 
 966 
 
 498 
 
 502 
 
 507 
 
 5" 
 
 516 
 
 520 
 
 525 
 
 529 
 
 534 
 
 5 l S 
 
 4 
 
 967 
 
 543 
 
 547 
 
 552 
 
 556 
 
 561 
 
 565 
 
 57o 
 
 574 
 
 579 
 
 583 
 
 4 
 
 968 
 
 588 
 
 592 
 
 597 
 
 601 
 
 605 
 
 610 
 
 614 
 
 619 
 
 623 
 
 628 
 
 4 
 
 969 
 
 632 
 
 637 
 
 641 
 
 646 
 
 650 
 
 655 
 
 659 
 
 664 
 
 668 
 
 673 
 
 4 
 
 970 
 
 677 
 
 682 
 
 686 
 
 691 
 
 695 
 
 700 
 
 704 
 
 709 
 
 7I § 
 
 717 
 
 4 
 
 971 
 
 722 
 
 726 
 
 73i 
 
 7 P 
 
 740 
 
 744 
 
 749 
 
 753 
 
 758 
 
 762 
 
 4 
 
 972 
 
 767 
 
 771 
 
 776 
 
 780 
 
 784 
 
 789 
 
 793 
 
 798 
 
 802 
 
 807 
 
 4 
 
 973 
 
 811 
 
 816 
 
 820 
 
 825 
 
 829 
 
 834 
 
 838 
 
 843 
 
 847 
 
 851 
 
 4 
 
 974 
 
 856 
 
 860 
 
 865 
 
 869 
 
 874 
 
 878 
 
 883 
 
 887 
 
 892 
 
 896 
 
 4 
 
 975 
 
 900 
 
 905 
 
 909 
 
 914 
 
 918 
 
 923 
 
 927 
 
 932 
 
 936 
 
 941 
 
 4 
 
 976 
 
 945 
 
 949 
 
 954 
 
 958 
 
 963 
 
 967 
 
 972 
 
 976 
 
 981 
 
 985 
 
 4 
 
 977 
 
 989 
 
 994 
 
 998 
 
 ♦003 
 
 ♦007 
 
 *OI2 
 
 *oi6 
 
 *02I 
 
 *025 
 
 ♦029 
 
 4 
 
 978 
 
 99034 
 
 038 
 
 043 
 
 047 
 
 052 
 
 O56 
 
 061 
 
 065 
 
 069 
 
 074 
 
 4 
 
 979 
 
 078 
 
 083 
 
 087 
 
 092 
 
 096 
 
 IOO 
 
 105 
 
 IO9 
 
 114 
 
 118 
 
 4 
 
 980 
 
 123 
 
 127 
 
 131 
 
 136 
 
 140 
 
 145 
 
 149 
 
 154 
 
 !58 
 
 162 
 
 4 
 
 981 
 
 167 
 
 171 
 
 176 
 
 180 
 
 185 
 
 189 
 
 193 
 
 I98 
 
 202 
 
 207 
 
 4 
 
 982 
 
 211 
 
 216 
 
 220 
 
 224 
 
 229 
 
 233 
 
 238 
 
 242 
 
 247 
 
 25 * 
 
 4 
 
 983 
 
 255 
 
 260 
 
 264 
 
 269 
 
 273 
 
 277 
 
 282 
 
 286 
 
 291 
 
 295 
 
 4 
 
 984 
 
 300 
 
 304 
 
 308 
 
 3*3 
 
 317 
 
 322 
 
 326 
 
 330 
 
 335 
 
 339 
 
 4 
 
 985 
 
 344 
 
 348 
 
 352 
 
 357 
 
 361 
 
 366 
 
 370 
 
 374 
 
 379 
 
 383 
 
 4 
 
 986 
 
 388 
 
 392 
 
 396 
 
 401 
 
 405 
 
 4IO 
 
 414 
 
 419 
 
 423 
 
 427 
 
 4 
 
 987 
 
 432 
 
 436 
 
 441 
 
 445 
 
 449 
 
 454 
 
 458 
 
 463 
 
 467 
 
 47 1 
 
 4 
 
 988 
 
 476 
 
 480 
 
 484 
 
 489 
 
 493 
 
 498 
 
 502 
 
 506 
 
 5" 
 
 5 X 5 
 
 4 
 
 989 
 
 520 
 
 524 
 
 528 
 
 533 
 
 537 
 
 542 
 
 546 
 
 550 
 
 555 
 
 559 
 
 4 
 
 990 
 
 564 
 
 568 
 
 572 
 
 577 
 
 58i 
 
 585 
 
 590 
 
 594 
 
 599 
 
 603 
 
 4 
 
 991 
 
 607 
 
 612 
 
 616 
 
 621 
 
 625 
 
 629 
 
 634 
 
 638 
 
 642 
 
 647 
 
 4 
 
 992 
 
 651 
 
 656 
 
 660 
 
 664 
 
 669 
 
 673 
 
 977 
 
 682 
 
 686 
 
 691 
 
 4 
 
 993 
 
 095 
 
 699 
 
 704 
 
 708 
 
 712 
 
 717 
 
 721 
 
 726 
 
 730 
 
 73 J 
 
 4 
 
 994 
 
 739 
 
 743 
 
 747 
 
 752 
 
 756 
 
 760 
 
 7 6 5 
 
 769 
 
 774 
 
 778 
 
 4 
 
 995 
 
 782 
 
 787 
 
 791 
 
 795 
 
 800 
 
 804 
 
 808 
 
 813 
 
 817 
 
 822 
 
 4 
 
 996 
 
 826 
 
 830 
 
 835 
 
 839 
 
 843 
 
 848 
 
 852 
 
 856 
 
 861 
 
 865 
 
 4 
 
 997 
 998 
 
 870 
 
 874 
 
 878 
 
 883 
 
 887 
 
 891 
 
 896 
 
 900 
 
 904 
 
 909 
 
 4 
 
 913 
 
 917 
 
 922 
 
 926 
 
 930 
 
 935 
 
 939 
 
 944 
 
 948 
 
 952 
 
 4 
 
 999 
 
 957 
 
 961 
 
 965 
 
 970 
 
 974 
 
 978 
 
 983 
 
 987 
 
 991 
 
 996 
 
 4 
 
 N. 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 D. 
 
TO THE TEACHER 
 
 Many colleges now require for entrance examination a 
 very elementary knowledge of the rudiments of the subject 
 of Graphs. It is the opinion of many good teachers that an 
 insight into Graphs is of considerable value to the pupil in 
 finding the roots of equations, especially equations of the 
 second degree and of degree higher than the second. All 
 agree that the study of Graphs tends to stimulate the 
 interest of the pupil in the work of finding the roots of 
 equations. 
 
 At the request of many teachers and superintendents this 
 short chapter has been added to this treatise on elementary 
 algebra. Five pages are devoted to giving the necessary 
 definitions and explanations of the subject, and showing 
 the pupil how to plot points. Four pages then treat of 
 solving linear equations, six pages of solving quadratics, 
 and one page of solving equations of degree higher than 
 the second. 
 
 A natural way is to study pages 409-417 after or with the 
 chapter on Simultaneous Simple Equations, and to study 
 pages 418-423 after or with the chapters on Quadratic 
 Equations and Simultaneous Quadratics. 
 
 408 
 
CHAPTER XXVII. 
 GRAPHS. 
 
 436. Graphs. Diagrams, called graphs, are often used to 
 show in a concise manner variations in temperature, in 
 population, in prices, etc., etc. 
 
 437. Variables and Constants. A number that, under the 
 conditions of the problem into which it enters, may take 
 different values is called a variable. 
 
 A number that, under the conditions of the problem into 
 which it enters, has a fixed value is called a constant. 
 
 Note. Variables are represented generally by the last letters of 
 the alphabet, x, y, z, etc. ; constants, by the Arabic numerals and 
 by the first letters of the alphabet, a, 6, c, etc. 
 
 438. Algebraic Functions. A function of a variable is an 
 expression that changes in value when the variable changes 
 in value. In general, any expression that involves a vari- 
 able is a function of that variable. If x is involved^only in 
 a finite number of powers and roots, the expression is called 
 an algebraic function of x. 
 
 An algebraic function of x is rational and integral as regards 
 x, if x is involved only in positive integral powers ; that is, in 
 powers and numerators, but not in roots or denominators. 
 
 Thus, x 2 , ^x' 2 + x, are algebraic functions of x; but a x , 
 
 3:3 + 4 1 x 
 >Tc are not algebraic functions of x. Of — , — » Vx, 2 x + a, 
 
 X 2 • X% X2 + a2 
 
 j ax 2 + 6x+c, the last three only are rational integral functions of x. 
 
 a+6 
 
 For brevity a function of x is represented by /(«), F(x) } 
 <l>(x), each of which is read function x. 
 
 409 
 
410 
 
 GRAPHS. 
 
 439. As an easy example we may illustrate by a graph 
 the changes in temperature for a day from 8 a.m. to 8 p.m. 
 
 The official temperatures for Boston, July 17, 1905, were 
 as follows: 8 a.m., 71°; 9 a.m., 72°; 10 a.m., 73°; 11 a.m., 
 77°; 12 m., 82°; 1 p.m., 85°; 2 p.m., 86°; 3 p.m., 88°; 4 p.m., 
 90°; 5 p.m., 89°; 6 p.m., 88°; 7 p.m., 86°; 8 p.m., 82°. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^ 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 9 
 
 J 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 „0 
 
 3 A 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 K 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ti 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 81 
 
 ) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 i 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ■1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 „ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 A 
 
 
 
 
 
 
 
 
 o 
 
 n 
 
 ° 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 } 
 
 
 ■i 
 
 1 
 
 n 
 
 i 
 
 I 
 
 i 
 
 ? 
 
 
 
 
 3 
 
 
 } 
 
 i 
 
 \ 
 
 t 
 
 
 ( 
 
 
 
 
 H 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Draw a horizontal line XX' and a line OF perpendicular to XX'. 
 Using any convenient units of length, lay off on XX' equal distances 
 to represent the hours and on OY equal distances to represent degrees 
 of temperature from 70° to 90°. At each point of division on XX' 
 draw a perpendicular of sufficient length to represent the temperature 
 at that hour. Through the upper ends of these perpendiculars draw 
 the line AB. This line, or graph, presents to the eye a complete view 
 of the changes in temperature for the day. 
 
GRAPHS. 
 
 411 
 
 440. Coordinates. Let XX' be a horizontal straight line, 
 and let YY' be a straight line perpendicular to the line 
 XX' at the point 0. Any point in the plane of the lines 
 XX' and YY' is determined by its distance and direction 
 from each of the perpendiculars XX' and YY'. 
 
 The distance of a point from YY' is measured from on 
 the line XX' and is called the abscissa of the point. The 
 distance of a point from XX' is measured from on the 
 line YY\ and is called the ordinate of the point. 
 
 Thus, the abscissa of Pi is OB h the ordinate of Pi is OA i ; 
 the abscissa of P 2 is 02? 2 , the ordinate of P 2 is OA 2 ; 
 the abscissa of P 3 is OB 3 , the ordinate of P 3 is (L4 3 ; 
 the abscissa of P 4 is OB±, the ordinate of P 4 is 0.4 4. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Y 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 A, 
 
 
 
 
 
 
 
 
 
 r 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 2 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ! 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 » 
 
 
 
 
 
 
 
 B, 
 
 
 
 
 
 
 
 
 
 ?4 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 b 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 Ri 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 * 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 H 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 P 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 The abscissa and the ordinate of a point are called the 
 coordinates of the point. The lines XX' and YY' are called 
 the axes of coordinates, or the axes of reference ; the line XX' 
 is called the axis of abscissas, or the axis of x\ and the line 
 YY' is called the axis of ordinates, or the axis of y. The 
 point is called the origin. 
 
412 GRAPHS. 
 
 In general, an abscissa is represented by x, and an ordi- 
 nate by y. The coordinates of a point whose abscissa is x and 
 ordinate y are written (x, y). In this notation the abscissa 
 is always written first and the ordinate second. 
 
 Thus, the point (4, 7) is the point whose abscissa is 4 and ordinate 7. 
 
 Abscissas measured to the right of YY' are called positive, 
 to the left of YY' are called negative ; ordinates measured 
 above XX' are called positive, below XX' are called negative. 
 
 Thus, in the figure on page 411 the point Pi is (8, 5), the point P 2 
 is (—6, 3), the point P 3 is (— 4, — 3), and the point P 4 is (5, — 4). 
 
 441. Quadrants. The axes of coordinates divide the plane 
 of the axes into four parts called quadrants. The quadrant 
 XOY is called Quadrant I, the quadrant X'OY is called 
 Quadrant II, the quadrant X'OY is called Quadrant III, 
 and the quadrant XO Y is called Quadrant IV. 
 
 Every point in Quadrant I has a positive abscissa and a 
 positive ordinate ; every point in Quadrant II has a negative 
 abscissa and a positive ordinate ; every point in Quadrant III 
 has a negative abscissa and a negative ordinate ; every point in 
 Quadrant IV has a positive abscissa and a negative ordinate. 
 
 Hence, the signs of the coordinates of a point show at a 
 glance in what quadrant the point is situated. 
 
 442. Plotting Points. It is evident that if the location of 
 a point is known, the coordinates of that point referred to 
 given axes may be found easily by measurement ; and if the 
 coordinates of a point are given, the point may be readily 
 constructed, or plotted. 
 
 Thus, a convenient length is taken as the unit, and the point P is 
 found by measurement to lie 2 units to the right of YY' and 4 units 
 below XX', and is, therefore, the point (2, — 4). 
 
 Again, to plot the point (—5, 2), a distance of 5 units is laid off on 
 XX' to the left from to JTi, and a distance of 2 units on YY' upwards 
 from O to K 2 . The intersection of the perpendiculars erected at K\ and 
 K2 determines the point K, which is the required point (— 5, 2). 
 
GRAPHS. 
 
 413 
 
 Coordinate paper is paper ruled in small squares. In 
 plotting points and graphs the student will find coordinate 
 paper of much help in giving accuracy and in saving time. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \j 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 F 
 
 < 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 M 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 ( 
 
 
 
 
 S 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 K 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 V 
 
 
 h 
 
 J 
 
 c 
 
 
 
 
 
 9 1 
 
 
 
 F 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 X' 
 
 
 
 
 bl 
 
 
 
 
 
 
 
 
 
 
 
 
 
 K 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 F 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 L 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 P 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 > 
 
 ' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Exercise 139. 
 
 1. In the figure on this page determine the coordinates 
 of the point B ; of M ; of N\ of R ; of S) of If; of L ; of 
 A j of >j of D; of (7. 
 
 2. What is the abscissa of a point on the axis of y ? 
 What is the ordinate of a point on the axis of x ? 
 
 3. Where must a point lie if its ordinate is zero ? if its 
 abscissa is zero ? if both abscissa and ordinate are zero ? 
 
 4. Plot the following points : (2, 5), (- 3, 6), (- 2, - 4), 
 (3, - 5), (7, 0), (- 5, 0), (0, 0), (0, - 3), (- 4, - 5), (7, 2). 
 
 5. In what quadrant does a point lie if its coordinates are 
 both positive ? if both are negative ? if the ordinate is pos- 
 itive and the abscissa negative ? if the abscissa is positive 
 and the ordinate negative ? 
 
414 GRAPHS. 
 
 6. Plot the points (-2, -8), (-1, -6), (0, -4), (1, -2), 
 (2, 0), (3, 2), (4, 4). Do these points lie in a straight line ? 
 Is the equation 2 cc — ?/ = 4 satisfied if the abscissas are sub- 
 stituted in turn for x, and the corresponding ordinates for y ? 
 
 443. Graph of a Function. Let f(x) be an algebraic function 
 of x, where x is a variable. If y =f(x), then ?/ is a new vari- 
 able connected with x by the relation of y =f(x). If /(#) is 
 rational and integral, it is evident that to every value of x 
 there corresponds one value, and only one value, of y. 
 
 If different values of x are laid off as abscissas, and the 
 corresponding values of f(x) as ordinates, a series of points 
 will be obtained. A line, straight or curved, may be drawn 
 through all these points. This line is called the graph of the 
 function f(x) ; it is also called the graph of the equation y = f(x) . 
 
 Plot the graph of the equation x — 2 y — 4 = 0. 
 Transpose, 2 y = x — 4. 
 
 05-4 
 
 The following table may be computed readily. 
 
 If x = 0, y = -2 
 x=-2, y = -S 
 s = -4, ?/ = - 4 
 
 If x = + 2, y = - 1 ; 
 
 x=+4, y= 0; 
 x = + 6, y = + 1 j 
 
 x = -6, y = -5. x = + 8, y = +2. 
 
 These points are plotted in the figure on page 415 and all lie on the 
 ►straight line A B. If, in the given equation x— 2y — 4=0, the abscissa 
 of any point in the line AB is substituted for x in the corresponding 
 ordinate for y, the equation is satisfied. The line AB extends indefi- 
 nitely in either direction and is the graph of the equation x— 2 y— 4=0. 
 
 If any two points of a straight line are known, the position 
 of the line is definitely determined. ■ 
 
 444. Linear Equations. The graph of every equation of 
 the form ax + by + c = is a straight line. For this reason 
 such an equation is often called a linear equation. 
 
GRAPHS. 
 
 415 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 > 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Exercise 140. 
 
 Plot the graph of the following equations by finding a 
 series of points: 
 
 1. Sx-2y = 6. 4. - x + 3y = 6. 
 
 2. 5x + 2y = 10. 5. 3x + 2y = 12. 
 
 3. 4# — y + 4 = 0. 6. # — 5 ?/ = 5. 
 
 Plot the graphs of the following equations by finding 
 the points in which the graphs cut the axes : 
 
 7. lx + 2y-lA = 0. 10. 4x + 3y + 12 = 0. 
 
 8. 5x- 32/- 15 = 0. 11. a; -8?/ + 8 = 0. 
 
 9. 3a; -4,7/ -24 = 0. 12. 5x + ±y + 30 = 0. 
 
 Plot the graphs of the following equations by finding any 
 two points : 
 
 13. x + y = 0. 15. x — 5y = 0. 17. 5x + 4:y = Q. 
 
 14. x — y = 0. 16. 2 x = 6 2/. 18. 7 x — 5 y = 0. 
 
416 
 
 GRAPHS. 
 
 19. In what respect do the equations of Examples 1-12 
 differ from the equations of Examples 13-18 ? 
 
 20. Does the graph of the equation ax ± by — pass 
 through the origin ? Why ? 
 
 21. The equation of the axis XX' is y — 0. What is 
 the equation of the axis YY'? 
 
 22. What is the position of a graph if its equation does 
 not contain x ? if its equation does not contain y ? 
 
 Plot the graph of : 
 
 23. 3x = 6. 25. x=-\. 27. 5# = 30. 
 
 24. 2 y = 5. 26. y = — f . 28. 6 y = — 42. 
 
 
 
 r 
 
 ^ f^ 
 
 
 ^* v *»»^ l ^ y^ 
 
 ^* s *"»««^ J 
 
 ■^^^ >' 
 
 "**■*■• ^K 
 
 /^-^.^ 
 
 ^_ o ^ --^ *■ 
 
 
 / 
 
 A 
 
 / 
 
 / 
 
 f 
 
 / 
 
 -i/V 
 
 1r t r- 
 
 " 
 
 
 445. Graph of the Solution of a Pair of Simultaneous Linear 
 Equations. In the figure the straight line AB is the graph 
 of the equation x -f- 3 y — 12 ; and the straight line CD is the 
 gr#ph of the equation 4 x — 3 y = 18. It is evident that 
 the coordinates of K, the point of intersection of the lines 
 AB and CD, must satisfy both equations. 
 
GRAPHS. 417 
 
 By solving the equations as simultaneous equations we 
 find that x = 6 and y = 2, which are the coordinates of K. 
 
 Hence, it is evident that it is possible by the use of graphs 
 to solve two simultaneous linear equations that contain only 
 two unknown numbers. In some cases exact values of the 
 unknown numbers may be found; in other cases only ap- 
 proximate values. The larger the scale used in plotting the 
 graphs, the closer will be the approximations obtained. 
 
 Exercise 141. 
 
 Find by graphs exact values of x and y in the following 
 equations and verify by solving the equations : 
 
 1. 2x-5y = 0\ 4. llx-2y = 21\ 
 4x + 2y = 24:J 2x + 4:y=-18J 
 
 2. 7x-2y = U\ 5. 5x + $y = 20\ 
 5x+ y = 10J 2x-3y = -23) 
 
 3. 5x + ±y = 301 6. 3x-f 4y = 30\ 
 
 x- y = - 3J ( 5x-6y = 12J 
 
 Find by graphs approximate values of x and y : 
 
 7. 4:x-5y = 10^ 9. 7x-2y*=U\ 
 2x + 3y= 9J 5x + 32/ = 15J 
 
 8. 8x+ y=2(Tl 10. 9x-4y = l$\ 
 2x-5y = 10) 2x + 5y = 20J 
 
 11. The graphs of the equations 2x + 3y=4:,2x— y=12, 
 and x -f 3 y = — 1 meet in a point. Are the equations simul- 
 taneous ? Give reason. 
 
 12. Do the graphs of the equations Ax — y = 2, x — 6y = 5, 
 and 3x + y = 10 meet in a point ? Are the equations simul- 
 taneous ? Are the equations inconsistent ? 
 
 13. Are the equations 2x — 3y = 5 and 2x — 3y = S 
 simultaneous ? What is shown by their graphs ? 
 
418 
 
 GRAPHS. 
 
 446. Graphs of Quadratic Equations. The graph of any given 
 quadratic equation in x and y may be drawn by the use of 
 the method shown in the solution of the following example. 
 
 Plot the graph of the function x 2 +- 3 x — 4. 
 
 If y = 0, x = + 1 or - 4 ; 
 ?/ = + l, x = + 1.19 or -4.19; 
 y= + 2, x = + 1.37 or -4.37; 
 y = + 3, x = + 1.54 or - 4.54 ; 
 y = + 4, x = + 1.70 or - 4.70 ; 
 y = + 6, x = + 2 or - 5 ; 
 y = + 8, x = + 2.27 or - 5.27. 
 
 it x 2 + 3 a 
 
 
 4 = y. 
 
 Then x = 
 
 y = 
 
 -<H, 
 
 X 
 
 = -1.5; 
 
 
 
 y = 
 
 -6, 
 
 X 
 
 = -1 
 
 or 
 
 -2; 
 
 y = 
 
 -5, 
 
 X 
 
 = -0.38 
 
 or 
 
 -2.62 
 
 y = 
 
 -4, 
 
 X 
 
 = 
 
 or 
 
 -3; 
 
 y = 
 
 — 3, 
 
 X 
 
 = + 0.30 
 
 or 
 
 -3.30 
 
 y = 
 
 -2, 
 
 X 
 
 = + 0.56 
 
 or 
 
 -3.56; 
 
 y = 
 
 -1, 
 
 X 
 
 = + 0.79 
 
 or 
 
 -3.79. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 A 
 
 
 
 
 
 V 
 
 
 
 B 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 J 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 I 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 1 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 J 
 
 
 
 
 
 
 C 
 
 * 
 
 
 
 o 
 
 
 D 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 V 
 
 
 I 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 -' 
 
 / 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 ^ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Plot the points found (-1.5, -6J), (-1, -6), (-2, -6), and so on. 
 Through these points with a free hand draw the smooth curve AB. The 
 curve AB is the graph of the function x 2 + 3 x - 4. This graph consists 
 of one symmetrical branch of infinite length and is called a parabola. For 
 values of y less than — 6£, the corresponding values of x are imaginary. 
 
 When y = 0, then x = 1 and - 4, the roots of the equation 
 x 2 - 3 x - 4 = 0. 
 
GRAPHS. 419 
 
 To solve an equation in x it is necessary only to find the 
 points in which the graph cuts the axis of x. The abscissas 
 of these points are the roots of the given equation. 
 
 447. A more rapid method of solving a quadratic "by the use 
 of graphs is shown in the solution of the following equation. 
 Solve the equation x 2 — x — 2 = 0. 
 
 Let x 2 =y, and put y for x 2 . Then the equation becomes y — x — 2 = 0. 
 
 The graph of the equation y — x — 2 = is the straight line AB, as 
 shown in the left-hand figure on page 420, and the graph of the equa- 
 tion x 2 = y is the parabola CD. 
 
 The abscissas of the intersections of AB and CD are 2 and — 1, which 
 are the roots of the given equation, as may be shown by solving it. 
 
 The great advantage of this method is that the same 
 parabola may be used in the solution of different equations 
 
 Exercise 142. 
 Plot the graphs of the following functions : 
 
 1. x 2 + 5 x + 4. 3. x 2 - 7 x -f 6. 5. 2 x 2 - 7 x + 5 
 
 2. x 2 + x-2. 4. x 2 -5x-\-6. 6. 3« 2 4-4a;--4 
 
 Find, by the method of graphs, the roots of : 
 
 7. x * _ x _ 6 = 0. 10. 7 x 2 + 14 x - 21 = 0. 
 
 8. 2 x 2 - 9 x + 9 = 0. 11. 4 x 2 - 12 x + 9 = 0. 
 
 9. 5z 2 -20 = 0. 12. 25x 2 + 60x + 36 = 
 
 Find approximately the roots of : 
 
 13. 3 x 2 - 2 x - 7 = 0. 15. 5 a 2 - 7 a; - 1 = 0. 
 
 14. 5 ar 2 - 3 x - 30 = 0. 16. 7 a 2 + 5 x - 31 = 0. 
 
 17. What is the nature of the roots of the equation 
 3rc 2 -r-2:E + 4 = 0? Construct the graph of the equation. 
 
 18. How does the graph of a quadratic show that the 
 roots are real and unequal ? real and equal ? imaginary ? 
 
420 
 
 GRAPES. 
 
 448. Solution of Simultaneous Quadratic Equations. In 
 
 general, the method of solving simultaneous linear equa- 
 tions (p. 416, § 445) should be followed. 
 
 Solve 
 
 * 2 + 2/ 2 = 251 
 4 x -f 3 y = 20 J 
 
 
 (1) 
 (2) 
 
 In (1), if x = 0, 
 
 x = ± 1, 
 x=±2, 
 
 y = ±5; x = ±3, 
 y = ±4.90; x = ± 4, 
 
 y =±4.58; x = ± 5, 
 
 2/ = ±4; 
 y = ±3; 
 
 y= o. 
 
 
 If x > + 5 or < — 5, the value of y is imaginary. 
 
 Equation (1) is symmetrical ; its graph also is symmetrical and is 
 the circle HK, as shown in the right-hand figure. The graph of equa- 
 tion (2) is found to be the straight line AB intersecting the circle at 
 the points -M"(lf, 4|) and N(5, 0). 
 
 Hence, the solution gives x = 5, y = 0orx= lg, y = 4f. 
 
 The straight line CD is the graph of 4 x -f 3y = 25. (3) 
 
 The solution of (1) and (3) gives the double solution x = 4, y — 3. 
 
 The straight line EF is the graph of 4 x + 3 y = 30. (4) 
 
 The solution of (1) and (4) gives imaginary roots, since the graphs 
 do not intersect. 
 
GRAPHS. 
 
 421 
 
 Exercise 143. 
 
 1. What does the right-hand figure on page 420 show about 
 the relation of AB, CD, and EF ? How do the coefficients 
 of x and y in equations (2), (3), and (4) show this ? Are the 
 graphs of ax -f by -f c = and ax -f- by -f d = parallel ? . 
 
 2. Write the equations of two parallel lines and con- 
 struct their graphs. 
 
 
 ' [_ 
 
 
 
 ::;:::::i:=::::5^:::;;:E;::;:;;: 
 
 EEEEEEEE^EEEEEE^EE=e£eEEEEEE 
 
 ^: :„*?^~ : 
 
 a 
 
 1 ^ ^ — i 
 
 
 3. If a straight line and a circle touch each other, how many 
 values has x ? how many has y ? How many values have x 
 and y when the line cuts the circle ? What is the nature of 
 the roots of two equations when the graphs do not intersect ? 
 
 Solve exactly or approximately by the method of graphs : 
 
 4. x 2 + if - 1G9 = 
 3x ~2y + 9 
 
 6. x 2 + y 2 = 10(n 
 5x -f y = 46 J 
 
 2} 
 
 a- 2 + y- 
 
 100\ 
 
 3 a; +4v/ ■ : 50 J 
 
 7. a- 2 4- y 2 = 1001 
 3x + 4y = 60 J 
 
422 
 
 GRAPHS. 
 
 8. Solve the equations of Example 7 as simultaneous 
 equations and explain why their graphs do not intersect. 
 
 9. The figure on page 421 shows the graph of the ellipse 
 4 x 2 + 9 y 2 = 288, and the graph of the parabola 3 y 2 = 8 x. 
 What roots satisfy these equations ? 
 
 10. The equation of the circle ax 2 + ay 2 = c differs in what 
 respect from the equation of the ellipse ax 2 + by 2 = c? What 
 is the shape of the ellipse when a and b differ greatly in value ? 
 when a and b are nearly equal ? when a and b are equal ? 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 f 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1b 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 0" 
 
 + m 
 
 r 
 
 V 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 
 
 V 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 / 
 
 r 
 
 
 
 \ 
 
 s. 
 
 
 \ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 — 
 
 f 
 
 
 
 
 
 s 
 
 ^ 
 
 \ 
 
 ■) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 A 
 
 \r 
 
 *>. 
 
 
 
 
 
 
 i 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 \ 
 
 
 \ 
 
 < 
 
 
 
 
 J 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 V 
 
 
 \ 
 
 
 
 > 
 
 f 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 s 
 
 
 =~ 
 
 y 
 
 ' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 r- 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 '' 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 11. The figure on this page shows the graph of the circle 
 x 2 + y 2 = 26, and the graph of the hyperbola #y = 5. What 
 are the coordinates of their points of intersection ? What 
 roots satisfy the equations ? 
 
 12. Solve the equations x 2 -f y 2 = 26 and xy = 5 as simul- 
 taneous quadratics and notice that the results are the answers 
 to Example 11. 
 
GRAPHS. 
 
 423 
 
 Solve by graphs : 
 
 13. x 2 + y 1 = 80 
 
 xy = 32 
 
 14. x 2 + i 
 
 xy 
 
 15. 
 3 
 
 } 
 
 2/ 2 =34| 
 
 y =15J 
 
 * 2 + 2/ 2 = 74 -^ 
 
 z 2 + 2/ 2 = 172j 
 
 16. 
 
 17. 
 
 ■>y 2 
 
 5x 2 + y 2 = S21\ 
 -196 a = 0J 
 
 ^ 2 + 62/ 2 = 791 
 a 2 - 4 y 2 = 89 J 
 
 :1) 
 
 18. Sx 2 -5y 2 
 xy 
 
 19. Explain the meaning of the imaginary roots of the 
 equations in Example 18. 
 
 Note. The equations in Example 18 are types of the two simple 
 hyperbola equations. The difference in form is due to the difference 
 in the position of the curves. 
 
 Determine by inspection the shape of the graph of : 
 
 20. 2x — lly=7. 23. x 2 + y 2 = 18. 26. 4z 2 +4y 2 =27 
 
 21. 5x 2 +8y 2 = 6. 24. x = 3y. 27. xy = 12. 
 
 22. y 2 = 8x. 25. 2x 2 = 7y. 28. 2x 2 -5y 2 =12 
 
 r ~ 
 
 
 
 :::::=:::::::-::::=:--:-:":^:-:: 
 
 [ 
 
 i 
 
 ^L 2 -^-- ° 4---* 
 
 === = = = = ==2 zf=: --^=== == 4- 
 
 ::::::::Vz:::--:h;:::-^:::::::: 
 
 ::: :::? =:: :=: "£-_::: : :: 
 
 :;=:::^::::::::::::::::::::::::-: 
 
 u '*■ - 
 
 
424 
 
 GRAPHS. 
 
 449. Graphs of Higher Equations. The graphs of equations 
 and functions of higher degree than the second may be plotted 
 by the method already shown (p. 418, § 446). 
 
 In general, the number of real roots of an equation in x 
 is equal to the number of times the graph cuts the axis of x. 
 If the graph is tangent to the axis of x, there is a double 
 root or a multiple root ; if the graph does not cut or touch 
 the axis of x, the roots are imaginary. 
 
 Plot the graph of the function x z 
 Put x 3 - x 2 - 16 x - 20 = y. 
 
 16 
 
 20. 
 
 x = + 6, 
 
 y = + 64 ; 
 
 If x = + 0.5, 
 
 y = -28.13 
 
 x = + 5.5, 
 
 y = + 28.13; 
 
 x = 0, 
 
 V =-20; 
 
 x = + 5, 
 
 y= 0; 
 
 x = -0.5, 
 
 ?/ = - 12.38 
 
 x = +4.5, 
 
 2/ = -21.13; 
 
 x = -l, 
 
 2/ = - 6; 
 
 x =+4, 
 
 V = - 86 ; 
 
 x=-1.5, 
 
 2/ = - 1.63 
 
 x = + 3.5, 
 
 y = - 45.38 ; 
 
 x = - 2, 
 
 2/= 0; 
 
 x = +3, 
 
 y = - 50 ; 
 
 x = -2.5, 
 
 2/ = - 1.88 
 
 x = +2.7, 
 
 y =; -60.81; 
 
 x = -3, 
 
 ? = - 8; 
 
 x = + 2.5, 
 
 y = - 50.63 ; 
 
 x = -3.5, 
 
 y = -19.13 
 
 x = +2, 
 
 2/=-48; 
 
 x = -4, 
 
 y =-36; 
 
 x= + 1.5, 
 
 2/ =-42.88; 
 
 x = -4.5, 
 
 2/ = - 59.38 
 
 x = +l, 
 
 y = - 36. 
 
 x = — 5, 
 
 y = - 90. 
 
 To make the figure compact use two spaces of the coordinate paper 
 for one unit of x, and one space for ten units of y. The curve CAEBD 
 (p. 423) is the graph of the function x 3 — x 2 — 16 x — 20. The graph shows 
 that the roots of the equation x 3 — x 2 — 16 x— 20=0 are 5, —2, and —2. 
 
 When x is greater than 6 the curve evidently extends indefinitely 
 above XX' ; when x is less than — 5, indefinitely below XX'. 
 
 To determine more accurately the shape of the curve, it is often 
 desirable to assume for x several values between two consecutive units. 
 
 Exercise 144. 
 
 Find by a graph the roots of ; 
 1. ^3 _ x 2 __ 12 x = 0. 2. 2 x s - x 2 - 26 a. + 40 = 0. 
 
 Find by a graph the number of real roots of • 
 3. x 9 - 8 = 0. 4. x s - 5 x 2 + 8 a + 14 = 0. 
 
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