LG 
 " 
 
 21) 7 bu 
 3 oplreberersrprtrsinrenystty Uitte 
 * sip peeacalatats peerage ret : 
 + ewes ios snsqpproreraselae’ (#038 TITIES 
 Peydiais: esa resbebywreenie hist Li Ls eal pacererrerr eer st: 
 OP st ES 
 

 
 
 
 
 
 
 
 ere Apsei git 
 
 LIBRARY _ 
 Bia.q 
 AR he 
 MATHEMATICS LIBRARY 
 
 
 
 
 
 
 
* Sey, 
 
 operty of the Math. Dept. 
 
 be left Return this book on or before the 
 Latest Date stamped below. 
 
 
 
 University of Illinois Library 
 
 
 
 
 
 
 APR 7 
 GAR - 9 wt 
 nEC23 R 
 
 L161—H41 
 
aa vhite 
 gd 
 
 fan as 
 
 ¥ 
 
 
 
Che Standard Sevics of fMathematics 
 
 
 
 THE 
 
 KSSENTIALS OF ALGEBRA 
 FOR SHCONDARY SCHOOLS 
 
 BY 
 
 ROBERT J.. ALEY, Pa.D. ra 
 
 PROFESSOR OF MATHEMATICS, INDIANA UNIVERSITY "¥ 
 
 AND @ 
 
 DAVID A. ROTHROCK, Pu: 
 
 ASSOCIATE PROFESSOR OF MATHEMATICS, INDIANA YNIVERSITY 
 
 
 
 A hen) 
 SILVER, BURDEFT AND COMPANY; 
 
 NEW YORKG “BOSTON CHICAGO 
 eG 
 Q 
 
 o 
 &, 
 

 
 Ye. CV 
 TOA EPEETOAS PAH RelAL 
 | VLECK 
 
 ait 
 
 Copyrieut, 1904, 
 
 By SILVER, BURDETT AND COMPANY. 
 
 
 
 ie! : at * Pa | ” 
 pry, i pat it ie ate ; 
 Ae \ary YY ae ove \ K ” a ; 
 ead y mi” em .! j uv a aes) ; ed 
 A q i 7 ead y enka + i 
 aw oa | ‘ Wyss’ 
 Mi < + - iy i m4 A 
 hak Wer a ois, Las an 
 : 7 at “ ony , a ire, 
 poly af ‘sp F : 20 
 = ty a 
 
 
 
 ~ 
 
 se 
 : 
 iB 
 1 
 ‘ 
 PT 
 : 
 t 
 Site Pil 
 a. : 
 tare 
 LE j 
 ' 
 - Fi 
 “1 - 
 Ki . 
 - i 
 iF 
 Me * 
 j ( ar 
 : De 
 . . 
 ral at a : 
 

 
 fa Xx w7 C6 uATHEMATICS LIBRARY 
 
 PREFACE. 
 
 In the preparation of this book the authors have made 
 an earnest effort to retain all the essentials of the older 
 Algebra text-books, and to introduce and properly empha- 
 size certain newer features which the mathematical studies 
 of the present demand. 
 
 The following are some of the special characteristics of 
 the book: 
 
 1. The Number System. ‘The number system is pre- 
 sented in the first chapter, and from the arithmetical 
 system extension is made to the algebraic number. system. 
 —. In this way the idea of negative number is introduced 
 — and the fundamental operations are explained. 
 ~~ 2. Factoring. This subject is treated with particular 
 3 fullness, and use is made of the factorial method wherever 
 _s-applicable in the study of Algebra. At the first reading, 
 
 Sections 79, 80, and 81, covering certain details of factor- 
 
 ing, may be omitted if thought desirable. The ordinary 
 
 student, however, should have no special difficulty in 
 mastering these sections. 
 
 3. The Graph. The work with graphs is made an in- 
 tegral part of the book. The graphs of simple and 
 quadratic equations are used freely to aid the pupil’s 
 understanding of the solutions involved. Graphic illus- 
 trations are given wherever it is thought they will make 
 
 the subject clearer. 
 iii 
 
 334.356 
 
lv PREFACE. 
 
 4. Type Forms. Type forms play an important part in 
 the study of Algebra. ‘The work of the student is greatly 
 simplified if he learns early in his course to recognize and 
 to understand these types. ‘Type forms are extensively 
 used in multiplication, division, factoring, and equations. 
 
 5. Exercises. ‘The exercises have been selected with a 
 view of clarifying the text and enforcing fundamental 
 _principles. ‘They are numerous, and are difficult enough 
 to call for effort on the part of the student. 
 
 It is believed that the book contains sufficient matter 
 to furnish a thorough training in the elements of Algebra 
 and to meet the entrance requirements of American 
 colleges. 
 
CHAPTER 
 
 TH 
 {Fi 
 i8ae 
 
 XVII. 
 XVIII. 
 348.4 
 
 CONTENTS. 
 
 INTRODUCTION F ; ‘ : 5 : ‘ ; 
 DEFINITIONS . : : , ; . : ; : 
 ADDITION AND SUBTRACTION. ; : A : 
 MULTIPLICATION AND DIVISION . ‘ : é : 
 IMPORTANT IDENTITIES : ; , : A : 
 FACTORING . ‘ ; : : : ri a : 
 Divisors AND MULTIPLES . : ; ; ; ; 
 FRACTIONS . ; : : , ; : : ; 
 EQUATIONS IN ONE VARIABLE . - 
 
 LINEAR EQUATIONS IN Two VARIABLES . ‘ : 
 SIMULTANEOUS EQUATIONS . 
 
 EVOLUTION . 2 Cie : 
 
 THEORY OF INDICES 
 
 Rapicats, SurRDs, AND IMAGINARIES 
 
 QUADRATIC EQUATIONS IN A SINGLE VARIABLE 
 SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS 
 RATIO, VARIATION, AND PROPORTION : ‘ 
 PERMUTATIONS AND COMBINATIONS . ; ‘ P 
 SERIES . : : ; : ‘ : : ‘ : 
 LOGARITHMS . : : - ‘ ‘ i : : 
 
 PAGE 
 
=e ' 
 “eS 4 Rll i si he al 
 7? is A sg is 
 
 ys: in “i { 
 | \ 
 
 
 
THE 
 ESSENTIALS OF ALGEBRA. 
 
 CHAPTER I. 
 INTRODUCTION. 
 
 1. The Integral Number System is that orderly succession 
 by ones which we first learn by counting. We are familiar 
 with it in the Arabic numeral form of 1, 2, 3, 4, 5, 6, T, 8, 
 Pave and so on. The characters 1, 2, 3, 4, etc., are 
 symbols of number, but we shall hereafter, by the use of a 
 common figure of speech, speak of them and other number 
 symbols as number. 
 
 2. Elementary Number Notions. We know that 3+4=7, 
 because by counting 3 and then 4 more we reach 7. This 
 may be seen by counting these groups, 
 
 @eo 6880806 @ 
 
 All the results of addition are primarily determined by 
 counting. In practice, a number of simple addition results 
 are determined by counting, and then these are made a 
 matter of memory. 3 x 4= 12, because by counting 3 
 groups of 4 each we reach 12. This is seen in the follow- 
 ing arrangement : 
 
 The truth of a multiplication table is also established by 
 counting. The number system shows that 34+4=4+3; 
 1 
 
2 THE ESSENTIALS OF ALGEBRA. 
 for by counting 3 and then 4 more, we reach the same 
 result as by counting 4 and then 3 more. 
 
 3 x4=4 x3 because 3 groups of 4 each meas the same 
 sum as 4 groups of 3 each. 
 
 Ie 
 Be uly 3 
 
 Illustrations enough have been given to show that the 
 integral number system is the real basis of the funda- 
 mental parts of arithmetic. 
 
 3. Fractions in the Number System. As long as no exact 
 measurements are needed, nor accurate divisions attempted, 
 the integral number system is sufficient. If a stick is 
 more than 9 inches and less than 10 inches in length, we 
 can not express its exact length by means of the integral 
 number system. A similar difficulty arises in attempting 
 to answer the question 8+ 3 = what? To answer all such 
 questions, fractions have been devised and made a part 
 of the number system. The addition of fractions to the 
 number system made possible many arithmetical opera- 
 tions which were before impossible. The field of arith- 
 metic was thus greatly enlarged. 
 
 4. Incommensurables in the Number System. When the 
 necessity for extracting roots arose in the development of 
 arithmetic, it was found that many roots could not be ex- 
 actly determined. For example, the square root of 2 lies 
 between 1 and 2, between 1.4 and 1.5, between 1.41 and 
 1.42, between 1.414 and 1.415, etc. We may extend this 
 
INTRODUCTION. 3 
 
 process of locating the square root of 2 between consecu- 
 tive numbers of the number system as far as we please, 
 but we can never find its exact value. Such numbers as 
 the square root of 2, and the square and cube roots of other 
 numbers which can not be exactly found, are called incom- 
 mensurable numbers or merely tncommensurables. Although 
 such numbers can not be exactly expressed, the number 
 system now includes them. 
 
 5. Numerical Arithmetic Complete. With the number 
 system so developed as to include integers, fractions, and 
 incommensurables, ordinary numerical arithmetic is com- 
 plete. This means that in performing the operations of 
 ordinary arithmetic no necessity arises for any other kind 
 of numbers. 
 
 6. Literal Arithmetic. In percentage we frequently 
 represent the base by 6, the rate per cent by 7, the per- 
 centage by p, the amount by a, and the difference by d. 
 When we do this, we can transform the rules for the cases 
 of percentage into the following forms : 
 
 Cen Ut. 
 (2) r=p+b. 
 (3) b=p+r. 
 
 (4) a=b+bx?r. 
 (ey ia D9, 
 
 The symbols 08, 7, p, a, and d may be considered as 
 particular numbers of the number system. When thought 
 of in this way, they are mere abbreviations of numbers. 
 Since they may be the abbreviations of any numbers 
 whatsoever, we may think of the symbols themselves as 
 
4 THE ESSENTIALS OF ALGEBRA. 
 
 numbers. When a symbol, such as any of the above, is 
 thought of in this way, it is called a general number. Such 
 a number is frequently called a literal number. These 
 symbols of general or literal numbers may have particular 
 numerical values assigned to them. In order to find 8% 
 of 250 we take form (1), on page 3, and put 250 instead 
 of 6, and .08 instead of 7. We then have 
 
 p= OX T= 200 vo —aue 
 
 7. Substitution. Zhe process of putting a particular num- 
 ber in the place of a general one ts called substitution. 
 
 By substitution all the results of general or literal 
 arithmetic become particular. The solution of a problem 
 in ordinary arithmetic is a mere matter of substituting 
 particular numbers for general ones in the proper literal 
 form, as is illustrated in the percentage problem of 
 Section 6. 
 
 The area of a rectangle is the product of its length and 
 width. If we represent area by a, length by J, and width 
 by w, we at once have the general form 
 
 a=Ilxw. 
 
 If we wish to find the area of a lot 66 feet long and 
 30 feet wide, we put 66 for 7 and 30 for w, and we have 
 
 a=lxw=66 x 30= 1980. 
 
 8. Algebraic Expression. Any combination of literal num- 
 bers or of literal and arithmetical numbers by means of any 
 or all of the signs of addition, subtraction, multiplication, 
 division, involution, and evolution 1s an algebraic expression. 
 
 a +y-—z is an algebraic expression and is read a plus 
 y minus z The algebraic expression a x b—e+d+4is 
 
 / 
 
INTRODUCTION, § 
 
 read a times 6 minus e divided by @ plus 4. The word 
 function is frequently used instead of expression. The 
 parts of an algebraic expression separated by either of 
 the signs + or — are called terms. In ax + by — cyz, there 
 are three terms; viz., ax, by, and cyz. 
 
 9. Signs used in Algebraic Expressions. ‘The signs +, 
 —, X, +, and +/ are used as in arithmetic. They denote 
 addition, subtraction, multiplication, division, and root 
 extraction, respectively. Multiplication is also indicated 
 by a dot (-), and by writing the characters adjacent to 
 each other. ax 6, a:b, and ab mean exactly the same 
 thing; viz., @ multiplied by 6. Between arithmetical 
 numbers also the signs x and - are used to denote multi- 
 plication. The multiplication of an arithmetical and literal 
 number or of two literal numbers is denoted by writing 
 them consecutively. 56 means 5x6. ab means a x 6. 
 6ab=6xaxb=6-a-b. The first form (646) is the 
 one generally used and is read szz ab. 
 
 There are five forms of division sign in general use, 
 6+ 2, 2)6, 6:2, §, and 6/2, all meaning exactly the 
 same thing; viz., 6 divided by 2. In algebra the 
 fourth and fifth forms are more frequently used than 
 the others. 
 
 Vz, Va, Vb are read square root of a, cube root of a, 
 and fourth root of 6, respectively. 
 
 a*, 68, z* are read a square, 6 cube, and x to the fourth 
 power, respectively. 
 
 a=aa; 6=bbb; 2t= rrr. 
 
 In the above expressions the 2, 3, and 4 are called 
 exponents. 
 
So 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 Read the following algebraic expressions: 
 1 a+b—c+4. 2. 4a—3b+<ab. 
 3. ab —4ac+3.ad —*. 
 ; is read b divided by c, or 6 over ec. 
 
 The last term of (3) should be read minus the quotient b divided 
 by c, or minus the fraction b over c. ° 
 
 4. 2a+b+5abe —A4 be. 6. 5a®—A aa? +Vab. 
 = 3 Baie: 
 5. a —4ar+vo0. 7. 6 abt — + Va 
 
 8. 5a ern, 
 y 
 
 / ee . 
 5a vu is read 5a times the square root of z. 
 
 9. nye +87 17 vay 10. Sat Vyz 24 8a Von, 
 
 * 10. Precedence of Signs in Algebraic Expressions. Jf only 
 the signs + and — occur in an algebraic expression, the 
 operations are to be performed in order from left to right. 
 
 For example: 8—4+4+3+42—5=4. 
 
 If only the signs x and + occur in an algebraic expres- 
 sion, the operations are to be performed in order from left 
 to right. 
 
 For example : 6+38x4+2=4. 
 
 If the signs +, —, X, and + occur in an aigebraie ex- 
 pression, the multiplications and divisions are first per- 
 formed, and then the additions and subtractions. 
 
INTRODUCTION. T 
 For example: 44+ 3x 2—12+6x342x4. 
 
 Performing the multiplications and divisions in the above, 
 we have 4+6—6+48. Now performing the additions and 
 subtraction, we get 12, which is the value of the expression. 
 
 * 11. Signs of Aggregation. ‘The signs of aggregation or 
 grouping are the parentheses ( ), braces { }, brackets [ ], 
 and vinculum or bar~——. Each of these signs indicates 
 that the expression within or under it must be treated as 
 a whole. (6c — ad)+ 6 means that the difference between 
 be and ad is to be divided by 6. The expressions (be — 
 ad)+6b, {be—adj+b, [be—ad]+6, and be—ad+6 all 
 mean precisely the same thing. ‘The four signs of aggre- 
 gation are all called by the general name parentheses. ‘The 
 different forms are necessary to avoid confusion when one 
 or more groups are included within another group. 
 
 
 
 For example: 5 x {12+(7 x64+2+4+[2x3+41])}. 
 
 This becomes 5 x $12 +(7 x 8+4+7)}, which in turn 
 becomes 5 x §12.+ 23, or 5x6=80. 
 
 Multiplication of a quantity within a parenthesis by any 
 quantity is indicated by writing the multiplier before 
 or after the parenthesis. 5(a+6) means 5x(a+6). 
 (a+6)5 means (a+ 6) x d. 
 
 * 12. Coefficient. In the expressions 5a, 32, and Ty, 5, 3 
 and 7 are the coefficients of a, x, and y, respectively. 
 
 In an indicated product any factor or factors may be 
 considered the named part, then all the other factors con- 
 stitute the coefficient. Thus, in 8 aay, 8 is the coefficient 
 of axy, 8 a is the coefficient of xy, and 8 az is the coefficient 
 
8 THE ESSENTIALS OF ALGEBRA. 
 
 of y. In the first case ary is the named part, in the second 
 xy, and the third y. 
 
 When no numerical coefficient precedes a literal expres- 
 sion, the coefficient 1 is understood. 
 
 Read the coefficients of y? in the following expressions: 
 
 $a, Says Lia, Ail Ua 
 
 % EXERCISES IN SUBSTITUTION. 
 
 Find the value of each of the following literal expressions, 
 in which a=4, P=2,¢=— 3, d=), 0—=6,y— yee 
 
 1. a+b—c+ ay. 
 
 Substituting the values given to the above letters, this expression 
 
 becomes 
 
 ‘In a similar manner determine the values of the following 
 expressions: 
 
 2 a+e2—yt+3d. oD Vatpete 
 
 3. ax+tbyte. 12. V6a+vVy. 
 
 4. «+y+a. 13. 5a+6b—4e. 
 
 5. ax? + by? +d. 14. 38a—4b406 
 
 ROUNE 15. 32—4y+10. 
 bd 16. («+ y)2+ (y+z2)a. 
 
 7. e+y+2+d. 17. abe + xyz. 
 
 8. x —y +4. x 18. ax —by+cz—8 abe. 
 
 9. w+22-+1. 19. + y+22—38 wyz. 
 
 10. a? +2 ay + 7%. 20. y +cu + (d—Va) +46. 
 
INTRODUCTION. 9 
 
 13; Use of General Number in Arithmetical Problems. All * 
 the problems of ordinary arithmetic may be made general 
 by the use of general or literal numbers in place of the 
 arithmetical numbers involved. 
 
 1. If John has 10 cents and Henry 12 cents, they together 
 have (10 + 12) cents. This becomes general by stating it thus: 
 if John has a cents and Henry 0 cents, they together have 
 (a + bd) cents. 
 
 2. If Mary is 10 years old and Susie is w years older, then 
 Susie is (10 +2) years old. 
 
 3. Ifa merchant sells a bushels of corn at b cents a bushel, 
 and c bushels of wheat at d cents a bushel, and divides the 
 money received equally among his e children, each one will 
 receive | (ab + cd) + e] cents. 
 
 4. A man has a cents and } dimes. How many cents 
 has he? 7 
 
 5. A man is a years old. His son is 4 as old. What is 
 the combined age of father and son ? 
 
 6. I traveled b miles at c cents a mile, and d miles at e 
 cents a mile. How far did I travel and what did it cost 
 me ? 
 
 7. A rectangle is 2 rods long and y rods wide. How many 
 acres does it contain ? 
 
 8. A farm a rods long and 0 rods wide is sold at ¢ dollars 
 per acre. Find the amount for which the farm sold. 
 
 9. How many hours will be required to travel # miles if 
 one third the distance be traveled at a miles per hour and the 
 remaining two thirds at } miles per hour ? 
 
 10. A man has # dollars in cash; 6 men owe him each y 
 dollars, c men owe him each z dollars. How much money 
 would he have if his collections were made ? 
 
10 THE ESSENTIALS OF ALGEBRA. 
 
 11. A rectangle is a rods long and b rods wide. Find the 
 length of its diagonal. 
 
 12. How many hours will be required for a train running 30 
 miles per hour to travel @ miles, if it makes n stops of 6 min- 
 utes each ? 
 
 13. Find the cost of excavating a basement a feet long, 
 b feet wide, and ¢ feet deep at a cents per cubic yard. 
 
 14. A man sold from a flock of n sheep; the mth part of 
 them for p dollars each. He then increased his flock by a sheep 
 and sold the whole lot at « dollars per head. How much did 
 _ he receive for the whole flock ? 
 
 14. Opposite Numbers. On the scale of a thermometer, 
 temperature is marked both ways from 0. On the centi- 
 gerade thermometer, temperatures above freezing 
 read from 0 up, and temperatures below freezing 
 read from 0 down. Longitude is measured both 
 east and west from a fixed 0 or prime meridian. 
 Latitude is measured both north and south from 
 the equator. We may consider direction along 
 a line to the right or to the left. Rotation may 
 be opposite to that of the hands of a clock, or 
 it may be clockwise. Numbers which in some 
 way indicate such opposites are called Opposite 
 Numbers. The need of opposite numbers becomes 
 apparent when we try to generalize the opera- 
 tions of arithmetic. a@—6 indicates the sub- 
 traction of 6 from a. 
 
 If a=10 and d= a a—b=10— 9=1, which 
 shows:that 9 is 1 less than 10. 
 
 If a=10 and 6=10, a—b= 10 —10 = 0, which shows 
 that 10 is 0 less than 10, or that 10 is equal to 10. 
 
 Centigrade Thermometer 
 
 
 
INTRODUCTION, Tel 
 
 If a=10 and 6=11, a—b=10—11, which, what- 
 ever it may be, ought to show that 11 is 1 greater than 
 10. 
 
 If a man has $500 and is in debt $400, he is worth 
 $ 500 — $400 = $100. 
 
 If a man has $500 and is in debt $500, he is worth 
 $500 — $500 = 0. | 
 
 If a man has $500 and is in debt $600, he is worth 
 $500 —$600. This statement harmonizes with the state- 
 ments made in the other two cases. What does it mean? 
 We may interpret it by saying that he owes $100 more 
 than he is worth, or that his liabilities exceed his assets by 
 $100, or that he is worth $100 less than nothing. 
 
 15. Negative Number. Such questions as the above are 
 answered by the extension of the number system so as to 
 include negative number. We may think of the arithmeti- 
 cal number system as starting at 0 and extending indefi- 
 nitely in a horizontal line to the right. It is an easy 
 matter to think of a similar system extending indefinitely 
 to the left from 0. These appear as follows: 
 
 200 .--100---60.--50--4382101234..-50---60---100---200 
 (ES EN ma a a SL a 
 
 This extension doubles the scope of the number system. 
 Integers, fractions, and incommensurables are all included 
 in the extension to the left. 
 
 16. Positive and Negative. That part of the number 
 system to the right of 0 is called positive. The positive 
 character of a number is indicated by the use of a + sign 
 before it. +4, +a, and +2? are positive numbers. In 
 
12 THE ESSENTIALS OF ALGEBRA. 
 
 practice the + sign is frequently omitted, so that the ab- 
 sence of a sign before a number shows that it is positive. 
 
 That part of the number system to the left of 0 is called 
 negative. ‘The negative character of a number is indicated 
 by the use of a — sign before it. — 6a, —a?,and — 112*y 
 are negative numbers. 
 
 The important idea in the two parts of the number sys- 
 tem is that of oppositeness. When anything is represented 
 by a positive number, its opposite is represented by a 
 negative number. If time A.D. is positive, then time B.C. 
 is negative. If distance to the right is positive, then dis- 
 tance to the left is negative. 
 
 ILLUSTRATIVE EXERCISES. 
 
 1. If two points are on the same meridian in latitude + 30° 
 and — 20°, respectively, how far apart are they? This means 
 that one is in north latitude 30°, and the other in south latitude 
 20°. They are evidently 30° + 20° = 50° apart. 
 
 Draw a diagram illustrating this. 
 
 2. On a certain day the lowest temperature recorded was 
 — 5° and the highest +12°. What was the difference in tem- 
 perature between the lowest and highest ? 
 
 3. A man was born in the year —31 and died in the year 
 +43. How old was he ? 
 
 4. A man travels +45 miles from A, and his friend travels 
 — 80 miles from A. How far apart are they ? 
 
 5. Two places are in —63° and +87° longitude, respectively. 
 How far apart are they ? 
 
 17. Extension of Meaning of Negative. In the above 
 exercises and illustrations we have thought of negative 
 number as beginning at 0 and extending in the opposite 
 direction from that of positive number, which also begins 
 
INTRODUCTION. 13 
 
 at 0. Beginning at zero is not a necessary part of the 
 meaning. The necessary part is that of oppositeness. 
 Number starting anywhere is negative if it denotes exten- 
 sion in the opposite direction to that of positive number. 
 
 A man travels east 12 miles, then west 4 miles, then east 11 miles, 
 then west 15 miles. If we select east as the positive direction, then 
 west is the negative direction. We may then say that a man travels 
 +12 miles, then — 4 miles, then +11 miles, and then — 15 miles; 
 as shown in the following diagram. 
 
 + 12 mi. 
 
 
 
 18. Double Use of the Signs + and —. The signs + and 
 — in algebra retain their arithmetical sense, and denote 
 addition and subtraction, respectively. When used in 
 this sense they are called signs of operation. The signs 
 + and — are also used to denote the quality of a number, 
 that is, to indicate that the number belongs to the posi- 
 tive or negative part of the number system. In this 
 sense, the signs + and — denote oppositeness, and are 
 called segns of quality. Smaller signs slightly elevated are 
 sometimes placed before a number to denote its quality. 
 
 Pemeraiiple: 75, 7D, ta, a, 7156, 718 ¢. 
 
 19. Algebraic Number. When the quality of a number 
 is considered, the number becomes algebraic. The signs of 
 algebraic number are + and —. The absolute value of a 
 number is its value without regard to quality. +a, +4, 
 +11, —T, are algebraic numbers. a, 6, 11, T, are their 
 absolute values. 
 
CHAPTER il. 
 DEFINITIONS. 
 
 20. Identity. In the equality 54—2xr=5-2 we have 
 merely the statement that 5 2 diminished by 2 2 becomes 3 z. 
 No question need be asked concerning the number or value 
 represented by z ‘The equality is true, whatever value x 
 may have. If x=1, the equality becomes 5x1—2x1 
 =$§xl. If27=5,it becomes5x5—2x5=3x5. te 
 be any number a, it becomes 6a —2a= 3a. 
 
 An equality, true for all values of the letters considered, 
 as called an identity. 
 
 In an identity both sides of the equality may be reduced 
 to the same form. 22+ 2ar+ y2—2ar=27+ y7 is an 
 identity because the left side becomes 27 + y? by uniting 
 2axand — 2 az. 
 
 21. Equation. ‘The equality x+5=7 is entirely dif- 
 ferent from that considered in the preceding Section. If 
 x be 1, the equality does not exist, for 1+ 3 is not equal 
 to 7. Neither does it exist if x be 3, for 8+ 8 18 note 
 If x be 4, the equality exists, for then we have 44+ 3=7, 
 a numerical identity. We see that the equality 7+ 3=7 
 restricts the value of x to 4. 
 
 An equality which contains one or more restricted letters 
 as called an equation. 
 14 
 
DEFINITIONS. NF 
 
 An identity is usually distinguished from an equation 
 by having its sign of equality written thus, =, while the 
 equation retains the sign =. The sign = is read “is 
 identical with” or “identically equals.” ‘The sign = is 
 read “is equal to” or “ equals.” 
 
 5a+3a= 8a is read 5a plus 8a is identical with 8 a. 
 3x%+4= 20 is read 3a plus 4 equals 20. 
 
 22. Variables. The letters of an equality which are re- 
 stricted in value are called variables. 
 
 They are generally, although not necessarily, represented by 
 the last letters of the alphabet. 
 
 In the equality « + y= 5, w and y are variables. 
 
 23. Constants. The letters of an equality which are not 
 restricted and all arithmetical numbers are called constants. 
 
 They are generally, although not necessarily, represented by 
 the first letters of the alphabet. 
 In the equality ax + by =c, the constants are a, b, and c. 
 
 EXERCISES. 
 
 Distinguish between equation and identity in the following 
 equalities; also point out the variables and constants. 
 
 1. 5a@—4e¢+4+a=2-2. 9. v—2ar+a74+2 ar=a7+22, 
 2. 8ytv—4y+6e=4y4+7u. 10. 82—52=12. 
 
 3. d4—5=—10. 1l. aw+ by=c. 
 4°124a—304+40=9a+4b. 12. 32+427=52422. 
 
 5. 4a+10=—18. 13. Te¢+a=12. 
 
 6. Zav+5ar—Sar=4ax, 14. ae+5arx=a’'x(8 — 2). 
 Toy fy 15, 15. 5u+9u—T x= 28. 
 
 8. 4ay+6ary —2 vy=8 wy. 
 
16 THE ESSENTIALS OF ALGEBRA. 
 
 24. The Root of an Equation. A value of the variable 
 which, when substituted for the variable, reduces an equation 
 to an ‘identity, ts called a root of the equation. 
 
 The equation 2 — 5 = 8 asks what number diminished 
 by 5equals 8. The answer, «= 13, is the root, for 13—5=8 
 is an identity. 
 
 The process of obtaining the value of the root is called 
 solving an equation. 
 
 25. Axioms. In solving an equation certain elementary 
 facts are taken for granted; that is, their truths are ac- 
 cepted without proof. Such self-evident truths are called 
 axioms. ‘The three following are of use to us at present: 
 
 (1) Numbers equal to the same number, or to equal num- 
 bers, are equal to each other. 
 
 Ex. 4+2=6,383+38=6; hence 4+2=—3+4. 
 
 If 3a—5=10, and 2644=10, then 83a—5=26+4+4. 
 
 (2) If equals be added to or subtracted from equals, the 
 results are equal. 
 
 Ex. 3+2=5; then8+2—2=5-—-2. 
 
 If a+b=e,thena+6b—6b=c—6. 
 
 If x+y=65,thnza+y+4=544. 
 
 (3) Tf equals be multiplied or divided by equals, the re- 
 sults are equal, 
 
 Ex. 1£=3+4H. 
 
 Multiply by 3, Ig X 8=3x3414x 3,orl0=9+4+1. 
 
 4z 20 
 
 If 4a = 20, then Se eee 
 4 4 
 
DEFINITIONS. aye 
 
 26. Solution of Exercises. The algebraic equation can be 
 used to advantage in the solution of many exercises found 
 in ordinary arithmetic. The process consists in first ex- 
 pressing the exercise as an algebraic equation, and then 
 applying the.axioms so as to find the root. This will be 
 illustrated in the following exercises. 
 
 EXERCISES. 
 
 1. A and B have $900; A has $100 more than twice what 
 B has. How much has each ? 
 
 SOLUTION. 
 
 Let x = B’s money. 
 22 + $100 = A’s money. 
 x +22 + $100 = both A’s and B’s money. 
 $900 = both A’s and B’s money. 
 By Axiom (1), z+22+100= 900. 
 By Axiom (2), +2 z=900—100 = 800 (subtracting 100 from equals). 
 oa = O00, 
 By Axiom (3), 2 = 2662, B’s money. 
 22+ 100 = 5334 + 100 = 6334, A’s money. 
 
 2. A and B have $1500; A has $300 less than 3 times B’s. 
 How much has each ? 
 
 3. What number added to twice itself will make 900 ? 
 
 4. The sum of two numbers is 84; the larger is 11 times the 
 smaller. What are the numbers ? 
 
 5. John has $300 more than Henry; they both have $2100. 
 How much has each ? 
 
 6: A house and lot cost $3700; if the house is worth $700 
 more than the lot, what is the value of each ? 
 
18 THE ESSENTIALS OF ALGEBRA. 
 
 7. Divide $720 among three men so that the second shall 
 have twice as much as the first, and the third 3 times as much 
 as the first. 
 
 8. Divide 440 into three parts so that the second. part’ shall 
 be 100 more than the first, and the third part as much as the 
 sum of the first and second parts. 
 
 9. A horse and carriage cost $288; the carriage cost 4 as 
 much as the horse. How much did each cost ? 
 
 SOLUTION. 
 Let x = cost of horse. 
 4a = cost of carriage. 
 x++#2z=cost of both. 
 $288 = cost of both. 
 By Axiom (1), 2+4#x2= 288. 
 By Axiom (2), 5x +42 = 288 x 45, multiplying by 5. 
 9x = 1440. 
 By Axiom (3), x = 1440 + 9 = $160, cost of horse. 
 #x=# of 160 = $128, cost of carriage. 
 10. What number increased by 2 of itself is 550 ? 
 11. One third of a number increased by 4 of the number 
 is 455. What is the number ? 
 12. A’s money is 2 of B’s money; together they have $1300. 
 How much has each ? 
 13. Four times a number increased by % of the number 
 is 475. What is the number ? 
 14. Two numbers added together make 80; the greater is 5 
 more than 4 times the lesser. What are the numbers ? 
 15. If to my age you add its half and its third and 50 years 
 more, the sum will be 3 times my age. What is my age? 
 
 16. If to the double of a number you add its half and 42 
 more, the sum will be 4 times the number. What is the 
 number ? 
 
DEFINITIONS. 19 
 
 17. One number is 3 of another; their sum is 156. What 
 are the numbers ? 
 
 1s. 7*@—52+112=390. Finda. 
 
 19. Divide the number 99 into three parts so that the first 
 shall be 2 times the second and 3 times the third. 
 
 20. A man bought two houses for $ 4400, paying 10 times as 
 much for one as for the other. What did each cost? 
 
 21. A man paid $700 more for one house than for another; 
 the cost of one being + of the cost of the other. What was the 
 cost of each ? 
 
 22. One seventh of a number exceeds } of it by 560. What 
 is the number ? 
 
 23. What number added to 2 of itself will make 1000 ? 
 
 24. What number diminished by # of itself will make 60 ? 
 
 25. What number increased by 4 and } of itself will make 
 
CHAPTER III. 
 ADDITION AND SUBTRACTION. 
 
 27. Arithmetical Addition. In elementary arithmetic, to 
 add two numbers, 4 and 5 for example, is to find in the 
 number system a number 9 by the process of counting, 
 first 4 and then 5 more. ‘The number so found is called 
 the swum, and the numbers added are called the addends. 
 The addition of any number of addends furnishes only an 
 extension of the above process of continuous counting. 
 If we think of number represented as heretofore upon a 
 scale, then addition may be represented as follows : 
 
 5 A 5) B 6 bs 
 
 Let us add 5, 3, and 6. 
 
 First, we count 5 from O, which takes us to A; then 3 
 more, which makes 8 and takes us to B; and finally 6 more, 
 which makes 14 and takes us to P. In this illustration O 
 is the zero point from which counting proceeds, and P is 
 the terminal point. Hence, the sum represents the counted 
 distance of the terminal point from zero. In practical 
 addition elementary sums are remembered, and thus the 
 actual counting is avoided. 
 
 28. Algebraic Addition. We have already seen that 
 ordinary algebra contains not only positive number (for- 
 ward counting), but also negative number (backward 
 
 20 
 
Oe 
 
 ADDITION AND SUBTRACTION, 21 
 
 counting). Part or all of the addends may therefore be 
 negative. Hence, the definition of algebraic addition 
 must be extended accordingly. 
 
 Algebraic addition is the process of finding a number 
 (sum) in the algebraic number system represented by the 
 terminal point reached by the successive forward and back- 
 ward countings indicated by the addends. 
 
 Thus to add 4, 5, — 6, and 2 is to count.4, then 5 more, 
 giving 9, then backward 6 to 3, then forward 2 to 5; 5 is 
 the terminal point of the successive countings and is the 
 sum of the given addends. 
 
 On the diagram the counting is from O to A, A to B, 
 Bto ©, Cto P. 
 To add 4, — 6, — 38, and +2 is to count 4, then back- 
 
 - ward 6 to — 2, then backward 3 more to — 5, and then 
 
 forward 2 to —3. The terminal point of the successive 
 countings is —3, which is therefore the sum of the addends 
 given. Show this by a diagram. 
 
 EXERCISES. 
 
 Find as above the sum in each of the following, and illustrate 
 by a diagram : 
 
 1. 4, 5, —3. 6. —8, —9, —4. 
 
 2. 6, —T7, —3. 7. 4, —8, +6, —5. 
 
 3. —4, 5, —1. 8. —6, +9, —10, +8. 
 4. 8, —9, +6, —4. 9. —4, —2, +8, +6. 
 5, 
 
 a es 10M Shee Omer 10 
 
Paya THE ESSENTIALS OF ALGEBRA. 
 
 29. Monomials and Like Monomials. Algebraic expressions 
 consisting of single terms are called monomials. 
 
 Monomials containing the same literal parts, each literal 
 part having the same exponent, are called like monomials. 
 
 62, —4ab, 3 xy, are monomials. 4a, 2a,and — 3a are like mono- 
 mials. To add 4a, 2a, and — 3a, is to count 4a, and then 2a to 6a, 
 
 and then backward 8a to 8a. 38a is the terminal point of the suc- 
 cessive countings, and is therefore the sum. 
 
 EXERCISES. 
 
 Add the following monomials: 
 4b, 5b, —2b, —3b. 5. ll abe, —10 abe, —4 abe. 
 30b, —6a*%b, —5a’bd. 6. 8a’, —52’, +727, —Oz-. 
 —5ay, Tay, —2 ay. 7. —2a’y, +4a°y, ay, —d ay. 
 6ax’, 3ax*, —10 aa’, +ax. 8 IYaxy, —Saxy, —Taxy, Lazy. 
 9. 3Vay, —2Vay, +6Vay, —V2y. 
 10. 4avz, +5avz, —9avz. 
 
 Pew N eT 
 
 30. The Commutative Law. We know that in arithmetic 
 8+4=443. This is known as the Commutative Law, 
 and means that the addends may be taken in any order. 
 The law is applicable to any number of addends. This 
 law holds in the extended number system of algebra. It 
 is algebraically stated as follows: 
 
 a+b=b+a. 
 
 The following diagrams show the truth of this law: 
 
ADDITION AND SUBTRACTION. 23 
 
 That a—6b=—6-+4a is shown as follows: 
 O a B 
 Y i =? 
 B “0 O 
 PRAMS iva Dare qa ee re Vig gi.) 
 +c 
 
 The distance from O to P is the same in each case. 
 
 31. The Associative Law. We know that in ordinary 
 arithmetic, in adding 2, 3, 4,5, we may, if we wish, first 
 add 3 and 4, and add their sum to 2, and then add 5. 
 We may, in fact, associate the addends in any manner we 
 choose. ‘This is known as the Assoczative Law. ‘This law 
 holds in the extended number system of algebra. It is 
 algebraically stated as follows: 
 
 a+tb+ec=a+(b+e). 
 
 This law, like the Commutative Law, applies to both 
 positive and negative numbers. 
 
 32. Addition of Monomials. Let us find the value of the 
 expression 
 Sa— 2a +47 — (a+ 82. 
 
 By applying the Commutative Law we may write this 
 expression in the equivalent form 32+427+82—22—1 72. 
 Now, by the Associative Law we may add all the positive 
 numbers into the one sum of 152, and all the negative 
 numbers into another sum of —92, and thus write the 
 expression in the equivalent form 15% — 92, which we at 
 
24 THE ESSENTIALS OF ALGEBRA. 
 
 once know to be 62. All this may be shown in the 
 following scheme: 
 8a—2r74+42e7-—T2r14+82 
 =3r+4xv+8x—2x—T2x, by Commutative Law. 
 = lbr—9za, by Associative Law. 
 Sad by adding. 
 The sum of a positive and negative number is equal to 
 
 the difference of their absolute values with the sign of 
 the greater prefixed; e.g. 6—4=+ 2, while 9-—13=—4. 
 
 Rute. To add like monomials, add the positive terms, 
 then add the negative terms ; to the difference of the absolute 
 values of the two sums prefix the sign of the greater. 
 
 EXERCISES. 
 
 1. Add 3a, —5a2, —9a, +112, —32, +72. 
 
 We may for convenience arrange the solution thus: 
 
 
 
 + 32 — 52 
 lice — 92 
 ie ae 3a 
 2Zlix —17e% 
 
 =172 
 
 + 4% 
 
 It will be noticed that we have arranged in the first column 
 all the positive nuinbers, and in the second all the negative 
 numbers. This is a mere matter of convenience. The stu- 
 dent should early accustom himself to pick out and unite the 
 positive numbers mentally, and likewise the negative numbers, 
 merely writing down the results. Unless the coefficients are 
 very large, a little practice will enable the student to do this 
 with accuracy and rapidity. 
 
ADDITION AND SUBTRACTION. 25 
 
 Add 5ayz, —11 xyz, 18 xyz, —4 xyz, —13 xyz. 
 Add 15 a’bc, 10 abc, —2a’bce, — 18 a’be. 
 
 Add 1027, —52’, —2a’, 1627, —82’. 
 
 . Add $ax, tax, —fax, —lLax, — az. 
 
 Mads V2 e,.—b5V22, —8V 2a, —V2 2, 6V2 a. 
 7. Add 4(a+6), —3(a+b), —4(a+)b), 2(a+0b). 
 
 Algebraic expressions containing like quantities within a paren- 
 thesis may be added as monomials. Thus, in Exercise 7, the quantity 
 a + 6 is common to each addend, hence the sum is — (a + b). 
 
 8. Add 5(@’+ 7’), —3(a’?+y"), T(a’+y’), —8(a’+y’). 
 9. Add t(ax+ by+2), —3(ax+ by+2), (av+ by+ 2). 
 lo. Add 3(@*—2y), —§@’*—2y), —4@’—2y). 
 
 a1. Add 3(Vz+Vy), —8(Va+vVy), 6(Va+vy). 
 
 oy? eT op? 2 
 12. Add eet) Al eiiatimto) eae) 
 
 33. Addition of Polynomials. An algebraic expression 
 consisting of more than one term is a polynomial. 
 
 If it has two terms, tt is called a binomial: and tf three, 
 a trinomeal. 
 
 A Aa Pw DW 
 
 If we desire the sum of 4a + 6 and 3a — 46, we may indicate that 
 sum thus: 4a +60+3a—4bD. 
 
 Now, by the Commutative Law, we may write this 4a+3a+b—44, 
 which by the Associative Law becomes 7 a—3b. 
 
 However many polynomials we might have to add, the 
 process would be an extension of the above. We have 
 then the following rule for adding polynomials: 
 
 RuuE. Unite the like terms of the various polynomials 
 into sums, and connect these by their a signs to form 
 the polynomial sum. 
 
26 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 1. Add 2ax4+5by+5cz, 4av—S5by, Tby—8 cz, 1lax— 
 4 by — 6 cz. 
 
 For convenience we may arrange these polynomials thus: 
 
 Zax+3 by+5cz 
 4 ax —5 by 
 T by —8 cz 
 
 11 aw — 4 by — 6 cz 
 1Tax+ by—4cz 
 
 It should be noticed that in the arrangement like terms have 
 been placed in columns. 
 
 2. Add 5a+5y+42, —3a—5y+z2, Tx—2y—z, 10%+y—2z. 
 
 3. Add 5ax—6 by+cz, Sax+by+5cz, —4ax+3 by—8 cz, 
 30 ax—3 by +3 ce. 
 
 4. Add 42°+3y'?—4, —30°+27?—6, 2? —y? +7. 
 
 5. Add 380° +4Y7+432?, —e?—47°4+222, 11 e—yY—-—2 
 
 6. Add 4ay—S8 yz2+7 2a, 4yz—Tay, 3yz+420, —4ayt+ 
 320 — 2 yz, — xy — 2H. 
 
 7. Add 4Vay+3Vyz+V2zx, —2Vyz2—3V2x, —2Vay— 
 BV y2 + -V 20. 
 
 8. Add 8abec—4ayz+Tlmn, —4lmn+12ayz, 5 ayz— 
 4abe+3lmn, —5 abe —T lmn. 
 
 34. Identity. The sum of the addends is identically 
 equal to the sum. 
 
 Thus, 8a+4xe—52x7=22. 
 
 Also, 2a+6+5a—4b—6a+4b=a+4+6. 
 
 Since an identity is true for all values of the letters involved, we 
 may make use of the identity existing between addends and sum to 
 
| ADDITION AND SUBTRACTION. Q7 
 
 verify our results in addition. If in the first illustration above we 
 make x = 1, it becomes 
 34+4—5=2, or 2=2. 
 
 If in the second illustration we put a=1 and b =1, it becomes 
 24+14+5-4-64+4=1+41, or2=2. 
 
 The use of 1 for each of the letters is convenient, but not necessary. 
 We may use any value whatever. Thus, in the second illustration 
 
 above, we may put Hee Sinviat Bele 6 
 The identity then becomes 
 6+5+415-20-—18+4+20=3+4+5, or 8=8. 
 
 EXERCISES. 
 
 Add the following quantities, and verify the results by sub- 
 stituting particular values for the letters used: 
 
 1. 35a+4)b), 5a—6b, Ta—4c, 5b+411¢. 
 4%—3y, 5a+3y, —Ta+y, 3u—2y. 
 2¢7—30, 50+40, 7T@7?—5B, —38a?— 02 
 5a +4ay+3y’?, 2e°—B5ayt+6y’, 32’°—8y’, 8 ay. 
 —dsaxr+4by+e, 5ax—6by—38c¢, aw—3by4+2e 
 12%@—3y+2, 6a¢—4y4+72z, —8x—38y4+382. 
 etyte2, sar+4y+522, —4e°—8y'?4+ 27. 
 w—3y, be+4y4+3, vty, y—swe. 
 8Va—4Vy, 5Va+8Vy, —6Va2+Vy. 
 
 10. 4a? —5b?+65b, 3a? +203—40b. 
 
 Nore. a? means Va; be means Vb. 
 
 11. 5Va—6Vb+¢, —3Va+4vb—4e, 2Va—3V0 + 2c, 
 N/a —38V/b4+3¢. 
 
 12. Saxy+4 byz4+ 6czu, 2 byz —5axy+ 6czx, —4axy + 2 byz 
 —Tczx, 2axy — byz — 7 cea. 
 
 oO YO ah wD 
 
28 THE ESSENTIALS OF ALGEBRA, 
 
 35. Subtraction. Subtraction is the process of finding 
 one addend, when the other addend and the sum of the twe 
 addends are given. 
 
 The given addend is called the subtrahend, the given 
 sum the minuend, and the addend which is to be found 
 the remainder or difference. 
 
 Referred to the number system, we may say that to 
 subtract a from 6 is to find the amount and direction of 
 counting necessary to pass from a to 6, both a and 6 begin- 
 ning at the zero point. 
 
 This is shown on our number system diagram as follows: 
 
 b 
 
 a. 
 
 If P is the terminal point of the subtrahend a and Q 
 that of the minuend 6, then the distance and direction 
 from P to Q is the result of subtracting a from 6. 
 
 36. Cases of Subtraction. A consideration of the defini- 
 tion of subtraction gives us the following four cases: 
 
 (1) If we subtract a positive number from a positive 
 number, the remainder is the arithmetical difference of the 
 absolute values, positive or negative, according as the 
 absolute value of the subtrahend is less or greater than 
 the absolute value of the minuend. 
 
 For example: 
 
 Ta—(+38a)=4a, Ta—(+10a)=— 38a. 
 
ADDITION AND SUBTRACTION, 29 
 
 (2) If we subtract a negative number from a positive 
 number, the remainder is the positive arithmetical sum of 
 the absolute values. 
 
 Forexample: 52—(—32)=8u. 
 
 (3) If we subtract a positive number from a negative 
 number, the remainder is the negative arithmetical sum of 
 the absolute values. 
 
 For example: —6ab—(+4ab)=-—10ad. 
 
 (4) If we subtract a negative number from a negative 
 number, the remainder is the positive or negative arith- 
 metical difference of the absolute values according as the 
 absolute value of the subtrahend is greater or less than 
 the absolute value of the minuend. 
 
 For example: 
 
 —Ta—(—I1l2)=42, —138¢%—-(—-9xv)=— 4-2. 
 
 The student should verify the truth of these cases by 
 testing them on the number system diagram. 
 
 EXERCISES. 
 
 From 192*y subtract 11 ay. 
 From 7ab subtract — dab. 
 From —Q9ayz subtract 12 xyz. — 
 From —17a subtract —15¢. 
 From —142? subtract — 23 27, 
 From 27 a’y subtract 43 a*y. 
 From 157’ subtract 15 7’. 
 
 From —7 aba subtract — 7 aba®. 
 From 17 Vzy subtract — 12 Vay. 
 
 See Se Sie ee Nac 
 
30 THE ESSENTIALS OF ALGEBRA. 
 
 10. From 11(a+ 5) subtract 15 (a +4 b). 
 
 11. From —6(e@+y) subtract 9(@ +7). 
 
 12. From —11(a’+ 0’) subtract — 7 (a? + 6’). 
 
 13. From 18 (ay + yz+ 2%) subtract — 7 (ay + yz + zz). 
 14. From 23(aa + by +c) subtract 40 (ax + by +c). 
 15. From —16(y’?+4 ax) subtract —21(y’+4 az). 
 
 37. Subtraction of Monomials and Polynomials. In the 
 cases considered in Section 86 it may be noticed that the 
 subtraction of any number is equivalent to the adding of 
 an equal opposite number. 
 
 Illustrations : 
 
 Ta—(+4a)= Ta+(—4a)=3a. 
 Ta—(—4a)= Ta+(+4a)=11a. 
 —Ta—(—4a)=—Ta+(+4a)=— 3a. 
 —Ta—(+4a)=—Ta+(—4a)=— Ila. 
 
 Hence, we have the following rule: 
 
 Rute. To subtract one number from another number, 
 change the sign of the subtrahend and proceed as in addition. 
 
 For example: The problem, from 18 ax*y take — 5 aa’y is 
 equivalent to this problem, to 18 aa’y add 5 ax*y. 
 
 In practice the change in sign should always be made men- 
 tally. 
 
 Rue. To subtract one polynomial from another, change 
 the sign of each term of the subtrahend and proceed as in 
 addition. 
 
 EXERCISES. 
 n bk From 9e°+ 38ay—117°4+7 
 Subtract 7a? —5ay—17y°+9 
 2e°+8ay+ b6y—2 
 
 
 
ADDITION AND SUBTRACTION. ot 
 
 In this we think of the 7 2? as negative, and add it to 92”, 
 giving 2a. The —5ay is thought of as positive, and added 
 to 3ay, giving + 8zy, and so on with the other terms of the 
 subtrahend. 
 
 2. From 9a? —11a?+5y subtract 14a°—7a’—8y. 
 
 From 11 at — 15 b* — 13 ab? subtract 7 a* + 108*— 3 a7b*. 
 From 4ay + 3a"? —11y’ subtract 7 vy —42?— 13 7°. 
 From 18 ax —14 by + 11c subtract 5 ax + 19 by —5e. 
 From 5 Va +8 Vy —134 subtract 8 Vx—5Vy+3a. 
 7. From 192°+72—5 subtract 42° + 11a? —8. 
 
 On See 
 
 Arranging for subtraction, we have 
 192 +Ta2—5 
 4771 oe —§ 
 15e¢ —1l1l¢?’4+Te+3 
 In the minuend there is no term in x’, or we may say there 
 is the term 0a. Hence the 112’ is to be subtracted from 0 2”, 
 and of course gives a remainder of —112a*. Inthe subtrahend 
 there is no term in 2, so there is nothing to subtract from 7 a, 
 or the remainder is 7 a. 
 
 
 
 8. From d2* —3 2° +-32-+6 subtract 6a* —32°+ 2? — 5. 
 9. From 4 277’? +6ay?—3a*y+y' subtract 3 y'+4 ay—6 ay. 
 10. From 3(a+6)—5(@#+y)+6c subtract 6(@+4+ y)— 
 5 (a+b) —5d. 
 11. From 14 -Vxv+13 y?—16 x subtract 15 y8+17 y—12 Vay. 
 12. From 4az+5y7?—62z+13 subtract 21417 «z—15awu 
 +11 y. 
 13. From 18 4? — 272° + 42 yz subtract 18 yz — 36 y? +17 2’. 
 14. From 36 a’ — 27 a®b —17 ab? + GO subtract 13 6°+ 17 ab? 
 — 37 ad. 
 
 15. From 137? +17 «wy — 5 subtract 18 y?—17 yz +32. 
 
O2 THE ESSENTIALS OF ALGEBRA. 
 
 38. Identity. Subtraction may be expressed as 
 Minuend — Subtrahend = Remainder. 
 
 This is an identity and will, therefore, be true for any 
 values of the letters involved. ‘This identical relation 
 furnishes a convenient method for verifying the results 
 of subtraction. 
 
 For example: Tab—14a?+ 110? Minuend. 
 4ab—11a2+4+ 19062 Subtrahend. 
 8ab— 8a%*— 806% Remainder. 
 
 If we put a =1, b =1, we get 
 (PUES SO Beaten o 
 4—1) 4°19 = 12 
 eg oe ee: 
 
 which shows our result to be true. 
 
 EXERCISES. 
 
 Perform the following subtractions and verify by putting 
 the letters each equal to 1: 
 
 1. From 9a?+ 11 ab—18a take 8 a?+14ab —9a. 
 
 From 3a—4y+7 take «—2y+8. 
 
 From 42° +57'+7 ay take 52’? —4ay4+3y?, 
 
 From 14 aw +12 by +7 take —3 ax + 2 by—5. 
 
 From 4(@+ y)+72—4 take 3@+y)—52—8. 
 6. From 16(a’?+ ay +’) +2 —w’ 
 
 take 12(a + ay+y?)-4245w". 
 7. From 2(b?—4 ac) + 2+ 7? take 5(6?—4 ac) + 52?—6 7. 
 8. From ax+ by+c take a'v+ dy +c’. 
 
 Gis Ph WN 
 
ADDITION AND SUBTRACTION. 33 
 
 9. From aa’ + 2 hay + by? +2 gx +2 fy 
 take ax? +2 h'xy + b'y?+2g'e@+2 flyt+ cl. 
 10. From 2+ 7’?+ aw + by take 32? +3 y+ le+ my. 
 11. From aV2+bVy-+c take a Va2+ b'Vy +c’. 
 12. From 5-V2? + 9? + 7-V/y? + 2 take 8-2? +? +8V y+ 2. 
 
 39. Removal of Parentheses. If we recall the principles 
 of addition and subtraction already developed, we can by 
 means of them remove parentheses preceded by the + or 
 — signs. 
 
 For ~a+(6—c+d)=a+b—c¢+d, 
 and a—-(6—e+d)=a—b+ec-—d. 
 
 Hence, a parenthesis preceded by a + sign may be omitted 
 without any change in the signs of the terms inclosed. A 
 parenthesis preceded by a — sign may be omitted if the sign 
 of each term within it is changed. 
 
 For example: | 
 
 ax+(3ab—4cy—3az)=axr+5ab—4 cy — 38 az. 
 ax —(3ab—4cy—8az)=ax—38ab+4cey4+ daz. 
 
 EXERCISES. 
 
 Remove the parentheses in the following and unite like terms: 
 1. 3a—(2a+4b—c). 2. 37—4y—(2e%+a—3y). 
 3. 5a’?—(b6+3a’) —4b—(2a°—38 D). 
 
 4. 3ax—[a?—3 ax + ay — (ay —5 aw +a’). 
 Remove the [ ] first, and we have 
 3 aa —a’?+3 ax —ay+ (ay —5 ax +a’). 
 Now since the parenthesis is preceded by a + sign it may 
 
 be omitted, and we have, when we unite the like terms, aa as 
 the result. 
 
34 THE ESSENTIALS OF ALGEBRA. 
 
 2a—8b—[7a—(4b—5a—b425)}. 
 
 3av—5 by —[7 by + 3 aw — (5 ax —3 by). 
 lla—(Ty—32)+5y—[8x—(4y+5a)]. 
 5ae—-{Ta—[3e—(5a—2e%—a)]t. 
 
 17a12e— fi5a—ll2+(8a—22)=—be— fat 
 
 10. 2~— {6y+[6z2—(@—y+2)—22]+d3y}. 
 
 In exercise 10 and the following put «=5, y =4, and z= 2, 
 
 and find the value of the expressions. 
 
 Ml. by—32+fa—y+22-—(8a42y—42)]. 
 
 12. 32—{2a—[52—(4y—Tx—2y4+2)—3a]+2 yf. 
 13. Qn+3y—(4y—z2+a) —[5e—(Ty4+82)]. 
 
 14. xy — (y2— 2a —2 xy — 2 YZ). 
 
 15. 38ay— {4e°—-y—[2Zay4+3a°—5y]—-CBx2’—-4y)}. 
 
 oO OND gw 
 
 
 
 40. Insertion of Parentheses. Many times it is just as 
 important to insert parentheses in an algebraical expres- 
 sion as it is to remove them. Evidently any number of 
 terms may be inclosed in a parenthesis without change if 
 the parenthesis is preceded by a + sign. Any number 
 of terms may be inclosed in a parenthesis preceded by a 
 — sign, if the sign of every term so inclosed is changed. 
 
 For example: 
 
 ax + by —cz+ab=axr-+ (by —cze+ab), 
 ax + by — ce+ab=ax—(— by + cz —ab). 
 
 The number in the parenthesis is thought of as one quan- 
 tity, and hence may be considered as one term. ‘This gives 
 an extension in meaning to the word term. In the illus- 
 trations just given the expressions on the left have four 
 terms, but those on the right have only two terms. 
 
ADDITION AND SUBTRACTION. oO 
 
 EXERCISES. 
 1. 3a?4+4ab—3c+d. Inclose the last three terms in a 
 parenthesis preceded by —. 
 
 2. ay+ by —cy—w«y. Inclose the last three terms in a 
 parenthesis preceded by —. 
 
 In each of the next four exercises inclose all the terms con- 
 taining @ in a parenthesis preceded by —. 
 3. 59—38xu+au—4abe. 4. Ty+d3z2+20—384u—5ypx. 
 5. 15—5ax+112—16 a2-+ 27a. 
 6. 27 y—30y 4+ 22ay—13b4 4 212%—50a4+16b. 
 In the following exercises put all the terms containing a in 
 a parenthesis preceded by —, and all the terms containing y 
 in a parenthesis preceded by —. 
 7. 15—8a+4y—S5av+T7ay+3ab—5 by. 
 Rearranging, we have 
 15—8a—5ax+3ab+4y4+T7 vy — 5 by. 
 Then 15—(8a+5ax%—3 ab) —(—4y—T ay +5 by). 
 8. 252+5y—824+7 ax—11 by+cew—38ac+11. 
 9. b°—ab+cy—2ac?—4b*y4+1la—4y. 
 10. 3%+2a—4y+5e2—6ab+3yz2—5au+11 by. 
 
 41. Adding and Subtracting with regard to a Named Letter. 
  Ta+5a—6a might be written (7 +5—6)a. So if we 
 had 7 az + 5 bx — 6 ex, we might write it (Ta +56 — 6e)z. 
 In this, z has been chosen as the element of likeness or 
 the denomination in the three terms. We have added the 
 coefficients, but in this case the coefficients are unlike and 
 we can only indicate the addition. We can subtract 5 aa? 
 from 7 ba? if we consider 2? as the element of likeness. 
 The remainder is (7 6 — 5a) 2%. 
 
36 THE ESSENTIALS OF ALGEBRA, 
 
 EXERCISES. 
 Unite with respect to x, 5ax +7 ay —3 xz. 
 Unite with respect to a, 3ax+4ay— daz. 
 Unite with respect to k, 4k +3 yk —12 zk. 
 Unite with respect to x and y, 4ax+6 ay —5 ba +12 cy. 
 
 YE ee Naa 
 
 Unite with respect to x and y, 
 3ax +4 by — (2 ax + 12 by). 
 6. Unite with respect to 2’, 
 ba + 16.07 — 12 be a 
 7. Unite with respect to a, y, 2, 
 3a+4y4+32—(5%—8y—T2). 
 (83—5)a+(44+8)y+(847)2=24412y7+4102. 
 8. Unite with respect to # and y, ? 
 da—2y4+7—(2a4+3y+4). 
 9. Unite with respect to a, y, and z, 
 —5a4+8y+724+ 6a—4y+4+2). 
 10. Unite with respect to the powers of a, 
 3802 —52?—8e4+ 72? —162+4 202°—212+6. 
 11. Unite with respect to the powers of w and y, 
 aa” + by’ + ca + dy +e — (a'x’ + bly’? + clu + d'y + e'). 
 12. Unite with respect to 2’, xy, and 7’, 
 4a? —3 ay +129? — (20? —4 ay +8 y?) + 16 ay. 
 13. Unite with respect to powers of k, 
 12¢+16yk—64%4122k4+ 807k? +12%°4+ 2%. 
 14. Unite with respect to a’, y’, 2°, and ay, 
 aa” + 2 hay + by? + cz? — (a'a? + 2 h'ay — b!y? + c'2?). 
 
ADDITION AND SUBTRACTION. Si 
 
 MISCELLANEOUS EXERCISES. 
 
 1. Aand B have $550; A has $100 more than twice as 
 much as B. How much has each? 
 
 2. Add 11 aa’?+138 ba’y+9 cay’?—5 dy’, 4 ba’y—38 aa’? +16 dy’, 
 31 ba’y — 24 cay”, and 10 dy’ —8 cay? + ax’. 
 
 3. Remove the parentheses and simplify, 
 
 13 2 —{8y—5a— (8e4+T7y)—2Zyi— (12%+8y422). 
 
 4. Unite with respect to a and y’, 
 
 aw — 5 a? + 34? — by? + ca? — dy’. 
 
 5. From 5a°— 6 a’b —7 ab?+11 0° take 100?'—3a'°+ 5 ab? 
 — 12a’), 
 
 6. By means of a diagram show that the sum of 8 and 
 —10 is —2. 
 
 7. By means of a diagram show that —10 taken from 8 
 leaves 18. 
 
 8. By means of a diagram show that 8 taken from — 10 
 leaves — 18. 
 
 9. With «=4 and y=—5, find the value of 
 Sy—j{sa—[4y4+2xe—(6e—3y—52a)}}. 
 10. With the same values of w and y find the value of 
 {8e—[3y+2e—(5y—382)]} (8e-—3y—2). 
 
 11. A father’s age is 10 years more than 3 times his son’s 
 age; the sum of their ages is 82 years. Find the age of each. 
 
 12. Add, 8y¥y4+3ay+112, Ty—138ay—212, —14y+ 
 10 ay +1227, and 2y—5 ay — 2 2. 
 Verify your result by putting y= 2 and x=1. 
 
 13. From ie ape ob 
 take 8a+ Sab—-1006+11 
 
 Verify your result by putting a=1 and b=2. 
 
 
 
38 THE ESSENTIALS OF ALGEBRA. 
 
 14. Add 8(a+b) —16(#—y) + 11Vaae, 5(@— y) + 2Vax— 
 7(a+b), and —38(a+b) +11(@—y) —14Vaz. (Regard the 
 quantities in each parenthesis as a single term.) 
 
 15. From 8(a?— 0b’) —17(«#+y) —11vV2'+/ 
 take 7 (a? — b*) — 20(@+ y) + A/a? + y? 
 
 16. Show by a diagram that the sum of 8, —6, —3, +2, 
 and —5 is —4. 
 
 17. Unite with respect to a and 2, 
 
 oa+8a—4a—Tx—ba+ bat cx — de. 
 
 18. 8%e+3y—(«+y)+2yz. Inclose all the terms after the 
 first in a parenthesis preceded by a — sign. 
 
 19. A farmer exchanged a bushels of wheat at 6 cents a bushel, 
 and ¢ bushels of corn at d cents a bushel, for a mowing machine 
 costing e dollars, and for calves at f dollars each. How many 
 calves did he get ? 
 
 20. John, James, and Henry together have $216. John 
 has one half as much as James, and Henry has as much as 
 both John and James. How much has each ? 
 
 21. Simplify 8a°4+3ay—4y+2ay+3y—Tw+4y7— 
 Q2ay—Ty+5e?+3ay—4y? +102” 
 22. Simplify 17-—134+44a?— #—(S6y+ 21¢+52— 11). 
 23. Ifa=—1, y=2, and z= 4, find the valine on 
 wye — fa? —[y? + 22 — (22 —3ay—4y)] + 227}. 
 24. With the same values of a, y, and 2, find the value of 
 Byz+{8a—[4y+2— (Say +4 yz) —8ayz]— 34h. 
 25. Add 3(a+y)— 4(a?—b)+ 3V/—P 
 —Vety +11 —b) —16VP SF 
 4(ea+y)— 8(@—d)— 4/75 
 B(a+ty)— 8(a@—b)— IVP —F 
 
CHAPTER IV. 
 MULTIPLICATION AND DIVISION. 
 MULTIPLICATION. 
 
 42. Integral Multiplication. ‘To multiply 7 by 8 may be 
 taken to mean that 7 is to be added 8 times. 
 
 TH+74+74+74+74+7+74+7=56. 
 
 In such an operation 7 is the multiplicand, 8 is the multt- 
 plier, and 56 is the product. Similarly, to multiply a by 6 
 is to find the sum of a added 6 times: 
 
 ata+ta+--(b times) = ab. 
 
 In this, a is the multiphcand, 6 the multiplier, and ab 
 (vead, a multiplied by 6) is the product. It is clear that 
 this view of multiplication has no meaning when the mul- 
 tiplier 6 is fractional or negative. We can not add a one 
 half times, or three fourths times, or three and a half 
 times; neither can we add a negative four times to obtain 
 the product of a by —4. All this shows that if we are to 
 have multiplication of algebraic numbers, we must extend 
 our definition. No change is made in the meaning of 
 multiplicand, multiplier, and product. 
 
 43. Multiplication Defined. Multiplication rs doing to the 
 multiplicand what has been done to unity to produce the 
 multiplier. 
 
 39 
 
40 THE ESSENTIALS OF ALGEBRA. 
 
 For example. To multiply 6 by 5 is to do to 6 what has 
 been done to unity to produce 5. Unity has been added five 
 times (1+1+1+1-+1=5) to make 5, hence we must add 6 
 five times to get the product 6x5. Or, we may say that 
 unity has been taken five times to produce 5, hence we must 
 take 6 five times to produce 6 x 5. The new definition is seen 
 to agree with the arithmetical notion of multiplying one integer 
 by another. ‘To multiply 6 by 2 is to do to 6 what has been 
 done to unity to produce 2. Unity has been divided into 3 
 equal parts and 2 of them taken to produce 2. Hence to 
 multiply 6 by 2 we must divide 6 into 3 equal parts and take 
 2 of the parts. It is thus seen that the new definition includes 
 multiplication of fractions. The application of the definition 
 may be seen by considering the following simple concrete 
 problem. 
 
 A water tank with a capacity of 1000 gallons contains at 
 the present time 600 gallons. Let us consider the following 
 four cases: 
 
 (1) If it have a supply pipe carrying 10 gallons an hour, 
 how much water will be poured into the tank in the next 
 84 hours ? 
 
 Evidently the amount poured in is 10 x 81. We must do to 10 
 what has been done to 1 to make 8}. To produce 84, 1 has been 
 added eight times, then separated into 2 equal parts and one of the 
 parts added. Hence we must add 10 eight times, giving 80, then 
 separate 10 into two equal parts of 5 each and add the 5 to 80, giving 
 85. Hence the amount poured in is 85 gallons. 
 
 (2) With the same rate of flow, how many gallons must be 
 added to find the amount in the tank 6 hours before the pres- 
 ent time ? 
 
 In this case the 6 hours is negative, as it is the opposite of the 
 time used in the preceding problem, which we considered positive. 
 Our problem now is to multiply 10 by — 6. We must do to 10 what 
 
MULTIPLICATION AND DIVISION. 41 
 
 has been done to 1 to produce — 6. — 6 is produced by subtracting 
 1 six times. Hence to multiply 10 by — 6, we must subtract 10 six 
 _ times, which gives — 60. Hence we must add — 60 gallons to 600 
 gallons to get the contents 6 hours before the present time. 
 
 (3) A discharge pipe is flowing at the rate of 5 gallons per 
 hour. How many gallons must be added to give the contents 
 10 hours from now ? 
 
 Since the inflow rate was positive, the outflow rate is negative. 
 Hence, our problem now is to multiply — 5 by 10. We must do to — 5 
 what has been done to 1 to produce 10; that is we must add — 5 ten 
 times, which gives — 50. So — 50 gallons is the amount to be added. 
 
 (4) The discharge pipe has been running at the rate of 6 
 gallons an hour for a number of hours. How many gallons 
 must be added to give the contents of the tank 8 hours ago? 
 
 In this case both the 6 and 8 are negative, and the problem is to 
 multiply — 6 by — 8. We must do to — 6 what has been done to 1 to 
 produce 8; that is, we must subtract —6 eight times, which gives 
 
 +48. So we must add 48 gallons to get the contents of the tank 8 
 hours ago. 
 
 The results of the four cases may be arranged thus: 
 10 x 8§ = + 85, 
 10 x —6= — 60, 
 —5x10= — 50, 
 (—6) x (—8)= +4 48. 
 Tf we should make our reasoning perfectly general by letting 
 
 the rate of flow be a and the number of hours be 8, then the 
 four cases would take this form: 
 
 oxo =-+ab, 
 ax (—b) = —ab, 
 —axb = — ab, 
 
 (—a) x (— 6) = + ab. 
 
49 THE ESSENTIALS OF ALGEBRA. 
 
 44. Law of Signs in Multiplication. From the definition 
 and the above considerations, we see that the product is 
 4- if the multiplier and multiplicand have like signs, and 
 it is — if they have unlike signs. 
 
 In multiplication, like signs in multiplicand and multiplier 
 give a positive product and unlike signs give a minus product. 
 
 45. Continued Products. Products produced by three or 
 more multiplications are called continued products. 
 
 If 6 boys buy 5 oranges each at 3 cents apiece, the total cost of the 
 oranges is 3x 5 x 6 cents. If 8 groups of 6 boys each should buy 
 oranges as above, the total cost of the oranges is 3 x 5 x 6 x 8 cents. 
 If instead of using arithmetical number we should use algebraic num- 
 ber, and say that there were d groups, c boys in a group, and each 
 boy bought 6 oranges at a cents apiece, then the total cost of the 
 oranges would be a x b x ¢ x d = abcd cents. 
 
 46. Factors. As in arithmetic, the numbers multiplied 
 together are called the factors of the product. 
 
 a, 6, and e are factors of abe. 
 
 5, a, 6, and e are factors of 5 abe. 
 
 In the latter case, 5 is considered a numerical multiplier 
 or coefficient, and is usually so designated, instead of being 
 called a factor. 
 
 EXERCISES. 
 
 Point out the factors and numerical multipliers in the follow- 
 irg products: 
 
 1. abex. 4. 11(22)(3y)z. 
 2. Sabdz. 5. (52) (4a) (82). 
 3.17 xX 9(8 2) =o) x92. 6. (Sy) (42) (2 x y). 
 
MULTIPLICATION AND DIVISION. 43 
 
 47. Signs of Continued Products. 
 (—3)x2x8 =—6x3 =— 18, 
 (—3)x(-—2)x3 =+6x3 =+18. 
 (—8)x(—2)x(—3)=+6x(—3)=— 18. 
 
 If we use algebraic instead of arithmetical number, the above 
 
 may be written: 
 
 (—a)xbxe =(—ab)xe = — abe. 
 (—a)x(—)b)xe =(+ab)xe = + abe. 
 (—a)x(—6)x(—e¢)=(+4 abd) x (— ¢) =— abe. 
 
 We see that one negative factor produces a negative 
 
 product, two negative factors a positive product, and three 
 negative factors a negative product. Hence, 
 
 Products resulting from an even number of negative fac- 
 tors are +, those resulting from an odd number of negative 
 factors are —. 
 
 EXERCISES. 
 Give the signs of the following products: 
 
 1. (—2a) x (5a)(—2). 
 (—a)(—b)(—0)(-@ 
 a(—b)(—0)d(—e) f(—9) (— A). 
 '(— a) (— b) (y*) (— 82). 
 17 (—5) (— 3) (— 2) (eyz)(—2)(—8). 
 48. Commutative Law of Factors. In multiplying to- 
 
 gether numbers in arithmetic, the factors may be taken in 
 any order without changing the product. 
 
 Hence |. SS OD: 
 eee a OX OX OS ox OS 8K Sx 2 
 meee x. ah xD Kh 2 = 30. 
 
 Oe SS 
 
44 THE ESSENTIALS. OF ALGEBRA. 
 
 If we construct a rectangle of length 5 and width 8, the 
 area is 5x38. If we construct another rectangle of length 
 3 and width 5, the area is 8 x 5. 3 
 
 qQ 
 
 These two rectangles are readily seen to be equal, for 
 the second one is merely the first one placed on end. 
 
 Hence, 5x8=8.x 5. 
 The reasoning would be exactly similar if we should use 
 the general rectangle whose length is a and width 6, and 
 the other general rectangle whose length is 6 and width a. 
 In this case our conclusion is 
 ake Osed. 
 
 The product is the same whatever the order of the factors. 
 
 This is called the Commutative Law of Factors. 
 
 This law holds for all algebraic numbers. 
 ax (— b)=(— b)a =— (ba) = — (ab) = — ab; 
 or, since —b=(—1)4, 
 ax(—b)=ax(—1) xb=(—1lab =— ab. 
 EXERCISES. 
 1. Commute the factors of 3 ay. 
 SRY = 3 YX SUSY = yd = YOM eee 
 
 2. Commute the factors of — abe. (—1 is a numerical 
 multiplier.) 
 
MULTIPLICATION AND DIVISION. 45 
 
 3. Commute the factors of ab(@+y). (The quantity in 
 parenthesis is a factor.) 
 ab(a + y)=a(@+y)b=(a+y)ab=(a+y)ba=b(x+y)a=ba(x+y). 
 4. Commute the factors of (a + b)c(x + 9). 
 5. Commute the factors of (a + b)(¢ + d)(e+/). 
 
 49. Associative Law of Factors. If 6 boys buy 4 oranges 
 each at 5 cents apiece, the total cost is 8 cents x 4 x 6. 
 We may think of this as each boy paying out 3 cents x 4 
 and the 6 boys as paying out (3 cents x4) x6; or we may 
 think of total number of oranges bought, 4 x 6, and then of 
 the cost of these as 38 cents x (4 x 6); or finally we may 
 think of the six boys as buying 1 orange each, at a cost 
 (8 cents x6), and then the total cost of 4 oranges each is 
 (5 cents x6) x4. Hence, we have 
 
 Ween (5 X4)x 6=3 x (4x 6)=—(8 x 6) x4. 
 If instead of the arithmetical numbers, 3, 4, and 6, we use 
 the algebraic numbers a, 6, and c, we have 
 
 eee a XO) xX C= a xX(b xX ¢)=— (a x ¢)x O. 
 
 The above is an illustration of the Associative Law of 
 Factors, which may be stated as follows : 
 
 The factors of a product may be grouped in any order. 
 Ge exo —(ab) X (cd) =a(bx e)d=a(bxexa). 
 8x(—8) x (—4) x(—2)=1[8x (—3) ]} x {(-4) 
 
 x (—2)}= etc. 
 
 50. Distributive Law of Factors. ‘The multiplication of 
 4465 by 7 may be indicated thus: (4+5) x7. The opera- 
 tion may be carried out in either of the following ways : 
 
 (4-5) x l= 9 x T= 63. 
 (44+5)7 =4x74+5xT=28+4 35= 63. 
 
46 THE ESSENTIALS OF ALGEBRA. 
 
 In the first instance the 4 and 5 have been combined 
 into 9, and the product of 9 by 7 taken. In the second 
 instance the 7 has been distributed as a multipler of the 
 terms 4 and 5 of the multiplicand, and the sum of the 
 separate products taken. If these arithmetical numbers 
 4,5, and 7 be replaced by the algebraic numbers a, 6, 
 and e, we have 
 
 (a+b)xe=axc+bxc=act+be. 
 
 The fact expressed by the algebraic identity 
 (a+b) xe=ac + be 
 
 is known as the Distributive Law. Since the multiplier 
 and multiplicand are commutative, 
 
 ex(a+b)=(a+ 56) x c=ace + be. 
 _ It may be noticed that the Distributive Law harmonizes, 
 as it should, with the definition of multiplication. To 
 multiply a+6 by eis todo to a+6 what has been done 
 to 1 to produce ec. This would certainly mean that we 
 are to do the same thing to a and 6 and add the results. 
 
 In distributing factors the law of signs must be observed. 
 
 (a—b)x5ce=5ae— 5be. 
 (— 2a? + 6) x(—38¢)=6 a@e—3 Be. 
 
 EXERCISES. 
 
 Distribute the following factors: 
 
 - (a+b—c)x 2d. 
 
 (2 eB i. te) C8 oh, 
 
 . 2at+ty—3s2)2a). 
 
 . 12 a —5 yy? + 15 ay) (4d). 
 
 . (15 wy — 15 yz + 12 az)(— 2 ad). 
 
 ao P&P WO N 
 
MULTIPLICATION AND DIVISION. 47 
 
 (ab —5 0? + 38¢ —a)j(dxy). 
 
 (ax — ba + cay)(— 3 mm). 
 
 (4 lx +5 my — 3 nz)(2 ad). 
 
 (— 6a + 5a’b — 38 ab? + b*)(— 3 2’). 
 10. (aa + bx + c)(— 2 yz). 
 
 oO NO 
 
 51. Index Law of Factors. Ina continued product, abed, 
 any two or more of the factors may become equal. ‘To 
 indicate the product when two or more factors have 
 become equal, a convenient notation has been devised. If 
 6 should become equal to a in the above product, we would 
 
 have aacd, and it would be written a2ed. The small 2 to the 
 right of the a and slightly elevated is called the exponent. 
 If 6 and ¢ each equal a, we have aaad =a'd. If 6b, c, and 
 d each equal a, we have aaaa=a'*. It should be noticed 
 that the exponents 2, 3, and * indicate in these cases the 
 number of times @ occurs in the respective products. 
 
 a-a=a’, read a square. 
 a-a-a=a’, read a cube. 
 a:-a:a-a=da', read a fourth power. 
 a-a-a--tomas=a", read a exponent m, or a mth. 
 It should be noted that a= al. 
 W=A+Aa+a-a-a. 
 Wxrr=Aa+-a-a+xe'a. 
 Write out the equivalents of 
 1) dz?= 
 (2) 5y = 
 (3) Tad-d-a:a°4:4x-¢x= 
 (4) a2?2(38 = 
 
 ©) 2-2-2.2-2-¢7-9-e-e-rs 
 
48 THE ESSENTIALS OF ALGEBRA. 
 
 @X@W=a-4-4X4:4=4-0:0:0-:4=0=0", 
 a" Xa"=(A-a+a- tOMAMS)(4+a:a- tO NAS) 
 =(a-a-a-+- to (m+n) a’s) 
 sete aes 
 The result CR 
 is the Index Law for positive integral exponents. It is read 
 a with exponent m multiplied by a with exponent n equals 
 a with exponent m+n. 
 ge. = ot? = 75, 
 y' z af” as Nek on y”. 
 
 EXERCISES. 
 
 a? - at. P= (a*- a?)(a*- a) =a" a, 
 8 
 
 ey P= 
 
 (ety) @ty-(@+y)P= 
 (a+ b)*- (a+8)%-(@+y)?-@+y)= 
 BT y+z-yt2)- (a—b)-@—b= 
 
 SOOO eae) aaa ee a Oo ear 
 oy) 
 bo 
 ou 
 oo 
 tc 
 _ 
 
 _ 
 9 
 
 52. The Multiplication of Any Number of Monomials. 
 
 RULES: 
 
 (1) Write the product of the numerical coefficients. 
 
 (2) Attach the literal factors of the product, observing the 
 Index Law for repeated factors. 
 
 (3) Prefix the proper sign, determined by the number of 
 negative factors, + if an even number, — tf an odd number. 
 
 These rules result from observing the laws already developed. 
 
MULTIPLICATION AND DIVISION. 49 
 
 Multiply together 3 ab, 4 ac, 5 bc, — 3 abe. 
 
 The indicated result is 
 
 3ab x 4ac x 5be x (—3abe) 
 =3.4-5-(—3)a-a-a-b-b-b-c-c-e (by Commutative Law) 
 
 =-—180a-a-a-b-b-b-c-c-c (by multiplying numerical 
 factors and observing law of signs) 
 
 = — 180 a*b*c3 (by Index Law) 
 
 The above has been developed by an application of the laws, but 
 the result is in exact accord with the rules on page 48. 
 
 (2a) x (— 8 ab) (— 2 bc) = + 12 abc. 
 EXERCISES. 
 
 Multiply together the following monomials: 
 
 1. 3ax,4ab, —S5aay. 5. —4lx’, —3 By’, —8 aty’. 
 
 2. 5a’x, 3 b’y, 2 a®a°y"*. 6. 2abc, 3 b’c®, — a*be’. 
 
 3. —4 mn, 2mn*, —S5m‘n’a?. 7. 3(a+b)*, 4y(a+b)’, 5(a+b)* 
 4. —axyz, Day’, —4 aye’. 8. 8 au’, 5(a+b)*, —2a%(a+b)?. 
 
 53. The Multiplication of a Polynomial by a Monomial. 
 RULES: 
 
 (1) Distribute the monomial as a multiplier of each term 
 of the multiplicand. 
 
 (2) Connect the results by the proper algebraic signs as 
 determined by the law of signs. 
 
 By the Distributive Law 
 (a+b) xec=ae+ be. 
 
 6 can be any number, as d+e; then (a+ b)c=ae + be 
 becomes 
 
 (a+d+e)e=ac+(d+e)c=ac+dc+ee. 
 In general, 
 
 (at+b+e4+d+-.-)ms=am+bm+em+dm-+-:-. 
 
50 
 
 L; 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 (5e+3y)x 2a=1020+6 ya=10 ar+ 6 ay. 
 
 It is customary to write the letters of a product in alphabetic 
 
 order. 
 Law. 
 
 ae ee Le 
 
 YH » eS 
 QO N HF © 
 
 14, 
 
 15. 
 16. 
 
 17. 
 
 This can always be done by an application of the Commutative 
 
 (8a+4b)x2a=6a'4+8 ab. 
 
 (83a?—4ab)x4ab= 
 
 Multiply 3 a°%—2a2?+3ay by 5 axy. 
 
 Multiply —3 a+ 4la’?—Sdlay by —2 Pa’. 
 
 (ax +3 ay—5 bey) x 8 axy = 3 way +9 aay? — 15 abcay’, 
 Multiply a?— ab + ab?— B® by —a%b®. 
 
 Multiply 3 at— 4a’ — 6 ab? +7 ab® by ab’. 
 
 Multiply a’?+6?+a’+y+2 by 6 aba”. 
 
 Multiply 5ab —12cd—3,fh by —6be. 
 
 Multiply —4 a’yz+7 ay’z—3 xyz? by — 5 ayz. 
 
 . Multiply 3 a’b? +7 b’c? —14 ca? by 8 a*b°c’. 
 . Multiply 3(@@+y)?+4(a+b)? by 2(@+y)(a+5). 
 
 B(w +9)? + 4(a +d) 
 2(a+ y)(a +d) 
 6a@+ya+b)+8@+y)(at oy 
 Multiply 
 (7+ yP—3a(e+yP?+4y(e+y’) by 5 ay(a’+ 7’). 
 Multiply (a—b)*—38 a(a—b)?+7 y(a—by* by 2 x’y(a—b)’. 
 Multiply 
 3(@+yyP+2a(e+y)/P?—38H(e+y) by —ab(a+y) 
 Multiply 
 
 5(a? + y)? — 38 a(a? + y)? —T atb(a? + y) by —4(a? + y)*. 
 
-_ MULTIPLICATION AND DIVISION. 51 
 
 18. Multiply 
 10(a? + 3 b?)* — 6(a? + 3 6)? +12(a@2 +3 b’) by —5(a?4 3 0?) 
 19. Multiply (a+6+c)?—3(a+b+c) by 5(a+b+¢)*. 
 20. Multiply 
 (au’ + bx +c)’ + 4(aa? + ba +c)? by — 5(aa’ + ba +e). 
 5%. Meaning of 0 x a and a x 0. 
 (6—c)a=ab—ae. 
 If 6=e, this becomes 
 (¢ —¢)a=ac— ae. 
 But e—c=0 and ac—ac=0. 
 Hence, Nie AUP 
 By the Commutative Law, 
 
 Oxa=ax0=0. 
 
 55. Multiplication of a Polynomial by a Polynomial. RULE. 
 To multiply a polynomial by a polynomial multiply each 
 term of the multiplicand by every term of the multiplier and 
 take the algebraic sum of the results. 
 
 This is a direct consequence of the Distributive Law. 
 Find the product of (a+6) by (@+y). 
 Let x+y be replaced by e. 
 
 Then (a+ 6) x(#+y)=(at6) x c=ae + be. 
 
 But c=r+y, 
 and so (at+b)(e@+y)=ac+be=a(rt+y)t+bet+y) 
 =axr+ ay + bx + by. 
 
 This product consists of the sum of the products of each 
 term of the multiplicand by each term of the multiplier. 
 
52 THE ESSENTIALS OF ALGEBRA. 
 
 If three polynomials are to be multiplied together, two 
 of them must be multiplied as above and their product 
 multiplied by the third. 
 
 Thus 
 
 (a+b)(a+ y)(r+8)=(ax + ay + bx + by) + 8) 
 (by multiplying factors one and two together). 
 (ax + ay + bx + by) (7 + 8)= are + ary + bra + bry 
 + asx + asy + bsx + bsy. 
 
 EXERCISES. 
 Find the product of the following polynomials: 
 1. GBa+bd)(a+b)=8¢0C+ab4+8ab4+0?=38a°?+4ab4 BD. 
 The distribution may be arranged conveniently as follows: 
 oa+ 0 
 a+ 0 
 30+ ab (the product of (8a-+ b) by a) 
 dab +b’ (the product of (8a +6) by 6) 
 3a? +4ab+ 6° (the total product) 
 This is called long multiplication, and should only be used 
 
 when the distribution can not easily be made in the straight 
 line form illustrated above. 
 
 2. (@+y)(@—y) =e 4 ay —ay— Y= ve — yy’ 
 
 3. (2e%+y)(2Qea—-y) =4e-y. 
 
 4. (4a4+38y)(4a—3y). 8. (x+ 5) (@— 2). 
 
 5. (au + by) (ax — by). 9. (y+ 4)(y’—3). 
 
 6. («+ 4)(@+4+ 2). 10. (ax? +b)(x+c). 
 7. (©—2)(@+ 38). 11. (8%+5) (4% + 2). 
 
 ee 
 
 12. (4a°+4+ 32°45 a—6) (82 —2a°?—324 2). 
 
MULTIPLICATION AND DIVISION. 53 
 
 Multiplicand and multiplier are both arranged according to 
 the powers of 2; that is, the exponents of a decrease uniformly 
 from left to right. The arrangement will be just as good if we 
 reverse both factors and have the powers of @ increase from 
 left to right. For multiplication we may arrange the work as 
 follows: 
 
 4a? +3a°+ 5a — 6 
 oa —2a— 3a-+ 2 
 120°+92°415e'—182? (product of the multiplicand by 32°) 
 
 —8x2°— 62*—102?+122’ (by — 2 2”) 
 —  —129'— 9o3—150°+182 (by —3 2) 
 | oe + 6274+102—12 (by 2) 
 
 
 
 oe ae oe — 292+ 3274+ 28e¢—12 (the total product) 
 
 The orderly arrangement according to powers of « merely 
 insures that the terms of the product will come in an orderly 
 way, thus making it easier to arrange like terms in columns 
 ready for adding. When arranged in this way the highest 
 power in the product appears first, and is the product of the 
 two highest terms in the factors. 
 
 —- 13. (4@—5a?+3a—6)(8—@+2a). 
 
 When these factors are arranged according to powers of a, 
 
 we have 
 (—5a?+4a°4+3a—6)(—a’?+2a+43). 
 
 Now multiply as in Exercise 12. 
 —14. (—6y+4+77’—12)(4—2y+4+7’). 
 When arranged, this exercise becomes 
 et YOY 12) GF + 0 yf! 2 y-+4). 
 
 Observe that in the multiplier the term 7? does not occur ; 
 in the arrangement according to y we write 0’. 
 
54 THE ESSENTIALS OF ALGEBRA. 
 
 15. (b¢—54+60?—3b+26*) (hh —264 2). 
 - 16. (@—Ta+120 + 6)(@—38a—5S). 
 
 ~17. (82°+4+120°—10%+4 4)(—2—5+4 82). 
 
 ~18. (5a°y? —6 xy +12 2°y? — 4) (8 —42%y? + yx’). 
 -19. (42—-4+462—52)(—42z+43). 
 
 ~ 20. (80?+4at+a+5a’°—4)(at—a@+a?—a+1). 
 al, (2084 3a2—42—1)(8a+44). 
 
 
 
 
 
 
 
 SOLUTIONS. 
 (1) | (2) 
 2¢ + S3e— 47 — 1 2+ 3— 4-71 
 ox + 4 o+ 4 
 6at+ 9a®—120°— 3a 64) 
 $a 42 a 16 eee § +12 =16—4 
 Gat t17e+ Oa—192_4 6417 +0 
 
 In solution (1) the multiplication is carried out in the usual 
 way, while in solution (2) merely the coefficients are used. 
 In the answer on the right the #’s should be inserted, beginning 
 with a, The method used on the right is called multiplying 
 by detached coefficients. It is a device which saves time by the 
 omission of all letters. 
 
 In using detached coefficients the following directions should 
 be observed : 
 
 (1) The multiplicand and multiplier must be arranged 
 according to the same letter. 
 
 (2) 0 must be used as the coefficient of every power of the 
 letter of arrangement which does not occur. 
 
 (3) The letter of arrangement is inserted in the product by 
 beginning at the left with a power equal to the sum of the 
 highest powers in the multiplicand and multiplier, and decreas- 
 ing uniformly to the right. 
 
 The following exercise illustrates this: 
 
MULTIPLICATION AND DIVISION. 50 
 
 22. (3a°+42°4+50°4 7x41) x (52°—32). 
 See 4 4 Ot he 
 oo — 3 
 1o6e-0 +20 +25. +35 + 5 
 “— 9 -—"0 —12 —15 —21 —3 
 P. 
 
 foe Ue lia + 25a 4+ 23 ae — 102 — 21a — Sa 
 
 The highest powers of « in the two factors are 5 and 3, so 
 the product begins with 2%. 
 In the following exercises use detached coefficients. 
 
 — 
 
 —23. (at—5a?+40—384+2a)(80—5a+2). 
 
 mated eo oa — 2a-+7)(a’+4a° —da’?+ 2a — 7). 
 ~ 25. (ay? —4 277? + 6 ay — 5) (a*y? + 4 wy? — 6 xy + 5). 
 
 — 26. (32% —Tae+4a—5)(22 —32+4 4). 
 
 — 27. (Ty —5y+3y—4A4y—8y+1). 
 
 — 28. (a*b®?— 6 ab + 7)(5 ab — 4 a°D"). 
 
 — 29. (4a‘b'— 7 — 6a’b? + 3 ab)(ab —5 + a’d’). 
 
 - 30. (807+ 4a— 5) (a’?+2+41)(@’—382x+3). 
 
 56. The Identity in Multiplication, The multiplicand 
 multiplied by the multiplier is identically equal to the 
 product. 
 
 Multiplicand x Multiplier = Product. 
 
 (a+b) xec=ac+t+ be. 
 
 Let a=b=c=1. 
 Meee Teed. vad sc dy 
 PCT Weer pad Me oe 
 
 aay, 
 A — we 
 
56 THE ESSENTIALS OF ALGEBRA. 
 
 This furnishes a convenient method of verifying the results 
 in multiplication, If in Exercise 22 above we put «=1, we 
 have 
 
 (8+4+547+4+1) x (6—3)=15411425+423—-10—21—3. 
 
 D0. eee 
 40=40. 
 
 This merely verifies the coefficients. In order to verify the 
 exponents, some value other than 1 would have to be sub- 
 stituted. In case more than one letter occurs, numerical values 
 must be given to each. 
 
 (a+y)(@—y)=e7—y’. 
 
 Let Ce DRT 99) ees 
 Then (142) —2)=1?— 2% 
 (3)(—1) =1-—-4 
 — > = — oe 
 we EXERCISES. 
 
 Perform the following multiplications and verify by means 
 of the identity : 
 -1. (+ ab)\(2a+43D). - 5. (5a°—S8ay42y’)(2a+3y). 
 2. (4a4+6b)2Qa—4b). -6. (8%+4a)(8a%—4a). 
 — 3. (724+ 3a+41)(a?—a#+1).- 7. (6 ab —7 cd) (5 ab 4 T cd). 
 ~ 4 (5ab +3 cd)(2ab—4cd). —8. (6a°y—4ay’+y") (8 cy—Ty’). 
 
 57. Involution. In multiplication when the factors are 
 alike, the operation is called involution, and the result a 
 
 Owe? 
 P aA-a-a=a’, 
 
 Qs a? - 9? =" 07 )8 == nd 1a 
 
 (a® . a® . a®) = (a°)® = aaa aaa aaa, 
 

 
 MULTIPLICATION AND DIVISION. 54 
 58. Meaning of (a’”)”. 
 (a”)” means a” -a™. a”... to n factors. 
 
 Since each of the n factors, a”, contains m a’s, there 
 are in the product mn a’s. 
 But a” means a-a-a-a-- to mn factors. 
 
 Hence, (ae er art, 
 More generally, (a7b')" = a™"b™. 
 The exponent n is distributive as to factors within the 
 p 
 parenthesis. oS 5 ae 
 ay")? = vy”. 
 
 The exponent is not distributive as to terms within 
 the parenthesis. 
 
 @+yP=@t+NetYNEety-. 
 (7 + y)? is not equal to 2? + y?. 
 
 The difference between the following forms should be 
 noted : 
 
 Ae na a’. a” = hn"). 
 (a)? aid ae. (a™" — gy, 
 EXERCISES. 
 
 Remove the ( ) and simplify: 
 . (a7)* (a®)? = a’ - ab =a". 
 
 (2) (@yyt 
 
 3? . (3), 
 
 (ae ys (aay, 
 -(@t+y)y - (8°)? « (27)? (a?) (a)? 
 (42) ty’) @+yy-@+y) 
 Roa = 2°. 26 = 218, 10. (ab’c)* + (a®be®)'. 
 
 Oo » © N HB 
 (ace st oO) 
 
08 THE ESSENTIALS OF ALGEBRA. 
 
 DIVISION. 
 
 59. Division Defined. — Division is the process of finding 
 one number, when the product of two numbers and one of 
 them are given. The given product is the dividend, the given 
 number the divisor, and the required number the quotient. 
 
 Division is the inverse of multiplication, the dividend 
 corresponding to the product, the divisor to the multiplier, 
 and the quotient to the multiplicand. 
 
 Since axb= ab, 
 
 ab+b=a. 
 
 60. Law of Signs in Division. 
 
 From multiplication we have 
 (+ a)(+ 6) =+ ab. 
 (+ a)(— 6) =— ab. 
 (—a)(+ 6) =— ab. 
 (—a)(—b6)=+ ab. 
 
 From the definition of division it follows that 
 +ab+(+b)=+a. 
 —ab+(—b)=+a. 
 —ab+(+b)=—a. 
 +ab+-(—b)=—a. 
 
 Like signs in dividend and divisor give a positive quotient, 
 
 and unlike signs give a negative quotient. 
 
 Gl. Index Law. We already know that 
 
 q'” x q” a qmtn, 
 
 Hence, qntn Pen qr ee a” — qQmtn-n, 
 Suppose mM+n= Dp, 
 then aP +a" = ap-", 
 
 More generally, aPb6%e" + a"bsct = q?-"b4-ser-t, 
 
MULTIPLICATION AND DIVISION. 59 
 
 For the present it is understood that the exponents of 
 the factors of the dividend are not less than the exponents 
 of the corresponding factors of the divisor. 
 
 The above considerations show that the exponents of the 
 Factors of the divisor are subtracted from the exponents of 
 the like factors of the dividend in order to obtain the expo- 
 nents of the factors of the quotient. 
 
 Illustrations: 
 
 eet iat 3. 
 
 (2) aP6-+ a3h5 = q0-3)6-5 = ib, 
 (3) atb°® + a®bect = abrct. 
 
 4) @+y%e+uy'+ ty E+uy=(@+yetw)® 
 
 62. Meaning of Z and a, 
 a 
 
 We know that ax0=0. 
 Hence, We 0. 
 a 
 (oii Agee EAH 
 Hence, ert 
 7 ks 
 But, by Index Law, a” = a" = a, 
 om 
 Now by Axiom 1, i ea 
 
 Any quantity with an exponent 0 ts equal to 1. 
 
 63. Division of One Monomial by Another Monomial. 
 RULES: 
 
 (1) Divide the numerical coefficients as in arithmetic. 
 
 (2) Attach the literal part determined by the Index Law. 
 (3) Prefix the proper sign determined by the law of signs. 
 
 Peer Onyer <= 3 72 —=:3.0%2, since 4 = 1. 
 
Fa 
 
 60 THE ESSENTIALS OF ALGEBRA, 
 
 a EXERCISES. 
 Divide: 
 
 21 a’a®y? by 7 ara’y. 
 
 72 atay’2" by 24 a®y’2™ 
 
 108.a7y*2? by 36 aty’2?, 
 
 81 a°b®cd? by 27 a°b*d. 
 
 144 a(x + y)’ by 48 (a+ yy 
 
 — 63 (a — b)’ (@ — y)' by 21 (a—b) *(w—y)?. 
 
 AB a*y?(a® + y?)? by —5 y (a? + y?) 
 
 3° atayz by 3? a’a*y. 
 
 — T*BaPyt(z— ax) by — T?Pa'(z— a). 
 
 3f- F(a — @)*(b — y)* by 3? - 5°(a — w)7(b — y)*. 
 
 A SS OY eh ae eee 
 
 = 
 9 
 
 64. Division of a Polynomial by a Monomial. From the 
 Distributive Law of Factors we know that 
 (a+b+ec) Xk = ak + bk + ck. 
 Hence, (ak+bk+ckh)+k=a+b+e, 
 
 which shows that # is distributed as a divisor to every 
 term of the dividend. 
 
 Rute. Divide each term of the polynomial by the mono- 
 mial and add the results. 
 
 10 a2? + 5 ax?y — 20 a8x*y? 
 5 ax 
 
 =2av+y—4ay?. 
 EXERCISES. 
 1. (12 a®y* — 8 atay?’ — 4 a’a*y*) + 4 a®y’. 
 
 2. (Qabcx* — 18 a7b*a* + 27 a®bc’ax) + 9 aba. 
 3. (380 a*y%z + 25 a*ytz? — 35 a*y'2”) + 5 a*y’e. 
 
MULTIPLICATION AND DIVISION. 61 
 
 4. (@+ y+ 5@+y)'—aw@t+y), (Regard x + y as a term.) 
 (@+y) 
 5. [4(a — b)* — 5a(a— 6)? + 11 ay (a — b)’] + (a — )*. 
 6. Divide 5 ay(a+y¥)*— 10 ay? (@ + y)? — 15 ay’ (@ + y) by 
 5 wy (w+ y/)*. 
 7. Divide 11 a*b?(a?+ a?) + 22 a% (2 + a’)? — 33 ab? (a? + a’)? 
 by 11 ab(@’+ a’). 
 8. Divide 
 T xyz (a? — b*)? + 21 ay’2? (a? — b?)" by —T7 ayz (a? — 0), 
 9. Divide 
 24 a*b’c (aw + b)* — 36 a’b*c’ (ax + b)’ by — 12 (a*b’c) (aw + b)*. 
 10. Divide 
 —33 (aa’+ba+c)?+44 at (aa’?+ba+c) by 11 (aa’?+ba+e)’. 
 
 65. Division of a Polynomial by a Polynomial. The divi- 
 dend and divisor should be arranged in descending or 
 ascending powers of some common leading letter. This 
 gives a quotient arranged with respect to the same letter. 
 
 The first term of the quotient is found by dividing the 
 first term of the dividend by the first term of the divisor. 
 The process is illustrated in the following solutions : 
 
 (1) Divide 
 
 2 —a*— 11234 1622-22-38 by 22-4243. 
 2—4¢+3)2°—2t—11234+162—-22—3(0°4+38 22?-22-1 
 
 —4e'+ 323 = 23(22?—4 2443) 
 8a*—14 23 +16 22-2 7—3=1st partial div. 
 32'—1243+ 922 =3 022-4443) 
 
 —223 +7a2?9-2¢—3=2d partial div. 
 
 —203 +8227-6a2 =—22(2?—42+3) 
 —z*+4x2—38=3d partial div. 
 —2°4+427—38=—-—1 (#?—42+4+3) 
 
62 THE ESSENTIALS OF ALGEBRA. 
 
 ‘This scheme of division is merely a separation of the 
 dividend into parts. In the example just solved we have 
 separated 2° — af — 112° + 162? — 2x —3 into these parts: 
 (—4 a'43 23) + (3 at—12 2?+9 27) 4+ (—2 23 +8 2-62) 
 
 +(—a?+4x—5). 
 Now, regarding the dividend in this separated form, we 
 have the division thus: 
 
 ge —4et+38e8  8at— 1228 + a? — 2224+ 822-62 
 
 w—47+4+38 7 4g lB we—4ateg 
 a oe ope teat == 7 oe ad e 
 (2) Divide at — 16 by 2+. 
 z+ 2)a* — 163 —2¢74+42-8 
 
 
 
 
 
 +e 
 
 
 
 
 
 g4+ 223 
 — 27° 
 — 222 — 427° 
 + 472 
 —8r—16 
 — 82—16 
 
 Divisions such as the above, which terminate without 
 any remainder, are called exact. 
 
 EXERCISES. 
 Divide: 
 1. «t—a®—92°+154%—12 by «—3. 
 2. #—5e—38x2+15 by «—5. 
 3. 60° +7e2—1824+5 by 24+ 5. 
 4. 2e*—9a*?+172a?— 14a by wv —2e. 
 5. Sat—6a0°+2a°4+14a—21 by a’— 2a438. 
 
MULTIPLICATION AND DIVISION. 63 
 
 a*+4a° +6070? +4 ab?+0' by a+2ab4+ 0% 
 wv’ —a® by a? — a’. 
 b’c®’ — a® by de® 4+ at. 
 m—3m'+3m?—1 by m?—1. 
 10. a” + 2a"b" +b" by a® +b" 
 11. a” —b™ by a®— 0" 
 12. a’"—b™ by a®— b”. 
 13. af —130?+472?—312+4 by 2 —6a+1. 
 14. «*—120°+ 54a?—108e%+481 by a’?—6249. 
 15. ab? —Sa’*b'cd + 5 aberd? — c'd® by ab — cd. 
 16. 12 a°y§ — 17 vty? +10 vy? —3 by 4a°y?— 3. 
 17. x°—10a*+40 2—80 a +4+80e—382 by a—4a44. 
 1s. 7+15¢—212°+18a?—4a* by T—62+4 2", 
 19. (a+ b)?—5(a+b)+4 by (a+b) —1. 
 20. («+y)?+7(e+y)—18 by (w+ y) 49. 
 21. (a+ 2)?—(a+a)—42 by (a+) —1. 
 22. (m+n)? —-T(m-+n)—44 by (m+n)+4. 
 23. (a+b)?—x* by (a+b) —«@. 
 24. («+y)>—(a+b)* by (@+y)— (a+D). | 
 G66. Detached Coefficients. When the dividend and divisor 
 are arranged in descending powers of some common letter, 
 the quotient is also thus arranged. We may then perform 
 the division by the use of the coefficients only. 
 (1) Divide 2*4+23— 327+ Tx—6 by 22-2742. 
 1—1+2)14+1-—-84+7-614+2-8 
 
 op Orga 
 
 
 
 
 
 (h=qheky 
 higeds 9 ha We 9 
 ya yee 
 
 —3+3—-—6 
 
64 THE ESSENTIALS OF ALGEBRA, 
 
 Since at-+22=.2?, we know that the quotient begins 
 with 27, and is 77+22—83. 
 
 In the use of detached coefficients all powers of the 
 letters from the highest to the lowest power must be 
 present in both divisor and dividend. If any powers are 
 absent, they must be inserted with zero coefficients. 
 
 (2) Divide 2—8 by x—2. 
 
 In the dividend neither z* nor 2 appears. We insert 
 them, writing the dividend 23+ 02?+ 02-8. 
 
 1—2)14+0+4+0-—8014244 
 
 
 
 1—2 v4+2x+4, quotient. 
 2+ 0 
 2—4 
 4—8 
 4—8 
 EXERCISES. 
 
 Divide, solving by detached coefficients : 
 w—5a’?+4 by «—1; by «+2. 
 at—Toe4+110°+T¢e%—12 by x—-1; by «—4. 
 e'—132°+ 36 by w+a—6. 
 
 at —18a°y? —175y* by wv — 257. 
 
 2m* —17 m®n + 31 mn? — 23 mn? +12 nt by 2m —Sn. 
 at — 256 by a? +16. 
 
 b§— 729 by b?— 9. 
 
 y®— 4096 by x? +8. 
 
 a+a*t+1 by at—a?+1. 
 
 16a*+4a?+1 by 4¢+4+2a+1. 
 
 . +4 by +2e42. 
 
 . 1864150°-170°—3 by 50?—46+438. 
 
 SO Cee a Oa UIE SE 
 
 Hoe 
 Nb fF Oo 
 
MULTIPLICATION AND DIVISION, 65 
 
 13. a°— 0 by a? +2 0 + 2ab?+ 5b. 
 14. 2+y by at — aty + a*y? — ay + y'. 
 15. 10 a*— 48 a*b + 26 a*b? + 240d? by —5a?+40ab4+30% 
 
 67. Inexact and Continued Division. 
 (1) Divide 27+ 1 by x+1. 
 : x+1)22+1(7—1 
 w+ ex 
 —z+l1 
 —x—l 
 2 
 In this example there is a remainder of 2, and the 
 division is inexact. In such examples the division should 
 continue until the largest exponent of the remainder is 
 less than the largest exponent of the divisor. 
 (2) Divide 1+ 2? by 1 +z. 
 1l+e2)14+21—24+2a2—228 
 
 bee oes 
 — pt 7 
 aT ee 7 
 2 x 
 2 a2 +- 273 
 — 273 
 — 248 2 at 
 
 2D at 
 
 dol 
 
 This division may end with two terms of the quotient 
 and the remainder 22”, or with three terms and the re- 
 mainder — 223, or with four terms and the remainder 2 2+. 
 Evidently the division might be continued to any number 
 of terms desired. When inexact division takes this form, 
 it is called continued division. 
 
66 THE ESSENTIALS OF ALGEBRA, 
 
 EXERCISES. 
 Find quotients and remainders: 
 1. Divide 2? —72?+11x%—7 by +3. 
 2. Divide + 72? —8%+413 by a? +3a—2, 
 Divide a—11@+ 21 by a +5. 
 4. Divide bt—146°+4 11 by 0? +9. 
 Divide a’a’? + 9 a’a’? — 7 by ax + 3. 
 
 sa 
 
 oO 
 
 In the next five exercises continue the division to four terms: 
 Divide 1 +a by 1 —@. 
 
 Divide 2+35a%—42° by 2— 2. 
 
 Divide 3 —6%+4 82? by 8+ 52a. 
 
 Divide 1 —17x%+4 132 — 82° by 1— 1624 527, 
 10. Divide 10 — 20a + 25 a*?— 31a by 2—5a. 
 
 © OND 
 
 68. The Identity in Division. 
 Dividend + Divisor = Quotient. 
 (—1)+(@—-l=274+er41. 
 
 This is true for all values of z. 
 Let z= 2, and we have 
 
 (23 —1)+(2-1)=274241, 
 
 (8 —1)+1 =142+1, 
 T+1 = 7, 
 ff = 7. 
 In this case, should we make «= 1, we get 0+ 0 on the 
 left of the sign =. 0-+0 is indeterminate. In using the 
 
 identity to verify divisions, avoid substitutions that will 
 produce this form. 
 
MULTIPLICATION AND DIVISION. 67 
 
 EXERCISES. 
 
 Divide and verify by substituting particular values: 
 
 1. 62 —177 + 241-16 by 2P—3/1+4+4. 
 2. y —1lby y—1. 
 
 3. 2 +1 by «+1. 
 
 4. a—120?+ 48a — 64 by a’ —8a+ 16. 
 5. 625 — 5002 4 1502? — 2022+ 2* by 5 —2. 
 6. 625 —2* by 254 2. 
 
 7. y'—dy — 154 by 7? — 14. 
 
 8. a‘b* + 4a°b? — 117 by a’b’ + 18. 
 
 9. 2° —y° by a’? — yx’. 
 
 10. at—4a?—34a°+76a+105 by a—T. 
 
 REVIEW EXERCISES. a 
 
 1. Find the value of 6 2? — 4 ay +12 77, whenv=4,y=—1. 
 
 2. Find the value of wz’? — 64y°+82—3 ayz, when «=0, 
 Boe = OD, 
 
 3. From 16 a? —4 7° + 12 2 — 14 a’y + 3 ay’ subtract 12 y+ 
 8a°+4 ay? —10 ay 4+ 11 2’. 
 
 4. Remove parentheses and unite like terms: 
 
 16 —§12a+[46—3c]+8—[8a—38(4—2))]}. 
 
 5. Remove parentheses and unite like terms: 
 —3a’+4[ay—ax(8a—4y)—3y(4ea4+2y)]—fa?+3(ay—y’)t. 
 
 6. Unite terms in a, y, z: - 
 
 ax + by + cz—4(— a'a + b'y —c') +3 (lx + my + nz). 
 
 7. Simplify «—4y—[z—y—(@+y—z)], and find value “ 
 Wie Ga yez = 1. 
 
68 THE ESSENTIALS OF ALGEBRA. 
 
 8. Simplify 4(a—5 {b— ct) —[8 b+ }2b—(ce—a)}], and a 
 find value when a=1, b=2, c=3. 
 
 9. Find value of Pr RTE lcd 6) | ze + V2? ey — vine 
 when wes) aye) ee, 
 
 3 Va? + aval ee 
 
 c+Y—z2 
 
 
 
 
 10. Find value of 
 when ¢=4, y=), 2= 6. 
 11. Remove parentheses and simplify: 
 a(y—2z) +y(z—2%) +2(@—y). 
 12. Remove parentheses and simplify: , 
 a? (y —2) + y? (2-2) +0? (2@—Y). 
 13. Multiply #+a+41 by #—a+1. 
 14. Multiply #’+7?+1—a—y-— xy by Petar 2: 
 15. Multiply 2’ — 47? by a + ay’. 
 16. Multiply 12a¢'—-3 03+ 102? -—5a2+4 by 322-2? +52 
 — 4, 
 
 17. Find the value of #—4a°+3a—5, (1) when x=2; 
 (2) when «= —1; (8) when «=0. 
 
 18. Find the remainder after dividing #—42a?+3a—5 
 (1) by «—2; (2) by. «+1; (8) by =. 
 
 Note that the remainders are the same as the results found in 
 Exercise 17. 
 
 19. Divide 2 a’ — 3 a‘b — 6 a°b + 18 a’b? — 6 ab® by 2a—3 b. 
 20. Divide 3 a*+140°+92+ 2 by 2?+5a+1. 
 
 21. Divide 2 a’°+ab—ac—30?’—4be—c by 2a+3b+¢e. 
 
 22. Divide a —(a+b)x+ ab by x—a. 
 
 . Divide a’ —(a + 6 + c) a’ +(ab + be + ca) « — abe by x—a. 
 . Divide #2—-U4+m+n)2+ (m+ mn+iln)z—lmn_ by 
 
 
 
 ee Pe (Ltn) z+ln 
 
 oe 
 

 
 MULTIPLICATION AND DIVISION. 69 
 
 . Multiply together («+ a") and (@ + a). 
 
 . Multiply 2” — y" by a +y". 
 
 . Multiply 30"%4 5a"+7 by 2a"—4a—3., 
 
 e Divide 3.47% +13 21415 a 4+ 9a by a+ Bar. 
 . Divide xt” + ya — yma" — y™™ 4 a" y" by a+ y”. 
 
 Multiply a?+0° by a—b, and divide the product by a+0. 
 
 . Multiply 3a?44a'—22" by 2x?—32'+2". 
 . Divide a” —b*”" by a"— bd”. 
 . Multiply 2? 4+ 3a"t!—5a* by a ?—2a +a". 
 
 Divide a” — by aim  ag3my™ gry” aeny” + y, 
 
 . Divide a®— b%" by a”— b*. 
 
 Divide (a + b)”" —«* by (a+ b)" +2”. 
 
 meeieioe (ye + y)*—1 by (7+ y)*—1. 
 (Vax +b)? + 4? by Vaa+b+y. 
 . Simplify @’—ayt+y)(@t+ty@e-yy+@e+y). 
 
 5(a® + a’b + ab? + b*) x 4(a—b)’?+ [2 (a? —b") x 10(a? +") ]. 
 
. CHAPTER V. 
 IMPORTANT IDENTITIES. 
 
 Many expressions in algebra appear in standard or type 
 forms. When this is the case, multiplications and divi- 
 sions can be performed mentally by. remembering certain 
 identities. 
 
 69. Multiplication Identities. 
 1. The Product of Two Binomials with a Common Term. 
 This form is given by 
 , (x + a)(x + 6) =x°?+ (a+ 6)x+ab, 
 or (a+ x)(6+x)=ab+(a+b)x+x°*. 
 (1) (2+ 10)(w+5)=274+ (10+ 5)z + 50 
 =27+ 152+ 50. 
 
 To find the product of «+10 and x + 5 it is only necessary 
 to see that a is 10 and } is 5 and make these substitutions. 
 
 (2) (e—7)(«@+8) =a°+(—7+8)a—56 
 =27’+a— 56. 
 
 In this a is —7, and D is 8. 
 
 (3) (542) (11 —a) =55+4 (11 —5)e—2, 
 
 In this exercise it is necessary to note the signs of the 2’s. 
 
 (4) (8e+y—5)(8e+y4+7)=B6e+y)?+2(8a+4+y) -385. 
 
 In this exercise the common part is 3@+y. 
 70 
 
IMPORTANT IDENTITIES. 71 
 
 EXAMPLES. 
 
 Write out the products of the following: 
 
 1. (w+ 10)(x — 2). 6. (aw —11)(ax + 6). 
 
 2. (3%+6)8e+1). 7. (3 xy — 5)(3 xy — 6). 
 3. (Ta—d)(Ta+A4). 8. (42a°+7)(4a°— 5d). 
 
 4. (564+2a)(54+6a). 9. (3—42y)(5 + 6 xy). 
 5. (2y—6)(2y+7). 10. (a+ b—6)(a+b+5). 
 
 ll. (€+60+4+7)(a+ b—8). 
 12. (©+2ab—3)(x+2ab+7). 
 13. j4—(@+2y)[ [5+ (@+2y)}. 
 14. [(a+))?—4a][(a+b)?+7 a]. 
 15. [Sab —(a—y)’|[5ab + (@— y)?]. 
 2. The Square of a Binomial Sum. 
 If, in the identity 
 (e+a)(e#+b)=224+(a+b)x+ab, 
 we let = tl, 
 it becomes (x +a)(«@4+a)=274+(at+a)zr + aa. 
 (x +a)?=x°+ 2ax+a’. 
 The square of the sum of two quantities is equal to the 
 
 sum of their squares increased by twice their product. 
 
 (Set+4yP=OrP?+2O6xndy+Gy” 
 
 =25 a+ 40 xy + 16 y?. 
 EXERCISES. 
 Write out the results in the following: 
 1. (~@+y). 2. (2%+a)* 3. (34%+4b)% 
 4. [(@+y+aP=(et+yy’+2aa+y+o. 
 5. [yt+t(a+b)]. 7. [((a—3)+5 yf. 
 6. [37+ (2a+c)}*. 8. [(a+b0)+(+y)}. 
 
re THE ESSENTIALS OF ALGEBRA. 
 
 3. The Square of a Binomial Difference. 
 If, in the identity | 
 (x#+a)?=2742ar+4 a’, 
 we change a to —a, it becomes 
 (x —a)?=x°—2ax+a’. 
 
 The square of the difference of two quantities ts equal ta 
 the sum of their squares diminished by twice their product. 
 
 (Za—y)y=(2a)?-2Cayt+y 
 =4a*—4dayt+y*. 
 EXERCISES. 
 
 Write out the results in the following: 
 
 1. (a—«@)*, 5. [(a+b) —ay]’. 
 
 2. a—y)* 6. [(Ba+y)—ab]. 
 
 3. (xy —4 b)*. 7. [(@+y)—(a+b)f. 
 4. (3 a’— by)’. 8. [((2a—y)—dsa’y/*. 
 
 4. The Product of the Sum and Difference of Two Quan- 
 
 teties. 
 
 If, in the identity 
 
 («+a)\(e#+b)=2?+ (a+b)r+ ab, 
 we let b=—Aa4, 
 it becomes (#@+a)(x—a) =2°+ (a—a)xu—aa. 
 (x + a) (x — a)= x°—a’. 
 
 The product of the sum and difference of two quantities 
 
 2s equal to the difference of their squares. 
 
 (1) (Tx—8y)\(T e438 y)=(T2)?— (8 y)?= 49 22— 9 of, 
 (2) (ax+by—c)(ar+ by + ¢)=(ar 4+ by)? — & 
 
+ 
 
 IMPORTANT IDENTITIES. 13 
 
 EXERCISES, 
 
 Write out the results in the following: 
 
 1. (@—y)(u+y). 2. (3a—b)(Ba+b). 
 
 3. (4x4 ab)(4 x%— ab). 
 ne: [(a+#)+a][(a+2)—a]=(a+aP—V=0+2 aa+a’—@ 
 
 = 2 aw +2". 
 
 . [@+2y)—#][(@+2y) +2]. 
 . (a? 4+ 2% 4+16)(a? + 2 a — 16). 
 - [@+b)—-@+y[@+o)+e+y]. 
 . (ax? + ba + ¢)(ax? + ba — c). 
 
 NO Oo 
 
 o 
 
 5. The Square of a Polynomial. 
 If, in the identity 
 (e+a)2=2°+2ar+ 
 we put a=Y +2, 
 it becomes (a@+y+2)?=274+2e(y+2)4+(y+2)%. 
 (xty4+zpexrt+y+ 2274 2x4 2xz+ 2yz. 
 This may easily be extended to include the square of a 
 
 polynomial of any number of terms, the result being that 
 
 The square of a polynomial equals the sum of the squares 
 of its terms increased by twice the product of each term by 
 every other term. : 
 
 (1) @t+y—avSr+y +a’ +2ay+22(—a)+2y(—a) 
 
 HErt+yt+art 2ay—2av—2Zay. 
 (2) (82+2a—4b) 
 = (3 a)?+ (2.a)?+(—4b)?+2(32)(2 a) +2(3 7)(—4b) 
 + 2(2.a)(— 4b) 
 =O +4¢°+ 1607 + 12 ar— 24 be —16 ab: 
 
74 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 Write out the results in the following: 
 
 1. (a—av—yy’. 6. (ec —y+t+a+b)* 
 
 2. (8x—y+d)?. 7. (m—n—p—q)’. 
 
 3. (2a4+3b—y)’. 8. (2a+y—sz+a)’. 
 4 (3a—5a—2y)* 9. (Ww—ab+2xe—s3y). 
 5. (2+5a—4))*. 10. (2x —y? + ay — a)’. 
 6. The Product of Three Binomials. 
 
 By actual multiplication, it is found that 
 (x +a) (x+6)(x+e)=x°4+ (a+6+4+e)x? 
 + (ab + bc + ca) x + abe. 
 
 The product is arranged according to the powers of the 
 common letter x The coefficient of 2? is the algebrate 
 sum of the second terms of the binomials, the coefficient 
 of 2 is the algebraic sum of their products in pairs, and the 
 term free from z is the product of the second terms of the 
 binomials. 
 
 (1) @+1)(@+ 2)(@+3) 
 
 =e+(14+2+43)e?+(1x241x 342 KS) ah ee 
 sagt 627-11 oe G, 
 (2) @+2)@—3)(a +4) 
 =a + (2—3+44)a? + [2(—3) +2 ‘i 3)(4) Je 
 + 2(—38)(4) 
 eo Boe 10 ods 
 
 EXERCISES. 
 Write out the results in the following: 
 1. (w+ c)(x + d)(a + e). 3. (6+ 2)(6—1)(6 +38). 
 + 2. (atajy(aty(ate). 4 (y—3)(y+2)(y—1). 
 
IMPORTANT IDENTITIES. T5 
 
 5. (m+5)(m — 4)(m — 3). 7. (a — d)(a — 3)(a? + 8). 
 6. (ay +2)(ay—T)(ay 1). & (4) + 11)(y'—7. 
 9 (@+y+5)(@+ty+3)@+y4+2). 
 10. (8a+y—2)(8ea+y—4)(8x+y+4+6). 
 ll. (aw+b+42)(av+b-+4 8) (ax+b—5). 
 12. (ax + ba + 4) (aa? + bx — 2) (aa? + bx —1). 
 
 £ T. The Cube of a Binomial. 
 If, in the identity 
 (e+a)(@+b)(@+c)=2+(a+b4e)2 
 + (ab + be + ca)x + abe, 
 we put 6 and ¢ each equal to a, it becomes 
 (7+a)(#+a)(e#+a)=22+ (a+a+a)z? 
 + (aa +aa+aa)zx+ aaa, 
 or (xa)'=x°480xr43a'%x+a° 
 =x*+a°+3ax(x+a). 
 If, in the above identity, we change a to — a, we have 
 (x —a)=x°—3ax?+3a°x— a? 
 = x°—a®—3ax(x—a). 
 The cube of a binomial is the cube of the first term plus 
 three times the algebraic product of the square of the first 
 term and the second term, plus three times the algebraic 
 
 product of the first term and the square of the second term, 
 plus the algebraic cube of the second term. 
 
 A) @+2yfP=e+3@)'2y+38@2 y+ y) 
 
 =0'4+6 ay+12 ay’?+8 7’. 
 Ba) 2 by ==(2 2)°+-3(2 a)*(—3 b) +-8(2 a)(—3 by 
 +(—3 b)? 
 
 =8 a'— 36 a*b+ 54 ab?—27 0°. 
 
76 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 Write out the results in the following: 
 
 1. (2a +50). 9. (4ab—5y)’. 
 2, (a—3d)*. 10. (5 48 =). 
 3. (8244) 2 
 11. [(@+y)—2)%. 
 
 4. (2xa—5y)* . 
 
 a 12. [(w—y) +a]®. 
 5. (3 a@ + ab)”. 13. [2(¢@+y)—3a)* 
 6. (2 mn — pq). \ 14. [ (aa a b) { Ny 
 7, SCD Seu 1s 15. [((a+b)+(#+y)]* 
 8. (6y—1)y. 16. [(ax +b) — (ca+d) 
 
 8. The Binomial Theorem. | 
 (a + 6)"=a"+ na" + nin 2) = q?26? ata ie —*) 5 ve 2 a°5: 
 
 f+ seeeeee tL nab-! a 6". 
 
 The above identity is known as the Binomial Theorem. 
 A general proof for it will not be given. For the present 
 we will limit the exponent n to integral values. 
 
 If we multiply both sides of the identity 
 
 (a+b6)=a?+ 3 a7b+3 ab?+ 6? by a+ 6, 
 we have (a+0)t=at+4 ab + 6 al? + 4 ab? + Ot. 
 
 If in the binomial theorem we make n = 4, we have 
 
 4-3 4-3-2 7 es es | 
 By at ah eee eee 33 
 (a+b)t=at+4a cA (a fy + Tone aa WT 
 
 =a'+ 4 ab + 6 ab? + 4 ab? + 6, 
 
 54 
 
 
 
 a result which agrees with that found by multiplying. 
 
IMPORTANT IDENTITIES. T7 
 
 If we multiply both sides of the identity 
 (a+ b)t=at+4a% + 6 a0? +4 ab? + Ot by a+, 
 we have 
 (a+ b)>=a° +5 ad + 10 a30? +10 a7? + 5 nee iP 
 If, in the binomial theorem, we make n = 5, we have 
 
 (a+b)S=a5+ 5 ath + a : 
 
 
 
 azb2 + - 
 
 = 05+ 5 a!6 +10 a362+ 10 7004.5 “BA. 
 a result which agrees with that found by multiplying. 
 _ The identity 
 (a+ 6)"=a"+ na” 6 ail aed ar a8 mn OS 2’) q” 68 
 
 as Suk dss - nao + b* 
 is often called the binomial expansion. 
 The following Laws should be observed in regard to 
 the Hxponents and Coefficients of the successive terms of 
 the binomial expansion. 
 
 (1) Law of Hxponents. The sum of the exponents of 
 a and 6 in any term is always n; the leading letter a ap- 
 pears in the first term with the exponent n which decreases 
 by unity in each succeeding term; the letter 6 appears in 
 the second term with the exponent 1 which increases by 
 unity in each succeeding term to the last term 6”. 
 
 (2) Law of Coefficients. If any term be taken, the 
 coefficient of the next succeeding term is obtained by 
 multiplying the coefficient of the given term by the ex- 
 ponent of the leading letter a, and dividing this product 
 by the number of the given term in the series. 
 
Tpaee THE ESSENTIALS OF ALGEBRA. 
 
 Thus, in 
 (a+ bys at + 4a + 6.00? +4 ado +O 
 the coefficient 6 in the third term is obtained by taking the 
 product of 4, the coefficient of the preceding term, by 
 the exponent of a, 3, giving 4 x 8, and dividing by 2, the 
 number of the term 4 a4 in the series. Hence, the coeffi- 
 cient of the third term is 4x 3+2=6. 
 
 Pascal’s Triangle. The coefficients of the terms in the 
 expansion of 
 (a+ b)!, (a+ 6)’, (a+b)%, (a+ b)4, ete., 
 
 may be arranged in a table forming what has been called 
 Pascal’s Triangle. ‘The arrangement follows. 
 
 Coefficients of (a+6)!are 1 1 
 
 Coefficients of (a+6)?are 1 2 1 
 Coefficients of (a+ 6)? are 1 3 8 1 
 Coefficients of (a+6)*are 1 4 6 4 1 
 Coefficients of (a+6)® are 1 5 10 10 5 1 
 
 etc. 
 
 Each number appears as the sum of the number im- 
 mediately above and the one to the left. Thus, the first 
 10 is the sum of 6 and 4; the second 10 is the sum of 
 4 and 6; the last 5 is the sum of 1 and 4. By this simple 
 arrangement the binomial coefficients for any power of 
 a+6 may be easily written out, provided we know the 
 coefficients of the expansion of a+6 for a power one 
 lower. Knowing 
 
 Lieb LOTR ecm 
 
IMPORTANT IDENTITIES. 79 
 
 to be the coefficients for (a+ 0)°, the coefficients for the 
 
 sixth power of a+ 6 are 
 
 1-6) ee Z0 eB 1 
 (1) (a+2b)=at+4a%(2b)+6a7(2b)? +4a(2b)*>+ (25) 
 = 0°48 ob - 24 0°b* ++ 32 ab? 16 bt 
 2) @2—y)s (22)? +5(2 a) —y) +10(2 2) — y)? 
 + 10(2 2)" —y)? +5 (2 @)(—y)*+(— yy 
 = 32 2° — 80 aty + 80 wy? — 40 a?y? + 10 ay* — 2. 
 
 EXERCISES. 
 
 1. Extend Pascal’s triangle to include the expansion of 
 
 (a+ b)”. 
 
 Write out the following expansions: 
 
 6 btbriniie 8 
 
 2. (a+b)*% (Use Paseal’s triangle.) i (« 1 s) 
 
 3. (x+y) 9. (8y—22)*. ~ 
 
 4. (m+n)*. 10. (2a +b)’. LG: , a cet 
 ae 17. e+ Y)+o/*. 
 
 : ep ll. (a@—3y)’. : 
 
 : ° 4 2 : A » 6 18. [ (a+b) —2x]* 
 
 6. (y+ 1)”. 12. (212+ a)* 19. [Sa—(a—y)]’. 
 
 7. (24+ a)%. 13. (w —2y)". Wee 22 
 
 8. (p— gq). 14, (a? — ¥)°. é Sey 
 
 EXERCISES. 
 
 By comparison with types write down the following products : 
 
 a 
 
 . (@+5)(@+3). 
 (x + 10)(x— 2). 
 
 . («—5)(a@+ 8). 
 
 . (ax + 3)(ax + 5) 
 
 &® BD N 
 
 5. (vy + a)(ay — 6). 
 
 6 
 7 
 8 
 
 6. (4a+2)(4a—5). 
 
 . (a? + 5) (a? — 10). 
 . (w+a43)\(a@+a—7). 
 
 wa 
 
80 THE ESSENTIALS OF ALGEBRA. 
 
 ey 
 
 10. 
 is ip 
 12. 
 
 13. 
 
 ey) 
 
 
 
 (3 + by)(4 + by). 14. (Sx+y)(Bux—y). 
 (7? +324 2)(@’+3a—5).. 15. (Qe—4y)\(2a+4y). 
 Baty)(B«a—2y). 16. (2a+y+8)2x2+y—8). 
 
 (a+b+12)(a+b—6). 
 y 17. (a+ y?— 4)(a? + y? +4). 
 25+ h)\(o—5—¥ . 
 2/\ 2 18. (av + by + c)(ax+ by — ¢). 
 19. (@’+1+42)(#?+1—2). 
 20. (ay + yz +2@ —a)(ay + yz +244). 
 
 Perform the operations indicated : 
 
 21. (wt+y+z)’. 27. (a+ y")?—4 ary”. 
 
 22. (8%+2y+2). 28. (v«+y+z2+w)’ 
 
 23. (aw+by+c)*. 29. («+3)(~+5)(@+6). 
 
 24. (x +3)?—122¢. 30. («+3)(a—3)(x+5)(a —5). 
 25. (w+3y—4)?+16(e+3 y). 31. («+3 a)(2?—3ax+9 a’). 
 
 36. 
 Si: 
 38. 
 39. 
 40. 
 41. 
 42. 
 43. 
 
 3xe+2 y+6)—24(3 2+4+2 7), 32. (~+35 yy)’. 
 33. («+ y)?—S vy(a@+y). 
 34. (a? +0°+C—ab—be—ca)(a+b+ ce). 
 35. (2° 4+4y4+1—2ay—2y—a2)(a+2y+1). 
 
 Show the truth of the following identities: 
 
 (w+ a)? — (@— a)’ = 4 a2. 
 
 (x + a)? — 4 ax = (a — a)’. 
 
 (P+ ay +yja-—ysv—y¥. 
 
 (P—wa +Pjeryav’t y. 
 
 (a + ay + xy? + 9°) (a — y) = at — of. 
 
 (2? — ay + ay? — y?)(x + y) = at — yf. 
 
 (a? + wy +o?) (a? — ay +o?) = xt + ay? + of, 
 
 (0? + y? +22 — avy —y2—2ze) (e+ y4+2=07°+ 742-8 awyz. 
 
IMPORTANT IDENTITIES. 81 
 
 70. Division Identities. From the relation of division 
 to multiplication, every multiplication identity gives rise te 
 at least two division identities. The following division 
 identities are of importance: 
 
 1. [x°+ (a+ 6)x+a6]+(x+a)=x+6. 
 2. (x°4+2ax+a’)+(x+a)=x+a. 
 3. (x4°—2ax +a’) +(x—a)=x—a. 
 4. (x°—a?)+(x—-a)=x+a. 
 oD (xe—y)+(x-yaxt+ayty’. 
 6. G+y)+ (x+y) =x x+y’. 
 7 xi—yie(x—-y)= x4 xytay' ty’. 
 It should be noted that in each of these identities, the 
 quotient might be the divisor, and the divisor the quotient. 
 (1) (#’—64)+(7%+8)=? 
 This is an example of type 4. By a comparison with that 
 type we see at once that the quotient is x— 8. 
 (2) (a@—11 4+ 30) + (#—6)=? 
 This is an example of type 1. 
 a=—6; ab=30; hence, b=—5. 
 The quotient is #— 5. 
 (3) 27 —a*)-+(8—a)=? 
 If we notice that 27 = 3°, we see that this is of the form of 
 
 type 5. 
 Hence, (27 —a®)+ (3 —a)=9+3a+a% 
 
 EXERCISES. 
 
 Perform the following divisions by comparison with 
 type forms: 
 1. (a?— 2a—63)+(a+7). 
 2. [(2 x)*— b*] + (40? — 0”). 
 
Oo ON AA Pw 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 (9+6a+a’)+(384a). 
 
 (8 a + b®) + (2a+5). 
 
 (81 a? — 25 b) + (9a+5 5B). 
 
 (1 —10a+4 25 a7) +(1—5a). 
 (y? + 11 y— 26) + (y+13). 
 
 [(a + b)? — 64] + (a+b—4). 
 
 [27 a — (@—y)*] +[8a—(a—y)]}, 
 [(a@+ y)?—4 ab?] + (@+y+2 ab). 
 (at — 14 a — 51) + (a — 17). 
 
 - [@+y)+11@+y) —60]+(@+y=—4). 
 - [et+y)’—16@+y)\a+b) +48 +b) ]+[@+y) 
 
 —4(a+b)]. 
 
 . (0 —9 x” — 112) + (a — 16). 
 . ("+16 3” + 64) + (y" +8). 
 
 [(a + 6)? —6 a(a+b) 4927] +(a+b—32). 
 
 . (a — 25 yy?) + (a + 5"). 
 
 [wt y)™— (a+ d)"] + [e+ y)"— (0 +)" I, 
 
 [ (ax + b)? + 12527] + (av + 6)? — 5x (ax + db) + 2527]. 
 [ (aa? + ba + c)? — (lx + m)*] + (aa? + bu +c+lex+n). 
 (y®" — a) + (y" — 20"), : 
 
 ) (y*™ eat a) 5 (y” hal a"). 
 
 
 
 [((V8a+y)?— a] +(V38e+y—a). 
 [( Vax + 6)? — 647] + (Var +b—4y). 
 
 : [(V aw + by)? + 125 a®] + (Vax + by +5 a’). 
 
CHAPTER VI. 
 FACTORING. 
 
 71. Products, Factors. Numbers which are multiplied 
 together to form a product are called factors of that product. 
 
 For example, in 5 x a x b=5atb, 5, a, 6 are factors of the product 
 5ab. Only expressions free from divisions and roots will be con- 
 sidered in factoring. In 7a(b+c)(x+y+2), 7 isa numerical factor 
 or numerical multiplier, a is a monomial factor, b + ¢ is a binomial factor, 
 and «+ y+ 2 1s a trinomial factor. 
 
 In multiplication, we have the factors given to find the 
 product; in factoring, the product is given to find the 
 factors. 
 
 72. The Degree and Number of Factors. The degree of an 
 algebrate monomial is the number of letters composing tt. 
 
 Thus, a%b is an expression of the third degree, being made up of the 
 product of axaxb. a*x*y is an expression of the fifth degree in 
 
 a, x and y; it is an expression of the second degree in a, also in z; 
 it is of the first degree in y. 
 
 The degree of an algebraic polynomial is the highest num- 
 ber of letters found in any term. 
 Thus, z?+ 3 2+ 4 is an expression of the second degree, containing 
 
 xxx in its highest term. az?+bz is an expression of the third degree 
 in a and z, but is an expression of the second degree in z. 
 
 The number of factors of an algebraic expression is not 
 
 greater than the degree of the expression. 
 83 
 
84 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 Determine the degree of the following with regard to all the 
 letters : 
 
 iT; 
 
 73. Monomial Factors. 
 may be written down by inspection. 
 
 Thus, 
 
 also, 
 
 oe + an. 
 
 2 UY +2. 
 
 3. wtoyt+y’. 
 4. 
 5 
 
 e4t3e°7+42-+ 16. 
 
 te? + be Le. 
 
 10. 
 
 = eS 
 
 e+ y+2—3 vyz. 
 ao? — a’. 
 
 v4? ata 
 abe-+- 74-0 =e 
 abe + b’ca + cab. 
 
 Factors of monomial expressions 
 
 abaz=a-a-b-a-a2; 
 
 DMPPeo.E EY + YY. 
 
 Here, 5 is not an algebraic factor in the sense of determin- 
 ing degree; it is a numerical multiplier. 
 
 Monomial factors contained in polynomials may be seen 
 as the inverse of the distributive law. 
 
 (a+6+c)m=am-+ bm-+ em. 
 
 Reverse this identity, and we have the factors of 
 
 am +bm-+em, namely m anda+6+e. 
 
 Factor: 
 
 Sela ee Oe ON fT 
 
 14 a*aty. 
 — 5 wy"s. 
 2 xyz, 
 
 § a?b?c°. 
 
 — 10 a®be. 
 12 ar ye", 
 
 EXERCISES. 
 
 as 
 
 4 abe! 
 
 8. 40°4+7Tv=a2(4e+7). 
 9. ax? +ay. 
 
 10. 
 11. 
 12. 
 
 3 ax + 6 aa 
 5a?+10ab+5 abe. 
 3 xy — 6 ay? +7 aeys 
 
FACTORING. 85 
 
 13. 3ma+4 ma? + 2 may. 17. 5a®yz + 30 a’y’z — 40 ayz’. 
 18. 7 a’a*yz +7 aay e +7 a®ay2’. 
 19. 8a@beVae+y—6ab'oVa—y. 
 20. 5 a’yzVax +b 
 
 16. Savy — 15 aay + 21 a®ay’. +10 ay22Vax—b. 
 
 14. —2lx+4ly+6 ly. 
 
 15. avy + aayt+ 3 azz’. 
 
 ~ 
 
 TTY Pits. 
 
 A great number of algebraic expressions may be fac- 
 tored by comparison with some known form of product. 
 The identities of the preceding chapter are reversible ; 
 when so written they become Types in factoring. 
 
 74. The Type x’ — a’, the Difference of Two Squares. This 
 expression is recognized as the product of «+a by x—a. 
 
 Hence, 
 x’ —a?=(x+a)(x—a). 
 
 The difference of two squares equals the product of the 
 sum and difference of the square roots of the two numbers. 
 Thus, a#—16=(4%#+4)(w#— 4); 
 also, (a+ b)?—16=(a+6+4)(a+6—4). 
 
 Factor: EXERCISES. 
 1. a’ — 40%. 6. 25 a27y? — 36 22. 
 oan — 9b’. 7. (e+y)— a. 
 3. a’b? — c’. 8 («+3y)— 2. 
 4. 16a°b?— 250’. 9. (3x—2y)?—(e+a)’. 
 5. 4—9 2”, 10. 4(x+2y)?—9(a+b)*. 
 
 M. et—Y=CW-Y)@+Y=e@+yYVNA-yCt+y. 
 
86 THE ESSENTIALS OF ALGEBRA. 
 
 In Exercise 11 we have two first-degree factors, «+ y and 
 a —y, also the second-degree factor, a?+y?. No factors of 
 a +4-y? can be found unless radicals be aes Such a 
 factor may be called irreducible. : 
 
 Factor : , 
 
 12. 42° — 16 (@—34)*. 14. (w+ y)*— 4 ay’. 
 13. 16—a*. 15. (8%+4y+45)—9. 
 16. (— Sor oy pe (w+ y)’. 
 
 17. (lv+ my +n)? — 4 (ax + by +c)”. 
 18. (aw + by)? —4 (lz + mw)’. 
 , 19. (aa)? — (b"y")?. 22. (x + 3)? — 
 20. 7" yy" — 2 4p", 23. (Sa+a)’?—9 Bb. 
 21. («@+1)?— a’. 24. («+a)*— (a+ 6)* 
 
 25. (@+3y)?—(S8x+y). 
 
 * 75, The Type x°+(a+6)x+a6. This expression is the 
 product obtained by multiplying +a by «+6. Hence, 
 x°+ (a+ 6)x + ab= (x +a)(x+ 5). 
 
 Examples belonging to this type assume the form 
 
 x’ + sx +p, | 
 
 where p is the algebraic product of two numbers, and s 
 is their algebraic sum. 
 
 If s and p be integers, the factors of p may sometimes 
 be found by inspection such that their sum shall be s. 
 
 Thus, to factor 22 + 62+ 8, we must find two factors of 
 8 whose sum is 6. These are seen to be 4 and 2. Hence, 
 
 v+6xr4+8=(1+4)(%+ 2). 
 
 To factor. a2?+ 10a — 24, we must find two factors of 
 
 — 24 whose sum is 10. These are 12 and —2. Hence, 
 
 a?+10a— 24=(a—2)(a4+12). 
 
FACTORING. 87 
 
 It should be observed that if in a®+sx+p, p be post- 
 tive, the factors of p chosen must be of like signs, and if s be 
 positive, both factors of p must be positive. If s be negative, 
 the two factors of p must be negative; if p be negative, one 
 factor of p must be positive and the other negative, and the 
 sign of 8 shows which is numerically the larger. 
 
 EXERCISES. 
 
 1. Factor 7? — 5y — 24. 
 
 Here the factors of —24 are —1, 24; —2, 12; —3, 8; 
 —4,6; and also these numbers with their signs changed. A 
 pair of factors must be chosen whose sum is — 5; this is seen 
 to be 3, —8. Hence, 
 
 f= Dy — 2A'=(y 4+ 8)(y — 8). 
 2. Factor (x + 2)?— 5(a + 2) — 14. 
 The two factors of —14, whose sum is — 5, are —7 and 2. 
 Hence, 
 (w+ 2)? —5(@+2)—14= (44+ 24 2)@+2-7) 
 = (x + 4) (a — d). 
 3. Factor (2?+ 32)? — 8 (a? + 3a) — 20. 
 The factors of —20, whose sum is — 8, are —10 and 2. 
 Hence, 
 (a? + 32)? — 8 (a? 4+ 32) — 20 = (a? 4 3a + 2) (24+ 3a%—10) 
 ={(@+1)(@+2)}{(@+5)@—2}. 
 
 Factor: 
 
 ara 3 ot 2. 9. v+5ay+6y". 
 
 5. 2—3a44+2. 10. v’y’— 3ay— 10. 
 
 6. a?+a2—2. 11. 143 a7 —102’y7, 
 
 7. v—x— 2. 12. 62° —5ay+y’. 
 
 8 a’o7’+ 5 az +- 6. 13. (w+ y)’+ 9(a@ + y) + 20. 
 
88 THE ESSENTIALS OF ALGEBRA, 
 
 14. (a+ 3b)?— (a+ 3b) — 20. 
 
 15. (67+ 35y)’+10(6%+43y) + 16. 
 16. (+ 4")?— 6 (a+ 9’) — 27. 
 
 17, (270+8y+4+5/4+9(2e243y+5) +18. 
 18. (@+ 5x)?+10(a’+ 52) + 24. 
 
 19, a'— 15 a+ 36. 
 
 20. (ax + by)?+ 8 (ax + by) +7. 
 
 21. (ax + by)?— (1+ m) (aw + by) + Im. 
 22. (a+ 6a)?+17 (a+ 62a) +72. 
 
 23. (a@’—5a+4)— (@’—5a+ 4) —2. 
 24. w"™—104"+ 16. 
 
 25. (ax)"+ (1+ m) (axv)”+ Im. 
 
 f 76. The Types x°+2ax-+a’ and x*°—2ax+a’*. These 
 two trinomials are perfect squares of 2+a and w—a, 
 respectively. 
 
 x 4+2ax+ a= (x+a)(x+a)=(x+a)’. 
 x°—2ax + a°= (x —a)(x—a)= (x—a)’. 
 The sum of the squares of two numbers, increased (or 
 
 diminished) by twice the product of the numbers, equals the 
 square of the sum (or difference) of the two numbers. 
 
 EXERCISES. 
 1. Factor 2? +62 + 9. 
 
 Here x and 9 are the squares of # and 3, and 6 @ is twice 
 the product of 3 and 2; hence, 
 
 ao +6e+9= (@+3)(@+ 38) = (@ + 38). 
 2. Factor 927+ 62-41. 
 9e°+6e4+1=(862+4+1)(8241)=(82+4+1)2 
 
FACTORING. 89 
 
 3. Factor 16 a? + 40 ab + 25 b2 
 
 16 a? + 40 ab + 250? = (4a)?+ 2(4.4)(5 bd) + (5 bY’ 
 =(4a+5b)(4a+45b)= 4a+4+ 5D)’ 
 
 4. Factor y? — 16 yz + 64 2”. 
 
 y —16 yz + 642 = y’? — 2(y) (82) + (82)? 
 
 Se 8 2) — 8) aa (Yin 8 2). 
 
 5. Factor (w? + 4a)? —4(a@’°4 4a) +4. 
 
 In this expression we may consider xz? + 4. as a single quantity. 
 
 (?+ 4a)? —4(a? +42) +4= (e+ 420) —2(a?+4 x) (2) + (2)? 
 = (a + 4a— 2)(e’ +42 — 2) 
 
 = (a + 4a — 2)’. 
 Factor: 
 
 6. «7 —S8x-+ 16. 
 7. 40°—12274+ 9. 
 8. ax? + 10 ary + 25 7”. 
 9. 49 a*b?— 14ab +1. 
 10. 100 — 20 ab + a?7b’. 
 ll. (aw + b)? +2c(aw+ b) +’. 
 12. (6e¢+4y)?’—6862+4+4y)4+9. 
 13. 40? 4 99? — 12 ay. 
 14. (ax + by + c)’?+ 8 (aw + by + c) + 16. 
 15. (2+ y’)? —2(a? + y?)2 + 2. 
 The next three exercises are squares of trinomials. See page 73, 5. 
 16. 74+ yY4+24+2 ay + 2yz2+ 2 2m. 
 Peo oe — 2 ab + 2 bc — Zea. 
 18. 40° + 9y +22412 ay + 6y2+4 20. 
 
90 THE ESSENTIALS OF ALGEBRA. 
 
 19. v7"4+12 2" + 36. 
 
 20. yi" —14 "4.49. 
 
 21. a” + 12 a"b™ + 36 0*. 
 
 22 (wt+y)"—G6a(a+y)"+9 a. 
 
 77. The Types x°?—y’ and x°+y’*. It has been shown by 
 actual multiplication that 
 YS (x — ye + ay ty); 
 and O+ p= (x+y) —xy+y’). 
 
 EXERCISES. 
 1. Factor 8 a? — 27 b*. 
 8 a? — 27 D8 = (2 a)’ — (8B)? 
 = (2a—30)[(2a)’?+ (2a) (86) + 6 b)*], 
 = (2a—5b)(40+4+6ab+49 0"). 
 2. Factor (a+ y’)? — 8 ay’, 
 (a? + 9%) — 8 ay = (a? + y?)8 — (2 ay)? 
 =(@'+y')—2ay]l[@+y) + @ + yx) 2a 
 + (2 xy)”] 
 =(@—y)[@ +9)? +2 ay (+ y’) + 4 xy’). 
 3. Factor (x + 5)? + 8 b% 
 (x+5)?+ 8b? = (+ 5)? + (2b) 
 =[(+5)+2b][(w~+ 5)? — (#4 5) 264 (2b)"] 
 =([%+5+426][(@+5)?—2b(@+ 5) +4 6" ]. 
 
 Factor: 
 4, 2° —8 Be, 8. («+ y)?—125 2a. 
 5S 0? a0 Tobe. 9. (8x2+4+4)+8 y% 
 6. a’x*® + 64. 10. (w+ 32+ 4)%— 64. 
 7. ofa — 27 yf, 1l. (av + by)? — ca’. 
 
 12. (8a%+4y)>+2a+y)% 
 
FACTORING. 91 
 
 Certain expressions may be transformed into the sum or 
 difference of two cubes, and the factors then found. 
 13. Factor «* + 7°. 
 In this case we may write 
 ab + y= (0°) + (yy 
 = [a*+y"] [@*)? — Gy’) + YY" 
 = [2+ y?] [at — vy? + y']. 
 14. Factor 2&—y*. 
 As in Exercise 13, we may write 
 gf — y= (a)? is (y?)3 
 = (x? iy y’) f(a")? ae ha an (y”)?] 
 = (#+y)(@—y) [t+ vy t+y*] Psst? 
 =@+y)@-y@t+ayty’) (a? —ay +47), 4, OG Fe 
 (See Exe ay f 42 page 80.) 4 
 
 
 
 Of course, the factors of 2° — 7° could hav en obtained by 
 comparing with the type 2 — a. Thus, Wy he 
 fi 
 
 eva @y— yy fe ly 
 ee ty)G@—y) gh! 
 =(@+y)@—a ty’) ery +2y +y%). 
 Other types of binomials, such as “at — bt, a? + b°, and so on, 
 
 may appear for factoring, but such special cases will not be 
 considered at this time.!'. © 
 
 | ” 
 
 Factor : | ; 
 15. Sa — b°. 18. a®u’+ (y+2)% 
 16. (ax)® + (by)®. | 19. (ax + by)* — (cz)% 
 17. 2° — 1. ; 20. 64 a°+ (bc)®. 
 
 21. [(w + Wii yy ls 
 27 ae + 3 ab? + b%)? + c%, : 
 
 3 
 
92 THE ESSENTIALS OF ALGEBRA. 
 
 78. The Type Ax°+Bxy+ Cy’ or Ax°+Bx+C. If the 
 first of these expressions is capable of separation into 
 factors free of radicals, it must be composed of two 
 binomials of the form az+ by and lx + my, where a, 6, J, 
 m are algebraic numbers, 2.e., they may be + or —, inte- 
 gral or fractional. 
 
 Multiplying these supposed factors together, we have 
 
 (ax + by) (le + my) = alz? + (am + 61) xy + bmy?. 
 Hence, to factor Az*+ Bry + Cy? we are to find four 
 numbers, a, 6, 1, m, such that 
 al= A, bm=(C, and am+6l/= B. 
 This method is illustrated by the following examples : 
 (1) Factor 62?+ 31 xy +35 y’. 
 The factors of 6 are 6, 1 and 3, 2; the factors of 35 are 35, . 
 1, and 7, 5. <A good plan is to arrange the letters thus, 
 was WiC DD 
 c+ y 
 Now attach the factors of 6 and 35 to a, y, respectively, as a 
 trial arrangement. Let us place them thus, 
 Sa+Ty |. 
 “2a+5y 
 The square terms appear correctly, 6 2’, 35 y, but the cross 
 
 products, 3x-5y and 2a“-7y, do not add so as to give 31 ay. 
 Hence, our trial arrangement is not correct. Let us try 
 
 2e+7T7y) 
 
 ou+5y) 
 This gives 6 a?+(2x54+3x7)ay+ 35 y’, the correct product. 
 Hence, 62°+4 3lay4+35y=(2247y)(82+5y). 
 
FACTORING. 93 
 
 The case in which y is equal to unity, giving Av’?+ Brt+ 0, 
 requires no special mention when the factors may be found by 
 inspection as above. 
 
 (2) Factor 3 a+ 16 ay +5 7’. 
 
 In this example we are to find the arrangements of the 
 factors of 3 and 5 with the letters x, y in 
 
 | a+y) 
 aty)’ 
 such that the sum of the cross products shall be 16 ay. 
 _ By trial, the arrangement is seen to be 
 eves 
 
 | det ¥ 
 
 Hence, 32°+16ay+5y=(w+5y\(Ba+y). 
 
 (3) Factor 152°+ 58 2+11. 
 
 The factors of 15 are 3, 5 and 1, 15; the factors of 11 are 1, 
 11. Our trial arrangement may be | 
 
 ye a iD 
 15%+ 1 
 But the cross products 11 x15+1 x1 do not give 58. Another 
 | trial may be 30411 
 5eo+t i |, 
 
 which gives for the middle term 3a+550=58 a. 
 Hence, 152°+58e%4+11=(82+11)(2+1). 
 
 Factor: EXERCISES. 
 
 1. 30? +7 ay+2y’. 4, 120° +2 vy —2 7’. 
 
 re 6 a? —5 ay —6 7’. 5. 3 ax’ + (9+ a) bx + 3 b* 
 3. 21¢’+3lay4+4y7". 6. 20 7° — 22 yz + 6 27, 
 
 Tek —2Gu2 — 8 2, 
 
O4 THE ESSENTIALS OF ALGEBRA. 
 
 79. The Type ax’?+6x-+ce. ‘This expression is called 
 the General Quadratic in a single variable x. For differ- 
 ent values of a, 6, e this expression represents every 
 quadratic that may be written. Thus, with a= 5, 6=6, 
 c= 2, we have 627+ 627+2; with a=4, 6=—5,c=—l, 
 we have 427— 5a — |, etc. 
 
 We shall illustrate the method of factoring the general 
 quadratic by a few special examples. 
 
 (1) Factor 2a? + 16a — 20. 
 By dividing out 2, we get 
 2x°+ 16% — 20 
 = 2 (a? + 8a — 10) 
 = 2[a°+8 «+16—10—16], adding and subtracting 16. 
 
 = 2[(a + 4)? — 26], rearranging terins. 
 26 may be written (/26)?, and we have 
 20 +162 — 20 
 
 = 2f (a 4+ 4)? —(V26)"), the difference of two squares. 
 =2[4+4+ V 26] -[e+4—~26], the product of sum 
 and difference. 
 
 Note. Adding 16 is called Completing the Square of the first two 
 terms. ‘To complete the square of x? + 2 mz we must add m?; 1.e., we 
 add the square of half the coefficient of x. To complete the square of 
 
 x? + 6 x2 we add (S)= 32. To complete the square of 2? + kx we add 
 
 Complete the square: 
 1. a’ + 52. 
 
 Here (3)? must be added. 
 If we wish the expression to remain unchanged, we must 
 also subtract (3)’. 
 
FACTORING. 95 
 
 Hence, e+ 5e=a+ 5a 4 25 — 25 
 
 
 
 a) 
 = (w+ $28. 
 2.  — 32. Y aed Wi ete a Sr 
 3. a +7 x. 8° 5a? +7 
 4. vw’ — 8x. 9. 3a’— 82. 
 5. 380°4+9x=3(a? +32). 10. 72? — 35a. 
 6. 52? — 25a. 11. 92?— 25 a. 
 (2) Factor 5a? 4 15a — 10. 
 527° +152—10 
 = 5[a?+ 3a — 2], dividing by 5. 
 = 5[a?+3a+ i — ($)?— 2], completing the square. 
 = 5[(x+ 3)? -— 2-2], rearranging. 
 =5[ee+ 471 
 = 5[ (a + 8)? — (V1)"], difference of two squares. 
 = 5[a +54 ek E pores Sl product of sum and 
 p; difference. 
 
 (3) Factor 42° + 6a + 2. 
 
 4o? +6242 
 = 4[a?+ 3a+41), dividing out 4. 
 =4[a’?+23e+ (3)?— (3)?+ 4], completing the square. 
 =4[(#+3)?— 3], rearranging. 
 =4[(@ + 3)’— (4)’], rearranging. 
 
 =4[4+2+41]-[v%+2—1], product of sum and difference. 
 
 =4[e%+1]-[v+4]. 
 
 According to the method here illustrated we may factor 
 the general quadratic. 
 
96 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 ax? + ba+e 
 = af 2 + es + al dividing out a. 
 a a 
 yale b \2 6 \2 
 =) 2 + —- 2+ | — ce , completing square. 
 a 2a Qa 
 ay b6\2 &—4ae si id 
 = al (x + ) — Pe combining terms. 
 = al (2 +5 ~)- “(ae 4 “) ‘| difference of two 
 4 at squares. 
 my Vb? — 4 ac b ia 
 =2+3,+~ G2] [7 +208 
 
 Hence, to factor any quadratic in which a, 6, ¢ have 
 been replaced by numerical values, we need merely to 
 replace those letters in the general factors above by the 
 special values found in the given example. 
 
 (4) Factor 3 a?+8 a+ 2. 
 
 Here a=3, b=8,c=2; hence, we have 
 
 
 
 
 
 
 
 
 
 
 
 3a + 82+ 2 
 
 8 , V64—4.3.- 8  ~V64—4-3-2 
 =3 Nest: g4 2. 
 
 Nae gone la 2am ] 
 
 on te 
 
 (5) Factor 5a?— 7a +4 3. 
 
 In this case a= 5, b=— 7, c=38; hence, 
 Dei eS 
 
 Ti eo ee —7 V(—7—4-5-3 
 
 =5 fs 
 
 Ate Bie aio al e+55 OE 
 
 
 
 
 
 v7 Th Gan Let ae ee 
 =s]e Tiyan | ets lle 10 Ge aT i 
 
FACTORING. 97 
 
 Note. The factors in this case, involving the square root of a 
 negative number are called imaginary. Such numbers as v — 11, 
 V —5) V— a, etc., do not belong to the algebraic number system ex- 
 plained in Chapter I. 
 
 (6) Factor — 32°+ 4a +4 2. 
 
 
 
 
 
 
 
 Here a =—3, b= 4, c= 2; hence, 
 —8attdet2=—afaty cit +2 
 2-(—3) 2.(— 3) 
 oe 
 ealT ee) ae) 
 
 coe a |S 24 us ry tien etl 
 agh Sb 
 u 
 
 =—3 [6-2-2 Vi) [2-245 1/40}. 
 jets EXERCISES. 
 226 7? — 25 a — 15. 6.0ll ae — 55a +99. 
 2. —22?4+142—10. 7, —427°—282+4+ 32 
 3. 32°7—52+4+9. 8. 62°—3027+4 48. 
 4. 7% —3524 49. 9. —82°+40 2-8. 
 5. —2’+11e¢—3. 1o. 10 x?— 702-4 20. 
 
 Il. FACTORS BY REARRANGEMENT AND GROUPING 
 OF TERMS. 
 
 The method of factoring certain algebraic expressions 
 will often be suggested by a proper rearrangement or 
 grouping of terms. Two general plans of grouping are 
 worthy of attention. 
 
98 THE ESSENTIALS OF ALGEBRA. 
 
 80. Grouping with Regard to the Descending Powers of 
 Some Letter. 
 (1) Factor 42° + 7 +244 2y +4224 2 yz. 
 If this example is not recognized as a perfect square, we 
 should proceed thus: 
 4 ary +244 vyt4 w242 ye=stet4a(yt2tyt+2 yz4+2 
 = 42°44 x(y+z)+(y+2)? 
 = (2x2+y+2). 
 (2) Factor 2? + 6ar—8 br +9 a?—240ab+16B% 
 Arrange with regard to the letter 2, giving 
 +6 ax—8be+9e@—24ab+16P 
 H=27+2e(3a—4b)+9e0—24ab4+160 
 =H=r+2e8a—46)+(8a—4b) 
 =(r+3a—4by. 
 (3) Ractar 4+T7axr+6 br+10 a? + 21 0b4+9 8% 
 Here we may arrange with regard to the letter a. 
 v4+T7axr+6b2+10a?+210ab+98P 
 =10a°?+Ta(#+3b)+2°+6b2495b? 
 =100°+7 a(x+3b)4+(e4+3 5)? 
 
 =(5a+2+3b)(2a+24+35). 
 “hae | EXERCISES. 
 1. 2¢0+3ar+2°4+7ab+50°+4 be. 
 2. 22° -8 +3 y+ bu—2 by +T ay. 
 3. 7 4+47+192—4ay—12 yz+6 22. 
 4. 4¢°4190?416°+4+12 ab—16ac— 24 be. 
 6. 22°—37—32—2y—52z+10 yz. 
 
FACTORING. 99 
 
 81. Grouping with Regard to Some Letter that Enters in 
 One Degree Only. By forming a product of a number of 
 factors, one of which contains a letter not found in the others, 
 we shall see the purpose and application of this method. 
 
 Let us multiply together the following: 
 
 (a#+353a+ b)\(7#+a4+™m). 
 The product is 
 e+4ar+3a*+be+ab+m(r+3a+b). 
 
 It will be noticed that m appears in but three terms, 
 and that these terms when collected constitute m times the 
 first factor of the product. Hence, if a literal expression 
 contain a single letter entering to a single degree only, the 
 coefficient of that letter contains a factor of the gwen expres- 
 sion, if the expression has any factors. 
 
 An example will illustrate this method. 
 
 Factor 32?+8ay+3ka+5ky+5y’. 
 
 Here k enters to a single degree. 
 
 Arrange with regard to k, and we have 
 
 8024+ 8ay+5y74+(8e+5y)k. 
 
 If the expression can be factored, 3%+5y must be one of 
 the factors; hence, 32?+8 xy +57? must containdx+5yasa 
 factor. 
 
 827 +4+8ay+5 743 ket bky=3v+8ay+5Y+k(8e+5y) 
 =(52+5y)(a#+y)+kh(5e+5y) 
 =(Se+5y)(x+y+h). 
 
 EXERCISES. 
 Factor: 1. 40?+4ay—357?+5ay—2 az. 
 2. Ptezr—4yz—ay—127y%. 4 wyt4ay42?—ar—20 a7 
 3. 4actbe—?412@—ab. 5. 142°47 pe—Sayt+py—y, 
 
100 THE ESSENTIALS OF ALGEBRA, 
 
 82. Binomial Factors by Trial. Let us divide 27+627+5 
 by x — m in the ordinary way. 
 x—-m)e+6e+o0(4+6+m, quotient. 
 x2 — mx 
 (6+m)x+5 
 (6 + m)x — m(6 + m) 
 m+6m + 5, remainder. 
 
 It will be noticed that the remainder found on dividing 
 a? +62+5 by «— mis precisely the value that the divi- 
 dend becomes when z has been replaced by m. Thus, if 
 in 2@+6a2+5, we put z=™m™, we get m+6m+5, the 
 same as the remainder above. 
 
 This remainder is, of course, true for any value of m. 
 We make a few special illustrations for various values of m. 
 
 (1) What is the remainder on dividing 
 7 x“ +6xe+5 by «—4? 
 
 Our result above shows the remainder to be the value of 
 v’+6ae+5 whenx=4; hence, the remainder after division is 
 
 44+6x4+5=45. 
 Verify by actual division. 
 (2) Find the remainder after division of 
 e’+6xe+5 by «—5, 
 Replace aw by 5, and the remainder is ' 
 °+6x5+5=60. 
 Verify by division. 
 (3) Find the remainder when 2? +6a-+5 is divided by a+ 5. 
 Here, m=—5; hence, the remainder found by dividing by 
 x+5is (= by 6 (= bya eb = 0) 
 
FACTORING. 101 
 
 The remainder being zero shows that # + 5 is an exact divisor 
 of 77+ 6 + 5. 
 (4) Find the remainder on dividing a —8#+12 by «—8. 
 The remainder is (8)? — 8(8) + 12 = 12. 
 Hence, « — 8 is not an exact divisor. 
 (5) Find the remainder on dividing x —8#+12 by «—6 
 The remainder is (6)? — 8(6) + 12 = 0. 
 Hence, x — 6 is an exact divisor, 7.e., a factor. 
 In factoring by trial, the number of trials is limited to the 
 number of divisors of the constant term. 
 (6) To find the factors of 2° — 7 a — 6. 
 The factors of 6 are +1, +2, + 3, + 6. 
 When divided by x—1 the remainderis 1— 7—6=—12. 
 When divided by «+1 the remainder is —1+ 7—6=0. 
 When divided by x—2 the remainderis 8—14—6=—12. 
 When divided by «+ 2 the remainder is —8+ 14—6=0. 
 When divided by x —3 the remainder is 27 —21—6=0. 
 No other divisors need be tried, for we already have the 
 
 three (© +1), (w+ 2), and (# — 3), and an expression of the 
 third degree can not have more than three factors. 
 
 EXERCISES. 
 Factor each of the following: 
 Lo — 2? —42+4 4. 
 Sea — 42 — 8. 
 ee oe — 13 2 — 15. e+ 15 2 — a2 — 15. 
 ag — 6 a + 11 « — 6. edo 47 & 60; 
 5. 2 —102°4+ 312—30. 10. 2—122°+4 482 — 64 
 
 gt 2'e* — 9'o — 18. 
 a +9 a7 26 o + 24. 
 
 oO ND 
 
102 
 
 26. 
 
 27. 
 
 a I 
 
 2. 
 
 THE ESSENTIALS OF ALGEBRA, 
 
 EXERCISES IN FACTORING. 
 
 bu? — b. 13. 27 — 642°. 
 10 em? — 40 c*. 14. 16 wy? — a. 
 a? + vy + wz + yz. 15. a — 20° SG, 
 lm + mr — lr — 1°, 16. m*—4 mn? + 4 nil, 
 20° +3ay—2az—3yz. 17. (a+b+c)?—(a—b—c)’. 
 §—5a+a’ 1s. 14a®—8a’—21a+4+12. 
 at + ay? — 72 v4. 19. p?—2pq+q?—7. 
 m—4m + 96. 20. 252?—10ay—92?+ y”. 
 hab? 2 a7b Has 21. 9a?— 3600+ o608 
 oe Aa 1: 22. a? -+-29a +120; 
 . 2P—8(m+ 1) 23. ot 412 27? — oo 
 . 64at — 81 4. 24. 247418743. 
 25 W(e@—y)+3aly—x) + 2(e—y). 
 pa te, 28... abc? — 2 a®b’c? + atd?c*. 
 BL afl 29. 9x°—6ay—1+y’. 
 m> + m> — m? — 1. 30. 646*-—127—1; 
 
 REVIEW EXERCISES. 
 
 Remove parentheses and simplify 
 2e—3sy—j{5ae—[8y+5a—(4a+y—38a—4y)]}. 
 
 Put «=5 and y=1 in the above exercise, and find the 
 
 value. 
 
 3. 
 
 From the sum of 3~a—8y+2z2 and 5y—Ta—3z2 take 
 
 their difference. 
 
 4. 
 5. 
 
 Multiply out (a* + 4) (a? + 2) (a? — 2). 
 Divide a" — Say + 3a"y— x? by a®—y. 
 
FACTORING. 103 
 
 6. Factor az’ + ba? —a— be. 
 7. Simplify 
 (a + a)? — (@—a)’?—[(@ + a)? + (a — a)(% — a) — x” J. 
 8. Divide a*+4y' by 2 —2ay42y’. 
 9. Two numbers differ by 17. One third of the smaller is 
 
 one greater than + ofthe larger. Whatarethenumbers? (Let 
 x= smaller, «+ 17 = larger.) 
 
 10. Divide (@+y*)(@—y’) by #—2a’y+ 2 ay’ — y'. 
 11. Factor «*—102?+ 9. 
 12. Add with respect to xv, ax +a(b—c)a—y'r+i11o. 
 13. Find the value of 
 
 w+ yz —yfa— (By +22) (5a+y—6)—2a—y}, 
 
 waen oO, y7—2, and z= 1. 
 
 14. Find the value of Spee when «=— i. 
 a 2) — x 
 
 15. (2a®— 32+ 5)? = what ? 
 16. Divide at? — gm yrth 4. gmtt yn — yfnt) py gm 4 yn, 
 17. Multiply a”*b + a7b"? by a™°b+ ab". 
 1s. Multiply (a*+1)(a"—1)(a"+4+2a"+1). 
 19. Verify a(a+1)(a+2)(a4+3)=(@+4+3a41)?-1. 
 20. Irom 
 (w@+y+2)(a+y—2) take a? —{y—[2y°—(—2ay+2)]}. 
 21. Find the sum, difference, product, and quotient of 
 Aa(y—z)ynr* and a(y—z)n"*, 
 22. Factor a®’—4a7+a-+6. 
 23. Divide 8a? —7°+2°+6ayz by y—2-—2. 
 24. From 4{(@—3y) —1(9y— 22) take 7, (74 —9y). 
 
 ivide 1a?—12a?+1q—1 Pipe se 
 25. Divide ta’—tia’?+14a— by ta—}. 
 
CHAPTER VII. 
 DIVISORS AND MULTIPLES. 
 
 83. Highest Common Factor. 
 
 A Common Factor of two or more numbers is a factor of 
 each of them. 
 
 a is a common factor of ax, a?y, and ab®. x—yisa 
 common factor of a7 — y? and 2 — y®. 
 
 The Highest Common Divisor of two or more numbers is 
 the product of all their common factors. 
 
 ax is the Highest Common Divisor (H.C.D.) of a®zy, 
 3 a222, and 5a2rz. a* is common, and so is z. ' 
 
 In arithmetic, the term Greatest Common Divisor is fre- 
 quently used. ‘This term is not applicable in algebra. In 
 the above example, a2 may or may not be greater than a. 
 If a is less than 1, then a? is less than a. Hence, in alge- 
 bra the term Highest Common Divisor is used. 
 
 84. Highest Common Divisor of Monomials. Rue. Zo 
 the Greatest Common Divisor of the numerical coefficients 
 affix each letter common to all the monomials, and to the 
 lowest power it occurs in any one of them. 
 
 S.aratyee?. 4 aaaeiegs fy age eee 
 
 The G. C. D. of 3, 4, and 5 is 1. 
 
 The common letters with the proper exponents are 
 a4, a3, ¥?, 2. . 
 
 Hence, the required H.C. D. is 1 a%a3y%z. 
 
 104 
 
DIVISORS AND MULT(PLES. 105 
 
 EXERCISES. 
 Find the H.C. D. of: 
 
 A afy’2, 8 xyz”, 12 abyz’. 
 
 5 abe’, 10 a*b®xay, 25 a7b*cx. 
 
 16 a®aFytz, 48 atba®y*zt, 36 a®btary"Z, 
 
 14 a®b?c*m?n?, 21 atb*m4, 42 a®b?mén’. 
 
 22 p®qta’y, 44 p*tg?a°y’, 66 pq®a’y®. 
 
 14 (a — b)x*y, 18 (a — b)°a*y’, 12 (a — b)Paty? 
 15 (@ — y)*2*, 21 (@— y)*2*, 33 (a — y)*2". 
 
 18 (a? — b*)ay, 27 (a? — b°)*2*y, 36 (a? — b°)?xty’. 
 
 oN OO PF oO D FP 
 
 85. Highest Common Divisor of Polynomials. The H.C.D. 
 of polynomials may be found by factoring. If each factor 
 is considered as a single quantity, the method is the same 
 as used in finding the H.C. D. of monomials. 
 
 (1) The H.C.D. of 2#’7—82+7, a?—1, 2?4+3a—4 is found 
 as follows: Bs et 7 aio Ter 1), 
 ae? — 1 = («+ 1)(@—1), 
 a +3e—4= (x4 + 4)(e—1). 
 It is seen at once that #—1 is the H.C.D. 
 (2) Find the H.C. D. of 
 e+5e—140, «e*—8a, 2t—42°+42, 
 w+ 5x2?—144%=2(x + 7)(a — 2), 
 ot — 8x = «(v’?+ 24+ 4)(x— 2), 
 et—4a° 44a? = a(x — 2)(x — 2). 
 
 Here we see that w(« — 2) is the H.C. D. 
 The H.C. D. is sometimes used in reducing fractions to their 
 lowest terms. 
 
106 
 
 \ 
 \ 
 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 Find the H.C. D. of: 
 
 1. 
 
 pon anh OD 
 
 ee ee ee 
 PON HF O 
 
 15. 
 
 ao? —y?, ow —Qay+y?, x? — ay. 
 
 a’? — b*?, a@—ab*, v?+2ab+0%. 
 
 a —y’, w+’, w+ xy, 
 
 e—Te+12, +2e2—15, 2’?—9. 
 w+8a+15, @—2a—35, a®?+3a—10. 
 b?—140+449, b?+b—56, b? —b— 42. 
 30°—12%+412, (w4— 2)8, 3a?—12. 
 a*b? — b*, ab? + 6°, ab — b?. 
 
 a — aty!, a8 (ay — 9?) 
 
 . (8—3 2"), x — avt— 62%. 
 
 . 0 —2 0° — 3504, a? — 25 ot, 
 
 + 2ey+y, e+ y7+3ay(e+y). 
 . m—3m—7T0, m?—11m?+10m. 
 
 x? — xy + x2 — yz, ry —y’. 
 a®’— 8, atb?—4a°b?, 4a?—16a+16. 
 
 86. Lowest Common Multiple. 
 
 When two algebraic expressions are so related that the 
 first is an exact divisor of the second, the second is said 
 to be a multiple of the first. 
 
 6 abcx is a multiple of 2 abc, because 2 abe is a divisor 
 
 of 6 a®bex. 
 
 A Common Multiple of two or more algebrate expressions 
 ws exactly divisible by each of them. 
 
 12 a®x37323 is a Common Multiple of 8 a®zry and 4 a?yz?. 
 
 The Lowest Common Multiple (L. C.M.) of two or more 
 algebraic expressions is the expression of lowest degree which 
 ws exactly divisible by each of them. 
 
DIVISORS AND MULTIPLES. 107 
 
 We use the term Lowest Common Multiple in algebra because we 
 are concerned about the degree and not about the numerical value. 
 
 a®z is the L.C. M. of a? and x; a‘z is also a multiple, and if @ is less 
 than 1, a*z is numerically less than a®z. 
 
 87. Lowest Common Multiple of Monomials. RuLE. To the 
 Lowest Common Multiple of the numerical coefficients affix 
 every letter found among the monomials, and to the highest 
 power it occurs in any of them. 
 
 3 ax, 5a*xy, 4ay*. By the above rule we write 60 a2ry4 
 at once as the L. C. M. of these expressions. 
 
 EXERCISES. 
 Find the L. C. M. of: 
 LZ ax, Sa'wy, 4 abey. 3. 8 pqr’, 24 n’q’r, 12 pg*r*. 
 we tao we, LO araty. 4. 10 Pm'n, 15 Pmn’®, 25 l*m?n. 
 
 5. 7 ab®xy, 14 a®b*y*, 21 ab*ary. 
 
 6. 3(a—)b)ay, 6 (a— b)*a*y, 12 (a — b) xy”. 
 
 7. 4(@— y)*ab, 5 (@— y)’a°b*, 10 (x — y)?(ab)*. 
 
 8. 40 (a®— «)*y’z, 60 (a? — x)*y2z*, 120 (a? — @)*y"2". 
 
 88. Lowest Common Multiple of Polynomials. ‘The L.C.M. 
 of polynomials may be found by factoring. Consider each 
 factor as a single quantity and proceed exactly as in the 
 case of monomials. 
 
 The L.C. M. of (@ — y)*, (28 — y®), a® — 6ay+5y* is 
 found as follows: 
 
 @—yY=@-WY@-y), 
 (P-YP)=@-WYeEtryt+y), 
 (?—6ry+5y*)=(@—-y)(a#—5y). 
 The L.C.M. is (a@—y)*Q@?+ay+y7)(@—5y). 
 
108 THE ESSENTIALS OF ALGEBRA. 
 
 In general the L.C.M. should be left in its factored form ; 
 that is, its factors should not be multiplied together. 
 
 The L.C. M. is used to a limited extent in reducing fractions to a 
 common denominator. 
 
 EXERCISES. 
 Find the L. C.M. of: 
 
 1. a— 0, @—2abd4+ 0, c—ab. 8. °—5r4+6, °+5r—24. 
 
 2. x? —16, 2?—9x2+ 20. 9. P+4a*, at—16 2%, 
 
 3. p?— 25, p? + p — 30. 10. Y —9y4+14, yr —4 y— 21. 
 iy pane Wr dues beth} 11, 8B, gt = 
 
 5. a@—4ab?, at—2a°*d. 12. 3 — 27 d’, ? —cd—6 a’. 
 
 6. 60, +3, &—Ssa. 13. 2—a, 4—2, 442°, 16-27 
 7. m’?+m+i1, m—1. 14.3406, 9—b*, 27-8. 
 
 15. «—3, +3, 2—627+9, +6249, 2?—9. 
 
CHAPTER VIII. 
 2 FRACTIONS. 
 
 89. Algebraic Fraction ; Numerator ; Denominator; Terms. 
 An algebraic fraction is an indicated division. 
 
 a+b, av + by, (a+6)+e. 
 It is usual to write these indicated divisions thus: 
 qa at} 
 he by p) 
 
 a. atl lw 
 ; is read a over 6, the fraction a over 4, or a divided 
 
 
 
 , or a/b, ax*/by, (a+ b)/e. 
 
 by 6. The preferred reading is a over 6. 
 The dividend its called the numerator, the divisor the 
 denominator, and the two together the terms of the fraction. 
 In the fraction ~, z is the numerator, y the denominator, 
 
 and a and y the terms of the fraction. 
 Any expression may be put into a fractional form by 
 writing it with a denominator 1; a= a py sree, 
 Since a fraction is an indicated division, we know that 
 - x 6 =a; for the fraction 7 may be regarded as the quo- 
 tient of a+ 6; but the quotient multipled by the divisor 
 _ equals the dividend. Hence, ; per Diets 
 109 
 
110 THE ESSENTIALS OF ALGEBRA, 
 
 90. The Sign of a Fraction. The sign of a fraction is 
 placed before the line which separates the numerator and 
 denominator. —* is read minus the fraction x over y- 
 
 y 
 
 Since a fraction is a quotient and the terms are dividend 
 and divisor, the sign of a fraction is determined precisely 
 as the sign of the quotient in division is determined. 
 
 sit n n n 
 ers == ras reas eaten =a a 
 +d d —d d +d do — d 
 
 A fraction preceded by a minus sign is equal to the same 
 fraction preceded by a plus sign, provided eather the numera- 
 tor or denominator be preceded by a negative sign. 
 
 a a — th, 
 Thus, Be Fee Nas atte os. 
 
 If the sign of either term of a fraction be changed, the sign 
 of the fraction ts changed. 
 
 Let : be a fraction. Change the sign of n, and it be- 
 
 comes hes a change the sign of d, and it becomes 
 eee ey 
 —d d 
 
 Tf the signs of both terms of a fraction are changed, the 
 sign of the fraction is unchanged. 
 
 If the signs of m and d are both changed, the fraction 
 ” becomes —~ =”. 
 d —d d | | 
 
 From the above it is evident that the value of a fraction 
 
 is unaltered by changing the signs of both terms, or by 
 
FRACTIONS. ph i 
 
 changing the sign of one term, provided in the latter case 
 
 the sign of the fraction is also changed. 
 ee 2h ine et 3 
 d —d’ —d- i 
 The line separating numerator and denominator acts as 
 a vinculum on both terms of the fraction. 
 ae means (x + y)+(a+b). 
 
 
 
 
 
 ab a = a a a 
 Se aa ————- = —————- @™ 
 Cc c a—2Z2 —a+2z 
 
 L— a 
 
 91. Law of Signs. If the terms of a fraction are made 
 up of factors, the signs of an even number of factors in 
 either or both terms may be changed without affecting 
 the sign of the fraction. If the signs of an odd number 
 of factors in either term are changed, the sign of that term 
 is changed, and hence the sign of the fraction is also 
 
 changed. a(— ©)(— y)(@) _ aryz 
 
 Cc Cc 
 
 (a—b)(b — e)(e — a) vs (6 —a)(b—e)(e— a) 
 ey 2 )(2— 2) (@—-y)(y-2)(@-2) 
 _ (b—a)(e — b)(a— @) 
 
 — (y—a)(y—2)(@— 2) 
 
 92. Reduction of Fractions. <A fraction is reduced when 
 its form is changed without changing its value. 
 
 Reduction of fractions depends upon the following 
 principle : 
 
 Multiplying or dividing both terms of a fraction by the 
 
 same number does not change its value. 
 
 
 
 
 
 
 
 
 
LIZ THE ESSENTIALS OF ALGEBRA. 
 Proof. Let an es 
 € 
 
 Then n= Ta, by multiplying both sides by d. 
 nxm=fdxm, by multiplying both sides by m. 
 
 
 
 
 
 
 
 
 
 
 
 : x teay, by dividing both sides by d x m. 
 xm / 
 Hence, a= 9 sei since both = ve 
 xe 
 Al n sie eae Mm 
 y d d+m 
 Proof. Let “=f. 
 roo e age 
 Then Wie FO. 
 n+m=fd+m, 
 NM 
 d-+m ee 
 Hence, 5 = 55 a 
 
 Fractions are reduced to. higher terms by multiplying 
 2, , 
 both terms by the same number; ~ = a (by multiplying 
 both terms by y7z). et! 
 
 Fractions are reduced to lower terms by dividing both 
 Vital foe 
 
 = -© (by dividing both 
 ahedy? ays § y dividing bo 
 
 
 
 terms by the same number ; 
 terms by a®2”). 
 
 A fraction is in its lowest terms when its terms contain 
 no common factors. 
 
 To reduce a fraction to its lowest terms, divide or cancel 
 all common factors out of its terms. 
 
~. 
 
 é 
 
 FRACTIONS. 113 
 
 @ (a? — 6?) a(a— by’ 
 
 by dividing out a?(a + b). 
 
 Selecting and canceling the common factors can gen- 
 erally be done mentally. 
 
 % 
 
 @ 
 
 Ho 6 Ve 2) 8) i 
 a A Bree) (ent 3) 
 
 (a — a) (a — b) 
 
 (aaa) 
 
 a— 3 
 a oes 
 
 
 
 (a—ax)(b—am) _ 
 
 22)(8— 2) erates 
 
 EXERCISES. 
 
 Cancel factors common to numerator and denominator in 
 the following: 
 
 di. 
 
 a‘b?a : 
 Te whe 
 3 aPa7y 
 15 a®ay? 
 AT ab?a? | 
 51 b*cx? 
 xt — ay? 
 gt — yf! 
 Dien 
 (a + 3) (@ — 2) 
 
 = 20.0 
 
 
 
 
 
 (a — 5) (w+ 5)a® 
 
 eee 1 ; 
 —627+9 
 
 a“? — 1 
 
 Fogg tN 
 
 ay — xy? 
 
 aty — wy? 
 
 
 
 aba 
 
 
 
 20. 
 
 21. 
 
 ax — bx 
 
 
 
 
 
 
 
 ah” 
 7. gp) 
 a? + aa 
 as oles a2 
 gen 
 a — 4? 
 BO oad yp 
 "8 13 
 at yt 
 (w+y)° 
 (a+ y)(@—y)(@—2) . 
 
 (vw — 2) (y— x) (—#—Y) 
 w+ 3a+2 
 ree 
 ao —5a+6. 
 a? 4+a—12 
 2 at + ty — 
 (w+ y)(@—y) 
 
114 THE ESSENTIALS OF ALGEBRA, 
 
 po ta eens a7, © 2 ae 
 
 (e+ y +2)" ee 
 so Bi sera ae) oa w+ 60% + 11a? 4 62% 
 20 + Oe (a + 1) (a+ 5a + 6) 
 ay a? —(a+b)x+ ad. 29 at — 60° + 11a? — 6a 
 ' gt — (b4+-0)a@ + be " (w — 8) (a —38@ + 2) 
 
 ao —Te+12 1+a+b+ab 
 x” — 9a + 20 (1 — a’) (1— Bb’) 
 
 sts a? + (a — b) a — ab “31 av —OP+0C+4+ 2ca 
 | #@—(b—c)x—be ' +b? £2 abe 
 
 25. 30. 
 
 
 
 93. Proper and Improper Fractions. A proper fraction is 
 one whose numerator is of lower degree in a named letter than 
 the denominator. 
 
 g4+2¢+5 
 g+327— Ta +4 
 ator is of second degree in 2, while its denominator is of 
 the third degree. 
 
 is a proper fraction because its numer- 
 
 An improper fraction is one whose numerator is of degree 
 equal to or greater than the denominator. 
 e+3sa%+5xr—4 
 An improper fraction may be reduced to an integral 
 expression and a proper fraction by dividing the numerator 
 by the denominator. 
 V+2¢r+1)e8+4+3e+5r—-4e4+1 
 B+2e4+e 
 
 is an improper fraction. 
 
 ve+4er—4 
 e+ 2er4+1 
 2x2—5 
 SHB ey Marae ee oe 92—5 
 
 Veghts Dae. e+ 20241 
 
FRACTIONS. 115 
 
 The process is similar to the arithmetical process of 
 reducing an improper fraction to a mixed number. 
 
 EXERCISES. 
 
 Reduce the following to integral or mixed expressions: 
 
 1. 
 
 94. 
 Since 
 
 
 
 
 
 74+ x C= te oe 12% 
 ————-e 5. —_——__---—_—____—————e 
 x e—4 
 e—de+tt e+ 2 avy + y? + 2? 
 eT erat 6) 
 aw (w+ y)" 
 3a? —5x+1 : aw — y —3 vy (a —y) 
 e+ 1 ; a ; 
 at — oF + a+ 5. 8 4at—6 0? + 12¢%+45_ 
 e+ ot x , a 1 
 4 
 Prove =) + a + a+ a 
 —2x 1—2 
 
 Tse? divide as in Exercise 9 to four terms. 
 x 
 
 
 
 } 1 66 6 66 6 6c 6 
 
 1—S2 
 
 Reduction of Fractions to a Common Denominator. 
 a fraction is not changed in value by multiplying 
 
 both of its terms by the same number, we may make the 
 denominator any number we please by properly selecting 
 
 our multiplier. ; and 2 may be made to have the com- 
 
 mon denominator 6d. 
 
 
 
 
 
 Cred Tad 
 6 bxd bd. 
 CEE S50 sear. 
 d dxb bd 
 
116 THE ESSENTIALS OF ALGEBRA. 
 
 The method of reducing fractions to a common denomi- 
 nator is the same as in arithmetic. It may be stated as 
 follows : 
 
 RuLE. Find acommon denominator, in general the lowest 
 common denominator (L.C.D.). Divide it by the denom- 
 inator of each fraction, and multiply the terms of the frac- 
 tion by the quotient. 
 
 (1) Reduce 5 
 the lowest common denominator. 
 
 The L. C. D. is (6 — #)(6 + 2). 
 
 L.C. D.+b—av=b+a, 
 L. C.D: +6+2=6—2. 
 
 to equivalent fractions having 
 
 
 
 
 
 and, 
 L b+ 
 
 
 
 Oey a(b + 2) 
 b—x (b—2x)(b+2) 
 Aoet c(b —& 
 ba (b—a)(b+2) 
 Ax an 
 
 (2) Reduce to equiv- 
 
 x 
 v—W (©—2)(@—3) @+2)@+3) 
 alent fractions having the lowest common denominator. 
 The L. C.D. is (@— 2)(% + 2) (@— 8) (@+ 8). 
 L. C. D. + (a — 4) = x” — 9, 
 L. C. D. + (@ — 2) (@ — 3) = (@ + 2) (« +3), 
 L. C. D. + (w@ + 2) (@+ 3) = (a — 2) (@— 3). 
 
 a(a?—9) x — Ox 
 @—D@—9) @—H@+D@—3)@+43) 
 4 w(x + 2) (a +3) 4 oF + 20 7? 4+ 242 
 
 (@—2) (@—3)(@+2)(@+3) @—2)(@+2)(@—3) @+3) 
 
 (2 — x) (@ — 2) (a@— 3) —¢+7¢7—16e+12 7 
 @+2)(@+3)\@—HE—3) @=2 CHEE) 
 
FRACTIONS. 117 
 
 EXERCISES. 
 
 Reduce to equivalent fractions having lowest common 
 denominators : 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Sem cee 5 3 4. 
 1 -, = ae d =» ee : 
 meu od. 3b ey C x+y ah x—y 
 2. 9 4 and I : 4. a pei and us 
 ey “x+y e¢ wt e+ 2 
 ee lan as OL tg, 
 e+l «—F «+2 x—2 
 6. ee aR sal , and 5D 
 e—4 “£42 e— 2 
 x oe 
 7 ——__—_—_ ae ea tae EE 
 x?7—62+8 ea e? —9 oa + 20 
 8. a Nye SUMS and i a a 
 e+be+6 +4243 v4+3e¢+2 
 a c 
 9. : 
 go? — y? ot yf? a? — avy” 
 2 2 
 ee and | 
 ey w—ay—2y" w+ sy —2y? 
 ee PP and 2%. 
 t—x 1+2* 1-2 b--s 
 12. 2 eet a and Rise Tey 
 eP4+5e+6 #+62+8 e4tT7e#+12 
 1 2 3 
 ee ed 
 a? — xy + y" x? + ey + y" xt + ay? + y* 
 a 3a 5 6 
 14. —— -— 15. ——__—_—_-. dt ee 
 b"c” Ortiemtl a" (a+b)? an x"? (a+b)* 
 
 95. Addition and Subtraction of Fractions. From division 
 we know that 
 Sree + d—e—f a 
 ee 9 Brgang 
 
 
 
118 THE ESSENTIALS OF ALGEBRA. 
 
 If we read this identity from the right, we have the 
 result of adding a number of fractions. Hence, we con- 
 clude that a, b_a+b 
 
 
 
 | ape ee NP 
 Ruue. Zo add or subtract fractions, reduce them to the 
 same denominator and then deal with the numerators accord- 
 ing to the rules for addition and subtraction, writing the final 
 result over the common denominator. 
 1 2, 32 
 
 a—x at+z a— 
 
 
 
 -== what? 
 A a 
 
 The common denominator is (a — a) (a+ @). 
 
 
 
 
 
 
 
 1 a+ 
 a—x @—2 
 2  2a—2¢ 
 ase a Cate oe 
 3D on 
 
 Fei’ Mier haem sin 
 Thus we have 
 
 a-nu (2a— 2a 3842  a+u+2a—22—32 3a—4e 
 
 
 
 
 
 
 
 
 
 ORs hat an Gs a” — x age 
 EXERCISES. 
 Combine and simplify : 
 1 et ey 10 
 x—1 BK 
 A: an 3 6 etat+ a 
 x+1 (x + a)? 
 5 a? Bias 
 3. — ° WON toe 
 x + 10 Mate 
 , 
 ie as ea, a ee 1 
 
 x+2 ota! acd jae 
 
FRACTIONS. 119 
 
 32+ bare ae A 6 + 41 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 j OT. he Te e+ a? e+ a’ 
 1 1 a b 3 ab 
 10. ; Pn i ea oe lea 
 Pe 1 Be Ceo a= 6 a 
 1) a Tat emer ED ea, 
 ee 2 a — 2 e=-9 e¢—5a+6 
 Mpoerd be? 1 4, fet Ge 8 a4 
 er (1)? e—1 @+20+1 2-1 («—1) 
 16 Le SO Ss ES eam eae 
 @—N)@—-)e+H @-HeEH 
 17. 2 2 a e+2 ; 
 (e+ 3)(@—2)(@+1) (#%+4)(a’—-a — 2) 
 it 1 
 ee 
 a) el a ci De 
 1 i 
 
 IPL) ot a ee alee 
 e—5e+6 w#-—624+8 
 
 * G—NU-) | W-)E-%)  E-9e—H) 
 
 96. Multiplication of Fractions. 
 
 To prove a Xess 
 bd bd 
 
 Let 5a Sy (1) 
 
 and < = f, (2) 
 Ait 2 
 
 Then 5 x Gali Sy 
 
 But from (1) a= 7,6, (5059 
 and from (2) Gas) othe (4) 
 
120 THE ESSENTIALS OF ALGEBRA. 
 
 Now multiplying Equation (8) by (4), member by . 
 
 member, ac =f fra. 
 = fifo by dividing both sides by 6d. 
 But 5 x ate 
 er re 
 Hence, b x ie bd 
 
 Rue. The product of two fractions is the product of the 
 numerators over the product of the denominators. 
 
 This covers all possible cases of multiplication involving 
 fractions, for all integers can be put in fractional form. 
 
 Dy ain ee te 
 b pe wa) 
 Ae ee 
 
 AEC leur ven. 
 
 Any number of fractions are multiplied together by 
 placing the product of all the numerators over the product 
 of all the denominators. 
 
 In all problems in multiplication of fractions free use of 
 cancellation should be made. 
 
 sai ene 
 a — Be xe—-y «uty 241 
 
 
 
 
 
 
 
 In this, if we cancel the common factors, our work 
 appears thus: 
 ria * Ly ee “P41 2+1 
 
 
 
FRACTIONS. 121 
 
 EXERCISES. 
 
 Perform the operations indicated, reducing the fractions to 
 lowest terms: 
 
 32 10y a a a? 
 1 Se pe er oe iin oe a See 
 FO 1 : 7. (= 1)x($-+1)= 5—1. 
 
 18 
 
 oy 1b 
 
 3, 3(@+y) , 6@—y)" 
 2@—y) I@+ry) 
 Sea ety. 
 
 o 
 
 © 
 TAT BN EG. oN 
 
 Sia Sia Sie SIs 
 | 
 Q}Se 
 oe 
 
 b@+yy @™a—y 
 12aa%y 14 dy) Pty) 4 
 16 bay? 6a = x(a@+y) 
 
 Sa(e+y) 2/x—y\ (¢ y 
 6. 22] erage Sharing 
 6b(a@—y)? 3\e+y inset 
 
 
 
 - 
 
 a 
 S!1S 
 
 a 
 
 L; 
 
 
 
 See 
 
 S 
 
 x 
 Re, 
 QO 1a 
 
 SS ae =, 
 tS 
 
 x 
 
 Fg 
 
 RAIS 
 
 2 
 
 is 
 
 x 
 
 Cees 
 
 QQ 
 Se 
 
 S 
 
 a 
 sie 
 MS” 
 
 = 
 Q 
 oe 
 aus 
 o1S 
 ——— 
 | 
 Mo 
 ols 
 ee EH 
 Qe 
 ees oP 
 + 
 Fe 
 Qa) 
 eae 
 ecaewttadl 
 a) 
 + 
 Qa 
 
 2 
 14. (E-2). 15. at;tt)(5-1). 
 
 Tepe ea: 4) x Soe Lie (Cancel common factors.) 
 e+i5ie+6 (e+ 1)(@+ 4) 
 
 w+ Te+l2 w+6a+5 | 
 oft 4a+3 ~2?4+9a4 20 
 
 uae aed me eat SD 
 e—Te2+12° 2? —62+5 
 
 e-E 1\? 5 (at— 25 e—1 
 19. . 
 oa Aaa ee) 
 or —4oa"+4 _ ym—1 
 Se as ae aa 
 
 17 
 
 18. 
 
 
 
 
 
 
 
 
 
132 THE ESSENTIALS OF ALGEBRA. 
 
 97. Division of Fractions. 
 
 To prove SSSR 8 
 Greed) 0 Wane 
 Let goty (1) 
 say (2) 
 Then Stat acti 
 Butfrom(1) a=f,d, 
 
 and from (2) Clea ifad- 
 
 
 
 , ES Rie ake 
 Co Wa Weta 
 ra zs = by multiplying both sides by a 
 els Oras b. 
 ay d ae ad aes d ae 
 os (bu WEbO We pee 5 
 Hence, ; “ . — 7 x <. 
 
 Rue. To divide one fraction by another, invert the divisor 
 and multiply. 
 
 Since all integers can be expressed in fractional form, 
 this rule suffices for all forms of division involving the 
 fraction. 
 
 
 
 (1) i Nios he a ee 
 y Ye wl yar aay 
 This is dividing a fraction by an integer. 
 9 .2_a4,@_a ¥ _ ay, 
 2) yf gh age (ee x 
 
 This is dividing an integer by a fraction. 
 
FRACTIONS. 
 
 
 
 
 
 
 
 Byes. 7:2 ae P Rekcaek £7 ey. 
 (8) Se B= x= 4-9). 
 
 ety w—-y wt+ty a- 
 
 123 
 
 In this case the common factors (a — b) and (#+y) are can- 
 ~eeled. The student should constantly be on the alert for 
 common factors and cancel them as soon as they appear. 
 
 EXERCISES. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 San, Gara ae +36 ab+30? 
 Baty 10 ay? -@+56 ° a +5ab 
 > Aara’ | 8 aa 6 —Ay oR cto 5 ary 
 "15 b%y? ~ B3by eae 20? — 8 ay 
 3, Le atylat A any? gq Sty . @ryy’ 
 " “Qatyd 3 atyz — @—y? (a —¥)? 
 pee et ves (2 t-Y/)" Poe alo hipy ems See 
 (e—y) w-—y v—y’ a? + ay + ye 
 eo .«—sy 
 Ly ae a Y. 
 (@+3y) a+3y 
 10 (a+a)(v—b) , #+(a—b)x—ab 
  @—a)(@+b)  2’—(a—b)x—ab 
 ee Die ee OC ase Oe 
 e—y* (a+y’) b°—(c+a)?> —a+b—e 
 
 98. The Complex Fraction. A fraction which has a frac- 
 tional expression for either or both of its terms is called a 
 
 complex fraction. 
 
 
 
 
 
 a lina 1 
 eas pal, ames Fe x+- 
 Bo 6x y 
 1,0 a+ 
 ae ey Y 
 
 are complex fractions. 
 ax 
 
 Since a fraction is an indicated division, a complex 
 fraction is simplified by performing the division indicated. 
 
124 
 
 
 
 
 
 
 
 
 
 
 
 a 
 URN wet. Oe, 0.400 
 terse pb \ic ane 
 a 
 aa al | Se ie 
 (2) ey ey wy wy _y—® 
 sa BES ytu ay 
 oy ty xy vy 
 bear xy 
 
 
 
 EXERCISES. 
 
 Simplify the following: 
 
 
 
 
 
 
 
 
 
 
 
 at 5 — +4 
 es 1+ - 1+2 
 sey y 
 " 6 
 e+——5 
 ee a 3 
 2," eee 
 oie 2a gl og: 
 Kae w+a_a—a 
 3. x Be A ak wa 
 ies aes ens Meg yee | 
 x %—-a «e+a 
 4 x+y a sy 
 a Op Oe 
 5 RecN kart 8! 
 a+y AFane he ral 
 
 
 
 
 
 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 bose 
 
 
 
 10. 
 
 11. 
 
 12. 
 
 wy y+e y+e 
 
 
 
 a+ 3 ey+ 27° 
 a Daiet 
 ety wx+2y 
 
 
 
 
 
CHAPTER IX. 
 EQUATIONS IN ONE VARIABLE. 
 
 99. Identity; Conditional Equation. Distinctions be- 
 tween an identity and a conditional equation have already 
 been made in Chapter II]. We have had illustrations 
 of the identity in all the fundamental operations and in 
 factoring. | 
 (w+ 2)r+4=277427+4, 
 and (e+1)(w@—5)=r7?—-427—-—5 
 
 are identities; that is, they are equalities that are true for 
 all values of x. tn 
 
 A conditional equation restricts the value of some one 
 letter, the letter so restricted being called the variable. 
 The other numbers of an equation are called constants. 
 
 (_) z«=5 restricts x to the value 5. 
 (2) 8x=18 restricts x to the value 6. 
 
 (8) ax=6 restricts x to the value us 
 a 
 (4) 2?=42 restricts 2 to the values 0 and 4. 
 
 Each of these equations becomes an identity for the 
 restricted value of the variable. 
 
 Thus, (1) becomes 5=5 for r=5, 
 
 (2) becomes 3 x 6=18 for x= 6, 
 125 
 
126 THE ESSENTIALS OF ALGEBRA. 
 
 (5) becomes a x ° = for re v 
 a a 
 
 (4) becomes 02=4>x 0 for 7=0; 
 and (4)? =4 x4 for z=4. 
 
 100. Root of an Equation. A value of the variable for 
 
 which the equation becomes an identity, is called a root of the 
 equation. 
 
 A root of an equation is frequently called a solution. 
 
 Any root of an equation when substituted for the vari- 
 able is said to satisfy the equation; that is, an equation 
 is satisfied by any value of the variable which reduces it 
 to an identity. 
 
 22+5=15 has 6 for a root. Substituting 6 for 2, 
 this equation becomes 2 x 6+ 3=15, which is an identity. 
 6 is the value of x which satisfies the equation 274+ 3=15. 
 
 101. Classes of Equations. A rational equation is one in 
 which the variable is free from radical signs. 
 
 8227—Tx+11=0 isa rational equation. 
 
 An irrational equation is one tn which the variable is 
 affected by a radical sign. 
 
 5 a? — 6Va2=12 is an irrational equation. 
 An integral equation is one in which the variable appears 
 only in the numerators of its terms. 
 Dee Ot ie if 
 
 3a? + 17 + — EST is an integral equation. 
 
 A fractional equation is one in which the variable appears 
 mn one or more denominators. 
 
EQUATIONS IN ONE VARIABLE. 127 
 
 (el ae 
 e xv—4t2r+1 
 
 Ba+ 
 
 is a fractional equation. 
 
 A linear equation is one in which the variable appears to 
 the first degree only. 
 
 32+ 7=11 isa linear equation. 
 
 A quadratic equation is one in which the highest power of 
 the variable is two. 
 
 522+ 7Tx2—16=0 is a quadratic equation. 
 
 A cubic equation is one in which the highest power of the 
 variable is three. 
 
 x —6a*+11x¢2—6=0 isa cubic equation. 
 
 It should be noticed that several of these terms may be 
 applied to the same equation. 
 
 322°—Ta+13=0 is a rational, integral, and quadratic 
 equation. 
 
 EXERCISES. 
 
 Classify the following equations : 
 
 1. av’?+b2+c=0. 5. +3 e—Te=l, 
 2. ax+b=0. 6. 3Vr=11—2. 
 
 3. ax’ + be? +cr+d=0. 7 8 5+28 4. 
 4. w+3e4+7=0. e e+1 
 
 102. Equivalent Equations. It will be noticed that the 
 equations ibys Sate 
 (2) 84—12=0 
 
 have the same root, viz., x = 4. 
 
128 THE ESSENTIALS OF ALGEBRA. 
 
 The equations 
 (3) 527 — 102 = 0°and 
 (4) 527-102 +6=6 
 are satisfied when x = 0 and whenzx=2. Such equations 
 
 as (1), (2), and (3), (4), having the same roots, are called 
 equivalent equations. 
 
 _Hquivalent equations are those having precisely the same 
 roots. 
 
 Determine which of the following equations are equiva- 
 lent: 
 (1) x—1=0, whose root is 1. 
 
 (2) 38x%—6=0, whose root is 2. 
 
 (8) 2¢%—2=0, whose root is 1. 
 
 (4) 22?—x=0, whose roots are 0 and 1. 
 (5) 2a—4=0, whose root is 2. 
 
 (6) z27—1=0, whose roots are 1 and —1. 
 (7) 832—6+4+5=5, whose root is 2. 
 
 (8) 2—a2—4=-—4, whose roots are 0 and 1. 
 
 (9) ar : = (0, whose root is 2. 
 
 2 
 (10) Aes : + 1=1, whose roots are 0 and 1. 
 (11) 27+3=4, whose roots are 1 and —1. 
 103. Solution of Linear Equations. 7 solve an equation 
 
 is to find all of tts roots. 
 
 The solution of an equation consists in deriving one or 
 more equivalent equations, the last of which gives the value 
 of the variable. 
 
EQUATIONS IN ONE VARIABLE. 129 
 
 Thus, to solve 
 (1) 82+4-—2=6+4 22, 
 
 we bring 22 to the first member of the equation and 
 (4 — 2) to the second member. This is done by subtract- 
 ing these quantities from both members and gives us 
 
 (2) 8a—247+4-—2-—(4—2)=64+22—24—(4-2), 
 which becomes, by omitting the terms that destroy each 
 other, 
 
 (3) 59-27-56 —4 + 2. 
 
 This process is called transposition. It is effected by 
 changing the sign of a term when it is moved from one 
 member of an equation to the other. 
 
 We next unite the terms in each member of Equation (3), 
 which makes 
 
 (4) oA, 
 
 This process is called combining terms. It is effected by 
 addition. The last equation expresses the value of the 
 
 variable 2 Hence, 4 is the root or solution of the given 
 equation. 
 
 It should be noticed that Equations (1), (2), (8), and 
 (4) are equivalent equations, each having the root 4. 
 
 If 4 is put for x in these equations, they become 
 
 (1) 8x4+4—2=6+2 x4, 
 
 (2) 83x4—-2x4+4-2-(4-2) 
 =64+2x4-—2x4-(4- 2}, 
 
 (3) 3x4-2x4=6-—4+2, 
 
 (4) 4. = 4, 
 
 Each of the above is an identity. 
 
130 THE ESSENTIALS OF ALGEBRA. 
 
 As another illustration let us solve 
 
 (1) 4¢4+2—T=227-8. 
 
 Transposing 2 and —7 to the second member and 22 
 to the first member, we have 
 
 (2) Ag —2og=—o- [ae 
 
 Combining, we have 
 
 (3) 2a=—3. 
 
 Dividing both terms of the equation by 2, we have 
 
 (4) B= — §. 
 
 The root or solution of Equation (1) is— 3. Equations 
 (1), (2), (8), and (4) are equivalent equations, each being 
 satisfied by the root — 3. 
 
 
 
 
 
 Putting v= — 3, they become 
 
 e 4(-D)+2-T=%A-H-8, 
 or —64+2—T=—3-—8. 
 
 @) -(--2(- = 247-8, 
 Or —6§6+38=—247—8 
 
 (3) 2(—3)=—8. 
 
 (4) —g=—8. 
 
 As a further illustration let us solve 
 
 () 328 46222! 418. 
 
 In order to get rid of the fractions in this equation, we 
 multiply both members of the equation by the Lowest Com- 
 mon Multiple of the denominators. Multiply both sides by 
 30, the L. C. M. of 2, 5, and 3; the result is 
 
 (2) 45¢—754+180=62—2027+4+10+4 3890. 
 
EQUATIONS IN ONE VARIABLE. 131 
 
 This process is called clearing of fractions. It is always 
 brought about by multiplying both members of the equa- 
 tion by the L. C. M. of the denominators. 
 
 (3) 452-6 7+ 20 x=75—180410+390, by transposing. 
 - 4) 59 = 290, by combining. 
 oo). Die ap by dividing. 
 5 is the root or solution of (1). 
 Equations (1), (2), (8), G), and (5) are equivalent 
 
 equations. 
 Show that each is satisfied by the root 5. 
 
 104. Rule for Solving a Linear Equation. 
 
 (1) Clear of fractions. 
 
 (2) By transposition bring all the terms containing the 
 variable to one member of the equation and all the constant 
 terms to the other member. 
 
 (3) Combine the terms of each:member of the SE by 
 addition. 
 
 (4) Divide both members of the equation by the coefficient 
 of the variable. 
 
 Steps (1) and (4) depend upon the axiom that multi- 
 plying or dividing equals by equals gives equals. Steps 
 (2) and (8) depend upon the axiom that increasing or 
 diminishing equals by equals gives equals. 
 
 EXERCISES. 
 Solve the following equations : 
 1, 82—5=19. 4. 7x—12+3=52+416—85. 
 
 2. (x—11 — 24. §. 12—4¢74+2=13—72+10. 
 3. 3a@—5=2+13. 6. 15—142—7=17—167—6. 
 
132 THE ESSENTIALS OF ALGEBRA. 
 
 7. Te—ll+42—-T=32—8. 
 
 8. 11—5e417-—38e7=18 —11 2428. 
 
 9. 5e«—16—6e—6=115—Ta—4a2—T. 
 
 10. 2~—224+7274+14=62-—8+4+42-4 42—5@2@ 
 
 11. 10(@— 3) =8(e—2). (Remove the parentheses first.) 
 12. 11(4ea—5)=7 (64%—5). 
 
 13. 3(@—2)4+2(2%—3)=3 (#@—4). 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 14. de@+1_x+12. 20. Ze—1 , 3a—b_ Oaye) 
 2 S o Z 4 
 5a—4 2443 . 32 oe ae £ 
 [2-0 2e+5 4ey—-2 , 8% 126 
 16. — . 22 7° 28— — Sie eis 
 3a ee oie °— 5 
 1. S20 ge owe a +1 2x—5 c_4 
 b 5 8 9 7 
 © o2 22 f= ee ee x—3 
 1s. —+—=6-—-——-1 24. a . 
 973 3 6 i 5 3 
 7 Amerie arte | etl 2a. x—9 
 19. —-—-=-+=-— 25. 4= 
 Pe Is: 103 9 fe a 8 
 26 221430288 19452, 
 
 27. 3(@ — 2) —2(¢#—5)+2e—20=17. 
 28. ax + ba=a’?+ 2ab+ G 
 (a+ b)a= (a? + 2ab+5). 
 _a@+2ab+0? 
 a+b : 
 “oa +d. 
 
 x 
 
EQUATIONS IN ONE VARIABLE. le 
 
 
 
 
 
 29. ax+.a’?= bx + 0’. 34. ce—3a=—c? — 9. 
 Sl. ag-+- ba =—a®-t b* 36. Spat 
 a 
 32. a’a—abe+ bae=a'?+ BF ; 
 H P 37. ole at Sh = 2 0. 
 33. av-+ b*= aq? — b*x— abe. atb a—b 
 
 38. (w—3)(@+5) —T = (w%+4)(x—8). 
 39. (24—5)’?4+4=(#—6)(4a—3). 
 40. 14—(2—2)’=5— (x#+3)(x— 2). 
 
 EXERCISES. 
 
 1. One half of A’s money is $35 more than B’s. They 
 together have $280. How much has each ? 
 
 SOLUTION. 
 
 Let x = B’s money. 
 xz + $35 = one half of A’s money. 
 2x +70 = A’s money. 
 x+22+ 70 = 280. 
 +27 = 280 — 70. 
 ge aa tA 
 x = 70 = B’s money. 
 22 + 70 = 210 = A’s money. 
 
 VERIFICATION. One half of $210 is $105, which is $35 more 
 than $70. Also, $70 + $210 = $280. 
 
 2. A has $10 more than 3 times as much as B, and they 
 together have $250. How much has each ? 
 
 3. Find two numbers whose sum is 81, such that one may 
 exceed 6 times the other by 4. 
 
134 THE ESSENTIALS OF ALGEBRA. 
 
 4. Divide 114 into three parts such that the first may exceed 
 the second by 15, and the third the first by 21. 
 
 5. Divide $176 among A, B, and C, so that B may have 
 $16 less than A, and $8 more than C. 
 
 6. Divide 440 into three parts such that the second is double 
 the first,increased by 10, and the third is the sum of the first 
 and second. 
 
 7. What two numbers have a sum of 861 and a difference 
 Ot 21 
 
 8. Find a number that exceeds 31 by the same amount that 
 2 of the number exceeds 1. 
 
 SOLUTION. 
 
 Let zx = the number. 
 
 x — 31 = the excess of the number over 31. 
 
 ; — 1= the excess of } of the number over 1. 
 
 By the conditions of the problem, these are the same; hence, 
 
 6x2 — 186 = 2 — 6. 
 62 —2z=— 186 —6. 
 
 5 « = 180. 
 x = 36. 
 VERIFICATION. 36 — 31 = 5. 
 1 of 86 —1=5 
 
 9. What number increased by + of itself and 80 is 30 more 
 than double itself ? 
 
 1o. Hight times the difference between the third and fourth 
 parts of a certain number is 40 less than the number. What 
 is the number ? 
 
EQUATIONS IN ONE VARIABLE. 135 
 
 11. If 10 be subtracted from a number, 4 the remainder 
 + 40 is 50 less than the number. What is the number ? 
 
 12. Find two consecutive numbers such that 4 of one plus + 
 of the other is 44. 
 
 SUGGESTION. Let x = one number, and 2 + 1 = the other number. 
 
 13. Find two consecutive numbers such that 4 their sum is 
 34 less than the larger one. 
 
 14. Find three consecutive numbers such that 1} the first 
 + 4 the second + } the third is 88. 
 
 15. In 10 years John will be twice as old as Henry was 
 10 years ago. John is 9 years older than Henry. Find their 
 ages now. 
 
 SOLUTION. 
 Let x = Henry’s age. 
 x + 9= John’s age. 
 zx +9 +10 = John’s age 10 years hence. 
 x — 10 = Henry’s age 10 years ago. 
 2+9+10=2(2—10). 
 24+9+10=22 — 20. 
 %z—2xr=—9-—10-— 20. 
 
 —xr=—— 39. 
 30, 
 zr+9= 48. 
 
 16. A man’s age plus that of his wife’s is 95 years; 40 years 
 ago he was twice as old as she was then. What are their ages 
 now ? 
 
 17. Eight years ago a father was 9 times as old as his son 
 was at that time; in 37 years the father will be 13 times as old 
 as the son is at that time. What are their ages now? 
 
 is. A man left + his estate to his son, + to a nephew, t toa 
 niece, and the remainder, amounting to $2600, to his wife. 
 What was the value of his estate ? 
 
136 THE ESSENTIALS OF ALGEBRA. 
 
 19. A house is sold for $2280. This is a gain of 14%. 
 What did the house cost ? 
 
 SOLUTION. 
 
 Let x = the cost of the house. 
 ii = gain. 
 x + pls x = 2280. 
 1002 + 147 = 228000. 
 1142 = 228000. 
 e200; 
 
 20. A horse sold at a loss of 7% brought $111.60. What 
 did the horse cost ? 
 
 21. A man invests } his capital at 4% and the remainder 
 at 5%. His income is $2800. What is his capital ? 
 
 22. What number must be added to each of the terms of the 
 fraction +4 to make it 24? 
 
 23. What number must be subtracted from both terms of 
 the fraction 42 to make it 2? ; 
 
 24. Divide $5600 into two parts such that the income from 
 one part at 3% may be equal to the income of the other part 
 at 4%. 
 
 25. Divide $760 among A, B, C, and D so that A and B 
 together shall receive $150, A and C together $190, and A and 
 D together, $ 580, 
 
 26. $7.20 is changed into 36 coins. Each coin is either a 
 dime or a quarter. How many of each are there ? 
 
 27. A bill of $10.20 is paid in an equal number of dimes, 
 quarters, and half dollars. How many of each are used ? 
 
 28. A man bought sheep at $4 a head, calves at $9, and 
 cows at $35. He bought twice as many calves as cows, and 
 
 twice as many sheep as calves. The cost of all the stock was 
 $690. How many head of each did he buy ? 
 
EQUATIONS IN ONE VARIABLE. 137 
 
 29, Find three consecutive numbers such that the sum of 
 the quotient of the first divided by 10, the second by 11, and 
 the third by 61, is 25, 
 
 30. Find three numbers such that the second is a times the 
 first, the third } times the second, and their sum ec. 
 
 31. One half of A’s money is equal to B’s, and five eighths 
 of B’s is equal to C’s; together they have $1450. How much 
 has each ? 
 
 32. A man walks out at the rate of 4 miles an hour, and rides 
 back at the rate of 10 miles an hour. How far can he go out if 
 he must make the round trip in 7 hours ? 
 
 33. A man sold 12 acres more than 1 of his farm, and had 
 2 acres less than 2 of it left. How many aeres had he? 
 
 34. A train leaves a station at 8 a.m. and rvuzs 30 miles an 
 hour. At 11 a.m. another train leaves in the same direction 
 running 45 miles an hour. When and where will it overtake 
 the first train ? 
 
 35. A and B are two towns 120 miles apart. A messenger 
 starts from A to B at 7 a.m. and travels 10 miles an hour. At 
 8 a.m. another messenger starts from B to A and travels 
 12 miles an hour. When and where will they meet ? 
 
 36. A man in traveling from New York to Buffalo, goes + as 
 far by boat as by train and 1, as far by carriage as by boat. 
 If the distance to Buffalo from New York be 490 miles, how 
 far does he travel in each conveyance ? 
 
 105. The Linear Type. Every linear equation in a single 
 variable may be reduced to the type form 
 
 ax+6=0. 
 
 In this form a and 8 represent any positive or negative 
 numbers whatever. 
 
138 THE ESSENTIALS OF ALGEBRA. 
 
 For example, S77 Ce eee 
 Clearing of fractions, 
 127-284 10%—-2=827-28+432. 
 Transposing all terms to the first member, 
 12%7+1027-—82-—3827—28 —24+258=0; 
 
 
 
 Collecting, IaH 20. 
 This is in the type form. 
 Comparing it with ax +b=0, we see that a= 11 and )=—2. 
 
 The solution of the type form azr+b=0 
 
 ‘ b 
 1S r=. 
 
 a 
 Hence, the solution of the above example is x = ?;- 
 Special roots of axw+b=0. 
 If 6=0, then the solution of az + 6=0 becomes 
 perme ef 
 a 
 If a=0 and 6 is not 0, then the solution of ar+6=0 
 
 pecomes 
 r= — 
 
 We have here a new form whose value we must in- 
 vestigate. jb 
 
 CER, 
 
 HP = +104, 
 += +1008, 
 + © = + 10008, 
 
 +b 
 + ___ 4 1900000008. 
 0000001 mY 
 
EQUATIONS IN ONE VARIABLE, 139 
 
 It appears that, as we decrease the denominator, the 
 value of the fraction increases. When the denominator 
 of the fraction is very small, the value of the fraction is 
 very large. When the denominator becomes 0, the value 
 of the fraction is large beyond measure. We express this 
 fact by saying that the value of the fraction is infinity. 
 The symbol for infinity is oo. 
 
 b 
 f= —-—-= — O, 
 
 0 
 Any number divided by 0 is equal to oo. 
 3 i ae: 10008 _ 
 
 oO, -=0o, ——_ = ©, ete. 
 
 ch area 0 
 
 If a= 0 and 6= 0, the solution of ax + 6=0 becomes 
 
 e=—- 
 
 0 
 Lae i 0 
 5-38 the symbol of indeterminateness. g may have 
 any value. 
 
 a — 
 (1) 
 
 part. 
 
 
 
 If in the above we put x = 1, it becomes 
 
 
 
 1-1 0 
 ———=1+41, or ~=2. 
 Reena 0 
 “ae 
 (2) cae lib ase 
 x—d 
 If in this we put x= 5, it becomes 
 25 — 25 
 
 
 
 0 
 (aS 5 5, ty a 10. 
 Rb aeY Nhe ai, 
 
140 THE ESSENTIALS OF ALGEBRA. 
 
 106. Equations of Second or Higher Degree which depend 
 upon the Linear Type. 
 
 If we have the equation 
 x—d2+6=0, 
 we may by factoring write it in the form 
 (x —3)\@—2)=0. 
 
 We know that if one factor of a product is 0, the 
 product is 0. The product (# — 3)(a@ — 2) may be 0 by 
 either factor being 0. Ifz«—38=0, then the product is 0, 
 or if r—2=0, then the product is 0; that is, the product 
 is 0 if z= 38, org = 2. 
 
 This is called equating the factors to 0. 
 
 A root is a value of the variable which satisfies the 
 equation. Hence, in the above equation, 3 is a root 
 because it satisfies the equation. 2 is likewise a root 
 because it also satisfies the equation. Therefore, the equa- 
 tion z2— 52+ 6=0 has the two roots x= 3 and z= 2. 
 
 An equation of higher degree than the first may be 
 solved by the linear type, provided, after all the terms 
 have been brought to one member, it may be factored into 
 linear factors. Each factor equated to 0 will give one root. 
 
 Hence, the number of roots is equal to the degree of the 
 equation. 
 
 EXERCISES. 
 1. vw —5a—24=0. 
 By factoring, this is written (# — 8)(a# + 3)=0. 
 Hence, ¢ + 8.=.0 oriaees, 
 also “e+-3=0orz=—3. 
 
 The two roots are 8 and — 3. 
 
EQUATIONS IN ONE VARIABLE. 141 
 
 2. vw —3x2—40=—0. 13. 1627— 25 =0. 
 
 3. v+624+8=0. 14. «7 —T2+10—0. 
 4. ¢4+1074+16=0. 15. 2 +3e2—10=—0. 
 5) a —d5x2—14=—0. 16. 2° +82+15=0. 
 6. # —162+448=0, 17. 4¢7—122%49=0. 
 7. @+4=42. 18. 2 —6'?+2ax+a’?—0. 
 8. 2? —1=—0. 19. 0? —12 2+ 35=—0. 
 9. vw —25=—0. 20. #’?— 2142+ 20=0. 
 10. 2 +11 = 36. 21. 2 + 282+ 75=0. 
 11. «?—16=0. 22. o —T.a= 98. 
 
 12. 2’—(a—b)’?=0. 23. wv +7T«x—98=0. 
 
 24. 32°+11¢—4=—0. 
 
 25. 12(¢+1)—3(@—1)4+2?-1=0. 
 
 26. (x — x)? — 22(a? x) + 40=0. 
 
 27. (@’ +3 2)?— 8(#’? +32) —20=0. 
 
 28. +5e7+62=0. 
 29. o°® —12 07+ 27 x=—0. 35.92 =o —2 ab) = 0. 
 30. (7+4)+(2 x—5)?=73. 36. 2-— 1327+ 36=0. 
 31. (2%—5)’—(24410)?=24. 37. (a@+4)?—-@ae—1ty=0. 
 32. (6%+4)—(B8a—8)’=0. 38 (@?+227)?—(@’—42)’?=0. 
 33. 7 —~=0. 39. 404-8 a —52’°=0. 
 34. aa’? —(b+c)?=0. 40. 2 —382+52—1=0. 
 
 107. Fractional Equations. Certain fractional equations 
 
 may be reduced to the linear form or to the form dis- 
 cussed in the last section. 
 
142 THE ESSENTIALS OF ALGEBRA. 
 
 Fractional equations are made integral by clearing of 
 Fractions. 
 
 The common multiple used in clearing of fractions will 
 contain the variable. It may give an integral equation 
 which is not equivalent to the given fractional equation. 
 
 (1) nae 
 
 x 
 
 Clearing of fractions by multiplying by 2, we have 
 
 2 — 1 = 0, 
 or (#¢—-1)@4+1)=0. 
 W hence, zx=1 and z=—1. 
 
 These roots both satisfy the given equation. 
 When z= 1, 1 becomes 1-F= 1—1=0. 
 x 
 
 When z=—1, path becomes -1-+ =-141=0. 
 x 
 
 — 
 
 If in clearing of fractions we multiply by 2%, the 
 resulting equation is 
 x—axr= 0, 
 or x(a —1)(x+1)=0. 
 Whence, z=0,2=1, andz=—l. 
 
 We now have three roots, two of which, 1 and —1, 
 satisfy the given equation, while the other one, 0, does not 
 satisfy it; for, when r=0, x seh becomes gi which 
 : x 0 
 is not equal to 0. 
 
 The root 0 which is here introduced by clearing of 
 fractions is called an extraneous root. 
 
 The root 0 occurs because we multiplied by a multiple 
 higher than the L.C. M. | 
 
EQUATIONS IN ONE VARIABLE. 143 
 
 In integral equations any multiple whatever of the denom- 
 enators may be used in clearing of fractions, but in frae- 
 tional equations the L. C. M. should always be used. | 
 
 
 
 (2) 54 ot a4, 
 
 §a7?—54+474+1=42'?'—4, 
 §at—4e2?4+7-—-54144=0. 
 
 
 
 
 
 
 
 v+a= 0. 
 a= 0Q0andz=—1. 
 When z=0, 
 5+ oti a4 becomes 5+ ot a4, or 5—1=4. 
 When z=—1, 
 Hee etl =4 becomes 5 + att a, or 5 +o=4. 
 a= —1 does not satisfy the equation and is therefore an 
 
 extraneous root. 
 
 The root —1 occurs in this solution because the fraction 
 a+1 
 g2—1 
 
 it to its lowest term the equation becomes 
 
 1 
 Dye me es 4 
 AVERT aT 
 
 
 
 was not reduced to its lowest terms. By reducing 
 
 fe —) +1 ss 47 — 4. 
 be 4e—5 2 1-4 — 0, 
 me 
 No extraneous root now appears. 
 
 Before beginning the solution of a fractional equation, all 
 fractions should be reduced to their lowest terms. The safe 
 
144 THE ESSENTIALS OF ALGEBRA, 
 
 plan in all fractional equations is to test every root, retain 
 only those roots that satisfy the equations, and reject all others 
 as extraneous. 
 
 
 
 
 
 
 
 EXERCISES. 
 ee ey 3 2) ae 
 ot DS wo 
 SL 3 2 
 2. 4 —=— —= 0. eee SS 
 ta oo il ean 
 Zz 1 1 
 5 aa ii aan 
 peer g SU IGG) a 
 
 6. The quotient of a number divided by 7,increased by the 
 quotient of 63 divided by the number,is 6. What is the number? 
 
 7. A number is increased by 82 and the sum divided by the 
 number; the quotient is ;/; of 1 more than the number. What 
 is the number ? 
 
 8. The sum of the squares of two consecutive numbers is 85. 
 What are the numbers ? 
 SOLUTION. 
 Let x = one of the numbers. 
 xz +1 =the other. 
 2 (el es, 
 z24724+227+1 = 85. 
 2277+227+1- 8-0. 
 227+227—84=0. 
 
 g+a—42=0, by dividing by 2. 
 (x +7)(x — 6) =0. 
 De arm] OTT ae 
 
 The numbers are 6 and 7 or —7 and — 6. 
 
 9. The sum of the squares of two consecutive numbers is 
 41. What are the numbers ? 
 10. ‘Two numbers differ by 5, and their squares differ by 105. 
 What are the numbers ? 
 11. Three times the product of two consecutive numbers 
 lacks 92 of being twice the sum of their squares. What are 
 the numbers ? 
 
 
 
EQUATIONS IN ONE VARIABLE. 1465 
 
 12. The area of a square field is doubled by increasing its 
 length 12 rods and its width 5 rods. What is the length of 
 
 one side of the field ? 
 SOLUTION. 
 
 Let x = one side of the field. 
 a? met deel) (tO). 
 222= 474+ 1724 60. 
 242 —2?-17x7-—60=0. 
 u2—17x—60=0. 
 (x — 20)(4 4+ 3) = 0. 
 x= 20 and x =— 5. 
 
 Both of these roots satisfy the equation, but only 20 can be used 
 in this problem, as it would not be possible to have a field one side of 
 which is — 3 rods in length. 
 
 13. The denominator of a fraction 1s 3 more than its numera- 
 
 tor. If7is added to each of its terms, the value of the fraction 
 is increased by ;3;. Find the fraction. 
 
 14. A can do a piece of work in 10 days, B in 8 days, and C 
 in 6 days. In how many days can they all do it working 
 
 together ? 
 SOLUTION. 
 
 Let 
 
 = the time required. 
 
 lI 
 
 part done in one day. 
 
 Slee; 
 
 II 
 
 part done in one day by A. 
 
 II 
 
 part done in one day by B. 
 = part done in one day by C. 
 
 = part done in one day by A, B, and C. 
 
 oe 
 CO | 
 4. 
 
 II 
 
 ole ole 
 
 on Ret 
 we pp Ole 
 
 12741 
 
 Spats 
 8S 8 & Ole QIK olHKolH 
 
 II 
 oO 
 
146 THE ESSENTIALS OF ALGEBRA. 
 
 15. A cistern has two pipes; one will fill it in 8 hours, and 
 the other in 12 hours. If both are open, how long will the 
 cistern be in filling ? cE 
 
 16. A cistern has three pipes; one will fill it in 12 hours, one 
 in 10 hours, and the other will empty it in 15 hours. If all 
 three are open, how long will the cistern be in filling ? 
 
 17. A number added to 22 times its reciprocal makes 13. 
 Find the number. 
 
 : is called the Reciprocal of x. 
 z 
 
 1s. A can do a piece of work in a days, B ean do it in Dd 
 days. In how many days can they together do the work? 
 
 19. The area of a square field is doubled by increasing its 
 length a rods and its width 6 rods. What is the length of one 
 side of the field ? 
 
 20. A fraction whose numerator is 3 less than its denom-: 
 inator added to its reciprocal gives 2;%. Find the fraction. 
 
CHAPTER X. 
 LINEAR EQUATIONS IN TWO VARIABLES. 
 
 108. Roots of a Linear Equation in Two Variables. The » 
 type form of the linear equation in a and y is 
 
 ax+by+ec=0. 
 
 a, 6b, and e may be any positive or negative numbers 
 whatever. 
 
 The equation 22+5y—10=0 is a special form of 
 ax + by+e=0; in which a= 2, b=5, and e= — 10. 
 
 If in 2~7+5y—10=0, we transpose 5 y—10 and di- 
 vide by 2, we have 
 pel) 0 Y, 
 
 2 
 
 x 
 
 The value of z depends upon y. It has one, and only 
 one, value for each value of y. 
 
 ia Uta 0. lfiy=—1, ¢= 4+. 
 yo, «x= 3. y= —2, c= 10. 
 =22=0. y= —3, 2 = 28, 
 y=3, c= — 3. yi 4 tee 1D: 
 at = — 5, y= — 9, w= 3B. 
 y= 9, r= — 4p. Yie= 10,92 = 20. 
 
 147 
 
148 THE ESSENTIALS OF ALGEBRA. 
 
 The equation 272+5y—10=0 is satisfied by r=5 
 and y = 0, for these values reduce the equation to 
 
 2x5+5x0—-10=0. 
 a= 3 and y=1 also satisfy the equation, for they reduce 
 Ey xbox oe 
 
 Therefore, z= 5, y= 0 and «= 3, y =1 are roots of the 
 equation 27a +o5dy—10= 0. 
 
 In the set of values of 2 and y above, every value of a 
 and the corresponding value of y constitute a root. The 
 number of such sets of values that are possible is evi- 
 dently unlimited. Besides the roots (5, 0), (3, 1), (0, 2), 
 Cr Bs 3), C= Dy 4), (Gatce D), (Jes val 1), ClO 2), and 
 GCG), — 3), any number more could be worked out at 
 pleasure. A root of an equation in two variables may 
 be written (m, n); m is the value of z, and. n is the 
 corresponding value of y, the two together constituting 
 a root. 
 
 109. Graph of the Linear Equation. ‘The codrdinate azes, 
 or lines of reference, are two hnes perpendicular to each 
 other. 
 
 The azis of abscissas, or x-line, is the horizontal line 
 
 WOK. 
 
 The axis of ordinates, or y-line, 
 
 is the vertical line Y’OY. 
 Abscissas, or x-distances, are al- 
 X ways measured parallel to the z-line. 
 
 They are positive when measured to 
 
 the right of the y-line and negative 
 
 Y’ when measured to the left of it. 
 Wia.-1: Ordinates, or y-distances, are always 
 
 
 
 
 
LINEAR EQUATIONS IN TWO VARIABLES. 149 
 
 measured parallel to the y-line. They are positive when 
 measured above the z-line and negative when measured 
 below it. 
 
 The coérdinates of a point are its x and y distances. 
 The w distance is the abscissa, and the y distance the 
 ordinate. ‘The codrdinates cf a point completely deter- 
 mine it with respect to the lines of reference. 
 
 A point is designated by its codrdinates written (m, 7). 
 This means that m is the abscissa and m the ordinate of 
 the point. 
 
 Y 
 
 P, (2,3) 
 
 
 
 P, (—5,—4) 
 
 Y 4 
 
 Higa: 
 
 The point P, or (2, 3) is located by measuring from O 
 to M, a distance of 2, and from M to P,, parallel to OY, 
 a distance of 8. The point P, or (—4, 2) is found by 
 measuring from O to N,a distance of 4, and then from 
 N tose: parallel to OY, a distance of 2. The abscissa in 
 this case is measured to the left of OY because it is nega- 
 
150 THE ESSENTIALS OF ALGEBRA, 
 
 tive. The location of the points P; or (—5 —4) and P, 
 or (6, — 2) is also shown on Figure 2. 
 
 BEC ns See: 
 Ch 
 SeEEPEE ai. 
 
 a a 
 
 
 
 Hd 
 Y 
 ~ 
 
 Suen ane 
 
 
 
 
 
 
 
 EXERCISES. 
 
 In Figure 3 the side of each small square is a unit or one. 
 
 1. Write the codrdinates of each lettered point. Thus, 
 A is (2, 2), & is (—6, —1). 
 
 2. Locate on Figure 3 the following points: (5, 1), (— 2, 1), 
 (3, satay (es 3, oe v (6, 1), i= 1, aoe Cass 2); (0, 3), E 3, 0), 
 (0, 0), (1d, 8), (23, 22), (— 3h 4 
 
 On Figure 4 are located the points (10, — 2), (74, — 1), 
 (6, 0), (24, 1), ©, 2), C— 24, 8), C— 5, 4), (— 7, 5). 
 
LINEAR EQUATIONS IN TWO VARIABLES. 151 
 
 It will be seen that these points are in a straight line. 
 The points located on Figure 4 are some of the roots of the 
 equation 22+ 5y—10=0, worked out in Section 108. 
 
 oe ee Ha a 
 ate 
 
 Estee: 
 a 
 fs 
 P| 
 im 
 ia 
 He 
 
 
 
 TAL__| 
 
 
 
 
 
 of 
 
 me MuaG 
 Sie 
 
 
 
 
 
 
 
 
 
 ay 
 ’ 
 
 
 
 The graph of an equation ts the line upon which are found 
 all the points indicated by its roots. 
 
 The line MN is the graph of the equation 
 
 The coordinates of every point upon this line satisfy 
 the equation 27+ 5y—10=0. The point P is (24, 1). 
 (24, 1) is a root ‘of 2x+-5y—10=0, for when x=24 and 
 y =1, the equation becomes 2 X 24+ 5X 1— 10=0, which 
 is an identity. 
 
 The graph ofx—2y=4. Herezr=4+42y. 
 
 @=—=4 when y=—0. a= 2 when y=—1. 
 
 ee Oowhen oy = 1. z= 0 when y=—2. 
 
 teow oen. i = 2. z=—2 when y=—83. 
 
Loy THE ESSENTIALS OF ALGEBRA. 
 
 The points represented by the roots above worked out 
 
 are (4, 0), (6, 1), (8, 2), @, —1), (0, —2), and (—2, —3) 
 
 and are shown on Figure 5. 
 
 y 
 FEEEE EEE eae 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Ber... 
 Plo ea 
 3 HERRERA 
 [eh ise, 
 
 
 
 ESS 
 =e 
 ae 
 
 Fia. 5. 
 The points tetera in Figure 5 lie upon the straight line 
 PQ, which is the graph of —2y=4. 
 The graph ofx+y+5=0. Herex=—(y+5). 
 e=—)d ~whene y= ae x=0 when y=— 6). 
 oe — 6 when aye: z=1 when y=—6: 
 x=—2 when y=—83. 
 
 _ The points represented by the above roots are (—5, 0), 
 (—6, 1), (—2, —3), (0, —5), and (1, —6). 
 
 Locating these points upon Figure 6, we find that they 
 all lie upon the straight line AS, which is the graph 
 of x+y+5=0. This line passes through the points 
 
LINEAR EQUATIONS IN TWO VARIABLES. 153 
 
 Ac—4, —1), B(—8, —2), and C(—1, —4). Verify that 
 
 these are roots of the equation. 
 
 aaa 
 
 R, 
 
 ——— 
 
 oo 
 
 (- 
 
 Spat 
 ae 
 Bava 
 
 bisaeules 
 EP 
 Alb 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 bad 
 Eames 
 te 
 a 
 sel 
 | 
 
 Pita at 
 
 EXERCISES. 
 Draw the graphs of the following equations: 
 1 3e—2y=6. 3. 24—5y=10. 5. e+y=4, 
 2. da—y=8. 4. ¢—y=4. 6. 24—3y=0. 
 The graph of every linear equation in two variables is a 
 straight line. 
 
 Since two points are sufficient to locate a straight line, 
 we need but two roots of an equation to draw its graph. 
 
154 THE ESSENTIALS OF ALGEBRA, 
 
 The graph of 844+ 5y=15. 
 Since 2=0 when y=8, one root is (0, 3)s 
 
 and since x=0d when y= 0, another root is (5, 0). 
 
 
 
 att 
 
 Fia. 7. 
 
 Locate these two points upon Figure 7, and draw through 
 them the straight line AB. ‘This line is the graph of 
 Sa+5y= 15. 
 
 In general, the most convenient pair of roots for deter- 
 mining the graph of an equation is found by making #=0 © 
 and solving for y, and then by making y= 0 and solving 
 for z. ‘These two roots give the points in which the line 
 cuts the codrdinate axes. 
 
 EXERCISES. 
 
 By the above method make the graphs of the following 
 equations : 
 
 1. 3x—2y=6. 4. Tx—y=T. 
 2. 4x—y=8. 5. 2 Tiere 
 3. 2y—5x=10. 6. 357+4y=12. 
 
LINEAR EQUATIONS IN TWO VARIABLES. 
 
 110. Graphs of Two Linear Equations upon the Same - 
 
 Diagram. 
 Graphs of x—y=6 and 2a+y=9. 
 
 
 
 
 
 
 
 
 
 
 
 PEN ae 
 
 “A 
 
 
 
 
 
 
 
 ea] Se 25 Ne 
 
 
 
 
 
 PAGS 
 
 
 
 ys : D 
 Fig. 8. 
 
 | 
 
 Pes | lee | Ne 
 
 155 
 
 SS 
 
 The graphs of these lines are AB and CD, respectively. 
 They intersect at the point P (5, —1). This point P hes 
 on both lines, and its codrdinates constitute a root of each 
 equation. By putting x=5 and y= —1, x—y=6 becomes 
 5—(—1)=6, and 22+y=9 becomes 2x5—1=9. This 
 
 verifies that (5, — 1) is a root of each equation. 
 
 Since two straight lines can intersect in but one point, 
 a pair of linear equations can have but one common root. 
 
156 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 By means of their graphs find the common root of each of 
 the following pairs of equations: 
 
 7 \Fescie, lat , {eA ae 
 laty=6d. et+y=—B5. 
 
 a ees a aan 24+4y=6, 
 x—2y=1. laty=1. 
 
 a raat te a (e—y=1, 
 e—2y=A4. (2e 4.2 ene 
 
 7. Graphs of e+y=1 and 2x%+2y=9. 
 
 Fig. 9. 
 
 The graphs of these equations are shown in Figure 9. They 
 are parallel lines, and so do not intersect. The two equations 
 have no common, finite root. It should be noticed that the 
 coefficients of x and y in the second equation are just double 
 the coefficients of # and y in the first equation. 
 
 { x—dsy=6, 3 xe+2y=6, 
 [3a—9y=9. , pea y = 2. 
 
CHAPTER XI. 
 SIMULTANEOUS EQUATIONS. 
 
 111. Definitions. Hqguations in two or more variables, 
 having the same solutions, are called equivalent when any 
 one of the equations may be changed to the exact form of the 
 others. 
 
 Thus, r+y=d 
 and 3824+3y=15 
 
 are equivalent equations; each has a root (1, 4), and the 
 first may be changed to the second by multiplying both 
 members by 3. 
 
 Equations not equivalent, but having the same solutions are 
 called simultaneous equations. 
 
 Thus, c+y=6 
 and. s2+y=12 
 
 are simultaneous, having the common solution r=3, y=3. 
 These two equations are not equivalent, since the first 
 can not be changed to the form of the second. 
 
 Two equations in two variables form a set of s¢multane- 
 ous equations; three equations in three variables form a 
 set of simultaneous equations, etc. 
 
 112. Elimination. To solve a set of simultaneous equa- 
 tions, we must so operate upon and combine the given 
 157 
 
158 THE ESSENTIALS OF ALGEBRA. 
 
 equations as to produce a single equation containing a 
 single variable. ‘The processes of obtaining such a single 
 equation are called elimination. , 
 
 In the operations of elimination the following principles 
 are to be carefully noted: . | 
 
 (1) For any expression in an equation an identical 
 expression may be substituted. 
 
 (2) When both members of an integral equation are 
 multiplied by an integral expression containing the vari- 
 able, the resulting equation is not equivalent to the 
 original equation. 
 
 For example: x+d3y=4. 
 
 Multiplying both sides by # — y, we have 
 
 (w@—y)@+3y)=4@—y), 
 an equation which not only has all the roots of a+5y=4, 
 but also all the roots of a—y=0. 
 
 (3) All the axioms heretofore given can be used in the 
 processes of elimination, with the single exception noted 
 above. 
 
 113. Elimination by Substitution. This method will be 
 understood by noting the following solution: 
 
 (1) a+2y=5, 
 
 te 52—2y=1. 
 By transposing the 2 y in Equation (1), we have 
 (3) e=d0—2y. 
 Substituting this value of z in Equation (2), 
 (4) 6(5 —2y)—-2y=1. 
 
SIMULTANEOUS EQUATIONS. 159 
 
 Removing the parentheses, 
 
 (5) 25—10y—2y=1. 
 (6) —10y—2y=1-—25, by transposing. 
 (7) —12y=-— 24, by collecting. 
 (8) gy aut by dividing. 
 Substituting this value of y in (8), 
 (9) = 5 —2(2). 
 (10) hoe 
 
 The root is (1, 2). 
 EXERCISES. 
 
 Solve the following equations by the method of substitution : 
 
 
 
 1 | a+ y=4, 4 v+4y=10, 
 ' (8a—4y=5. | ~(38e— y=12. 
 Beets, 11, AEE OE ms seh 
 22+ y=9.  (8a4+2y=T7 
 Soy — 2, ne rye 
 | w«+3y=114. ise homers 
 2 
 lx 9» 
 6a+ y= 20. Outen 
 5a+6y=16, 1) Bp A 
 —8e+38y=— 18. eet egy=li 
 
 © exh 
 
 Construct graphs for Exercises 4, 5, 6, and 7 
 
160 THE ESSENTIALS OF ALGEBRA. 
 
 114. Elimination by Comparison. ‘I‘he following problem ~ 
 will illustrate the method : | 
 
 
 
 (2) 4vu— y=5. 
 By transposing 8 y and dividing by 5, we get from (1) 
 19—3 
 (3) ta ee 
 
 By transposing — y and dividing by 4, we get from (2) 
 
 5 
 
 (4) ime te 
 
 Equating the two values of 2 given by (8) and (4), 
 
 19—3 5 
 
 (5) a ee, 
 (6) 76 —12 y= 25+ 5y, by clearing of fractions. 
 (7) —12y—S5y=25— 76, by transposing. 
 (8) —1l7y=-—4dl, by collecting. 
 (9) y= 3. 
 
 Substituting this value of y in (3), 
 
 (10) pa PU eXt a2, 
 
 The root is (2, 3). 
 EXERCISES. 
 
 Solve by the method of comparison : 
 
 a (ie eae f4et+ y = 3, 
 (2Qa+ y=3. "(| 2Qe4+8y=4. 
 - eee pa tay F (3a—4y=4, 
 
 “x—dsy=—A4, |2a+5y=10. 
 
SIMULTANEOUS EQUATIONS. 161 
 
 : lai (—4¢e—Zy=1, 
 (8e2+7y=21. ~(— w+5y=6. 
 aoe” i a 
 lat2y=b. l5a+ Piezo 
 ama 4 ores Wie 
 L38a+2y=12. (dsat4y=4m. 
 
 Construct graphs for Exercises 3, 5, 7, and 9. 
 
 115. Elimination by Addition or Subtraction. ‘he two 
 problems here solved will illustrate the method. 
 
 ei jeer +o Y= 0, 
 
 Bical 
 eo. es Ta Oy =. 
 
 We first make the coefficients of the y’s have the same 
 absolute value. This is done by multiplying Equation (1) 
 by 2 and Equation (2) by 3, thus giving us 
 
 (8) 4x7+6y= 10, 
 (4) 2l2—6y= 165. 
 Adding Equations (3) and (4), 
 
 (5) 25 2 = 25. 
 
 (6) goerai li 
 Substituting this value of a in (1), 
 gi) 2XL+ oY = 0. 
 
 (8) 38y=5—-—2=3, 
 (9) y=. 
 
 The root is (1, 1). 
 
162 THE ESSENTIALS OF ALGEBRA. 
 (1) 24+Ty=38. 
 (2) 84+4y=81. 
 
 We can make the coefficients of the a’s alike by multiply- 
 ing Equation (1) by 3 and Equation (2) by 2, thus giving 
 
 2. Solve 
 
 (3) 624+ 21ly=114, 
 
 (4) 62+ 8 y= 62. 
 Subtracting Equation (4) from (8), 
 
 (5) LS a ae 
 
 (6) y= 4. 
 
 Substituting this value of y in Equation (1), 
 Gh) 24+7-4=388. 
 
 (8) 2x2= 38 — 28 = 10. 
 
 (9) area lta 
 
 The root is (5, 4). 
 
 The method of elimination generally used is that of 
 addition or subtraction. ‘The method by comparison is 
 merely a disguised form of eliminating by subtraction. 
 The particular method to be used must be determined by 
 a careful inspection of the problem. 
 
 116. Some Illustrative Examples. 
 
 ( Dey 
 i a i — 4, 
 ib) 413 
 
 Here there is no need of clearing of fractions. By addi- 
 tion, y is eliminated, and 
 
SIMULTANEOUS EQUATIONS. 163 
 
 a2 
 
 3 —= 6. 
 om ; 
 (4) 3x2= 24, 
 (5) oa; 
 Substituting 8 for z in (1), we have 
 gy 
 oy —+“=4, 
 (6) ats 
 (7) pat 2=2. 
 (8) y = 0. 
 The root is (8, 4). 
 Bee 17 
 1) - a? 
 . @) x ss y 24 
 iy Gagan 
 mony 24. 
 In problems of this form never clear of fractions. 
 (3) pee Ae Bh by multiplying (2) by 2. 
 Gey 24 aa 
 ee aye 
 (4) anor by adding (1) and (3). 
 (5) sr pkcgeaad Wop tyre 
 (6) emia fe 
 Substituting 8 for 2 in (1), 
 pera LT 
 1) g ty bt 
 rus all YE ear ee ME 
 ©) y 248 8 
 @) y= 6. 
 
 The root is (8, 6). 
 
164 THE ESSENTIALS OF ALGEBRA. 
 (lez — 37 = 10: 
 
 In this form, in which one equation has the variable with 
 a coefficient 1, use the method of substitution. 
 
 From (1), 
 
 (3) x=3y+10. 
 
 Substituting the value of x in (2), 
 (4) 88 y410)+5y=2. 
 (5) 9y+380+5y=2. 
 
 (6) 14 y= — 28, transposing and combining. 
 (8) x= —64+10=4, from (3). 
 
 The root is (4, — 2). 
 
 
 
 G) oe 
 - BY y 
 2 UI Tene 
 (2) a a 
 
 In examples of this form it is best to clear of fractions. 
 
 (8) 14+2274+3y=3z2, by clearing (1) of fractions. 
 
 (4) —x+8y=-—1, by transposing and collecting. 
 (5) xet+y=5, by clearing (2) of fractions. 
 (6) 4y=4, by adding (4) and (5). 
 (T) y=1. 
 
 GS) ake x=4, by substituting 1 for y in G). 
 
 The root is (4, 1). 
 
SIMULTANEOUS EQUATIONS. 165 
 
 EXERCISES. 
 
 Solve the following simultaneous equations: 
 
 
 
 eas y =, 2.3 TH 
 : ae ala? 
 (2e%—5y=12, [~+-=22. 
 : x 
 2 { de +8y=13, a4 <6, 
 eee 8 — 71. 11 ‘ 
 A (fe—ty=1, o%—-—-=3 
 (ga+tozy=4 
 [e+7=a 
 ( aes , ue 
 4x —_=6, 12. Y 
 5 7 
 : kx —— =) 
 Beye 2. 35 | 
 fnan 
 A an ay : 735° : 
 ie) 18 eo 
 Bie b;. Hb y 
 ax + by == 4m besa eased 
 2 p. a 1 ee Tt 
 ' n+y ™m 
 F mie? 15 (3ae—Ty=0), 
 ee 7 | lge+hy=7. 
 é My ae (ax — by=0, 
 ug me+ ny =q. 
 9. Shak 2 . f mar + ny = me’, 
 Dany. bnew + my =n. 
 
166 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 
 
 CEE ee oe v—a , y—b 
 1s. - —= —Ty— =(, 19. — “~—1[= e 
 8 at 4 3e—T Yy—di es ; < ; 
 
 syt+Ta+1 2y—32+8_5 
 re aH 
 5 3 
 20. 
 Sy—Tex+10_ sy+24+6_9 
 3 5) ; 
 
 Construct graphs for Exercises 2 and 6. 
 
 EXERCISES. 
 
 1. The sum of two numbers is 32, and one number is 3 
 times the other. What are the numbers ? 
 
 SOLUTION. 
 
 Let x and y be the numbers. 
 
 Then (1)! eee, 
 and (2) xz=3y, by the conditions of the problem. 
 (3) 8y+y=32. 
 (4) 4 y= 32. 
 (5) y =8. 
 (6) ey 
 
 2. Eight apples and 5 oranges cost 31 cents, and 5 apples 
 and 10 oranges cost 40 cents. What is the cost of 1 apple and 
 of 1 orange ? 
 
 3. Three bushels of wheat cost 15 cents more than 5 
 bushels of corn, and 2 bushels of wheat and 1 bushel of corn 
 together cost $2.05. What is the price per bushel of each ? 
 
 4. A fraction is equal to 3. If both of its terms are 
 increased by 12, the value is then 4. Find the fraction. 
 
 (Let x = numerator, y = denominator, ~ = fraction.) 
 
 5. Find a fraction such that if 1 is added to the numerator 
 it becomes 4, and if 5 is added to the denominator it becomes }. 
 
SIMULTANEOUS EQUATIONS. 167 
 
 6. The sum of two numbers is 75. The larger contains the 
 smaller 5 times, with a remainder of 3. Find the number. 
 
 7. There are two numbers; 3 times the first is 8 more than 
 the second, and their difference is 42. Find them. 
 
 8. A man spent $225 for sheep at $3.50 a head and calves 
 at $10 a head. He bought 42 head inall. How many of each 
 did he buy ? 
 
 9. Ten years ago a father was 5 times as old as his son. 
 Twenty years hence he will be twice as old. What are the 
 present ages of each ? 
 
 10. A said to B, “Give me $60, and I shall have twice as 
 much as you.” B said to A, “Give me $90, and I shall have 
 as much as you.” How much had each ? 
 
 11. Find two numbers such that + the first and + the second 
 is 36, and } the first and + the second is 18.. 
 
 12. There are two numbers such that if each is increased by 
 5, the sums are in the ratio 5 and 11, and if each number be 
 decreased by 15, the remainders are in the ratio 1 and 7. Find 
 the numbers. 
 
 13. A farmer has two horses and an $18 saddle. If the 
 saddle is put on the cheaper horse, the horse and saddle are 
 worth 2 of the better horse. The better horse and saddle lack 
 $12 of being worth twice as much as the cheaper horse. What 
 is the value of each horse ? 
 
 14. If the greater of two sums be multiplied by 5 and the 
 lesser by 7, the sum of the products is 140. If the greater be 
 divided by 7 and the lesser by 5, the difference of the quotients 
 is 0. Find the numbers. 
 
 15. There are two numbers which differ by 11. One sixth 
 of the larger is 1 more than + of the smaller. Find the 
 numbers. 
 
 o 
 
168 THE ESSENTIALS OF ALGEBRA. 
 
 16. A number consists of two digits whose sum is 18; if 27 
 be added to the number, the order of the digits is changed. 
 
 Find the number. 
 SOLUTION. 
 
 Let x = units’ digit. 
 y = tens’ digit. 
 10 y + x = the number. 
 
 Then (1) Zit iz ls, 
 and (2) 10y+2+27=102+~y, by conditions of problem. 
 (3) 9y—-I9x=— 27. 
 (4) y—-r=—3. 
 (5) 24 = 10; by adding (1) and (4). 
 (6) y = 5. 
 (7) alias from (1). 
 
 107 + x = 58, the number. 
 
 17. If to a certain number of two digits the tens’ digit be 
 added, the sum is 80. If the units’ digit be subtracted, the 
 remainder is 70. Find the number. 
 
 18. A number is composed of two digits whose sum is 13. 
 If their order is inverted, the new number is 4 less than double 
 the original number. Find the number. 
 
 19. A sum of money was divided equally among a certain 
 number of people. If there had been 3 persons more, the share 
 of each would have been $2 less; but if there had been 2 per- 
 sons fewer, the share of each would have been $2 more. How 
 many persons were there, and what was the share of each? 
 
 20. A lost 3 of his money and then borrowed 4 of B’s money, 
 when he had $12. At first A had 2 as much as B. Find 
 how much each had at first. 
 
 21. The sum of a number of two digits and the number 
 formed by reversing the order of the digits is 110. The differ- 
 ence of the digits is 8. Find the number. 
 
SIMULTANEOUS EQUATIONS. 169 
 
 22. A man has a certain number of silver dollars and quar- 
 ters. He notices that if his dollars were quarters and his 
 quarters dollars he would have $22.50 more than he now has. 
 He also notices that if his dollars were dimes and his quarters 
 half dollars, he would have $1 more than he now has. How 
 much money has he ? 
 
 23. In a certain school 4 of the number of boys is equal to 
 4 of the number of girls; twice the whole number of pupils in 
 the school is 100 more than 3 times the number of girls. 
 How many pupils in the school ? ; 
 
 24. A and B are 45 miles apart. If they walk in the same 
 direction, A overtakes B in 45 hours. If they walk toward 
 each other, they meet in 5 hours. Find their rates of walking. 
 
 25. If the first of two numbers be divided by 12 and the 
 second by 15, the sum of the quotients is 12; if the first be 
 divided by 4 and the second by 3, the difference of the 
 quotients is 12. What are the numbers ? 
 
 26. Find two numbers such that the sum of their reciprocals 
 is 25, and the difference of their reciprocals is ;35. 
 
 27. If the base of a rectangle be increased by 6 feet and 
 the altitude by 4 feet, the area is increased by 216 square feet. 
 If the base be decreased by 4 feet, and the altitude increased 
 by 4 feet, the rectangle becomes a square. Find the base and 
 
 altitude of the rectangle. 
 
 28. If B loans A $500, A will then have 3 times as much 
 money as B has left; but if A loans B $ 200, B will have twice 
 as much money as A has left. How much money has each ? 
 
 29. A sum of money on interest amounted to $ 824 in 9 months 
 and to $840 in 15 months. Find the principal and the rate. 
 
 30. If the greater of two numbers be divided by the less, 
 the quotient is 1, with a remainder of 8; if 4 times the less be 
 divided by the greater, the quotient is 2 with a remainder of 
 22. What are the numbers ? 
 
170 THE ESSENTIALS OF ALGEBRA. 
 
 31. Sixty workmen, consisting of men and boys, did a piece 
 of work in 5 days and received for it $430. The men were 
 paid $1.75 a day, and the boys 80 cents a day. How many 
 men and how many boys were there ? 
 
 32. Find a fraction such that when the numerator is trebled 
 and the denominator decreased by 4 the value becomes 3, and 
 when the denominator is trebled and the numerator increased 
 by 2 the value becomes 4. 
 
 33. In 10 years I will be 5 times as old as my son was 5 
 years ago, and 2 years ago I was twice as old as my son will 
 be 4 years hence. Find my age and that of my son. 
 
 34. The lengths of two ropes are as 4: 5, and when 20 feet 
 is cut from each rope the remainders are as 3:4. Find the 
 lengths of the ropes. 
 
 35. A river flows 3 miles an hour; a boat going down the 
 river passes a certain point in 12 seconds and in going up it 
 takes 18 seconds. Find the speed of the boat in still water 
 and the length of the boat. 
 
 117. System of Linear Equations with Three or More 
 Variables. We have seen that in order to solve linear 
 equations with two variables, we must have a set of two 
 independent equations; in like manner, when we have 
 three variables, we must have a set of three independent 
 equations ; when four variables, we must have a set of 
 four equations, etc. 
 
 The method of solving a problem in three variables 
 will be understood by noting the following solutions: 
 
 (1) e-2y+2=1, 
 ai (2) 3a+y—z2=4, 
 (3) 2a+ty+2=12. 
 By looking at this problem we see that the z’s can be 
 easily eliminated. 
 
SIMULTANEOUS EQUATIONS. WG 
 
 (4) 4x —y=5, by adding (1) and (2). 
 
 (5) o2+2y= 16, by adding (2) and (8). 
 
 (6) 8x—2y= 10, by multiplying (4) by 2. 
 
 (7) bo == 20, by adding () and (6). 
 
 (8) Nee ee 
 
 (9) - y=, by substituting 2 for x in (4). 
 (10) ae, 
 
 by substituting 2 for x, 3 for y in (1). 
 ‘The root is (2, 8, 5). 
 (1) 4a—3y4+22=8, 
 ay bes 64+3y4+32=T, 
 (8) 2x—6y+52=4. 
 By adding (1) and (2), we eliminate y, and have 
 
 1072+ 52=10, 
 or (4) Phe en ae ie ay 
 
 Multiplying (1) by 2 and subtracting (3), we have 
 (5) G27 2, 
 
 Adding (4) and (5), 
 (6) pO ta4. 
 (7) t=. 
 
 Substituting x= 4 in (4), 
 
 (8) | Li a2. 
 (9) z=. 
 Substituting ¢=4 and z= 1 in (1), 
 
 (10) 2—3y+4+2=3. 
 (11) —3y=-1. 
 (12) Y= 
 
 The root is G, 4, 1). 
 
172 
 
 (Do not clear of fractions in Exercise 
 
 ( 
 
 
 
 = eee er ee ee 
 J 
 
 RliR SiR Sik 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 “x +2=11, 
 
 y+z=6, 
 2ae+y = 25. 
 22+32= 54, 
 Ty+5z=106, 
 de+ 5y= 76. 
 
 10. 
 
 EXERCISES. 
 
 
 
 2 83 
 £ ¥ B 
 «| 3 
 20. aye 
 ey OZ 
 nt nn" 
 mM n 
 Phy Zz 
 7 
 Yo 
 Gar : 
 5 q 
 OP gaa . 
 8. po 2 ee 
 ey) Daa 
 ae 
 9, 2 i+2=5, 
 tf See 
 Bo 3 
 D-) 
 
 e+ytz+w=10, 
 2e—y+32+w=13, 
 
 “e+ 3y—2z2—-—w=—3, 
 (y —2y4+32—-2w=—2. 
 
SIMULTANEOUS EQUATIONS. 173 
 
 11. Three men have together $750; 4 of A’s and 4 of C’s 
 is equal to 4 of B’s; twice A’s is $150 more than both B’s 
 and C’s. How much money has each? 
 
 12. The sum of three numbers taken two and two are 68, 
 94, and 62, respectively. Find the numbers. 
 
 13. There are three numbers such that the first with $ the 
 second is 68; 4 the first with 2 the third is 73; and the second 
 with + the third is 90. Find the numbers. 
 
 14. A number consists of three digits whose sum is 17. The 
 hundreds’ digit is 3 times the tens’ digit. If the order of the 
 digits be reversed, the number is increased by 297. Find the 
 number. 
 
 15. A and B can doa piece of work in 6 days, B and C in 
 74 days, and C and A in 10 days. In how many days can each 
 do the work separately ? 
 
 16. A cistern has three pipes A, B, and C. If A and B fill 
 while C empties, the cistern will be filled in 60 minutes. If A 
 and C fill while B empties, the cistern will be filled in 24 minutes. 
 If B and C fill while A empties, the cistern will be filled in 120 
 minutes. In what time could each pipe fill it alone ? 
 
 17. There are three numbers whose sum is 113; 4 the sec- 
 ond is 2 more than } the third; 2 of the first lacks 3 of being 
 4 the second. Find the numbers. 
 
 18. Separate the number 180 into three parts, such that the 
 second divided by the first equals 2, the third divided by the 
 second equals 3, and the first divided by the third equals 4. 
 
 19. A, B, and C have certain sums of money. If A gives 
 B $100, they will have the same amount; if C gives A $200, 
 he will have as much left as A and B together then have; if 
 B’s money were doubled and <A’s increased by $100, they 
 would then have together as much as C. What sum has each? 
 
CHAPTER XII. 
 EVOLUTION. 
 
 118. Definitions. Square Root. One of the two equal 
 factors of a number is called its square root. 
 
 25 = 5 x. 5, hence 5 is a square root of 25. 
 
 Cube Root. One of the three equal factors of a number 
 is called its cube root. 
 
 64=4x 4 x 4, hence 4 is a cube root of 64. 
 
 nth Root. One of the nm equal factors of a number is 
 called its nth Root. 
 
 a“=axaxaxa-+- to n factors, hence a is an nth 
 root ola”: 
 
 Evolution. ‘The process of finding roots is called evolu- 
 tion. It is the inverse of ¢nvolution. 
 
 From the definition we see that the square root of at is. 
 
 4 
 $ — a, the cube root of a” is a =a, and the nth root 
 
 a 
 
 Of a” is he 
 
 The Radical Sign. When the sign -+/ is placed before a 
 number, a root is to be extracted. The number written over 
 the radical sign is called the index, and denotes what root 
 is to be extracted. Thus, V16 = 4, V125 =5, V16=2, 
 V32 = 2, Va" =a; the indices 2, 3, 4, 5, n, denote, respec- 
 tively, the 2d, 5d, 4th, 5th, and nth roots of the number 
 before which the radical sign (,/) is placed. The square 
 root sign (/) is usually written -/. 
 
 174 
 
EVOLUTION. 175 
 
 The Radicand. The number whose root is to be ex- 
 tracted is called the radicand. Thus, in V25, V125, 25 
 and 125 are radicands. 
 
 The root of an expression of two or more terms is 
 denoted by the radical sign in connection with the vinculum 
 or parenthesis. 
 
 Vat+2ab+0 and /(a?+2ab + 62) each denote the 
 square root of a? + 2 ab + 0%. 
 
 119. Even and Odd Roots. An even root has an even 
 index, an odd root an odd index. 
 
 Ae V5, are even roots. V8, V32, Vals, are odd 
 roots. 
 
 An even number of equal positive or negative factors 
 multiplied together gives a positive product. Hence, only 
 positive numbers can have real even roots. An even root 
 of a negative number is called an imaginary number. All 
 other numbers are real. All numbers are either real or 
 amaginary. 
 
 5 Wt, V1, a, 6, are real numbers. V—1, V—5, 
 
 Ai a 6 r : ° 
 V—16, V— a’, are imaginary numbers. 
 
 
 
 120. The Law of Signs. (1) An even root of a positive 
 number is either + or —. 
 V81=+9, for (+9)x(+9)=81 
 and (—9)x(—9)= 81. 
 V a2 i td 
 (2) A negative number has no real even roots. 
 (3) An odd root of a negative number is —. 
 
 peep - 8, for (—3)(—3)(— 8) = 27. Wa B2= ~—2: 
 
 
 
 
 
176 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 Extract the indicated roots: 
 
 
 
 
 
 
 
 
 
 
 
 1. V25@0=+5a. 11. 625 a™b*c2, 
 2. Valle = + @bVe. 12. 1000 a®b%, 
 3 V16 abil 13. V400 a®)¥!8, 
 4. V/169 xyz, 14. bec 
 (w+ y)’ 
 SAO Ty 15. Ve 343 DPC, 
 6. V—8 abi = — 2 abc’. 16 «| SL 
 7. Va—4ab+4b'=+ (a—2b). Vag a®btc® 
 8. Va?+6ax+9 x 17. V16 a®p%e8, 
 Ay Aap 18. \/— 32 aay, 
 ae . 19. v/aPb%c®, 
 0° New 20, /— bE, 
 
 SQUARE ROOT. 
 
 121. The Square Root of a Polynomial. The square root 
 of a polynomial is found by the reversal of the method 
 used in eae a polynomial. 
 
 (A+ B)*= A?+2 AB+ Bis the general type form of 
 the square of - sum of any two quantities, and is the. 
 type form used in the reversal process. 
 
 By comparing any perfect square, whose root consists of 
 two terms, with this type, its root may be easily determined. 
 
 get 10 9 + 25. 
 
 This may be written 27+ 22-5 +4 52. 
 It is at once seen that 4d=z and B=5. Hence, the 
 square root of z74+10%+25isv+5 
 
EVOLUTION. BEG 
 
 The same method may often be applied to polynomials. 
 whose roots have three terms. 
 e+y?+9+2ay+6u+6y. 
 Arrange this in type form, according to a. 
 P+ 2a(yt+3)+y+6y+9. 
 This may be further arranged 
 w+ 2a(yt+3)\t+(y +3), 
 and the square root is at once seen to be 7+ y +3. 
 
 EXERCISES. 
 
 Find the square root of the following by comparing with 
 the type form: 
 1. 2 +1624 64. 
 2. 9a°+24x%+4+16. (Write (8 2)?+ 2(8 2)4416.) 
 a 2 — 19%. 81. 5. 1627+ 56 vy + 49 7’. 
 4. xv —10 ay + 25 y’. 6. a’a’+ 2 abuy + by’, 
 7. P42 eyt+yY +242 2042 yz. 
 8. w4+9yV4t2+6ay4+22e4+ 6 yz. 
 5. a—G6bc+ 0? +9 ce —2 ab+6ca. 
 lo. 42°4+7°+92—4 ay —6 yz4+12 2. 
 MW. a?+0?4+ 4+’? 4+ 2ab4+ 2 bc4+ 2ca+2ad+2bd+2 cd. 
 12. 2 —6ay+9y+16+8 a¢— 24 y. 
 
 122. Formal Extraction of Square Root. When the root 
 can not be easily determined by inspection, we reverse 
 
 the type form. A2+2AB+ B?| A+B 
 A2 
 
 2A+B|2AB+ B 
 2AB+ B 
 
178 THE ESSENTIALS OF ALGEBRA. 
 
 The first term of the root A is the square root of A?. 
 The second term, B, is contained in 2 AB, and is found | 
 from it by dividing by 2A. 2A is called the trial divisor. 
 (2 A+ B) is the complete divisor. When this is multi- 
 plied by B, the result is 2 AB+ B%, which is the part of 
 the square remaining after A? is subtracted. 
 
 (1) Find the square root of 36 #?— 144 xy 4 144 7’. 
 
 36 a — 144 ay + 144 9? |6a—12y 
 36 2 
 
 122—12y| — ldtay+ 1447 
 — 144 ay + 1447? 
 
 Here the square root of 3627 is 6x. 122 is the trial 
 divisor, which divided into — 144 2y gives —12y as the © 
 next term of the root. The complete divisor is 127-12 y, 
 which multiplied by — 12y gives — 144 2y + 144 7°, the 
 remaining part of the square after 362? is subtracted. 
 62—12y is the required square root. ‘This method is 
 easily extended, as the following example will show : 
 
 (2) Find the square root of 24+ 62?+a?—247+16. 
 a+ 6 a8 4+22—247+16|a?+30—4 
 
 a! 
 Q2432/623+ 2 
 § a +9 a2 
 —8 7?7—24 7416 
 —8 2?— 2947416 
 
 
 
 
 Oia +6 7 —4 
 
 Note that the polynomial is arranged according to 
 powers of 2. 
 
 The square root of 2* is z. The trial divisor is 2 2%, 
 which divided into 62° gives 3a, the next term of the 
 
EVOLUTION. 179 
 
 root. The complete divisor is 22?+32. When this is 
 multiplied by 32 it gives 623+927, which leaves, when 
 subtracted from the remaining part of the square, 
 —82?—247416. (274372) is now regarded as the first 
 part of the root, giving a trial divisor of 222+ 6a. This 
 gives —4 as the next term of the root. ‘The complete 
 divisor is 22?+6a—4. When this is multiplied by — 4 
 it gives —8 a? — 244+ 16, which is the part of the square 
 remaining. The root is 27+ 3a —4. 
 
 123. Rule for Extracting the Square Root: 
 
 (1) Arrange the terms with respect to the powers of some 
 letter. 
 
 (2) Extract the square root of the first term, place its root 
 as the first term of the root sought, and take its square from 
 the given polynomial. 
 
 (3) Double the root already found for a trial divisor, 
 divide the first term of the remainder by the trial divisor, 
 placing the quotient as the neat term of the root, and also 
 annexing it to the trial divisor to form the complete divisor. 
 
 (4) Multiply the complete divisor by the last term of the 
 root, and take the product from the first remainder. 
 
 (5) Continue this process until the other terms of the root 
 
 are found. 
 EXERCISES. 
 
 Extract the square root of the following: 
 1. af 14 974+ 42° + 20 2 + 25. 
 —423 +100? —1224+9. 
 Aat + 25 a? +12 a + 24a + 16. 
 mee alee oF 1 AD ot 1 Oe 38 a 21 a. 
 » 4at +h oy? + y' + 4 aby + 2 wy’. 
 
 a - oO N 
 
180 THE ESSENTIALS OF ALGEBRA. 
 
 9 xt 4 37 ay? + 4 yt — 30 ay — 20 anf. 
 1—2e—6y4+7+6ay+9 yy’. 
 
 Aa? — 12 2y +16024+9 924162 —24 ye. 
 4b +4¢7— 123-694 oe 
 
 ht 
 
 7 
 
 o ON 
 
 y' 2 by? 4 b°y? 3 
 10. = — + —_+ b 
 Fait sa as Oe 
 
 124. Inexact Square Roots. ‘The following example will 
 sufficiently show the method. 
 Extract the square root of 1+3 2 to four terms. 
 
 1+8e¢|l+3e—-feP+ite 
 
 1 
 2+3713824 
 382+ 32 
 
 24+3x2—32" 
 
 
 
 bole 
 
 co Be 
 qv 
 — a2? 22934 81 at 
 
 24+3u—3 27+ 22 a3 | At 
 
 
 
 
 
 EXERCISES. 
 
 Extract the square root of the following to four terms: 
 
 oe bee § 1+597- 37 
 2.1+42+827 | 6. 1— Ae oe 
 3) er, 7. 1-387 +5ee 2 
 4, 144427, 8. 1t+u+e?+a?+ or 
 
 125. Square Root of Arithmetical Number. The following 
 principles may be easily verified by the student: 
 
EVOLUTION. 181 
 
 (1) The square of a number of one digit consists of one or 
 two digits. — 
 
 (2) The square of a number of two digits consists of three 
 or four digits. 
 
 (3) The square of a number of three digits consists of five 
 or six digits; and so on. 
 
 From these principles we can at once tell how many 
 digits in the square root of a given number. The square 
 root of 390625 will consist of three digits. 
 
 The number of digits in the root may be indicated by 
 separating the given number into groups of two figures 
 each, beginning at the right. The left group may contain 
 either one or two digits. 8906 25 and 1 98 21. 
 
 Now solve by using the same method as in algebraic 
 problems. 
 
 Extract the square root of 390625. 
 
 89 06 25 | 600 + 2045 
 
 36 00 00 
 1200, trial divisor 
 20 
 1220, complete divisor 
 1240, trial divisor 
 5 
 1245, complete divisor 
 
 
 
 It is usual to put the work in the following shorter form : 
 39 06 25 | 625 
 36 
 122 | 3 06 
 
 1245 
 
 
 
182 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 Find: 1. 104976. 2. \/ 278784. 3. VbT121, 
 4. 61504. 5. 235225. 6. 1092025. 
 
 126. Roots of Decimals. Group both ways from the deci- 
 mal point, and solve exactly as in whole numbers. 
 
 36 74.87 63 64. ‘This shows the root to be made of two 
 whole number places and three decimal places. 
 
 EXERCISES. 
 Find: 
 1 W18'3184, 3. /.133225. 5. +/1.110916. 
 2. 8.6436. 4. >/.00300304. 6. VE: 
 
 If the result of Exercise 6 is desired to three decimals, we may 
 write it- V2.000000, and then proceed as in the above exercises. 
 
 7. ~/5 to three decimals. 8. \/11 to three decimals. 
 
 
 
 127. Cube Root of Polynomials. ‘The type form is 
 (A + B)® = A? 4+ 3 A2B + 3 AB? + B. 
 
 All polynomials in this form may have their cube roots 
 written by inspection. a%+62%y+122y?+8y? may be 
 written, 2+ 3a%(2y)+3a2 y)?+C2y). 
 
 The cube root is readily seen to be 7+ 2y. 
 
 By exactly reversing the type form we can extract the 
 cube root of any polynomial which is a perfect cube. 
 
 A4+3A°B+3AB+ B|A+B 
 AB 
 
 3A24+3AB+4 B?|3 A2B+3 AB? + B 
 3 A2B +3 AB2+ Be 
 The term B of the root is found in 3 A?B and is obtained 
 
 by dividing 3 A?B by 3 A?. This shows 3 A? as the trial 
 divisor, and 3 4243 AB+ B? as the complete divisor. 
 
 
 
1838 
 
 EVOLUTION. 
 
 Extract cube root of 
 
 27 #® — 108 2° + 198 at — 208 28 + 18207-4824 8. 
 
 
 
 IOSTAIP 939[du109 ‘¢-+2 FZ—-t QO+ 
 ef TL—# LO =et(* F—e GE te F—e BE 
 S+e BF—@ TEL +e FEL st FG | —_ TOSTAT [UMD ge Qh+ ee GLH LE =e Fe? GE 
 eC FQ — FEL to QOL — | AOSTAIp oqopdutod {we OT + et 9G —¥% LG 10 
 (et A) + @ FD Et 1] 
 IOSTATP [LMI] 5% 15 =eGr GE 
 
 SL BP—et GEL tet FFL —¥ FS 
 
 
 
 el 806 — 3% SGI Hc% SOT — 
 gt LG 
 S+X QF— wv TET Heh 80G — 3 BEL TeX SOT — 9% LZ 
 
 
 
 "NOILATOG 
 
184 THE ESSENTIALS OF ALGEBRA. 
 
 128. Rule for Extraction of Cube Root. 
 
 (1) Arrange with respect to some letter. 
 
 (2) Hetract the cube root of the first term for the first 
 term of the root, and subtract the cube frem the polynomial. 
 
 (3) Use three times the square of the root found for a 
 trial divisor, and by dividing the remainder by this divisor 
 get the second term of the root. | 
 
 (4) Add to trial divisor three times the product of the first 
 part of the root and the part of the root last found and the 
 square of the root last found. 
 
 (5) Subtract the product of the complete divisor and the 
 part of the root last found from the remainder of the 
 polynomial. 
 
 (6) Repeat this process until the root has been completely 
 determined. 
 
 EXERCISES. 
 
 Extract the cube root of: 
 
 et hart 12¢45. 
 
 ee Ba 6 ot 7 of 6 ee 
 
 of — 3 ay + 6 at? — 7 oP + 6 x7 — Say 
 Ve+3¢0(6+1)4+3a(041)?+60430?4+3641. 
 a@+3a%+60?+ 12004 300?+0' + 1240-12026 
 27 a — 64 y® — 108 a?y + 144 ay’. 
 
 e° — 9 o' +. 42 ot = 117 oF + 210 a? — 22 eee 
 
 8 a®b* — 36 a°d® + 78 att — 99 a®b® + 78 a7b? == 36 aa 
 
 OA) Poe WN 
 
 129. Extraction of Cube Root of Arithmetical Numbers. The 
 following principles may be verified as in square root: 
 
 (1) The cube of a number of one digit consists of one, two, 
 or three digits. 
 
EVOLUTION. 185 
 
 (2) The cube of a number of two digits consists of four, 
 jive, or six digits. 
 
 (3) The cube of a number of three digits consists of seven, 
 eight, or nine digits ; and so on. 
 
 From these principles we can at once tell how many 
 digits in the cube root of a given number. 
 
 If the given number is separated into groups of three 
 figures each, each group will correspond to a digit of the 
 root. | 
 
 For example: 95 256 152 263. There are four digits in 
 the cube root of this number. ‘The group at the left may 
 contain one, two, or three digits. 3365 791 and 871 625. 
 
 An example will illustrate the method of solution and 
 show that the same plan is followed as in the extraction 
 of the cube root of polynomials. 
 
 Extract the cube root of 262144. | 
 
 2+ 3 ab+ 3 ab? + 6 = 262 144| 60+4=6 
 ae = 603 = 216 000 
 (8 a2 + 3ab4+0)b= 46 144 
 
 ego 5 x 607 = 10800 
 +3ab=3x60x4= 720 
 + §2=— 42 = 16 
 (3 ab + 3.ab + 67)6 = 11536 x 4= 46 144 
 The above shows the similarity to the general type. 
 In practice the solution should appear as follows: 
 
 262 144 | 64 
 63 = 216 
 Trial divisor 6? x 300 =10800| 46 144 
 Cite eB = -. 720 
 42 — 16 
 11536 | 46 144 
 
 
 
 
 
 
 
 
 Complete divisor 
 
 
 
186 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 Extract the cube root of each of the following numbers: 
 1. 4913. 3. 753571. 5. 2628072. 
 2. 300763. 4. 614125. 6. 1.728. 
 
 In Exercise 6, group both ways from the decimal point. If neces- 
 sary, annex ciphers to fill the last group. If the root is not exact, by 
 annexing ciphers the result may be carried to any desired number of 
 decimal places. 
 
 VAD UREN stone Bat 9. .0081 to three decimals. 
 
 8. 3 to three decimals. 10. 2.05 to three decimals. 
 
CEAPTER XIII. 
 THEORY OF INDICES. 
 
 130. The Index Laws. ‘The following laws for integral 
 exponents have already been proved: 
 
 rie a” x a” =— qztn. ) 
 
 2. a” +a" =a”, when m is greater than a. 
 8. (a”)” — qi, 
 4. (a x ce a” ~x 6”. 
 ee eo 
 3 b a bn 
 _ m 
 6. Va" =a", when m is divisible by a. 
 
 The laws of algebra should be general. Let us assume 
 that the above index laws hold true for all values of m 
 and ”, and find consistent meanings for certain special 
 forms, viz., a, a~”, ip 
 
 Any number affected by an exponent is called the base with 
 respect to that exponent. 
 
 In a”, ais the base. In (@+y)”, 2+ y is the base. 
 
 131. The Form a®. The second index law is 
 a™ au a” —_ Fy hha 
 Let m =n, and this becomes 
 a® + a" = a, 
 But aa" = 1. 
 Hence, a =a" +a"=1. 
 187 
 
188 THE ESSENTIALS OF ALGEBRA. 
 
 Any quantity with an exponent zero is equal to unity. 
 10? 
 
 
 
 2—1, for 2+2=1; 10> Se 
 102 
 ee Ct Pp (e+5)=1 
 
 132. The Form a—". The first index law states that 
 q”™ x a” — qt, 
 Let us assume that this holds for all values of m and n. 
 Let m= —n, and we have 
 
 OOK = 0 ee 
 ae ey 
 Hence, a~" =—, by division. 
 a 
 
 a~” means the reciprocal of a”. 
 
 1 if ie 
 o°=—-—; 10 1=—- 7? = —___—_; 
 3 10) ed Oe ee 
 ag t= £ ; ax *y™ = —; 8 a7* = —- 
 xv xy" qa” 
 
 A factor may be removed from the numerator to the denom- 
 nator, or vice versa, if at the same time the sign of its exponent 
 be changed. 
 
 
 
 B2p-By2 —52z2y2 «2 x Gte2e- 1-2 __ de 
 Biker Nn igh 9 | Su ng: ty 2p a 
 p 
 133. The Form a’. We know that (a”)"=a™. 
 PR Phi arn 
 Put m = and n=q. Then (a) =(a)!=a¢=a" 
 
 P 
 Now extract the gth root of both sides of (a%)?’=a?, and 
 
 Dp Pp 
 we have az=~V/a?; that is, ay means the qth root of the pth 
 power of a. 
 The numerator of a fractional exponent means the power 
 of the base, and the denominator the root to be extracted. 
 
THEORY OF INDICES. 189 
 
 To extract the qth root of a”, divide the exponent p by q. 
 
 p p ie 
 The gth root of a? is az, or a7 =Va?. 
 at xat=aitt=al, 
 
 axa= (a2)? 
 
 nh. 
 COM eam he 
 of: ed 
 a=Va. 
 Bee eh at We. 
 
 We find that by assuming the generality of the endex 
 laws we have consistent meanings for zero, negative, and 
 fractional exponents. We will hereafter use the six index 
 laws of page 187 for all values of m and n. 
 
 
 
 
 
 EXERCISES. 
 0) ec ee 
 ae Oa 
 Bas - mt. git ge ae gece a Op Se | z 
 2. —jui 4 12T2-T—He 
 aot = ee 
 2 
 
 
 
 
 
 Simplify the following by making all exponents + after 
 combining like numbers: 
 
 — 3 '— 1 Lye | be pbs 
 fe VawVvb ‘ ; 32a by dy ; 
 4 N/a /b . 32 a byt zt 
 
 52x 63 x 10° 
 
 
 
 : Vv «art 5aty\ 
 —3 +3 =a OF K ‘ 
 aa x WD? 
 2 eS ae: ees 
 Vaxby 9. (a *bic!) + (a *b*c 8). 
 
 10. (a-2b-"c~?) . (2b) . (adic). 
 
190 THE ESSENTIALS OF ALGEBRA. 
 
 Remove the parentheses and simplify : 
 
 11, $[(ay2 ts 12. ([§ (aw + by)?t4]%)2, 
 1 
 
 y 
 
 14. (@’—y)?, 15. (7 — yy. 
 
 13, (wf yaa ¢Qaytpytaat 42h 
 y 
 
 It is to be noted that while (a2b2)? = 0, (a2 4. b2)2 does not 
 equala+b6. An exponent is distributive to the factors, but not 
 to the terms of an expression. 
 
 (a? +b)? = a +2 02d? +B. 
 (a? + b?)2 = V/a? + b?. 
 Perform the operations indicated in the following. 
 16. (w+ y")(@—y""’). 
 17. (@ +e+1+a'+4a-)(4—a"), 
 18. (a —y"*) + (#—y}). 
 19. (@ —2a+a°— 2a 1)(2? a +2"), 
 20. (a ?+4+ 2a%b 14 0~*)(a1+4 bd). 
 21. (a°bscta?)? a (ai bt c3 x )8, 
 22, (a 2b) cmt g 3) + (a3 b3ety-2) 1, 
 23. (2 abe? +. 8 a~‘h-¢2) + (6a °be), 
 24, (9 avi —16a°b%) + (3 aby3 + 4 abi). 
 25. (2 aty t+ y a ary ee 
 
CHAPTER XIV. 
 RADICALS, SURDS, AND IMAGINARIES. 
 
 134. Definitions. 
 
 Radicals. A root indicated by means of a radical sign 
 is called a radical. ) 
 
 As noted in Chapter XII the quantity under the radical 
 sign is called the radicand. 
 
 VI, V5, Va, V28 + y? are radicals, and 7, 5, a, a + ¥3 
 are the radicands. 
 
 These radicals may also be expressed in equivalent 
 expressions by means of fractional exponents. ‘Thus, 
 
 V7 = 7, 4/5 = 58, Va = a’, Ve + 3 = (23 + yy. 
 The laws of algebra apply to radicals, since radical signs 
 
 may be replaced by exponents. All the laws of exponents 
 hold for radicals. 
 
 Thus, Vab = Vavo, 
 1 
 for Vab = (ys = aibr =Va Vo. 
 This law may be extended to any number of factors. 
 Vabe = VavobVe. 
 Commensurable Numbers. A commensurable number 
 is one whose value may be expressed as a fraction with 
 
 integral terms. 
 Thus, 49 = 4 
 
 
 
 2 = 98, is a commensurable number. 
 191 
 
 gi 
 
192 THE ESSENTIALS OF -ALGEBRA. 
 
 Incommensurable Numbers. A number which can not be 
 expressed as a fraction with integral terms is called an 
 incommensurable. 
 
 Thus, V2 = 1.4142... is inecommensurable. 
 
 Surds. A surd is an incommensurable root of a com- 
 mensurable number. 
 
 Thus, V2 is a surd, for it is an incommensurable root of 
 a commensurable number. 2 is commensurable, but V2 
 is 1.4142..., an incommensurable number. 
 
 \ 1 + -vV2 is not a surd in the sense of the above defini- 
 tion, for 1 + V2 = 2.4142... is itself an incommensurable 
 number. 
 
 Examples of Surds: V3, V2, V5, V7, Va. The latter 
 expression, Va, is a surd if a be not a perfect square. 
 
 A surd is always a radical, but a radical is not always 
 a surd. 
 
 V5, Vat, V21, V16, V22y8 are radicals, but V16, V22y 
 
 are not surds. 
 
 135. Surds Expressed Graphically. Many surds may be 
 expressed graphically. In doing this, use is made of the 
 Pythagorean proposition. Jn a right triangle the square 
 of the hypotenuse is ia to the sum of the squares of the 
 two legs. 
 
 If AB = BC=1, 
 then AC= V1? 4 12= V2. 
 
RADICALS, SURDS, AND IMAGINARIES. 193 
 
 If DEH=2and EF =1, 
 then DF=V+4 2 = V5. 
 
 If X is a cube whose volume is 5, 
 
 then’ KL = V5. 
 
 
 
 EXERCISES. 
 
 Represent graphically 
 dae |) 2. V3. SOuNALS. 4, V/34. 5. V6. 
 
 136. Surd Forms. Radicals whose radicands are alge- 
 braic numbers are generally considered swrds unless the 
 radicand is the nth power of an algebraic number, » being 
 the index of the root. 
 
 Vat 4, Va + 6)%, ae talk ay are surds. 
 
 V(a—x)3, Vata, V2 +22ry +7? are radicals, but not 
 surds. 
 
 These latter expressions are frequently spoken of as 
 being in the surd form. 
 
 137. Irrational Numbers. An expression involving a 
 surd or surds is an irrational. 
 
 5+V6, 34+V2—V5 are trrationals. 
 
194 THE ESSENTIALS OF ALGEBRA. 
 
 138. Kinds of Surds. ‘The order of a surd is denoted by 
 the index of the root. V5, V4, Vz are surds of second, 
 third, and nth orders respectively. Surds of the second 
 order are generally called quadratic surds, those of the 
 third order ecubie surds, etc. 
 
 Surds are of the same order when they have the same 
 root index. 16, Vz, Va+6 are surds of the same 
 order. 
 
 A monomial surd consists of a single surd term. 
 
 A binomial surd consists of two surd terms, or a surd 
 and a rational term. V5 +-~V72, 5+-V10 are binomial 
 surds. 
 
 A trinomial surd consists of three terms, two of whick 
 at least are surds. 3+V2—V5 and V3 4+~V2—4V5 
 are trinomial surds. 
 
 A mixed surd consists of a rational factor and a surd 
 factor. 5V3, 4V2, aVx +y are mixed surds. 
 
 A surd is in its semplest form when the root index is 
 the smallest possible and the radicand the simplest possi- 
 ble integral expression. 
 
 V2T=V9x38=V9x V3=3V3 (simplest form). 
 /36= V6 (simplest form). 
 
 Noe ye =x Vab = 2 - Vab. 
 
 Similar surds are re La when reduced to their 
 simplest form, have the same surd factor. 
 
 5V4, 8V4 are similar surds. 
 
 In this chapter only quadratic surds will be considered. 
 
 
 
 
 
 139. Transformation of Quadratic Surds. <A _ rational 
 quantity may be put in quadratic surd form by squaring 
 
RADICALS, SURDS, AND IMAGINARIES. 195 
 
 it and indicating its square root. Thus, 5 = V5? =V25. 
 This is readily extended to the case of reducing a mixed 
 surd to an entire surd. 
 
 56V38 =V25 x 8=V71b; 4V2=VI16 x 2 = vB2. 
 
 A quadratic surd is reduced to its simplest form by | 
 factormg the radicand and removing to the left of the 
 radical one of each pair of equal factors. 
 
 (1) alin 2 ed = 
 (2) Simplify V75 + V243 —-V108 — V27. 
 V5 = VEX S =5V3. 
 V243 = V2 x B= OV. 
 V108 = V6? x 3 = 6V3. 
 VIT=V 3x 8=3V3. 
 The expression now is 
 
 5V3 + 9V3 — 6V3 — 3V3 =5Vv3. 
 
 EXERCISES. 
 , Simplify the following surd expressions: 
 fey oO 4/5 + 6-8. 
 2. V124+V27 —V75. 
 3, 50 —V32 4+ 2V18—5V8. — 
 4. V3004+-V108 — -V243. 
 5. 3-Va8y! — 2V ay? + 13-V a7. 
 6. 25 a®b3 —-V/81 ab +-V144 a'd*. 
 7. V8(@+ y)? —5V27 (e+ 2ay + y) +-V12(@+ y\(at+y). 
 8. Vaety +a) Vo avy +2 +3VI6W +5! 
 
 
 
196 THE ESSENTIALS OF ALGEBRA. 
 
 V125 + -V.245 —-V/320 + -V405 — V720. 
 
 10. V82— 24 97+18 2 — V2 o— 1207+ 18 2. 
 
 11. V8 m3 — 16 mn + 8 mn?—-V2 m3+ 4 mn + 2 mn? 
 12. («—y)V84V32+ 6 ay + 3y?— («+ y)V108. 
 13. V4a2+4a%y 4+ V4 ay? +4 9. 
 
 14. 59 a’) +27 a? +7 aV25 64 75. 
 
 15. V48 ay + yV75 2+ V3 2@—9 yy 
 
 16. V75 a& — V3 a3 + 27 ab? — 18 ab + V27 ab? 
 
 17. V5(a—b)?—V 20 a? +40 ab +20 b?+ V20 a?—40 ab+ 20 b2 
 1s. 99 — -V176 + V539 + 4/275. 
 
 19. VW52—38V117 + 5V 1578. 
 
 20. (a+b) V(a— b+ 9) + (a—b) VE ya bp 
 +(8a—4b)V8a+4b)(e+y). 
 
 so 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 140. Product of Quadratic Surds. Products of surds obey 
 the following law: os sy 7) 
 (1) V3iT x V388 = 38V3 x 4V9 
 =12V3xv2 
 —12V6. 
 (2) V3 (a+b) x V2(a—b)= V6 (a2 — 8). 
 (3) V150 + V48 =5 V6 + 4V3 = §V8 = §V2. 
 
 EXERCISES. 
 Multiply the following surds: 
 
 1. 20 x V/80. 3. 5V3 x V48 x V75 x V'15. 
 2. V32 x V200 x V50. 4. 3Vaxr x V16 aba x V48 aa’, 
 
RADICALS, SURDS, AND IMAGINARIES. 197 
 
 . Ve@e—yyPx Ve@— yy. 
 . V60 a?y? x W135 ay x V36 aty®. 
 . V8(a—bY x V2(a—b) x V6 (a— bY. 
 8 Vb(e—y) x V20(@+y) x V3 (e— ye +ay +). 
 Simplify : 
 V5 aa? x V72 a+ y) x V2(a—y) 
 V25 wat x V82a(@—yra@ty) 
 10. (32 + V48) + (V2 + V3). 
 
 No o 
 
 2, 
 
 
 
 
 
 141. Multiplication of Polynomial Surd Expressions. Such 
 expressions as 
 (a+vVb+ Ve) x(a—Vb4+Ve) 
 are multiplied together in the same way as_ integral 
 expressions, the extended law for multiplication of radi- 
 cals being observed. 
 
 (1) a vb +) Ve 
 Fee MN Ds tt ENG 
 ataVvb+ ave 
 Say b ab — V/ be 
 eat + Vbe+te 
 a? Pare —b +e 
 
 (2) +v3) x (V2— V5)=v24+V6—V5—VI15. 
 
 EXERCISES. 
 Multiply the following: 
 1. (1+-V3) x (1—-v3). 5. (3V2 —4)(8V2 4-4). 
 2. (2—¥V5) x (2+ V5). G0 V,0 — 2a hi 
 3. (V24+V3)(V2— V3). 7. .(V2-+- V5). 
 4. (2V38—1)2V3+1). 8. (V2+V3—1)% 
 
198 THE ESSENTIALS OF ALGEBRA. 
 
 9. (2V3+1)(V3 + 2). 
 
 10. (V5+V3+4)(V5—V3—4), 
 
 1. (2V3—3V5+1)(5V3 +2V5—1). 
 
 12. (4-7 — IV 6) Ta 6 
 
 13. (V3+1)(V9—V3+1). 
 
 14. (Vat Vb— Ve)(Va— Vb + Vo). 
 
 15. (2Va—3Vy —5-V2z)(2Va+ 8Vy —5-V2). 
 VbB-V7T_ , 2V54+2V7 
 
 2V3—-4V5 8V346V5 
 
 142. Conjugate Binomial Surds. 
 Two quadratic binomial surds differing only in the sign 
 of a surd term are called conjugate surds. 
 Va+tvb and Va—-vob, or 4+ Vy and 2—Vy, 
 are conjugate binomial surds. 
 (Va+vb) x (Va—Vb)=Vai—-VP=a-5. 
 (x+Vy) xX (@—-Vy)=2—-VyP= 2 —y. 
 
 The product of two conjugate binomial surds is rational. 
 
 143. Rationalizing Factors. When the product of two 
 surd expressions is rational, one expression is said to be the 
 rationalizing factor of the other. 
 
 For example: Va— -Vé is the rationalizing factor of 
 Va+vb, because (Va+Vb)(Va—Vb)=a—8, a rational 
 result. 
 
 a+vVy is the rationalizing factor of 2— Vy, because 
 (@—Vy)@+Vy)=v—y. 
 
 V5 is the rationalizing factor of V5, because V5 x V5=5. 
 
RADICALS, SURDS, AND IMAGINARIES. 199 
 
 Suppose the value of is required. It will 
 
 re 
 V5—VB 
 evidently be very much simpler if we first rationalize the 
 denominator. 
 
 
 
 
 
 
 
 V5. - V5(V5+-V3) _5+V15_5+4+V15- 
 V5—V3° (Vb—V3)(Vo+V38) 5-38 
 pe V2 VT V2) V2) VIS 
 meee 69 i -Vvav2. 2 
 EXERCISES. 
 
 Rationalize the denominators of the following fractions: 
 
 
 
 
 
 
 
 1. ah 5. fal ee —s 9. 2 a V3. 
 v2 42/3 ay ie 
 2. L+v2. 6. pee Vie —* 10. 3 ds v5, 
 V3 Bo 1G hee sss 
 eg, SVE nn, 442 V3. 
 2V2 4—3~V7 V5 —V3 
 4 fp ae Taye = aoe 
 13. 5+8V5 | a7, Lt 8ve—1. 
 2V5—V3 1—3Vae—-1 
 14 84+2V2+V3_ 18. OV GAEN Fs 
 2V543V7 (@+y)+V2ay 
 15: Set ilp aby a—Vayt+y. 
 Va+ Vb Vae—wVy 
 Tee PVE 20, Ve EN 2 ae 
 
 2Va+4vb ' Va—2b—-V2a—b 
 
200 THE ESSENTIALS OF ALGEBRA. 
 
 144. Rationalizing a Trinomial Surd. A trinomial surd 
 expression may be rationalized by two operations. 
 
 (1) Rationalize Va+Vb+~Ve. First multiply by 
 Va+tvb—vVe. This gives a+6b—c+2Vab. Now 
 multiply bya+6—e—2-Vab. This gives (a+b—c)?—4 ab, 
 a rational expression. Hence the rationalizing factor of 
 Vat+vb+Ve is (Va+vVb —Ve) (a +6 —e — 2 Vab). 
 
 (2) To rationalize V2+V38+~V5, we multiply by 
 (V2+V38 —V5)(2+38—5—2-V6). This will give a 
 product 24. 
 
 EXERCISES. 
 
 Find the rationalizing factors of the following: 
 
 Dee tny Cerys 5 VW1i0=V2se 
 Den) SaNy & ANT 6. Va+vVore 
 
 SINS +V 24/0. 7, 1—2 Voeee ee 
 Bey Gey ols 8. 2Va—V2b4+3Ve. 
 
 145. Rational Numbers and Surds. 
 
 THeorem I. Jf Vax and Vy be surds, then Vx can not 
 equala+~Vy, where a is rational. 
 
 For, assuming Vz=a+-Vy, and squaring, we have 
 e=a+y+2a Vy; 
 
 x—-@—y _ 
 or oS Gir 
 
 But the left member of this supposed equality is rational, 
 and therefore can not equal the surd Vy. 
 
 Hence, Vaetat+vy. 
 
 The sign + is read, is not equal. 
 
RADICALS, SURDS, AND IMAGINARIES. 201 
 
 THEOREM IT. ais a +Va= b +Vy, where a and 6b are 
 rational, and if Vx and Vy are surds, then a=b and x= y. 
 
 The proof of this theorem is similar to that of Theorem I. 
 
 We have te Vie 
 
 Squaring, (a—6)?4+2(a—b)Vr+ae=y. 
 
 Transposing, — x+y —(a—b)?=2(a—6b) Vz. 
 
 Here, a rational number, —x + y —(a— 6)?%, is equal to 
 2(a—b)timesasurd. But this can not be true, except in 
 the case a=0b; but when a=6 the original equality says 
 that x= y. 
 
 146. Square Root of a Binomial Surd. ‘The square root of 
 certain binomial surds may be extracted. 
 
 (1) Find the square root of 5 + 2 V6. 
 
 Let V54+2V6=Vr+ Vy. 
 - Then §+2V6=ex+y4+2Vz2y, by squaring. 
 
 ey =O, ry ==, by Theorem II. 
 
 The question now is to find two numbers whose sum is 
 5, and whose product is 6. ‘These are seen to be 2 and 3. 
 
 Hence, V5429V6= V3 4 V3. 
 (2) Find the square root of 8 — 2 -V15. 
 As in (1), let © V8 — 2/15 = Vz— Vy. 
 
 Then 8—2Vib=2+y—2Vz2y. 
 r+y=8, vy=15, 
 then sie py 
 
 Hence, V8—2V15 = V5— V3. 
 
202 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 Extract the square root of the following: 
 
 Tent Oise won 7.31.2-e a 
 
 Qa iee wey oo 8. 1824742, 
 Beeliletaoeny Og 9. 49 +125. 
 
 4. T—V40=7—2-V10. 10, ¥—-2V yee 
 
 5, 1642/55. 11; 87 — 12Ra 
 
 6. at+b—2Vab. 12. at+b+c+2Vac-+ be. 
 
 147. Imaginaries. An imaginary number has already 
 been defined as the even root of a negative number. 
 
 We shall have occasion to use only the square root of 
 negative numbers. 
 
 V/—5, V—10, VEo16 =4V— 1 eee 
 
 are imaginaries. 
 
 It is to be noted that every square root of a negative 
 number may be expressed as a real number multiplied by 
 the square root of —1. Thus, 
 
 V—16=V16 V—1=4V_-1; V—5=VvV5 V-=—1; 
 V—@=VaiV—1=avV—1; V—a=Va V1. 
 
 The V—1 is usually denoted by 7, and is called the 
 
 imaginary unit. 
 
 148. Some Properties of / =V—1. 
 
 (1) i =V—1. 
 
 (2) 2=(V—1)?=~-1. 
 
 (3) &®=(V—1)8=71() = -7. 
 
RADICALS, SURDS, AND IMAGINARIES. 203 
 
 C4) = (2)? =1. 
 (5) Bait) at. 
 
 (6) 6-1) =2= —1. 
 (T) Tat) =-2. 
 (8) B=(i)2=1. 
 
 (9) =i) =1; ete. 
 
 This table shows that the powers of ¢ repeat the values 
 V—1, —1, —V-—1, 1, in cycles of four. 
 
 149. Operations with Imaginaries. All the operations 
 possible with surds are also possible with imaginaries. 
 The properties of 2 must be observed. 
 
 (1) V—36 +V—81+v-— 100 
 
 = V36(— 1) + -V81(— 1) + V100(— 1) 
 =6V—14+9V—1+10V—-1 
 =67+97+102 
 e200. 
 (2) V—48 + V— 75 +V— 248 
 = V16(— 3) +-V25(— 8) + V81(— 8) , 
 = 4iv8 4 5iV3 + 9iv8 
 SS B/S. 
 (8) V—5xV—T=Vbixvit 
 =V5VT ii 
 = V35? 
 = — V35. 
 
204 THE ESSENTIALS OF ALGEBRA. 
 (4) (V—84V—2)(V—8-—V—2) 
 =(iV3 +iv2)(iv8 —tv2) 
 = 12(V3 + V2)(V8 — v2) 
 
 . = 12(3 — 2) 
 = —1(3—2) 
 Lies Boia, 
 (5) 8+iv2)?=94 6iV2472-2 
 —-9+6V2i-2 
 —7T+6V24. 
 5 4V37 
 CDE TAY Ey 
 
 To rationalize and make real the denominator, multiply 
 both terms of the fraction by 5 + -V37. 
 
 Oe — ee 
 
 (5 —V3i) (6 +V3t) 25 — 37? 25 —(— 3) 
 _ 224+10V32 
 er anes 
 _ 114+ 5V3¢ 
 14 
 EXERCISES. 
 2, V—18+V¥—128 — VS bt 
 2. (31+V21) x5V2i=? ¢ StV=5_» 
 3. (4—V3iy=? 38—V—5 
 4, (V—34 V2) (V—243)=? 7 22a 
 PEED Su le: 34+ 2V3% 
 V3—-V—2 8. V—75 +V—25 =? 
 
 (Make denominator real and 
 rational.) 
 
RADICALS, SURDS, AND IMAGINARIES. 205 
 
 9. (av —a2+yV— y) (av —2—yV—y) =? 
 
 
 
 Tig =» sir Rice ae 
 a— bi V/—5x 
 
 
 
 hg BSL L234 
 2 — i WS eel ee 
 “(Oia | Cormar 
 
 150. Graphical Representation of Imaginaries. Complex 
 Numbers. We are accustomed to represent real numbers 
 upon a straight line, the positive numbers in one direction 
 and the negative numbers in the opposite direction. 
 
 Let OA =1, 
 OA' =—1, 
 1x2=—1. 
 
 
 
 Hence, multiplying 1 by 2? turns it so as to reverse it in 
 direction. OA is turned about O to the position O0A' when 
 it is multiplied by 7?; that is, OA is turned through a half 
 circle. From this we conclude that, when OA is multi- 
 plied by 2, it will be turned half as far; that is, to the 
 position OB. If OA is multiplied by 23, it will be turned 
 to the position OB’. It is customary to regard BB’ as 
 the line of imaginaries. 
 
 A number like 3+ 27 is called a complex number. The 
 type form of such a number is 
 
 a + 67. 
 
2.06 THE ESSENTIALS OF ALGEBRA. 
 
 The following diagram shows how complex numbers are 
 graphically represented. 
 
 (P,) -2+22 (P) 3424 
 @ e 
 
 
 
 EXERCISES. 
 
 Locate on a diagram the following complex numbers: 
 
 1. 447. 4 —347. 
 2. —3— 31. 5 — 34. 
 3. 2—5%. 44¢ 
 
CHAPTER XY. 
 QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 
 
 151. Definition. An equation in which the highest power 
 of the variable is two is called a quadratie equation. 
 
 Thus, 327+427+5=0, 2-—5274+1=0, az’?+6=), 
 are quadratics. 
 
 152. Type. Every quadratic in a single variable may 
 be reduced to the type ax?+ 6x+e=0, where a may be 
 any quantity except 0, 6 may be any quantity, and e may 
 be any quantity. | 
 
 3a2—4a2+5=0 is a special form of the type, in which 
 a=3, 6=—4, ande=5. 
 
 EXERCISES. 
 
 Reduce the following equations to the form of the type 
 ax’ + 6x +c=0, and indicate the particular values of a, 6, and 
 c in each case: 
 
 
 
 
 
 
 
 IS a = o — 5. ee ae 
 ( Ppa 
 2. 6a2=2—27. 
 : , 5 t—4_7a°-3 
 (Clear of fractions.) Mice Tinka ° 
 2e—T 2—27 1 3 
 3 ———_ = —_—_- 6. x©+-=5—--.- 
 3 5 % 
 
 207 
 
208 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 xv? — 4. 3  8e+1 
 7. —— =2 =, . ——=22. 
 oq ala 9 z 22 
 8. iO a Hag) 10. da+1 2042. 
 e+ 2 e+1 e24+2 
 
 153. The Pure Quadratic. If 6—0, the general quad- 
 ratic ax?+6x+e=0 becomes ax?+e¢=0, an equation 
 sometimes called a pure quadratic. 
 
 The solutions of ax?+e¢=0 are easily found. 
 
 Transposing e, ax*=—e. 
 Dividing by a, a= —*. 
 a 
 
 Extracting the square root, z= a 
 a 
 
 If a and ¢ have like signs, the roots +{—-— are both 
 a 
 
 imaginary ; if a and e have unlike signs, the roots are both 
 real. =e 
 Thus, ose = 0 has the roots r= oe ee vee are 
 
 imaginary; 827—5=0 has the roots 7= ssf , which are 
 real. 
 
 The roots are real and rational if a and ec have unlike 
 
 e Cc e e . 
 signs, and — is a perfect square; the roots are irrational 
 a 
 
 C .e 
 when — is not a perfect square. 
 a 
 
 Thus, 
 
 322—5=0 hastherootsra= + , which are irrational ; 
 
 Se 
 
 | 
 
 | 2) 
 
 Pe +3, which are rational. 
 
 3a2—27T=0 has the rootsz = wp 
 
 Se) 
 
~ 
 
 QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 209 
 
 EXERCISES. 
 
 Reduce the following equations to the form of the type: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 | ax’+c=0. 
 
 Mee fe 0. Gee eee 
 7 3 32 4x4+5 
 (Clear of fractions.) 2 S2t+4_ 5a—3 
 
 G—-3 _ ea pat Se oi 4 
 
 5 U+9 1 P 
 ee er g, =" =" _. 
 
 3. = = 
 7 142 ; 2 eae 
 
 ” ia © 9. e,4_5 
 ed @+3 hia el 
 ne ee 6 v 10. Sal ea be % ° 
 
 72 11 x 3u+7 
 
 EXERCISES. 
 
 Solve the following equations : 
 
 ie 7e?—112=0. 
 First SoLuTIon. 
 
 Transposing, ee 12, 
 
 Dividing by 7, re 1G: 
 
 Extracting the square root, x= +4, rational roots. 
 
 SECOND SOLUTION. 
 
 The general pure quadratic 
 ax?+¢c=0 ‘(uh 
 has the solutions r= Ly = “ 
 
 Comparing 7 2? — 112 — 0 with this general equation, 
 ant. eo = = 112, 
 
 Ni 1155 fii2 
 
210 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 c—3 12 6.00 = eee 
 2. se . YP cig =-—2. 
 5 ees te a 
 x — Doe goo Oe 49 
 BA ————— eo . 38e __3a+30 
 7 14 2% 8. 9b ee 5 
 4. (w©+3)(@—3) =2a?— 45, 
 5. 3¢—27=721. 9. of 4+ 6 ee 
 4 3 3 
 6 OX Se! 
 es eee 10. —aa’?+25b=0. 
 
 154. Solution of the General Quadratic, ax?+ 6x +¢=0. 
 
 Dividing through by a, we have 
 
 b 
 v2+—-—x rin ett 
 a 
 my « ~ Cc 
 lransposing, e+ -_7 = 
 a a 
 
 2 
 Adding + (the square of half the coefficient of 2) to ° 
 t 
 
 both sides, we have 
 b b? igh en 
 7 aes ee 
 —4 ae 
 4 a? 
 This is called completing the square, because it always 
 makes the first member of the equation a square. 
 Iixtracting the square root of both sides, 
 
 b Nees: 4 ac 
 
 C--— = 
 2a 2a 
 arising: Vh2— 4 ae 
 ransposing, Gaon A” ae a 
 2a a 
 —b+v82—4 ae ac 
 
 2a 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 211 
 
 _ 
 
 This result shows that the quadratic ax’ + 6x +¢=0 has 
 two roots; namely, 
 
 ya b+ ve— Fae 
 2a 
 
 a Pen vO pe aC: 
 2a 
 
 This is the solution of the general quadratic and may 
 be taken as the type solution for all quadratics. 
 
 Solve 327°+827+5=0. 
 First SoLurion. 
 Dividing by 3, a+ $r+3=0. 
 
 Transposing, a+ bas — 
 
 color 
 
 Completing the square, 
 
 at +4. 8 ax + 16 = 16 — 
 
 
 
 Extracting the square root, ° 
 
 =— land —8. 
 Verifying the Solution: Put x=—1 in the equation 
 3a°+82+5=0 and 
 
 pea yt8 (1) b= 8 — 845 — 0. 
 
 Put s=— 3 and 
 
 3(— H+ 8(-H4+5=%— 4045 = 4h 40 4 = 0, 
 
 3 3 
 Replacing the variable of an equation by a root reduces 
 
 the equation to an identity. This process is called verify- 
 ing the solution, 
 
212 THE ESSENTIALS OF ALGEBRA. 
 
 SECOND SOLUTION. 
 Comparing 322+ 82+5= 0 with az7+ br+e¢=0, 
 a=8, b=8, and c= 5d. 
 Let us substitute these values in the type solution, viz., 
 
 —btive?—4ae 
 = raat PEREIRA Se ara | 
 Qa 
 
 = 8 4 vi64— 60 
 a | 
 _—8+v4 
 Seren: 
 —8+2 
 6 
 It will be noticed that the substitution in the type solu- 
 tion gives results identical with those obtained by carrying 
 out all the steps of dividing, transposing, completing the 
 square, extracting the root, and transposing. 
 
 and we have L£= 
 
 
 
 =-—iand —§. 
 
 EXERCISES. 
 
 Solve the following equations by both processes in the order 
 of the above illustration and verify your results ; 
 
 1, 22?7—52=3. 10. 11 2? =: 39 2 4+ 20. 
 2. 42°4+7x¢= 16. ll. 622+72—20=0. 
 3. da’ =19 e+ 14. 12. 6a°—47 e+4+77=0. 
 2 5 o* —12 = 4a. 13. 3¢7+427+10=0. 
 G 
 
 . 63"+¢—156=0, : 
 
 ene 5+ 18. 2 
 
 15. 3%——=1. 
 ve Si a ie 2 
 &. 150?—8= 37a. 1s. 1+ f2=—3 28 
 9 777+7=502. 17, ve’ +1=-—2.9¢24 
 
 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 
 
 ‘ 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 3 8 12 
 18. “a nee =—-e 20. ie | es ne 
 e—-2 #e2+2 5 x re 
 21 —38 80 
 [| . a 2 —— =0 
 5x +20 20—5a ee 
 x x—3 x 2 os at ae 
 22. -— ig el eee SS ee 
 e—3 x bake x 3 
 23. 2+2ba+c=0. 
 24, 2 —(m+n)e+mn=0. 
 25. (a—b)x*—br=a. 
 26. (a*— b*)x’ — (a’+ b*)~r+ab=0. 
 7 
 b—a b+a V—x’ 
 28. Ve—5ae= eo 
 Ve—5e2 
 29. am poo 
 Vie 827+ 5 
 30. Vae+ca= gene —— 
 aa” +- cx 
 
 155. The Double Root. 
 
 Solve 
 
 Comparing with the type az? + br+ ¢=0, we have 
 
 Hence, 
 
 9227-2074 50=0. 
 
 a= 2, 6=— 20, e= 50. 
 20 + -V400 — 400 
 a 
 4 
 _ 20+V0 
 ea. 
 =5+0. 
 
 5) 
 
 od 
 
 13 
 
914 THE ESSENTIALS OF ALGEBRA. 
 
 In this case the roots are 5+ 0 and 5—0; that is, 5 in 
 each case. Hence, the equation has two equal roots. In 
 the above example 5 is said to be a double root. ‘The 
 quantity under the radical, 62—4 ae, is 400 — 400 or 0. 
 Hence, the condition for a double root, or two coincident 
 
 roots, 18 
 ; b?— 4ac=0, 
 where a, 6, e¢ are the coefficients in the general quadratic 
 
 az? + br+c=0. 
 
 EXERCISES. 
 
 Solve the following equations, noting those which have double 
 (coincident) roots : 
 
 1. wo —4e44=0. 6. #—le+7=0 
 2. 4e¢?7+424+1=0. 7. 5° —4 9 aw 
 
 6 HS SUF et elt). 8. 2527+ 302+9=0 
 4. 4e?—127+4+9=0. 9. fv’+ie«+1=0 
 5. 8e?+42+1=0. 10. —3e¢’+42+:=0. 
 
 156. Irrational Roots. 
 Solve 3822? -—-927+4+2=0. 
 Comparing with the type 
 
 a+ br +e=0, 
 
 we have a= 3,6=—9, e= 2. 
 Hence, r= 94 V81— 24 
 6 
 _9-v51 
 
 6 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 215 
 
 Here it will be noticed that the roots are irrational. 
 The quantity under the radical, 6? — 4 ae, is 81 — 24 or 57, 
 and is not a perfect square. ‘he roots are conjugate surds. 
 
 EXERCISES. 
 
 Solve the following equations : 
 
 
 
 
 
 
 
 
 
 1. 42? +2—1=0. 5. Bgl Vee Wad 
 3 5 
 2. 3x —Te#+3=0. 6. (v—1)(a—2) =1. 
 Se te — 7 — 0. 7. 807 —21=202. 
 oe: ike ees 
 4. 52° —32—3=—0. BOGS pate ae 
 > Siig see ea licas see 
 2 D 2 \ 
 10. 9 m'n*a? — n? = 6 m?n®a. 
 157. Complex Roots. 
 Solve 627 —Ta+3=0. 
 Comparing with the type 
 ax? + br +e¢=0, 
 we have a=d,6=—T,e¢= 38. 
 T+V19 — 60 
 ae Pt Vette 00 
 10 
 _T+v—11 
 10 
 _T+vil-2 
 LO aie 
 
 In this case the roots are imaginary. The quantity 
 under the radical, 6? — 4aec, is 49 — 60 or — 11, which is 
 negative and therefore its square root is imaginary. ‘The 
 roots’in this case are conjugate complex numbers. 
 
216 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 Solve the following equations : 
 1. #—52+8 —0. § —327+11382=20. 
 2. 22°+924+11=—0. 6. 2? + .54¢74+ .3=0. 
 3. 3a?—10724+9=0. 7 — 32° 4+.82 —10= 0) 
 4. Te—l11v7+8=0. 8 
 
 158. The Discriminant. The solution of the general 
 quadratic ax2 + bx +¢=0is 
 
 Pater seh A Nee et 
 2a 
 
 From the examples just given we see that the character 
 of the roots is determined by the quantity under the radi- 
 cal. This quantity, 62 — 4 ae, is called the discriminant of 
 the quadratic. 
 
 159. Some Conclusions. 
 
 (1) When 62 —4ac =a square, the roots are real, rational, 
 and different. 
 
 (2) When 62—4.ac=0, the roots are real, rational, and 
 equal. 
 
 (3) When 62—4ac =a positive number not a square, the 
 roots are real and conjugate surds. 
 
 (4) When 62—4ac=a negative number, the roots are 
 conjugate complex numbers. 
 
 (1) 4a -—Tr+38=0. 
 The discriminant 62 — 4ae= 49 — 48 = 1. 
 
 Hence, the roots of this equation are real, rational, and 
 different. 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 217 
 
 (2) 2a7?-—4742=0. 
 
 The discriminant b?— 4 ace=16—-16=0. 
 
 Hence, the roots of this equation are real, rational, and 
 equal. 
 
 (8) 5a74+82—2=0. 
 
 The discriminant 6? — 4 ac = 64 + 40 = 104. 
 
 Since 104 is not a square, the roots are conjugate surds. 
 
 (4) Tz?-—5274+1=0. 
 
 The discriminant 0? — 4 ac = 25 — 28 = — 3. 
 
 Since the square root of — 3 is imaginary, the roots are 
 conjugate complex numbers. 
 
 EXERCISES. 
 
 By means of the discriminant tell what kind of roots belong 
 to each of the following equations: 
 
 a2 
 
 a » 2 
 
 e—d5ae—9—0. 6. 
 (7 +32—1=—0. 7. 
 5e7+9e2+11=0. 8. 
 62?—-Ta+$=0. 9. 
 9e?—1827+4=0. 10. 
 
 EXERCISES. 
 
 ax’ + 2aux--(a—4)=0. 
 1le’?—32e4+7,=90. 
 e—3ae+$=0, 
 4¢+132+11=0. 
 ax’? +5e2—1=0. 
 
 Solve the following quadratics by comparison with the solu- 
 tions of the type: 
 
 i 
 
 ee i net 
 
 $e +5a2—3=0. a 
 327—62+1=9. 8. 
 —227+82+6=—0. 9. 
 6a? +a—2=—0. 10. 
 24¢7+142—5=0. 11. 
 
 xv” — 55a —.065 = 0. 12. 
 
 5 a? — 500 = 0. 
 627+ 82=0. 
 92? — 252+ 50=0. 
 
 307+ 212—5=0. 
 7e?’—424+3=0. 
 
918 THE ESSENTIALS OF ALGEBRA. 
 
 13. 1a°’-—6a%—13=0. 15. lla’—2a4+7=0. 
 = (), 16. jv —5248=0. 
 
 160. Graphical Solution of the Quadratic. The general 
 quadratic ax ti bx t¢e= 0 
 is equivalent to the two equations 
 
 Y= ur, 
 ay+ber+e=0. 
 
 For if in the second of these, we put the value of y from 
 the first, we get the quadratic aa*+ br +¢=0. We know 
 that y + bx + ¢=0 has a straight line for its graph. (See 
 page 1538.) 
 
 Let us see what the graph of y = 27 is. 
 
 y can not be negative, because a square can not be 
 negative. 
 
 Solving for 2, we have 
 
 
 
 
 
 
 Y a=tvy. 
 ageoy) Tk y=0, «=0. 
 y=1, x=+1and —1. 
 
 (4,16) 2, w=+1.414 and —1.414. 
 
 3, v©=+1.7382 and —1.782. 
 y=4, x=+2 and —2. 
 etc. etc. 
 
 (314,121%) 
 
 (3,9) 
 Representing these points with refer- 
 
 (24,6'4) ence to the codrdinate axes and draw- 
 ing a smooth curve through them, 
 (1,24) we have the curve of the adjacent 
 - figure, which is the graph of y= 2%. 
 This curve is a y-parabola and is 
 the same for every quadratic. 
 

 
 QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 219 
 
 The line az+ by +e¢=0 can only be specifically repre- 
 sented, when we give particular values to a, 6, and e. 
 
 If the quadratic becomes particular, then the lne is 
 specific, and we can readily draw its graph. 
 
 Determine by means of graphs the roots of 
 
 v—x—2=0. 
 This equation is equivalent to 
 
 y = x", 
 eo, 
 In this a=1,6=—1, ande=—2. 
 In the equation y—x—-2=), 
 when y= 0, x=— 2, and when z= 0, y=2. 
 
 Hence, the x-intercept is — 2 and the y-intercept is 2. 
 
 Now, drawing the graph upon the same diagram that 
 contains the y-parabola, we get the result Y 
 shown in the adjacent figure. 
 
 It is seen that the line cuts the y-pa- 
 rabola at P and Q: The z of P is OY, 
 which is — 1, and the 2 of Y is ON, which 
 is’ 2. 
 
 The roots of the quadratic 72—2—2 
 = 0 are —1 and 2. 
 
 Hence, we see that the intersections of 
 the line az + by+e= 0and the y-parabola 
 y =x" have for their abscissas the roots of the quadratic 
 ax? + 6x+ec=0. 
 
 
 
 161. Graphical Solution of the Pure Quadratic. If the 
 quadratic is az?-+¢=0, the two equations to which it is 
 equivalent are y = 2, 
 
 ay+e=0. 
 
220 THE ESSENTIALS OF ALGEBRA. 
 
 Here again we have the same y-parabola. The line ay + 
 c=0 is parallel to the z-axis, and so the abscissas of the 
 two points of intersection will be equal 
 
 in value but opposite in sign. 
 Determine by means of graphs the 
 
 P Q roots of Pee Pig ee) 
 The equivalent equations are 
 (ia 
 The: graphs show the two roots to be 
 +2and —2. 
 162. Graphical Solution in Case of Equal Roots. 
 
 Construct graphs showing the roots of 
 
 v—47+4=0. 
 The equivalent equations are P 
 oa 
 y—4x24+4=0. 
 
 In this case the line just touches the 
 parabola at P, whose abscissa is 2. 
 
 The quadratic z7—4 7+4=0 has equal 
 roots, each of them being 2. 
 
 163. Graphical Representation in Case of Imaginary Roots. 
 
 Construct graphs showing that 27—2a2+5=0 has imagi- 
 nary roots. 
 
 The equivalent equations are 
 
 eink 
 
 y—224+5=0. 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 221 
 
 This equation has imaginary roots. The line 
 y—2x4%+5=0 
 does not touch the parabola y = 2%. 
 
 164. Certain Conclusions. 
 
 (1) In the case of real different roots, 
 the line ay+x+e=0 cuts the parabola 
 y = x" in two places. 
 (2) In the case of real equal roots, the 
 line ay+bx+ec=0 touches the parabola 
 y=x", but does not cut it. 
 (3) In the case of imaginary roots, the line ay+bz+c=0 
 is entirely outside the parabola y = 2. 
 
 EXERCISES. 
 
 Using the parabola in the book, or a similar one of your 
 own construction, solve by means of graphs the following 
 equations : 
 
 1. 2¢+4¢—-1=0. 
 
 Norr. The line in this case is2y+2-—1=0. 
 fA yf aards 
 ee se 
 
 gz and y intercepts are 1 and }. 
 
 Merely lay a ruler across the parabola so as to 
 make these intercepts, and note the abscissas of 
 the points of intersection with the parabola. 
 
 2. @—S32+2=0. 6. 52?—6a—8=0. 
 3. 4e¢7+4274+1=0. 7, #—Azgt4d=0. 
 
 aoa O80 4-0 = 0, 8. 2307-524 3=.0, 
 5. 102°—32—4=0. 9, 27—424+3=0. 
 
ph THE ESSENTIALS OF ALGEBRA. 
 
 165. Relations among the Roots and Coefficients of a 
 Quadratic. The roots of aaz2+bx+c=0 are 
 
 _ —b+VP—4a0 fae 
 AE, 
 
 ray ly clase mad) tac 
 2a 
 
 Here we use « (alpha) and ® (beta) to represent the 
 two roots.. 
 Adding these two roots we have 
 —~btVRPo dae 2b = Vee 
 
 1 eo ai a 
 
 Multiplying together these two roots, we have 
 
 yaw alge ou kabey Ady, 2 2 
 of = (= +5" =a\ 6—vb? at) 
 
 2a 2a 
 VC eine: Cokes 
 4 a? 402 a 
 If the coefficient of 27 in the general quadratic 
 av?+ br+e= 0 
 
 be made unity by dividing by a, the equation takes the 
 form 
 
 b é 
 g+—71+—-=0,. 
 nie 
 The sum of the roots e+ B=— a and the product of 
 | a 
 the roots «8 = <. 
 Hence, if a quadratic be written so that the coefficient of 
 x is unity, the coefficient of x is the negative of the sum 
 
 of the roots, and the constant term is the product of the roots. 
 This has already been seen in factoring, page 86. 
 
 \ 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 223 
 The roots of the equation 25a’—15a%+2=0 are 2 and t. 
 When this equation is divided by 25, it becomes 
 
 o—se+7,=0. 
 
 The sum of the roots is 2+ 4= 3, and their product is 2. 
 This sum and product are, respectively, the negative coefti- 
 cient of # and the constant term. 
 
 166. Formation of Equations with Given Roots. If the 
 roots of an equation are 2 and 1, what is the equation ? 
 Here 
 
 and 
 
 Colbo 
 ~ 
 
 | 
 
 | 
 coi bo ke 
 
 g 
 
 on 
 
 jes teae = 
 I| 
 
 or 38a2—5xr4+2=0. 
 In general, 227—(a+8)x+e8=0 is the quadratic 
 equation whose roots are « and £. 
 EXERCISES. 
 
 Make equations which have the following roots: 
 
 1. 3, —4; —5 4; 3,4. 6. 3—iV2, 3+iv2. 
 2. 1 +~/5, 1 Lin/5. wh 3 — 5,V5, 8+ 5,5. 
 
 e354, 8— bi. 
 
 8. a+bi, a— bi. 
 ae es 9. 314 5Vmi, 31—5V mi. 
 
 5. 5, — Ad. lo. V5+3V3, V5 —8v3. 
 
 
 
224 THE ESSENTIALS OF ALGEBRA. 
 
 167. Generalized Quadratic. aX?4+6X+c=0. 
 The above is a quadratic in X. Its roots are 
 —b+VP—4ae 
 2a 
 
 X may be any algebraic expression. Whenever an 
 equation may be arranged like the above type, it is said 
 to be of the quadratic form. 
 
 C1) 324+797+4=0. 
 
 Put 2? = X, and the equation becomes 
 
 38X274+7X+4=0, 
 
 xX = — 
 
 
 
 
 
 whence X= sieve. 
 =—l and — 4. 
 
 But X = 7, 
 hence, a= —1and — 4, 
 and g=+2and +2V32. 
 
 The roots of 304+ T72+4=0 
 are Tivewi tore V3 4. 
 
 (2) (a? — 5)? — T(a@#— 5) 4+ 12=0. 
 
 Put 2 — 5 = X, and the equation becomes 
 X?—-7X+12=0. 
 CXa Ay XE SNe Ol 
 
 X = 4 and 3. 
 Hence, z?—5=4 and 3. 
 z2=9 and 8. 
 
 e=+3and +2v2. 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 225 
 
 (3) 202 +5 Vo? —5274+3=1074+ 6. 
 ‘Transposing, 222—102+ 5 Va?—52+3-—6=0. 
 By adding 6 and subtracting 6, we have 
 9a? —107+64+5V22?—52+3-—6-—6=0, 
 or, 0 A(a@®' — 524+ 8)4+5V22?—524+8-—12=0. 
 Put Va? —52+3=_X, and the equation becomes 
 2X24+5X—12=0, or (2X —3)(X + 4)=0. 
 X =— 4 and 3. 
 Since X = V22— 52+ 3, we have 
 Ve—5x2+3=—4 and 3. 
 a—5x2+3=16 and 2. 
 v—d5xe—13=0 and 2—52+3=0. 
 
 _5+V25 + 52 54VIT 
 
 b) 
 
 
 
 2 2 
 Fav 8) ba! 
 and ee ee ea a 
 
 The roots are 5tViT and. 5+ V22. 
 2 2 
 EXERCISES. 
 
 Solve the following equations: 
 
 1 (4+32)4+38V#+3e—4=0. 
 (a? — 2)? — 5 (a? 22) +6=0. 
 e828 oD & 0. 
 Vb 02+ 404+3(5a2+40) = 24, 
 (80+5)+4V32+54+7=0. 
 
 ole died ae 
 
226 THE ESSENTIALS OF ALGEBRA. 
 
 6. (2+5)+3(2+7)-4=0 
 x 2 
 97. 8(22+4)—4V2e+441=0, 
 8. 28+727—18=—0. 
 9, w+4e?4+4+4+5(a°+2)—6=0. 
 lo. 28— 5a? = 36. 
 11. v'—92? = 400. 
 p i 2a—84+V2e—5=—1. 
 13, x—5—V22—11=8. 
 14. po 0 ea On/m one eee 
 
 
 
 15. PtatvVvetet5 =25. 
 16. +504+4=5V2e'?4+5a+4+ 28. 
 17, @4+5e—Ve+5a+14= 42. 
 ia 3 VatT_ Vets. 
 
 Ve4-3 V2 
 19. V2i—3 Va =40. 
 20. 22°9—324+6V22—324+2=14. 
 
 
 
 
 
 EXERCISES. 
 1. One half a number plus the square of the number is 150. 
 Find the number. 
 
 2. The sum of two numbers is 15 and their product is 56. 
 Find the numbers. 
 
 3. Find two numbers whose sum is 80 and whose product 
 is 216. 
 
 4. Separate 41 into two parts such that the product of the 
 part is 530. 
 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 227 
 
 5. Two numbers differ by 3, and the sum of their squares 
 is 225. Find the numbers. 
 
 6. The sum of the squares of three consecutive numbers is 
 434. Find the numbers. 
 
 7. A rectangular lot is 32 feet longer than it is wide. It 
 contains 13200 square feet. What are the dimensions ? 
 
 8. A man buys a certain number of chickens for $6. If 
 he had paid 10 cents apiece more for each, he would have 
 gotten 5 fewer for his money. How many chickens did he 
 buy ? 
 
 9. Find a number such that if its square be diminished by 
 1, } of the remainder is 18 more than 10 times the number. 
 
 10. There are two numbers whose difference is 8. If 540 
 is divided by each of these numbers, the difference of the 
 quotients is 18. Find the numbers. 
 
 11. One side of a rectangle is 7 feet longer than the other 
 and its diagonal is 13 feet. Find the area. 
 
 12. The difference of the reciprocals of two consecutive 
 
 numbers is ;34,5- Find the numbers. 
 
 13. The difference between the reciprocals, two consecutive 
 
 odd numbers, is x25. Find the numbers. 
 
 14. By increasing his speed 1 mile an hour a man finds that 
 he takes 5 hours less than usual to walk 60 miles. What is 
 his ordinary rate ? 
 
 15. The larger of two pipes will fill a cistern in 6 minutes 
 less time than the smaller. When both pipes are open, the 
 cistern is filled in 131 minutes. Find the time required by 
 each pipe to fill the cistern. 
 
 16. A and B have a distance of 150 miles to travel. B 
 starts 10 hours before A and arrives 10 hours after A. A 
 travels 2 miles an hour faster than B. What is the rate of 
 each per hour ? 
 
ete maine THE ESSENTIALS OF ALGEBRA. 
 
 17. A rectangular field is 4 times as long as it is wide. If 
 the width is increased 20 rods, its area is doubled. Find the 
 area of the field. 
 
 18. What number increased by 4 and squared is equal 4 of 
 itself increased by 10 and squared? 
 
 19. A man sold a horse for $144, thereby gaining as many 
 per cent as the horse cost him dollars. What was the cost of 
 the horse ? 
 
 20. A boat’s crew can row 9 miles down a river and back 
 in 4 hours. The rate of rowing in still water is double the 
 rate of the current. Find the rate of rowing and the rate of 
 the current. 
 
 21. The hypotenuse of a right-angled triangle is 5 feet 
 longer than the base and 10 feet longer than the perpendicular. 
 Find the sides of the triangle. 
 
 22. The sum of two numbers is a and their product is B. 
 What are the numbers ? 
 
 23. The perimeter of a rectangular field is 168 rds. and 
 its areais 9 A. Find the length of the sides. 
 
 24. Two men start at the same time from the vertex of a 
 right angle and walk along its sides at the rate of 3 and 4 miles 
 per hour, respectively. In how many hours are they 50 miles 
 apart ? 
 
CHAP THR @X0V I: 
 SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 
 
 168. When simultaneous equations involve quadratics, 
 they must be solved by methods which depend upon the 
 form of the equations. Various methods are shown 
 under the following cases: 
 
 GASH 1. 
 
 169. A Linear and a Quadratic. A simultaneous set of 
 this kind may always be solved. The equations are of the 
 forms, 
 
 (1) az?+ bry +cy=d, 
 
 (2) le+ my = k. 
 
 (3) se nai, from (2). 
 
 (4) Cae Amu Tey Ea") y +cy2=d, 
 
 by substituting in (1). 
 
 (5) Cam? — blin + cl?) y? + (blk — 2. akm)y + ak? — di? = 0, 
 by rearranging (4). 
 
 Equation (5) is a quadratic in y, and therefore has tivo 
 roots. The substitution of these roots in (8) will give 
 two values of x. Hence, the set of equations has two 
 
 roots, and only two. 
 229 ° 
 
230 THE ESSENTIALS OF ALGEBRA. 
 
 In the above we have used general equations and we 
 have found that the solution depends upon a quadratic in 
 one variable. Such a quadratic can always be solved. 
 . Hence the simultaneous set can always be solved. 
 
 I. Solve the following equations: 
 
 (1) a? + y? = 25, aa} 
 (2) Ty—a2= 26. 
 (3) z= Ty — 26, from (2). 
 
 (4) 49y?— 350 y+ 625+ y?= 25, by substituting in (1). 
 (5) 50 y2— 3850 y + 600 = 0. 
 
 (6) y2—Ty+12=0, by dividing (5) by 50. 
 EO) (y—3)(y—4) =), by factoring (6). 
 (8) y = 9 and 4. 
 Co) 2a=7Tx38—25 and 7 x 4— 25, 
 by substituting in (8). 
 (10) x=—4and 3 
 
 The roots are (— 4, 3) and (3, 4). 
 
 Care should be given to the proper association of the 
 values of x and y. It should be remembered that a root 
 is a properly associated value of 2 and of y. 
 
 The graphs of 27 + y? = 25 and T y—# =25. 
 From 2* + y* = 25 we have 
 y= +V 25 — 2%. 
 When z= 0, y=+5 and —5; 
 (0, 5), (0, —5) are roots. 
 When g=+1l and —l,y= 2/6 and —2V6; 
 (1, 2V6), (1, —2V6), (—1, 2 V6), (—1, —2-V6) are roots. 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 231 
 
 When «= 2 and —2, y=V21 and —v21; 
 (2, V21), (2, —-V21), (—2, V21), (—2, — V21) are roots. 
 When «=3 and —3, y=4 and —4; 
 (3, 4), (8, —4), (—3, 4), (—3, —4) are roots. 
 When z= 4 and —4, y=3 and —3; 
 (4, 3), (4, —3), (—4, 3), (—4, —3) are roots. 
 When x=5 and — 5, y= 0 and 0; 
 (5, 0) and (—5, 0) are roots. 
 Locating these roots and drawing a curve through them, 
 we find the graph to be a circle. 
 The graph of Ty—xv=25 is a 
 straight line. It is the line PQ. 
 This line cuts the circle in the two 
 points Pand QY. The codrdinates of 
 the two points P and Q, where the 
 line cuts the circle, are (—4, 3) and 
 (3, 4). These are the two roots of 
 the given equations. That they should be the roots 
 
 appears from the fact that they are the only two points 
 whose coordinates are the same for the line and the circle. 
 
 
 
 II. Solve the following equations: 
 
 (1) x + y? = 25. | 
 
 (2) Sa+4y= 25. 
 
 (3) a SAY, from (2). 
 ioe 2 
 
 (4) wee TN IO + y* = 25, by substituting in (1). 
 
 (5) 625—200 y4+16 42+9 y? = 225. 
 (6) 25 y? — 200 y + 400 = 0. 
 
22, THE ESSENTIALS OF ALGEBRA, 
 
 
 
 (8) Y-HY-4)=9. 
 
 (9) 3 y =4 and 4, 
 (10) R20 a: hd ea oe 
 3 3 
 (11) x= 3 and 3. 
 
 The roots are (3, 4) and (3, 4). 
 ‘These roots are the same. Equations (1) and (2) are 
 sald to have a double root. 
 
 The graphs of 224 y?=25 and 82+4y=25. 
 a+ y%=25isthesamecircleasin(1). 32+4y= 26 is 
 a straight line. ‘The graphs are 
 shown in the adjacent figure. 
 In this case the line PQ, which 
 is the graph of 3872+4y= 25, 
 p does not cut the circle, but just 
 touches it at the point Y. The 
 coordinates of the point @ are 
 (3, 4). (8,4) is one of the two 
 equal roots of the given equa- 
 tions. In the case of equal roots the graph of the linear 
 equation just touches the graph of the quadratic equation. 
 
 &) 
 
 III. Solve the following equations: 
 
 (1) | w+ y2= 25. 
 (2) x+y=10. 
 
 (4) 100—20 y+y2+y2= 25. 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 233 
 
 (5) 2y2— 2 y+75=0. 
 20 + V400 — 600 
 6 piesa Nee OY 
 (6) y 4 
 _ 20+10V—2 
 4 
 T10fhV =2 -1045V2 
 Seana a Fe 
 10+5V27 
 (T) Ba a 
 _10F5V27, 
 2 
 
 The roots are 
 
 10—5V2¢ 104+5V2: J 104+5V2¢ 10—5V2: 
 Mee Kg ae OLR 
 
 
 
 These roots are both ¢maginary. 
 
 The graphs of 27+ y?= 25 
 and z+y=10 are shown in 
 the adjacent figure. 
 Inthis case the line PQ, which e 
 isthe graphof w+ y=10, neither 
 euts nor touches the circle. Q_ 
 In the case of imaginary roots 
 the graph of the linear equation 
 neither cuts nor touches the 
 graph of the quadratic. 
 
 170. Graph of the Quadratic in x and y. From the pre- 
 ceding discussion it must not be inferred that the graph 
 
234 . THE ESSENTIALS OF ALGEBRA. 
 
 of the quadratic equation in x and y is always a circle, 
 It may be any one of the following curves : 
 
 OC )C FZ- 
 
 Circle Ellipse Parabola Hyperbola 
 Graph of 427249 y*= 36. 
 
 From this 
 
 
 
 When 2=0, y=2 and — 2; (0, 2), (0, —2) are roots. 
 When z=1 and —1, pearey and — 4/2; 
 
 CL4V2), (1, —4V 2), (= ey 2)nCea ~4V3) are roots. 
 When z=2 and —2, y=8V5 and — aV5; 
 
 (2, 2V5), (2, —2V5), (—2, V5), (—2, ~3V5) are roots. 
 When a=3 and —3, y=0; (8, 0), (—3, 0) are roots. 
 
 Locating these points and drawing a curve through 
 them, we have an ellipse like the following. 
 
 | 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 285 
 
 EXERCISES. 
 
 Solve the following sets of equations: 
 
 e +y =5, (s%x—2y=8, 
 a? + y? = 25. oil 4 wy = 32. 
 meee = (eat2y=9, 
 (2a +3y =6. eee ay 
 3. eee . fa+3y'=27, 
 
 aed ~ ( aty=10. 
 4. ala of ay? = 26, 
 ay = 50. 2. Ale y 296 
 5. ae 
 ay = 27. 13 ae: at Bae 
 o—y" = 16; 
 x —y =4, 
 6. ; os 
 ae sarubre 
 Layty’=9 
 , ee 
 , ry == 0 e+2y=7, 
 15.,3 10 
 5 vw? — ay =2 
 
 Construct graphs for Exercises 1, 2, 3, 5, 13. 
 
 CASE II. 
 
 171. Both Equations Quadratic of the Form ax’ + by’=c. 
 When the equations are of this form, one of the variables 
 may be eliminated as in simultaneous equations of the first 
 degree, and the resulting equation is a pure quadratic in 
 the other variable. 
 
236 THE ESSENTIALS OF ALGEBRA. 
 
 I. Solve 
 C1) rat yt see 
 (2) 427+ 25 y*? = 100. 
 Multiplying (1) by 4 and subtracting from (2), 
 
 (3) 21 y = 36. 
 (5) yotviz=+2vi. 
 (6) 2+(+vi2)?=16, substituting in (1). 
 (7) a= 16—-1= 1), 
 19 
 8 r= t—" 
 *) V1 
 The roots are (- 10 9 ve) Ge 24/8 =) 09 24/3) 
 V7 v/ 
 (= Semeny Vere 24/2 >) There are, as in all solutions under this 
 
 case, a roots. 
 
 Graphs of a+ y¥2=16 and 427+ 25y2=100. The 
 graph of the first equa- 
 tion is a circle and of the 
 second an ellipse. ‘They 
 are shown in the figure. 
 
 The graphs of the two 
 equations intersect in 
 the four points P, Q, R, 
 and S. The codrdinates 
 of these four points are 
 the four roots of the set 
 of equations. 
 
 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 237 
 
 I. Solve | Srey ay 
 
 (2) 4a? +4 25 y? = 100. 
 Eliminating z* as in Example I, 
 (3) ge U, 
 (4) y= 0. 
 (5) y=0 and 0. 
 (6) x? + 0? = 25. 
 (7) e410. 
 
 The roots are (5, 0), (5, 0), (— 5, 0), (— 5, 0). The 
 
 set of equations has two pairs of double roots. 
 
 Graphs of 2+ y?=25 and 4274+257?=100. The 
 graphs are a circle and ellipse. They are shown in the 
 adjacent figure. 
 
 The graphs of the two equations do 
 not intersect, but they touch each 
 other at the points Q@ and P. The 
 coordinates of these points are the 
 roots of the equations. As in Case I, 
 when the graphs touch, the codrdi- 
 nates of the points where they touch 
 are double roots. 
 
 Chyiat 4? =I, 
 
 (2) 422+ 25 y= 100. 
 Eliminating 2? as in Example I, 
 
 (3) Zyae 96: 
 
 (4) y= Ft 
 
 (5) y=tV3=24V4. 
 
 les Solve 
 
238 THE ESSENTIALS OF ALGEBRA. 
 
 (6) e+ (tVipt= 1 
 v= 1-§t=— Hf 
 
 The roots are (6V+¢, 4/2), (5Vti, —4Vv2), 
 (—5V1i, 4V2), (—5V4t, —4-V2). 
 
 These roots are all imaginary. 
 
 Graphs of 2+ y2=1 and 427+ 25 y?=100 are a circle 
 and an ellipse, respec- 
 tively. They are shown 
 in the figure. 
 
 The graphs of the two 
 equations neither inter- 
 sect nor touch. The 
 
 circle is entirely within the ellipse. As in Case I, the 
 roots are imaginary. 
 
 EXERCISES. 
 
 Solve the following sets of equations : — 
 
 a+ y?= 18, 5 {| v+37r=52, 
 | (Qe + y=. ' (2a? 4+5y?=19, 
 2 Pe hedg 
 ieee ae 6 1 ee 
 LC — a . 
 e+ 2y?= 41. ae oe 
 Ct ae 
 3 (ety =%, BAA 
 (3.024 44° = 180. ae ve 
 | A 9 ein 
 (raat ae ty ia esis 
 +24? = 107. 3 Fi ae 
 
 Construct graphs for Exercises 1, 6, 8. 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS, 239 
 
 Case III. 
 
 172. Both Equations Homogeneous in the Part involving the 
 Variables. ‘he equations are of the form 
 ax* + bery+ey2=hk. 
 The first step of the solution is the elimination of the 
 constant terms of the two equations. 
 
 
 
 1) 222-—8 a 2, 
 I. Solve , Me ee 
 (2) 2a%-—3 y?= 838. 
 Multiply (1) by 8, (2) by 2, and subtract. 
 (3) 2a7—9ry+9y?=0. 
 (4) Qar-—sy@—3y)=0. 
 (5) ° ara Y. 
 (6) B= FY: 
 (7) w= By. 
 Substituting z= 3 y in (2), 
 BY 2 
 (8) o(-# : i} res 8. 
 2 
 (9) ane ee 3 ye — oe 
 By" 
 (10) 5 ny 
 (11) y" = 2. 
 (12) y=+v2. 
 (18) a= 8(+2)=+8V2 
 Substituting 2=3 y in (2) 
 18 472-38 y=3 
 I= 5 
 eos 
 y= +}Vv5. 
 
 a= 8(+1V5)= +3V5. 
 
240 THE ESSENTIALS OF ALGEBRA. 
 
 The roots are 
 (8V2, V2), (—8v2, —V2), GV5,4V5), (—2v5, —4FVb). 
 In this case there will always be four roots. ‘There 
 may be one or two pairs of double roots. Two or four of 
 
 the roots may be imaginary. 
 In this case the equations may be of the form 
 
 ax* + bry + cy* = dx or ey. 
 They are solved precisely as the above. 
 (1) 2227-8ay+y7=2y, 
 (2) 2279-38 y=3y. 
 
 Eliminate the right-hand members by multiplymg (1) 
 by 3, (2) by 2, and subtracting, 
 
 II. Solve | 
 
 (3) 2a7—-9ayt+9 y=. 
 (4) (2e—By)\(@—3y)=0. 
 (5) a=yand dy. 
 Substituting z= 3 y in (2), 
 
 (6) 2g yy? —- 8 y= By. 
 
 T BY By, ya—-2y=0 2)=0 
 (7) 9 8 Y 49 = 0 
 (8) y= 0 and 2. 
 
 (9) x= 0 and 8. 
 Substituting z= 3 y in (2), 
 (10) 9y—sy¥=3y. 
 CLL) 6 y= 3 y. 
 
 (12) y(2y—1)=9. 
 (138) y = 0 and 4. 
 
 (14) a= 0 and g. 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 241 
 
 The roots are (0, 0), (0, 0)), (8, 2), (3, 4). (0, 0) is 
 a double root. 
 
 In each of these examples, after finding the value of 2 in 
 terms of y, the substitution might have been made in the 
 first equation instead of the second. ‘The second was 
 selected because it was of simpler form than the first. 
 
 EXERCISES. 
 Solve the following sets of equations: 
 
 ee - (2a°—ay+y7’?=16, 
 eb = 5. a + ay +2 y? = 44. 
 
 w+ 2 ey—y’=T, a Deed ie? 
 22° —38ay—y =—1. y? + xy = 60. 
 {sy—5x%=70, 
 ly? — 3 ay = 10. 
 
 la? + my’? =n, 
 
 ev’ —y’? = 3, 10. , 
 4. be ax + by? =. 
 —2Zaeyt+2y=2. 
 
 (4? +4y=—13 —4 ay, 
 
 — a i Oe 
 ca = 41, 8 2? —12 ay=11—8 77 
 
 >= =37—y' 
 sey 12 4247 12. : te 
 
 aw 
 2 
 32° —9 xy —y’=119. (a 
 
 + = 73 — 2. 
 
 CASE IV. 
 173. When Both Equations are Symmetrical in x and y. 
 
 Hquations are symmetrical in x and y when the interchange 
 of these letters does not change the equations. 
 
242 THE ESSENTIALS OF ALGEBRA. 
 
 Examples: (1) 2—-382y+y?=27T. Interchange 2 and y, 
 and we have y? — 3 yx+ 2°= 27, which is the same as (1). 
 
 (2) v—2ayt+y*=16, 
 
 (3) a+ y* = 12, 
 
 (4) i tLe 
 and (5) xe+y=6, 
 
 are all symmetrical equations. 
 
 I. Solve | SAE TA a 
 (2) 1A bie 
 Add 2 zy = 24 to (1), and we have 
 (3) + 2ay+y*%= 49, 
 (4) a+y=x+T, from (38). 
 Subtract 2 cy = 24 from (1), and we have 
 (5) ew—2eyt+y=l, 
 (6) e—y=H+1. 
 
 From (4) and (6) we have, by adding and subtracting, 
 z=4, 3, —4, —3, 
 y =3, 4, — 3, —4. 
 The roots are (4, 3), (8, 4), (— 4, — 3), (— 3, —4). 
 When both equations are general quadratics, the solu- 
 
 tion depends upon a cubic or quartic. The investigation 
 of such equations is beyond the compass of this book. 
 
 (CA) #f+2ey-—y/=T, 
 12) e—2y>+y =2, 
 Solving (2) for 2, 
 
 (3) t= $V242 Py. 
 
 II. Solve 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 2483 
 
 Substituting in (1), 
 
 
 
 (4) 242y—yt2yV2+2y—y—y=T. 
 (5) t2yV24+2P—y=5—y ty. 
 Squaring, 
 
 (6) 8 y?2+8 yt—4 p= 254 y'—9 y?—2 yP +10 y. 
 (7) Tyt—2y +17 Y—10 y= 25. 
 
 This equation is a quartic in y, and unless it breaks up 
 into factors of degree not higher than two it can not be 
 solved by our present methods. 
 
 Graphs of 27+ y¥?=25 and zy = 12. 
 
 The graph of 27+ y? = 25 is a circle. 
 
 ey = 12. 
 
 When r= 0, y=; (0, &) is a root. 
 
 When z=+1and —1, y=12 and — 12; 
 C, 12), (—1, —12) are roots. 
 
 When 2=+2 and — 2, y=6 and —6; 
 (2, 6), (— 2, — 6) are roots. 
 
 When z=+3 and —3, y=4 and —4; 
 (3, 4), (-- 38, — 4) are roots. 
 
 When «=4 and —4, y=3 and — 3; 
 (4, 3), (— 4, — 3) are roots. 
 
 When z=5 and —5, y= 22 and — 22; 
 (5, 22), (— 5, — 22) are roots. 
 
 When x=6 and —6, y=2 and — 2; 
 (6, 2), (—6, — 2) are roots. 
 
244 THE ESSENTIALS OF ALGEBRA. 
 
 When z=7 and —T, y=16 and — 12; 
 
 (7, 18), (— T, — 18) are roots. 
 When a= 8 and — 8, y=1) and —13; 
 
 (8, 14), (— 8, —14) are roots. 
 When «= 9 and — 9, y=14 and — 14; 
 
 (9, 14), C(— 9, — 14) are roots. 
 When a= 12 and —12, y=1 and —1; 
 
 CL2t 1: (— 12, —1) are roots. 
 When a2 = 24 and — 24, y=4 and —3; 
 
 (24, 3), (— 24, — 4) are roots. 
 
 The graph of zy = 12 
 is the adjacent hyper- 
 bola intersecting the 
 graph of 22+ y?=5 in 
 the points P, Q, AR, and 
 iS, whose coordinates are 
 the four roots of the 
 equations, 
 
 In — solving 
 this example, 
 we get the 
 equations «+y=+T7 
 and <—y=2 1 ie 
 graph of x+y=7 is 
 the line PQ, and that 
 of «+y=-—T is the 
 line RS. The graph of 
 x—y=1 is the line QS, and that of z—y=—1 is the 
 line PA. ‘These four lines intersect in the four points 
 
 
 
 
 
 
 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 245 
 
 Pro h, and ‘S. 
 
 The four lines have precisely the same 
 
 intersections as the circle and hyperbola. ‘This is why the 
 
 set 
 
 is equivalent to the set 
 
 as was used in the solution. 
 
 2 y = + 1, 
 
 a + y* = 25, 
 ig) fend 8 
 
 EXERCISES. 
 
 Solve the following sets of equations: 
 
 far + y= 13, 
 ar = 6, 
 a’ + y? = 34, 
 xy = 16. 
 
 xy = 24. 
 
 | — 
 
 al 
 
 | habe 
 9. | 
 
 ing) ie 
 
 | a y 
 
 co 
 
 ve Y 
 lo. | 
 
 roe ea 
 
 [a y 
 
 ae 5 
 ae Ey 6 
 
 Ei Sige aie 
 
 Lae, eae 
 zo [3 o+-5y = 2 ay, 
 { xy = 15 
 “e Donna 
 x+y =90 
 
 er Yy=aA, 
 14. 
 
 fe = 1(a’?— 6’) 
 
 Construct graphs for Exer- 
 cises 4 and 7. 
 
246 THE ESSENTIALS OF ALGEBRA. 
 
 174. Special Methods; Higher Degrees. Simultaneous 
 equations of higher degree than the second can frequently 
 be solved by special methods. ‘This is particularly true 
 when they are symmetrical. 
 
 A few of the special methods will be illustrated. In 
 such problems the student is expected to devise his own 
 methods. 
 
 I. Solve 
 
 (1)+(2) =) at + ay + oy? + cy + yf = 211. 
 (2)4=(4) at —4a8y + 62%y?-— 4ay24 yt =1. 
 (8)—(4), (5) 5aty — 52%? + Say? = 210. 
 (6) a8y — ay? + vy? = 42. 
 CT) xy(a? + y*) — ay? = 42. 
 From (2), a —2ay + y2=1., 
 e+ y=2ey+1. 
 Substituting in (7), 
 cy(2xy +1)— ay? = 42. 
 ary? + xy —42=0. 
 (cy + T) (zy — 6)=9. 
 sy= 6 and <= (i 
 
 6 ix 
 t= = and =——s 
 Y Y 
 
 Substituting in (2), 
 
 Daay arte and = a 
 y y 
 
 6 ay = ee 
 Pt Yee yy ee 
 
 (y—2)(y+3)=9, ae —-14+V-—27 
 y=2and —3; 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 247 
 
 may, 
 The corresponding values of z are 3, — 2, Lea 
 
 ye 
 The roots are (8, 2), (— 2, — 3), 
 ee Oe eas 87) ft nN OF 1 <1). 
 7 0 er, are | F D 9 D) 
 oo bees 
 2 ee Aa 
 II. Solve aes 
 |@) peed a3: 
 Oe 1 oat 
 Lees be -9 
 2) = (3) —— — + — = -— 
 CYE@) a5 tia- 4 
 ‘5 
 ()-@)=(4) ==9 
 ry 4 
 1 2 eee 
 1 i Dt = 
 
 y CF, 
 Combining (2) and (6) < addition, 
 
 1 3 
 —* and Dr 
 z=4and — 2. 
 y=% and —4. 
 III. Solve - ne ees: 
 Shaye yf == (. 
 Put L=U+, 
 Y=U—v 
 Moe a Ad 
 te 
 
 at + yt = 2ut +12 u%?2420t= 641. 
 
248 THE ESSENTIALS OF ALGEBRA. 
 Putting in the value of », 
 2 ut + 12? (42) + 2(42)? = 641. 
 2u4t + 147 wv? — 2727 = 0, 
 
 4 
 e 7 paar 
 g=utv=d, 2, 5 + a 
 T — 303 
 y=u-—v=—2, —5, =p hae 
 a+ y*= 10, 
 
 IV. Solve es 
 (2) vy—ax—y=-1. 
 
 Multiply (2) by 2 and add to (1), 
 
 (3) e+ 2ayt+y2?—-2e+y)=8. 
 (4) (c+ yy? 2(e@ty)—8=0. 
 (5) (e@+y-—)N@tyt+2)=0. 
 (6) x+y=4or —2. 
 
 From (2) by substituting the value of x +4 y, 
 (7) ry = 9d, or — 3. 
 
 Multiply (7) by 2 and subtract from (1), 
 (8) | 2 —Qay +y2=A4, or 16. 
 
 (9) a—y=+2, or +4. 
 
 Combine (6) and (9), 
 
 The roots are (8, 1), (1, 3), (1, —3), (—3, 1)- 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 249 
 
 EXERCISES. 
 
 Solve the following sets of equations: 
 
 Pee 1 11 ees LS apg 
 ah y * Lay =12. 
 ee a ei Reb Mec 
 : as? e—y=1, 
 cy —y = 4, dal Carty = 82, 
 = ees, | if “e+y=4, 
 ied ss (a? +24? = BA, 
 a ol ) | xy + y” = 35. 
 a+ y=. ( 304+ 5y°=17, 
 a? — y= 19, a Fs foe: ee Lp 
 5. | es ame 
 e—y=1 28 o 
 a+ yi = 91, LeMah ean 
 6. | ee 
 e+y=7. 
 aw + 4? = 28, 17. ie pres 
 ae 2M Rage 
 ( xy + ay? = 30, sm J x 4-4? = 37, 
 | aty=b. le+y+ay=13. 
 pierre ay, he 
 x—y=1. 
 — 2 —s = 
 10 { MN +3 @—N=18, 20 | ae 35, 
 a+ y =T. (@+y)(@ +y") = 65. 
 EXERCISES. 
 
 1. Find two numbers whose difference is 5 and the differ- 
 ence of whose squares is 145. 
 
 2. The difference of two numbers multiplied by the greater 
 
 = 100, but multiphed by the less = 84. Find the numbers. 
 
250 THE ESSENTIALS OF ALGEBRA. 
 
 3. The sum of two numbers is 7, and the sum of their cubes 
 is 91. Find the numbers. 
 
 4, The product of the sum and difference of two numbers 
 is 96, and the sum of their squares is 146. Find the numbers. 
 
 5. The sum of two numbers multiplied by their product is 
 120; and their difference multiplied by their product is 30. 
 Find the numbers. 
 
 6. The difference of two numbers is 3, and the difference of 
 their cubes is 117. Find the numbers. 
 
 7. The sum of the areas of two square fields is 2500 square 
 rods; the sides of the fields are to each other as 5 to 4. Find 
 the area of each field. 
 
 8. If the length and width of a rectangular field are each 
 increased 10 rods, the area is increased 5 acres. But if the 
 dimensions are each decreased 10 rods, the area will be 24 
 acres. Find the dimensions of the field. 
 
 9. The diagonal of a rectangle is 180 feet; the length 
 of the rectangle is 22 times the width. Find the dimen- 
 sions of the rectangle. 
 
 10. Find two numbers such that their product is 16 times 
 their difference, and one of the numbers is double the other. 
 
 11. A rectangular lot containing 13200 square feet is sur- — 
 rounded by a walk 6 feet wide. The walk contains 3336 
 square feet. Find the dimensions of the lot. 
 
 12. In acertain number of two digits the sum of the squares 
 of the digits is one more than twice their product, and the dif- 
 ference of the squares of the digits is 7. Find the number. 
 
 13. The fore wheel of a carriage makes 12 revolutions more 
 than the hind wheel in going 240 yards; but if the circumfer- 
 ence of each wheel is increased 1 yard, then the fore wheel will 
 make only 8 revolutions more than the hind wheel in the 
 same distance. Find the circumference of each wheel. 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 251 
 
 14. If a man had worked 5 days less and had received $1 
 a day less, he would have earned $30. If he had worked 10 
 days less, and had received $2 a day more, he would have 
 earned $50. How many days did he work, and what were 
 his wages a day ? 
 
 15. Ifthe numerator of a fraction be increased by 3 and the 
 denominator be decreased by 3, the resulting fraction is the 
 reciprocal of the first. If 47 be added to the fraction, the sum 
 is } the reciprocal of the fraction. Find the fraction. 
 
 EXERCISES— MISCELLANEOUS. 
 1. Extract the square root of 2’y? — ary? —(4a—Ly)a’y 
 +2ay+4a' 
 2. Find the roots by factoring: 
 (a) # —Tx=390. 
 (6) #+72=60. 
 (c) y—9ay+20a=0. 
 (d) By+4(2y—3)—39=0. 
 (e) y—(e—a(e—b) = (a—D)y. 
 3. Determine whether 1, —1, 2, or any one of them, is a 
 root of 9a’°—3a=2. 
 
 4. Make an equation whose roots are 3+ -V7 and 3— v7. 
 5. Simplify 8-V3 + 13-243 — 5-121 +.4-V27. 
 
 6. Multiply (@+Va7) by a? —Va~. 
 
 seeLE SV Dia. 
 
 2V5 —3-V6 
 34+V—4 
 6—V—16 
 
 7, Rationalize the denominator of 
 
 8. Rationalize the denominator of 
 
 1 
 3 
 
 9. Divide a—b by a’ — 3. 
 
Oo THE ESSENTIALS OF ALGEBRA. 
 
 Seon 
 v42%=8, 
 vo sy 
 
 10. Solve ee 
 Ge oe 4, 
 Ey, 
 
 11. Solve de?—2x+c=0. 
 
 12. By means of the discriminant, tell what kinds of roots 
 each of the following equations has: 
 
 (a) 3a°—5a+2=0. (c) 5a°—5a+10=0 
 
 (b) 22?+112—10=0. (d) —a’+4e+4+2=0. 
 
 13. Find two consecutive numbers whose product is 1260. 
 
 14. A number consisting of two digits which differ by 3, is 
 6 less than 7 times the sum of the digits. Find the number. 
 
 15. What value must a have to make the roots of 52?—112 
 +a=0 equal ? 
 2 ey, 
 
 ——-=0, 
 a1 fd eed Find « and y. 
 9 
 +55} 
 “Ber ir: Find « and y. 
 dy—v7+3=0. 
 
 18. Two trains start at the same time to go 320 miles. One 
 goes 8 miles an hour faster than the other and reaches its 
 destination 2 hours sooner than the other. Find the rate of 
 each train. 
 e+2 “+3 
 
 eo 1 
 =p oy 
 PINE at eel 
 
 
 
 
 
 
 
 19. Solve 
 
 20. Express 27 ys 4 2aiy7# without negative or fractional 
 exponents. 
 
 21. 120*—172°4+6=0. Find z. 
 22. y°—3d ye == O6..q Ing: 
 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 253 
 
 23. A number consists of two digits. If the order of the 
 digits be changed, the sum of the new and original number is 
 77 and their product is 1300. Find the number. 
 
 24. 36074 29 ax+5a°?=0. Find z. 
 25. Make an equation whose roots are 
 26. V2 +27 =8. Find 2. 
 erie oO ee 0 —C. 
 b+ta w-—a b+c we-—Cc 
 28. Find the roots of (y—2)(y? —12 y + 20)(y—1) =0. 
 
 29. DV — 2? x 8V— 7 x 2V—9 ce? x 8 — 48? = what? 
 
 30. Multiply V3 —V2—vVy by V3+ Va— Vy. 
 
 31. Perform the indicated operations and simplify the 
 
 result: 
 Ve= (aa 
 a Gee 
 
 5 
 
 32. Square ai + bt — 3, 
 33. Solve 11 x—11= ae 
 
 a+v2 and a—Vv2 
 3 3 
 
 27; Find a. 
 
 
 
 
 
 
 
 
 
 
 
 34. Solve 22—352+4+22°=0. 
 a ws 5. 2a+3 
 —5 2a—-3 
 a+5 a= ba 
 2a+3 Seen 
 36. A path around the outside of a rectangular garden is 6 
 feet wide and 4224 square feet in area. The area of the gar- 
 den is 28000 square feet. Find the dimensions of the garden. 
 
 
 
 
 
 35. Simplify — 
 
 
 
 
 
 Byipsinplity 22. 
 
954 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 
 
 A Bs Ob 
 aa Solve eee A 
 Ohh Se OR es ed 6 
 3 2 ace tie der 2 a2 — 14 
 39. vipers are Nae ’ Find @# and y. 
 
 2y—dSay+ouv= 6. 
 
 —6 2 y—1 2 
 40. Solve &— a : 
 Abr AD RMON pe aE So 
 
 41. Solve (a? — b’)(a—1) =2 a(a’ + 0”). 
 
 42. Solve s]2—4{2—a(2—- 2) | | as. 
 
 43. What must be the value of x in order that 
 may equal —1? 
 
 44. Find the value of at—5a*—122?—13%—T7, when 
 g=—i(14+~V—8). 
 
 45. Two rectangular fields each contain 10 acres. The 
 
 perimeter of one is + longer than that of the other. One of the 
 fields is asquare. What are the dimensions of each field ? 
 
 
 
 
 
 
 
 
 
 (w+ 3)? 
 32°+9a—5 
 
 46. If ab+bce+ca=0, prove that 
 (a) (a+b+cPr=e?40?+e%. 
 (6) (a+b+cPf=e04+?+¢—3 abe. 
 (c) (@+b+c)=ai+b*+ct—4abe(a+b+0). 
 
CHAPTER XVII. 
 
 RATIO, VARIATION, AND PROPORTION, 
 I RATIO. 
 
 175. The ratio of a quantity A to a quantity B is the 
 quotient of A by B. 
 
 This quotient may be written in any one of the forms, 
 A + B, 4 A/B, or A:B, each of which is read, the 
 ratio A to B. 
 
 176. Ratio can exist only between two abstract num- 
 bers, or between two concrete numbers of the same kind. 
 The ratio 5 to 7, or ? , has a meaning, so does the ratio 
 6 bushels to 15 bushels, but not so with 6 bushels to 15 
 
 inches. Ratio merely expresses the part one magnitude is 
 of another. 
 
 177. The terms of a ratio are the numbers compared, 
 the numerator being called the antecedent, the denominator 
 the consequent. 
 
 178. The ratio of antecedent to consequent is called a 
 direct ratio; the ratio of consequent to antecedent is called 
 an inverse ratio. 
 
 Thus, 14: 28 is direct, while 28:14 is its inverse ratio. 
 
 b is the inverse of ©. 
 a b 
 
 255 
 
250 THE ESSENTIALS OF ALGEBRA. 
 
 179. A compound ratio is the product of two or more 
 single ratios. 
 
 Thus, 5 x is the compound ratio of the single 
 
 bh gh 
 <a *f 
 ratios, ~ “ # 
 
 5 b’ Teck 
 
 180. Laws of Ratios. Since a ratio is a fraction, the 
 operations which may be performed upon fractions may 
 likewise be performed upon ratios. Below are enumerated 
 the more‘important laws relating to ratios: 
 
 (1) A ratio is unchanged by multiplying or dividing 
 both antecedent and consequent by the same number. 
 
 MeO oe 00-2 ee mA A-+-m 
 § 8x4 849° Bo mB Boome 
 
 
 
 
 
 
 
 This law shows that the ratio of two concrete magni- 
 tudes of the same denomination is independent of the unit 
 
 of measurement. ‘The ratio of 2 miles to 5 miles is 53 the 
 
 ratio of 2 miles, expressed in feet, to 5 miles, expressed in 
 2x 5280 2 
 5 x 5280. 5 
 numerator and denominator of a ratio merely changes the 
 denomination of the terms, if they be considered as con- 
 crete quantity, the ratio of the two magnitudes remaining 
 unchanged. 
 
 (2) A ratio is changed by extracting the same root of 
 each term of the ratio, or by raising each to the same 
 power. 
 
 This law is true except in the case when antecedent and 
 
 feet, is 
 
 
 
 The introduction of a multipher in 
 
 A VA 
 consequent are equal. If pa then will TR = /m, and 
 
RATIO, VARIATION, AND PROPORTION. 257 
 
 k a 
 se =m*; but Vm#m, and m‘*#m. Hence, the ratio has 
 been changed in all cases except that in which m= 1. 
 
 (3) A ratio is changed by performing unlike operations 
 upon antecedent and consequent. 
 
 ees es 
 
 — + ———, in which the antecedent has been multipled 
 
 AE re 
 A mA mA (m\A 
 by 2 and the consequent by 3. —+——, for ——= (“) at 
 sf : 7 Bb 7 aB nB 1} BD 
 (4) A ratio is changed by adding the same quantity to 
 antecedent and consequent, except when the ratio is unity. 
 
 BoA BALL oe 
 
 7 ORI apr ey igs 
 
 
 
 The truth of this law may be easily proven. Take the 
 A+” 
 B+a 
 
 
 
 : A 
 two fractions, 2 and , where x is any number what- 
 
 ever. By division, 
 A+z A, x(B—A) 
 
 B+« B BCB+2) 
 Alte 
 Brow 
 (5) A ratio is made more nearly equal to unity by 
 adding any positive number to each of its terms. 
 
 
 
 
 
 which shows that oe except when A = B. 
 
 Let 4 be any ratio, and z any positive number. Then 
 
 
 
 
 
 A+a _ ve ee 
 
 B+o2@ Bia 
 
 and Ay why eres 
 B B 
 
 The first of these two differences is seen to be smaller 
 than the second. Why? Hence, the truth of the law 
 is established. 
 
ih) THE ESSENTIALS OF ALGEBRA. 
 
 
 
 (1) Compare = and 
 
 
 
 
 
 
 
 2_20 243 5 _ 86, 
 TALT0 SU oe LOM 
 2 
 = is more nearly 1 than = 
 (2) Compare > and ote. 
 DUS BOU Di toes en 
 8 21° 344 7 21 
 30 
 
 ae 
 nA is more nearly 1 than a 
 
 os 
 
 From these illustrations we may see that if a ratio be 
 less than unity, the addition of the same positive number 
 to the antecedent and consequent increases its value toward 
 unity; and if the ratio be greater than unity, the addi- 
 tion of the same positive quantity to both antecedent and 
 consequent diminishes the ratio toward unity. 
 
 181. The terms ratio of less inequality, ratio of equality, 
 and ratio of greater inequality are sometimes employed to 
 describe ratios less than unity, ratios equal to unity, and 
 ratios greater than unity, respectively. 
 
 182. Limit. The result shown above, 
 
 wales a clas ps Par ee oe) 
 B4+z Bax 
 
 
 
 
 
 A+z 
 
 
 
 indicates that the difference between and unity 
 
 is a fraction 
 
 
 
 whose value may be made as small 
 
 as we please by making @ sufficiently large. Hence, the 
 
RATIO, VARIATION, AND PROPORTION. 259 
 
 A+z 
 
 value of the ratio -, as x becomes infinitely large, 
 x 
 
 
 
 approaches unity, which is called the lamit of the ratio. 
 The value that any algebraic expression continually 
 approaches -but never reaches is called its limit. 
 
 EXERCISES. 
 
 Write in their simplest forms the ratios of: 
 
 
 
 1. 625 to 125. 4. 2 —(y+z2) to ~+y+z. 
 
 2. 4802 to 120 2”, 5. 2—y to a+tay+y’. 
 
 3. 2432? toxr+3. 6. (‘St 1) A a 
 4 ab 
 
 SUGGESTION. pee (et 8) x, 
 
 r+3 r+3 
 7. a@—12a+ 20 to a—10. 
 8 627+ 23 ax +20 a? to 32+4a4. 
 9. wt +ary’?+ y* to v—ayt+y’. 
 10. (a+ 7’)? —4 a°y* to (a? — y’)’. 
 Write the compound ratios of the ratios: 
 11.35 to 5 and 10 to 15. 
 12. ~+y to «—y and w—y7’ to (#+ ¥)*. 
 13. 25 to «* and 32’ to 50. 
 14. a? — 270° to (a—35b)? and a—3b to &+3ab4+90". 
 15. (©+1)?: @’+2a+41), (#’+1):(#+1), 
 
 and (a—1) : (#’—a+1). 
 Find the value of x for which the ratio of: 
 16. 128 to 2 is 2. 1602+) 60,4 — 1 18.7. 
 Te O20 LO 2 18 0. 19500 + 1207) to: 27-Eb- is. 3. 
 
 20. (7+ 4):(8e%41)=4. 
 
260 THE ESSENTIALS OF ALGEBRA. 
 
 Arrange the following ratios in descending order of magni- 
 tude : 
 
 
 
 21 9 21 AS er I Yeah EF 8 4 1 
 © 2:02 PS22 9 '2 GF. 92 Fee D4) oO iy 2 eee 
 22 ats “a+ @-+3) aie 
 
 
 
 
 
 
 
 
 
 
 
 II]. VARIATION. 
 
 183. The term variation has little use in ordinary alge- 
 bra, but its use is so frequent in physics that a brief treat- 
 ment of the subject will be introduced here. 
 
 In physics we say “the weight of a uniform mass varies 
 as the volume.” ‘This means that if W is the weight, and 
 V the volume, then is W=k x V, where k is a constant, the 
 weight of a unit volume of any given substance. 
 
 In mensuration the circumference varies as the diameter. 
 This means, that if ( be the circumference and D the 
 diameter of any circle, then will 
 
 C=kx D, 
 k being a fixed constant for all circles. ‘This constant is 
 
 usually denoted by the Greek letter 7; its numerical value 
 is an incommensurable number, 3.14159 «--ccccceeeeeseee : 
 
 184. In general a variable y is said to vary as another 
 variable 2, when 
 “= a constant. 
 x 
 
 The phrase, y varies as x, is sometimes written 
 Yer, 
 
 but is to be interpreted to mean 
 wi 
 
 x 
 
 i NOT eae, 
 
 From this definition we see that a variation as here con- 
 sidered is equivalent to an equation. 
 
RATIO, VARIATION, AND PROPORTION. 261 
 
 185. Variations may be classified as follows: 
 (1) Direct. y varies directly as x, when 
 y=kx, k=a constant. 
 
 The circumference of a circle varies directly as the 
 
 radius. 
 C=kR, where k= 27. 
 
 (2) Inverse. y varies inversely as x, when 
 
 y=: 
 es 
 The volume of a gas varies inversely as the pressure, 
 
 k 
 V=—, where V= volume and n= pressure. 
 n 
 
 (5) Joint. y varies jointly with xv and z, when 
 Dire abe 
 The weight of a rectangular parallelopiped of metal of 
 unit height varies as the product of the length by the 
 
 width, 
 Veer delle xan). 
 
 k = weight of unit volume of the substance, / = length, and 
 6 = width of the rectangular solid. 
 
 (4) Quadratic. y varies as the square of 2 when 
 URW Le 
 
 An example of such variation is found in the law of 
 falling bodies; 7.e., the space fallen through by any body 
 starting from rest equals a constant times the square of 
 the time expressed in seconds. 
 
 w= ki?, k=tg9, g = 32 feet, 2 inches. 
 
 S'= space described, t = time in seconds. 
 
262 THE ESSENTIALS OF ALGEBRA. 
 
 (5) Direct and Inverse. y varies directly as x and in- 
 
 versely as 2 when | 
 
 = k-. 
 
 Yer 
 
 An example of this form of variation is found in 
 
 Newton’s Law of gravitation. If M, m, be the masses of 
 
 two attracting bodies, D their distance apart, and G the 
 force of gravitation, then 
 
 Mx m 
 G = k——— 
 Dp 
 EXERCISES. 
 
 1. If yoo, and y=b when «=a, find the value of y when 
 
 2 pint 
 SOLUTION. 
 
 If yea, thenisy=kx. Buty =b when x=a; 
 b 
 bence, b=ka, or k=-- 
 a 
 a 2 for any value of zx. 
 
 b 
 Hence, y=—-c when x= c. 
 a 
 
 2. If you, and if y=10 when x=2, find the value of 9 
 when «=12. 
 
 3. If you, and if «=16 when y= 64, find the value of a 
 when y= 15. 
 
 4. The circumference of a circle varies as the radius 
 (Cx). If C=3.1416 when R=4, find the circumference 
 of a circle whose radius is 12. 
 
 5. If excy and wz, prove ~ «4. 
 
 Mies 
 
 6. If xxy and vet, prove avcyt. 
 
RATIO. VARIATION, AND PROPORTION. 263 
 
 7. If xw«y, prove that zy”. | 
 8. If wxy and zcy, prove that (a — 2°) oy’. 
 
 9. If y varies inversely as a’, and if y=16 when «= 4, find 
 2 when y= 10. 
 
 10. The volume of a sphere varies as the cube of its radius. 
 If the volume of a sphere whose radius is 3 be 113.1, find 
 the volume of a sphere whose radius is 20. 
 
 Ill. PROPORTION. 
 
 186. Proportion. The equality of two ratios is called 
 a proportion. 
 
 Thus, a=< is a proportion. 
 
 Various forms have been employed in writing a propor- 
 tion, the following being the ones more frequently used: 
 
 foe C.D), are Balad. 
 i eee: 
 A+B=C+D, eat 
 
 Each form is read A is to Bas Cis to D. 
 
 187. Proportionals. The terms of the two ratios are 
 called proportionals. 
 
 In the proportion A: B=C: D, the terms A and D are 
 called extremes, the terms Band Care called means, of the 
 proportion. 
 
 In the proportion A: B= B: D, B is called the mean 
 _ proportional to A and D; D is called a third proportional 
 to A and B. 
 
 188. Theory of Proportion; Theorems. A Theorem is a 
 statement of a truth to be proved. 
 A Corollary is a truth derived from the proof of a theorem. 
 
264 THE ESSENTIALS OF ALGEBRA. 
 
 The following theorems apply to proportions in which 
 the terms of each ratio are considered abstract numbers. 
 
 THEOREM I. In any proportion the product of the ex- 
 tremes equals the product of the means. 
 
 Given A: B=C: D, or, more simply, 
 AEC. 
 Bed: 
 hens A= Bes WW hye 
 
 CoROLLARY. The mean proportional to two numbers 
 equals the square root of their product. 
 
 This corollary results from the above theorem by letting 
 
 C= 8, a=, or AD= B*, whence B=VAD. 
 
 THEOREM II. Jf the product of two numbers equals the 
 product of two other numbers, then either pair may be taken 
 as extremes, and the other pair as means, of a proportion. 
 
 (Inverse of Theorem 1.) 
 Given AD=BC. 
 
 Divideshy 7 ale 4-5 
 Divide by CD, (2) fas. 
 Divide by AG, (3) 4 = 7m 
 
 In each of the proportions (1), (2), (8), we have taken 
 one pair of factors, A, D, or B, C, as extremes, the other 
 as means. 
 
 ——— eS eee 
 
RATIO, VARIATION, AND PROPORTION. 265 
 
 THEOREM III... Jf four numbers be in proportion, they 
 are in proportion by inversion. 
 
 Expressing this theorem algebraically, 
 
 D 
 
 = = oe from which B =i. 
 
 tee D Ae 
 
 Proof is left to the student. Result is easily shown ° 
 true from Theorem I. 
 
 THEOREM IV. Jf four numbers be in proportion, they 
 will be in proportion by alternation ; that is, the first ts to 
 the third as the second is to the fourth. | 
 
 Pigepraically, if A: B=C:D, then.is A:C=B: D. 
 
 See (2) under Theorem II. 
 
 THEOREM V. Jf four numbers are in proportion, they 
 are in proportion when taken by composition; that is, the 
 sum of the first and second is to the second as the sum 
 of the third and fourth is to the fourth. 
 
 This theorem stated algebraically is, 
 A Sault Ee a Ee 
 
 if — = fa then is 
 
 B D B D 
 Proof is easily derived by adding 1 to each member 
 of the given proportion and reducing each member to a 
 fractional form. 
 
 
 
 
 
 THEOREM VI. In a series of equal ratios the sum of 
 all the antecedents is to the sum of all the consequents as 
 any antecedent is to its consequent. 
 
 Proof. Let the equal ratios be 
 — Ss ea as - =" #** =7, 
 
 where ¢ is the common value of the ratios. 
 
266 THE ESSENTIALS OF ALGEBRA. 
 
 Then oar, or A=r xB, 
 f=, or C= nxil; 
 Ser, or H=r KF, 
 amr, OF = cae 
 
 Adding the equalities, 
 
 
 
 A+C+H+G+--=7x(B+D+F+7aR 
 ry A+C+H#H+G+4+>. AL 
 
 H 9 —-- > = —  — See, 
 BE D+F tps 1p 
 
 I Eee aK! 
 
 then is 
 1 §+14+35+44+6 8+4 54146 
 
 —_ rd 
 
 3° 1643494129518 9419" [oe 
 
 and so on. 
 
 
 
 EXERCISES. 
 
 Find the value of the variable for which each of the following 
 proportions is true: 
 
 LY O20 95-40: 
 ee Dt 0. 520s 
 edd = D2 a. 
 (8a+4):(@+5)=(5e%+1): Ga—4). 
 (4e2—3):(2e+1)=(7 7-4): (844 2). 
 
 6. Find a fourth proportional to 12, 16, and 40; also to a, 
 b, and c. 
 
 Ses 
 
RATIO, VARIATION, AND PROPORTION. 267 
 
 7. Find a mean proportional to 16 and 49; also tol and m. 
 8. Find a third proportional to 25 and 35; also to a and «wz. 
 
 Are the following proportions true for all. values of the 
 letters: 
 
 9. 9—2):(842)=(8x—2"): 4? 
 
 2 
 10. (Fat): 2ae—N =a): dee? 
 ll. (w+ y)?—2): (e@+y+2)=(@+y—2): a? 
 etme « 
 
 Sa yaa 2 + .w, show a= 
 lyt+mw y 
 
 
 
 et Zz 
 suGGESTION, Let —~=7r, —=r, then x=ry, z=rw, also v= Iry, 
 v 
 
 a 
 mz=mrw. Add, lx + mz=r(my + mz), ete. 
 
 2 2 1 re gl A 
 13. If -=—, show ———.~=—=-—. 
 w yw. yw. yy 
 14. The rates of walking of two travelers are to each other 
 asatob. If one walk c miles in a given time, how far does the 
 other walk in the same time ? 
 
 15. The rear wheel of a wagon is a feet in circumference, 
 the fore wheel is } feet in circumference. How often does the 
 fore wheel rotate while the rear wheel makes m revolutions ? 
 
CHAPTER XVIII. 
 PERMUTATIONS AND COMBINATIONS. 
 I. PERMUTATIONS. 
 
 189. This subject can best be understood by introduc- 
 tion through a few concrete examples. 
 
 (1) How many different numbers of two digits each 
 can be formed by using in every way any two of the five 
 digits %, 0, seo.) ore 
 
 By writing any one digit first and each of the remaining 
 four digits after it, we have the following five rows, each 
 composed of four numbers: 
 
 56, 57, 58, 59, 
 65, 67, 68, 69, 
 195, 76, Tek 19; 
 85, 86, oh SU. 
 95, 96, 91, 98. 
 
 In all there are 5 x 4= 20 different numbers. 
 
 (2) How many different numbers of two digits each 
 can be formed by using in every way any two of the four 
 digits 0, 0.) 7,10 ¢ 
 
 Here we select any one of the four digits as the first, 
 and place after it successively every one of the remaining 
 three digits. 
 
 268 
 
PERMUTATIONS AND COMBINATIONS. 269 
 
 - This gives the following numbers: 
 
 56, 57, 58, 
 65, 67, 68, 
 795, 76, 78, 
 85, 86, 87. 
 
 In all there are four selections of the first digit, and four 
 less one selections of the second, giving 4x 38=12 different 
 numbers. 
 
 (3) How many different numbers of three digits each 
 can be formed by using in every way any three of the five 
 PIPING t 4 Oy OD? 
 
 The first digit can be any one of the five; hence, the 
 first place of each number can be filled in five different 
 ways. Four digits remain to fill the other two places of 
 each number. But we have just seen that two digits can 
 be selected from four in 4 x 3=12 ways. Hence, with 
 each of the five selections of the first digit, can be placed 
 twelve selections of the digits filling the two remaining 
 places. Hence, there are 5x4x3=60 different numbers. 
 
 190. Definitions. (1) The number of ways of selecting 
 three things from a collection of five things is called the 
 permutations of five things taken three at a time. 
 
 (2) The number of ways of selecting r objects from a 
 collection of n distinct objects, regard being had for the order 
 of selection, is called the permutations of n things taken r 
 at a time. 
 
 In this general case, may be any number of objects, 
 and r may be any integral number from 1 to n. 
 
270 THE ESSENTIALS OF ALGEBRA. 
 
 191. Symbol. Instead of writing the permutations of 
 n things taken r at a time, the symbol ,P, is generally 
 used. 
 
 Illustrations. (1) ,P,=5 x 4, the permutations of five 
 things taken two at a time. | 
 
 (2) ,P, =4 x 3, the permutations of four things taken 
 two at a time. 
 
 (3) ,P3,=10x9x8, the permutations of ten things 
 taken three at a time. 
 
 192. Hzamples: Let the pupil construct tables, if 
 necessary, to verify the following results : 
 
 Clit ley eens 
 
 Aig Pes: 
 
 (3) Fe Baa eg Ame ce 
 
 (4)° Pe xt aes 
 
 COs) Aiea. 
 
 CDi): a Fg aha 
 
 (Ty Pep es) 
 
 (8) Pp AX oer 
 
 (9)\ SP as 5x4 Xo eee 
 
 CLOT Rebs Lue 
 The above examples indicate that there is a law govern- 
 
 ing the formation of permutations. It will be noted that 
 the number of factors giving the permutations in each 
 case equals the number of objects in each selection ; the 
 
 highest factor is the number to be permuted, and each 
 succeeding factor is one less than the preceding. This 
 
PERMUTATIONS AND COMBINATIONS, Atel | 
 
 observation should lead one to some conclusion regarding 
 the value of the general symbol 
 
 atone 
 We should expect the number of permutations of » things 
 
 taken 7 at a time to be expressed by a product of 7 of the 
 natural numbers beginning with n. Hence, we should find 
 
 A Oe 
 
 (2p aoe n(n — 1). 
 
 (8) ,P3,=n(n—1)(m— 2). 
 
 (4) ,P,=n(rn—1)(n—-2)™m-—8). 
 
 (5) ,P,=n(n— 1)(n— 2)(n— 3) (m— 4). 
 
 (r) ,P,=n(n—1)(n—2)(r— 3) « (n—r+1). 
 
 193. Value of ,P,. ‘I’o determine the number of permu- 
 tations of n things taken 7 at a time, we may proceed as 
 follows: 
 
 (1) Let the n distinct things be represented by n letters 
 of the alphabet. 
 
 (2) Select any one letter to stand first in a set of words 
 of two.letters each. Then there would remain n — 1 letters 
 to fill the second place; but the first letter may be selected 
 in nm ways, and with each of these selections any one of 
 the n — 1 remaining letters may be placed. 
 
 (3) Hence, for the number of permutations of n things 
 taken 2 at a time we have 
 
 anal): 
 
 (4) Let the first two letters of a three-lettered word be 
 selected from the n letters; this selection can be made in 
 
O12 THE ESSENTIALS OF ALGEBRA. 
 
 n(n —1) ways, as shown in (3) above. Now we may 
 select any one of the remaining n— 2 letters to fill the 
 third place. 
 
 (5) Hence, the formation of a three-lettered word from 
 n letters can be accomplished in 
 
 nes =n(n—1)(m— 2) ways. 
 
 (6) In a similar manner we may show that 
 
 ne ,=nPs x (n—3)=n(n—1)(n—2)m— 8), 
 
 »P,=nP, Xx n—4)=n(n—1)(n— 2)(n— 8) (n— 4), 
 
 and in general 
 per pee Sheet aD 
 =n(n—1)(n—2)(n— 38) + (m—r+1). 
 
 This general result may not be understood at the first 
 
 reading of this subject, but its truth may be assumed until 
 the pupil has had more experience. | 
 
 CoroLuARY I. Whenr =n, the general formula becomes 
 nen =n(n—1)(n—2)(m—3)4x38x2xI, 
 
 a result easily remembered. 
 
 194. The Factorial Symbol. In the value of ,P,, above, 
 we have the product of the natural numbers from 1 to n. 
 This product is often spoken of as “factorial n,” and is 
 
 written for brevity 
 [n, or nl. 
 
 Either |n or n! is to be read factorial n, and means 
 the product 
 
 n(n —1)(n—2)(n—38)+5x4x38x2x1. 
 
PERMUTATIONS AND COMBINATIONS. ie 
 
 | EXERCISES. 
 Find values for: 
 
 
 
 
 
 
 
 
 
 
 
 
 
 sae eee 16 Ae 
 [5 5, [ie 8. [25 e 
 [6 [8 [8 Bale 
 2. B | 
 | 10 6 as 9 ee 
 Spel abe 115 |4 oh 4s 
 fate [15 [4 i 
 20 20 8 {10 
 4. Eu zs ue 10. es 
 
 
 
 
 
 [16 x [2 [10 [12 [15 
 
 195. Coro.tLARY II. When objects are permuted all 
 together, but are not all different, the number of distinct 
 permutations is given by ,P,,+|s, where s is the number 
 of objects which are alike. 
 
 Illustration. Required the number of different numbers 
 obtainable from the five digits 5, 6, 6, 7, 8, taking five 
 at a time. 
 
 If all digits be different, the number of selections would 
 clearly be ,P;=|5. But the two sizes, when permuted, 
 give no new numbers; hence, all the permutations of the 
 two sixes, z.e. |2, must be excluded (divided out) from 
 the total. 
 
 
 
 Pepe 2 Xt x8 x12 5 
 Sa Peis 
 EXERCISES. 
 
 1. How many different numbers of three digits can be made 
 from 1, 2, 3, 4, 5, 6? 
 
 2. How many different permutations can be made by taking 
 4 of the letters of the word working ? By taking all of them ? 
 
274 THE ESSENTIALS OF ALGEBRA. 
 
 3. Find the value of i.P33 1;£s3 »Ls 
 
 4. How many permutations can be made from the 26 letters 
 of the alphabet, taking + at a time ? 
 
 5. How many six-place numbers can be formed from the 
 Arabic numerals? (Include 0.) 
 
 6. In how many ways can a class of 6 be seated in a row of 
 6 chairs ? 
 
 7. In how many ways can the front row of 6 chairs be 
 filled from a class of 20? 
 
 8. How many different permutations can be made from the 
 lette1s of the word Indiana? Mississippi ? 
 
 9. How many even numbers of 6 places can be formed from 
 the digits 1, 3, 4, 5, 7, and 9? 
 
 10. How many numbers between 50,000 and 60,000 can be 
 formed from the digits 3, 4, 5, 6, 7 ? 
 
 11. In how many ways can 10 books be arranged on a shelf 
 provided 2 particular books are always to be at the ends of 
 the shelf ? 
 
 12. In how many ways can 12 balls be arranged, if 5 are 
 red, 4 white, and 5 blue ? 
 
 Il. COMBINATIONS. 
 
 196. (1) How many products of two factors each can 
 be made from the five digits 5, 6, 7, 8, 9? 
 
 We have seen that the number of ways of selecting two 
 things out of five is the permutations of five things taken 
 twoatatime. But in the case of products, 5x 6=6 x 5; 
 hence, each arrangement of two digits is the result of a 
 permutation of two things taken two at a time. These 
 must all be excluded. Hence the number of products is 
 
 5 x4+|(2=5x4+2=10. 
 
PERMUTATIONS AND COMBINATIONS. Bes 
 
 (2) How many products of three factors each can be 
 made from the five digits 5, 6, 7, 8, 9? 
 
 Since any three factors may be arranged in 3 x 2 x 1 
 different ways, each arrangement giving the same product, 
 we shall have to divide out |8 of the permutations of 
 five things taken three ata time. Hence, the total number 
 of different products is 
 
 ox4x3+|3= 10. 
 
 197. Definition. (1) The number of ways of selecting 
 three things from a group of five, no regard being had for 
 ‘he order of selection, is called the combinations of five things 
 saken three at a time. 
 
 (2) In general, the number of ways of selecting r things 
 from a group of n things, no regard being had for the order 
 of selection, is called the combinations of n things taken r 
 at a time. 
 
 198. Symbol. Instead of the phrase, combination of n 
 things taken r at a time, the symbol ,C, is usually em- 
 ployed. 
 
 Thus, ,C, is read, the combinations of five things taken 
 two ata time; ,,(, is read, the combinations of ten things 
 taken four at a time, ete. 
 
 199. Relation between ,C, and ,P,. It is easy to see that 
 if we select from a given number of things any specified 
 number, and do this in every possible way, having no 
 regard to the order of selection, and then permute all the 
 objects in each group in every way, we shall have the total 
 permutations of the n things taken r at atime. ‘The selec- 
 
276 THE ESSENTIALS OF ALGEBRA. 
 
 tions of the groups are the combinations, and the objects 
 of each group are permuted r at a time; hence, 
 os 4 
 AY edema b MEN ay EPe 
 OMY ROM 
 MW aU. x 7. 
 
 er, al, n(n) n= 2) Gree 
 n~7 |” |” 
 
 A second form for ,C,, may be had by multiplying the 
 numerator and denominator of the fraction on the right 
 
 
 
 
 
 
 
 
 
 
 
 by |jn—vr. This multiplier makes the numerator |n, and 
 the symbol ,C;, becomes 
 cee 
 “" |r |n— 7 
 12 |12 
 Cc = = e 
 aie Khe ei ee 
 From the first form 
 | yy _12-11-10-9.8-7-6 
 1277 [7 
 
 12-11 10-9. Sy aeeeee 
 
 
 
 Notr. When r=n, the second factor of the denominator |n — 7 
 becomes |0, a symbol whose value is to be taken as unity. 
 
 To show 
 jo =1, 
 we take the equality [m =m x |m —1, 
 and put = 
 
PERMUTATIONS AND COMBINATIONS. pag 
 
 then (has deta; 
 and as [1 = 1, |O must be 1. 
 oe |O =~ i 
 The fo in Lie ee 
 [2 [therm 
 
 is mor? easi 7 remembered than the form 
 
 C, =m = UG — 2)ur ea —r +1) 
 a 9 
 
 but the latter is especially useful in many applications. 
 
 
 
 n 
 
 EXERCISES. 
 
 1. How many combinations can be made from 9 things 3 
 atatime? 5 ata time? 
 
 2. Find the values of Cy, Co, 12C3. 
 
 -3. In a meeting of 20 people, in how many ways can a 
 committee of 5 be selected ? 
 4, A school is composed of 19 boys and 25 girls. In how 
 many ways can a committee consisting of 1 boy and i girl 
 be selected ? 
 
 5. From the above school, how many comm.ttees consisting 
 of 2 boys and 1 girl can be selected ? 
 
 6. From 15 persons, how many committees of 5 can be 
 formed, provided one particular person is to be a member of 
 every committee ? 
 
 7. If out of 9 candidates there are to be 5 officers elected, 
 how many different tickets can be formed ? 
 
 8. From 4 vowels and 8 consonants, in how many ways can 
 5 letters be chosen, provided exactly 2 of them are vowels ? 
 Provided at least 2 of them are vowels ? 
 
 9. How many even numbers of 4 places can be formed from 
 the digits 1, 2, 3, 4, 5, 6, 7, 8? 
 
 Recs. C1; find n. 
 
CHAPTER XIX. 
 SERIES. 
 
 200. General Definitions. (1) Any set of numbers is an 
 array, Or Succession. 
 
 (2) A series is a succession of numbers arranged ac- 
 cording to some law. 
 
 Thus, 1, 2, 3, 4, 5, 6, ---, is a series, the law of formation 
 being that any number is to be had from the preceding 
 by adding 1. 
 
 We may also define a sertes as a succession of numbers, 
 the knowledge of two or more successive ones being suffi- 
 cient to determine all. 
 
 Thus, 5, 7, 9, 11, ---, form a series, since by inspection 
 of any two we see their difference to be 2; hence, any 
 number of the series may be had from the preceding by 
 adding 2. 
 
 (8) The numbers forming a series are called the terms 
 of the series. 
 EXERCISES. 
 What law of formation exists in each of the following? 
 1, 2, 4, 6, 8, 10, ---. 3. 4, 1, §, 2, §,.3, £, See 
 
 2. 5, 10, 20, 40, 80, ---. 4. 3, 3, 2, 3, Be, +. 
 278 
 
  — ae 
 
th i ti 
 
 SERIES. 279 
 Bani 1 a, Oy 1, Os 22% 
 6. a, a+d, a+2d, a+3d, »-, a+ (n—1)d. 
 7. a, ar, ar*, ar, «+, ar", 
 gs. 5, —15, 45, —135, 405, — 1215, 
 y oe x ag a 
 
 Ben ey (8, 
 
 nig ta tat 
 
 ee 8. 
 
 (4) When the number of terms of a series ts finite, the 
 serves is called a finite series. 
 
 Thus, 2, 5, 8, 11, 14, is a finite series. 
 (5) When the number of terms is infinitely great, the series 
 ws called an infinite series. 
 
 Thus, if a series be formed by making any term the 
 half of the preceding term and this process be continued 
 indefinitely, as, 
 
 
 
 a ws 1 1 EE a eee 
 see eS S716) Ba. "2 
 we have an infinite series. 
 
 (6) If the sum of n terms of an infinite series can be 
 shown to approach some finite number as n 1s made to 
 approach infinity, the. series is called convergent. If this 
 sum can not be shown to approach some finite quantity, the 
 series is called divergent. 
 
 In the discussion of the subject of series, we shall 
 examine only three special forms, the arzthmetical series 
 the geometrical series, and the binomial series. 
 
280 THE ESSENTIALS OF ALGEBRA 
 
 Y. ARITHMETICAL SERIES (ArirumMeticaLt ProGREssion). 
 
 201. Definition. An arithmetical series 1s a series in 
 which the difference of any two successive terms 28 a constant. 
 
 Illustrations. (1) 5, 9, 18, 17, 21, ---, is an arithmetical 
 series, since the difference of any two successive terms is 4. 
 
 (2) 8, 8.5, 4, 4.5, 5, 5.5, +--+, is arithmetical, since the 
 difference is .5. 
 
 (3) a,at+d,a+2d,a+5d,a+4d, --,a+(m—1)d, is 
 arithmetical, since the difference of any two succes- 
 sive terms is d. In this illustration @ is taken as any 
 algebraic number, commensurable or not, and d likewise 
 as any algebraic number. 
 
 202. Notation. We shall denote by a the first term of 
 any arithmetical series, by d the constant difference (com. 
 mon difference) between any two successive terms, by / 
 the last term or nth term, and by S' the sum of n terms of 
 the series. 
 
 203. Fundamental Formulas. 
 (1) l=a+(m—1)d. 
 
 (2) Sat xn 
 
 = FAt aE xn, 
 
 In these two relations five letters are involved, any two 
 of which may be unknown. 
 
 The first of the above formulas is easily seen to be true 
 from the manner of formation of the general arithmetical 
 series shown in illustration (8) above. 
 
SERIES. 281 
 
 To derive formula (2), we write the series 
 Ssat(atd)+(a+2d)+- panko! 
 Then reverse the series, 
 S=1+d—d)+(U—-—2d)+--+(a+2d)+(a+d)+a, 
 and add the two equalities, giving 
 28=(a+)+(a+)D+@4+)D+(a4+)+ 
 +(a+l)+(a+l) 
 = n(a+1), since there are n terms in the series. 
 
 i genGek: l 
 2 
 The second form of formula (2) is derived by replacing 
 1 by its value from formula (1). 
 
 xn. 
 
 
 
 204. Arithmetical Mean. If a, b,c be three successive 
 terms forming an arithmetical series, 6 is called the arith- 
 metical mean of a and e. 
 
 =4(a+e), for by the definition of the arithmetical 
 series, 
 
 b—a=c—); 
 (transposing), 2b6=a+e, 
 or b=4(a+e). 
 
 205. Arithmetical Means. In an arithmetical series a, 4, 
 ce, d, e, f,--:,l, the terms J, ¢, d, e, f, +++, are called the arith- 
 metical means ota and J. 
 
 206. To insert k arithmetical means between any two 
 numbers. 
 
 Let a and 6 be any two numbers. After * means have 
 been inserted, the whole series will consist of # + 2 terms. 
 
282 THE ESSENTIALS OF ALGEBRA. 
 
 Hence, 6 is the last or (k + 2)” term of an arithmetical 
 series of which a is the first, and d an unknown common 
 difference. 
 
 
 
 Hence, b=a+(k+2-l1)d 
 =a+(k+1)d, 
 Bs eed 
 Ray 
 
 The common difference d being known, the series may 
 be easily written thus: 
 
 9) Be 
 a 44 20 a) 
 
 bis eG; 
 k+l eee 
 
 a, 
 
 
 
 207. An arithmetical series is determined when two of its 
 terms are known. 
 
 Let a be the kth and 6 be the mth term of an arith- 
 metical series. Let a be the first term and y the common 
 difference. 
 
 Then a=x+(k—-1)y, 
 
 b=x+(m—1)y. 
 By subtraction 
 b—a=(m—k)y, 
 b— 
 
 or v= oe (the common difference), 
 m — 
 
 
 
 by Ge a\| _(m—l)a—(k-1)@ 
 and x=a—(k—1) {= iy [ = 
 The first term 2 and the common difference y being 
 known in terms of a, 6, &, m, the series may be written 
 down. 
 Determine the arithmetical series in which the 5th term 
 ig 17, and the 12th term is 88. 
 
SERIES. 283 
 
 SOLUTION. 
 Let x = first term, and let y = common difference. 
 
 Then =a-+ (n—1)d becomes respectively, 
 17=x+(5-1)y, 
 aus (12—-1)y. 
 By subtraction, Bl =i y, OY ¥ = 3; 
 when Ep ria 
 Then the series is 5, 8, 11, 14, 17, 20, ---, 35, 38. 
 
 EXERCISES. 
 Find the 18th term of 2, 5, 7, 10, ete. 
 Sum 4, 7, 10, etc., to 9 terms. 
 Insert 5 arithmetical means between 10 and 34. 
 
 PO B 
 
 Find the 15th term of an arithmetical series whose 2d 
 and 8th terms are 9 and 21, respectively. 
 
 5. Which term of the series 1, 6, 11, 16 is 96? 
 
 6. Find the sum of the natural numbers from 91 to 187. 
 
 7. Show that if any four numbers are in arithmetical pro- 
 gression, the sum of the Ist and 4th is the same as the sum of 
 the 2d and 3d. 
 
 8s. Find the 18th term of 27, 21, 15, 9, ete. 
 
 9. Find the sum of 12 terms of 3, 44, 6, 74, ete. 
 
 10. How many terms of 1, 2, 3, 4, etc., will make 465 ? 
 
 11. How many terms of 7, 11, 15, 19, etc., will make 297 ? 
 
 12. How many strokes does a clock strike in 12 hours ? 
 
 13. Find the sum of all the even numbers from 100 to 200 
 inclusive. 
 
 14. Find the sum of all the numbers from 48 to 135 inclu- 
 sive which are divisible by 3. 
 
 15. What debt could be paid in a year by the payment of 
 10 ¢ the 1st week, 40¢ the 2d week, 70 ¢ the 3d week, etc.? 
 
284 THE ESSENTIALS OF ALGEBRA. 
 
 16. Determine the series whose 10th term is 51 and whose 
 20th term is 101. 
 
 17. Determine a series whose 15th term is 0 and whose 31st 
 term is 64. 
 
 18. Find the sum of all numbers from 105 to 361 inclusive, 
 which, when divided by 4, leave a remainder of 1. 
 
 Il. GEOMETRICAL SERIES (GEOMETRICAL PROGRESSION). 
 
 208. Definitions. A geometrical series is a series in which 
 the ratio of any two successive terms ts a constant. 
 
 Illustrations. (1) 2, 4, 8, 16, 32, is a geometrical series 
 in which the ratio 16+8=8+4=32+16=2 is a constant. 
 
 (2) J, 4, 4, a gh +) 18 a geometrical series with ratio 
 equal to 4. 
 
 (3) 1, x, 2*, 23, at, ---, is a geometrical series with ratio 
 equal to z. 
 
 (4) a, ar, ar’, ar, art, ---, ar""|, is a geometrical series 
 in which the ratio is 7. 
 
 209. Notation. ‘The illustration (4) above suggests a 
 notation for the geometrical series. 
 (1) a = first term. 
 (2) r =constant ratio. 
 (3) J = last term, or nth term. 
 (4) S=sum of n terms. 
 210. Formulas. 
 CO eat oe 
 
 (2) S=a("=7)=a(5=") 
 
 (3) S= j “_, when r <1, and n=a, 
 —r 
 
 
 
 
 
 
 
SERIES. 285 
 
 The first of these formulas results from the law of forma- 
 tion of the series as indicated in illustration (4) above. 
 The second formula we may derive as follows: 
 
 GQ) S=a+ar+ar4+ ar+---+ ar? + art 
 Multiply by 7, 
 
 (2) Sr=ar + ar? + are +--+ +ar"* + ar”! + ar". 
 Subtracting (2) from (1), 
 
 (3) S—Sr=a-— ar". 
 
 (4) SQ-r)=ad—>2). 
 
 (5) ieee x cz “). 
 
 
 
 —?r 
 Another method of derivation is worthy of attention. 
 By actual division we know that 
 
 
 
 
 
 
 
 ats: 
 1 pee te 71 72, 
 
 l-—r 
 
 Seid 
 
 oe (a pte 
 1l—r 
 
 eu, 5 
 
 I Peltr4r4r4+7, 
 1l—r 
 
 and so on; for the general case, 
 
 Foal rr rt oe bm 
 4 8 eet i 
 
 Now by writing the value of S again, 
 S=a+ar + ar? + ar? + art + ee + ar™? + ar, 
 and factoring out a from each term on the right, we have 
 S=a(l trp ret ro foe for? rr), 
 The value in this bracket is the same as the value of 
 
 above; hence, Pa 
 Sean. ( ) 
 
 et 2” 
 
 
 
 ee et 
 
 
 
 me fe 
 
 
 
286 THE ESSENTIALS OF ALGEBRA. 
 
 211. Sum to Infinity when r<1. If the ratio 7 be less 
 
 : a 
 than unity, r°<1,and whenn=o,7r"=0. Hence;s= f 
 
 | ; —r 
 when r< ll, andn=o 
 
 
 
 The sign = is read approaches. 
 Illustration. Find the sum of 1+4+4+ $4 ete. 
 
 1 
 i ae 
 
 
 
 1 
 Here, a=1,r= = and S'= 
 _ 
 
 212. The Geometrical Mean. Jf a, 6, c, be three successive 
 terms of a geometrical series, then 6 is equal to the square 
 root of the product of a by c, and is the geometrical mean of 
 a and e. 
 
 By the definition of a-geometrical series, 
 b 
 
 c — 
 Fare. whence, 6? = ac, or b = Vae. 
 a 
 
 213. Geometrical Means. Jn a geometrical series the 
 terms lying between any two terms are called the geometrical 
 means of those two terms. } 
 
 Thus, 5, 10, 20, 40, 80, 160, are six terms of a geometrical 
 series with ratio 2. The terms 10, 20, 40, 80, are the four 
 geometrical means between 5 and 160. 
 
 214. Insertion of Geometrical Means. Any number of geo- 
 metrical means may be inserted between any two numbers. 
 
 Proof. Let aand 6b be any two numbers, and let k be 
 the number of means to be inserted. Then 6 is the 
 (k+2)” term of a geometrical series, whose first term 
 isa. Ifr be the unknown ratio, 
 
 Writ ipl hinge cmengs Or 
 
 ip eee slog 
 sae" 
 
SERIES. ei ORF 
 
 Insert five geometrical means between 128 and 2. 
 These means may be written down if we know the value 
 of the ratio. ‘This is given by 
 
 pat N{5, where = 5, b= 128, a=2, 
 a 
 " AL Vv 64 = 2. 
 
 Hence, the series is 2, 4, 8, 16, 82, 64, 128, and the 
 means are 4, 8, 16, 32, 64. 
 
 215. A Geometrical Series Known. <A geometrical series 
 is known when any two terms are known. 
 
 Proof. Leta be the &th, and 7 the mth terms of a geo- 
 metrical series; and let 7 be the unknown ratio, with zx 
 as first term. 
 
 Then a= zr'-}; 
 and Pere. 
 
 These two equations are sufficient to determine the first 
 term x and the ratio r. 
 
 ‘ ps ey 1 
 By division —— =-, 
 Co ee 
 or prern. &, 
 a 
 m—k l 
 (1) ° 7 = i, 
 
 
 
 Equations (2) and (1) give the’ first term and ratis, 
 respectively, of the required series. 
 
288 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES. 
 
 1. Find the 10th term of 1, 2, 4, 8, etc. 
 2. Find the sum of the 10 terms in (1) above. 
 
 3. The 5d and 6th terins of a geometrical series are 27 and 
 729. Find the 8th term and the sum of the 8 terms. 
 
 4, Find the sum of 10 terms of 1, 4, 4, 4, ete. 
 5. Sum 3, — 3’, 3°, — 3‘ to 8 terms. 
 
 6. A house with 8 windows was sold for $1 for the 1st 
 window, $2 for the 2d, $4 for the 5d, ete. What was received 
 for the house ? 
 
 7. If you receive $5 Jan. 1, $10 Feb. 1, $20 March 1, and 
 so on for each month of the year, what is the total amount you 
 will receive during the year ? 
 
 
 
 8. Find the sum to infinity of 535, 735, zara ete. 
 9. Find the sum to infinity of 1, 4, 4, 34, ete. 
 
 10. The arithmetical mean of two numbers is 13 and their 
 geometrical mean is 12. Find the numbers. 
 
 11. Show that the series of alternate terms of a geometrical 
 series is also a geometrical series. 
 
 12. If every term of a geometrical series is divided by the 
 same quantity, the quotients form a geometrical series. 
 
 13. The reciprocals of the terms of a geometrical series form 
 a geometrical series. 
 
 14. The difference between the 1st and 4th of four numbers 
 in geometrical progression is 208, and between the 2d and 3d 
 is 48. Find the numbers. 
 
 15. The sum of 3 numbers in geometrical progression is 14 
 and the sum of their reciprocals is 7. Find the numbers. 
 
a 
 
 SERIES. 2Q8y 
 
 Ill. BINOMIAL SERIES. 
 216. Definition. The series defined by 
 
 (a + b)® =a" tna + ey LD gr-2h2 pose 
 
 4 a ae 3) oe (M=P ED) nrg. 
 T 
 is called the benomial series. 
 Nore. In higher algebra it is shown that this series gives a true 
 
 value of (a+6)” for all values of a and b provided n is an integer. It 
 also defines (a +b)" properly when n is a negative number, or a 
 
 fraction, provided ® bea proper fraction. The general proof of these 
 a 
 
 assertions will not be attempted in this development. 
 
 EXERCISES. 
 
 ieee inthis n= 30, ¢— 1, b'= 2. 
 
 
 
 (1+ 2)=143004 0%) wg AXES a® + ete. 
 
 2. Write out the first 6 terms of (1+2)"; (1—y)"; (w+y)” 
 3. Find the coefficient of 2” in (1+). (Referring to the 
 binomial series, we see that 7 = 15.) 
 20 0 CS Co ara a Ga 
 — 3)(—3—1)(—3— 2) 
 a rca v ea ni (— 2) 
 
 
 
 =14+-32+4+627 + 102°- ete. 
 5. Write out the first 5 terms of 
 
 (a) A—2)7; (&) +2); © G—-a)?; @ A+ay. 
 
290 THE ESSENTIALS OF ALGEBRA. 
 
 EXERCISES— MISCELLANEOUS. 
 1. Expand (w#+2y)°; (2a—y)*®; (2a—38b)). 
 2. Find the sum of 8 terms of 1, 2a, 42%, «+. 
 3. Find the sum of 30 terms of 7, 11, 15, «+». 
 4. Find the 6th term of (1—2a)~*. 
 
 5. How many arithmetical means must be inserted between 
 10 and 40 so that the sum of the series may be 275? 
 
 6. Divide 26 into three parts which are in geometrical pro- 
 gression, and such that when 4 is added to the second part, the 
 three parts are in arithmetical progression. 
 
 7. Insert a geometrical mean between 5 and 45. Explain 
 the double sign of the result. 
 
 8. Sum to infinity 3, 2, 38, +. 
 9. Find the coefficient of ay’ in (24—3 y)™ 
 10. Find the coefficient cf a in (1 — 2a)-?, 
 
 11. In how many ways can 9 persons be selected from a 
 party of 21 people? 
 
 12. How many commiitees consisting of 4 men and 3 women 
 can be formed from 12 men and 10 women? 
 
 13. Each member of a baseball nine, except pitcher and 
 catcher, can play in any position. In how many ways can the 
 team be arranged upon the field ? 
 
 14. In how many ways may a baseball nine be selected from 
 16 candidates, if 2 are pitchers, 3 are catchers, and the remain- 
 der can play in any position ? 
 
 15. Expand (4 — 23 to five terms, and thus get an approxi- 
 mate value of 2. 
 
SERIES. 291 
 
 16. Expand (100-1)? to five terms, and thus get an approxi- 
 mate value of 99. 
 
 a 6 
 17. Expand (1 +0) . 
 
 1s. Find the sum of all the even numbers between 205 and 
 341. 
 19. Find the sum to infinity of 1 ++ +5 t + —— 1 — -[- ove, 
 V2 2/2 
 _ 20. The sum of the first three terms of a geometrical series 
 is 21, and the sum of their squares is 273. Find the series. 
 
 21. A debt is to be paid by 10 payments which form an 
 arithmetical progression. The third payment is $220, and the 
 seventh is $360. Find the last payment and the total debt. 
 
 22. How many consecutive odd numbers beginning with 11 
 must be taken to make a sum of 759? 
 
 23. Prove that the squares of the terms of a geometrical 
 series also form a geometrical series. 
 
 24. (Va+1—Va—1)'= what? 
 25. Expand (1+2+4 27). 
 
CHAPTER XX. 
 LOGARITHMS. 
 
 217. Definition. The exponent with which a given num- 
 ber a must be affected in order to produce another number N 
 ts called the logarithm of N to the base a. 
 
 Hxamples : 
 
 (1) The logarithm of 32 to the base 2 is 5, for 2°=382. 
 (2) The logarithm of 125 to the base 5 is 3, for 5°=125. 
 (3) The logarithm of 100 to the base 10 is 2, for 107= 100. 
 
 218. Notation. log 
 
 oa 
 
 N means the logarithm of WV to the 
 base a. 
 The two equations, 
 
 a*= N and x=log, NV are’ equivalent. 
 
 219. System of Logarithms. The logarithms of all 
 numbers to any given base constitute a system of loga- 
 rithms. Any positive number other than unity may be 
 used as the base. Hence, there may be an infinite number 
 of logarithmic systems. ‘There are only two systems in 
 common use, the Napiertan and the Common. ‘The Napier- 
 ian system uses the base e (2.7182818). This system is 
 used in analytical and theoretical discussions and is but 
 rarely used in caleulation. The Common or Briggs system 
 uses the base 10. ‘This system is the one in general use. 
 
 292 
 
LOGARITHMS. 293 
 
 220. Properties of Logarithms. 
 (1) The logarithm of 1 is 0. 
 
 Since a°=1 for all finite values of a, log 1=0 for all 
 bases. 
 
 (2) The logarithm of the base itself is unity. 
 
 Since tea, lov a = 1. 
 
 (3) The logarithm of a product is equal to the sum of the 
 logarithms of its factors. 
 
 Let F, and F, be the factors, and put #, = a" and F, = a’. 
 Pieter, and y—log, i. Fx £,=ata =a. 
 log, (Ff, x £,) =2+ y= log, Ff, + log, J. 
 
 Similarly, log, (f’, x F, x #3) =log, #, + log, F,+log, F;. 
 
 The property is true for any number of factors. 
 
 Ex. log,, 210 = log,, (2x 3 x0 x 7) =log,, 2 + log,, 3 
 + logy 5 + log, 7. 
 
 Ss 
 
 (4) The logarithm of a quotient is equal to the logarithm 
 of the dividend minus the logarithm of the divisor. 
 
 Let #, and F, be dividend and divisor respectively and 
 represent them as in (3). 
 
 > 
 
 log, (3) =z—y=log, lf, —log, F,. 
 
 BD semen eS hast ~) 
 Ex. logy 13 a log 49 35 logy 13: 
 
 ah pee eT. ou tes 
 _ logy, 5 + logy 7 logy 2 logy 2—log,, 3. 
 
294 THE ESSENTIALS OF ALGEBRA. 
 
 (5) The logarithm of the nth power of a number is equal 
 to n times the logarithm of the number. n may be integral 
 or fractional. 
 
 If WV” be the given power, 
 put IViexnges 
 then, Nee 
 
 log, NV" = nz = nlog, WV, since — oma 
 Ex: log,,310% =o lop nolo. 
 logy) V125 = logy 125% = 4 logy 125. 
 
 221. Characteristic and Mantissa. 
 The equation for Common Logarithms is 107 = WN. 
 This equation shows that the value of 2 (the common 
 logarithm) will not in general be integral and that it may 
 be positive or negative. 
 $756 > 10°sand <a10 
 
 Hence, log 8756 = 3 + a fraction. 
 058 > 10-2 and 10 3 
 Hence, log .053 = — 2 +4 a fraction. 
 
 The Characteristic is the integral part of a logarithm. 
 The Mantissa is the fractional part of a logarithm. 
 
 222. To find the Characteristic : 
 (1) Of the logarithm of a number greater than unity. 
 
 LOO ein 
 0} ech) 
 1.07100 
 103 = 1000 
 
 10" = 1 with n ciphers annexed. 
 
LOGARITHMS. 295 
 
 A study of the above table shows that the logarithm of 
 a number between 1 and 10 is 0+a fraction; between 
 10 and 100, 1+ a fraction; between 100 and 1000, 2+a 
 fraction; and so on. 
 
 If V is a number whose integral part has n digits, we 
 
 “may write N a 190@-D+ a fraction 
 Hence, log V = (n—1)+ a fraction. 
 
 The characteristic of the logarithm of a number greater 
 than unity ts one less than the number of digits in its 
 entegral part. 
 
 Ex. log 8542 = 3 + a fraction. 
 log 96.54 = 1 + a fraction. 
 
 (2) Of the logarithm of a decimal fraction. 
 
 Aer ete 
 tulle Or 4 
 OU tree UEs 
 S000 Peers 1. OQ 
 
 .(n ciphers) 1 = 10-@+» 
 
 An inspection of the above table shows that the logarithm 
 of anumber between 1 and .1 is —1+ a fraction; between 
 -Land .01 is —2+a fraction; between .01 and .001 is —3+ 
 -afraction; between .001 and .0001 is —4+a fraction; of any 
 decimal beginning with n ciphers is —(n+1)+a fraction. 
 
 Let V = a decimal fraction beginning with n ciphers. 
 
 Then, N = 10-@+) +2 fraction, 
 log MV =— (n+ 1) + a fraction. 
 
296 THE ESSENTIALS OF ALGEBRA. 
 
 The characteristic of the logarithm of a decimal fraction is 
 one more than the number of ciphers immediately after the 
 decimal point, and is negative. 
 
 Ex. log .0856 = — 2 + a fraction. 
 log .00058 = — 4 + a fraction. 
 
 EXERCISES, 
 
 Determine the characteristic of the logarithm of each 
 of the folowing numbers: 
 
 Dips rs ire AL MAS Lovo 7. .0000031. 
 2. 95.64. Breas 8. 231.416. 
 3, aD: 625.005 (0 9. (63454. 
 
 10. 1375.603. 
 
 223. The mantissa depends upon the sequence of digits, 
 and not upon the position of the decimal point. 
 The logarithm of 4880 is 3.6415. 
 10345 = 4380 or log 4380 = 3.6415. 
 
 Dividing in succession by 10, 
 1025 — 488. or. Jop 436 =e 
 1015 — 43:8 or log 43.8 == JeGci. 
 10°45 = 4.38 or log 4.38= 0.6415. 
 10-145 = 4388 or log .438=—1+4.6415. 
 10-2+-6415 — 0438 or log .0488 =— 2+ .6415. 
 - 10-8+ 6415 — .00438 or log .00438 = — 3 + .6415. 
 
 This table makes it evident that the mantissa does not 
 change so long as the sequence of digits is unchanged. 
 
LOGARITHMS. 297 
 
 EXERCISES. 
 log 456 = 2.6590; what is log 456000 ? 
 log .987 = —1+4.9943; what is log 9870? 
 log 57.9 = 1.7627; what is log .000579? 
 log 3210 = 3.5065; what is log 3.21? 
 log .0705 =— 2+ 8482; what is log 7050000? 
 log 7.350 is .8663; what is log 73500 ? 
 log 2 is .8010; find the log 2%, 24, 2°. 
 log 3 is .4771; find log 3%, 3%. 
 From examples 7 and 8 find the log 6; log 24. 
 log 5 is .6990; find log 25, log 125. 
 From data above given find log 20, log 75. 
 . log Tis .8451; find log 49, mg 343. 
 Find log 42. 
 
 Find log 140. 
 reed a x oF 
 
 Oe Oe Se eee 
 
 KH YH eH eS eH 
 Piet OTe RE tN 
 
 mn 
 
 Find log 
 
 USE OF TABLES. 
 
 224. Approximate Arithmetical Calculations involving 
 multiplications, divisions, raising to powers, and extraction 
 of roots may be greatly abridged by the use of Logarith- 
 mic Tables. Such tables have been constructed showing 
 the Common Logarithms of numbers from 1 to 10,000. 
 We give a four-place table showing the mantissas of the 
 logarithms of numbers from 1 to 1000. 
 
THE ESSENTIALS 
 
 OF ALGEBRA. 
 
 
 
 
 
 O 
 
 1 
 
 Det 
 
 
 
 3 
 
 
 
 4 
 
 
 
 5 6 7 8 9 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 0000 0043 0086 0128 0170 | 0212 0253 0294 0834 0874 
 11 | 0414 0453 0492 0531 0569 | 0607 0645 0682 0719 0755 
 2 | O792 0828 0864 0899 0934 | 0969 1004 1038 1072 1106 
 13 | 1189 1173 1206 1239 1271 | 1303. 13337 “Jae; eae 
 14 | 1461 1492 15293 1553 1584 | 1614 1644 1673 "u70sea7ee 
 15| 1761 1790 1818 1847 1875 | 1903 1931 1959 1987 2014 
 16 | 2041 2068. 2005 2122 2148 | 2175 22901 2997) 9953) 99070 
 17 | 2304 2330 ~ 2355 2380 2405 | 2480 2455 2480 2504 9599 
 18 | 2553 2577 2601 2625 2648 | 2672 2695 2718 2742 2765 
 19 | 2788 2810 2833 2856 2878 ; 2900 29238 2945 929067  ~90R9 
 20; 3010 3032 3054 38075 3096 | 3118 3139 3160 3181 3201 
 Y1 | 3222 3243 3263 3284 3304 | 3324 3845 3365 3385 3404 
 29 | 3494 3444 3464 3483 4502 | 3592 3541 3560 3579 3598 
 23 | 3617 3636 3655 3674 3692 | 3711 3729 3747 VaiGeuneres 
 24 | 3802 3820 3838 3856 5874 | 3892 3909 3927 3945 3962 
 25 | 3979 38997 4014 4031 4048 | 4065 4082 4099 4116 4133 
 26 | 4150 4166 4183 4200 4216 | 4232 4249 4265 4281 4298 
 97 | 4314 4330 4346 43862 4878 | 4893 4409 4425 4440 4456 
 98 | 4472 4487 4502 4518 453: 4548 4564 4579 4594 4609 
 99 | 4624 4639 4654 4669 4683 | 4698 4713 4728 4742 4757 
 30 | 4771 4786 4800 4814 4829 | 4843 4857 4871 4886 4900 
 3 4914 4928 4942. 4955 4969 | 4983 4997 5011 5024 5038 
 32 | 5051 5065 5079 5092 5105 | 5119 5132 5145 5159 5172 
 33. |°5185 5198 5211 5224 52937 | 5250 5263 5276 5289 5302 
 34 | 5315 5328 5340 5353 5366 | 5378 5391 5403 6416 5428 
 35 | 5441 5453 5465 5478 5490 | 5502 5514 5527 5539 5551 
 36 | 5563 5575 5587 5599 5611 | 5623 5635 5647 5658 5670 
 37 | 5682 5694 5705 5717 5729 | 5740 5752 5763 5775 5786 
 38 | 5798 5809 5821 5832 5843. | .5855 5866 5877 5888 5899 
 39 | 5911 5922 5983 5944 5955 | 5966 5977 5988 5999 6010: 
 40 | 6021 6031 6042 6053 6064 | 6075 6085 6096 6107 6117 
 41 | 6128 6138 6149 6160 6170 | 6180 6191 6201 6212 6222 
 2 | 6232 6243 6253 6263 6274 | 6284 6294 6304 6314 6825 
 3 | 6835 6345 6355 6365 6375 | 6385 6395 6405 6415 6425 
 44 | 64385 6444 6454 6464 6474 | 6484 6493 6503 6513 6522 
 45 | 6532 6542 6551 6561 6571 | 6580 6590 6599 6609 6618 
 46 | 6628 6637 6646 6656 6665 | 6675 6684 6693 6702 6712 
 47 | 6721 673 6739 6749 6758 6767 6776 6785 6794 6803 
 48 | 6812 6821 6830 6839 6848 | 6857 6866 6875 6884 6893 
 49 | 6902 6911 6920 6928 6937 | 6946 6955 6964 6972 6981 
 50 | 6990 6998 7007 7016 7024 | 7033 7042 7050 7059 ‘7067 
 51 | 7076 7084 7098 7101 7110 | 7118 7196 “V13h Jide 
 yn Aaa MrGiaeh. WON” <item 7202 7210 7218 7226. ssieee 
 53. | 7243 7251 . 7259 7267 7275 | 7284 7292 7300 . 7308 ue 
 54 | 7324 7332 7340 7348 7356 | 7364 7372 7380.) 7oSeeue 
 N O 1 2 3 4 5 6 v4 8 9 
 
 
 
 
 
 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 LOGARITHMS. 299 
 
 N]/] O 1 2 3 4 5 6 7 8 9 
 
 55 | 7404 7412 7419 7427 7435 (443 7451 7459 7466. 7474 
 56 7482 7490 7497 7505- 7513 TOLD, Des) TbS6r' (ozo UL DDL 
 oT 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 
 58 7634 7642 7649 7657 T664 7672 T679 7686 7694 £7701 
 59 T1709 7716 7723) T7310 1738 | T1745 «T1752. «7760)=— TT6T)~—s T7174 
 60 | 7782 7789 7796 7803 7810 (818 7825. 7832) 7839 7846 
 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 
 62 7924 7931 ° 7988 7945 7952 7959 7966 7973 7980 T7987 
 63 7993 8000 8007 8014 8021 8028 80385 8041 8048 8055 
 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 
 65] 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 
 66 8195 8202 8209 8215 3292, 228), (S200 |) S241 8248 8254 
 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 
 68 8395 8331 83838 8344 8351 8357 83863 8370 8376 £8382 
 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 
 7O)} 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 
 V1 8513 80919 8525 8531 853 8543 8549 8555 8561 £8567 
 72 8573 8579 8585 # £8591 597 8603 8609 8615 £8621 8627 
 73 8633 $8639 8645 8651 8657 8663 8669 8675 8681 8686 
 74 8692 8698 8704 8710 8716 SLO hk IEOloon (OlOo. 6 O40 
 75 )| 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 
 76 8808 8814 8820 8825 8531 8837 8842 8848 8854 8859 
 Lit 8865 8871 8876 8882 8887 88938 8899 8904 8910 8915 
 78 8921 8927 89382 8938 8943 8949 $8954 8960 8965 8971 
 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 
 80 | 9031 90386 9042 9047 9053 9058 9063 9069 9074 9079 
 sl 9085 9090 9096 9101 9106 9112 117 9122 9128 9135 
 82 O1388 °° 9143 9149 9154 9159 9165 9170 9175 9180 9186 
 83 9191 9196 9201 9206 9212 OO Ty: 92225 9227) 19232. 9235 
 84 0243 9248 9253 9258 9263 9269 9274 9279 9284 9289 
 85 | 9294 9299 9304 9309 9315 9320 93825 9330 93835 9340 
 86 | 9345 9350 9355 9360 9365 | 9370 9375 9380 9385 9390 
 87 | 9395 9400 9405 9410 9415 | 9420 9425 9430 9435 9440 
 88 | 9445 9450 9455 9460 9465 | 9469 9474 9479 9484 9489 
 g9 | 9494 9499 9504 9509 9513 | 9518 9523 9528 9533 9538 
 90 | 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 
 Ol 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 
 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 
 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 
 94. 9731 97386 9741 9745 £9750 9754 9759 9763 9768 9773 
 951 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 
 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 
 O7 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 
 98 9912. 9917 9921 9926 9930 9934 9939 9943 9948 9952 
 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 
 N O 1 2 3 4 5 6 7 8 9 
 
300 THE ESSENTIALS OF ALGEBRA. 
 
 225. To find the Logarithm of a Given Number. The 
 logarithm of any number consisting of three consecutive 
 digits has its mantissa given in the table, and its charac- 
 teristic is found by Art. 222. 
 
 Hxamples: 
 
 (1) Find log 358. } 
 
 The first two digits, 35, are found in the column marked 
 NV, and the third digit, 8, is found in the top row. 
 
 In the row opposite 35, and in the column under 8, we 
 find the number 5539, which is the mantissa of the logarithm 
 of 858. By Art. 222, the characteristic is 2. 
 
 Hence, log 308 = 2.5539, 
 
 (2) Find log 755. 
 In the row opposite 75, and in the column under 3, we 
 find 8768. The characteristic is 2. 
 
 Hence, log 753 = 2.8768. 
 
 (3) Find log 7.53. 
 The mantissa is the same as for 753, and the character- 
 
 Hence, log 7.55 = 0.8768. 
 
 (4) Find log .0758. 
 
 The mantissa is the same as in (2) and (8), and the 
 characteristic is — 2. 
 
 Hence, log .0753 = — 2+ .8768. 
 
 This is usually written 2.8768. 
 The negative characteristic is frequently made positive 
 by adding and subtracting 10. 
 
 Thus, log .0753 = 2.8768 = 8.8768 — 10. 
 
LOGARITHMS. 301 
 
 The characteristic may be retained as a negative, but 
 the mantissa is usually written positive, and is not sub- 
 tracted from the negative characteristic. 
 
 (5) Find log 3965. 
 
 The mantissa for 396 is .5977 and for 397 it is .5988. 
 The difference is .0011. The mantissa for 3965 is approxi- 
 mately .5977 + .5 x .0011 = .59825. Since we are to re- 
 tain but four places, this would be written .5983. The 
 characteristic is 3. 
 
 Henee, log 8965 = 3.5983. 
 
 By means of such approximations as above shown, the 
 accompanying table may be used to find quite accurately 
 the logarithms of numbers to 10,000. 
 
 (6) Find log 87.55. 
 
 The mantissa for 875 is .9420, and for 876 it is .9425. 
 The difference is .0005. ‘The mantissa for 8755 is approxi- 
 mately .9420 + .3 x .0006=.9422. The characteristic is 1. 
 
 Hence, log 87.53 = 1.9422. 
 
 226. To find the Number corresponding to a Given Loga- 
 rithm (Antilogarithm ). 
 
 Hxamples : 
 
 (1) Find the number whose logarithm is 2.6877. Find 
 in the table the given mantissa or the one nearest to it, 
 but less. | 
 
 5877 is found in the row opposite 38 and in the column 
 under 7. Hence, the three digits of the number are 587. 
 Since the characteristic is 2, the number is 387. 
 
 The number whose logarithm is 2.5877 is 387. This is 
 frequently written log~! 2.5877 = 387. 
 
302 THE ESSENTIALS OF ALGEBRA. 
 
 (2) Find the number whose logarithm is 1.8963. 
 
 In the table the nearest mantissa to 8963 is 8960. 8960 
 corresponds to the sequence of digits 787. ‘The difference 
 between 8960 and the given mantissa is 8. The difference 
 between 8960 and the following mantissa in the table is 5. 
 3+5=.6. The next digit in the sequence is 6. The 
 sequence of digits corresponding to the mantissa 8963 is 
 7876. ‘The characteristic 1 shows that there must be two 
 digits in the integral part. 
 
 Therefore, log-) 1.8963/==(3eia. 
 
 (3) Find the number whose logarithm is 3.5896. 
 
 The mantissa in the table next below 5896 is 5888, 
 which corresponds to the sequence of digits 388. The 
 difference between 5888 and the given mantissa is 8. The 
 difference between 5888 and the following mantissa in 
 the table is 11. 8+11=.7. This gives 7 as the fourth 
 digit of the sequence. Since the characteristic is — 3, 
 the first significant figure of the required number is in 
 the third decimal place. 
 
 Therefore, log! 3.5896 = .003887. 
 
 EXERCISES. 
 
 Find from the table the logarithms of each of the 
 following numbers: 
 
 Load. 5, 320.3, 9. .00056. 
 2. 984. 6. 948.6. 10. .5236. 
 3. 506. Be lkaavilie 11. 3.1416. 
 4. 1345. 8. .0679. 12) 2 
 
LOGARITHMS. 303 
 
 13, 1.40. 16. 3. 19. .3964.. 
 14. 122. 17. 984}. 20. .005876. 
 15. 37}. ; 18.79.0906. 
 
 Find the number corresponding to each of the following 
 logarithms : 
 
 21. 2.8136. 28. 3.2084. 35. 9.4669 — 10. 
 22, 1.9145. 29. 0.3756. 36, 1.4784. 
 
 23. 2.8904. 30. 1.5967. ST leon, 
 eae, (001, Bie pio, ag. 97/38945'— 10. 
 rime tna yee g28eo1 584: 39. 9.4896 — 10. 
 26. 0.4472. 3405, 5922. 40. 8.5986 —10. 
 OTe ako iO. 34. 8.6493 — 10. 
 
 227. To perform Arithmetical Operations by the Use of 
 Logarithms. 
 
 Examples : 
 (1) Find the value of 469 x 541 x .0878. 
 By Art. 220 (3), the logarithm of a product is equal to 
 the sum of the logarithms of the factors. 
 log 469= 2.6712 
 log 541= 2.7332 
 lope 0818 =~ 8.5775 — 10 
 log of product = 13.9819 — 10 = 3.9819 
 
 The product = log? 3.9819 = 9592. 
 
304 THE ESSENTIALS OF ALGEBRA. 
 
 5237 x Gi9dee 0059 
 
 By orinditherealie at 
 C2) HANG he alte Ob Gas ne 
 
 
 
 By Art. 220 (4), the logarithm of a quotient is equal to 
 the logarithm of the dividend minus the logarithm of the 
 
 divisor. 
 169025 (et oe 
 
 log 6.94= 0.8414 
 log .0059= 7.7709—10 
 
 Adding, log of numer- = 12.3314—10=12.3314—-10 (1) 
 ator (dividend) 
 
 log 497 = 2.6964 
 log .7648 = 9.8836 — 10 
 lop oF = 0S1G58 
 Adding, log of de- | 
 aay = 13.7159 —10 =i 9 
 nominator (divisor ) j : (4) 
 
 Subtracting (2) from (1), log of quotient = 9.5555 — 10 
 
 
 
 The quotient = log~!(9.5555 — 10) = 0.85983. 
 
 V578 x .038575 x V38.7 
 292 x V.875 x V8.58 
 
 
 
 
 
 (3) Find the value of 
 
 We apply the principles of Art. 220. 
 
 log V578 = 4 log 578=4 x2.7619= 1.3809 
 log .03575= 8.5533—10 
 log V38.7 =1 log 88.7=1 x 1.5877= 0.5292 
 Adding, log of numerator = 10.4634—10 (1) 
 
 
 
LOGARITHMS. 308 
 
 log 29? = 2 log 29 = 2 x 1.4624=2.9248 
 log V.375 = 4 log .875 
 
 = 4x (9.5740 —10) =4.7870—5 
 
 log V3.58 = d1og 3.58 =1 x 0.5539 = 0.1846 
 
 
 
 Adding, log of denominator =7.8964—5 
 = 2.8964 (2) 
 Subtracting from log of dividend 10.4634 —10 (1) 
 log of divisor 2.8964 (2) 
 
 
 
 log of quotient = 7.5670 — 10 
 
 The quotient = log !¢7.5670—10)= .00569. 
 
 EXERCISES, 
 
 By the use of logarithms evaluate the expressions given 
 
 below : 
 
 a 
 
 nt ee 
 
 6. 
 
 487 x 394 x 3.75. 
 
 57.84 x 589 x 427.5 x 52.36. 
 oTd4 x 69.54 x 16.66 x .000584. 
 17.58 x 125.6 « .00825 x 2.65. 
 1.578 x 3584 x 27.5 x .398. 
 
 3857 x 47.65 x 394, 
 467 x 89.84 x 3.875 
 
 
 
 7, BA416 x 13.6% x 87.5 9 -V946 x 39.872 x V4.749 
 
 57.6 x 39.57 
 .0236 x 34.63 
 
 
 
 V84.65 x 8.75% x 39.46 
 10. V658 x 47.688 x V380.57. 
 
 
 
 " 48.94 x 3.92. 4982 x .0388 x .00051 
 
 228. Exponential Equations. Hquations in which the 
 unknown quantity occurs as an exponent are called expo- 
 nential equations. ‘They are easily solved by logarithms. 
 
306 THE ESSENTIALS OF ALGEBRA. 
 
 Ex. Find the value of a in 267 = 137. 
 Taking the logarithms of both members of the equation, 
 ars log 25° = log 137. 
 x log 25 = log 187. 
 
 Plog oie od 
 
 
 
 Se isa oa sara 1.5285. 
 EXERCISES, 
 Solve the following equations : 
 Lael 6. Cleats": 
 Cue BX e100, 7... (00 aes 
 oy daub ap 8. (OC) * aa 
 Seo te Ope 9. 108 e*— ieee 
 Se ee 10. 3" 'X 574 
 
 229. Compound Interest. The computation of com- 
 pound interest may be easily performed by the use of 
 logarithms. : 
 
 Let P=principal loaned. 
 # = rate of interest. 
 n = time in years. 
 A,, = compound amount for n years. 
 Then A,=P+ PR=P(1+ £R), amount for one year. 
 A,=PA1+Rh)+P14+ 8)R=PA+ &)?, 
 
 amount for two years. 
 
 A, = P(1+ R&)", amount for n years. 
 
LOGARITHMS. 307 
 
 If the interest be compounded semi-annually, the com- 
 pound amount is R\2 
 a=r(te¥ 
 
 If compounded quarterly, it is 
 
 A, 2 PU i zy". 
 
 Examples: 
 
 (1) Find the amount of $850, compounded annually 
 for 8 years, at 5%. : 
 Ag= S00 (1+ .05)? = 350(1.05)° 
 
 log 850 = 2.9294 
 8 log (1.05) = 0.1696 
 log A, = 3.0990 
 
 A, = log! 3.0990 = $ 1256. 
 
 
 
 (2) Find the amount of $598, compounded semi- 
 annually for 6 years, at 4%. 
 
 PVs bet @ Deg be, 
 log 598 = 2.7767 
 12 log 1.02 = 0.1032 
 log A, = 2.8799 
 A, = log™1 2.8799 = $758.40. 
 
 
 
 (3) In how many years will $596, compounded annually 
 at 4%, amount to $978 ? | 
 From the general formula 
 Ae Cl fh), 
 
 we have GLE i, t= oe 
 
308 THE ESSENTIALS OF ALGEBRA. 
 
 n log (1+ &)= log A, — log P 
 " lop A; loge 
 log(1+ &) 
 
 Substituting the values of A, P, and R, we have 
 _ log 978 — log 596 
 log (1.04) 
 
 2.9903 — 2.7752 
 __ 2:9903 — 2.71702 _ 49 6 
 0.0170 PA 
 
 EXERCISES. 
 
 1. Find the amount of $1565, compounded annually 
 for 10 years, at 44%. 
 2. Find the amount of $1968, ere seml- 
 
 annually for 12 years, at 4%. 
 
 3. In what time will $965 amount to $1500 when 
 compounded annually at 33% ? : 
 
 4. What principal, compounded annually for 20 years _ 
 at 34%, will amount to $2500 ? 
 
 5. Find the time in which a given sum will double 
 itself when compounded annually at 3%. 
 
 6. Find the compound amount of $1 at 4%’for one 
 century. 
 
 7. Find the rate at which a given sum will double itself 
 in 20 years, if compounded annually. 
 
 8. Find the amount of $364, compounded annually for 
 12 yr. 5 mo. 10 da., at 5%. 
 
INDEX. 
 
 Abscissas, axis of, 148, 149. 
 
 Addition, 20-28, 35; identities, 26, 
 27; of fractions, 117, 118; in 
 elimination, 161. 
 
 Algebraic expression, 4; signs used 
 in, 5, 6; addition, 20, 21; sub- 
 traction, 28, 29; fraction, 109. 
 
 Arithmetic, numerical, 3; literal, 3, 
 4; addition in, 20; square root 
 in, 180, 181; cube root in, 184. 
 
 Arithmetical mean, 281. 
 
 Arithmetical means, 281. 
 
 Arithmetical series (progression), 
 280. 
 
 Associative law, definition of, 23; 
 of factors, 45. 
 
 Axioms, 16, 
 
 Binomial, definition of, 25 ; identi- 
 ties, 70, 71, 74, 75; theorem, 76 ; 
 factors, 100; series, 289. 
 
 Coefficients, definition of, 7, 8; de- 
 tached, 63; law of, 77; of a 
 quadratic, 222. 
 
 Combinations, examples of, 274; 
 definition of, 275; symbol of, 
 275. 
 
 Commutative law, definition of, 22 ; 
 of factors, 43. . 
 
 Constants, 15, 125. 
 
 Coordinate axes, 148, 149. 
 
 Cube root, 182, 184. 
 
 Distributive law, of factors, 45. 
 Division, 58-67 ; identities, 66, 81; 
 of fractions, 122. 
 
 Elimination, definition of, 157; in 
 substitution, 158; by compari- 
 son, 160 ; by addition or subtrac- 
 tion, 161; illustrative examples 
 in, 162-164, 
 
 Equation, definition of, 14; sign of 
 15 ; root of, 16, 126 ; conditional, 
 125; linear, 125, 147, 148, 170, 
 229; of second or higher degree, 
 140; fractional, 141, 148; inte- 
 gral, 143; simultaneous, 157- 
 173 ; quadratic, 207-228 ; homo- 
 geneous, 239; symmetrical, 241 ; 
 in higher degrees, 246. 
 
 Evolution, 174-186. 
 
 Exponents, definition of, 5; law of, 
 77; integral, 187. 
 
 Factoring, iii, 83-103. 
 
 Factors, definition of, 42, 83; laws 
 governing, 43-51; degree and 
 number of, 48; monomial, 84; 
 by rearrangement and grouping, 
 97-99; binomial, 100; rational- 
 izing, 198. 
 
 Fractions, in number system, 2; 
 algebraic, 109; signs of, 110, 
 111 ; reduction of, 111, 115, 116; 
 proper and improper, 114 ; addi- 
 
 309 
 
310 
 
 tion and subtraction of, 117, 118 ; 
 multiplication of, 119, 120; divi- 
 sion of, 122 ; complex, 128. 
 
 Geometrical series, 284-287. 
 
 Graph, the, iii; of the linear equa- 
 tion, 148; of two linear equa- 
 tions, 155; of surds, 192, 193 ; 
 of the quadratic, 218-221. 
 
 Highest common divisor, 104, 105. 
 Highest common factor, 104. 
 
 Identity, definition of, 14, 125; sign 
 of, 15; in addition, 26, 27; in 
 subtraction, 82; in multiplica- 
 tion, 55, 70-78; in division, 66, 
 81. 
 
 Imaginaries, 202; operations with, 
 203; graph of, 205. 
 
 Index laws, 47, 58, 59, 187. 
 
 Involution, 56. 
 
 Logarithms, 292. 
 Lowest common multiple, 106, 143. 
 
 Monomials, definitions of, 22; addi- 
 tion of, 28, 24; subtraction of, 
 30; multiplication of, 48, 49; 
 division of, 59, 60; H.C.D. of, 
 104; L.C. M. of, 107. 
 
 Multiplication, 89-57; of monomi- 
 als, 48; of polynomials, 49, 51; 
 identities in, 55, 70-73 ; of frac- 
 tions, 119, 120. 
 
 Notation, 280, 284. 
 
 Number system, the, iii, 1-3. 
 
 Numbers, incommensurable, 3, 192 ; 
 general, 4, 9; literal, 4; oppo- 
 site, 10; negative, 11; positive 
 
 INDEX. 
 
 and negative, 11, 12, 18; alge- 
 braic, 18; commensurable, 191 ; 
 irrational, 198; rational, 200; 
 complex, 205; ratio of, 255, 
 
 Ordinates, axis of, 148, 149. 
 
 Parentheses, 7; removal of, 33; in- 
 sertion of, 34. 
 
 Pascal’s triangle, 78, 79. 
 
 Perinutations, examples of, 268, 
 270 ; definitions of, 269; symbol} 
 of, 270; factorial symbol of, 272. 
 
 Polynomials, definition of, 25; ad. 
 dition of, 25; subtraction of, 50; 
 multiplication of, 49, 51; divi« 
 sion of, 60, 61; square of, 73¢ 
 H. C.D. of, 105) EG aie 
 107 ; square root of, 176; cube 
 root of, 182, 184. 
 
 Proportion, definition of, 263; theo 
 rems of, 263-265. 
 
 Quadratic, the pure, 208; see Equa . 
 tions, 
 
 Radicals, definition of, 191. 
 
 Ratio, definition of, 255; terms of, 
 255; compound, 256; laws of, 
 256; limit of, 258. 
 
 Root, of equations, 16, 126, 140, 
 147, 222; even and odd, 175; of 
 decimals, 182; the double, 213; 
 irrational, 214 ; complex, 215. 
 
 Signs, algebraic, 5, 6, 138; of aggre. 
 gation, 7; in addition, 20, 21; 
 in subtraction, 28, 29; in multi. 
 plication, 42, 43; in division, 58 ; 
 in factoring, 87; in fractions, 
 
INDEX. | 
 
 110, 111; radical sign, 174; in 
 roots, 175. 
 Square root, 174, 176, 177, 179-181. 
 Substitution, 4, 8; in elimination, 
 158. 
 Subtraction, 28-38; of fractions, 
 117, 118; in elimination, 161. 
 Surds, definition of, 192; expressed 
 graphically, 192; forms, 193; 
 kinds of, 194; quadratic, 194, 
 196; polynomial, 197 ; conjugate 
 binomial, 198; trinomial, 200; 
 and rational numbers, 200 ; 
 square root of binomial, 201. 
 
 Trinomial, definition of, 25. 
 Type forms, iv; in multiplication, 
 
 51, 57, 70-76; in division, 59, 
 81;-in factoring, 85, 86, 88, 92, 
 94 ; in linear equation, 137, 147; 
 in square root, 176; in cube 
 root, 182; in index laws, 187, 
 188 ; in quadratic equations, 207, 
 208, 209, 211, 216; in permuta- 
 tions, 271 ; in combinations, 275, 
 276; in arithmetical series, 280, 
 281; in geometrical series, 284, 
 285. 
 
 Variable, definition of, 15; equation 
 
 in one, 125-146; linear equa- 
 tions in two, 147-156 ; quadratic 
 equations in a single, 207-228. 
 
 Variation, 260-263. 
 

 
 
 
 
 
oF 
 
 “iptays 
 
 oh a alll allt 
 
 ANSWERS. 
 
 1 
 
 ee Oe 
 
 Page 8. 
 24, 4. 22. 130 17. 2. 
 29. 8. 39. 13. 20. 18. 52-72, 
 Tee 9. 49. 14. 7. 19. 10387. 
 151. 10. 49. 15. 24. 20. 193. 
 123. 11. 123. 16. 136. 
 
 Page 9. 
 (10 b + a) cents. 10. (x + by + cz) dollars. 
 
 : vs years. 11. Va? + b? rods. 
 
 : x , nb 
 (6+d) miles. (bc+de) cents. An. & + =a) Hours. 
 
 . © acres. abex 
 160 13. 27 cents. 
 abe 
 
 fo COUATS, np ah fe 
 160 14. et ( ay a |} dollar 
 
 x 2% 
 — + — | hours. 
 i a 4 3 ) 
 Page 17. 
 
 . $1050, A; $450, B. 5. $1200, John ; $900, Henry. 
 300. 6. $2200, house ; $1500, lot. 
 (aire 7. $120, $240, $360. 
 
 . 60, 160, 220. 15. 428 years. 21. $3500, $2800. 
 330. 16. 28. 22. 31,360. 
 
 . 780. 175 60,,.96. 23. 600. 
 
 . $520, A; $780, B. 18. 30. 24. 300. 
 
 mel 00. 19. 54, 27, 18. 25. 60. 
 
 BUNS 20. $4000, $ 400. 
 
 Page 22. 
 4b. 3. 0. 5. —8abe. 7. — 2aty. 9. 6V xy. 
 — 8 ab. 4. 0. 6. 10 x?. 8. axy. 10. 0. 
 
fa 
 
 oA FS 
 
 ee 
 
 Si sae ane aed 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 Page 25. 
 — 5 xyz. 6. 0. 
 5 a2be. 8. «2 + y?. 
 Lies, 9. 0. 
 — ax. 10. 4(a?—2y). 
 Page 26. 
 19%—38y+22. 6. 
 Tax —5by+4cz. vf 
 242+ 4 y? — 3. 8. 
 
 13 a? — y2 4+ 4 2%. 
 Page 27. 
 
 Page 29. 
 8 x2y. 6. — 16 ay. 11. —15(@+y). 
 12 ab. 7) 0) 12. — 4(a? + dB). 
 — 21 xyz. 8.10. a 13. 25(ay + yz + 22). 
 dg, 9. 29V xy. 14. —17(ax+ by+c). 
 hs 10. —4(a+ bd). 15. 5(y? +4 az). 
 Page 31. 
 —5ai—4474 18 y. 9. 10 xy? + 2 xy? — 3 wy — 2 yf. 
 4 at — 25 bt — 10 a?b?. 10. 8(a+ b)—-11@+y)+6ce+5. 
 —Sxy + 7x2 + 2 2. 11. 26Va2y —2y3— 162 —17y. 
 165 ax — 33 by + 16 ¢. 12. 194ax-+5y?— 23 ez—-l1ly—8. 
 —8V2+ 18Vy — 164. 13. 54 y3 — 44 23 + 24 yz. 
 —e—e7?4+34411. 14. 36 a2 + 10 a2b — 34 ab? — 12 6. 
 
 ll. Vx+vy. 
 
 — 8ay + 2yz + 18 2x. 
 2Vxy — 2V yz —V 2x. 
 —6abe+4+ 18 cyz — lmn. 
 
 No answers given, as student is to verify his results. 
 
 15. 1822+ 17 ay +17 yz — 18 y? — 37, 
 
 Page 32. 
 
 No answers given, as student is to verify his results. 
 
 pak ap inact 
 
 Page 33. 
 a—4b+4+¢e. 7. Wat+2y. 12. —8., 
 x—y-—a. 8. 10x%—18a. 18. 21: 
 —20b. 9 6x%—5a. 14. 46. 
 —10a+40b. 10. — 25. 15. 22. 
 
 5 ax — 15 by. We eu 
 
i 
 
 ANSWERS. 3 
 
 Page 35. 
 1. 3a2?-—(—4abd+3c-—d). 
 2. ay—(— by+ cy + ay). 
 38. 5—(d8%-—ar+4abz). 
 4. Ty+382—(—24+382x+4 dyn). 
 5. 18 — (Sax — 11%+ 16 ax) + 274. 
 6. 27 y — d0y — (— 22 zy 4+ 15 bx — 2127) — 80a4 165. 
 8. 11 —(8ac—7axr) — (11 by—5y) 4+ 254-344 cu. 
 9. b?-—(ab+2ac?—1la)— (4 b*¢y —cy+4y). 
 10. 5%+5x2—(2ab+4+6ab+ bax) — (14y — 3 yz — 11 by). 
 Page 36. 
 (6a+7y—382z)z. 3. (444+ 3y—122)k. 
 (8x%+4y—5z)a. 4. (4a—5b)x4+ (6a4120c)y. 
 5. (8a—2a)x4+ (40—12b)y; ax —8 by. 
 . (44 16a — 12 bd) x2. 9. —2x%+4y+4 8z. 
 . 2©—dby4+ 3d. 10, 5 3? 422 4 37 eH 6. 
 
 . (a—a')e?+(b-—)b')y+(c-—c)x+(d-ad)y+e-el. 
 . 247417 ay 4+ 4 y?. 
 . 644+ (16y4122)k4 (82? + 12) k? + Kk’. 
 
 (a—a@')v?+ (2h—-2h')cy + (64 0!) y? 4+ (ce — c') 2. 
 
 Page 37. 
 $400, A; $150, B. 15. (a2—b2)4+3(a+y)—15 V 22+ 92. 
 9 ax3+48 bx2y—23 cay? +21 dy? 1%. (1—b)a+(—44+60+4+c¢-—d)z. 
 18 22—Ty—6%2. 18. 8%—[—38y4+ (#+y)—2 yz]. 
 (a—5+c)+ (3—b—d)y’. 19 ab + cd — LOW Bh al Gad! 
 8 a3 + 6 ab — 12 ab? + 6. 100 f 
 Zi. 20. $36, John; $72, James; $108, 
 462. Henry. 
 . 64 yr., father; 18 yr., son. 21. 16274 6ay — 8 y?. 
 By — 5 xy. 22. 28 —34%-—27-4y'. 
 38a—15ab+76-—11. 23. — 5. 
 —2(a+ b) Vaz. 24. 79. 
 25. 5(a +y) — 4(a? — b) — 26 Vy? — 
 Page 46. 
 2ad + 2 bd — 2 cd. 3. 4axr+2 ay —6 az. 
 —6ad —9 bd + 12 cd. 4. 48 bx? — 20 by? + 52 bay. 
 
 5. — 30 abry + 26 abyz — 24 abzez. 
 6. 5abxy — 25 bey + 15 cay — 5 axy. 
 
oo 
 
 we ae 2 ge 
 
 ray 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 % —Samxz? + 3 bmx? — 3 cmzy. 
 8. 8ablx + 10 abmy — 6 abnz. 
 9. 18 a3a3 — 15 aba? + 9 ab2x3 — 3 b328, 
 10. — 2 ax’y?z — 2 buy?z — 2 cy?z. 
 Page 48. 
 . wa, 6. 29.39. 611, 9. (a+b)#(4+y)2, 
 ys. 7. q%qltyl8, 10. 79(y+z)4(a—b)}6 
 2 ey ttyl, 8. (7+ y)7!. 
 Page 49. 
 . — 60 ardbx2y. 4. 20 xty%z4. 7. 60 y(a + b)?. 
 . 80 a5b2x3y5, 5. — 96 ltxby’. 8. — 80 atz*(a+b)> 
 . 40 mnsad, 6. — 6 adic. 
 Page 50. 
 . 12 a3b — 16 a®d?2. 5. 6 l4xt — 8 [8x5 4 10 arty, 
 
 . 15 akx2y — 10 a2a3y +15 ax?y2, 7. — a'b? + ath? — a3b4* + a?d®. 
 8. 3 a5b3 — 4 atb* — 6 a3b® + 7 a?b?. 
 9. 6 a®bx? + 6 ab3a? + 6 abet + 6 abaty? + 6 abxrz?. 
 10. — 30 abc + 72 be?d + 18 befh. 
 11. 20 x3y2z2 — 35 w2y3z? + 15 x2y2z8. 
 12. 24 atbh+c? + 56 a2btct — 112 atb2ct. 
 14. 5 xy(a? + y?)4 — 15 a2y (a? + y?)3 + 20 xy?(a? + y?)2. 
 15. 2x2y(a — b)® — 6 wy(a — b)® + 14 wy?(a — D)t. 
 16. — 38 a3b?(~ + y)® — 2 ab?(x% + y)® + 38 a3b4(x + y)?. 
 17. — 20(42 + y)® + 12 a3(a? + y)* + 28 ath (a? + y)3. 
 18. — 50(a? + 3 b?)7 + 30(a? -+ 3 b2)5 — 60(a? + 3 b?)4. 
 19. 5(a+60+c)'—15(a+ 54 6). 
 20. — 5(ax? + bx + c)* — 20(ax? + bx + €)3. 
 
 Page 52. 
 16 «2 — 9 y?, 8. 22+32— 10. 
 a2x2 — b2y2, 9. yt + y? — 12. 
 x + Sgt 8. 10. ax? + acx? + bx + be. 
 2+ 4 = 6, ll. 12224 262+ 10. 
 
 5a5 — 14 a4 — 10 a3 + 24 a2 —3 a — 18. 
 14. 7647 y5 — 8 yt — 22 y3 + 40 y? — 48. 
 15. 66+ 464 — 1163+ 138 624+ 46 — 10. 
 16. a + 9 at — 48 a? — 38.4? +17a— 30. 
 
sole os 
 
 ao Femi ipeee gal oh 
 
 ri acct oi ak ado Ae 
 
 ANSWERS. 5 
 
 19s 3 oc + 12 ot + 91 ze? — lida? + 82 2 — 20. 
 
 18. 5 xy — 8 wy5 — 54 wtyt + 35 x3y8 4 52 x2y2 — 18 xy — 12. 
 19. 42—102* —17 22+ 8422+ 2-12. 
 
 20. 4a2+ a7’ +2a8—a—8at+7a®—2a24+5a—-—4. 
 
 93. 3a° + 7 a> — 33 at + 39 a? — 29 a2 +19 a —6. 
 
 24. a — 16 a°+ 40 a — 41 at + 76 a? — 74 a2 4 28a — 49. 
 25. xoy® — 16 xty* + 48 x3y3 — 76 x2y? + 60 xy — 25. 
 
 36. 6 2' — 925 — 295 + 21 xt — 20 28 — 22 22 4+ 31 x — 20. 
 27. 28y7 — 20 y® — 56 y5 + 59 y* — 21 8? — 24 y? + B5y — 4, 
 28. — 4 a®b® + 29 atht — 28 a8b? — 30 a2b? + 35 ab. 
 
 29. 4a7b7 — 2 a®b® — 17 atb*t — 13 a3b? + 33 a2b? — 22 ab 4+ 35. 
 30; 327° — 275 — 102+ + 14234 42?— 122 — 15. 
 
 Page 56. 
 
 No answers given, as student is to verify his results, 
 
 Page 57. 
 16 al2y4, 4, 410 718/24, 7. qidxl4. 9. (a2 + y2)M, 
 (a+ y)% 6. 34, 8. 3! . 2472, 10. abc, 
 Page 60. 
 3 aay. 5. 3a(x+y). 9. 7 lyt(z2—<2). 
 3 a2xz. 6. —3(a—b)(a—-y)>. 10. 3?(a — 2). 
 BY. 7. —9 x3y, 
 3 bed. 8. 38a2z. 
 3y — 2 axy? — ax, 3. Bay + 5 y?z —T yz. 
 cx’ — 2 abu? + 8 a2c?, 4. («+ y)?+ 5(4+ y)— a. 
 
 5. 4(a—b)—52+4 11 ay(a—b)?. 
 6. (e+ y)?—2xy(et+ y)—3y(a+t y)%. 
 7. ab +2 a(x? + a?) — 3 b?(xz? + a?). 
 8. — «2 —3 yz(a? — b?)%, 9. —2a+38 Bct(ax + b)3. 
 10. —3+ 4 a*(ax? + ba + c)3. 
 
 Page 62. 
 x3 +2 12 —3 1+-4. 8. bc? — at. 15. @2b?—2 abed+c?d?. 
 x8 — 3. 9. m*—2m?+4+1. 16. 3aty* —2 x2y?+1. 
 $22—44+1. 10. a” +. b". 17. 2?—6224+1227—8, 
 227—52+4+7. ll. a + 6, 18), 1-8 a= x2, 
 3 a? — 7. 12. a" + arb" + 5, 19. (a+ b)—4. 
 a2 + 2ab + b2. 138. 2? —72+4. 20. (x+y) — 2. 
 
 xt + xa? + at. 14. «27-6249. 21. (a+24)+6. 
 
THE ESSENTIALS OF ALGEBRA. 
 
 22. (m+n) —11. 
 23. (a+b)?+ (a+ b)e4+ 2%. 
 24. (w+ y)?+(a+y)(a +b) + (at b)% 
 
 
 
 Page 64. 
 e+a2?#—4x4—4; c2—22? 8. y6 —8 yt + 64 y? — 512, 
 —x+2, 9 at+a*41. 
 v8 —602+524+12; 2-32? 10. 4a?-—2a+1. 
 —-2£+3. ll. «2? —22+4+2. 
 x? —a2— 6. 12. 3b—1. 
 2 + 7 y?. 13. a? — 2 a?b + 2 ab? — 6. 
 m3 — 7 m?n + 5 mn? — 4 ni. 14. «+y. 
 a2 — 16. 15. —2a?+ 8ab. 
 644+ 9624 81, 
 Page 66. 
 x? — 10% +4 41, rem. — 150. 8. 1—1L a+ 72 x — 395 23, rem. 
 x2 +44 —10, rem. 30x — 7. 1gy® at, 
 a? — 16 a+ 80, rem. — 379. 9. 1—x—8 «2? — 181 23, rem. 
 b2 — 23, rem. 218. 2056 at + 655 2°. 
 ax? +6 ax — 18, rem. 47. 10. 5+ 22 me aie ay rem. 
 14+2%+207+42 48, rem. 2 xt, 1255 4 
 1424-22 —143, rem. —i ct. 
 
 Page 67. 
 
 No answers given, as student is to verify his results. 
 
 Page 67. Review Exercises. 
 124. 3. 823 — 16 y3 + 23 — 4 ay — ay? 
 488. 4. —4—4a4+20643¢e. 
 — 16 4? — 31 xy — 21 y”. 
 (a+4a4+3l)¢+(b—-4b'4+38m)y4+(c4+8n)24+4e. 
 
 24—-2y—22; —2. Lit: 
 3a—25b+421¢; 16. 12. y2z — xy. 
 5. 13. 2+ 4241. 
 3 V41 —4 V3. 14. «2 + y84+1-—32y. 
 3 15. xt — 4 xy? 4+ a3y? — 4 yt, 
 36 x7 — 2126 + 98 25 — 88 xt + 79 «3 — 69 a2 + 402 — 16. 
 —7; —18; —5. 20. 3a2—a4 + 2. 
 Cat cede pedo, 21. a—b—c. 
 
 a*—38a?b+2 ab. 22. «—b. 
 
ANSWERS. i 
 
 23. 22 —(b+c)x+4 be. 25. atl + quar + ax + antl, 
 24. z2—™m. 26. x2" — yn, 
 27. 6 a — 12 grt) 4 42n — 20 unt] — yn — 284 — 21. 
 28. 3a" 4+ 4a"-14 3 yr”, 
 29. om — ym 1. 
 30. a? —2 a2b + 2 ab? — DB. 
 31. 6x — 12 x74 — 2 xr — opt — arptr + 10 gat, 
 32. qn + arbn + h2n, 
 833. w2nt2 +- gent] — 10 92" + 13 72n—-1 — 5 92n-2, 
 34. x — y”, 88. (Vax+b)?—y(Vax+ b) + 2. 
 35. ai” + a2nbin + D8, 39. (x — y)%. 
 36. (a+ bd)” — x. 40. a—b. 
 37. (a2 + yy) + (a2? + y)" 4+ 1. 
 Page ‘71. 
 1. 42+8% — 20. 8. 1644+ 8 x? — 35. 
 2. 947+ 21446. 9. 94+ 6 xy — 24 xy?. 
 3. 1902 —7a—20. 10. (a+ b)? —(a + b)— 30. 
 4. 254+ 40a+412 a2, 11. (a+ 6)? -—(a+4 5) — 56. 
 §. 4y2?+2y — 42. 12. (x+2ab)?+4(%+2 ab) — 21. 
 6. ax? — 5 ax — 66. 13. 20 -—(#+ 2y)—(%+ 2y)?. 
 7. 9 0?y? — 33 xy + 30. 14. (a+ b)4+ 3u(a + bd)? — 28 x2. 
 15. 15 a7b?—2 ab(x—y)?—(x-y)!. 
 1. 27+ 2axy+4+ y?. 5. y2+2y(a+6)+(a+ bd)?. 
 
 2. 407+ 4axr +4 a’. 6. 927+627(2a+c¢)4+(2a+¢)% 
 3. 92? + 24 bx + 16 52. 7. (a—38)?4+10y(a — 3) + 26 y?. 
 8. (a@+b)?+2(a+d(@+y+(at+y) 
 
 Page ,72. 
 
 1. a? —2ax+ x2. 4. Dat —6 arby + by. 
 
 2. 9a27—6Bay y?. 5. (a+ db)? —2 xy(a + bd) + x2y?. 
 3. «?y2 — 8 bry + 16 D2. 6. (8 a+y)?—2 ab(8x+y) +4752. 
 7 Ce+y)?—2(@+y)(a+ b)+(a + dD)? 
 
 8. (2x—y)?—6 xy(2u —y)+ 9 xty?, 
 Page 73. 
 1. x? — y?. 6. v#+ 443+ 4272 — 256. 
 2. 9a? — bd. 7 a+2ab+4+ b?— 4? —2 ay — y2. 
 3. 16 42 — a2b2. 8. a*xt + 2 abu + bx? — c%. 
 5. 4ay + 4 y?. 
 
OIA Th oD 
 
 es 
 o © 
 
 on 69 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 Page 74. 
 
 a+ a+ y2—2axr—2ay+2 xy. 
 
 9 a2 + y2 + D2 —6 ay + 6 ba — 2 by. 
 4249624 y2+ 12ab —4 ay — 6 by. 
 
 9 a2 + 25 a2 + 4 y? — 30 aw — 12 ay + 20 xy. 
 
 4+9a2+ 1662+ 12a— 16) — 24 ab. 
 
 et y2+a2+ bh? —-2ay+2axn+2be —2ay —2 by +2 ab. 
 
 m? + n?+ p? + q2?—2mn—2mp —-2mq+2np+2 nq +2 pq. 
 4e274+ y2 4922+ a? 4+ 4 ay —12 42+ 4 a0 —6 yz + 2 ay — 6 az. 
 a* + ab? ++ 427249 y?—2 0b +4 a*x —6 ay—4 abx+6 aby—12 xy. 
 4u2+ yt 4+ o2y2? + a —4 ay? 4+ 4027 —4 an —2 xy3 4+ 2 ay? —2 ary. 
 
 1. w+(c+d-+ e)x? + (ced + de + ec)zx + ede. 
 
 2 @8+(a+yt+ za? + (ay t+ ye + ex)at xyz. 
 b3+467+ 6-6. 6. «xy? — 6 xy? — 9 ay + 14. 
 y>—2y?—5y4+ 6. 7. 2 — 49 x? + 190. 
 m3 — 2m? — 23 m + 60. 8. 7° — 93 y? + 308. 
 
 9. (x+y)? + 10(~+ y)? + 31(4% + y)+ 30. 
 
 10. (8x%+ y)? — 28182 4+ y)+ 48. 
 
 11. (ax + b)? + 5(ax + b)? — 84(ax + b)— 80. 
 
 12. (ax? + bx)? + (ax? + bx)? — 10(ax? + bx) + 8. 
 
 Page 76. 
 
 1. 8a3+ 12a7b + 6 ab? + 6. 3. 27 43 + 10822 + 144% 4 64. 
 
 15. 
 16. 
 
 a? — 9 a2b + 27 ab? — 27 b3. 4. 8 «?—60 x2y+ 150 xy? —125 y3 
 5. 27 43 + 27 abx? + 9 a®b2x + a3, 
 6. 8 mn? — 12 m2n’pq + 6 mnp2q2 — p3q?. 
 12628 —7524+15a2—1. » 8. 216, — 1842-0 oe 
 9. 64 a3b3 — 240 a2b?y + 800 aby? — 125 y3, 
 PE Ory Olek 
 1l. (a+ y)? — 6(@ + y)? + 12(4 + y) — 8. 
 12. (x—y)?4+ da(a— y)? + 8a2(z — y) + a. 
 18. 8(2 + y)? — 36.a(a + y + 54 a2(a + y) — 27 a8. 
 14. (ax + 6)? + 3c(ax + b)? + 3c2(ax 4+ b)4 c3. 
 (a+b)? + 3(a+ b)*(e+ y+ 38(a+ d)a@+y)*+(@+y)* 
 (ax + b)® — 3(ax + b)?(cx + d)+ 3(ax + b)(cx + d)? —(ca + d)3. 
 
 
 
 Page 79. 
 2. a®& + 6a5d + 15 ath? + 20 a3b3 + 15 a2b4 + 6 abd + 5S, 
 3. at + 7 axoy + 21 wy? + 85 xty? + 35 x3yt 4+ 21 x?2y5 4+ Tay? + y7. 
 
ANSWERS. 9 
 
 m8 + 8 mn + 28 mon? + 56 mon? + 70 mint + 56 m3n5 + 28 m?2ns 
 +8mn' + n8, 
 
 . 14924 867? + 84273 4+ 126 x4 + 12625 4+ 84278 4+ 386274 9 784 2, 
 
 y + 10 79 + 45 8 + 120 y7 + 210 95 4+ 252 y5 + 210 yt + 120 73 
 + 45 y?+ 10y 4 1.- 
 
 . 162! + 82 ax? + 24 a?2x? + 8 a3x 4+ at. 
 
 . p’—8 p'q+28 p'g?—56 pq? + 70 p4q* — 56 p2q° + 28 p?g® —8 pg? + q°. 
 . Bl yt — 216 y3x + 216 y22? — 96 yx? + 16 x4. 
 
 . 824° + 80 atd + 80 a3b? + 40 a2d3 + 10 abt + DS. 
 
 © — 15 a4y + 90 x3 y? — 270 xy? + 405 ayt — 243 y?, 
 (21 x)® + 6(21 x)°a + 15(21 &) 4a? + 20(21 w)3a3 + 15(21 x) 24 
 + 6(21a)a° + a®. 
 xvl2 — 12 xy + 60 x8y? — 160 xy? + 240 xty? — 192 x2y5 4+ 64 7/8. 
 — 9 xl6y2 4 36 altyt — 84 wl2y6 + 126 xly8 — 126 x8yl) 4+ 84 8712 
 — 36 xtyl4 + 9 x2yl6 — 18, 
 
 » + 4aly + Tady? + 7 ary? + 3 axtyt + ey? + ye wy? + ys ay" 
 
 + ghz ys 
 
 . dx? — 7-38. 4 aby 4 21-35. 42x°y? — 35. 34. 48aty3 4 35. 38. 4434 
 
 — 21-32. 45a7y5 + 7-3 . 46ry6 — 47y7, 
 (a + y)* + 12(4 + y)3 + 54(a + y)? + 108(% + y) 4+ 81. 
 (a+ b)*—82a(a+ b)? + 2422(a + b)? — 3223(a + Db) + 1624. 
 
 . 245 4° — 405 at(a — y) + 270 a (a — y)? — 90 a2 (a — y)8 
 
 + 1l5a(e — y)* —(% — y)?. 
 
 ie 4. 5 y+5 oer 32 xy? | 16a2y* 64ay? | 64 y8 
 55 5S 52. 33 5.33 5-34 36 
 . 2+8r7415. 4 a2x?+ 8ar4+ 15. 7. of — 5 27*— 50. 
 
 v24+8x%—20. 5. x2y2+(a—b)xy—ab. 8. (w4+a)*—4(x7+a)—21. 
 v+3e2—40. 6. 1622—122—10. 9. 12 +7 by + by?. 
 
 (x?+5 27)?—3(x?4+3 2) —10. 15. 427 — 16 y?. 
 
 . 92? — 3xy — 2 y?. 16. (2% +4 y)? — 64. 
 (a + b)? + 6(a + b)— 72. 17. (a? + y?)? — 16. 
 (x ~5)2— 18. (ax + by)? — c?. 
 
 19. 2* + 27+ 1. 
 
 . 9x? — y?, 20. (xy + yz + 2x)? — a2, 
 
 21. 224+ 4724+ 224 2xey+2yz24+ 2 eu. 
 
 22. 902+ 4y24 224 1l2ay4+ 6924+ 4 yz.. 
 23. a2x?2 + b?y2 + c2 + 2 abay + 2 ace + 2 bey. 
 24. x27—-6x4+ 9. 
 
 25. 474+ 9y?+16+4 82+ 24y + Bay. 
 
10 
 
 wo © © 29 
 woes © 
 
 Sen oe a ea 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 26. 942+ 4y?4+ 36 — 36a — 24y 4 122y. 
 Dra gn — 2 ynynr + yn, 
 28. x2 4+ yy? +224 w?+ 2ry+ 202+ 2aw + 2yz2+ 2yw + 2ew. 
 
 
 
 
 
 
 
 . 2+ 14424 638% +4 90. 33. «3 4+ 93, 
 et — 3427 + 225, 34. a? 4+ 6? 4+ 3 — 8 abe. 
 
 . 02 + 27 a8. 35. 2+ 8y3+ 1-—62y. 
 
 » 8+ 9x7y + 27 xy? + 27 y?. 
 
 Page 81. 
 a— 9. » 18 eR 
 442 + 62, 15. y"+ 8. 
 3+ a. 16. a+b—3%. 
 4a?—2ab+ 6. 17. vw" — 5y", 
 9a — 5 db? 18. («+ y)"+(a+ db), 
 1— da. 19. av+b+ 52. 
 y — 2. 20. ax? + b4 +¢— lx —m. 
 
 . (4+ b)2+4(a+ b)+ 16. 21. y2” 4 yranr + 12”, 
 9a2?+3a(e—y)+(e—y)% = 22. _-¥yBr 4 yPrgnr 4 yrg2m 4 xn, 
 (1+ y)—2ab. 238. V3z+y4+ 4. 
 
 ny a? 24. (Var+by+4yVar+b4+16y2 
 epee Lbs 25. (Vaxr-+by)? —5a2Vax + by+ 
 
 . ©+y—12(a+5b). 25 at. 
 
 Page 84 
 a(a + y). 15. axy(l+at+say). 
 3 ax(1+2 ax). 16. 38axy(a—54+4 7 ay?). 
 
 . 5a(a4+264 be). 17. S5axyz(a? + 6 xy — 8 2). 
 
 . ey(d5 —Caxry +7 xy?). 18. 7 a®xyz(x? + y? + 27). 
 
 . m“(3+4mxr4+2y). 19. 3abc(aVe+y —2bVx%—y). 
 21(—24+2Wy+4+38y). 20. 5ayz(aVar+b+2yVax—b). 
 
 Page 85. 
 (a+2b)(a—2b). 5. (24+82)(2-—382). 
 
 (2a+3b)(2a—38b). 6. (day + 62w)(S5ay —6 zw). 
 (ab + c)(ab — Cc). 7 («tytaj)(x+y—a). 
 (4ab+5c)(4ab—5c). 8. («+ 3y+2)(4+3y—2). 
 (8%—2y+z2+a)(8x—2y—z2-a4). 
 [2(@ + 2y)+ 3(a + b)][2(# +2y)— 38(a + d)]. 
 
 [2”%+ 4(x% — 34) ][2 a — 4(x — 84)]. 
 
 (4 + a*)(24+ a)(2— a). 16. (8%+4y+4+8)(82+4y4 2). 
 (x2 + 4 xy + y?) (xa? + y?). 16. (—2%+38y—5)(—4x2+y—8S). 
 
 a 
 
ANSWERS. 11 
 
 (lx + my+n-+2 an +2 by+2c)(le+my+n—2 ax —2 by —2¢). 
 (ax + by + 2lz+2 mw)(ac + by — 2 lz—2 mw). 
 
 ~ (arn + Ory”) (Caran — Bry”), 21. («#+a+1)(e«—a+1). 
 
 ~ (ary 4+ Zw") (ary” — 2"), = 22. (44+ 2y4+38)(e-—2y438). 
 . (84+38b+a)\8x—8b4 a). 
 
 . [(@+a)?+(a+ b)?7]@+2a+b0)(4—b). 
 
 . 8a+yy-—*). 
 
 Page 87. 
 
 . («© +1)(@+42). 15. (644+ 3y+2)6%+3y+8). 
 (x —1)(@ — 2). 16. (x2 + y? — 9) (a? + y? + 8). 
 (a+2)(a—1). 17. (2x+38y+4+8)(2044+3y+41)). 
 (* —2)(%+1). 18. («+1)(@+2)(@+8)(4%+4 4). 
 (ax + 2)(ax + 3). 19. (7©+38)(%@—38)(“#+2)(@ — 2). 
 
 -. (@+2y)(*4#+3y). 20. (ax + by +7) (ax + by + 1). 
 (xy — 5) (xy + 2). 21. (ax + by —1)(ax + by — m). 
 
 . (1+ 5ay)(1 —2 ay). 22. (1+ 38)#(a + 2)(a + 4). 
 
 . B«a-y)2Qae-y). 23. (427-— 54+ 2)(a7-54+4 5). 
 
 . (©+y¥4+4)(4+y4+ 5). 24. (x — 8)(a" — 2). 
 
 . (@+3860—5)(a+3b+44). 25. [(ax)" + I][(ax)" +m]. 
 
 Page 89. 
 (x — 4). 15. (a? 4+ y? — 2?)?. 
 (2% — 3)? : 16. (x+y+2z)?. 
 
 - (ax + 5y)?. 17. (a—b—c)? or (—a+6+40c)?, 
 (7 ab — 1)?. 18. (244+ 3y+4 2)? 
 
 . (10 — ab)?. 19. (a + 6)?. 
 
 . (ae +04 c)?. 20. (y*” — 7)?. 
 
 . (8a+4y—3)2. 21. (a" + 6 b*)?2, 
 
 . (22-38 y)?. 22. [((a@+y)"—saj 
 
 . (ax + by+e4 4)? 
 
 Page 90. 
 
 4. («© —2b)(a? + 2 bx + 4 0?). 
 5. (2a—38b))(4a2+6ab+4 9b). 
 6. (ax + 4)(a2x? —4ax+4 16). 
 7. (ax —3 yz) (a2? + 3 axyz + 9 y?z?). 
 8. («+y—S5z)[(e@t+y)?+52(e+y) + 25 27]. 
 9. (84+442y)[(82+4)2?-2y(82+4)+4y?]. 
 10. «(a+ 3)[(22?+ 3844 4)?4+4(@?4 82+ 4) 4 16]. 
 11. (ax + by — cz)[(ax + by)? + cz (ax + by) + c?z?). 
 12. 5@+y)[Gx+4y)?—-Be+4y)(2e+y)+ 2@et+y)?). 
 
12 THE ESSENTIALS OF ALGEBRA. 
 
 15. (Qa — b?)(4 a2 4 2 ab? 4- D). 
 
 16. (ax? + by?) (atx* — a?b2a?y? + bty*). 
 
 17. (w—1)(#4+ 1) (a? —a44+ 1) (2? 4+ 441). 
 
 18. [a%x? + (y+ z)*][atat — ata? (y + 2)? + (y + 2)4]. 
 
 19. [(ax + by)? — c*z*][(ax + by)* + cz? (ax + by)? + ctz4]. 
 20. (4 a? + bc?) (16 at — 4 a?b?c? + dtc). 
 
 21. 4 xy (8 a? + y?) (a? + 3 y?). 
 
 22. [(a+ b)?+ c][(a+b)*—c(a+ bd)? 4 c4]. | 
 
 
 
 
 
 
 
 
 
 Page 93. 
 1 (8a+y)(e*+2y). 5. (ax +3b)(6x44+ 5). 
 2. (84+2y)(2x—33y). 6. 25y—32)(2y—~2). 
 3. (7xe+y)(8x%+4y). 7% 2(4y+2)8y—42). 
 4. 2sx—y)(2x2+y). 
 Page 97. 
 1) Ba ed al) (eee 
 ore. 89 2 oat 
 9 —2(a— 24 aR) - 7-3). 
 2 2 2 2 
 3 (2-245) (eB), 
 6 6 6 6 
 
 4 1 Cae ee (x-38-¥=*). 
 2 2 aie? 2 
 
 eee (2 rele xe (« 4" sp a 
 
 
 
 
 
 
 
 2 2 2 
 6. 11(2—34+ YH) (2-8 - v 
 2 2 2 2 
 7. —4(x%+8)(x—1). 
 2 2 2 £ 
 9. -8(2 84 21) (2-332) 
 2 2 2 
 10. 10(2 244) (x -7_ A), 
 2 2 2 2 
 Page 98. 
 
 l. (w@+a4+3b)(4+2a+4 Db). 38. (x —2y+32)% 
 2. (t+3y+b)\(2e4+y-—D). 4. (2a+3b—4c)% 
 5. (2a4-—S38y+z)(a@+y—382). 
 
ge alae ool ea! 
 
 a 
 
 ll Dg 
 
 a a 
 OP WON OO SAD AP ww Pr 
 
 - (@—2)(a+1)(a—-1). 
 
 ime Sas Sk 
 
 ANSWERS. 
 
 Page 99. 
 8. (4a+6)(8a—b+40). 
 4. (x+4a)(4—5a+y). 
 
 (2%—-—5y)\(2x+7T7y-— a). 
 (x—4y)(e+3y+4+2). 
 
 13 
 
 5. @aty@r—y tp). 
 
 Page 101. 
 6. (x+2)(@ — 5) (x + 8). 
 7 (© +2)(%+ 8)(@4 4). 
 8. (7x—1)\(@+ 1)(~+4+ 15). 
 9. (x+3)(%+4)(~+ 5). 
 10. (7—4)(a—4)(x—4) or (x—4)3. 
 
 (7 —1)(% — 2)(%4 2). 
 (w — 2)(% + 2)(x% 4+ 2). 
 («© +1)(%—38)(#+4 5). 
 (% —1)(% — 2)(a% — 8). 
 (2 — 2)(% — 8)(a% — 5). 
 
 Page 102. 
 b(a@ + 1)(#— 1). 16. 
 10c(m + 2c)(m — 2¢). VG: 
 (x+z2)(~1+y). 18. 
 (m—r)(l+ 7). 19. 
 (7 —2z)(2x+4 3y). 20 
 (38 — a)(2— a). 21 
 (x? + 9 y*) (a? — 8 y?), 22. 
 (m — 12)(m + 8). 23 
 a(a + 6)?. 24 
 (8% +1)(%+4+1). 25. 
 
 2(7+4(m+1)][7—4(m+1)]. 926. 
 (8 a? + Dy?) (8 a? — Dy?). 
 (8 —4%7)(9 + 12474 16274). = 27. 
 
 e(4y+nr)(4y—~2). 
 
 28. 
 29. 
 
 (m? — 2 n2/?)2, 
 4a(b+¢). 
 
 (7a — 4)(2a? — 3). 
 Rwy Ei Ci 0 ioe 1). 
 
 . (84%Ay7¥4+82)(54—-—y-—382Z). 
 . 9(a — 2b)?. 
 
 (a + 5)(a + 24). 
 
 . (77+ 11)(% + 8)(% — 38). 
 . 0(2141)(414+1). 
 
 (% — y)(a — 2) (a — 1). 
 
 Beeb 4% 
 ate )(s *) 
 (m?+1)(m?+m+1)(m—1). 
 
 a*b*c?(1 — ac)?. 
 (a—y+1)(e—y-1). 
 
 30. (1662+1)(2b4+1)(2b—1). 
 
 Page 102. Review. 
 
 x—-5y. 10. 
 0. 1) 
 6x—l6y+4z2, or —14x%4 12. 
 10 y — 62. 13. 
 a’ — 16. 14. 
 x" — 2ary + y?. 15. 
 (a + bx) (x+1)(@—-1). 
 2 ax — x?. 16. 
 x2 + 2ay + 2 y 17. 
 
 18, 35. 
 
 08 + 2 x2y +2 xy? + y, 
 
 (a —1)(%+1)(x—8)(x + 3). 
 
 (a8 + ab —ac—y?+ 11)z. 
 
 — 421. 
 
 54, 
 
 4q64+9 22 + 25 —12 a3x+20 a3 
 — 30%. 
 
 gmt l _ yrtl, 
 
 q2m+l1p2 4 ambnts + qmt4pn-1 + 
 azbn, 
 
14 
 
 18. 
 20. 
 21. 
 
 dha on pe 
 
 phate bee el er 
 
 — 
 
 a mE 5g Pek al li ecole 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 ain + 2q38 —2aqr—1. 
 
 0. 
 
 22. 
 23. 
 
 5a(y—z)™-4*; Sa(y—z)m-*; 24. 
 
 4 x2 (y—z)4m-8 : 4, 
 
 4 yz. 
 5 a?b, 
 
 , Ar ee. 
 
 x— Y. 
 a+ b. 
 x+y. 
 x— 3. 
 a+. 
 
 12 a®baty. 
 
 plo alr, 
 
 24 p2q3r2, 
 
 a(a — b)*(a + Db). 
 
 (7 — 4)(% + 4)(@ — 5). 
 (p — 5)(p + 5)(p + 6). 
 (LEB) 0) 79: 
 
 25. 
 
 (a+ 1)(a—2)(a — 38). 
 —y?—ye— 22-2 xy—447242 x2, 
 4(8a2—19y). 
 
 b@—jath. 
 
 Page 105. 
 
 4. 7 ab?m3. 
 
 7. 3(x — y)32°, 
 
 5. 22 p2q2x3y. 8. 9(a? — b?)2xy. 
 6. 2(a — b)?x?y. 
 Page 106. 
 6. b—7. 11. s4(4-F Oe 
 7. «© — 2. 12. (x+y). 
 8. b. 13. m — 10. 
 9. x2(4—y). 14. x—y. 
 10. x3(% — 3). 15. a—2. 
 Page 107. 
 
 4. 150 l*m3n’3. 
 5. 21 a®bta ry’. 
 
 a3(a —2b)(a +2). 
 
 6 a?(a? + 3)(a? — 38). 
 (m — 1)(m? + m+ 1). 
 (r —2)(r—- 8)(r+4+ 8). 
 
 a 
 
 dbx 
 ay 
 
 
 
 
 
 
 
 
 
 7. 20(@ — y)8a3b3. 
 8. 120(a? — x)4y?z2. 
 492, 
 
 e(1+42z)(1-—472). 
 
 (y —2)(y —TD(y + 8). 
 
 (a 4+- 8 b®)(a — 2b). 
 
 (ec —3d)(?+3cd+ 9d?) 
 (c+2d). 
 
 (16 — zw). 
 
 (8 + b)(8 — b)(9+ 3b + B?). 
 
 (x — 3)?(% + 3)2 
 
 
 
 x 
 
 6. 12(a — b)2x 
 Page 108. 
 9. 
 10. 
 11. 
 12. 
 13. 
 14. 
 15. 
 Page 113. 
 Bye 
 b 
 g 
 b 
 D etLD: 
 Sie 
 r—a 
 
 9 @teyty, 
 
 a+y 
 
 . Nocom. factors. 
 1 
 
 +1 
 
15. 
 
 16. 
 
 ANSWERS. 15 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1 ; 18. —1. 91. 2% —Yy. 
 Oy +P 19, 242. ny 
 Sha Ale x—l 99. Tty—%, 
 x? — xy + y? 90, 2=2. xe+ty+z2 
 ee ay + yt x + 4 grat 
 (x+y)? x 
 hath pe vai Tif jens nL SSR 
 %—C xr—y (1 —a)(1 — dD) 
 %— 3 28. «x. 31. ie UEC, 
 x—5d 3 a+b—ce 
 x+a 
 r+e 
 Page 115. 
 x+1. 7. x2 —Qary 4+ y?. 
 width, Gh fe 10 eo Gy ee 
 9 ev2+ae+1 
 ee a TON en ee oe 
 a2 + 3445 ages 
 % — 24+ ——_—_—_.. , 48 
 e+ 24+ x Il. 1+ 22 4+ ot + 2h + : 
 e+ 3a. 1 — x? 
 gz 9 ome De 
 Jaypee 12. Be ea ae a a 
 Page 117. 
 3bc 6ac 2be bac 3ab_ 
 3abe Babe Babe Babe Babe 
 yaty) (e+ y) Yee: 
 cy(et+y) ey(at+y) xy(et+y) 
 3(z@—y)i 4(e+y). 
 gz — 2’ 2 — y? 
 a(x + 1)(z + 2) ba(xe + 2) co{e+1) 
 
 a(x +1)(2+2)’ x(x+1)(@4+2)’ x(x4+1)(x +2) 
 a(a—1)(a2—-4) b(a+1)(a2?—4) c(a?—1)(a@—2) d(a?—1)(a+2) 
 (2-1) (a?—4)’ (a?—1)(a?—4)’ (4?—1)(a?—4)’ (#?—1)(a?—-4) 
 Be eee -—-6e ba + 10% 
 
 ’ 
 
 
 
 g?—4’ g2—4 2 —4 
 
 un? —5x 322—6%2 , 
 (9) 4) (2 — 5)’ “(4 — 2) (@ — 4)(2 — 5) 
 
 e-+-1 e+2 2+3 f 
 (%+2)(%+3)(x+1)’ (%+2) (a wee eG (+2) (*#+38)(*+1) 
 ax? + axy? bx ca? + cy? 
 
 x(at — ys)" x(at — yt) x(a — yt) 
 
10. 
 
 12. 
 
 14. 
 
 16. 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 xy — 4 xy? x2y? + ry? — 2 yt xt — x8y — 2 xy? 
 (a? — 7) (a? — 497)’ (7 — YP) @?—4y)’ @—Y)\@—4~) 
 
 14-2442 2742238424 32-323 52 1—2%4+22?— 208+ 2* 
 
 
 
 ’ ’ 
 
 1—< 1a7t 6 leat 1—z4 
 5 2+20 64+18 7x“2+14 ; 
 (a+2)(a+8)(a+4)’ (%+2)(%+38)(2+4)’ (2+2)(@+8)(a44) 
 tee 2x”7—2ay+2y? 3 i 
 + oy? 4 y?? at + oy? + y4 > ot 4 a2y2 + yt 
 
 abe ae 15 5(a + b) O ame 
 
 prtlem+1? patlem+l : a"(a + b)*’ a"(a + b)4 
 Page 118. 
 
 re tk 9. 322+ 3% —3 3. 50%. 4. 2 ; 
 
 a—l x-+1 z+ 10 x+2 
 8 + x 6. + 3a +3a% + a2+%—-—a ”. 8 
 x—2 (x +a)? a 
 ee 10 ee 19. 
 z+a 2 —1 (2—1)(a+1)2 
 2x7*—62%+3 wu, (2?+6, 13. a? — bab — 6°) 
 x? + a8 at — 4 a* — b? 
 
 227+182+4 16 : 17. 3827+182+6 ’ 
 (a — 2) (« — 8)(@ + 8) C+) + Date 
 Maa taa ta ac IP 18. —2y ‘ 
 
 et —247+1 (xty+z2)(x+y—zZ)(@—y— 2 
 __ 40(5%+4) 19, — wl Eee 
 
 (x + 1)?2(@ — 1)? (x — 2)(a — 8) (a — 4) 
 
 11 —32 ; 90. ry + y2+ 2x — 2 — y? — 
 (x —1)(@—2)@ +4) (x—y)(y —2)(2-”) 
 , Hage 121. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 cy 10 (ae Oa 16. &+ 3. 
 
 ety  b2 hage ee 
 
 b(% — y) | 1 b2c7d? Lik 
 
 a?(x + y) "  @6 1890: 
 
 7 by*(a?—ay + y?). 12. a? 4904 y* 19. «+1. 
 
 4 ax y? x 90 (4-2) Ge Ty 
 13. @ 4 or (y™ + 1)(a" + 2) 
 
 
 
 
 
 
 
 5 
 
 g 
 
 Ne) 
 o 
 o 
 co 
 Q 
 Cy 
 
ry 
 S 
 
 10. 
 
 (OAD TP © 0 
 
 OIA AP WN 
 
 
 
 
 
 ANSWERS. 17 
 
 
 
 
 
 
 
 Page 123. 
 y. 1 1 a 
 ar ; 3. ai 4. qo yt uz ye 5. b 
 10 by 
 2 x? — 16 xy + 327? oe LE en ap a 
 ' Oat — 30 x3y + 25 xy? x2(8% — 5 y)? 
 Reem re to. 1° ion 1. 1 19.4, 
 (x+y)? : “—y 
 Page 124, 
 Le Bak 10 eerie 
 y g, 2(@— 8), + day + 2y? 
 £-+ a. aid 1, 10%=3, 
 22-24 4 2ax_. dxz—1 
 art " gt 4 @? 19, 2*—2, 
 e2+1 8. x—y. 2xz-1 
 (a+ y)? 9. b+4 
 y b+2 
 Page 1381. 
 8 i Eos 21. 4,7. 32. a+. 
 5 12. 10. 22. 3. 33. a— b. 
 9 13. 0. 23. 7. 34. c43. 
 10 14. 3. 24. 13. 35. 3m+15n. 
 3 15. 41. 25. 17. 36. ab(b+ a). 
 13 16. 43. 26. 5. 37. b? — a’, 
 11. i owe & ry Ee oe 38. — 12, 
 41, 18. 34. 29. —(a+)). 39. — 14. 
 13. 19. —6. 80. a2?+ab+b% 40. 4. 
 10}. 20. 5 31. a?—ab+ b 
 Page 133. 
 $190, A; $60, B. 11. 130. 
 70, 11. 12. 75, 76. 
 36, 21, 57. 13. 100, 101. 
 $72, A; $56, B; $48, C. 14. 80, 81, 82. 
 70, 150, 220. 16. 50 years, man ; 45 years, wife. 
 541, 320. 17. 35 years, father ; 11 years, son. 
 100. 18. $6300. 
 120. 20. $120. 
 
18 
 
 21. 
 22. 
 23. 
 24. 
 25. 
 
 26. 
 27. 
 28. 
 29. 
 
 G0 et 63 Se 8S 
 
 os Ne Pn cone tag 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 $ 60,000 30. Cc ac ; abe. 
 37. l+a+ab 14+a+ab 1+a+ab 
 iA 31. $800, A; $400, B; $250, C. 
 $ 38200, $2800. 32. 20 miles. 
 
 $80, A; $70, B; $110, C; 33. 80 acres. 
 
 $500, D. 34. 5 p.m., 270 miles. 
 12 dimes, 24 quarters. 35. 1 p.m., 60 miles from A. 
 12, 10 miles by carriage. 
 
 
 
 10 cows, 20 calves, 40 sheep. 36. 160 miles by boat. 
 
 120, 121, 122. | 520 miles by train. 
 Page 141. 
 
 8, — 5. 15. 2, —5 28. 0, —2, —3. 
 
 —2, —4 16. —3, —5 29. 0, 3, 9. 
 
 —2,—8 17. 8, 3. 30. 4, — 3. 
 
 Sat i 18. —a+b, —a-—b. 31. —1}3. 
 
 412, 1926557, 32. 4, —6. 
 
 2, 2. 20. 1, 20. 33. 3, — 3. 
 
 1, —1. 21. — 3, — 25. 34 6+ Ce ue. 
 
 bibs py Rane bisa. ie a 
 
 5, — 5. 23. — 14, 7. 35. a+b, —(a+b) 
 
 4, —4, 24. 1, —4. 36. 2, —2, 3, —3 
 
 a—b,-—a-+ob. 25. —2, —7. 37. 3, — qb. 
 
 3, — i. 26. —1, 2, — 4, 5. 38. 0, 0, 1. 
 
 2, 5. 27. —1, —2,—5,2. 39. 0,0, §, —4. 
 40.01, 1.12 
 Page 144. 
 
 3, —3. 7. 41, — 28. 15. 44 hours. 
 
 1,-—1 9. 4,5; —5, —4. 16. 8+ hours. 
 
 4, WR temas $ Lis 2.0re 
 
 5. 11. 9,10; —9, —10. 18. ab days. 
 
 14. 13 Si met MI a+b 
 
 ai hav oe 4 
 
 2 Bde AL tOuper OF (Factoring by Art. 79.) 20. 2. 
 Page 159. 
 
 (8, 1) 4. (3, 2). 7. (4:5, 195). 9. (6, 10). 
 
 (4, 1) a Ca 8. (144, 14). 10. (6, 15). 
 
 1,4 6. (—5, 0). 
 

 
 
 
 
 
 ANSWERS. 19 
 Page 160. 
 1 (2) = 1). 5. (129, 214). 8. (—}, 14). 
 2. (17, #4). 6. (* a+b rea), 9. (%, h). 
 Soe (85 0): 3 3 Tle (0 m)s 
 4. (49, 2). 7. (4, 0). 
 Page 165. 
 Lie yds By (2, 10): Sa (4e1). 
 2. (4, — 4). 6. (5, 2). 9. (#8, — 29). 
 3. (— #, 2). 7 ae lc—ap\ 10. (4, 1). 
 4. (48,, 234). ' \am—odl ee ae ae 
 12. Ga bm km + ae 16. ( bq ne ag i: 
 In+km ak—lUb mb+an mb+ an 
 13. & — bP b? — a? ie 17. (e + mn + nt — mt). 
 ac—b* b(c—a) m+n m+n 
 14. (2n—m,2m—N). 18. (5, — 4). 
 15. (7, 8). 19 (= ="). 
 20. (2, 5). “\ab- a — > 
 Page 166. 
 2. 2 cents, 3 cents. 20. $24, A; $36, B. 
 3. 80 cents, 45 cents. 21. 19 or 91. 
 _ 4. 36: 22. 10 dollars, 40 quarters, $20. 
 5. 7% 23. 200 boys, 300 girls. 
 6. 12, 63. 24. 5miles, A’s rate; 4 miles, B’s 
 weace, 61. rate. 
 8. 30 sheep, 12 calves. 25. 108, 45. 
 9. 60 years, 20 years. SG 118.13. 
 10. $540, A ; $360, B. 27. 24 feet, 16 feet. 
 11. 36, 54. 28. $760, A; $920, B. 
 12. 20, 50. 29. $800, 4%. 
 13. $180, $330. 30. 27, 19. 
 14. 14, 10. 31. 40 men, 20 boys. 
 15. 25, 36. 32. i. 
 Lig. 33. 40 years, 15 years. 
 18. 49. 34. 80 feet, 100 feet. 
 19. 12 persons, $10 each. 35. 15 miles, 3164 feet. 
 Page 172. 
 ae. 10.70, 1). 3. (1, —1, 0). 5. (4, t, 4). 
 2. .(12,-8, 10). 4. (14, 10, —8). 6. (G4, 4, 4)- 
 
20) THE ESSENTIALS OF ALGEBRA. 
 
 mr nr ir 9. (6, 4, 3). 12. 18, 50, 44. 
 ( 2” a 10. (1, 2, 3, 4). 13. 50, 54, 72. 
 (22 24 2n 11. $300, A; $300, 14. 629 
 2 ight Le B; $150, C. 
 
 15. 15 days, A; 10 days, B; 30 days, C. 
 
 16. 547 ininutes, A; 80 minutes, B; 40 minutes, C. 
 17. 24, 38, 51. 
 
 18. 20, 40, 120. 
 
 19. $500, A; $800, B; $1200, C. 
 
 Page 176. 
 3. +: 4 a3bc?. 11. + 25 a'pl5c, 16 9 x2yt 
 4. +13 xyz. 12. 10 abc’. ~ 16 a'b2e3 
 5. 12 xy". 13. + 20 ad8c9, 17. +2 atb®c?. 
 8. +(a@+32). 144 12 ab. 18. — 2 a>xoy. 
 Oe Date, Ni eaey 19. + a2bic?, 
 io. + &. 15. — 7 ade. 20. — a%bic8, 
 cd 
 Page 177. 
 1 +(@+ 8). §. +(42%+4+7y). 9 +(a—6+4+3¢). 
 2. +(8%+4 4). 6. + (ax + by). 10. +(2%—y+32). 
 3. +(@-—9). 7% t(e+ty+2). ll. +(a+b+c¢c+d). 
 4. +(%—5y). 8. +(%+3y+4+2). 12. +(~7—38y+4). 
 Page 179. 
 (The double sign is not written, but is understood.) 
 eae b, 5. 202+ xy + y?. 9. 2+ 3a ae 
 Ais ais 2 2 2 
 eee 2x +3. 6. 32 5 ary +2 y?. 10. Y + by +2. 
 3. 2274+ 327 +4. 7 l—-ax—8y. 3 2 
 4. B+60e7+5x44+1. 8 24-—8y+42. 
 Page 180. 
 1. 14 boat t dat . 1 oo 
 221 + 2a 2 at A ge, 2 
 3. l—4xr—f2?- 223, if aa 2 2 4+ fF 
 4. 14+4%4+4 327 — 3 2. 3 92 a 
 a E x + —— g 
 5. 1+ 3% — 4b a? + 195 23. +5 nF 8 "1G 
 
 Page 182. Section 125. 
 
 1. 324. 3. 239. 5. 485. 
 2. 528. 4. 248. 6. 1045. 
 
ANSWERS. 
 
 Section 126. 
 
 21 
 
 
 
 
 
 4.28. 8. .865. 5. 1.054. 4. 2.236: 
 2.94. 4. .0548. 6. 1.414. 8. 3.316. 
 Page 184. 
 x + 2. 4. a+b+1. % 227—32+ 5. 
 e2toa+i. §. a@+6 42. 8. 2a7b? —8ab+2. 
 x? — xy + y?. 6. 3a—4y. 
 Page 186. 
 Lis 3. 91. 5. 138. cp RUE 9. .2008. 
 67. 4. 85. Gi 12: 8. 1.442. 10. 1.270. 
 Page 189. 
 at y* 10.7 1; 
 cs ales 1 
 4 b6 11. = 
 5x 10% gre: % 
 2 yes BE 
 62 b3 
 1 ee Be lat | 
 al wa oy SW 14.0% = 
 a*b2xy®, 9. ai2p%cl2 y2 yt 
 eee Q1. a 3b 26348, 
 y? eames $44 25 
 x? — y~?, Soir airUr Ce 
 3 2-2 __ 7-8 
 = a e rae 23. 4 ch + 4 aqbe? 
 et—3 4344 2?—6243-—2271. 24. 323y3 —4ab3 
 
 y? 
 Dm +- e- 2-2 + yt. 
 
 a343 a-2b-143 a-1b-24 b-3, 25. 
 Page 195. 
 4V3. PEN Sep 3y/2, 4. 7V3. 
 
 (3 ay? — 2 ay + 13 y?) Va. 6. (5ab — 9 4+ 12 a2b?) Vab. 
 (ho GP ESA ee 
 
 8. [ax2(y +2) —38a% + 12(y4 2)]Vy +z. 
 
 
 
 V5. 16. «v3 2. 
 av2 x. 16. (4a+6b)V3a. 
 (m —38n)V2m. 17. (a.— 6 b)V5b. 
 —(4%+6y)Vv3. 18. 26V11. 
 . (2242y)Vx + y. 19. 48V13. 
 . avd +3. 20. (11 a2 — 1802) Vx 4+ y. 
 
Sree awe 
 
 10. 
 
 11. 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 
 
 
 
 
 
 
 
 Page 196. 
 40. | 7. 6(a— b)*Va—0. 
 400 V2. 8. 10(@ —y)2(a@4+ y) V3(a3— 9). 
 900V5. 9 BV2(¢ +9). 
 48 ax? Vat 2V5(« — y) 
 (@ —y)?Va —y. 10. 4. 
 540 wty?. 
 Page 197, 
 —2 2-1 8 -1 4 11 5.2 6. 56—12V3. 
 7 +2V10. i 12. 42. 
 6 + 2V6 — 2V2 —2V3. 13. 8V3 +1. 
 84+5V3, _ 14. a—b—c+2Vvbe. 
 —14—8v3. 15. 4% —10Vxz + 252 —-9y. 
 BV 8 bVb 1116 ay 16. 
 Page 199. 
 v10. 4 2V2—-2. y —12V6 + 9v42 
 2 gecvaal aa 47 
 Ee pynnucriie US 8. —(3 + 2V2). 
 v 
 eres 6. 5V2 4+ 4V3. 9, oe ss 
 74+3V5. 19, 6V8—8V15 + 3V2—4V10. 
 2 105 
 2V5 + 2V8 + VI5 +38. 13. ve oe ALLIS ft 
 ( 
 14, 9V74 6V14 + 8V21 — 6V5.— 4V10 — 2V15, 
 43 
 15, 2Va—3Vb—4Vab + 4a, 
 ; a—b 
 16, 2Ve—5Vab —4Vb —2b +34, 
 : 2a—8b 
 17. 6V~a—1+4+94%-8 
 10 —92 
 ig, @tyve+(et+y)Vy —2v2y —yv2 0, 
 x? + y? 
 
 19, tVe+ yVvy. 
 %—y 
 90. - (24 —3b+4+2V2@—3 ab + ey) 
 (a+ b) 
 
 
 
ae! m oN 
 
 rr 
 ee ee OOS 
 
 i 
 
 ANSWERS. eS: 
 
 Page 200. 
 
 . 14+Vv2—v3) v2. 8. (V8 + V2 —V5) V6. 
 
 Dewi 2 V7 )(1 —2°V 15). 4.0 (V6 + V2 11) (1. V2); 
 HCV 10i4--V2 — V3)(6 — 2 V6). 
 
 6. (Va—Vb—c)(a— b— c+ 2c Vb). 
 
 %. (142 V2 —3 V3)(6-V6 + 17). 
 
 8. (2Va+V2b—38vc)(4a—2b —9ce—6 V2 bc). 
 
 
 
 Page 202. 
 V3 4 V7. 5. V11+V5. 9. 243 V5. 
 vi — v5. 6. Va—vb. LOSE yt 
 V3 + V8. 7. V10—v2. ll. 8V7—2 V6. 
 V5 — v2. 8. -V74 V6. 12. Vat+b+Vv6e. 
 Page 204. 
 LO V24. g 243 V5i, 190, Va +2 abi, 
 . —15 V2 —10. . ae ae We 
 13 —8 V3i. ee ee eV t, Livia 2. 
 —~V64+3 V2 7 1, AO +4b45, 
 +(24+3V3)i 8 V3. aa ayers. 
 . 8+v6i. 9. y> — a3, 
 Page 210. 
 +6 ih AES v651 
 Page 212 
 8, —1. 12. 43, 7. 23. —bive?—c. 
 3, —3. 13. —4, — 42. 24. m, n. 
 3 il. 14. 3 =I. 25. oe —1. 
 7 15. 2, 6. a—b 
 oy) — 3 16. — 3, — 3 Sap re bere, 
 9, —4. 17. — 3, —%. MCRL Lars 
 Bae 18. 3, — 1A, 27. 3, Fe 1. 
 See € 19 2ee 28. 3, 2. 
 ea 20. 2, 6. 29. 3, 5. 
 4, — 5, 215.110; 99, poe VE +4 ab, 
 4,—3 22. +6. 2a 
 
i" 
 
 ve 
 
 a i 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 
 
 
 
 Page 214. 
 
 Bio 3. —4,-h.) 5) —}, 1. Fee 9..° +8) 8: 
 L$) 48 6,402: —3, —%. 10. 3, —4. 
 Page 215, 
 
 Bg bs BAVA if 4 3 + V69_ 7, S£VO6T. 
 8 \ipaed 0 d 4 
 T4+VB ee aut eee 
 Buen ; 5 g, 43+ 11299. 
 a 75 
 = NLeE Via 6. 3 v5, m+Vm2+1 
 4 2 10. 
 3m2n 
 Page 216. 
 5+ ivi. g Stiv2, 5, Stiv7l, 
 2 3 6 
 —9+iv7 g Ut iv103, 6. 
 4 14 ; 2 
 "7. 8 4 £1.08, 8. o + tv 23 iV23. 
 Ae) 
 Page 217. First Exercise, no answer. 
 Second Exercise. 
 —~5+4+vV6i 7. +10. 154+ V117 
 pata: “alin ieh 19) eee 
 6 8. 0, — j. 4 
 34+ V6. 25 + 5iv47. 181s 
 3 18 14. 1, —1. 
 24+V7. 10, ~2ZLt ¥501, 15, Lt 2iv2, 
 i, — 3. Gir + 80 eee 
 1, —4§, wu, 2£tv17, 16, Do£9V33, 
 ey AG, 7 2 | 
 Page 223. 
 
 w—{e—-2=0, 2+2%—20=0, 22?7—1824+1=0, 
 
 6. zc? -—62+11=0. 
 
 7. 14422 — 864”%+4 1171 =0. 
 8. x7-—2ax+a?+b?=0. 
 9. 27-—61lx+ 912+ 25m =0. 
 10. a? —2V5ax + 48 = 0, 
 
 7 — 245 —4= 0. 
 v*—6x+34=—0. 
 
 . de?7—-4xe—-55=0. 
 

 
 
 
 25 
 
 ANSWERS. 
 Page 225. 
 1 ~8tVB -38tV13_ 11. 4+ 5, 4 4% 
 2 4 2 19.80%: 
 Beate Wd N/S 13. 18. 
 [pie 2 cas ais 14.14V4 ¥ 2Vv3i. 
 ota — 6 + 2V'89 | Tyee: —145Vv5_ 
 iy 15 2 
 5. =Aa tive, 16. 4, —9, — SE VOL, 
 bi 2S, Lats, Vio 410, =a 
 7 — 33, —3. 18. 9. 
 8. +37, + V2. 19. 16, 5V5. 
 9. +2V2i, +i. , 384+V505 
 3/= M4 20. 2, Oe) 
 10. 8vV3; 227-4. 4 
 Page 226. 
 Peele teil 2k., 14. 4 mi. 
 SS. 15. 24 min., 30 min. 
 3. 18, 12. 16. A’s, 5 mi.; B’s, 3 mi. 
 4. 30; 11. 172 10;A; 
 a . mh i na We = 04 ve Wee 
 : ’ 9 495 to, — 14, — 11. 
 7. 100 ft. by 132 ft. Le Vins eee 
 8. 20. 20. 6 mi., 3 mi. 
 9. 15. 21. 15 ft., 20 ft., 25 ft. 
 1Oredy 20) 3-20, — 12. 99, et Va'—4ab a~Va—4 ab 
 11. 60 sq. ft. 2 ; 2 
 12. 39,40; — 40, — 39. 23. 24 rd. by 60 rd. 
 Pit a) 23. — 21; 24. 10 hr. 
 Page 235. 
 1, *(0, 5),-(6, 0). Sin GL 8) Cae) 
 2. (0, 2), (3, 0). On" 452), Co hi ©) 
 Patan). (—6, — 8). LONE 72 10): 
 4. (7, 8), (8, 7). uw. (ee 4v3E ets! : 
 Barto ),4--15, — 9). AIS Oca i. Vaid in 
 fee my, C6, == 9). P25 (105 )5 C1 Ea): 
 7. (2,3), C1, 6). 
 
pb oe 
 
 a. 
 2. 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 13, (6, 8), (4, 0). 
 
 ra! (-} 1 2evR), (-} 1 —). 
 3 6 3 6 
 
 15. (— 3, 5), (—7, 7). 
 
 Page 238. 
 
 (2, 3), 2; —3), (— 2, 3), (= —2, —35). 
 
 (3, 4), (5, —4), Co 3,4) a, a). eae ~ ou 
 (4V5i, V105), (4V5i, —V105), (—4V57, V105), 
 (—4V51, — V105). 
 
 (3, 7), (3, Ei), (—3, 7), (Oy ale 
 
 (-V203 i, V85), (V203 i, —V85), (—V 2087, V85), 
 (— V203 i, — V85). 
 
 (6,2); (6, 9), (5) 9) (bao 
 
 (6V 3h 2V$r), (6V 3 -2V ’ (-6V3;, 2V35), 
 C= 6V sr, —2V3,). 
 
 (6V43, 6V 2,1), (6V$8, —6V 2,1), (6V43, 6VE, HO; 
 (—6V$3, —6V 7%, 1). 
 
 
 
 
 
 
 
 Page 241. 
 
 (2,1), (—2, —1), (1, 2), (-1, —2). 
 
 (Vii, Sawa 4) CWA, 3Vi i), (SV, 5Vae)s 
 ez —8V4h, —5V 4). 
 
 (3, 4), (—8, —4), (7V$, 2V4), (-7V§, -—2V9). 
 2, 1), (—2, —1)s (4V4, v4); (—4v4, — v4). 
 (8V3, v3), (—3 v3, —v3), (4, 5), (—4, —~ 0) 
 (5, —L); (-5, 1), (3, =); (—3, 4). 
 
 (2,4), (2; —4), (V2, 3V2), (1 ee 
 (4°) (46): 
 
 Cd, 5), Ce 1, =); (4V42i, V#2 4), (—4vV42i, —v}2%). 
 
 lem — bn, [an — Gf 
 Ay ———— ————. 
 * Fee oe am — Ib 
 
 : G; 3) (—-$, —- » (4 3), Gey 3 ’ — $). 
 
 (3, 5), (—3, 165; (8, 5), Ce 8, —5). 
 
 Page 245. 
 
 4. (28))-( 2) — 2h eye Cee 
 2. (8, 5), (8; — By (5) 3)) (oem 
 
OTSws ooo ie 
 
 ANSWERS. QT 
 
 
 
 Br (o8)..(8, 3). 8. (8, 5), (4 10). 
 4. (3,3), (3, 8). 9. Gb, Gd- 
 5. (7, 2), (—2, —7). 10. (4, 4). 
 6 (4, 5), (5):8). Tete 8) oe). 
 Pans, —3). 1246, 3):2(57 3): 
 
 13. (10, 20), (20, 10). 
 
 a+b a—b a—b a+b 
 eae) a) 
 Page 249, 
 
 (4, 3), (—4%, a) 4. (2, 1), (1, 2). 7. (3, 1), (1, 3). 
 (4, 2), (5, 4). 5. (3, 2), (—2, —3). 8. (3, 2), (2, 3). 
 (1, 2), (14, 4). 6. (4, 3), (3, 4). 9. (38, 2), (—4, —4). 
 
 fee ey 1). (8,4), (4, 3); (—6, —2), (—2; —6). 
 PeewotehVe1t\ | flaVoleealev sii 
 
 (3, 2), (—2; —3), (ee ey ey Se ok 
 
 fetta), (24 57,2 — 51), (2— 57, 24.5%). 
 
 (2, 5), (—2, —5), (8V3, 7V4), (—8 V4, —TV}). 
 
 (2, 1), (— 2, —1), es 4), — 3, ey t). 
 
 (3, 9), (9, 8). ‘VW. Cio ZY (-4.\ 20-0); 
 (1, 6), (6, 1), (Met BN, (25M Sev, 
 2 2 2 
 (4, 2), (—2, — 4). 20. (2, 3), (, 2). 
 Vie: 6. 5; 2. 11. 66 ft. by 200 ft. 
 25, 21. 7. 900sq.rd.,1600sq.rd. 12. 48 or 34, 
 3,4 8. 50 rd. by 20 rd. 13. 4 yd., 5 yd. 
 6, 11. 9. 50 ft. by 120 ft. 14. 20 da., $3. 
 5, 3 10. 8, 16. 15.44, 
 Page 251. 
 cy —iay —2a?. 
 - (10, By (— 12, 5), (4a, 5a), (8, —34), (-—c+a,c¢—Db). 
 “2 —62x+4 2. 10. (— rary 
 137 V3 — 56. 11 1+v1—4ced 
 a8 + o2Va-1 — 1 — Va-8, ; od 
 10 +. 8V30 — 3V6 —2V5. 13. (35.and 86) (— 86. and — 85). 
 34 14. 85. 
 5 + 127 15. 6,,. 
 26 , 1G WiK6 A). Ge Gin 4). 
 ad + add} + 03. 17. (3, 2), (— 3, 
 
re 
 
 ay Pd heer 
 
 
 
 
 
 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 52 mi., 40 mi. oa ere 5a 
 OF Do 4’ G+ 
 
 =the VSB 25. Ont 6 ak URateee 
 Hy ae 26. 16. 
 ion gE a 27. bd. 
 
 Vy ys 28. 2,1, 2, 10. 
 
 +iv3, +4. 29. 450 abay. ei 
 
 30. 8+y-H = 2V8¢g. 
 
 
 
 V(11)?, 4. us 
 é: ) 31. 4 b. 
 . 26, 52. a+b 
 4 3 2.3 2 5 8 5 
 82. a + b2 +064 2a8bt — 2a8c? —2dtc?. 
 11+ 4V6. 40. 2, 5. 
 35 + V1049 41, CHP 4V2 a +204 
 Land , a? — b» 
 @2@ — 15)(2a +38) 42. 144. py 
 Ta(a — 3) 43 —15+V161. 
 140 ft. by 200 ft. 8 
 —2y. 44. 0. 
 +S, + 3, 45. 40rd. by 40rd. 20rd. by 80rd. 
 (1; 3)6 =) 8): | 
 Page 259. 
 $, 6: a—b ll. 2. 16. 4+ 8. 
 4 4 ab hb Awe by ei 
 ¥ 7 a-—2 13 hou Lesa: 
 a2, 8. 224+ 5a. 2 7 19:1 om 
 x—-Y-—2. 9 at+ay+y*% 14 1. 20. 7. 
 x—Y. LOWE. 15. x«—1. 
 Page 262. 
 60. 3. 33. 4. 75.3984. 9. + V10. 10. 33511.1. 
 Page 266. 
 113 5. + aN ene 1 8. 49, 22, 14. ac 
 37° 6. 53! b 9.88; b 
 + 15, anit ae 10. Yes. 15, i, 
 — $$. EE Nha 11. No. ‘True b 
 forz=1. 
 Page 273. Section 194. 
 120. 3. 1260. 5. 12,870. 7. 13992. 9. 91,390. 
 6. 4. 58,140. 6. 2. 8. 300. 10. +45. 
 
ANSWERS. Se RD AL, 
 
 Section 195. 
 
 17.120. 5. 75,600. 9. 120. 
 2. 840, 5040. 6. 720. 10. 24. 
 
 3. 3360; 57,120; 1,860,480. Mae BI GON 1 ZU: il. 80,640. 
 4. 358,800. 8. 630; 34,650. 12. 27,720. 
 Page 2777. 
 
 184-126. 3. 15,504. 5. 4275. rf Pyee ba 9. 840. 
 
 2. 210; 220; 220. 4. 475. 6. 1001. 8. 336; 456. 10. 16. 
 
 1. 53. 4, 365. 
 2. 144. 5. 20. 
 3. 14, 18, 22, 26, 30. 
 
 \6. 6, 11, 16, 21, etc. 
 
 Reo ka 
 2. 1023. 
 
 8. 6561, 9840. 6. 
 
 = 
 
 Page 283. 
 6. 13, 483. 10. 30. 13. 7650. 
 8. —75. A Eo Oly 14. 2745. 
 9. 135. 12. 78. 15. $405. 
 
 772 50) 268, 482d) ot) 18015, 146: 
 
 Page 288. 
 4043, 7%. $20,475. 10. 8, 18. 
 — 4920. 8. 5. 14. 8, 24, 72, 216. 
 $255. 9. 14. 15. 2, :4, 8. 
 
 Page 289. 
 
 (1+18a% + 153 #2 + 816 23 + 3060 2 + 8568 a5+ 
 2. | 1—21y + 210 y? — 1330 y? + 5985 yt — 20349 y+ 
 (25 4+ 25 x24y + 300 x23y? 4+ 2300 x22y3 + 12650 a21y4 + 53130 129yo+ 
 
 1x2x3x.-.-15 
 
 30 x 29 x ---16 ‘ 
 
 ((a) 1+%+27+ 23+ xt+ 
 
 (0) 1444-42? + apa — hy ott 
 | (cy) Lt gat yiat + Hh? + pag att 
 
 (a) 1— at §a2— 490 + Hg ot 
 
 — 
 
 Page 290. 
 
 64 28 — 192 xy + 240 aty2 — 160 x3y3 + 60 x2yt — 12 xy® + yw, 
 
 x + 10 vty + 40 wy? + 80 xy? + 80 ryt + 32 ¥?. 
 * | 128 a? — 1344 abd + 6048 a5b? — 15120 a'b? + 22680 abt 
 
 [ §  _ 90412 a2b® + 10206 ab® + 2187 b7. 
 
co 
 
 THE ESSENTIALS OF ALGEBRA. 
 
 
 
 256 x8 — 1 6. 2, 6, 18 11. 293,930. 
 2”%—1 7 +16. 12. 59,400. 
 1950. 8. 27. 13. 5040. 
 672 x. 9. — 252 x 32 x 243. 14. 1980. 
 saa 10. £433. 
 18. 2h ere te — rot 
 
 
 
 = Se 2 oe 2) eee 
 16. 10 20 E000 — 1800000 1280000000° 
 
 17. 1+6y%+15%+20% v2 +15 a 46 ania 
 $ 
 
 20. 
 25. 
 
 rm 
 
 i 
 = 
 
 Ene er 
 
 ge ep LT le lee od a 
 
 b2 b2 Vb B3 
 18,564. 21. $ 465 ; $3075. 
 A TEEY 22. 23. 
 Lote 6. 24. 8a*—S8ava?—1—4. 
 1+65%+ 152724 30234 4524 + 51 2 4+ 45 xo “ 30 x7 + 1528 
 + 5x9 + x, 
 Page 296. 
 2. Sopa 5. 0. U8! 9. 5 
 1. Alene 6.) = 3. 8. 2. 10. 3 
 Page 297. 
 5.6590. 6. 4.8663. 11. 1.3010; 1.8751. 
 3.9943. 7. .9030; 1.2040; 1.5050. 12. 1.6902; 2.5353. 
 a At 7627 8. .9542 ; 1.4313. 18. 1.6232. 
 0.5065. 9. 1781 ; 1.3801. 14. 2.1461. 
 6.8482. 10. 1.3980; 2.0970. 15. 1.6620. 
 Page 302. 
 2.7292. 11. 0.4971. 21. 651. 31. .592. 
 2.9930. 12. 0.43848. 22. 82.13. 32. .0144. 
 2.7042. 13. 0.1504. OS hay: 33. .00391. 
 3.1287. 14. 1.0969. 24. 5890. 34. .0446. 
 1.3075. 15. 1.5740. 25. 14.8. 35. .293. 
 2.07 Tl: 16. 1.5740. 26. 2.80. 36. .30087. 
 0.1976. 17. 2.9931. 27. 143.74. 37. .33392. 
 2.8319. 18. 0.7321 28. 1812.9. 38. .00248, 
 4.7482. 19. 1.5981. 29. 2.3747. 39. .30871. 
 1.7190. 20. 3.7691. 30. 39.509. 40. .039682. 
 
oS 
 
 _= 
 
 719500. 
 
 762400000. 
 
 254. 
 
 4.39. 
 2.52. 
 
 $ 2429.40. 
 $ 3165.70. 
 
 ANSWERS. 
 Page 305. 
 4. 19.014. 7. 9.556. 
 5. 61900. 8. 1138.05. 
 6G. 1004.6. 
 Page 306. 
 3.7 622.23 Tae 2. 
 1.59 6. 4.54 8. 3.59. 
 Page 308. 
 3. 12.86 yr. 5. 23.51 yr. 
 4. $1258.60. 6. $50.12. 
 
 31 
 
 9. 8.9975. 
 10. 1221100. 
 
 7. 3.55%. 
 8. $668.17. 
 

 

 

 
iil 
 
 URBANA 
 
 C001 
 
 oS 
 = 
 7 — 
 = 
 Li. 
 i=) 
 > 
 = 
 w 
 c 
 Lu 
 = 
 
 wahet Hp fg wr twt 4 
 hase eve gee eters 
 Se rerrrpiichmoe wae? 
 
 THE ESSENTIALS OF ALGEBRA NEW YORK 
 
 512.9AL2E