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L/i> 



AD 



y 



ALGEBRA FOR BEGINNERS. 



BY 



JAMES LOUDON, M.A. 

PKOFESSOR OF MATHKMATICS AND NATURAL PIULOSOPHV, 
l'NIVF.R.SITY COI.LKUF,, TOUONTO. 



TOROXTO : 
COPP, CLARK & CO., FRONT STREET. 

1 8 7 (5 . 






4^CQ 



PEE FACE 



V 



Be,.nners, for who„. this book is intended, are recon.- 

mended to work according to the rules printed in itahcs, and 

to om.t, on a first reading, the articles in sn.all tvpe. Those 

^sho desire only a practical acquaintance with the method. 

of working, may also omit Chapters X and XXI. 

The author will feel obliged to tearhnrc f,..- o / 

^'tot-ti lo leaciieis for any suggestion. 

they may have lo communicate 



SKPlKAIlJiiK. 1S75. 



CONTENTS. 



PACiK. 

I. Introduction r 

II. Addition 8 

III. Subtraction i6 

IV. The use of Double Signs and Brackets 20 

V. Multiplicatio' 23 

VI. Division 30 

VII. Examples involving the application of the first 

four Rules 39 

VI II. Simple Equations 42 

IX. Problems 49 

X. Particular results in Multiplication and Division 55 

XI. Involution and Evolution 61 

XII. The Highest Common Measure 72 

XIII. The Lowest Common Multiple 84 

XIV. Fractions 88 

XV. Simple Equations, continued 105 

XVI. Problems, continued 108 

XVI I. Quadratic Equations 113 

XVI 1 1. Problems , 119 

XIX. Simultaneous Equations 123 

XX. Problems 129 

XXI. Exponential Notation 1 33 

Answers 141 



I 
■I 

1 



I 



I 
8 



ALGEBRA. 



■•o*- 



CHAPTER I. 



f 



I 



INTRODUCTION. 

1. The operations of addition and subtraction, which in 
Arithmetic are stated in words, are denoted in Algebra by 
the signs + and — , respectively. 

Thus, add together 12, 6, and 6, is expressed 12 + 5 + 6 ; 
from 15 tahe 9, is expressed 15 — 9 ; add together 3, h, 5^ and 
from the sum lake i and 3^, is expressed 3 + 2 + 5^—^—31; 
and so on. 

2. The sign + is called the plus sign, and the sign - the 
minus sign. 

Thus, 24 + 2 - 15, is read 24 plus 2 minus 15. 

Exercise I. 

Employ plus and minus signs to express the following 
ope ;ations in Algebraical language : — 

(1.) Add together 56, 10 and 15; 10, 12 and |; 2, land f ; 

6> 10* 12> «*"^ 3« 

(2.) From 29 take 15 ; from 2^ take 1^ ; from 2*5 take 1*6. 

(3.) From the sum of 11, 35 and 6, take 17; from the sum 
of 81 and 75 take 69 and 42. 

9 



INTRODUCTION. 



(4.) To the difference between 15 and 7 add the sum of 
8 and 9. 

(5.) To the difToi'cnce between 28 and 16 add the difference 
between 10 and 4. 

3. The signs + and — are also used to denote that the 
quantities l)efore which they are written are, respectively, 
io he added to and subtracted from some quantity not neces- 
sarily expressed. 

Thus, +4 denotes a number 4 to he added to some number; 
—5 denotes a number 5 to be subtracted from some number. 

4. Quantities to be added are called positive quantities, 
and quantities to be subtracted negative quantities. 

Thus, +2, +7, +^, are positive quantities; and —3, — 1, 
— I, are negative quantities. 

5. As an illustration of positive and negative quantities, 
we may take the following examples : — 

(i.) A man has a certain amount of cash in hand ; he owes 
$150 to one man, and $280 to another ; and there are owing 
to him the several sums of $100, $210, and $120. Now, in 
order to determine what the man is worth, these several 
sums are to be considered in connection with the cash in 
hand ; $150 and $280 are evidently to he subtracted, and $100, 
$210, and $120, to be added. The former, therefore, may be 
de?ioted by -150, -280, and the latter by +100, +210, 
+ 120. In other words, in the process of finding out how a 
man's business stands, the sign + may be used to denote his 
assets, and the sign — his liabilities. 

(ii.) The mercury in a thermometer rises and falls in con- 
sequence of changes in the temperature, and the amounts of 
these variations are expressed in degrees. In order to deter- 
mine the reading of the thermometer, some of these variations 
must be added to, and others subtracted from, the reading 
before the variations took place. Suppose, for example, that 
during the day there take place a rise of 5°, a fall of 7°, a 



INTRODUCTION. 



I 



rise of 12°, and a fall of 8° ; 5° and 12° are to lie added to, 
and 7° and 8° subtracted from, the reading of the morning. 
The forjner, therefore, may be denoted by +5°, + 12°, and 
the latter by —7°, —8°. If the degrees are measured from a 
zero point, distances above may therefore be denoted by + , 
and distances below by — . Thus +20° means 20° above 
zero, and —5° means 6° below zero. 

6. In like manner the signs + and — may generally bo 
employed to denote the two relations of contrariety which 
magnitudes of the same kind may bear to one another in 
some defined respect. Thus, if money received be denoted 
by + , money paid away will be denoted by -- ; if distances 
walked in one direction be denoted by + , distances walked 
in the contrary direction will be denoted by — ; if + denote 
games won, — will denote games lost ; and so on. 

Exercise II. 

(1.) A owes B $60, B owes C $30, and C owes A $20; 
employ the signs + and — to denote the assets and liabilities 
of A, B, C. 

(2.) B and C owe A $20 each, C and A owe B $30 each, 
and A and B owe C $40 each ; express in Algebraical lan- 
guage the assets and liabilities of A, B, C. 

(3.) A pays B $10, B pays C $7, and C pays A $4 ; express 
Algebraically the amounts paid and received by A, B, C. 

(4.) Denote the following variations in the thermometer ; 
a fall of 2°, a rise of 5°, a fall of 3°. 

(5.) Denote that the thermometer rises and falls alternately 
1° per hour for 5 hours. 

(6.) Denote that the thermometer £Eillsand rises alternately 
2° i)er hour for 4 hours. 

(7.) Denote the following readings: 25° above zero, T 
below zero. 

7. In Algebra the numerical values of quantities are denoted 

B 



INTRODUCTION. 



by (1) figures, as in Arithmetic; (2) tho letters of the 
alphabet, either alone or iu combination. 

Thus, if there lie three points. A, B, C, situated in that 
order on a right line, the distances between A and B, and 
between B and C, if unknown or variable, may be denoted by 
o and h feet, respectively ; and, consequently, the distance 
between A and C will be denoted by a + 6 feet. If lie 
between A and B, and a and h denote, as before, the lengths 
AB, BC, the length AC will be denoted by a—h feet. 

Again, if x denote tho length, and y the breadth, in feet, of 
a room, the dimensions of a room 8 feet longer and 5 feet 
narrower will be a? +3 and y— 5 feet, respectively. 

8. In Arithmetical operations which are performed with 
figure symbols alone, all mention of the unit is suppressed, 
the results of these operations being true, whatever be the 
unit in view. Thus, 10 and 5 added together make 15, 
whether the suppressed unit be a pound, a gallon, or an inch. 
Algebraical symbols have a still greater generality. Not only 
is the unit suppressed, but the number of units is not assigned, 
as in Arithmetic. Thus a may represent any number referred 
to any unit. Operations performed in Algebraical symbols 
will, therefore, give results which are true for any numerical 
values which may be assigned to the symbols. 

9. Tho product of symbols which denote numbers, is repre- 
sented by writing them down in a horizontal line one after 
another, in any order, with or without the multiplication 
sign X , or dot . , between them. 

Thus a6, 6a, «•&, b'aaxb,bx a, all denote the product of a 
and b; ubc, a-b-c, axbxc, the product of a, h, c; and 
so on. 

Figure symbols are written first in order ; thus, da, 6ab, 
6abc, f«. When there are two figure symbols, the sign x 
only must be used between them. 

Thus, the product of 5 and 6a is denoted by 6 x 6a, and 
not by 5-6a, or 56a, whose values avejive decimal six times o, 
waA. fifty-six times a, respectively. 



.«f. 



INTRODUCTION. 



\ 



When thero arc throe or more figure symbols, either the 
sign X or • must bo used. 

Thus the product of 2, 5 and 7 is denoted by 2-5 -7, or 
2x5x7. 

10. The symbol a^ stands for im ; a^ for naa ; a* for aaaa ; 
and so on. «^ is read a s(/Hared, or a to the power of 2 ; u^ is 
read a cnbnl^ or a to the power of ^\ a* is read « to the power oj 
4 ; and so on. 

11. Tlio power to wliich a letter is raised is called the 
iiiikx or exponent of that power. 

Thus, 1 is the index or exponent of a ; 2 of tt^ j 3 of a*, 

12. The quotient of one quantity, n, divided by another, &, 
is denoted by either of the forms, « -f- i or ^ . 

Thus the quotient of 2a divided by 36c is denoted by 
2a-r-36cor by|;i. 

13. Any object or result of an Algebraical operation is 
called an Algebraical quantity or eapression. 

14. It is to be observed that the numerical value of an ex- 
pression which is not fractional in form may be a fraction, 
and the numerical value of an expression which is fractional 
in form may be an integer. 

Thus, if to a we assign the value 2, and to 6 the value ^, 

tho value of =- will be 2 -f- 2, or 4 ; whilst if the values of a 


and 6 be ^ and i, respectively, the value of ah will be i x i, 
or i, and the value of ^ will be J -r i, or 2. 

15. The symbol = stands for is or are equal to, and is 
written between the quantities whose equality it is desired to 
express. 

Thus a = 3 denotes that the value of a is 3. 



INTRODUCTJON. 



Exercise III. 

If rt = 1, ft = 2, c = 3, ^ = 4, e = 5, a; = 0, find the values of the 
following expressions : — 

(1.) a + 2i + 8c + 4'i. (2.) 13a-4/> + 5c. (3.) ah\de. 

(4.) ah-^'k—cx. (5.) 4a&c, SictZ, Gfifca;. 

(6.) a2 + 6-+c2. (7.) ^.d'^-W-cK (8.) 5rt(^+4/>3-cU 

a t7 



(9.) 



h''~e 



(^^•) ^f + f- 



X, I V ah ad bx 
^ de be ce' 



a' 



3 2b Sd'e 



(^^•) :x^-Q,.2w + 



yZ;^""3c2(^'^8a^6** 



(13.) If h = c=d=a, find what each of the following be- 
comes in terms of a : 

2bca,3a%2a^ + dahc-\-4:b^cd. 

16. When two or more quantities are multiplied together, 
each is said to be a factor of the product. 

Thus, a and 6 are factors of ah ; 3 and a^ are factors of 3a'^; 
and 6, h, ar d c are factors of 5&c. 

17. One factor of a quantity is said to be a coefficient of the 
remaining factor, and is said to be a litend or a cmtrical 
coefficient according as it involves letters or not. 

Thus, in 3^ and fa''' the numerical coefficients of x and a^ 
are 3 and *, respectively ; in ax^ and 3cd the literal coefficients 
of x'^ and d are a and 3c, respectively. 

Also, since a; = 1 x x, the coefficient of x in the quantity x 
is 1. 

The sign + or — when it precedes a quantity is also a sign 
of the coefficient. 

Thus, the coefficient of a; in +803 is +3, of x^ in — 5aas* is 
—5a, and of dz^ in —^cdz^ is — fc. 



i 



INTRODUCTION. 



I 



f 



I 



In the case of fractional numerical coefficients the letter 
symbols are sometimes written with the numerator ; thus, 

18. Quantities are said to be like or unlike according as they 
involve the same or different combiuatiors of letters. 

Thus, +5a, —7a are like quantities ; and so also are +Qx-7/, 
—bx^y; —2a, — a^ are unlike quantities. 



Exercise IV. 
(1.) Name the coefficient of x in 2aj, ^ax, fx, lex, 4ta%x, 

-^(XX, 

(2.) Name the coefficient of a in —a, +3a, —fa, +2ab, 
— 5ax2. 

(3.) Name the coefficient of x^ in -\-x^, -x^, -3ax\ -\-^dx\ 
(4) Name the coefficient of xym ^xy, H-Sa'^a;?/, — faxyz. 

(5.) Name the numerical coefficients in ^, ^, -|, +?^' 

5.15 

(6.) Name the niunerical coefficients in a:, — «/, —2x2, -.3^^^ 

(7.) Name the like quantities among 2x, ax, x, 3ca;l 

(8.) Name the like quantities among — a^, +2a2a;, +|a-', 
— a'flj. 

(9.) Name the like quantities among —Za^x, ax\ ahx^. 



I 



C 8 ) 



CHAPTER II. 



ADDITION. 



I. Figure Symbols. 

19. In order to explain the meaning of Algebraical addition, 
we shall, in the first instance, suppose the numerical values 
of the quantities to be represented by figure symbols, as in 
Arithmetic. We shall consider in order the addition of 

(i.) Positive Quantities. 
(ii.) Negative Quantities. 
(iii.) Positive and Negative Quantities. 

20. (i.) Positive Quantities. 

We have seen that a positive quantity, as + 5, represents a 
quantity 5 to be added to some number which may or may not 
be known or expressed. The sum of any numher of positive 
quantities is denoted hy writing them in a row with their signs 
between them, or by a positive quantity ivhose numerical value is 
their Arithmetical sum. 

Thus the sum of +4, +2, and +10 is +4+2 + 10,or +16; 
that is to say, the addition of 4 and 2 and 10 to a number 
is equivalent to the addition of 16 to that number. 

In like manner, the sum of +2, +f, +-*- and +5 is 

+ 2+1 + 1 + 5 =+8/^. 
The Algebraical statement 

+5 + 6 + H3^ = +14f 



ADDITION. 



may therefore be read the addition of^^^,\ and 3^ to a nurnber 
is equivalent to the addition of 14| to that number, 

21. (ii.) Negative Quantities. 

Since —4 denotes a quantity 4 to he subtracted from some 
number, when there are several negative quantities, as —4, 
—7, —8, denoting that they are all to be subtracted from 
some number, the operation may be denoted by writing them 
in a row, thus —4—7—8, or by a negative quantity, —19, 
whose numerical value is liheir Arithmetical sum ; that is to 
say, 

-4_7«8 = -19. 

In Algebra —4— 7— 8, or —19, is called the sum of —4:, —7, 
and —8. Thus the sum of —1, —10, — |, and — f is 

-1-10-t-f = -12ii. 

It follows therefore that the Algebraical sum of any number 
of negative quantities is a negative quantity whose numerical 
value is their Arithmetical sum. 

It will be observed that, instead of saying to subtract 12, we 
may say in Algebra to add —12. The Algebraical statement 

-2-7-10 =-19 

may therefore be read, in Arithmetical language, the subtrac' 
tion of 2, 7, and 10 from a number is equivalent to the subtrac- 
tion of Id ; or, in Algebraical language, the addition of —2,-7, 
and —10 ^0 a number is equivalent to the addition of —19. 

As an illustration of the foregoing phraseology we may 
take the following example : Suppose a man's gains to be 
denoted by +, and his losses by — ; then the statement 

-200-60-500 = -760 

may be read, if a dollar is the unit understood, the sum of a 
loss of 200 dollars^ a loss of 60 dollars, and a loss of 500 dollars 
is equivalent to a loss of 760 dollars. It may also be read, the 
subtraction of a gain of 200 dollars, a gain of 60 dollars, and a 
gain of 600 dollars is equivalent to the subtraction of a gain 
of 7QQ dcUars. 



J, ,1 



lO 



ADDITION. 



22. (iii.) Positive and Negative Quantities. 

Since +5 denotes a number 5 to he added, and —2a number 
2 to be subtracted, the performance of both these operations 
may be denoted by +5—2; and since the performance of 
these two operations is equivalent to addir.g 3, we may write 

+ 5-2 = +3. 

In Algebra +5—2, or +3, is called the sum of +5 and 
-2. 

Thus, +7-5, or +2, is the sum of +7 and — 5 ; +f — i, 
or +i, is the sum of +? and — i. 

Again, since to add 2 to a number and then subtract 5 
is equivalent to subtracting 3, we may write 

+ 2-5= -3; 

that is, in Algebraical language, to add +2 and —5 to a 
number is equivalent to adding —3. 

The statement 

+7-5+2= +4 

may therefore be read, in Arithmetical language, to add 7 to, 
then subtract 5 frort\ and finally add 2 to a number is equi- 
valent to adding 4 ; or, in Algebraical language, the sum of 
+ 7, —5, and +2 is +4. 

So also the statement 

-3+10-15= -8 

may be read, in Arithmetical language, to subtract 3 from, 
then add 10 to, and finally subtract 15 from a number is equi- 
valent to subtracting 8 ; or, in Algebraical language, the sum 
of —3, +10, and —15 is equal to —8. 

As an illustration of the foregoing phraseology, we may 
again take the case of a man's gains and losses. Thus the 
statement 

+ 10-8= +2 

may be read, if a dollar is the unit understood, a gain of 
10 dollars and a loss of 8 dollars are equivalent to a gain of 
2 dollars. So also the statement 



ADDITION. 



II 



-25 + 20= -5 

may be read a loss of 25 dollars and a gain of 20 dollars are 
equivalent to a loss of 5 dollars. 

23. From the preceding cases we can deduce the follow- 
ing rule for finding the sum of any positive and negative 
numbers. 

(i.) When the signs are all alike— i^'mc^ their Arithmetical 
sum, and 'prefix the common sign, 

(ii.) When the signs are different— i^«i(Z the numericcd 
difference between the Arithmetical sum of the positives and 
the Arithmetical sum of the negatives, and prefix the sign of the 
numerically greater sum. 

Examples. 

(1.) The sum of +4, +3, +h, and +7 is 

+4+3 + 1 + 7= +14^. 
(2.) The sum of -5, -12, -f, and -3 is 

-5-12-^-3= -20|. 

(3.) The sum of +4, -2, -3, +5, and +7 is 

+4-2-3 + 5 + 7= +4+5+7-2-3 
= +16-5 
= +11. 

Here +16 is the sum of the positives, and -5 of the 
negatives ; 11 is the numerical difference between these sums, 
and has the sign of the numerically greater + 16. 

(4.) The sum of +i, -2, -^, +1, and -f is 

+ i-2-^ + l-|= +i + l-.2-|-| 

— 4- 5 _ l_3 

= -2. 

Here +f is the sum of the positives, and —^-^ of the 
negatives ; 2 is the numerical difference between these sums, 
and has the sign of the numerically greater — ^^. 



12 



ADDITION. 



Exercise V. 
Find the sum of 

(1.) + 2, + 5, +18. (2.) +i +3, +f. 

(3.) _ 8, -13, - 7. (4) -i -1, -|. 

(5.) -rl8, -13. (6.) -26, +20. 

(7.) +i -1. (8.) -t, +1. (9.) +2-5, -3-2. 

(10.) +2, -3, +12. (11.) -3, +4, -6, +7. 

(12.) +5,-8, -12, +3. (13.) +^, -1, +1, -f 
(14.) ~|, -1, +2, -1 +i-. 
(15.) +2-58, -3-26, +1-089, -0-067. 

II. Letter Symbols. 

24. (i.) Like Quantities. 

We shall now show how to find the sum of quantities 
whose numerical values are represented by letters. When 
these quantities are like quantities, their sum is obtained by 
the rule — 

The sum of any numher of like quantities is a like quantity 
vjhose coefficient is the sum of the several coefficients. 

Examples. 

(1.) The sum of + 2r<, + 5a, and + 10a is 
+ 2a + 5a + 10a = + 17a. 

Hero + 17 is the sum of +2, +5, and + 10. 

(2.) The sum of -3c, -10c, and -12c ie 
_3c;_l0c-12c=-25c. 

Here —25 is the sum of —3, —10, and —12. 

(3.) The sum of + 8a;, — 12a5, and + Ix is 
+ 8a;-12x + 7.x=+3.x. 

Here +3 is the sum of +8, —12, and +7. 

(4.) The sum of —10a?, +12a5, and —6a; is 

— 10a; + 12a;— 5a= — 3a;. 
Here —3 fs the sum of —10, +12, and —5. 



Fine 

(1.) 
(4.) 
(6.) 
(9.) 

(11-: 
(13. 

(15.: 

25. ( 

The 

writing 

I between 

I Thus 

— 5y is 

+3&, ai 

I —2a; a: 

-3c. 

26. I 
+ or - 

Thufi 



ADDITION. 



13 



Exercise VI. 
Find the sum of 

(1.) +a, +2a. (2.) +3a, +6a, +7a. (3.) -a, -4a 
(4) -2a, -6a, -5a. (5.) +5x2, ^3^2^ 

(6.) -a^ +4a^ (7.) +4a, -7a. (8.) -2c, +5c, +7c. 
(9.) -10c, +8c, -3c. (10.) +a;^ -7a;2, +3a;2, -a;'^. 

(11.) -2a&, +lla&, +a6, -3a&. (12.) -ia, +|<i. 

(13.) +|a2, -fa2. (14) _2a, +^a, -a. 

(15.) -^-a,—^, +|<i, —2a. 

25. (ii.) Unlike Quantities. 

Tlie sum of any number of unlike quantities is denoted hy 
writing them in a row in any order, with their proper signs 
between them ; and each quantity is called a term of the sum. 

Thus the sum of +2a and +36 is +2a+36; of +Sx and 
-5y is +3x-52/; of -W md -26^ is -ia2_2&2; of +2a, 
+ 36, and — 5c is + 2-* + 36 — 5c. The terms of — 2ic + 6y are 
— 2a3 and +5y ,• and of — 6a+76— 3c are —6a, +76, and 
—3c. 

26. If a quantity contains no parts connected by the sign 
+ or — it is called a mononomial. 

Thus +2x, — 3a6, +6x^2/, are mononomials. 

27. When a quantity consists of two terms it is called a 
binomial expression; when it consists of three terms it is 
called a trinomial expression ; and generally when it consists 
of several terms it is called a polynomial, or multinomial 
expression. 

Thus +2a— 36 is a binomial, and +a— 26 + 3c a trinomial 
expression. 

28. The sign of a mononomial, or of the first term of a 
I polynomial, if it is positive, is generally omitted. 

Thus 2x^ stands for +2a;2, and a— 6 + c for +a— 6 + c. 

29. Like terms when they occur must be added together. 
I The operation may be conducted by arranging the several 



14 



ADDITION. 



quantities in rows under eiich other, so that like terms shall 
stand in the same column. 

Exam'phs. 

(1.) Find the sum of 2a+3i and 5a— 26. 

2a + 3& 
5r^26 

Here la is the sum of %x and 5a, and +& of +36 and —26. 
(2.) I'ind the sum of xy—^x and x—^xy. 

—^£-\-xy 
x—'^xy 

^bx^lxy. 

Here — 5rc is the sum of —6^- and x, and —Ixy of -^-xy 
and —^xy. 

(3.) Find the sum of 3 + a; + a7?/-8.T^ -3a;?/ + 2- 6a;, and 
4!r?/+fl32-|.l. 

3+ a; + cry -8x2 

2— 6.r— 3a;y 

1 + 4^xy + re'' 

(d-Zx^^xy-lx^. 

Here the sum of 1, 2, 3 in the iirst column is 6; of +ir, 
-6x in the second is -5a:; of ■\-xy, -3.r?/, +4a?/ in the 
third is ^2xy ; and of -Sa;^, +a;2 in the fourth is -7x^ 

(4.) Add together ha-^b-ic, ^J + ic + ^a, Ac-ia-i6. 

— ^-T^ + k 

Here -j-^a is the sum of ^a, ia, and -ia ; - JL6 and + ^^c 
are the sums of the quantities in the second and third 
columns, respectively. 



ADDITION. 



15 



Exercise VII. 



Find the sum of 



(1.) 2a, -3&. (2.) -oj, +3y. (3.) -2aj, -3y, -2. 

(4.) 3a, 2.x, -57/. (5.) 4, -.r, %j. (6.) ««, -.52, i. 

(7.) a, -' \ 3c, -c?. (8.) -a:, 2?/, -2, 1. 

(9.) a-3&,8a + &. (10.) -2a'''+&c, 3a2-56c. 

(11.) 3a-5?^ -4a + 2&, 5a-6&. 
(12.) a-.3&, 26-5c-, 4c-3a. 
(13.) 4x— 3?/+22, — 3;c + ?/— 42, a;— 4y + 2. 
(14.) a-a5 + 3, 5a + 2a5-5, -2a + 7. 
(15.) 2«-7, 5a+4, -6:c + 3. 
(16.) a-2&+3c, &-2c+3a, c-2a+3?). 
(17.) a;-2y+33-l,2ir + 3-42, 52/-2 + 7^. 
(18.) a;2 + 2aa; + a^2a;2-2a^£c2-2aa; + a2 
(19.) 3aS+a''^&-2a&2+&3^ 3a&2-2a26+a3, a%^al^^W. 

(20.)a-?> + |, a + &-|, &-a + |. 

'^"■'•^2 3^4' 2 3^1' 2 2^4' 
(22.) a+56-c, 2a-4&-c, |-a+|. 



( i6 ) 



CHAPTER III. 



SUBTRACTION. 



fM ■■ 



h-}> 






30. The Alyehraical difference between one quantity and 
anotlier is the quantity wliicli added Algebraically to the 
latter will produce the former. 

Thus the difference between 2 and —4 is the quantity 
whicli added to —4 will produce 2; the difference between 
—5a and 2a is the quantity which added to 2a will produce 
—6a ; and the difference between 3.r and 2a;— 5 is the quan- 
tity which added to 2x—6 will produce '6x. 

31. The quantity to bo diminished is called the minuend, 
and the quantity to be subtracted the subtrahend. 

82. The difference between two quantities is found by the 
rule — 

Add the first quantity to the second with its sign or signs 
changed. 

The reason for tliis rule will appear from the following 



Examples. 

(1.) From 5 take 3. 

Here the difference is the sum of 5 and —3=5—3=2, 
because 2 added to 3 makes 5. 

(2.) From 7 take -4. 

Here the difference is 7 -r 4=11, because the sum of 11 and 
-4=11-4=7. 



SUBTRACTION, 



17 



i 



(3.) From -6 take -4 

Here the diflference = —6+4=— 2, because the sum of 
-2 and -4= -2-4= -6. 

(4.) From -8 take 5. 

The difference =— 8— 5=— 13, because the sum of —13 
and5=-13 + 5=-8. 

(5.) From 2a take —3a. 

The difference =2a+3a=5a, because the sum of 5a and 
— 3rt = 5a — 3a =2a. 

(6.) From —hx take 4. 

The difference =— 5a;— 4, because the sum of —5a;— 4 and 
4 is —5a;. 

(7.) From 2a take -3a + 2&. 
The difference = the sum of 

2a and 3a— 2&=2a+3a— 2&=5a— 2&, 
Iv^cause the sum of 5a— 2& and —3a + 26 is 2a. 

The operation of changing signs and adding may be per- 
formed mentally, and the difference exhibited as in the fol- 
lowing examples, in which the minuend and subtrahend are 
written in rows with the difference underneath. In this 
arrangement the first row is equal to the sum of the second 
and third rows. 

(8.) From Zx^-^y take a;^— 5?/. 

3x2+2/ 

2x2 +6y 
Here Sx^ is the sum of 3x2 ^nd ^^-^ and +6y of +2/ and 

(9.) From 5a+3&— c take a— 6+ 3c. 

5a + 3?>— c 



a 



- &+3c 



4a+46— 4c 



]8 



SUBTRACTION. 



Hero 4a is tho sum of 5a and —a; +4Jof +36 and +&; 
and — 4c of — c and —3c. 

33. From tho foregoing examples it appears that, in Alge- 
braical language, to subtract a positive ijuitntity is eciuivalcnt 
to adding a negdiive ; and to sidUract a negative (juantity is 
equivalent to adding a positive. 

This phraseology may bo illustrated by taking the case of 
a man's gains and lasses to he denoted by + and — , resi)ee- 
tively. Thus, to subtract a gain of 10 dollars is equivalent to 
adding a loss of 10 dollars ; and to subtract a loss of 25 dollars 
is equivalent to adding a gain of 25 dollars. 

Moreover, if a man gains a dollars and loses h dollars, we 
say, in Arithmetical language, either that he gains a—h dollare, 
if a is greater than b, or loses b—a dollars, if b is greater than 
n. Either of these phrases may be employed indifferently if 
we agree that a gain of —c dollars means a loss of c dollars, 
and that a loss of —c dollars means a gain ofc dollars. 

Thus if a man gains 10 dollars and loses 5 dollars, we may 
cither say that ho gained 10—6, or 5 dollars, or that he lost 
5— 10, or —5 dollars. Again, if ho gains 8 dollars and loses 
12 dollars, we may either say that he gained 8—12, or —4 
dollars, or that he lost 12—8, or 4 dollars. 






Exercise YIII. 

(1.) From 1 take -3. 
(2.) From 1 take -1. 
(3.) From —5 take 4. 
(4.) From 12 take 15. 
(5.) From -3 take -8. 
(6.) From -8 take -5. 
(7.) From 4-56 take -6*04. 
(8.) From -I'Oi take 235. 
(9.) From -432 take -2-16. 
(10.) From -1089 take -0*123. 



SUBTRACTION, 



19 






(11.) From 2a take 8a. 
(12.) From —5a; take 2aj. 
(13.) From Sa^ take -2a^ 
(14.) From —3c take —5c. 

(15.) From 2a take h 

(16.) From -a' take 2a». 

(17.) From bx^ take -Gail 

(18.) From la take 4a-&. 

(19.) From a + a; take a— a;. 

(20.) From 5a-2a;+3 take 2a-a;-l. 

(21.) From 3a2-4a& + 62 ^ake SaHaJ-i*. 

(22.) From aa;— 4&y + 3c2 take 2ax-^hy—cz, 

(23.) From -4a+6-l take 2a-a; + 3. 

(24.) From 12a;2-5a; + l take 7a;3-16a;2+l. 

(26.) From^+3/+| take |+|-;3. 
(26.) From|o+6-^ take a+|6-^. 



o2 



( 20 ) 



CHAPTER IV. 



TEE USE OF DOUBLE SIGNS AND OF BRACKETS. 

34. The operations of addition and subtraction of positive 
and negative quantities may also be denoted by the use of + 
for the former operation and — for the latter. 

Thus instead of saying add together 2a and —Sh, we may 
employ the notation 2a -\ — 'Sh, the equivalent of which is 
of course 2a— db. So also tJw sum of —5a^, +Sb, and —2c 
may be written — 5cr+ +3iH 2c, which is equivalent to 

— 5a'-^ + 3&— 2c; and the difference between 6a; and —7y may 
be expressed hx ly, the equivalent of which is hx-\-ly. 

Accordmg to this notation, therefore, 2a'^H — 3&+ +2c 

means the sum of2a^, — 3&, and +2c; —60;+ +83/ H 1 the 

sum of —5a?, +Sy, and —1; 7a \-2b the difference between 

la and +2'>; —2x -5 the difference between —2x and —5; 

4a'^ 7 the difference between 4a- and —7. 

35. When any of the quantities before which the double 
signs are to be used contains more terms than one, it must 
be enclosed in a bracket; thus + (2a— 3&), + (—a; + 4), 

- (.r-5), -(-2 + aHa;). 

Thus a+(3i— c) denotes the sum of a and 36— c ; 2x 
+ (4y-2) + (-2^ + 1) the sum of 2x, 4y-2, and -22 + 1; 
2a2_j_ (3a2+4) the difference between 2a^-b and Sa-+4:; 
and 2a+ {b—1) — (c+4) the sum of 2ct and &— 1 IckSs c+4. 

Exercise IX. 

Retaining the given quantities, denote by using the double 
signs and brackets (when necessary) in the following opera- 
tions : — 



i 



USE OF DOUBLE SIGNS AND OF BRACKETS. 21 



(1.) The sum of 2x2, _i. 3^^^ __^y^ __^.^ 2(t, -3&,+4c. 
(2.) From 2c. take —6a. 
(3.) From -Stake +5^. 
(4.) From the siim of 2a and — 3& take +7. 
(5.) From the sum of 5 and +a; take —3a. 
(6.) The sum of 5a and &— 4. 
(7.) The sum of —a and —6+5. 
(8.) The sum of a— 4 and 2&— c. 
(9.) The sum of x^ and 2?/ + 5 less 2?. 
(10.) The sum of a— 1 and 3& + 5 less —3c. 
(11.) The sum of x, 2*2-1, and-3a:2-8. 
(12.) The diflference between da^ and V'—c. 
(13.) The diflference between a^^^ and — 2&+3. 
(14.) The diflference between 2a— 5 and a2>_2a+3. 

(15.) From the sum of a+&+c and a—h—c take — ct 
+26-3c. 

86. Double signs may be equivalently replaced by single 
ones by the rule : — 

Like signs produce + , and unlike signs — ; that is 
+ + = -- = +, 

Thus a++5=a + 5; 2x a=2x+a; 3+— 4g=3— 4c; 

c— +2a=c— 2a. 



37. Expressions may be cleared of brackets by the rule : — 

The sign + he/ore a bracket does not change the signs ivithin^ 
whilst the sign — changes every sign within. 

Thus 4 + (&-c)=4 + 6-c, 

2a+(— x + c— 2(f)=2a— cc+c— 26^, 
4a2_i_(26' + c)=4a2-l-262-c, 
3a;-(-4?/ + 5)=3«;+42/-5, 
03— (?/— 2)-f4— (— 3?/ + a;)=£c— y + «+4 + 3?/— .r. 



22 USE OF DOUBLE SIGNS AND OF BRACKETS. 



EXEBGISE X. 

Eeplace the double signs \y single ones in the expres- 
sions:— 



(1.) 2a++3&+-c. 
(3.) a;2+-4a;+-l. 

Clear of brackets : — 
(5.) 8a^.-(J+c). 
(7.) 8a-(-2&+30. 
(9.) (»+5-(2-42/)+8. 
(11.) 3a;2-l-(-aj+4)+2iB-(sc2-.6). 



(2.) ah—+hc c. 

(4.) 5x'-\--Sx' ?a;-+8. 

(6.) 8a-(5-c). 

(8.) 2a-l + (&-6)+c. 

(10.) a + (&-c)-(a-c). 



( 23 ) 



CHAPTER V. 



MULTIPLICATION, 



38. When it is desired to denote the operation of multiply- 
ing several expressions together so as to exhibit the various 
factors, we enclose each in a bracket and write them together 
in a row in any order. 

Thus ( + 2a) (— 3&) denotes the product of -I- 2a and — 3& ; 
(2a -1) l-h) the product of 2a--l and -&; (x'-d) (2a; + o) 
the product of 05^—3 and 2x + 5; and («— 1) (a:; + 2) (2x—b) 
the product of a;— 1, a; + 2, and 2ic— 5. 

39. Each of the quantities so enclosed in brackets is called 
a, factor of the product. 

Thus —2a, a^— 1, and 2a— 3 are the factors of (—2a) («"— 1) 
(2a-3). 

40. When a factor is mononomial, it is called a simple 
factor ; otherwise a compound factor. 

Thus in the preceding example —2a is a simple factor, and 
a^—1 and 2a— 3 compound factors. 

41. In the case of a simple factor the bracket may be 
omitted (i.) if the simple factor is written in the first place ; 
(ii.) if the sign of the simple factor is not expressed. 

Thus we generally write —2a (a3— 1), a:(a;''— 2), (a— &)a;, 
(a— Z>) (c^d)x. 

If the sign of a simple factor is expressed, the bracket 
may be replaced by a multiplication sign, x . 



24 



MUL TIPLICA TION. 



Thus 



-2^ X +3A = -2r(( + 36), 

3ic X — 5?/ = 8.t(— 5y), 

(a-1) X -2tr = (rt-l) (-2^2). 



JC?/^^. 



Exercise XI. 

Express in Algebraic<al language, retaining the given factors 
and using brackets wlion necessary : — 

(1.) The products of a-1, 2a2-3; -2+a, -3-a2; x-h, 

-2x + 7. 
(2.) The products of -%.i^, 6^-1; a^-l, -3a; 5x, 

—a;- + 3. 
(3.) The products of \, jk-I ; \, 2i»-3; -f, x^-S. 
(4.) The products of — oo?, a:— 1, ^7 + 2; a;-— 4, +5x, 2a + 3. 
(5.) The products of +8a5, —hyjXy—\\ —la, ab—'d, +86. 

42. The mode of performing the operation of multiplication 
whereby products are expressed as mononomials or poly- 
nomials will now be explained. It is convenient to make 
three cases : 

I. The Multiplication of Sijiple Factors. 
II. The Multiplication of a Simple and a Compound 

Factor. 
III. The Multiplication of Compound Factors. 

43. I. The product of two simple factors is obtained by 
the following rule : — 

(i.) The siffn of the 'product is obtained by the rule of signs : 
like signs produce -\- , and unlike signs —. 

Thus the signs of +2a ( + 36), -2a (-3&), +2a (-3&), 
—2a ( + 36) are, respectively, +, +, — , — . 

(ii.) The numerical coefficient of the product is the product of 
the numerical coefficients of the factors. 

Thus the numerical value of the coefficient of the product 
of ~2a and +36c is 2 x 3 - 6. 

(iii.) The literal part of the product is the prodmt of the 
literal parts of the factors (9). 



MUL riPLICA TION, 



25 



Thus tho literal part of the product of — 3aj and ^ifz is 



%yh. 



(iv.) The 'p oditct of two powers of the same letter is a power 
whose index is the sum of the ii. dices of the factors. 



Thus 






1 1 






Examples, 

(1.) The product of +2a and -36= + -2x3a6=-6a&. 
(2.) The product of -5a and 2d=-5x2ad=-10ad. 
(3.) The product of -^a and +3&'-=- + ^ x3a62=-.|a6='. 

(4.) -a?<-3c) = 3a&c= + 3a&c. 

(5.) -ia^( + ^Z>)=- + ^ • ia2&=— ia^ft. 
(6.) -2x(-f3a;-)=-+2x3x. a;2=:-6a;3. 

(7.) - 6ftX- Sit «) = 6x3 a3. «*= + 18al 

(8.) +2a2&c3(-5«6V)= + -2 x 5 a • a^ . j . j2 . ^ . c*- 



Exercise XII. 
Find the product of 
(1.) +3^, -26; -«, +5c; -2a2, -36; 5x, -6y. 
(2.) 2a6, -7c2; -4a^ +56c; -2iB, S?/^; -6, -8a. 

(3.) -ix, 3^; |a, -26; -hx, -\y; 2a\ -^. 

(4.) 2^2/^ —3a;-?/ ; —ax^y, —Zxy^ ; |-a6^c, — ^a^6c^ 

^-Ks a __2a^ ah a-W _%c,ij^ 3x^2/' . _jxx^ _J)Oiy^ 
^^ 2' "3 * 5' IT ' "5"' "8" ' T' 3 • 

44. II. The product of a simple and a compound factor is the 
Algebraical sum of the products of the simple factor and the 
several terms of the compound factor. 



26 



MULTIPLICA TION. 



Thus the product of 2a and 3ft— 5c+<^ is the sum of the 
products of 2a, 3&; 2a, —5c; and 2a, +c?; and is therefore 
equal to 6a6— 10ac+2a(^. 

The work may be arranged as in the following 

Examples^ 
(1.) Find the product of —803 a;ttd 2?/'*— 403^—5, 

-3a; 

— 6rt2/^ + 12a3'^y + 16a;. 

(2.) Find the product of 2x'^—^-\-\ and ^^^xy, 

2x2-ia; + i 

— fa;y 



Fs'^y* 



(3.) Clear of brackets the expression —2a (Sa"— 5a+l). 
As this expression denotes the product of —2a and 
Sa^— 5a + l, it is equivalent to — GaHlOa^— 2a. 

Exercise XIII. 
Find the product of 

(1.) 4a-36 + c, -2x; 3a;2-2x + l, 4?/; 2a&-3c, -d 

(2.) a;2«2x--5, 3^; 2rt2-3a + 7, -a»; a;'''-aa; + 2a2, -4aa;. 

(3.) 2x-\-y--^z,%iyz; la^-ah^ + ah, -4ub. 

(4.) a + ^ft-fc, 2a5; 2a2-3a+4, -|a; ^x^''ax-{-ia^,^ax. 



^rs 4a 2ah ^ ,^ 2.k ^ , 3aa; ,. 
W-) g — -g- +1,-15; _- 1+__,A| 



/X'» 



Clear of brackets : — 

(G.) 5(^2-a^-f-4); -2(a-a& + 3); -a(a2- 2aa; + l). 
(7.) (a-a;)a;; (a^h + c)c; (-!-{■ ab-Sa^)bab. 

(8.) |'(6a''-9a + 12);- |'(-10 + 2x-15x«). 

45. in. The product of two compound factors is the Alge- 
hraiml sum of the products of one factor and the several terms 
of the other. 



MUL TIP Lie A TION, 



2^ 



I 



Thus the product of 2a— 36 and 4c +5^/ is the sum of the 
products 2a— 3?', 4c and 2a— 36, +5<?, and is, therefore, equal 
to 8ac-126c + 10a(^-156d 

The work may be conveniently arranged as in the following 

Examphs. 
(1.) Multiply 2a;2-3a;+5 by 4a;-7. 

2ic2-3a!+6 
4a; -7 
8a;»-12a;H20a; 

-14a;H21a;-35 

8a;»-26xH4ia;-35. 

Here ^-llx^^-'ildx is the product of 4(r and Ix^—^x-^-^-^ 
*-14icH21a5— 35 is the product of —7 and 2ar^— 3a;+5; and 
the sum of these two partial products, which are arranged so 
that like terms stand in the same column, is the product 
required. 

It will be observed that in the foregoing process we work 
from left to right, and not from right to left, as in the cor- 
responding Arithmetical operation. 

(2.) Multiply 2a-6 by c-M, 

2a-6 



2ac-^hc 



"Qad-^-Shd 



2ac •^hc— Gad + dbd. 

In this case, as there are no like terms in the partial pro- 
ducts, the second is placed entirely to the right of the first. 

(3.) Multiply l-2x+3x- by 4(r-5a;2+2. 

It will be found most convenient in this and similar 
examples to arrange the given factors according to ascending 
or descending powers of x ; that is, so that the exponents of 
tlie successive terms shall continually increase or decrease, 
in the former arrangement the numeral stands first, in the 
latter last. 



I.. 



28 



MULTJPLICA TION. 



iii 



m 



■i 



In the present case let them be arranged in the order 

l-2£c + 3x=^ 
2+4x-5»2 

2-4r+6x2 

+ 4:03-8** +12^3 

-5a;- + lQx-"-15a;* 
2 -7ar^ + 22x3- 15a;*. 

(4.) Multiply 2a2-a& + Z>2 ^^y aHa&-36^ 
2a2-a& + 62 

2a4-a3i+^r2 

-6a'6H3a63-354 

2aHa'6-6a''^6H4aft«-36^ 

Here the factors are arranged according to descending 
powers of a and ascending powers of h. 






Exercise XIV. 
Multiply together 

(1.) 2*-3, .a-{-4; 4aH-5, -.r + l; 2-3x, l+cc. 

(2.) a;^-2,2.r-l; 1-a-, 2 + 3^2; \^x\^-x^, 

(3.) 2a--ti + 4, 3a~2; !-« + «=, 1 + a; l + a + a^^ 1-a. 

(4.) « + i, 1)1— n ; a-{-l)'-^c,7n-\-2n ; 2))i—ii, 2m -{-n, 

(5.) fi;?/ + u.^a;y/-a'2; x + o:y-y\ ar-'ly. 

(6.) 2«-^-5« + l,aH3^i-4; a + 26-3c, a-26 + 3c. 

(7.) 3a + 2i-c,o^~2/^ + 8c,- 3xH2x2/ + 2/', «'-2a'y + 8?/2. 

(8.) a;2-i,.r2-.>; « + ! ^-i ; 2a-^, a + i. 

(9.) «;2_^x + l,2^— i; 3*--|:« + i, 3x— ^ 

aO.) ^+|-2,|-3y; ^'-2x + i3^-|. 



a^ 2a 



a- 



2a 



Ui.^g— 3- + ^' 2 ^'3~'"-^' 






MULTIPLICA TION, 



29 



(12.) c»H2/^-a;y+a'+2/-l»«'+2/-l- 
(13.) aH&Hc''— &c-ca— a&, a+&+c. 

46. The <product of three expressions is found hy muUijplying 
thejproduct of two of them by the third. 



Examples. 



(1) 

Here 
(2.) 



Multiply together 2a;, -Sx% ^ixy^z. 
2a;(-3x22/)=-6x' 

Multiply together 



c^y ; and —Qx^y{'-ixyh)= +|a;y2;. 



X 



-L aj-2, a;-3. 



X 
X 



-I 
-2 



ay^—x 



~2ig+2 
a^-dx+2 

{B8-3x2+2a; 

-3a;'^+9a;~6 
ar'-6x2 + lla;-6. 



EXEBCISE XV. 



Multiply together 
(1.) -3a2, +2a% -6ab^; Ix, -^x', +|(b*; 



-S:x^y, 



—\y\ ^yz. 



(2.) -'2x,Zxy,^x'-hy; 2a&, -^^a^, 2a2-3a& + l. 

(3.) 2a;-3,4a; + l, a;-2. 

(4.) a;2_a; + l, a;-l, x + 1. 

(5.) a;2+2a£C+a2, a;2-2aa;+a2, a;H2a2a;2+a*. 



" > 



( 30 ) 



CHArTEB VI. 



DIVISION. 






MJi 



47. Division being the inverse of multiplication, it follows 
that, when two factors are nmltiplied together, either factor 
will be the quotient of the product divided by the other. 

Thus, since +a(— 26) =—2ab, +a is the quotient of — 2a6 
divided by —28, and — 2& is the quotient of — 2a6 divided by 

Again, since •~2x^ (x^'-4a}-\-d)=^2x^-\-Sx^—6x^, it follows 
that 05^—403 + 3 is the quotient of —2x^-\-Sx^—Qx'^ divided 
by —203^, and — 2a;^ is the quotient of — 223^+80^—6052 divided 
by oj^— 403 + 3. 

48. "When it is desired to denote that one quantity is to 
be divided by another, they are enclosed in brackets and 
written in a row with the sign ~ between them ; or the 
second quantity is written below the first with a line between 
them. 



Thus (-2a;2)-f-(+3!r), or 



-2ar^ 
+Sx' 



denotes that — 203^* is to 



be divided by +dx; (3a:2-2a5+6)-f-(aJ-4), or ?^-^-±_5. 

03—4 

that Sas^— 2a3+5 is to be divided by as— 4. 

49. The first or upper quantity is called the dividend, and 
the second or lower the divisor. 

50. The bracket is generally omitted in the case of a 
mononomial. 

Thus (-203) -r (-32^) is written -2a;-7--32/; (2a;2-l) 
-r(4x) is written (2x2-1) ~ 4a; ; (-3x3)-;- (2a- 1) |g 
written -3a!3-r(2!»-l). 



DIVISION. 



3« 



EZBBOISE XVI. 

Betaining the given quantities and employing brackets only 
when necessary, express in Algebraical language — 
(1.) Divide 2a-5 by -3a; da^-Sa + l by 3a-4. 
(2.) Divide 2a by -36; -a;^ by +2a;; 3a; by 2a. 
(3.) Divide4a;'^by 2a5— 5; —aa;'* by oc- a. 

51. The mode of performing the operation of division 
whereby quotients are expressed as mononomials or polyno- 
mials will now be explained. We shall consider in order 
three cases : 

I. When the Dividend and Divisor are Mononomials. 
II. When the Divisor only is a Mononomial. 
III. When the Dividend and Divisor are Polynomials. 

52. I. The quotient of one mononomial divided by another 
is obtained by the following rule : 

(i.) The sign of the quotient is obtained by the rule of signs : 
like signs produce +, and unlike signs — . 

Thus the signs of -^2ah-^Scd, Sx^-7'-2x, 4a2-i--2a, 
—bx^y-r- ■\-xyy are, respectively, +, +, — , — . 

This rule follows from the rule of signs in multiplication; 
thus, since +a (+i) = +a&, it follows that 

+ a6. 
+ a' 

So also from the equivalent forms — a ( + &) = — o&, — a (— &) 
= + a6, +a (— &) =— a6, we deduce 

— a& , 7 +a6 , —at- , 
—a —a +a 

(ii.) The numerical coefficient^ without regard to sign^ of the 
quotient is obtained by dividing the numerical coefficient of the 
dividend by the numerical coefficient of the divisor. 

Thus the numerical coefficient of 12a^-^Sa is 12-7-3=4, of 
2a^-T-Sx is f, and of la^^fa is I -~|=|. 

(iii.) The literal part of the quotient is obtained by dividing 
the literal part of the dividend by the literal part of the divisor 
(12). 



•= + b. 






If 

I'll 



1: 



32 



DIVISION. 



Thus the literal part of the quotient of 2a2 -f-Jc is ~. In 

be 

applying this rule the fractional form of expressing the 

quotient should always bo used. 

(iv.) The (jHothnt of two poirem of the same letter ia a power 
inJaise index in the di'jference 0/ the indices of dividend and 
diiHHor. 

The reason for this rule will he evident from the following 
examples : — 



From (10) 



a^ aaa 



a a 

a" aaaaa 



■ aa =a2— 0^3-1 . 



=iaa=a^=a'~^; 



a" aaa 
of aaaaaaa 



=aaa = a^ssa^"* : 



a 



aaaa 



ir 
.i 



So likewise 



a" 



±=a^-^=a\ 
a^ 



53. Since the quotient of any quantity divided by itself ia 

1 , this rule can be applied when the indices of the dividend 

and divisor are equal, if the zero power of a letter is considered 

' a^ " d~ 

eiual to unity. Thus -n=l ; find, by the rule, --zra^-s—fjO. 

a^ a^ 

therefore a°=l. Whenever, therefore, by applying the pre- 
ceding rule, we get such symbols as x^, y^, c", we must 
replace each of them by 1. 

Examples. 

(1.) -.i2-~+6=-V=-2; 

-15~-10=+i^=+f; 
1 

4.IJ: #— _=* — -_2 



f 



(2.) -6aS-3/. = -«. ?=-^*. 

u 



(1- 



(2. 



(3 



DIVISION. 



n 



^=:rt"^ 



(3.) -2a6-r--5c = +f^^=+|^. 

c 6c 

(5.) -4a6-f--7a2= + |a°-!^=+-fa3. 

(6.) 2tt263c» -r 3a6c2 = ^a'-'-i ft^-i cO-2 ^ 1^52^5^ 

Exercise XVII. 
Divide 

(1.) -16 by +4 ; 20 by -4 ; -f by -|; +5 by -f, 
(2.) -a by +2a;; Sa^ by -26; -Gajy by -3a. 

(3.) ^ by -I; -' 



^ oy g, ^ oy -^. 



'A " 3 

(4.) 2a» by a^\ 3a* by 3a; 8iB« by 2a;*. 
(5.) -4a26 by 2a& ; lOa^JSc* by 2a6c. 
(6.) a^y^ by — 3aa;2/; —ZQ^xy^ by -"Baxy. 

54. II. T/je quotient of a polynomial divided hy a mono- 
tiomial is the Algebraical sum of the quotients of the several 
terms of the former divided by the latter. 

Thus the quotient of Sx^—Qx^-\-Sx divided by —2x is the 
sum of the quotients 3x'-^ — 2x, —Gx'^-i — 2x, +8x-z — 2a5 ; 
and is therefore equal to — fa;"+3a3— 4. 

The work may be arranged as in the following 

Examples. 

(1.) Divide Hx^-^x^+^x by -2x. 

-•2x)S x^-4:X^+2 x 
-4xH2iB-l. 
(2.) Divide 6a»6-10a26='+2a& by 5ab. 
5a6)5aV-10a26H2a6 
a2- 2db^ + 1 

(3.) Collect coefficients of x in 2aa;— S*. 






"it 






II 



34 



DIVISION, 



As this means that 2aa;— Sec is to be expressed as the pro- 
duct of two factors one of which shall be x, we have merely 
to divide the given quantity by x to get the other factor. 

Thus 2aa5-3x=(2a-3)a5. 

(4.) Collect coefficients of cc^ in ^ax^—bx^+x^. 

By dividing the given quantity by x'^ we get 

ba3?—hx^ + flj2=(5a— 6 + 1) x^. 



«t 






Exercise XVIII. 
Divide 

(1.) 10a-15&+20by -5; -4aa! + 12--8a2by -4. 

(2.) 4a2a;-3aHa'by-a2; 12x3-6!B2+9a; by 3x. 

(3.) 3a''-12aH15a2by -S^^. 2a;32/-6xV+8iC2/* by 2a;«^. 

(4.) 2a^6c— 3a6^c+a&c2 by a&c. 

(5.) 20a26c2-15a&V+5afeby -5a6. 

(6.) Collect coefficients of x in ax—hx, 2ax—cx+x. 

(7.) Collect coefficients of as?/ in 4ixy—axy, Sx^y—xy^. 

55. Since by (54) ^^ 
x-1 



X 



X 



it follows that 
likewise 



g .=^-i and by (44) Ka'-l)-^-^, 
and K^~l) ^J*® equivalent forms. So 



2a;-3 



i(2aj-3), 

56. III. When the dividend and divisor are polynomials, 
the quotient is obtained by the following rule : — 

(i.) Arrange both dividend and divisor according to ascending 
or descending powers of some letter. 

(ii.) The first term of the quotient is found by dividing the 
leading term of the dividend by the leading term of the divisor. 
The product of the divisor and the first term of the quotient is 
subtracted from the dividend, giving the first difference. 



ro- 
ely 



DIVISION. 



35 



(iii.) Tlie second term of the quotient is found hy dividing the 
leading term of the first difference hy the leading term of the 
divisor. The product of the divisor and the second term of the 
quotient is subtracted from the first difference, giving the second 
difference in the process. 

And so on until the last difference is zero. 



y- 



Examples. 

(1.) Divide Qx^-hx'-'^x^-^ by 2f3!r. 

3ic + 2) 6a;3-5.x2-3x + 2(2x2-3a;+l 
605^ + 4tx^ 

-9jr;^-"-3x + 2 
— 9£c^— 6a3 

" 3;rT2 

3x+2 



— 1. 

a* 

So 



iais, 

iing 

J the 
isor. 
nt i& 



Here the divisor is first arranged, like the dividend, ac- 
cording to descending powers of x. The first term of the 
quotient 2x^ is obtained by dividing 605^, the leading term of 
the dividend, by 'dx, the leading term of the divisor. The 
product of the divisor 3«+2 and 2x'^, which is 6a5^+4a5^, is 
then subtracted from the dividend, giving — Qx'^— 3£c+2, the 
first difference. The second term of the quotient —^x is 
obtained by dividing —9*^, the leading term of the first 
difference, by 3a5, the leading term of the divisor. The pro- 
duct of the divisor .ind —^x, which is — 9a5^— 605, is then 
subtracted from the first difference, giving 3x+2, the second 
difference. The third term of the quotient +1 is obtained 
by dividing 3x, the leading term of the second difference, by 
3cc, the leading term of the divisor. The product of the 
divisor and +1, which is 3a3 + 2, is then subtracted from the 
second difference, giving the last difference zero ; and thus 
the process ends. 

The latter terms of the differences need not be expressed 
until the corresponding like terms in the par^i il products 
are to be subtracted from them; thus in the following ex- 

D 2 



36 



DIVISION. 



I % 



!ir 



If 



It i 



ample — ITx + G is not expressed in the first difference, nor 
+ 6 in the second. 

(2.) Divide 6x4^5a;3 + 6x--17^ + 6 by 2.c-l. 

2x-l)6xH5x3 + 6.z;'^-17u; + 6C3x3 + 4x2 + 5x-6 

-12aj + 6 

(3.) Divide l-2x3+&;'' by l-2ic+a;2. 

l-2x + x2)l-2ie3-ha:«(l + 2a; + 3x2 + 2x3 + a;* 

l-2x+u;^ 

2a;-x2-2:^3 
2ic— 4a;^ + 2x* 

3x2— 4a;^ 
3.c2_5a;=' + 3x* 

2x3-3x4 

2x3-4.r* + 2xS 

a;'* — ■ia'-'' 4- x* 



Here +x' is not expressed until we reach the last dif- 
ference. 

(4.) Divide 2a5-6a3& + 13a262-6a63-3 1<& by 2a-3&. 

Here wo shall arrange the dividend and divisor according 
to descending powers of a. 

2a-36)2a5-3a*6-6a36 + 18a262-6a&3(a4-3a26+2a6''^ 

2a°-3a*6 

-6a36 + 13a2&2 
-6a36+ 9a262 



• ^ 



4a=*62-.6a6' 
4a262-6a63 



DIVISION. 



37 



Exercise XIX. 
Divide 

(1.) a;2-7a! + 12 by a;-3; and 3x2 + 7a; + 2 by a; + 2. 
(2.) a;2-4x-5 by a;-5; and 4x2-9 by 2x + 3. 
(3.) 6x2-5x-6 by 2x-3 ; and 9x'- 18x2 + 26^-24 by 
3x-4. 

(4.) x3-4x2 + 6x-2 by x2-3x+2; and cc^ + x^+l by 

x'^ + X + 1. 
(5.) 4x4-15x3 + 14x2-6x + l by 4x2-3x + l. 
(6.) X*— 1 by X— 1; and x^ + 1 by x + 1. 
(7.) x2— 2x?/ + 2/'^ by x—y ; and x^-^-if by x + y. 
(8.) aHa'&H&^bya2-a& + &2^ 
(9.) 20x2 + 9x?/-12x- 182/2 + 92/ by 4x-32/. 
(10.) x6-2ax* + 3a2x3-3a3x2 + 2a*x-a5 ijy x3-ax2+a2x'-a8, 
(11.) x^-^x^y—xy^-\-y^ by x2 + x2/ + ?/2. 

(12.) |a*X-\fa3x2 +|a2^3 +_3_o^a;4_^5 l^y |.f^3__A,j2^^1a;3^ 



57. When the division is not exact, the last difiference is 
called the remainder. In this case the product of the quotient 
and divisor added to the remainder will be equal to the 
dividend. 

Example. 

Find the quotient and remainder in dividing 10x3+ 7x'* 
-8x-2by 2x + 3. 

2x + 3)10x3+7x2~8x-2(5x2-4x+2 
10x3 + 15x2 

-8x2-8x 
-8x2- 12x 

4x-2 
4x+6 



-8 



Quotient =5x2— 4x + 2. 
Remainder = — 8. 



I i 



38 



DIVISION. 



Exercise XX. 

Find the quotient and remainder in dividing 
(1.) 4x3-4a;H8a;+2by2a5+l. 

(4.) 2a5«-2x* + 9x3 by 2a;' + a? + 1. 
(6.) 2a;*x2x*+5a;3 by a;»+a^+»+l. 






I 



H 



( 39 ) 



CHAPTEE VII. 

EXAMPLES INVOLVING THE APPLICATION OF THE 

FIEST FOUB BULE8. 



58. In the following examples some of the given quantities 
are expressed by letter symbols, and the object of the exercises 
is to express in like manner other quantities which by the 
conditions of the question are related to the former. When 
a doubt exists as to the manner of solving a question, it will 
be well to substitute numbers for letters in order to see what 
operations ought to be performed in the given symbols. 

59. The sign .*. will be used to mean hence, or therefore, and 
the sign '.* siuce^ or because. 

Exumples, 

(1.) I buy goods for 2a+Sb—c dollars, and sell themfo^r 
4a— 6 + 2c dollars ; what do I gain ? 

4a— 6+ 2c 

2a-H36-c 

2a-4&+3c 

.*. the gain, which is the selling price less the cost price, is 
2a— 4&+ 3c dollars. 

(2.) A man has 3*^^ + Tas + 2 dollars and spends aj +2 of them 
per day ; how long will his money last ? 

a;+2)3xH7a;+2(3a;+l 

»+2 
»+2 



40 



EXAMPLES OF THE FIRST FOUR RULES. 









.*. the required number of days, which is equal to the num- 
ber of times the amount of his daily expenses is contained in 
the amount he possesses, is Sx- + 1. 

(3.) A man walks x miles in y hours : at what rate per hour 
does he walk ; how far will he walk in 5 hours ; and how long 
"will he be in walking 12 miles ? 

*.* he walks x miles in y hours, 
„ ^ „ „ 1 hour, 

and 1 mile in ^ hours. 



• • » 



X 



Again, 



X 



he walks - miles in 1 hour, 

y 



» 



>j 



5x 

y 



,, 5 hours; 



and 



he takes ^ hours to walk 1 mile, 

X 



if i> 



12y 

X 



» » i> 



12 miles. 



or OuC 

The answers are, therefore, - miles, — miles, and 

y y » 



12^/ 



hours. 



Exercise XXI. 



(1.) A man walks x, +a, and x—2a miles in tb'^ same 
direction ; how far does he walk altogether? 

(2.) A man has 100 dollars, and owes 50— x dollars; what 
is he worth ? 

(3.) A man walks a-\-h miles and returns a—h miles; how 
far is he from the starting point ? 

(4.) What is the area of a room a? +2/ feet long and x—y feet 
broad ? 

(5.) A man walks x miles at a miles an hour ; how long is 
he on the road ? 

(6.) A has X dollars, B 50?/ cents, and C 76z cents ; how 
many dollars have A, B, and C together ? 



EXAMPLES OF THE FIRST FOUR RULES. 41 



(7.) A has X pounds, B has y shillings, and C z pence ; how 
many pounds have A, B, and C together ? 

(8.) A spends a dollars in x days ; in how many days will 
he spend 10 dollars? 

(9.) How many square yards in a floor which is a feet by 
X feet ? 

(10.) What is the cost in dollars of painting a floor aj-hy feet 
by 05— y ^66* at x'^-k-y^ cents per square foot? 

(11.) A owns a acres, B 6 acres, and C 5 acres less than one- 
half of A's and one-third of C's together; what is the whole 
amount possessed by A, B, and C ? 

(12.) A owns a+ 6 acres, B a— & acres, C half as much as A, 
and D half as much as B ; how much more do A and C own 
than B and D? 

(13.) A walks a miles in t hours, and B half as far again in 
the same time ; how far will B walk in 10 hours? 

(14.) A walks 10 miles in x houis ; how long will he be in 
walking a miles? 

(15.) A spends at the rate of x dollars a day for a days, a 
dollar a day less for twice that time, and a dollar a day 
more for three times that time; how much does he spend 
altogether? 



c 42 ) 



CHAPTER VIII. 



SIMPLE EQUATIONS. 



I 

I i; 



V 



60. An equation is the statement of the equality of diiferent 
quantities, and these quantities are called the equation's 
members or sides. 

Thus 205+3=7 is an equation whose sides are 2a? + 3 and 
7, and a;*— 5a; +6=0 is an equation whose sides are a;'''— 5a+6 
andO. 

61. An identity is the statement of the equality of two like 
or different forms of the same quantity. 

Thus 2a + &=2a + 6, 2a; + 3a;=5a;, x^-bx-\-6=(x-'2X«:-S), 
are identities. 

62. In the case of an identity the equality holds for all 
values of the quantities involved, whereas in an equation the 
equality does not exist except for a limited number of values 
of the quantities involved. 

Thus the statement a;^+2a;+l= (a; + l)^ holds no matter 
what X is; but 5a;— 3=7 holds only when a;=2,and a;^+6=5a; 
only when a;=2, or a;=3. 

63. A symbol to which a particular value or values must 
be assigned in order that the statement contained in an equa- 
tion may be true is called an unknown quantity. 

Thus the unknown quantity in 6a;— 6=9 is x, and in 
22/2-2/=8 is y. 

The letters a, &, c, Z, m, ?^, ^, q, r are generally used to 
denote quantities which are supposed to be known, and x,y, z 
those which are for the time unknown. 



SIMPLE EQUATIONS. 



43 



-2/,» 



&i:. Quantities which on being substituted for the unknown 
reduce the equation to an identity are said to satisfy the 
equation, and are called its roots. 

Thus 5 is a root of 2ac— 3=7, because 5 when substituted 
for X reduces the equation to the identity 10—3=7. So 2 
and 3 are the roots of a;'' + 6=5a;, because when either is sub- 
stituted for X the equation is satisfied. 

65. The determination of the root or roots is called the 
solution of the equation. 

66. An equation is said to be reduced to its simplest form 
when its members consist of a series of mononomials involving 
positive integral powers only of the unknown. 

Thus5a;-8=0, a;2-5a; + 8=0, 2a;H6a;=7, a;^- 6*2= 7a; -8 
are in their simplest forms. 

67. Equations when reduced to their simplest forms are 
classed according to their order or degree. 

68. Simple equations, or those < f the first degree, are those 
in which the highest power of the unknown quantity is the 
first; as, for example, 2£c=5, 5a5— 8=0, 305—7=0. 

69. Quadratic equations, or those of the second degree, are 

those in which the highest power of the unknown quantity 

is the second; as, for example, a;^— 2xH-3=0, 2a;2=9, 
4a;2_3=10a;. 

70. Equations of the third and fourth degrees arc called 
cubic and biquadratic equations, respectively ; thus r' + 2a;=10 
is a cubic, and a;^— 2a;^=10a;— 5 a biquadratic. 

71. It is proved in works on the Theory of Equations that 
the number of the roots of an equation is equal to its degree ; 
so that a simple equation has one root, a quadratic two roots, 
a cubic three roots, and so on. 

72. In order to solve an equation it is generally necessary 
to reduce it by one or both of the following processes : — 

I. Transposition of Terms. 
II. Clearing op Fractions. 



44 



SIMPLE EQUATIONS. 



%\ 



These operations will be illustrated by applying them in 
order to the solution of simple equations. 

I. Transposition of Terms. 

73. If an equation contains no fractions it may be solved by 
transposition of termSf which consists in taking the unknown 
quantities to one side of the equation and the known to the 
otheVy the signs of the quantities which are so transposed being 
changed. 

Thus, if the equation is 4a; +5=10, by subtracting 6 from 
each side we get 

4a; +5-5=10-5, 
or 4a;=5; 

and so any quantity may be transposed from one side to the 
other by changing its sign. 

Examples. 
(1.) Solve 5a;+15=25. 

Transposing + 15 we get 

5a;=25-15=10. 

The value of x is then found by dividing both sides by its 
coefficient 5. 

/. x=2. 

(2.) Solve 8a;-4=2a; + 20. 
Transposing —4, 8a!;=2a5 + 20 + 4. 
Transposing 2x, 8a; — 205 =20 + 4 ; 

6a;=24. 

:.x =4. 

(3.) Solve 10 + 2(6a;-l) =32-3(a;-4). 

Clearing of brackets, ^; 

10 + 12X— 2=32-3a; + 12. 
Transposing 10, —2, —3a;, i 

12a; + 3a;=32+12-10+2; [ 

15x=36. 

• «._ 3.6 —i)2. 



"^ 



SIMPLE EQUATIONS. 



45 



(4.) Solve 3(a;*+2x) + 13 = 3a;2-7+4(3a;-l). 
Clearing of brackets, 

3a;H6a; + 13 = 3a;2-7+12aj-4. 
Transposing +13, 3a;'», +12a?, 

3a!2-3a;2+6a;-12a; = -13-7-4; 



Dividing by— 6, 



- 6x = -24. 

03 = 4. 



When, as in this case, the same quantity is common to 

t)oth sides, it may be struck out without actually transposing ; 

thus 

5a;2-6a; + 7=8a; + 5!»2_10 

becomes — 6a3 + 7=8a;— 10. 

(5.) Solve ax-\-'b=c. 

Transposing +&, aa5=c— 6. 

Dividing by a, a;=£rL. 

a 

(6.) Solve ax-\-'b—cx-\-d. 

Transposing +&, ex, 

ax—cx = d'-'b. 

Collecting coeflBcients of oj, 

(a— c).r = c?— 5. 
Dividing by a— c, 

a—c 

(7.) Solve a(a;— &) = &(a; + a)— c. 

Clearing of brackets, 

ax—ab = hx-\-ab—c. 

Transposing — a&, hx, 

ax—bx = ab-\-ab—c. 

Collecting coefficients of a;, 

(«— &)a; = 2a&— c. 
Dividing by a— &, 



a;= 



2a6— c 



4« 



SIMPLE EQUATIONS. 



Exercise XXII. 

(1.) 3-|-a;=6. (2.) «-6 = 4. (3.) a; + 5 = 12. 

(4.) a; + 9=4. (5.) 2x-l=3. (6.) 5a;+4 = 29. 

(7.) 4-3a;-5. (8.) l-aj=6. (9.) 3=6-2». 

(10.) 2a; + 3 = .'r + 5. (11.) 5i»-2-2x + 7. 

(12.) .'c+4 = 18-4x-4. (13.) 2r+3 = 3x-4. 
(14.) 16-2«=46-5a;. (15.) 3(a:-l)+4 = 4(4-aj), 

(16.) 5-3(4-2a;)+4(3-4a!)=0. 
(17.) a!-l-2(a;-2) + 3(£c-3) = 6. 
(18.) 6(c»-.5)+2(a;-3)-(»>-l)=9. 
(19.) 2(x-.2)-3(«-3) + 4(c«-4)~5(a;~5) = 0. 
(20.) 4(a;-ll)-7(x-12) = 6-(a;-8). 
(21.) «=2a-». (22.) 2a-3a;=8a-5a;. 

(23.) a;-2&=2a~a;. (24.) a + o:-& = a + 6. 

(25.) aas— a&— ac=0. '26.) ax—a = b-—bx. 

(27.) ax-a^=hx^b\ (28.) «(«-&) = c(a3-a). 



I 



II. Clearing of Fractions. 

74. If an equation contains fractions, it may be reduced to 
a form capable of solution by transposition, by multiplying 
both sides of the equatio7i by the L. C M. of all the denominators 
of the fractions. 

In the following examples numerical denominators only 
will be considered. 



(1.) Solve 



XXX 






2 3 5 

Multiplying by 30, the l.c.m. of 2, 3, 5, 

15x — lOx = 6x — 30. 

Transposing, 15a; -- lO.r — 6x = — 30 ; 

-a; =-30. 
.*. x = 30. 

(2.) Solve ^^^^&« + l?^ 

A o o 

Multiplying by 24, the l.c.m. of 2, 3, 8, 

12(a!-l) + 8(2a; + 3)=:3(6.K+19); 



SIMPLE EQUATIONS, 



47 



whence on clearing of brackets and transposing we get 
It must be observed that ^=i(a;-l), ?^= i(2a; + 3), 



3 



6x4-19 
fl-'^d y =J(6a? + 19); and therefore the brackets must be 

supplied in the first step since the numerators become bino- 
mial factors. 



(3.) Solve 



8-^:=l + ^+2^(j 



2 • 3 

Multiplying by 6, the l.o.m. of 2 and 3, 
■18~3(a;-l) + 2(aj + 2) = 0; 
whence on clearing of brackets and transposing we get 

a5=25. 



m 



X X 



Exercise XXIII. 

X , X 



(l.)i-i=3. (2.) ^+^=7. (3.) 1-1+1=10. 



(4.) ^+^=10. 



2 3 '4' 
(5.) ^+1=20-^9 



(6.) ^+^=4-^. (7.) ^ + 2=^-K«^ + l). 

(8.) 2(a)-l)-K2a;-9) = K17-2x). 

(9.) §^=4-^^-12 (10.) |(x-4) + K«'-6)=2x-15. 
(11.) 3-|cc=l-J^(7a;-18). 
(12.) 6(x-l)-l=f(5-2x) + .3(x + l)+4a;. 

(13.) x+^^—^^-9h=0. 

(14.) 4.+2,V+^-^^-^-^=^+M. 
/iK\ * + 4 3a;--2 , 1 x—1 

\^'^') — o -ITT- +4 = 



12 



3 • 



..n. S-2x 6x 5_3(2x + 6) 2x 

^,10.; -^-+7—7 14- -y 









48 



SIMPLE EQUATIONS. 



,.^.?>x-\ 13-;r 7.T_^ll.x-|-33 n 

(1 i .) —^ >.- —-rr+ ^. =^- 



5 






6 



^^'•^2 3'T 5- + " 9"-^* 
(20.) |(x-.8)-'^--*-g=0. 



ih 



•I 



( 49 ) 



11 



CHAPTER IX. 



PROBLEMS, 



75. When a question is assigned for solution the unknown 
quantity, or number, is generally involved in the various 
conditions which are proposed for its determination. The 
expression of these conditions in Algebraical language leads 
to an equation, the solution of which will be the solution of 
the question. 

76. In some cases, although there are more unknowns than 
one, they are related to each other in such a manner that 
when one is determined the others become immediately 
known. In such cases the unknowns can be expressed in 
terms of one unknown. 

Thus, if the sum of two unknowns is equal to 8, we may 
denote one of them by x and the other by 8— jc, or one of 
them by 4: + a3 and the other by 4— a?,- if the greater of two 
unknowns exceeds the less by 3, the former may be denoted 
by X, and the latter by jc— 3 ; if there be two numbers of which 
one exceeds 4 times the other by 7, the former may be denoted 
by 4a3 + 7, and the latter by x. 

In like manner, if there be three unknowns, of which the 
first exceeds the second by 3, and the second exceeds the 
third by 5, the first may be denoted by a?, the second by 03—3, 
and the third by as— 8. 

77. The following examples will illustrate the method of 
solving problems by means of simple equations : — 

(1.) What number exceeds its fifth part by 20 ? 

Let X be the required number. 

E 



50 



PROBLEMS. 



1^' 

m 



Then its fifth part =. | ; and by the condition of the 
question 

5 

/. x=25. 

(2.) The Slim of two numbers is 71, and their difference 43 : 
find them. 
Let x be the greater number. 

Then .a— 43 is the less : and since their sum is 71, we have 

a; + a;-43=71. 
/. a; =57, the greater; 
and a; -43 = 14, the less. 

This question may also be solved as follows: 

Let X be one number, the greater suppose. 

Then 71 -a; is the less i and since their difference is 43, we 

have „^ ^ Ao 

a;-(71-x)=43. 

.*. a; =57, 
and71-aj=14. 

(3 ) A boy is one-third the age of his father, and has a 
brother one-sixth of his own age; the ages of all three 
amount to 50 years. Find the oge of each. 

Let the boy's age z=x years. 

Then the father's age —'6x years, 

And the brother's age =| years. 



% 



\ 



And by the condition of the question 



o 
/. .x=12, 
3a;=36, 

^-2. 

Fractions may be avoided by supposing the ages of boy, 
father, and brother to be 6a?, 18a7, x years, respectively. 



PROBLEMS. 



5' 



5: 



re 



we 



IS a 
iree 



(4.) A and B start from two places, 90 miles apart, at the 
same moment, A walking i miles per hour, and B 5 ; when 
will they meet, and how far will each have walked ? 

Let the time of meeting be x hours after starting. 
Then A will have walked 4a5 miles, and B 5x miles ; and 
since the sum of these two distances is 90 miles, 

4a; + 5x=90. 
/. aj=10. 

/. 405=40, and 5x=50, are the distances in miles walked by 
A and B, respectively. 

(5.) How much tea at 90 cents per lb. must be mixed with 
50 lbs. at $1"20, that the mixture may be sold at $1*10 ? 

Let X = the number of lbs. at 90 cents, the value of which 
will be '900; dollars. 

Then, since there will be a? + 50 lbs. in the mixture, its 
value will be 1*10 (cc + 50) dollars ; and since the value of the 
50 lbs. at $1'20 is 60 dollars, we have 

•90x + 60=l-10(£c + 50). 

Multiplying by 100, 

90a; + 6000=110(x- + 50). 
.*. 3J=25. 



boy, 



Exercise XXIV. 

(1.) Divide 25 into two such parts that 6 times the greater 
exceeds twice the less by 70. 

(2.) Divide 135 into two parts such that one shall be |- the 
other. 

(3.) The sum of two numbers is 37 and their difference 3 : 
find them. 

(4.) A fish weighed 71bs. and half its weight : how much 
did it weigh ? 

(5.) At a meeting 43 members were present, and the motion 
was carried by 9 : how many voted on each side ? 

E 2 



5« 



PROBLEMS. 



Xh 



(6.) Divide 326 into two parts, such that f of the one shall 
be equal to the other diminished by 7. 

(7.) What is the number whose 4:th and 5th parts added 
together make 2i ? 

(8.) Forty-two years hence a boy will be 7 times as old as he 
was 6 years ago : how old is he ? 

(9.) A father is 57 years old, his son 13 : when will the 
father be 3 times as old as his son ? 

(10.) I have made 164 runs at cricket this season in 12 
innings: how many must I make in my next innings to 
average 14? 

(11.) My grandfather told me 10 years ago that he was 7 
times as old as myself; I am now 18: how old is my 
grandfather ? 

(12.) If in a theatre f of the seats are in the pit, ^ in the 
lower gallery, \ in the upper, and there are 50 reserved seats, 
how many are there altogether ? 

(13.) After losing \ of our men by sickness, and 210 killed 
and wounded, the regiment was reduced by \ : how many 
men did the regiment originally contain ? 

(14.) In a certain examination f of a boy's marks were gained 
by translation, ^ by mathematics, and -^ by Latin prose : he 
also obtained 1 mark for French. How many marks did he 
obtain for each subject ? 

(15.) Two men receive the same sum ; but if one were to 
receive 15 shillings more, and the other 9 shillings less, the 
one would receive 3 times as much as the other. What sum 
did they receive ? 

(16.) A and B begin trade, A with 3 times as much stock 
as B. They each gain £50, and then 3 times A's stock is 
exactly equal to 7 times B's. What were their original stocks? 

(17.) One-tenth of a rod is coloured red, one-twentieth 
orange, one-thirtieth yellow, one-fortieth green, one-fiftieth 



PROBLEMS, 



53 



•e to 

the 

sum 



blue, oue-sixtieth indigo, and the remainder, which is 302 
inches long, white : what is its length ? 

(18.) Find three numbers whose sum is 37, such that the 
greater exceeds the second by 7, and the second exceeds the 
third by 8 

(19.) Find a number such that if 5, 11, jmd 17 be suc- 
cessively subtracted from it, the sum of the third, fourth, 
and sixth parts of the respective results shall be equal to 19. 

(20.) How much wheat at 44s. a quarter must be mixed 
with 120 quarters at 60s. that the mixture may be sold for 
50s. a quarter ? 

(21.) How many lbs. of tea at 2s. 6d. per lb. must be mixed 
with 18 lbs. at 5s. per lb. that the mixture may be sold for 
4s. per lb. '? 

(22.) How much sugar at 4id. per lb. must be mixed with 
50 lbs. at 6^d. per lb., that the mixture may be worth 5d. 
per lb. ? 

(23.) A bag contains a certain number of sovereigns, twice 
as many shillings, and three times as many pence ; and the 
whole sum is £267 ; find the number of sovereigns, shillings, 
and pence. 

(24.) I wish to divide £5 4s. into the same number of 
crowns, florins, and shillings ; how many coins must I have 
of each sort ? 

(25.) A person gets an income of £550 a year from a 
capital of £13,000, part of which produces 5 per cent, and 
part 4 per cent. : what are the amounts producing 5 and 4 per 
cent., respectively ? 

(26.) I invest £800, partly at 4| per cent., and partly at 5^ 
per cent. ; my income is £39 10s. : what are the sums invested 
at 4 J and 62 per cent., respectively ? 

(27.) A garrison consists of 2600 men, of whom there are 
9 times as many infantry and 3 times as many artillery as 
there are cavalry : how many men are there of each ? 

(28.) My grand&ther's age is 5 times my own ; if I had 



.1 



54 



PROBLEMS. 



U\ 



been born 100 years ago, I should have been born 15 years 
before my grandfather : how old am I ? 

(29.) There is a number consisting of two figures of which 
the figure in the unit's place is 3 times that in the ten's; 
if 36 be added, the sum is expressed by the digits reversed : 
what is the number ? 

(30.) A miner works for 6 weeks (exclusive of Sundays), 
his wages being it the rate of 24s. per week, but he is to 
forfeit Is. besides his pay for each day that he is absent ; at 
the end of the time he receives 4 guineas : how many days 
was he absent ? 

(31.) A contractor finds that if he pays his workmen 
2s. 6d. per day, he will gain 10s. per day on the job ; if he 
pays them 3s. a day, he will lose 18s. : how many workmen 
are there, and what does the contractor receive per day? 

(32.) An officer on drawing up his men in a solid square 
finds he has 34 men to spare, but increasing the side by 1 
man he wants 39 to make up the square : how many men 
had he ? 

(33.) If the mean velocity of a cannon-ball at effective 
ranges is 1430 feet per second, and that of sound 1100 feet, 
how far is a soldier from a fort who hears the report of a gun 
1% of a second after he is hit ? 

(34.) An army in a defeat loses one-sixth of its number in 
killed and wounded, and 4O0O prisoners. It is reinforced by 
3000 men ; but retreats, losing a fourth of its number in 
doing so. There remain 18,000 men. What was the 
original force ? 

(35.) Su])pose the distance between London and Edinburgh 
is 360 miles, and that one traveller starts from Edinburgh 
and travels at the rate of 10 miles an hour, while another 
starts at the same time from London and travels at the rate of 
8 miles an hour : it is required to know where they will meet. 

(36.) There are two places 154 miles apart, from which 
two persons start at the same time with a design to meet ; 
one travels at the rate of 3 miles in 2 hours, and the other at 
the rate of 5 miles in 4 hours : when will they meet ? 



( 55 ) 



CHAPTER X. 

PARTICULAR RESULTS IN MULTIPLICATION AND 

DIVISION. 

78 There are several results in multiplication and division 
which should be committed to memory, as they enable us to 
dispense with the labour of performing the operations. The 
following cases occur most frequently. 

I. Since by actual multiplication 
(a + by=a^ + b^ + 2ah, 

&c. = &c. 

we can hence write down the square of a polynomial bv the 
rule: ^ 

The square of a polynomial is equal to the sum of the squares 
of the several terms and twice the sum of the products of every 
tivo terms. ^ 

Thus in the last example a\ +&2, a,c\ are, respectively, the 
squares of a, +&, -c; +2a& is twice the product of a and 
+ &, -2ac IS twice the product of a and -c, and -2ic is twice 
the product of +& and — c. 

In taking the products of the terms, two and two, it will 
be found most convenient to take in order the products of 
the first term and every term that follows it, then the pro- 
ducts of the second term and every term that follows it, and 
«o on, if there be more terms than three. 



m^ 



56 



PARTICULAR RESULTS IN 



Examples. 

(1.) (a + 2ie)2=a2+4a!2+4aa;. 
Here +4aa3 is twice the product of a and +2a;. 
(2.) (2a-5£c)*=4a2 + 25a;2-20aaJ. 

Here +250?^ is the square of —6x, and — 20aaj is twice the 
product of 2a and —6x. 

(3.) (2ar2-3a;+4)2=4a;* + 9a;2 + 16-12a;3 + 16»2_24a; 

=4x^-12a;H25!»'^-24x + 16. 

Here — 12a5^ is twice the product of 203^ and —3a!, +16a!r' of 
2x^ and 4, and —24a; of —Sx and +4. Like terms are added 
together and the terms are arranged according to descending 
powers of x. 

(4.) 992 =(100-1)2=10000+1-200=9801. 



Exercise XXV. 

Write down the squares of 
(1.) 03—1, x+a, x—5, x-\-d. 
(2.) 2a; + l, Sx-1, 2x+3, 3a;-2. 
(3.) x"-a, 2xy + l, Sx^-'2a, aa;2-46. 
(4.) 05— 2/ + Z, 2oj + 32/— s, x—2y^^z, 2a5— 42/ + 1. 
(6.) 2a2 + a + 3, 3a2-4a + l, a2-2a-4. 
(6.) Find the squares of 49, 98 and 995. 

79. II. Since (a +6) (a— 6) =a^^b^, it follows that the pro- 
duct of the sum and difference of two quantities is equal to the 
difference of their squares. 

Thus (2xVSy)(2x-Sy) =^x^^Qy^; 

(a2+l)(a2_l) =a4_l; 

(50)2+42/) (5a;2-42/) ='26x*-16y^; 
(2ar»+a*)(2a;3-a^) =4:x^^a^; 

501 X 499 =(500 + 1) (500-1) 
=50(^-1 
=249999. 



MULTIPLICA T20N AND DIVISION. 



57 



Exercise XXVI. 

Write down the products of 
(1.) oj-l, x + 1; a+3, a-3; 2+03, 2-a;. 
(2.) 2j5 + 1, 2x-l; 5a+2, 5a-2; 4a3+a, 4a;-a. 
(3.) aHa;,a^--a5; aHl,a'-l; a'* + a;2, aO^^jZ^ 
(4.) 3aH26, 3a2-2&; 4a»+2a;2, 4a3-2x2; 7a*-5a3, 
7a*+5a8. 

(5.) Find the products of 48, 52; 95, 105; 695, 705. 
80. III. Since by actual multiplication 

Id'^W-k-ab) (a-&) =a3-63, 

it follows that the sum of the squares less the product of two 
quantities multiplied by their sum is equal to the sum of their 
9ubes. 

In the latter identity the two quantities are a and —h ; 
the sum of their squares less their product is, therefore, 

^2+62 ah=a'^-\-b^-\-ab; and, since the cube of —6 is —6', 

the sum of their cubes is a^—b^. 

Examples. 
(I.) (x^-x + l) (x + 1) =x^-\-l. 
Here the two quantities are x and 1. 
(2.) (x^+x+1) (x-l)=x^-l. 
Here the two quantities are x and —1. 
(3.) (4a;2-2x + l)(2«+l)=8x» + l. 
Here the two quantities are 2x and 1, the cubes of which 
are 8x^ and 1. 

(4.) {x'--a^x' + a')(x^-\-a^)=x''-\-a\ 

Here tlie two quantities are x^ and a^, the cubes of which 

are .x^' and a*^. 

(5.) (4:x^-\-6x^ij-\-di/) (2x^-Si/) =8x^-27 f. 

Here the two quantities are 2x^ and —dy, the cubes of 
which are Sx*' and —27y\ 



58 



PARTICULAR RESULTS IN 



\ ' 



EXEUCISE XXVII. 

Write down the products of 

(1.) m^"mn-\-ii^,in-^u; 2^"-^Pl-^Q^>P-'Q' 

(2.) m^-m + l, ?/i + l; l + q-{-r,l-fj. 

(3.) x^-Bx + 9,x + 3', a2+4a-l-lG,«-4. 

(4.) 4a2-2a + 1, 2a + 1 ; leas^ + 4ax + a-, ^x—a. 

(5.) ^a'-Qab + W, 2a + 36; 9xH 15x7 + 25^/2, 3a;- S^/. 

(6.) a;< + a;2 + l, a;2-l; (c^-aV+a*, ccHa^ 

81. IV. By actual division it can be shown that the sum of 
any the same odd powers of two quantities is exactly divisible by 
the sum of the quantities. 

Thus, ^±i' =1, 

x+y 

t±y'=x'-xy + y\ 
x-\ry 

J-Ji-J- = 03* — xhj + x^y^ — xy^ + y^, 

&c.=&c. 

It will be observed that the signs of the quotient are 
alternately + and — , and that the successive powers of x 
are in descending whilst those of y are in ascending order. 

82. V. The difference of any the same odd powers of two 
quantities is exactly divisible by the difference between the 
quantities. 



Thus, 



— X, 

x—y 



_ — ^=zx^-\-xy-\-y'^, 
x-y 



x—y 



ic* + x^y + x^^y"^ + xy^ + y* 



&c.=&c. 






h 



MULTIPLJCATJON AND DIVISION, 



59 






Hero tlie signs of the quotient are all + . 

It may be noted that this case is included in the preceding 
(81) by supj)osing the two quantities to be x and — y. Thus 
the sum of the cubes of x and — ?/ is a;^— t/^ which is exactly 
divisible by their sum x—y. In fact the formulas of (82) 
are deducible from those of (81) by substituting —yioxy in 
the latter. 

83. YI. The dljference hetvjeen any the same even powers of 
tioo quantities is exactly divisible by the sum of the quantities 
and also by their difference. 



Thus (i.) 






y, 



x + y 

— ^ =jf-j'y-\rxy^^y^, 
x-{-y 



03' 



P — lfi 



yi. =x'^—'ji*y + x^y^—a^y^-\-xy*--y^f 



(ii.) 



x-i-y 
&c. =&c. 

x-y 

-^^ =x^-{-x^y + xy^-\-y^f 

x-y 



/yjC _^ yd 

iZ. = x** + 05*2/ + x^y^ + xy^ + xy* + y'^, 

x-y 

&c.=&c. 

It will be observed that when the divisor isx—y, the signs 
of the quotient are all + ; and when the divisor i&x + y the 
signs of the quotient are alternately + and — . It should 
also be noted that the formulas (ii.) are deducible from (i.) by 
substituting —y for y in the latter. 



Exercise XXVIII. 

Write down the quotients of 
(1.) x^+1 and 03^ + 1 divided by a; + 1. 



6o 



PARTICULAR RESULTS. 



II ! 



i: 



(2.) a;3_i and ;r'-l divided by x^l. 

(3.) ^2-1 and «<-! divided by « + !. 

(4.) x^-l and a;*-l divided by a;-l. 

(5.) 4a2-9i2 divided by 2a + 36. 

(6.) 9j^«-4a2 divided by 3.x3-2a. * 

(7.) ia'-a;« divided by i«Ha;3. 

(8.) Find what the quotient of x^ + if divided by aj+y 
becomes when (i.) a;=2a, y=36 ; (ii.) x=a\ y=2. 

(9.) Find what the quotient of x^^y^ divided by x-^y 
becomes when (i.) a;=3a, y=6 ; (ii.) aj=2a2, y =36. 



84. 
expr( 
of pc 
expm^ 

85. 
the p 
the p 
with 
of the 

Thi 



i; I 



Ii i: t 



( 6l ) 



CHAPTER XL 



3y a;+y 
jy x—y 



INVOLUTION AND EVOLUTION, 

84. The process by which the powers of quantities are 
expressed as mononomials is called Involution, The powei-fl 
of polynomials when so expressed are said to be developed, or 
expanded. 

85. We have already explained the notation for denoting 
the powers of a single symbol, as a, as, y. In all other cases 
the power of a quantity is denoted by enclosing it in brackets 
with the number indicating the power above and to the right 
of the bracket. 

Thus (—2a)'' denotes the third power of —2a; {a^lif the 
square of d^h\ {a%c^y the fourth power of a%c^\ (a— by the 
cube of a— 5; (x"— 2x+3/ the fifth power of a;'^— 2x+3. 

86. The same notation is used for denoting powers of 
powers of a quantity, brackets of different shapes being em- 
ployed when necessary. 

Thus (a^y denotes the square of a'; {(— 2icy)'^p the 
cube of (-2«2/)2 ; {(x^-S)'}^ the fourth power of (x^-bf. 



Exercise XXIX. 

Retaining the given quantities, denote 
(1.) The cubes of -a, 2x, Zxy\ 'la^Wc. 
(2.) The squares of 2a-l, a-& + l, x^—l. 
(3.) The squares of (xO^ (-2a)3, {4mx)\ (3a%c*y. 
(4.) The cubes of (a-6)3, (jb^-I)^, {9^^3x+2y, 



62 



INVOLUTION AND EVOLUTION. 



(5.) The squares of the cubes of x, — 2vC, aP-h^ a;— a, 

(6.) The cubes of the squares of —a, x^, 4x— 1, as^— a'*. 

87. A power of a power of a quantity is expressed as a 
power of that quantity according to the rule 



Thus, 



a ) —CI . 



88. So also 
Thus, 



(a"'¥c'^y=ia""''hP"-ci\ &G, 
(abhy=a^b'V-'. 



89. A rule has already been given in Art. 78 for expanding 
the square of any polynomial. The expansion may also be 
effected as in the following examples, in which the various 
parts are arranged in rows. Jn the first row occurs the square 
of the first term of the given qucDittfy ; in the second row the 
^>roduct of twice the first term added to the second and the 
second; in the third row the product of twice the first term 
added to twice the second term added to the third term and the 
third ; and so on. 



Mf 



Examples. 

(1.) (a + 6)2=a2 

+ (2a + 6)&=&c. 

(2.) (&-c)^=62 

+ (26-f)(-c')=&c. 



INVOLUTION AND EVOLUTION. 



^^h, jB— a, 


(3.) (af&+c)2=a2 




+ (2a + &)6 


f^a\ 


+ (2a + 26 + c)c=:&c. 


ised as a 1 


(4.) (a-5-.c)2=a2 


i 


+ (2a-.&X~5) 


1 


+ (2a-2&-c)(-c)=&c. 


1 


(5.) (a2~& + c2-(^)2=((^'2)2 


1 


+ (2«'-&)(-&) 


1 


+ (2a2-.2& + c2)c2 


J 


+ (2a2-2& + 2c2-t^)(-c?v.==&C. 



63 



panding 
also be 
Various 

(3 square 
row the 
and the 
^st term 
and the 



Exercise XXX. 

Express as powers or products of powers 

(1.) (x')\ (2xy, (x^)3, (2x3)8, (3x2)*. 
(2.) {ax'f, («V)2, (ah;h/)\ 
(3.) {ahcj, {aV)\ (2a^6V)3. 
(4.) {x-ifzy, (a6V)«. 

Expand 
(5.) (x + l)2, (2^-3)2, {x^^bf, {x^-^a'f. 
(6.) (a;2 + 2.c + 3)2, (a;2_3x + 4)^ (2x«-x«+5)2. 

90. Pligher powers of polynomials are developed by tlio 
Bmoraial and Multinomial Theorems, the explanation of which 
may be found in more advanced works. 

91. The process by which the roots of quantities are deter 
mmed is called Evolution., 

92. The nih root of a quantity is denoted by writino- the 
quantity under the sign V" , the line above being 
sometimes replaced by brackets enclosing the given quantity. 

Thus, a/2«. denotes the square root of 2 / ; 



K,.2 



J> 



5r«2j 



^«^ „ cube 

^^^^'+3 „ fourth „ xH3; 
^(«2-2a + 3) denotes the lifth root of «'^-2r< + 3. 



!! 



li 1! 



64 



INVOLUTION AND EVOLUTION. 



93. The mth root of the wth root of a quantity is denoted 
by writing the quantity under the sign "v ^ . 

Thus, V V2a denotes the cube root of V2a ; 
V^4^6^ „ fourth „ >^5^; 



V Vo;*— ir'+l „ cube 



» 



94. The mth root of the wth root of a quantity is expressei 
as a root of that quantity by the rule 



Thus, 



3 



VVa = v^(i; V V23D = '>i/2«; 



95. The reason of this rule will appear from the following case : — 
Let s/ i>Ja-s.x, Then on cubing both sides ot this equation we have 

Squaring a=a?«. 

Extracting the sixth root, 

Exercise XXXI. 

Retaining the given quantities, denote 
(1.) The square roots of 2aj, ax^^ 03^—1, a;^— 3a;+4. 
(2.) The cube roots of — a;^ 3a^ a - &, (a^— 3a+4). 
(3.) The square roots of the cube roots of 3aa;, a;— 1, 

(4.) The fourth roots of the cube roots of 2, 3a;— 1, 
2a*-aH3. 

Express as roots of the quantities under the double sign 

(5.) ^/^!a, y/I^, \/V3a, \/,yi^. 



IN VOL VTION AND E VOL UTION. 



65 






(6.) \/-^x^-l, V^'-^'Jar^-O, V^^x«-6;k* + 7. 

96. Since the square of a quantity is equal to the square of 
the same quantity with its sign or signs changed, it follows 
that there will be two square roots (if there be any), one 
being derived from the other by a change of signs. 

Thus, since ( + fr)^=(— a)-= + rt'-^, it follows that the square 
root of +«- is +a or — tt. These two roots may be repre- 
sented by the symbol ±« (read jilas or minus a) ; so that wo 
have ^a^=±a; ^x*^±x^; /^dx^=±3x. 

Again, since by Art. 78 

&c.=&c., 
it follows that 

/^a'—Saft + &''=«—&, or — a + 6, 
= ±(a-b); 
v'CaH&^ + c^— 2t«6 + 2ac— 2&c)=:a— /) + c, or — (i + &— c 

= ±(a-& + c). 

In the following examples we shall only determine thtict 
square root of a polynomial whose leading term is +, the 
other being derivable by a mere change of signs. 

97. Since \^a;'^"'-=a;'», it follows that 

j^ X ^^X" f \X ^^X , j^ X =: X \ 

where it will be observed the index of the root is one-half the 
index of the given power. 

98. The square root of a polynomial can generally be found 
by the following rule. 

(i.) Ari'd'iuje ih*' f/iiX'n ijiiantitu uccordiHg to afictniding or 
tiescettdhvj jiowers 0/ f^oiiie letter. 

(ii.) The first ternt of the root is the square root of the leading 
tor lit of the given (jnuhtit^,froui wJtich <7.s sqiatre is suhtractcdy 
leaving the first difference. 

(iii,; The frst divisor is fn'ire the first term <f the root added 

W 



66 



INVOLUTION AND EVOLUTION. 



to the second term. The second term is the quotient of the 
leading term of tlie first difference divided hy the leading term 
of the first divisor. The product of the first divisor and the 
second term of the root is subtracted from the first difference, 
leaving the second difference. 

(iv.) The second divisor is twice the sum of the first and 
second terms of the root added to the third term. The third, 
term is the quotient of the leading term of the second difference 
divided hy the leading term of the second divisor. The product 
of the second divisor and the third term of the root is then sub- 
tracted fro7n the second difference, leaving the third difference. 

The process is thus continued until the difference is zero. 



Examples. 

(1.) Find the square root of 9fl5^— 12x+4:. 

9a;2-12x+4(3x-2 
9x^ 

6a;-2)-12x+4 
-12a;+4 

Here the first term 3a; is the square root of the leading 
term of the given quantity, from which its square 9x^ is sub- 
tracted, leaving — 12ic+4. The leading term of the first 
divisor is 2 x Sx=6x. This is divided into — 12j;, the leading 
term of the first difference, giving —2, the second term of 
the root, which is also the second term of the first divisor. 
The product of 6x—2 and —2 is subtracted from the first 
difference, leaving remainder zero. The root is therefore 
3a;-2. 

(2.) Find the square root of 4«^— 12*H5a?2 + 6x + l. 

4^ *__ 

4x2 -.3^)« 12x3 + 5^ij + 6x + 1 

''12>:^-\-9x' 

4a;2-(k-l)-4xH 6a; + 1 
-4^H6,r + l 



INVOLUTION AND EVOLUTION. 



67 



of the 
ig term 
md the 
ference, 

rst and 
\e third 
•fftrence 
product 
hen siih' 
fence. 

s zero. 



leading 

is sub- 

lie first 

lleading 

term of 

livisor. 

lie first 

lerefore 



The first two terms, 2x^ and — 3x, are found as in Ex. 1. 
The first two terms of the second divisor = 2 (2/j^ — Sx) 
= ^x^—6x, the leading term of which is divided into —4:x^, 
the leading term of the second difference, giving— 1, the third 
term of the root, which is also the third term of the second 
divisor. The product of 405^ — 6aj — 1 and — 1 is then sub- 
tracted from the second diff'erence, leaving remainder zero. 

The root is therefore 2x^-^3x—l. 

The latter terms of the differences need not be expressed 
until the corresponding terms in the partial products are to 
be subtracted ; thus in the foregoing example 602 + 1 might 
have been omitted from the first difference. 

(3.) Find the square root of x^^ix^y -\-10x^y^—12xy^ -\-9y\ 
x*-4.<€hj + lOxV- 12a.y + 92/* {x^-^^xy + 3y^ 



^Ix^ — 2x2/) — 4x^2/ + lOx'^2/'^ 
—4x^2/+ ^^y^ 
2a;2-4^^H^^^2^~6«V-12a;2/H 92/* 
6a;y-12x2/^4-92/* 

In this example the given quantity is arranged according 
to descending powers of x, and the first two terms only of 
the first difference are expressed. 

99. The reason for the rule given in the preceding Article will 
appear from the following method of considering the last example. 
The given quantity is thei*e seen to be equal to 



a?» 






that is, to 



^i2x"-^2xy)(i-2xy) 
-ti2x^-Axy+'5y')Sy^ 

which by Art. 89 is equal to {x^ — 2xy-\-oy^y. 

Now, since a?' = (a--)-, the first term, x^, of the root is the square root 
of X*, the leading term of the given quantity. Also since —4tx^y — 
2x\—2xy), it follows that the second term —2xy is the quotient of 

F 2 



•:H 



68 



INVOLUTION AND EVOLUTION. 



— 4a;'y, the leading term of the first difference, divided by 2j;-, the 
leading term of the first divisor. Again, since 6a;-//- = 2x^(3,7^), it 
follows that the thii'd term ^xj^ is the quotient of 6,/;-^-, the leading 
term of the second difference, divided by 1x^^ the leading term of the 
second divisor. 

100. When the process for extracting tho square root is 
applied to a quantity which is not an exact square, a result 
is obtained the square of which added to the last difference is 
equal to the given quantity. 

Example, 

Find three terms of the square root of 1— 2a;. 

x^ 
1— 2a5(l—a;— o" 

_1_^ 

2-!c)-2a; 

-2x+a;2 



|-2a;-|]- 



X' 



-ajHccH?- 



— 05^ — 



X* 



: r,ti' I 



The 






dd 



03"— ^ 



square root is, therefore, l^x—% and remainder 
Hence 



(l-.-fj-.?--^ = l-%c. 



EXERCISIS XXXII. 



Find the square roots of 
(1.) ^ti*h\ 25x2//6, ^UY^ 
(3.) a6x2-36.»-H9. 



X 



(5.) a^H^-f,^. 



(2.) 16^H40a; + 25. 
(4.) 1 + 6a; + 9x2. 

(6.) x2~7x + ?. 



INVOLUTION AND EVOLUTION. 



69 



2.C-, the 

3/y2), it 

leading 

I of the 

root is 
- result 
•ence is 



linder 



(7.) 4a;2-Aa: + yi^. (8.) 4a;'-'-12x2/ + V. 

(9.) a;H4a;3 + 6a;2 + 4x + l. (10.) «* + 2xH3a;2+2.x + l. 
(11.) .x*-4a;3?/ + 6.xV-4a:2/H?/. 
(12.) 4,x«-4a;^-ll** + 14.«3 + 5ir-^-12.« + 4. 
(13.) Extract to three terms the square root of 1 + ic. 

101. The method of extracting the square root of a numerical 
quantity is founded on the Algebraical process, as will appear by- 
comparing the examples giv^n below. We shall first show how the 
number of figures in the root is determined by dividing the given 
quantity into periods. 



Since ^\-\, ^lOO^lO, ^10000=100, ^1000000=: 1000, &c., 
it follows that the square root of a number between 1 and 100, that is, 
containing 1 or 2 figures, lies between 1 and 10, and therefore contains 
1 figure ; the square root of a number between 100 and 10000, that is 
containing 3 or 4 figures, lies between 10 and 100, and therefore con- 
tains 2 figures ; so likewise the squai'e root of a number containing 
5 or 6 figures contains 3 figures ; and so on. If, therefore, we divide 
a number into periods of 2 figures each, beginning at the units, the 
number of such periods, whether complete or not, will be the number 
of figures in the root. 

In Arithmetic if the root is a+64-c4- &C'> 

2a is called the p'st trial-dmsorj 
2a +26 „ second „ 

2a + 26+ 2c „ thinl 

find so on. 



j> 



»> 



Instead of obtaining the divisors as in the previous examples, we may 
form them as below, where it will be observed that the sum of a 
divisor and its last term or digit gives the next trial-divisor. 

Examples. 

(1.) a a2+2a6+62+2ac+26c+c2(a+6+c 

a a^ 

2a+b ) 2a6+62+2ac+26c+c2 

6 2a6+62 

2a+26+c ) 2flrc+26c+c2 

2ac+26c+c2 



)--t i 



70 



INVOLUTION AND EVOLUTION. 



300 

300 


10'69'29 (300 + 20+7 
9 00 00 


600 + 20 

20 


) 169 29 
124 00 


600+40+7 


) 45 29 
45 29 



In the numericiil example, since the given number contains 3 periods, 
the root will contain 3 figures. The leading figure of the root, 
which is also the number of hundreds, will be 3, since the given 
number lies between 90000 = 300^ and 160000 = 400=. The second 
figure of the root, which is also the number of tens, is obtained by- 
dividing the first remainder 16929 by the first trial-divisor 600. The 
third figure of the root, which is the number of units, is obtained 
by dividing the second remainder 4529 by the second trial-divisor 
640. 

Omitting all unnecessary figures, we may arrange and describe the 
work as follows: — 



3 
8 

2 
647 



10'69'29 (327 
9 

)T69 
124 

)4529 
4529 



li 



The leading figure of the root is 3, the squai-e of which is the 
greatest square number under 10, tiie first period ; the square of 3 is 
subtracted from the first period, and to the remainder is annexed 
the second period 69 to form the first dividend 169. The first 
figure of the root is doubled to give the first trial-divisor 6, the 
division of which into 16 indicates the second figure of the root- 
The second figure of the root is annexed to the trial-divisor to form 
the first divisor 62, which is multiplied by the second figure of 
the root, and the product is subtracted from 109. To the remainder 
is annexed the third period to form the second dividend 4529. 
Under the divisor 62 is written its last <ligit, and the sum forms 
the second trial-divisor 64. The third and last figure of the root 
is 7, because when annexed to the trial-divisor to form the 



INVOLUTION AND EVOLUTION, 



n 



7TLT'""''' '^' ^''^'''' '^ '^' '-^ 7 is equal 



dividend 
(2.) 



2 
2 

44 
4 

481 
1 

4825 
5 

48306 



to 4529, the last 



5'83'51'23'36 (24156 

4 

T83 
176 



751 

481 

27023 
24125 

28^836 
289836 



ir 



( 72 ) 



CHAPTER XII. 



TEE HIGHEST COMMON MEASURE, 



^.| 



102. A QUANTITY is Said to be of so many di'mensions, in any 
letter, as are indicated by the index of the liighest power of 
that Ujtter involved in it. 

Tlids 'dx^—2x+4tis of 2 dimensions in x; 3?/ + 2?/— 5isof 
4 dimensions in y ; and ax^—hx^ + cis of 3 dimensions in x. 

103. A whole exvression, or quantity ^ is one which involves 
no fractional forms. 

Thus 3x^, —^xy, 2a;*— 3x4-4, are whole expressions, as are 
also all positive and negative integers. 

104. When two or more whole expressions are multiplied 
together, each is said to be a measure of the product, and the 
product is said to be a multiple of each factor. 

Thus 1, 4, a, and h are measures of 4a6; 1, 3, x and cc+1 
are measures of 3cc" + 3ji;; 5, a;-, x—\, and y'^-^l are measures 
of 5x2 (^_i) (r-1). 

105. It must be carefully observed that the terms measure 
and multiple are to be used only in connection witli whole 
expressions. In order, therefore, to obtain a multiple of a 
quantity it must be multiplied by another whole quantity ; and 
to obtain a measure of a quantity it must be divided by a 
whole quantity, the quotient also being a whole quantity. 

Thus the terms measure and multiple cannot bo us- d in 

connection with facc^, a:;^— _ + 3, because they involve tractions; 
whilst 1, 3, a, a;, x^, and « — 1 are measures of 3c<x^(a;— 1), 



THE HIGHEST COMMON MEASURE. 



73 



because the quotient of the latter divided by oach of tho 
former is a whole quantity. 

lOG. The hlyhei't mean/i.re of a quantity is tho quantity 
divided by + 1 or —1, that divisor being taken which will 
make tho first term of tho quotient positive. 

Thus the highest measure of — 4x- is 4:x^, oti>x—7 is 5;c—7, 
and of —2x~+ x—S is 2.c^—x-\-3. 

107. The Joivest mnUqjIe of a quantity is the quantity mul- 
tiplied by +1 or —1, that multiplier being taken which will 
make the first term of the product positive. 

Thus the lowest multiple of i2x— 3 is 2x— 3, and of 

—flc^ + aj— 5 is x'^—x + 6. 

108. When one quantity is a measure of two or more others, 
it is said to be a common measure of those quantities. 

Thus 2x is a common measure of 4x^ and 2.^'-— 4x, and x—1 
is a common measure of '2x—2, x^—^.*: + 1, and x^—1. 

109. The highest common measure of two or more quantities 
is the common measure of highest dimensions and greatest 
numerical coefficient or coefficients. 

Thus the common measures of li;^ and Qxhj are 

1, 2, X, 2x, 2x% 

of which the last is called the highest common measure 

(h.o.m.) ; the common measures of 4:{x'^—T) and 6(£c— 1)^ 

are 

1, 2, x-1, 2(x-l), 

of which the last is the h.c.m. 

110. We shall consider the process for finding the h.c.m. 
in the three following cases, namely — I. When one of the 
quantities is a mononomial. II. When the two quantities are 
liolynomials, neither of which has a mononomial measure. 
III. When the two quantities are polynomials, one or both 
of which have nioiionomiai measures. 

111. I. (i.) When the given quantities consist of two or 
more mononomials, their h.c.m. is the product of the G.C.M. 



74 



THE HIGHEST COMMON MEASURE. 



of the numerical coefficients and the highest power or powers 
common to the several (jiven quantities. 

Examples. 

(1.) Find the h.cm. of ISa&^x^ and Iha^'W. 
Here the g.cm. of 18 and 15 is 3; 
the highest power of a common to both is a ; 

ft it ^ f> M " » 

and there is no power of x common to both ; 

.'. H.O.M. is 3aP. 
(2.) Find the H.O.M. of 12xyz\ IQxYz^ and 28x*yz\ 
The G.CM. of 12, 16, and 28 is 4; 
the highest power of x common to the three quantities is x"^ ; 



It 


i» 


y 


a 


it 


» 


» 


z 


it 


M 



y ; 

i 

.*. H.O.M. is ^a?yz^. 

112. (ii.) The h.cm. of a mononomial and a polyn^ni •' 
is the H.O.M. of the mononomial and the h.cm. of the several 
terms of the polynomial ; and may be found by the following 
rule : — 

Express the polynomial as a product one factor of which is the 
H.CM. of its several terms: the H.CM. of this simple factor and 
the given mononomial will he the h.cm. required. 

Examples. 
(1.) Find the h.cm. of &aW and 8o:'W-VlaW. 

Here ^a^h^-VlaW^^a^W (2&-3a), where 4^262 is the h.cm. 
of U%^ and \2a%^; and the h.cm. of ^^h' and &aW is 2ai^ 
the H.CM. required. 

(2.) Find the h.cm. of 15x7/V and lQx'y'^z--l^xhfz^ 
+20a;Vs^ 

Here lOxhfz^ - Ibxhfz^ + IQxhfz^ = 6xYzX'2x - 3?/ + 4^), 
where 5a;2?/V jg the h.cm. of lOxYz^, I5:>y^y^z^, and 20.c:^y^z^ ; 






THE HIGHEST COMMON MEASURE. 



7$ 



and the h.c.m. of bxhfz^ and li^xifr} is 6xyh'\ the h.o.m. 
required. 



Exercise XXXIII. 



Find the h.c.m. of 



(1.) 12rt&2 and IGa^fts. (2.) 15a^6 and 20ah\ 

(3.) da.7i/ and SQxyz. (4.) 'Jax\y^ and 15a^.r2;. 

(5.) 4:2a^x^y and 35a^ic'^2/'*' (^O «&^cw'y and Sa'^twy'^i^. 

(7.) 8a*&, 12a»6^ and IGa^i^. 

(8.) 30a«<2/^ 426a;32/3, and IQcxY- 

(9.) 4a&2 and Ua^hx-^SahY 
(10.) 10tt&2cand30a»6H45a2i*. 
(11.) 10a&2a;y and 4:2a¥cx~70h*ci/. 
(12.) Swv^t^; and 12uhv^-24:u^viv^+S6uHo*. 

113. II. When two polynomials, neither of which contains 
a mononomial measure (other than unity), involve powers of 
a single letter, their h.c.m. can be obtained by the following 
rule : — 

(i.) Having arranged the given quantities according to descend^ 
ing powers^ choose that one which is not of lower dimensions than 
the other as divisor. 

(ii.) Divide this into the other multiplied hy the least number 
ivhich will make its leading term a mtdtiple of the leading term 
of the divisor. When this numher is unity, actual multiplication 
may he dispensed with. 

(iii.) Divide the first difference hy the highest mononomial 
measure contained in it. When this measure is +1, actual 
division may he dispensed with. 

(iv.) Bepeat the steps (i.), (ii.), (iii.), ivith respect to this last 
quotient (or diff'erence) and the first divisor; and so on, until there 
is no diff'erence. 

The last divisor ivill he the H.C.M. required. 



i 



76 



THE HIGHEST COMMON MEASURE. 



,.* 



It will be observed that no fractions occur in the process, 
and that the leading signs of all divisors are made positive. 

Examples. 
(1.) Find the h.o.m. of 2x2-7a; + 5 and Sx^-Tac+i. 

2 



2x2-7x+5)6x'^-14a;+ 8(3 
6a;2-21x + 15 

7)7^£T 

X- 1 

Here since the dimensions of the two given quantities are 
equal, either one may be made the divisor. 205^—70; + 5 being 
taken as divisor, Sx^— 7^ + 4 is multiplied by 2 in order to 
make the leading term Gx^ a multiple of the leading term 1x^ 
of the divisor. The first difference 7i»--7 is divided by its 
highest mononomial measure 7. 

In the next step x—1 and 2x^—lx-\-b are to be treated in 
the same manner as the given quantities. 

a;-l)2a;2_7a; + 5(2a; 

2x2-2x 

—5) -5a? + 5 

03 — 1 

The leading term of ^x^—lx-\r^ is a multiple of the leading 
term of u:— 1, and therefore the multiplication of the former 
by 1 is omitted. The difference — 5» + 5 is divided by its 
highest mononomial measure —5. 

In the next step the quotient x—1 and divisor x—\ are to 
be treated as the previous quotient and divisor were. 

a;~l)a;-l(l 

OJ-l 



Th( 

X — 

oft 
the 



THE HIGHEST COMMON MEASURE. 



77 



process, 
sitive. 



;ities are 
- 5 being 
order to 
^erm 2a5^ 
d by its 



eated in 



I leading 
former 
by its 

1 are to 



The process thus terminates and the h.c.m. is the last divisor 
a;— 1. 

Whenever as in the second step the difference is a multiple 
of the divisor, the division may be continued and the work of 
the last step avoided. Thus 

a;-l)2a;2-7a; + 5(2a;-5 

2x2 -2x- 









— 5x+5 
-5x + 5 






le whole work 


may 


be arranged j 


JB follows : 








3x2 


-7x+ 4 
2 






2ic2. 


-7aj+5) 


Gx'^- 
6x2. 


-14x+ 8(3 
-21x + 15 

7)7x- 7 








:=a;~l« 




X- 1)2x2 
2x2 


-7x+5(2x- 
-2x 


-5 


H.O.M. 




-5x+5 

— 5x + 5 





(2.) Find the h.c.m. of x2+2x— 3 and xH5x + 6. 

a;2+2x-3)x2 + 5x + 6(l 

X- + 2X-8 

~3)3x + 9 

^'+3)x2 + 2x~3(x-l 
X" + 3x 

H.C.M.=:X + 3. — X — 3 



Here the multiplication of x2 + 5.r + 6 by unity is unneces- 
isary. The other steps are similar to Ex. 1. 

(3.) Find the h.c.m. or2x''-7x-2 and 2':^-x-G. 



78 



"//£ HIGHEST COMMON MEASURE, 



2a;2-a;-6)2x3--7a'-2(a;, 1 



2i«;^— x"— 6. 



X 



X 



^--x-^ 



2x2-2x-4: 



2x2- 



X- 



■6 



-l)-a: + 2 



X 



--2)2x2-cc-.6(2x+3 

2x^—4:0; 



fi.o.M.=:a;— 2. 



3x--6 
Sic— 6 



Here 2x'^— cc— 6 is used as divisor in the second step, the 
dimensions of the first difference being 2. The partial 
quotients x, 1 of 2x^— 7x— 2 and 2x^—'2x—4: divided by 
2x2— cc— 6 ^j,Q separated by a comma to distinguish them 
from parts of an ordinary quotisnt. 

(4.) Find the h.o.m. of 

4x2 + 3x-10 and dx^+Tx^-Sx-lS. 

^^■{■3x-^10)ix'-\-7x'- 3x-15(x + l 
4a;H3x-2-10x 

4a;2 + 7a;— 15 
4lx^-\-3x-10 

Sic^)4a;H3x-10(a; + 2 



H.C.M,=:4x — 5. 



8.-10 
8r-10 



In this example there is no necessity to introduce or 
suppress any mononomial factors. 

114. The process of the foregoing examples will frequently 
enable us to find the h.c.m. of polynomials involving powers 
of several letters, as in the following 



bep, the 

partial 

cled by 

li them 






2 



ice or 



lently 
kowers 



THE HIGHEST COMMON MEASURE. 



79 



Exam.'ple. 
Find the h.c.m. of %^-^xy~6if and 3x2—40:^+2/'^. 



2 



_6x2 + 3.i7/-%2 
-lly)-ll.rv/ + 112/2 



H.C.M. =£13 — ?/• 



cc — 7/) 2«;2 -^xij— 2>if- (2x + By 
'dxy-^'dif 



Here the mononomial factor — lly is suppressed. 

115. The reason for the rule in Art. 113 will appear from the follow- 
ing proposition and its application in the next Art. 

V/licn one quantdij is a mea'^ure of two others, it will measure the sum 
and difference of any multiples of them. 

Let the quantities be A, L\ C : and let A measure B and C, so that 
JB = inA, Cz:=7iA, where m and 7i are wliole quantities. 

Take any multiples pB, qC of B,C, where p and q are any whole 
quantities whatsoever. Then, since ^9^ =/wi^4, qC=qnA, 

pBdzqC = p}nA±qnA = (pm±iqn)A. 

. pB±.qC , 

A 

that is, A is a measure of j)BdzqC, the sum or difference of any 
multiples of B and C\ because the ([uotiont pia-±.qn is a whole 
quantity. 

Thus, *Ja'-, which is a measure of (j.r-' ;ind 8,r-y, will measure 
(U"'( - -la) - ^jrii{ - ox), (Jx^-f y.r-//, ^ix^ - 8x'-//(4-.i7/), &ic. 

11(3. Sui)pose, now, that A and J! denote two ]iolynomials (as in 
Art. ll'i), neither of which contains a mononomial measure other than 
unity ; and let the dimensions ot' A be not greater than the dimensions 
of B. Divide A into // n)ultiplied by a mouonomiid whole quantity a, 
which makes its lirst term a multiple of the lirst term of ^4 ; and 



8o 



THE HIGHEST COMMON MEASURE. 



divide the difference G by the highest mononomial meatiure which it 
contains, and lot the quotient be D. 

B 

A)aBib 
64 

c)C 

D 

Now, C being equal to aB—bA, or the diffbrence of two multiples 
of A and B, is a multiple of all the common measures of A and B, and 
therefore of their H.c.M. 

Again, every common measure of C and A is a measure of C+bA, or 
ctB, and therefore of B, because A has no mononomial measure. 

Hence the H.C.M. of A and B is the same as the H.c.M* of ^1 and C, 
which is the same as the n.C.M. of A and D, because A has no mono- 
nomial measure. 

The problem is thus reduced to finding the H.c.M. of A and 1), 

These two quantities, A and I), are then treated in precisely the 
same manner as A and B ; and the process is continued until it 
tciminates as /> " ws, when the last divisor, F (suppose), is a measure 
of the last divide. 1 Q. 

P)Q(r 
rP 



The problem is thus finally reduced to finding the n.c.M. of P and 
Q. This is evidently P. 

Hence the last divisor in the above procef^ will be the H.C.M. 
required. 



PJXERCISE XXXIV. 

luiid the H.c.M. of 

(1.) 3.r2 + 2.t;-'il ?vA .^r>;2-fl3.r-6. 
(t2.) 2.tH;r-auivi8:<;-.-l^i 1. 
(3.) a;2-5.r + 6 and .'.'^-f;:^:-! 9. 
(4.) i>jHi0a; + 21 and x2-2.r-15. 



THE HIGimST CPMMOJV MEASURE. 



8i 



v'hich it 



lUltiples 
i i)\ and 

'+6^, or 

« 

A and C, 
lO mono- 

sely the 
until it 
measure 



f P and 



H.C.M. 



(5.) 2xHa;--15and2a;^-19a; + 35. 

(6.) x2-4a; + 3 and 4«3-9a;2- 15^ + 18. 

(7.) ct" + 10.rj + 25 and x^ + 15^;2 + 75 ,. ^ i25. 

(8.) ;/;3-6a-- + ll*-6 and a;3-it;^-Ux- + 24. 

(9.) ;t3-3x2-9x + 27 and 3,c3-a;a-27.« + 9. 
(10.) 3.>;2-22x + 32 and x^ -\\x" ^■Z'lx-'l^, 
(11.) 7a)2 - 12x + 5 and Ix? + .*- - Sa? + 5. 
(12.) 5.xH 2x^-15*;-. 6 and 7x^— 4cb2_21xH-12. 
(13.) 2x3 + 9^-2 ^ 4,^, __ 15 j^nd 4a;3 + 8^2 + 3x + 20. 

(14.) a;3-6x2 + lla5-6 and a;4-2x3-13a;2 + 14a; + 24. 
(15.) a;*-2x2 + l and a;*-4x3 + 6x2-4x + l. 
(16.) x^-^xy— Ykf and x^ — bxy + 6?/-. 
(17.) 2x2+3x2/ + ?/2and3x2+2x2/-2/'^. 

(18.) £c^ + a;22/ + ^2/ + 2/'^ ^Ji<i *''■~■2/^• 
(19.) 5*2 + 26jry + 33?/2 and 7x^ + 19i»2/ - Qtf. 
(20.) 3a;*-a;V-2i/^ and2x-* + 3x3?/-2xV-3a;2/». 

117. III. The H.o.M. of two polynomials involving mono- 
nomial measures is found as follows : 

Express each polynomial as the product of a moiioiioimal and 
a polynomial luhich contains no mononomial measure. Then 
the H.C.M. required is the product of the H.CM. of the rnono- 
Qiomial factors and the H.C.M. of the polynomial factors. 

Example. 

Fin( '. the h.c.m. of 

'6x'^y + l%9y + ^xhj and Qx^y^—Qx^y^ — l^xy"^. 

Here the given quantities arc equal to 

4^2^(2x2 + 3,/j + l) and ijxif(^-x^'^. 

The H.C.M. of 4^7/ and (Sxy^ is %ry ; and the h.c.m. of 
2*^ + 3x- + 1 and x- - x - 2 is u; + 1. 

.*. the H.C.M. re(iuired is %icy{x + 1). 

118. The H.C.M. of three polynomials is the H.o.jr. of any 
one and of the H.c.31. of the other two. 

o 



82 



THE h 'CHEST COMMON MEASURE. 






Example. 

Find the h.c.m. of x^-l, ccH 2^2-3, and 2x^ + 2x3 + 3a; + 3. 

The H.C.M. of x^—1 and a;^ + 2x^—3 is cc^— 1; and the h.c.m. 
of x^— 1 and 2x'* + 2;*3 + 3x + 3 is as + l, the h.c.m. required. 

Exercise XXXV. 
Find the h.c.m. of 
(1.) 12aa;^-27aa;2and2a2ic3+aV-3a2a;. 
(2.) 10x2+40^ + 30 and 4a;3-16a;2-84a;. 
(3.) 2x«-6x3-4x2and3x*-3x3-12x. 
(4) 2x2+x-3, x2-l, and x'-^^.x-h. 
{L) 6x2-a;-2, 21x2-17x + 2, and 15x2+5x-10. 

119. When all the component factors of the given quan- 
tities are known or can be determined, the h.cm. may be 
found by the rule of Art. 111. 



Examples. 
(1.) Find the h.c.m. of 

4(x-l)2(a; + 2)3 and 6(x-l)3(a;+2). 
The g.o.m. of 4 and 6 is 2 ; 
the highest power of x— 1 common to both is (x— 1)* 



)} 



X 



+ 2 



}) 



03 + 2. 



the h.c.m. required is 2(x--l)2(ic + 2). 



(2.) Find the h.c.m. of 

By Art. 80, Sn^x(x'--l) = H(fx(x-tXuHx-hi); 
Pj Art. 79, 12r/,TV-l) = 12ax'-^(.t;2-l)(a;^ + l) 

-I2r/x2(x-l)(x + l)(x2 + l); 



^;^fJ^fffffSrC0MM0J^ MEASURE. 

Now the O.O.M. of 8, 12, and 20 is 4 
the highest power of 









and the other factors a. + l, ^2^'i^ ^2 



IS a; 



a common to the three quantities 

X 

x—\ 

a:— 1; 

. +, + »'• + 1 arc not common 

. . tlie H.C.M. required is 4ax(x--l). 



Exercise XXXVI. 
Find the h.o.m. of 

(2.) 6a2(^+2Xa.-3)and8a^(«:-3)0 
(3.) ax2_2aa; + a and 2a^x'^^2a'^ 



X 



+3). 



(4.) a;2-l,a;3+l 



and a?*— 1. 



C5.)a;+2,a;2_4,andx3+8 

(60 3^3_8i,«,2^6^^9^^^^2^3__ 



18a7. 



I 



1 11 



g2 



li I' 



( 84 ) 



i. 






1- 
\ 



CHAPTER XIII. 
THE LOWEST COMMON MULTIPLE. 

120. When one quantity is a mnltiplc of two or more 
others, it is said to be a common multiple of those quantities. 

Thus l%t^ is a common multiple of 2.c and 3*^; and 
Xhx(x—V) of 3, 5, 15.^- and a; — 1. 

121. The lowest common midtiple (l.c.m.) of two or more 
quantities is the common multiple of lowest dimensions and 
least numerical coefficient or coefficients. 

Thus of the follo\*ing common multiples of 1x and Sx^, 

namely, 

&x\ 12^2, 18x2, 

6x^ 12xS 18xS &c., 
the first 6x2 ^g q^\\q^ the lowest. 

122. The L.C.M. of two quantities is found by the following 
rule : — 

(i.) If they contain no common measure except unity, their 
L.C.M. is their product. 

Thus, tho L.C.M. of 4x and Tab is 28a/;x. 

(ii.) If they contain a common measure, their L.c.m. is c^jual 
to one of the given qtiUiitifies multijilied by tjie quotient (f the 
other dlnlded liy their H.c.M, 

It will be generally found most convenient to expresK the 
L.C.M. as the product of several fuciors. 



THE LOWEST COMMON MULTIPLE. 



85 



(1.) Find the l.c.m. of ^.rhj and Sixif, 

The H.o.M. of these quantities is '6xy, 

.-. by the rule, l.c.m. J^f^- x Sixy'^X^xhf. 

'6xy 

(2.) Find the l.c.m. of 2x2-7a: + 5 and Zx^'-lx^^, 



The H.c.M. is found to be x—\\ and since 
= 2x-5, the L.C.M. will be (2a;— 6) (^x'-lx-Y^). 



2a;2-7a; + 5 



X' 



123. The following is the proof of the rule given in the preceding 
Article : — 

Let the two quantities be denoted by A and B^ and their ii.c.M. by 
C; and let A — aC^ B — bC, where a and h are whole expressions which 
have no common measure except unity. 

Then, since the L.C.M. of a and b is ab, the L.C.M. of aC( = Ji) and 

bC( = B) IS abC= —— =-^= -^.B=-^.A. 

124. The L.C.M. of three quantities is the L.C.M. of any one 
and the l.c.m. of the remaining two. 

The L.C.M. (if four quantities is the L.C.M. of any one and the 
L.C.M. (f the remaining three. 

And so on. 

J/Jxample. 

Find the l.c.m. olUx^ifz, Gxyz^, and lOx^yz^. 

The L.C.M. of 3x'i/'z and 6.///^!- js Gx'VV; and the l.cm. of 

QxYz^ and lOx'vjz^ is SOx'-y'z^ 

125. When alx the component factors of the given quan- 
tities are known or can be found, their l.c.m. may also be 
obtained hy multiplyimj the l.c.m. of the numerical factors by 
the Jiiyhest j)oiuer or 'powers of the several factors that occur in 
the fjiven quantities. 



86 



THE LOWEST COMMON MULTIPLE. 



I- 



I f' 



! 



.^ 



d 



Examples. 
(1.) Find the l.c.m. of 6x^^22^ i.x?u\ and 8xYz. 

The L.C.M. of 6, 4, and 8 is 24 ; the highest power of x 
which occurs amongst the factors of the given quantities is 
£c* ; and the highest powers of y and z are, respectively, y^ 
and z\ 

Therefore the l.c.m. required is 2ix* ijh^. 

(2.) Find the l.c.m. of 15«&(a— ft), 21a(a + bXa—h), and 

The l.c.m. of 15, 21, and 35 is 105; the highest powers of 
a,h,a—h, a + b, wliich occur amongst the given quantities, 
are, respectively, a, U^, a—b,a-\-b. 

Therefore the l.c.m. required is 105ahXa—b){a-\-b). 

(3.) Find the l.c.m. of x^-^l, a;^— 1, and cc^+l. 

Here x^-l = (x + l)(x^l); 

a;3-l = (a;-l)(,;:2 + aj + l); 
x'^+l = (x-\-lXx'-x-\-l). 

/. the L.O.M. = (cc + l)(x - l)(a:;2 + a; + l)(a;2-a; + 1) 
= lx'-lXx'+x^ + l). 

Exercise XXXVU. 

Find the l.c.m. of 

(1.) Sahx, 2bxy. (2.) Sa^xy, Uax% (3.) a¥, bc\ ca\ 

(4.) Sa%c, 12ah% 24Mhc\ 

(5.) Uhcu^, lQcav\ 20ab7v'-, 40a2'>V. 

(6.) x^-7x-\-12,x^-^x-G. (7.) 2x'-5cr-3,4a;2+4« + l. 

(8.) 3a;^-ll« + 6,2a;2-7a; + 3. 

(9.) cc^— 4aa;^ + 5a2£c— 2a^, o;'-?-^^*— 4a'. 
(10.) 8(a;2-l), 12(x-l)2. 



.0^ 



TJIEJ^ESrcOMj^O^ MULTIPLE. 

(13.) a^-{.a%, ab"b\ a^-~b'i 

(14.) 2x(:c^ + a; + i)^3^,3_3^^^a_^ 



«7 



(15 



/^ +r,i>'-?^i>H«3^ 



(16.) 



r -I,i>*-1,7>«-.1. 



(17.) (a-&) (a-,)^ (j.,.>> ^j_^^^ ^^^^^ 

(18.) 8a«6(a«6), 12a6(6-a), 3(a^-J2), W(j, J^,^^ 




IMAGE EVALUATION 
TEST TARGET (MT 3) 



<- 

s^^ 







^ 



'^ 



1.0 



I.I 



■SO 






^ 1^ 

1^ 12.2 



1^ 



2.0 



m 





F' ¥' r 




< 


6" 


► 






7 




^ 



^ %> V 
\y^ 
> 








^;7* 



Hiotographic 

Sciences 

Corporation 




33 WEST MAIN STREET 

WEBSTER, N.Y. 14S80 

(716) C '2-4503 






^^i^" 
^ 






^ 







u:^ 



rU 



'V 



It 



ill 



( 88 ) 



CHAPTER XIV. 



FBACTI0N8. 



hi 



126. When one quantity is not exactly divisible by another, 
the quotient is represented by writing them in the form of a 
fraction. 

Thus, — ^ denotes the quotient of —2 divided by +3; 
n?- the quotient of -2a divided by -^Sx; .'^"■^ ^ the 

quotient of a;— 1 divided by a;'*— 3a; +4. 

127. A fraction is not altered in value by multiplying 
or dividing the numerator and denominator by the same 
quantity. 

-2_-2x3_--6 ±6 _ +^6 X -4 _ -24 
Thus, ^3 - ^3^3 - ^9; _8 " -8x -4 "" +82^ 

-20 _ -20-j — 5 _ +4, -4 _ -4xf __ -^f . 



-25- -25-r— 5~ +5' +5 



+ 5x| 



-^'■r 



a ac —ax ^ 
h ~~ he ~ ~-hx ' 
x-l __ (x-l)(.r + l) 
2:^-3 ~ (2x-3)(x + l) 



a;2-l 
2x2-a;-3' 



128. The statement in the preceding Article depends on the two 
following propositions : — 

I. 77ie numerical value of a fraction is unaltered by multiplying or 
dividing its numerator and denominator by Vie sam^ quantity, 

(i.) Let a, 6, m, be integers. Then, since ^ denotes a of the 6 parts 
into which a unit is divided, it follows that 



FRACTIONS. 



89 



b ' mb 



a 

b 



ma 



. (1) 
. (2) 
. (3) 



Before considering the case where a, 6, m are fractional, we shall 
show how the operations of multiplication and division of numerical 
fractions must be performed. 



mc 



Let Of, 6, c, c?, m be integers. Then, since — =m, it follows that 





But by (3), 



a mo a , ma 
-.— =-xm=-- 
h c b 



ma _mao 
1 'be 

, a mc_mac 
"^ c bo 



. (4) 



Hence, if - be multiplied by an integer in a fractional form, the 
b 

product L a fraction whose numerator is the product of the numerators 
and denominator the product of the denominators. If, now, we wish 

ft o o 

to find the product of - and -, where - is not equal to an integer, the 
*^ b d a 

operation of multiplication must be in accordance with this rule ; for 

any application of the term multiplication to cases where its primaiy 

meaning (which is repetition) does not apply, must not be inconsistent 

with the cases where it does apply. 



, a c _ac 
"VlTbd 



(5) 



Again, since diyision is the inverse of multiplication, both in its 
primary and extended applications, it follows from (5) that 



ac 



a 



bd d b 
__acd 
'~bcd 

__ac d 
~bd' c 



by (3) 
by (5) 






^i; 



i \ 












I '9' 



M'l 



90 



FRACTIONS. 



Also 



ac ,a 
hdrV 



_c 
'd 

_abc 
abd 

bd a 



by (3) 
by (5) 



* /I 

Hence it follows that the quotient of the fraction - divided by 

b 



-. is equal to the product of - and — : that is, 
d b 

a.c a d_ad 
6 



d 



_a _ 
b c he 



. (6) 



(ii.) Now, let a, 6, m be fractional, which will include the case 
where some of them may be integers. 



Let 



Then 



y u d 



X 



z yz 



z 
u 



And 



mh' 



c 
c 



X ex 



cz 



by (5) 



d u du 

__cxdu 
czdy 



by (6) 



=^ by (3) 

yz 



••T 



.Tna 
mb' 



II. If, in the fraction -, a and 6 denote positive and negative 

quantities, the sign of the fraction depends on the signs of the nu- 
merator and denominator. These signs will be either like or unlike, 
and on multiplying or dividing by a positive or negative quantity they 
will still be like or unlike, and the sign of the fraction will therefore 
remain unchanged. 









FRACTIONS. 



91 



ivided by 



the case 



negative 

the uu- 
r unlike, 
tity they 
therefore 



For example, multiplying by — 1, 

I 2 — 2 2 

— „ = — - = +- , a positive fraction ; 
+ — o o 

_i_3 _3 q 

— - = — _ = - a negative fraction. 

— 4 +4 4 

129. A fraction which involves fractional coefficients in the 
terms of the numerator and denominator can always be 
reduced to one whose numerator and denominator are whole 
expressions by multiplying the numerator and denominator 
by the l.o.m. of the several denominators. 

Thus, n^ii_3j ■j ^ is reduced by multiplying nxmierator 

and denominator by 12, the l.c.m. of 2, 4, and 6, to the 

. . _ 12x-6 

equivalent from 24^2^:y^^2- 

130. A fraction is said to be in lowest terms when its 
numerator and denominator contain no common measure, 
except unity. Hence a fraction is reduced to lowest terms hy 
dividing its num>erator and denominator hy their H.C.M. 

Examples. 

(1.) Eeduce " lOa^S'-^y/ *^ lo^^st terms. 
The H.O.M. of SaWx and Ua^b^y is 4a^h\ 

Sa^b^x _%x 
•'• UaWy "" 3ay* 

■x^—1 
(2.) Reduce -3-r-i to lowest terms. 

The H.O.M. = 03 + 1. 

x^—\ x—\ 

^2^~^Tl* 



ceHl ~ 



as' 



2^2 I g^ 2 

(3.) Eeduce 03,2^53; 7 9 to lowest terms. 

The H.O.M. = 2x— 1. 

2^3^-2 _ a^-f^ 
•*• 2a;2-5;r + 2 "■ x-2' 



! 



i, ' 



92 



FRACTIONS. 



m 



!-M 

I, 

I 

I. 



'I 






Exercise XXXVIII. 
Reduce to lowest terms — 






(5.) 
(8.) 
(11.) 
(13.) 



6a2Z>2 



(3.) 



Qx^yT? 



Sxy'^^ 



O" 






a»-l 

^-7«J-10 

3 + l%+3a;2 
S-\-Sx-3x^' 



(6.) 



OJ + l 



C7) -"^ 



(9.) 



x^—y^ 



a^—y^' 



(10.) 



jc"— .4 
a;-3 



(12.) 



a;2-4x + 3' 

a;H7a; + 12 
a;2-ic-20' 



n4N g;'^-3ag-70 



'Ti^ cc^— 6a;— 9 
^ '^ a;* + 3x3-9a;-9' 



(16.) 



12x^-15x+3 



6*3_6x2+2x-2* 



(17.) 



a;^— 4a3V 



x3-6a;'^^ + 12ay-82/3 



nS) «'^— 3a;'^y + 3a;;/''— y ^ 
x^—x^y-'xy^-^y^ 



131. Fractions ate said to be like or unlike, according as thej 
have the same or diflferent denominators. 

2a 



Thus -, - are hke fractions, as are also 



SiC 



X X 



a;2--r a;=^-l' '3o 



2a 

jr- are unlike fractions. 

Sx 

132. Unlike fractions are reduced to like fractions by mul- 
tiplying the numerator a:>d denominator of eadi fraction by the 
quotient of the L.O.M. of the several denominators divided by 
lis deno7ninator. 

The common denominator will accordingly be the l.c.m. 
of the several denominators. 

Examples. 

(1.) Reduce p^, -^ to like fractions. 
00 4» 

The L.C.M. of 3& and U k i%d. 



1 



6ah 
4x+3* 



3 
-2* 

ig as thej 



SiC 



2a^ 
^-1' 3»/ 



s 6y mul- 
lon by the 
ivided by 

;he L.c.M. 



FRACTIONS. 



93 



The multiplier for the first fraction is -w. -^l\ 

•* Wh~md' 

The multiplier for thti second fraction is -j-^- = 36 ; 

. 3c__ %c 
" 4a5 126c^' 



(2.) Eeduce 



a—b ' 6— a 



to like fractions. 



The L.o.M. of a—b and b—a is a--6, the quotient cf which 
divided by 6— a is — 1 ; 

6— a a— 6* 



(3.) Eeduce 



1 x-l xj-2^ 

x-r x^+x+v x^-i 



to like fractions. 



The L.c.M. of the denominators is x^—l, the quotients of 
which divided by x-1, jrHas + l, jk»-1 are, respectively, 
tr/^ + cc + l, x—1, 1. 

a;-l __ a;'^-2a; + l 
aHaJ + l a^'-l ' 
as— 2»'^_£c— 2ar 
£C^— 1 os^— 1' 



EXEECISB XXXIX. 

Eeduce to like fractions : — 
1 2 



(1.) 



a 



^^•^ i^ ' ~ab' 



(3.) '^ , -5^ , 

cc xy xyz 



(5.) i , J- . A. 

a —a ax- — xy ««;«/ 



(4.)". ^ 



a 



4 

la 



(6.) - - , -^-y . 



(7.) - 



x-\-l u;+3 



W --r, 



3— a; 



x 



-r i-x- 



(9.) 



a 



•a 



a—b^ b—a 



m 



94 



FRACTIONS. 



1.1 



(10.) 
(12.) 
(13.) 
(14.) 



4 3^ 1_ 
2 



x^\ 



3aJ 



(^^•) a^^-x + l'a^ + l'x^+l* 
3 1 



(a: + 2) (x-lj ' (1-u^) (2-x) ' (a— 2) (x + 2) 

1 _ __ 1. _ 1 

(6_a) (c-ii) ' («-/>) (c-&) ' (a-c) (6-c)' 

1^ 1 1 

a((0—b) (.c— a) ' h(h—aj (x—h) ' a&x'* 



i ^t 



Addition and Subtraction. 

133. The operations of addition and subtraction of frac 
tions may be denoted by the signs + and — respectively. 

(c-2 



a 



-2& 
;2-3 
-2b 



Thus, the sum of , -^ , ajid may be denoted by 



_a , — 'J<> , a?— 2. 
x-l x^-3 dx-^' 

and ^Ac diference between -_,^ , a?^c? v^— * by 

Sx"^—! oa;--4 



4a5 



a;-l 



3a;''^— i 5a;— 4* 
Thus also +? denotes the fraction ? ^o be added to, and 



— - the fraction - to be subtracted from some quantity not 
expressed. 

184. Sums and differences of fractions when expressed as 
single fractions are said to be simplified, the operation being 
IDerformed according to the three following rules : — 

(i.) The sum of any number of UJce fractions is a like fraction 
ivhose numerator is the sum of their numerators. 

Thus — i- + ~^ + ^'^— ^ = 4--a; + 2a;— 3_a? + 1 
i»— 1 x—1 OS— 1 a;— 1 cc— 1' 

(ii.) The difference between two like fractions is a like fraction 
whose numerator is the difference between their numerators. 



FRACTIONS, 



3a; 



af frac- 
vely. 

loted by 



to, and 
ity not 



ssed as 
11 being 

''raction 



raction 






95 



(1.) 

(2.) 



Examples, 

5a;~4 _ 6 _ 5a;-4- 6_ 5a;~10 
2x-3 2a;-3 2a;-3' 2aj-3* 

4ag --2|K_4a: + 2x_ 6aj 



03' 



It will be observed that the — before the second fraction 
changes the sign of — 2ic. 

^ '^ " X+l X^ + 1 X^ + 1 

_Sx^-x-]-6-x^-{-Qx + S 

_.2x'^ + 5x + 13 

a^ + 1 " 

In this case the — before the second fraction changes the 
signs of all the terms of a;^— 6a3— 8. 

It appears from the preceding rules that 

_ a—b _ — a + 6 
c c 

and conversely ; in other words, the suhtraction of a fraction 
is equivalent to the addition (fa like fraction, whose numerator 
is the numerator of the former icith its sif/n or signs changed. 



Thus, 



a;2-2a: + 3 , -a-2 + 2x-3 



+ 



x^-l 
-2^5 



X^—1 

_ 5 
3a;2-7* 



2^-b 



(iii.) Addition and subtraction of unlike fractions are per- 
formed by reducing the unlike to like fractions and proceeding 
as above (i.), (ii.). 



Examples. 



(1.) Simplify .- + - + 



be ca ab 



>h 






III 



||.' 



)|i*'" 



|t|H 



i;. 



■#' 



96 



FRACTIONS. 



einco the l.o.m. of he, ca, ah is ahc, this sum is equal to 



a 

abc ubc ^ abc 



"^ nh,.— 



a^\■h■\^c 



ubc 



2r* 



+ i-^. - l_a2 



(2.) Simplify 1^^ . i_^^ 

Tlie L.c.M. of tho denominators is l--a\ 



1 



2« 



•'• 1 + a "^ 1-a "l-a^ - l-a2 + i_^2 -JZ:^ 



1— « 1+a 2a 
1— a + l + a— 2a 



- l-a" 

_2--2a 
~ l-a2 

2 

~ 1+a 

(3.) Simplify ^^—y - ^^::;^ - (^3^- 

The L.O.M. of the denominators is (x-\-yy (x—yY 

. 1 1 1_ 

" (x+yf y'-x^ (x-yy 

__ (j^—yf y'^-^x^ (?i-\-yf 

- (a;2-2/2)2 - (^2_2/2)2 - (052-2^2)2 

_ x'^—^xy^y'^—y^-\-x^ ~'X^'-^lxy'^i^ 

x^—Axy—y'^^ 
~~ (x^—y^y 



Simplify — 



EXEBCISE XL. 



(^•) 2a"'"(/ 



X 



/ox- 1 2 



FR^CT/ONS, 



97 



1 2 



2 8a 1 



Wx~3^' (S-^^+^^-i^'* <6-) 



a;— 1 ^—8 

2a5 ~ a" ' 



2a-3fe 3a--26 a-36 
<.y-; a~6 a + 6 



a+6 a-& 
(8-) ^3^6+^6- 

(10.) i:i^.+ii:^- 



(^^•-^ a'*"a + 6 + a*+a6 ' ^^^'J a"2x-l""4x2-l * 

^^^•''a;2 + a; + l+a;2-a;+I ^^*'^ 2"(»-l)"2(a; + l)"a;«' 

(■"•^0 ^c^iy'-^^'iy ■*■ x^^^p ■ 
.... 2 1_ a;+3 

. 2y' . 



^ -^ a:;=' + a;?/+2/^^a;'*— as*/ + «/'*'»* +33 V+!/ 
/iQ\__^""l JL_ g ^ + 1 1 

2 2 1 

(^^•) (¥:^"XiT2)~5^)(a;-2)''"(x-2)(a;+2)' 

ah ac 

(^0-) (a-6)(6-c),'''(a-cXc-6)' 

V^J-O (a;-6XaJ-c)'*"(a;-cXiC-a)+(a;-a)(x-6) 



(ic— a) (as— ft) (as— c) * 



OS— a 



cc— 6 



as- c 



(22.) (fc-aXc-«)"*"(a~&Xc-«»)'''(«^0(^-^c)' 

(^^•) ("^:^(^:::^+(6-«x^-c)+((;-6Xc-a) ' 

135. A mixed quantity is the sum of a whole expression and 
a fraction; as, for example, 

6 „ 1 3a; _ as— 1 






FRACTIONS. 



136. A mixed quantity may he expressed as a fraction by 
considering a whole expression as a fraction whose denominator 
is unity ; and conversely, a fraction may be expressed as a 
mixed quantity when part of its numerator is a multiple of its 
denominator. 



liii 



ii i 



[11 






Examples, 

,- . _ 1 2a; 1 2aa; + l 
(1.) 2x+-=-j- + -=— - . 

, 1 a;— 1 1 a^ 

(3.) p-+i-»"+i=,;=-TT — r=x^+i- 

a;'— 1 
It will be observed here that —a^ + 1 = —(a;''— 1)= — -| — 

(4.) Express ^.^ as a mixed quantity. 

On dividing 2a5*— 3a;+4: by ac+1 we get quotient 2a;— 6 and 
remainder 9. 

x-\rl x + 1 

(5.) Express ^±^^^^ as a mixed quantity. 

In this case the quotient is a? +2, and remainder — aj+1. 

. a;8 + a;2_2a;+3_^ . o. -fc + 3^ 

*, 5 x-r^-t-A — i . 

a;''— a; + l US'*— a; + l 



= a; + 2- 



a;-l 



a;'^— a; + l ' 



Here the + before the fraction is changed to — by changinp: 
at the same time the signs of all the terms in the numerator 
-a;+l. 



iion by 
minator 
ed a8 a 
)h of its 






—5 and 



FRACTIONS. 



99 



Simplify — 
(1.) 1 + 5. 

(4) ^+a-l 

X 



EXEBCISE XII. 



X 



(2.) ^,-1. 
(5.) ??-o+l. 

X 



(3.) 2-?. 
a 

(6.) 2+?-i. 



(7)8- 






03* 



05 — 1 

(12.) 4a; -1- 



(10.)-^, -a. 



(11.) 2- 



a + 6' 



2a; + l 



a; + 3 
Express as mixed quantities — 



(13.) xHiry + t/'^-^"^'. 



(14.) 
(17.) 
(19.) 
(22.) 



a 



2a;2-aj + 3 



X 

a;' + 3a; + 4 
a;«-3 a;H2 



(15.) ^-i. 
a; 

(18.) 
(20.) ^^-6 



(16.)-^ 
4-3:^ + 6a:'' 



2-3a; 



3x ' 



'Sx 



(21 >i ^ + ''^'' 



(23.) 



6a:'— 4a; + 5 
2a;=^-a; + r' 



(24.) 



jc*— a; + 5 



^\ 



B + 1. 



langmg 
nerator 



Multiplication. 

137. To denote that two or more fractions, or a fraction and 
a mononomial whole quantity, are to be multiplied together, 
they are written in a row with the multiplication sign x , or 
• (dot), between them. 

Thus ? X -,, or V * 4 denotes that ~ is to be multiplied by 
d a b 

c, x-l 20^-3 jej^oteg the product of '^"- and ?^'"?. 



d' a' 



x^-l 



X--1 



^^^ X — 3aaj denotes the product of ^. - and — 3«a:. 
b 

h2 



100 



FRACTIONS, 



1^ 



K j 



Hi ^ 



lit * 



■S v- 



Sums and differences when they are factors must bo 
enclosed in brackets. 

Thus, _„A_/a'^— aaj+cc^ denotes the product of -3^ and 



a^— ic^ 



a^— as' 



a^— ax+a;2; _^(a;+^) denotes the product of -^ — and 
cc+aX XI flj + a 

0; + ^; -^(^+?-i\ denotes the product of 4^:? and 
X 05-^— 1\2 X a2V cc*— 1 

138. Multiplication of fractions is performed according to 
the following rule : — 

The product cf any number of fractions is a fraction whose 
numerator is the product of their numerators and denominator 
the product of their denominators. 

A whole quantity is to be considered as a fraction whose 
denominator is u iiity ; and sums and differences of fractions 
and mixed quantities must first be reduced to fractions. 



(1)^ 



x^—a 



Examples. 

ax _(x^—a~)ax_x—a 
2ax x + a 2ax(x-^a) 2 

Factors common to any numerator and any denominator 
may be struck out. Thus ax which is common to the first 
denominator and second numerator, and x+a which is com- 
mon to the first numerator and second denominator, may 
be struck out before multiplying, and the result will be in 
lowest terms. When possible, therefore, the component 
factors of the several numerators and denominators should 
be obtained. 

(^\ 4a^— 4aa;^ hc-\-hx_4ia(a+x)(a-^x') ^ &(c + cc) 
^ "^ Sbc^-Sbx^ ' o^^ac" 35(c + a;) (c-a;) * a(a-x) 

__A(a+x) 
3(c-a;)' 



FRACTIONS., 



lOI 



Here a,a^Xy b, and c+a; are struck out, being common to 



the numerators and denominators. 



(3.) 



X 



(a; + 6)(a; + 7) 



(a;2-49) = 



X 



(x + 6)(a; + 7) 
x(x-7) 



(x + 7)(x^7) 



\ x^—i)x^ + l x^—1 x^ + 1 



xHl 



x-l 



(x-tl)(x-l) JcHl 

1 



x+r 



ExEBCiSE XLU. 



Simplify — 

yo\ b^c <?a o^}> 

KP'J ~~~-\ • — o • — 5« 

2/2S 2i» ^32/ 



^2 



a 



c?-W 



(5.) . — ^. 



.ON a- <>- c- 

oc ca ao 

(4.) '^ y 

(6.) 



1—03 ' l + ai* 



(7.) 
(9.) 



^S + p • ^2^2,2 

1 ac + l a rl 



^°'^ -(7-6)3 --^3 • 



05 + 1 0)2 + 1 a;*+l 

(11.) /I+I+IW 

\x y zj 

a3.)(..:.«i)(«-g. 

(17.) (t,— ■^-+i)^ 
\x* xy y^Jx 



(10.) 



x^—l x^—\ a;*— 1 



x^y^ 



(12.) -^(^-«\ . 

(14.) 5;+* f(-5 1-). 

ae.) ^-(&+^Ui— ^V 

6x\ <* / \ a + xj 

(18.) (l±'+^\4dt' . 



I- • * 









1 



i ■ 



w 



it : 



102 



FRACTIONS. 



Division. 

139. To denote that one fraction is to be divided by another, 
they are written in a row with the sign -f- between them. 
The same notation is employed when the dividend or divisor 
is a mononomial whole quantity 

Sums and differences when they are the object of division 
must be enclosed in brackets. 

Thus, --f-- denotes that - is to be divided by - ; 



4a 
a 



|4-(2a-l) 



4a 

,2 



a 



»> 



ii 



X 

y 



it 



X' 



n 



ii 



a 



Ii 



it 



a 



ii 



Ii 



3c; 

2a-l; 
3a; 



X 

2a 

X 



.2__: 



a 



The same thing may also be denoted by writing the 
quantities which are the objects of division in the form of 
a fraction. 

2^ a 

Thus, i==27j-^a* 2^n=6-^-(2a-l); 
a 

a. — •*' 



X 



140. When the product of two quantities is unity, each is 
said to be the reciprocal of the other. 

Thus, since axl=l , ? . ^ = 1 , ^^ . ?£r§_i :. r^, 
lows that 






f'R ACTIONS. 



103 



1 .. 



a 
a 
h 

2x-3 



is the reciprocal of a , and 



ft 



)} 



tt 



»i 



b 
a' 

2x-S 
x-l 



» 



■5 » 



1. 

a 
a . 
a " b' 
2.r-3 x-1 



a of 
b 



n 



x-1 " 2a:-3' 



141. Division of fractions is performed according to the 
following rule : — 

The quotient of one fraction divided hy another is the product 
of the former and the reciprocal if the latter, 

A whole quantity is to be considered as a fraction whose 
denominator is unity ; and sums and differences of fractions 
and mixed quantities must first be reduced to fractions. 

Examples, 
^ ''' bxy'' ' luxhj 5xy^ ' 2ab^~ 'ET' 



(2.) 



a' 



^y 5xy^ 

a^ + ax _ a^ 



by 

a^-}-ax + x^ 



a^—a^ ' a^^ax+x^ a-'—x^ ' a^ + ax 



. a^ 

(a— a;) (a^ + a.x + a;^) 
a 

a^—x^' 



a^-\-ax-\-x^ 
a(ci-\-xj 



Wh) ' \a b) ab" ' ~ab 



_bx-\-ay ab 

ab bx—ay 
_bx-\-ay 
bx—ay 






_y^—xy-\-x^ 



x^y"^ 



ix+y)(x^-xy-\-y^) 






■M 



1 1 * 



I 



I 



aV(a;+2/)' 



ini 



i •■■■■'• 

II : 






I' 






W': 



104 



FRACTIONS. 



EXBBOISE XUn. 



Simplify— 

(3.) — , -^ -ir- 
^ ' cp + l • as— 1 



(9.) 






a; 



y 



(11.) ^-r^+y. 

05—2^ as+y 
^ - ^ - 

i+6 S"*'G 






a a 



(6.) 



(4.) -J^-:--^-. 
a+cc • a— as 



... /I 1\ . 1 .g. ia^h a^^\a^ 



6 



a+ac a— a; 



/in\ a—x'_a±x 



(12.) ij^ 1 



(B 9? i9 



«? 



(14.) 



8~10 10~li} 

S+io 10+12 



( I05 ) 



CHAPTER XV. 



SIMPLE EQUATIONS (continued). 



I ,• 






142. We shall give in this chapter some examples of equa- 
tions involving fractions with literal denominators. Such 
equations may be cleared of fractions by the rule already 
given in Art. 74. In some cases before applying this rule 
it will be found more advantageous to simplify parts of the 
equation separately. 



(1.) Solve 



Examples, 
x—a x~-b 



b ~ a 
Multiplying by ah, the l.c.m. of the denominators, we get 

(x—a)a = (x—h)b. 
Clearing of brackets and transposing, 

ax—bx^a^-^hK 
Collecting coefficients of x, 

(a-'b)x = a^—b\ 
Dividing by a—h, 



(2.) Solve 



x = r- = a+&. 

a—b 



3a;-l 4a;-2 



Multiplying by the l.c.m. 6 (2x— 1) (3c5— 2), 
6(3a;-lX3x-2) . 6(4a;-2X2a!-l)=(2x-l)(3a!-2). 



io6 



SIMPLE EQUATIONS. 






m 



w 



n-T I 






Clearing of brackets, 

54a;2-54:a;H-12-48a;2+48x-12=6x2-7x+2. 
Transposing, 

54x2-48x2-6a;2-54a;+48a;+7a;=2-12+12, 

.*. a;=2. 

a;— l__a;— 2_a7— 4__aj— 5 
03—2 as— 3 as— 5 as— 6* 



(3.) Solve 



Simplifying the sides separately, 

(sc-1) (a;-3)-(a;-2y_ (a;-4)((g-6)-(a;-5)a 
(a;_2)(a;-3) (!B-6)(a;-6) 

Clearing the numerators of brackets, 

-1_ ^ -1 

(a;-.2Ki»-3) (ir-5)(a;-6)* 

Multiplying by the l.o.m. (as— 2) (ai— 3) (as— 5) (as— 6), clearing 
of brackets, and solving, 

as =4. 



Solve- 



EXEBOISE XLIV. 



a\ 12 1 _ij 
•-^¥"^12^-^*' 



(2.) 



(3.) 
(5.) 

(7.) 



16 



27 



3aj-4 5a;-6* 

a;-l_7a;-21 

a;_2 7a;-26' 

_^+2 = -^. 

a; + l as + 2 



(4.) .- 



42 _ 35 
a;-2 a;-3* 

45 57 



^QN 5as— 3 __ 2as + 3 _j^ 
(ll.)^=:i+«^~^ 



2a; + 3 4a;-5* 

C6 "i 2as— 6 _ 2 as— 5 
'^ '^ 3x-8 3^^* 

xo N 2as— 3 , 2as— 1_Q 
^^'^ 2HHhl + 2^+3~''' 

/I n\ Gas + 13 2as_ 8.r + 5 



2a;-5 



(12.) 



g!-14 _ 2a;-29 _ 1 
a? 2aj-20 2!b' 



(13.) 



(2aj+3)(a!-5) (3x-2) (jb-11) ' 



SIMPLE EQUATIONS. 



107 



X 



-^ = 0. 



/15) JP a;--l_g;— 3_ a;— 4- 

''^ a-l a;-2 a;-4 a;-5' 

(16.) zr"' 



6— a? 6— a; 4— a;~ 

Uo-; — T~ + = r + -. 

a, a 

(20.)^-^+^=!. 

(22.) ^— ^ -L^'"^ - ^"^^ I ^jj. 



(17.) 2-5= c. 
a b 

(19.) 5+»=5+*. 
(21.) A-.«=j2-a2. 



( io8 ) 



-it-- 






1' 



I'm'* 



i ■■ 



I':. *^ 



CHAPTEE XVI. 

PROBLEMS (continued). 

143. We shall give in this chapter some examples whicl 
are more difficult than those in Chapter IX. 

Examjples, 

(1.) If A can perform a given work in 60 days, and B in 4C 
days, in how mauy days will A and B, working together, be 
able to perform it ? 

Let w denote the work to bo done, and x the required 
number of days. Then 

w 
amount of work done by A in one day=gQ» 

B 

AandB 



$$ 



» 



li 



W 

io 






M 



f9 



t> 



X 



w w w 



1 1 



Divide by «;,- = gQ+|^- 

Multiply by 120a;, 120=2x + 3a;. 

.*. a:!=2-l. 

(2.) At what time between 5 and 6 is the minute hand of 
a watch 5 minute divisions behind the hour hand ? 



PROBLEMS, 



109 



les whicl 



id B in 4C 
;ether, be 

required 



Lot X =5= the number of minute divisions between the hour 
hand and 5; then 5— a; = the number of minute divisions 
between the minute hand and 5. But the number of minuto 
divisions between 12 and 5 is 25 ; therefore the number of 
minute divisions between 12 and the minute hand is 25— 
(5-a;)=20 + a;. 

The hour hand thus moves over x minute divisions while 
the minute hand moves over 20 + a;; and since the latL.r 
moves 12 times faster than the former, it follows that 

20 + a:=12a;. 

• T — l-fi- 

• > •*' — "'■ 1 T* 

Hence tho required time is 12!r=21^Y minutes past 5. 

(3.) A grocer bought 200 lbs. of tea and 1000 lbs. of sugar, 
the price of the sugar being \ of that of the tea. He sold the 
tea at a profit of 40 per cent., and the sugar at a loss of 2i 
per cent., gaining on the whole $4550. "What were his buy- 
ing and selling prices ? 
Let the cost price of the sugar per lb. =x dollars. 

•*• a ii tea „ =033 „ 

Then „ „ 1000 lbs. of sugar = 1000a; dollars. 

„ „ 200 „ tea =1200a! „ 

40 
The profit on the tea-TQ^* 1200a; =480a; 

2J 
The loss „ sugar = jQQ- 1000x = 25a; 

/. 480a;~25x=45-50 
a;=-10 

.". the buying price of sugar is 10 cents, and the selling 
price 91 cents per lb. ; the buying price of tea is 60 cents, 
and the selling price 84 centa per lb. 



( ■■ 
; 



te hand of 



Exercise XLV. 

(1.) I arrange 1024 men 8 deep in a hollow square : how 
many men will there be in each outer face ? 



no 



PROBLEMS. 






If 



!!l- 



Ik 






(2.) A regiment containing 700 men is formed into a hollow 
square 5 ranks deep : how many men are there in the front 
rank ? 

(3.) A man has a number of cents which he tries to 
arrange in the form of a square ; on the first attempt he has 
180 over ; when he increases the side of the square by 3 cents 
he has only 31 over. How many cents has he ? 

(4.) On a side of cricket consisting of 11 men, one-third 
more were bowled than run out, and 3 times as many run out 
as stumped ; two were caught out. How many were bowled 
and run out, respectively ? 

(5.) Water expands 10 per cent, when it turns to ice. How 
much per cent, does ice contract when it turns to water ? 

(6.) A manufacturer adds to the cost price of goods 20 per 
cent, of it to give the selling price ; afterwards, to eflfect a 
rapid sale, he deducts from the selling price of each article a 
discount of 10 per cent., and then obtains on each article a 
profit of 8 shillings. What was the cost price of each article ? 

(7.) A person invests £14,970 in the purchase of 3 per 
cents, at 90 and 3r per cents, at 97. His total income being 
£500, how much of each stock did he buy ? 

(8.) A and B join capital for a commercial enterprise, B 
contributing £250 more than A. If their profits amount to 
10 per cent, on their joint capital, B's share of them is 12 per 
cent, on A's capital. How much does each contribute ? 

*(9.) In a concert room 800 persons are seated on benches 
of equal length. If there were 20 fewer benches, it would be 
necessary that two persons more should sit on each bench. 
Find the number of benches. 

*(10.) A man travelled 105 miles, and then found that if he 
had not travelled so fast by 2 miles an hour, he would have 
been 6 hours longer in performing the journey. Determine 
his rate of travelling. 

(11.) An express train running from liondon to "Wake- 
field (a distance of 180 miles) travels half as fast again as an 



These questions belong to Exercise XLVIII. 



PROBLEMS. 



Ill 



a hollow 
he front 

tries to 
)t ho has 
^ 3 cents 

rac-third 
J run out 
e bowled 

ice. How 
,tcr? 

ds 20 per 

) efifect a 

article a 

article a 

article ? 

of 3 per 
ne beincr 

rprise, B 
nount to 
is 12 per 
e? 

benches 
ivould be 
h bench. 

hat if he 
uld have 
etermine 

o Wake- 
tiin AS an 



ordinary train, and performs the distance in two hours less 
time ; find the rates of travelling. 

(12.) A can do half as much work as B, B can do half as 
much as C, and together they can complete a piece of work 
in 24 days; in what time could each alone complete the 
work ? 

(13.) Three persons can together complete a piece of work 
in GO days ; and it is found that the first does ;I of what the 
second does, and the second % of what the third does : in what 
time could each alone complete the work ? 

(14.) What is the first time after 7 o'clock when the hour 
and minute hands of a watch are exactly opposite ? 

(15.) The hour is between 2 and 3 o'clock, and the minute 
hand is in advance of the hour hand by 14i minute spaces of 
the dial. What o'clock is it ? 

(16.) At what time between 3 and 4 o'clock is one hand of 
a watch exactly in the direction of the other hand produced ? 

(17.) The hands of a watch are at right angles to each 
other at 3 o'clock : when are they next at right angles ? 

(18.) How much water must be mixed with 60 gallons of 
spirit which cost £1 per gallon, that on selling the mixture at 
22s. per gallon a gain of £17 may be made ? 

(19.) How much water must be mixed with 80 gallons of 
spirit bought at 15s. per gallon, so that on selling the mix- 
ture at 12s. per gallon there may be a profit of 10 per cent, 
on the outlay ? 

(20.) If 16 oz. of sea-water contain 0'8 oz. of salt, how 
much pure water must be added that 16 oz. of the mixture 
may contain only 0*1 oz. of salt ? 

(21.) I have a bar of metal containing 80 per cent, pure 
gold, which weighs 30 grains : how much must I add to this 
of metal containing 90 per cent, pure gold, in order that the 
mixture may contain 87 per cent. ? 

(22.) How much silver must I add to 2 lbs. 6 oz. of an 



1 ■. 



l!, 



112 



PROBLEMS. 






iii> 



oUoy of silver and gold containing 91*7 per cent, of pure 
gold, in order that the mixture may contain 84 per cent, of 
gold? 

(23.) A person started at a certain pace to walk to a rail- 
way station 3 miles off, intending to arrive at a certain time ; 
but, after walking a mile, ho was detained 10 minutes, and 
was in consequence obliged to walk the rest of the way a milo 
an hour faster. At what pace did he start ? 

(24.) A person started at the rate of 3 miles an hour to 
walk to a railway station in order to catch a train, but after 
he had walked \ of the distance he was detained 15 minutes, 
and was obliged in consequence to walk the rest of the way 
at the rate of 4 miles an hour. How far off was the station ? 

(25.) A wins the 200 yard race in 28i seconds, B the con- 
solation stakes (same distance) in 30 seconds: how many 
yards ought A to give B in a handicap ? 

(26.) A wins a mile race with B in 5' 19". B runs at a 
uniform pace all the way ; A runs at \^ of B*s pace for the 
greater part of the distance, and then doubles his pace, win- 
ning by a second : how far did A run before changing his 
pace ? 

(27.) A boy swam half a mile down a stream in 10 minutes ; 
without the aid of the stream it would have taken him a 
quarter of an hour. What was the rate of the stream per 
hour ; and how long would it take him to return against it ? 

(28. A contractor undertook to build a house in 21 days, 
and engaged 15 men to do the work. But after 10 days he 
found it necessary to engage 10 men more, and then he accom- 
plished the work one day too soon. How many days behind- 
hand would he have been if he had not engaged the 10 
additional men ? 

(29.) Two crews row a match over a four-mile course; one 
pulls 42 strokes a minute, the other 38, and the latter does 
the distance in 25 minutes; supposing both crews to row 
uniformly, and 40 strokes of the former to be equivalent to 
36 of the latter, find the position of the losing boat at the end 
of the race. 



( "3 ) 



CHAPTEB XVn. 
QUADRATIC EQUATIONS. 

144. Wb have already defined Quadratic Equations in 
Art. 69. They are further called adfeded or pure, according 
as the term involving the first power of the unknown quan- 
tity does, or does not, appear. 

Thus, 4«;2-5x + 7=0, «;2+6x~3=0, a:^-3x=0,areadfected 
quadratics; 3a;2-8=0, 0^24-6=0, are pure quadratics. 

145. I. Pure Quadratics are solved bi/ transposition of 
terms and extraction of tlie square root. 

Examples. 
(1.) Solve jb2-4=0. 

Transposing, a?=L 

Since the square root of a positive quantity is either + or 
— , we have, extracting the square root, 

a:=±2. 
Thus the two roots are +2, —2. 

(2.) Solve 0^+6= Var'-ie. 
Clearing of fractions, 

3a;H15=10ay»-48. 
Transposing and dividing by —7, 



X 



^=9. 



.-. X =±3. 
Thus the roots are +3 and -3. 



I 



114 



QUADRATIC EQUATIONS. 



if'i:i; 

! 



!!'■« lil,; 



*il 



n Li's' 



146. n. Adfected Quadratics may be solved by one of 
the following three rules : — 

(i.) When tlie equation is in the form of the product of two 
factors, each containing the unknown, equated to zero, the solu- 
tion is effected by equating to zero each factor in turn. 

Examples. 

(1.) Solve (x-i-l) (2a;-3)=0. 

Since, when the product of two factors vanishes, one or 
other must be zero, we have 

either a; +1=0, and /. a;=— 1; 

or 2a5— 3=0, and .*. a;=f. 

Thus the two roots are —1 and f. 

(2.) Solve a;2-5aj=0. 

Factoring, x{x—b)=0. * 

:. either a;=0, 
or cc— 5=0, and /. £c=5. 

Thus the roots are and 6. 

Whenever, as in this case, the terms of an equation are 
divisible by the unknown x, we can infer that one root is 
zero. 

(3.) Solve (2x-5) (ax-43}) =0. 
Here either 2x— 5=0, and .*. a;=f ; 

46 



or aaj— 4&=0, and /. a;= 



a 



ib 

Thus the roots are f and -- 

■* a 



Q UADRA TIC EQUA TIONS. 



"5 



•y one of 

\ct of two 
the solu- 



one or 



;ion are 
root is 



Exercise XL VI. 
(1.) a)2„36=o. (2.) 5a;2=45. 

(4.) 2(a;2-7)+3(x2-ll)=33. 
(5)- K«^'H4) + Ka^H3) = a;2 + l. 



(3.) I =27. 



,(.. _6 7_ 



(7.) x'=Sx. (8.) 2+6^^=0. (^') i(x-'-4x)=6x. 

(11.) 4x2+1=0. 



(10.) x^-'~=0. 



(12.) a;2— ^=2a:2+iK. (13.) (x-S)(x-5)=0. 

(14.) (a5 + 5)(a;-7)=0. (15.) (a; + l)(a; + 3)=0. 

(16.) (2i»-l)(3x-4)=0. (17.) (3x-5)(2a; + 7)=0. 

(18.) (5u; + 6)(6a; + 7)=0. (Id.) (ax^b){cx-{-d)=0. 
(20.) a;^— aaj=a£c--a5'^. 

147. When tlie quadratic is not in a lorm adapted for 
applying rule (i.), it may be solved by either of the following 
rules : — 

(ii.) Having transposed the unknowns separately to one side, 
make the coefficient ofs? unity by division (if necessary). Then 
add the square of^ the coefficient of x, and the solution ?s effected 
hy the extraction of the sqtcare root of both sides. 

148. (iii.) Having cleared the equation of fractions (if neces- 
sary), and transposed the unknoums separately to one side, mid- 
tiply both sides by 4 times the coefficient of x^, and add the 
square of the coefficient of x. The solution is then effected by the 
extraction of the .iquare root of both sides. 

Examples. 
(1.) Solve a;2- 12a; + 35=0. 

By rule (ii.), transposing, 

a;2-12r=-35. 

i2 



- p. 4 






kiu* 



'*!' 



I 



I 



I 



[^'■1' ! 



ii6 



QUADRATIC EQUATIONS. 



Adding the square of one-lialf 12, 

a;2-.12^ + 6"=:36-35=l. 

Extracting the square root, 
a;-6=±l; 

that is, a:— 6=1, and /. a; =7, 
or 05— 6= —1, and ,*. a;=5. 

Thus the roots are 7 and 5. 
(2.) Solve 2a;2+5;«-3=0. 
By rule (ii.), transposing and dividing by 2, 

Adding the square of one-half -|, 

«.2 1 5^ I /5\2 — 2 5 _1_3— 4.9. 

X -\r-^x-t Vi^ --^6^^2--l6• 
Extracting the 6quar(3 root. 



»+^=±I. 



• • flj— 



-5±7 



4 =¥ or -3. 



Thus the roots are \ and— 3. 



(3.) Solve a;2+i)a;+g'=0. 

By (ii.), transposing, 

03^+^03=— g'. 

Adding the square of one-half^, 
Extracting the square root, 



QUADRATIC EQUATIONS. 



Thus the roots are 

2 2 

(4.) Solve 2a;2 + 5a;=3 by rule (iii.). 

Multiplying by 4x2=8, 

16x2+40x=24. 

Adding the square of 5, 

Vox' + 40a; + 5^=25 + 24=49. 

Extracting the square root, 

4ir+5=-fc7. 
/. a;=2, or — 3. 

(5.) Solve aic2+6ic+c=0. 

By (iii.), transposing, 

aa)2+5x=— c. 
Multiplying by 4a and adding W, 

Extracting the square root, 

2aa; + &=±V62— 4ac. 



j^^-6±V6''-4ac 



2a 



Thus the roots are 

" 2a" 



and 



— &-V&^— 4ac 

2tt 



117 



i 



Exercise XLVn. 



(1.) (B2-6ir+8 = 0. 
(3.) a;2+4a;-21==0. 

(5.) l-a:^=|. 



(2.) x2«4a,«6=o. 
(4.) 2*2_5^^2=0. 
(6.) ^=x^+-^. 



^ -i 



ili 






f 






1^' 



lis 



QUADRATIC EQUATIONS. 



(7.) 4a;=^-4a;=15. 
(9.) |'-a.=12. 



1-a; 2- 



03 



(8.) 6a;2-lla;+4=0. 
(10.) a;+l=" 



(14.) 



£C 



a; + 60 3x-5* 



(17.) a;=5— i-. 
a? — o 

(19.) ?±i_5fe±|)=8. (20.) 

(21.) (a;-3)2-3(x-2)(a;-7)=21. 
(22.) »2-(a+6)a;+a6=0. 



y-ip s 2a7— 3 a;— 3 _t 
'^ "^--^ 2^+l"3^T2-^ 

(18.) ^-2xz:^=f . 



2a;-5 3x--2 



a; + 2 



4— a; 



1,# 






( "9 ) 



CHAPTER XVm. 



PROBLEMS. 



149. When the Algebraical statement of a problem leads to 
a quadratic equation, the unknown quantity will be one of 
the roots. In some cases either root may be taken, but it 
will generally be found that one of the roots must be rejected 
as being inconsistent with the conditions of the particular 
question proposed. 

Examples. 

(1.) A person laid out a certain sum of money in goods 
which he sold again for $24, and lost as much per cent, as he 
laid out. Find out how much he laid out. 

Let X = number of dollars laid out. 

.'.05—24:= „ „ lost. 

But the loss is also x per cent, of x =— x»= ^ 



100 



100' 



X" 



^x-IL 



" 100 

a;2-100a;=-2400. 
03=40 or 60. 

The amount laid out was, therefore, $40 or $60. Thus 
both roots satisfy the conditions of the problem. 

(2.) A person buys a certain number of shares for as many 
dollars per share as he buys shares; after they have risen as 
many cents per share as he has shares, he sells and gains 
$100. How many shares did he buy ? 



I20 



PROBLEMS, 






Hi** 






#; 



Let X = the number of sLares bought, the price of which 
at X dollars per share is x^ dollars. 

The rise being x cents or ^ dollars, the price for which 

he afterwards sells the x shares at a^+^Tw^ dollars per share is 
(ic+=^W But the gain is $100. 

a;2=10000. 
/. X = + 100, or -100. 

As the negative root would not answer the conditions of 
the problem, it must be rejected. The answer is, therefore, 
100. 

Exercise XL VIII. 

(1.) A rectangular room which contains 1800 square feet is 
twice as long as it is broad : find its dimensions. 

(2.) Divide 20 into two parts whose product shall be 91. 

(3.) Find a number whose square increased by 20 is 12 
times as great as the number itself. 

(4.) Divide 15 into two parts such that their product shall 
be 4 times their difference. 

(5.) By what number must I divide 24 in order that the 
sum of the divisor and quotient may be 10 ? 

(6.) Find three consecutive numbers such that the square 
of the greater shall be equal to the sum of the squares of the 
other two. 

(7.) A ladder 34 feet long just reached a window of a house, 
when placed in such a position that the height of the window 
above the ground exceeded the distance of the foot of the 
ladder from the wall by 14 feet. Find the height of the 
window. 






PROBLEMS. 



121 



(8.) A horse is sold for £24, and the immber expressing 
the profit per cent, also expresses the cost price of the horse : 
what did he cost ? 

(9.) An article is sold for £9 at a loss of as much per cent, 
as it is worth. Find its value. 

(10.) A and B start together for a walk of 10 miles; A 
walks 1-2 miles an hour faster than B, and arrives 1^ hours 
sooner than he does : at wl)at rate did each walk ? 

(11.) After selling a part of an estate, and the same part of 
the remainder, I find I have left nine- tenths of the part first 
sold : what part did I sell at first ? 

(12.) An uncle leaves 14,000 dollars among his nephews 
and nieces, but 3 of them having died in his lifetime, the 
others received 600 dollars apiece more : how many nephews 
and nieces were there originally ? 

(13.) A number is composed of two digits, the first of 
which exceeds the second by unity, and the number itself 
falls short of the sum of the squares of its digits by 26. What 
is the number ? 

(14.) The sides of a rectangle are 12 and 20 feet : what is 
the breadth of the border which must be added all round 
that the whole area may be 384 square feet ? 

(15.) One hundred and ten bushels of coals arc distributed 
among a certain number of poor persons; if each had 
received one bushel more, then he would have received as 
many bushels as there were persons. How many persons 
were there ? 

(16.) A sum of £23 is divided among a certain number of 
persons ; if each one had received 3 shillings more, he would 
have received as mary shillings as there were persons. How 
many persons were tnere ? 

(17.) A company at an inn had £7 4s. to pay, but before 
the bill was settled 3 of them left the room, and then those 
who remained had 4s. apiece more to pay than before; of 
how many did the company consist ? 



■ I 



'' i 



122 



PROBLEMS. 



, ii 









(18.) A person rents a certain nuu* jer of acres of pasture 
land for £70; he keeps 8 acres in his own possession, and 
sublets the remainder at 5s. an acre more than he gave, and 
thus covers his rent and has £2 over. How many acres were 
there? 

(19.) An oflBcer can form the men in his battalion into a 
solid square, and also into a hollow square 12 deep ; if the 
front in the latter formation exceed the front in the former 
by 3^ find tho number of men in the battalion. 






t ,, 



li'^ii 



of pasture 
ission, and 
gave, and 
acres were 



( 123 ) 



lion into a 
3p ; if the 
ihe former 



CHAPTER XIX. 
SIMULTANEOUS EQUATIONS, 

150. If two unknowns are to be determined, there must 
be two independent equations. These equations are called 
simultaneous equations, because the same values of the un- 
knowns X and y must be substituted in both equations. 
Thus if 

2x^y = 9, 

the only values which satisfy both these equations at the 
same time are x=7,y=6. 

151. It must be borne in mind that there is an infinite 
number of values which will satisfy either equation sepa- 
rately. 

Thus, in the equation 203— 2/ =9, 

ifa;= 1, 2--2/=9, and .'. 2/=- 7; 
if 05= 2, 4— y=9, and /. 2/=— 5; 
if a:=10, 20-2/=9, and .*. y= 11 ; 
and so on. 

152. If three unknowns are to be determined, there must 
bo three independent equations ; and generally the number of 
unknowns must be the same as tho number of independent 
equations connecting them. 

153. The solution of simultaneous equations is effected by 
deducing from them other equations, each of which involves 
one unknown. This process is called elimination, and may be 
conducted according to one of the following methods :~ 



I 



m 






124 



SIMULTANEOUS EQUATIONS. 



I. Substitution. 
II. Comparison. 
III. Cross Multiplication. 

I. Method of Substitution. 

154. Tins method consists in finding from one equation the 
value of one unknown in terms of the other, and substituting the 
value so found in the second equation, which is thereby reduced to 
a simple equation in one unknown. 

For convenience of reference the given equations and others 
which arise in the process of solution are numbered (1), (2), 
(3), &c. 

Example. 

Solve x-\'y=d .... (1), 

2aj+2/ = 4 .... (2). 

From (1) we find 2/ = 3— a; .... (3). 
Substituting this value of y in (2), 

2a; + 3-a;=4:. 

Substituting this value of x in (3), 

2/=3-l=2. 
Thus the solution is x~l. y=2. 



II. Method of Comparison. 

155. This method consists in finding from each of the pro- 
posed equations the value of one and the same unknown i7i terms 
of the other f and equating the values so found. 



Solve 



Example. 

7a:~32/ = 19 .... (1), 
4a;+7y=37 .... (2). 



lation the 
tuting the 
'educed to 

Qd others 
^ (1), (2), 



ne pro" 
n terms 



SIMULTANEOUS EQUATIONS. 



125 



From (1) we find 



y 



__7i»-19 



• . (3); 



and from (2) 



y 7- .... (4). 



Equating these values of ?/, 

7aj-19_37-4a; 
3 T~' 

,\ a;=4. 
Substituting this value of x in (3) 

Thus the solution is a;=4, y=6. 

III. Method of Cross Multiplication. 

156. T/u-s ^^e^7,of^ co«,s/sf.s in Tnulfiplyhifj the qiven equations 
{reduced to the form ax + by=c) Inj such quantities as will 
render the coefficients of the same nuhioum mtmerically equal 
By adding or suhtracting the equations so found, ive obtain a 
simple equation in one unknoivn. 



Examples. 
(1.) Solve lx-2y= 5 . . . 

13«+4?/-30 . . . 
Multiplying (1) by 4 and (2) by 9, 

28x-36v/= 20 . . . 

117a^ + 36;y=270 . . . 
Adding (3) and (4), 

145ir = 290, 

Again, multiplying (1) by 13 and (2) by 7, 
91X-117?/- 65 .... (5), 
91as-+ 28^=210 .... (6). 



(1), 
(2). 

(3), 



c- 



\^ 



■ 

f! 



126 



SIMULTANEOUS EQUATIONS. 



Subtracting (5) from (6), 

145?/ = 145, 

/. 2/ = l. 
Thus the solution is x = %y = \. 



\ (2.) Solve 8a; + 25?/ = 9 . . . . 


(1). 


^ |, 12a;-10iy = 4 .... 


(2). 


,!|"' Multiplying (1) by 2 and (2) by 5, 




[ 16x + 502/ = 18 . . . 


. (3), 


;Li. 60a;-50?/ = 20 . . . 


.(4). 


Adding (3) and (4), 




76a; = 38, 




1, * rf — 1 

1 ..»«'— a • 




Again, multiplying (1) by 3 and (2) by 2, 


1 i 24a; + 75?/ = 27 . . . 


. (5), 


j ; 24a;-20//= 8 . . . 


.(6). 


Subtracting (6) from (')> 




! ; 95?/ = 19, 

.1 




V • ?y-i 




! • • 2/ - 6' 




-i^ Thus tho solution is x=\, ij=^. 





I*'^- 



m 



Exercise XLIX. 

(1.) 4x + 2/=ll, a; + 4?/=14. 
(2.) 2x + 32,'=21, 3a; + 5,?/=34. 
(3.) 3x=:23-2?/, 10 + 2.t=5?/. 

^^•^ 2 3' 32 

^^■^5^6~2^'^' 3 *'10 4* 

(6.) 3a;-22/ = 3(6-a;), 3(4a;-3?/) = 72/. 

(7.) 7(a)-l)=3(2/+8), ^+?=5^. 



SIMULTANEOUS EQUATIONS. 



127 



^ON 2a;—;/ 3 3y « 

(SO "1-^-2 = -4 -^-2, a7+y = 8. 

(10.) 2a;-2^3=5fr2 g.,_^~5_7?/-7 
5 2 ' *^ 3 2~' 

(ll.)T^(aj + ll)+Ky-4)=:a;-7,-Ka; + 5)-.K2/-7) = 3y-a:. 
(12.) a;-24 = |+16, Ka'+y)+x=3(2y-a;) + 105. 

(13.) K3^-72/)=K2a;+2/ + l), 8-K«'-2/) = 6. 
(14.) ^^±^=l + 2(2a;-6?/ + l) a;_,, 

(15.) a; 4- 2/ = a, a; -2/ = 6. 
(16.) aa; + a2/ = a2+6^ a;=a. 

(17.)^ + |=l,c.+y=c. 

157. When there are three simultaneous equations containing 
three unhnoivns, the solution is effected hy eliminating one of the 
unknoivns hetiveen the first and second equations, and also he- 
twecn the first and third, or second and third. Tu-o equations 
are thus obtained involving two unknoivns, ivhich may he found 
hy the methods already exi>lained. The value of the third un- 
knoivn may then he found hy substitution. 

Example, 

Soke 2.r-32/+ 2= 1 ... . (1), 

305-52/ + 4^= 3 . . . . (2), 

4x + 22/-32=13 .... (3). 
Multiplying (1) by 3 and (2) by 2, 

6a;- 92/ + 32=3 .... (4), 

6a;-102/ + 82=6 .... (5). 



>}' 



128 



SIMULTANEOUS EQUATIONS. 



if :■■ 



Subtracting (5) fr^^m (4), 

2/--5,i=— 3 .... (6). 

Thus a- is eliminated from (1) and (2). 
Again, multiplying (1) by 2, 

4a— •62/ + 22= 2 . . . . (7), 
and 4a; + 2y-32=13 .... (3). 

Subtracting (7) from (3), 

8?/-5^= 11 ... . (8). 

And y—5z=-'3 .... (6). 

From (6) and (8) we find y=2, 2=1. 

Substituting these values of y and z in (1), (2), or (3) 
we get a; =3. 

Thus the solution is x=3, y=% 2=1. 



n. 



EXEKCISE L. 

(1.) x + By-^2z=ll, 2x + y-i-3z=U, Sx -{-^y + z=ll. 
(2.) a; + 2^ + 37;=13, 2.b + 3?/ + 2=13, 3x-{-y + 2z=10. 
(3.) 2x-{-Sy-'4:Z=10, 3x-4:y + 2z=6, 4:X-9aj+Sz=21. 
(4.) lOx-2?/ +42=10, 3a; + 52/+32=20, a;+32/-2.2=21. 
(5.) 3^ + 2^=13, 3?/+22=8, 32+2x=9. 

(6.) ^-f 1+1=3, 4:x + 5y-{-6z=77,z+x=2y. 
4 5 D 

(7.) 3a;-2.y=6, 3y-22=5, 32-2a;=-2." 

(8.) |^(«-l)-2/=35, %-5z=43, x+y-{-z=60. 

(9.) 2/— 2 + 3=0, z;--iC=-5, a3 + 2/=6. 
(10.) 2,' + 2=a, z + x=h, x-\-y=c. 



( 129 ) 



CHAPTER XX. 



!), or (3) 



LI. 
10. 
=21. 
2=21. 



PROBLEMS. 

158. In the following problems the various unknowns are 
expressed m terms of separate and distinct symbols Td th^ 

ot equations. If two symbols x and y be employed thA 

tit and 1 *'' -"f ^^". "^^* '"^^ two'indepentn tiua! 
tions, and three mdependent equations will be required to 
determine three unknowns, x, y, and z, ^ *"* 

Examples^ 
(1.) A fraction becomes equal to 1 when 1 is added fn fli« 

rni^rir Lot* '''^" '^^-^^ *^^-^- 

Let - be the fraction. 

y 

Then by the conditions of the question, 



a+l_ 



y 

X 



1, 



the solution of which is x=5, y=.Q, 
Hence the fraction is |. 

^^«Z^^? ^^""^ numbers such that the sum of the first 
one-fifth the secondhand one-tenth the third, shall be equal 

K 



130 



PROBLEMS, 



li 'I 



H : 



to 4; the sum of one-half the first, the second, and one- 
tenth the third equal to 7 ; and the sum of one-half the first, 
one-fifth the second, and the third equal to I'i. 

Let aj= the first, y=. the second, and 2= the third number. 

Then by the conditions of the question, 

2 5 

Multiply these severally by 10, 

10a;+ %j-\- 2= 40 ... . (1), 

5x + 10y-f 2~ 70 . . . . (2), 

hx\ 2?/-|-10;^=120 .... (3), 

2x(2)-(l), 182/-h2=100 .... (4), 

(3)-~(2), -87/ + 92= 50 ... . (5>, 

9x(4)-(5), 170y/=850. 

.•.?/=5. 

Therefore from (4) 2=100-18?/=10 ; 

and from (1) a; =2. 

The required numbers are thus 2, 5, and 10. 



Exercise LI. 

(1.) One of the digits of a number is greater by 5 than the 
other. When the digits are inverted, the number becomes 
I of the original number. Find the digits. 

(2.) In a division the majority was 162, which was i\ of 
the whole number of votes ; how many voted on each side ? 

(3.) The sum of two digits is 9. Six times one of the 
numbers they form is equal to 5 times the other number, 
i'ind the digits. 



PROBLEMS. 



131 



md one- 
tho first, 

lumber. 



than tlio 
becomes 

as -i\ of 
side? 

of the 
Qumber. 



(4.) If the numei^ator and denominator of a fraction be 
each increased by 3, the fraction becomes 2 ; if each be in- 
creased by 11, it becomes f . Find the fraction. 

(5.) A number consists of two digits whose sum is 12, and 
Btich that, if the digits be reversed in order, the number pro- 
duced will be less by 36. Find the number. 

(6.) Three towns A, B, and C are at the angles of a 
triangle. From A to through B, the distance is 82 miles ; 
from B to A through C, is 97 miles; and from C to B 
through A, is 89 miles. Find the direct distances between 
the townSk 

(7.) The diameter of a five-franc piece is 37 milUmetres^ 
and of a two-franc piece is 27 milUmetres. Thirty pieces laid 
in contact in a straight line measure one metre exactly. How 
many of each kind are there ? 

(8.) At a contested election there are two members to bo 
returned and three candidates. A, B, C. A obtains 2112 
votes, B 1974, C 1866. Now 170 voted for B and C, 1500 for 
C and A, 316 for A and B. How many plumped for A, B, C, 
respectively ? 

(9.) A boat goes up stream 30 miles and down stream 44 
miles in 10 hour >. Again, it goes up stream 40 miles and 



down stream 55 .irui>k 
stream and boat. 



in 13 hours. Find the rates of the 



(10.) At a contested election there are two members to be 
returned, and three candidates, A, B, C. A obtains 1056 
votes, B 987, and C 933. Now 85 voted for B and C, 744 for 
B only, 98 for C only. How many voted for C and A, how 
many for A and B, how many for A only ? 

(11.) Seventeen gold coins, all of equal value, and as many 
silver coins, all of equal value, are placed in a row at random. 
A is to have one-half of the row, B the other half. A's share 
is found to include seven gold coins, and the value of it is 
£6. The value of B's share is £6 15s. Find the value of 
each gold and silver coin. 

k2 



it 



V 



»32 



PROBLEMS. 









(12. 



road 



and 



)m A to D passes through 
cessively. The distance between A and B is six miles greater 
than that between C and D, the distance between A and C 
is ^g of a mile short of being half as great again as that 
between B atd D, and the point half-way from A to D is 
between B and C half a mile from B. Determine the dis- 
tances between A and B, B and C, C and D. 

(13.) Fifteen octavos and twelve duodecimo volumes are 
arranged on a table, occupying the whole of it. After six of 
the octavos and four of the duodecimos are removed, only | 
of the table is occupied. How many duodecimos only, or 
octavos only, might be arranged similarly on the table ? 

(14.) Three thalers are worth \d. more than 11 francs. 
Five francs are worth \d. more than 2 fijrins. One thaler is 
worth %l. more than a franc and a florin together. Find the 
value of each coin in English money. 

(15.) Six Prussian poimds weigh \ oz. more than 5 Austrian 
pounds. Twenty-five Austrian pounds weigh \ oz. more than 
14 kilogrammes. One kilogramme weighs 1 oz, less than the 
sum of the weights of a Prussian and an Austrian pound. 
Find the number of ounces in each foreign measure of 
weight. 

(16.) A person walks from A to B, a distance of 9i^ miles, 
in 2 hours and 52 minutes, and returns in 2 hours and 44 
minutes, his rates of walking up hill, down hill, and on the 
level being 3, 3*, and 31- miles an hour, respectively. Find 
the length of level ground between A and B. 



( 133 ) 



CHAPTER XXI * 



EXPONENTIAL NOTATION. 



u 



159. Although the notation adopted in the preceding pages 
is sufficient for the purposes of the operations herein treated 
of, yet it is found expedient, before proceeding farther, to 
employ another notation to express roots, powers of roots, 
and their reciprocals. This notation, which consists in em- 
ploying fractional exponents instead of radical signs and 
integral exponents, and negative exponents instead of reci- 
procal forms, possesses the great advantage of reducing to a 
few uniform laws the operations of Multiplication, Division, 
Involution and Evolution, with respect to powers, roots, 
and powers of roots, of a quantity, and their reciprocals. 

160. The exponential notation consists in writing 



and 



a" instead of y^a'", 
1 



a-P 



Thus, according to this notation. 



a^=^a, 



a^=^a\ 



a^=ya», 



a 


1 




a 


■3 1 






-8 

a = 


1 


a 


■^ 1 


1 




a 


•f_ 


_1 


1 





* This Chapter may be omitted by those who do not intend to read 
more advanced works on Algebra, 



% 

« ! 



1 1 



:l 



jr,:i 



134 



EXPONENTIAL NOTATION, 



-3 2 

2a =^,' 









161. When the exponential notation is employed, the 
quantity is said to be raised to the -power indicated by tho 
exponent. 



Thus 



a^ is read a to the 'power ^ ; 



a 






it 



if 



i» 



a 



X 



X 



it 



n 



a 



a 


f; 


a 


-3; 


a 


-|. 



The term poiver in this extended Algebraical sense thus 
includes the terms power, root, root of a power, reciprocal of a 
power, reciprocal of a root, reciprocal of a root of a power, as 
used in tho ordinary or Arithmetical sense. 



U 






i:: 



EXEBOISE LII. 

Express in the Arithmetical notation— 

,_ ^ .1. A 2. 3. 
(1.) a^, a«, «», a2. 

(2.) x-^, x-\ x-'^'^. 
(3.) m 3, n '^, p *. 
(4.) 2a*, 3a:- 2, Gm'^. 
Express in the Exponential notation — 
(5.) ^x, ^m, ^n. 

X a'^ a'' a* 
(7.) ^'] ^^% V^, 

^^'^ ^x ^x j;/^' 



EXPONENTIAL NOTATION. 



135 



(9.) 



sd, the 
I by tho 



ise thus 
)coX of a 
oweif as 



2 3 10 

m rr p^ 



(10.) -!,. 4. ' 



^x ^x ^x^' 

162. The utility of employing the exponential notation will 
be exhibited in the statement of the three following rules, 
which are usually called Index Laws. 

I. The product of any two powers of the same quantity is a 
power whose exponent is tlie Algebraic sum of the exponents of the 
factors. 

Since the product of a quantity and its reciprocals 1, this 
rule cannot be applied when the exponents of the two factors 
are numerically equal and of opposite signs, unless tlie zero 
power of a quantity be considered=l. 

Examples. 
(1.) a'^ ' a3 = a2 3=^6. 

^8 8. 3 + 6. XS 

(2.) a* • a*^=a'^ ^=a^^, 
(3.) a^ .a-i=a3-i=a2. 
(4.) a-2 . a-3=a-2-3=ra-6. 

-^ ^ —A 1 — .' i 

(5.) a- a •^=a '■'=a'\ 

5. -3 5_8 JL 

(6.) a^ • a * = a« * = ai2. 
(7.) a- a-i=ai-i=a°=l. 

(8.) a^ ' a"2=ra" = l. 

This law may also be expressed briefly as follows :— 

where m and n are any quantities whatsoever, positive or ne^Of 
Uve^ integral or fractiotial, including zero ifsP=l. 






136 



EXPONENTIAL NOTATION. 



I' 



■'%''■ 



m' 



% 









m 



163. Proofs of the preceding rule will be exhibited in the 
following particular cases. 

(1.) If m and n be positive integers, 

a"* • a.^-:^aaa . . . . (w factors) xaaa . . . . (w. factors), 

= aaa .... (w+n factors), 
_ „♦»+» 

(2.) aa . a8=Va ^a^^~Q^ ^a2 = ^a'^=o« = a2^3. 

Here v^ = >v/^> because each of them when multiplied 
by itself six times produces a*. So also ^a = y~€?-\ and 
generally, if m, %, j9, are integers, 

because each of these quantities when multiplied by itself «p 
times produces a'"^. 

(3.) If m, n^p^qhe positive integers, then 



= ^a»»«+p», by Ex. 1, 



m p 



(4.)a».a-»~=^=a->=a'^-». 



a* a 



Exercise LIII. 

Find the products of 

(1.) 2a;, Sec"; x^, 405*"; dac", a;^™, 

(2.) a;2, 2a;*; 3a; s, 2x2; 6a;*, 5a;^. 

(3.) as", aj*; 2a;2, Sec"; a;2^ a;}f. 



EXPONENTIAL NOTATION. 



>37 



(4.) 2a», a- a ; a-\ Sa^; 5a, ^a'\ 

(5.) Q^, a" 3; 2a*, a"«; a, a'^. 

(6.) aa, a"3; a«, a~^; a^", a"^. 

(7.) a», a-3; a», a-; 2a, 3a-»; wa^, Ma-«. 

164. n. When one power of a quantity is divided hy another y 
the quotient is a power whose exponent is the Algebraic difference 
letween the exponents of the dividend and divisor. 

Since the quotient is=l when the dividend and divisor 
are the same, this rule cannot be applied in the case of equal 
powers unless the zero power of a quantity he considered=.l. 



a 



(1.) -3=a«-s=a8. 

2 

(3.) -3= a =a • 
a 

(5.) ''=|=a-^t=aA. 



Examples. 

(2.) «^=at-t=a*. 



ai 4-i A 



ai 



(4.) — r— a =a^ 



a 



a" 



(6.) -,=a»-fi=aO=l. 






This rule may also be expressed briefly as follows :— 



a 
a" 



TO 



-Qfn^n 



where m and n are any quantities ivhatsoever, positive or nega- 
tive, integral or fractional^ including zero ifaP=l. 

165. The proof of this rule rests on that of the preceding. 
For since a"*-". «"=«"', it follows that 

a"* 



il 



If ¥' 



1*1 



I3S 



EXPONENTIAL NOTATION. 



Also, since t^ =«"•*, the second rule may be considered to 
be included under the first. Thus 



a" 



W y^— M __ /»W1"»II 



ssa™. a "=«' 



.■T 



'iSf' 



Divide 



Exercise LIV. 



(1.) a^"* by a*"; a"" by a*. 



1. 
3 . 



2 

.3 



(2.) or by a^ ; a by o^ 

(3.) a;^ by a; ; ^ by a*. 
(4.) 01? by a;-^ ; x^ by a;'^. 
(6.) a;-^ by a;-2 ; a;-^ by as'". 

(6.) a;^ by a;'^; a;^ bya;-^ 

(7.) a;"" by aj-^"; a; 2 by a?"". 

166. m. The power of a power of a quantity is a power 
whose exponent is the product of the numbers expressing those 
powers. In other words, 

where m and n are any quantities whatsoever^ positive or nega- 
tive j integral or fractional, including zero ifaP=X. 



Examples. 
(1.) (a^y=a^; {a'y=--a'^. 

(2.) (a^)2=a«; (««> = «*. 
(3.) (a"^)a = a«; (a^y = a^. 

(4.) (a-l)2=a-2. (^-2)3^^-6^ 

(5.) (a-i)-2=a2; (a-3)-4=ai2. 
(60 ia^r^ = a~^; (a^y^z=a'K 



EXPONENTIAL NOTATION. 



139 



167. Proofs of the preceding rule will be exhibited in the 
following particular cases : — 

(1.) («")=^=a2 . a2 . a2=««=a3'<2. 

(2.) (a3)2=a'3 • a3=a^"'3=a« 

(3.) If n be a positive integer, 



(»"»)•»= a*" • a" 



— /ym+w-f 



=«•»". 



. to n factors, 

to n terms 



(4.) If ^ and i? be positive integers, 

(a")«=a«, 

because each of these quantities when raised to the ^th power 
produces a"^. 

EXEEOISE LV. 

Express the following as powers of a : — 

(1.) (aO^(«^y;W^ 

(a-i)2; (a-2)3. (^-3)4^ 
(a2)-3; (a-2)-3; (a-3)-*. 



(2.) 
(8.) 

(4.) 

(5.) 
(6.) 



(a3)2; (0^4)3 J (a"4)3. 



-tx-1 



(c.-^-^ ; (a-t)-^ ; (a-t)-!. 



168. The Index Laws (I., IL, III.) are thus seen to be true 
on the assumption that 



-1' 1 



for all values of m and w, positive or negative, integral or 
fractional, including zero. 

Instead, however, of treating the subject of the Index Laws 
and notation as in the preceding Articles, we may proceed as 
follows : — 

If m and n be positive integers, it may bo proved, as is done 
in Arts. 163, 165, 167, that 






i I 



n 



if 



h" ■ 



140 



EXPONENTIAL NOTATION. 



L 

II. 
III. 



a 



m 



^n—^m+n^ 



^ sa"'-", w being greater than n. 



a 



(a"')'»=a»'*. 



These laws, which are thus proved to hold in the particular 
case where m and n are positive integers, are then assumed to 
be true when m and w are any quantities whatsoever, posi- 
tive or negative, integral or fractional, including zero ; and 
from this extension of these laws we deduce that 



a" must=v a"*, a" 


'^=0^, and a°=l. 


Thus, by I., 




But 


a/ a. /s/a =a. 




/. a^=^a. 


By m., 


(afy=:a\ 


But 


(*/'^y=a\ 




.-. a^=Va\ 


ByL, 


a* . a^=a\ 


But 


a' X l=a'. 




/. a«=l. 


ByL, 


a3 . a-3=a»=l. 


But 


a^ 




• rt-3 — -^ 



a*" 



169. From the remarks of this chapter it will be thus seen 

that, although there is no absolute necessity for using such 

3. 4, —5. 
symbols as a*, a" , a 3, still their introduction gives rise to a 

uniformity in certain Algebraical processes ; and the number 

of rules which otherwise would be required to meet the 

different cases that arise in those operations thus becomes 

largely reduced. 



( 141 ) 



ANSWERS. 



1. 

(1.) 5r 10+15; 10+12+f ; 2+i+f ; a+^+|i+^. 

(2.)i 5; 2i-li; 2-5-1-6. 

(3.) 11 + 35 + 6-17; 81+76-69-4a 
(4) 15-7+8 + 9. 

(50 28-16 + 10-4. 

U. 



(1.) A's -60, +20; B's +60, -30; C's +30, -20. 
(2.) A's +20, +20, -30,-40; B's -20, +30, +30, 
C's -20, -30, +40, +40. 

(3.) A -10, +4; B -7, +10; -4, +7. 

(4.) -2°, +5°, -3°. 

(6.) +1°, -P, +P, -P, +P. 

(6.) -2°, +2°, -2^ +2°. 

(7.) +25°, -7°. 



m. 



-40; 



(1.) 30. 


(2.) 20. 


C3.) 22. 


(4.) 22. 


(5.) 24, 120, 0. 


(6.) 14. 


(7.) 7. 


(8.) 16. 


(9.) ^' 



El 



14^ 



ANSWERS. 



(iO.) h (11.) M. 

(13.) 2a^3a^2(i2 ^3^3 +4^4^ 






IV. 

(1.) 2, 3a, I, &c, 4a*-^6, k. 

(2.) -], +8, -f, +26, -5i»2. 

(3.) +1, -1, -8a, +|c(£. 

(4.) -1, +3a^ -faz. 

(6.) 1, -1, -% -3, +i -f. 

(8.) -a2, +|a^ +2a2a;, ^a^x. 

(9.) -Sa^^a;, +2a*£c; aos^, ^ax\ 



(5.)^>|,4, +f, 
(7.) 2£c, oj. 



'f* 



if 






(1.) +25. 
(4.) -21^. 
(7.) +^. 
(10.) +1L 

(13.) -H. 



(1.) V6a. 
(4.) -13a. 
(7.) -3a. 
(10.) -4x2. 
(13.) -bV*'- 



(1.) 2ct-3&. 

(4.) 3a + 2a;— 5!/. 

(6.) d'-^li'^h 

(8.) -«;+22/-2 + l. 



V. 




(2.) +43^. 


(8.) -28. 


(5.) +5. 


(6.) -6. 


(8.) -aV 


(9.) -0-7. 


(11.) +2. 


(12.) -12. 


(H.) +1/5. 


(15.) +0*342. 


VI. 




(2.) +15a. 


(3.) -5a. 


(5.) +2x=». 


(6.) +3a2. 


(8.) +10c. 


(9.) -5c. 


(11.) -*-7a&. 


(12.) +> 


(14.) -fa. 


(15.) -f|a. 


VII. 




I.) -a; + 82/. 


(3.) -f:!;z;-3?/-2'. 


(5.) 


4-a) +2^. 


(7.) 


a-2ft + 3c-(?. 


(9.) 


4a-26. 



(10.) a2-45c. 
(12.) -2a-&-c. 
(14.) 4a + 07 + 5. 
(16.) 2(1 + 26+ 2c. 
(18.) 4a;l 



X 



ANSWERS. 



M3 



(21.) |+i%2/+i^^. 



(11.) 4a-9&. 
(13.) 2.x-.6y-2!. 
(15.) 5a— 4x. 
(17.) lOx + Sy-.-, 
(19.) 4a' +46^ (20.) a + &. 

(22.) 2a+|5-|c. 



(1.) 4. 
(4.) -3. 
(7.) 10-6. 
(10.) -0-96G. 

(13.) lOa^ 

(16.) -W. 

(19.) 2«. 

(22.) — aa;— 5&2/+4:C2. 

(24.) -7ccH2az;2-5x. 

(26.) ia+i&--ic. 



(2.) 2. 
(5.) 6. 
(8.) -3-39* 
(11.) -a. 

(14.) 2:;. 

(17.) \\x\ 
(20.) 3a-cc+4. 



(3.) -9. 
(6.) ~3. 
(9.) -21G. 
(12.) -7a;. 

(15.) 2a-|. 

(180 3a + &. 
(21.) -5a& + 26=^, 



(23.) -6a + &+«-4. 
(25.) 1+14^. 



IX. 



(1.) 2*2+-l; 3a; + -42^+-5; 2a+-.36++4c. 



(2.) 2a- -5a. 
(4.) 2a+-8/>-+7. 
(6.) 5a + (6-4). 
(8.) a-4 + (26-r). 



(3.) ~6-+5x. 

(5.) 5++a: 3a. 

(7.) -a +(-6 + 5). 
(9.) a;2 + (2i/ + 5)-;^. 



(10.) a-l + (36 + 5)--3c-. 

(11.) ic + (2x''-l) + (-3x^-8). (12.) 4a2-(62-c). 

(13.) «2 + 4-(-26 + 3). (14.) 2a-5-(a2-2a-+ 3). 

(15.) a + 6 + c + (a-6-c)— (-a + 26-3c). 



■I 



144 



ANSIVERS, 



J*' 



(1.) 2a+36-(;. 
(3.) x'^^x-l. 
(5.) 8a— ft— c. 
(7.) 8a+26-3c 
(9.) a;+ll+4y. 



(10.) h. 



XI. 



(2.) o&— 6c+c. 
(4.) 5a;»-3a;2+7xB-8. 
(6.) 8a-6+c. 
(8.) 2a-6 + 6+c. 
(11.) 2aiH3a;. 



<■ 1 

1 


(1.) 


(2.) 


|i 


(3.) 


11 


(4.) 



(1.) (a-l)(2a2-3); (-2+a)(-3-a2); (a5-5)(-2x+7): 

(2.) -2a2(62«i); (a2-l)(_3a); 5a;(~a;H3). 

(3.) K^'-l); K2^-3); -1(0)^-5). 

(4.) -5<a;-l)(a; + 2); (a;«-4)( + 5iB)(2a+3). 

(50 +8<-52/)(a:y-l); -7a(a6-3)(+86). 



xn. 



— 6a&; — 5ac; +6a*6; — SOajy. 
— 14a&c''; — 20a'*6c; — 16a3yz; +48a. 
-fa^?^; -fa&; +iajy; -fa^J. 
— 6a5^«/^; +3aa"''«/''; — 4a'fe^c*. 



w 



a^b* 



(5.) -g ; 15 ; ~" 



"20"' ''""12^ 



xm. 

(1.) -8aa; + 66aj-2cx; 1203=2/ -8a;y+43^; -2a6c?-|-3c(f. 
(2.) 3x»-6»2-15i»; -2a'''+3a*-7a^ -4aa;H4aV-8a»a;. 
(3.) 4a;V + 2^y'^^-6ic^2^ -28a362+ 4^253 _ 4^2 j2^ 

(4.) 2fi26 + a?>2-fci&c; -a3 + fa2-2a; iaa;^- ia*^;* +|a'a;. 
(5.) -12a + 10a6-15; ^-^ax^^\^ax-\-^%\ 
(6.) 5x'*-5a; + 20; -2a+2a6-6; -a'+2a'»aj-a. 
(7.) ax -a;*; ac -&c + c*; -Saft + Sa^i^-lSaSj, 
(8.) 4a3-6a2+8a; +2x'-§a;H3a;». 



ANSWERS. 



»4S 



XIV. 
(1.) 2a;2+6a;-12; -^x^^x+h; 2-!c-.3a;2. 
(2.) 2a;«-a;2_4^^2; 2-2x+3a:2-.3a)«; 3+3x2-a;3-.a;« 
(3.) 6a3-7a2+14a-8; l+a^; i_^3^ 
(4.) am-an ^bm-hn; am +hm^cm + 2an+ 2bn^2cn' 

(5.) a;y-x*; a;3+a,3y-a;2^2_2a;y-2a;2/H2y3. 

(6.) 2aHa3-22a2+23a-4; a2-4&H126c-9c2. 

(7.) 3a|-4^&+8ac-46H8&c-3c2; 3a.*-4a;3y+6xV+4a;y» 

(8.) a)*-fa.Hi; a^^a-i; 2a='-i 

(9.) 2^3-|x2+|ia.-i; 9a.«-|a;2^.|^_^^ 

(11.) ia^-^H|a-l. (12.) a;«+3a:y-2a:+y3-2y+l. 
(13.) a»-3a6c + 63+c3. 

XV. 
(1.) +30a«J»; -J^a;7. ^i3^y,^',^ 

(2.) -24a;V+30a;V; -2a'>6+3a*62-a85. 

(3.) 8x3-26x«-17a.+6. (4.) a.*.a:3+^_i, 

(6.) a;8-2aV+a8. 

XVI. 

(1.) (2a-6)-r-.3a; (4a2..,3a + l)-f.(3a-4). 
(2.) 2a-f— 3&; -a^-j- + 2x'; 3a;-r-2a. 
(3.) 4a;2-j-(2c-5); -aaj2_j>(a;-a). 



xvn. 

(1) -4; -6; +f ; -V. (2.) ^±. -?^' 

. "^ 2a;' 26 



+ 



2«y 



a 



146 



ANSWERS. 






(3-) -|; 



6«a5. 



W 



56' 
(5.) -2a; 6a6V. 



14a2- 



(4) 2a'; a»; 4a^. 
(6.) -ia;?/'; +fay. 



xvm. 

(1.) -2a+36-4; aa;--3+2a2. 

(2.) -4a;+3-a; 4a;2_ar + 3. 

(3.) -aH4a-5; a;2-3a;i/+4/. 

(4) 2a-36+c. (5.) -4ac2+3&c'-l. 

(6.) (a— 6)0?, (2a— c+l)a;. (7.) (4— a)a;s', (3a3— y)a;y. 

XIX. 

(1.) a!-4; 3x+l. (2.) «+!; 2a;-3. 
(3.) 3a5+2; 3a:2-2(»+6. 

(4.) a;-l; oj^-jb+I. (5.) x^-Zx->^\, 
(6.) a;Ha;Ha;+l; (c^-ac'+cc^-aj + l. 

(7.) a;-3/; x^-xy-\-y\ (8.) aHaJ+ft^. 

(9.) 5a; + 6y-3. (10.) aj^-aaj+a''. 

(11.) a;2-2i»2/+2/'. (12.) faa;-2a;2. 

XX. 

(1.) ar2-3a5+ V , - f . (2.) x-a, '2a\ 

(3.) !r2_aa;+aa, -2a3. (4.) a-2-a;-f4, -ac-4. 

(5.) 2a;2+3, -5a;2-3a;-3. 





XXI. 


(1.) 3a3— amiles. 


(2.) 50 +x dollars. 


(3.) 26 miles. 


(4.) x^—y^ square feet 


(5.) - hours. 
a 


(6.) .+|+3^ 


<7-) -+^+4- 


/Q \ lOoj 

(8.) ^ . 







ANSWERS, 


1 

147 




(9.) |. 


(!«•) 1^00- 


100* 




(11.) ^+f -5. 


(12.) 3& acres. 




(13.) ^^ miles. 

V 


(14.) 1^ hours. 




(15.) 6aa;+ a dollars. 








XXII. 




L. 


(1.) 2. 


(2.) 10. 


(3.) 7. 


•y)xy. 


(4.) -5. 


(5.) 2. 


(6.) 5. 




(7.) "h 


(8.) -5. 


(9.) f . 1 




(10.) 2. 


(11.) 3. 


(12.) 2. j 




(13.) 7. 


(14.) 10. 


(15.) 2f 1 




(16.) i. 


(17.) 6. 


(18.) 61 1 




(19.) 7. 


(20.) 13. 


(21.) «. f 




(22.) 3a. 


(23.) a +6. 


(24.) %. 1 


I 


(25.) 6+c. 


(26.) 1. 


(27.) a2+a& + &^ 1 


1 




(2«.) '^*-^'. 


•1 


I 




a— c 


i; 
1 


1 




XXIII. 


s 


1 


(1.) 12. 


(2.) 12. 


(3.) 24. 1 
(6.)6A. ! 


1 


(4.) 30. 


(5.) 23i. 


^^^1 


(7.) 6. 


(8.) 3. 


(9.) 3. 

" i 


I 


(10.) 9. 


(11.) 4. 


(12.) 5. ' 


I 


(13.) 3i. 


(14.) -2. 


(15.) 33. 


I 


(16.) 2i. 


(17.) 2. 


(18.) 2. 


1 


(19.) 120. 


(20.) 13. 




1 




XXIV. 


{ 


1 


(1.) 15 and 10. 


(2.) 60 and 75. 


(3.) 20 and 17. 


1 


(4.) 14 lbs. 


(5.) 23, 17. 


(6.) 181 and 145. 
l2 



148 



ANSWERS, 



(7.) 5. (8.) 14 years. (9.) In 9 years. 

(10.) 18. (11.) 66 years. (12.) 400. 

(13.) 700. (14.) 30 for translation, 

5 for mathematics. 
4 for Latin prose. 

(15.) 21 shillings. (16.) A's £800, B's £100. 

(17.) 400 inches. (18.) 18, 11 and 8. 

(19.) 35. (20.) 200 quarters. (21.) 12 lbs. 

(22.) 150 lbs. (23.) 240 sovereigns, (24.) 13. 

480 shillings, 
720 pence. 
(25.) £3000 at 5 per cent., (26.) £450 at 4i per cent. 
£10,000 at 4 per cent. £350 at 5i per cent. 

(27.) 1800 infantry, (28.) 17 years. 

600 artillery, 

200 cavalry. (29.) 26. (30.) 12. 

(31.) 56 workmen; 150 shillings. (32.) 1330. 

(33.) 4290 feet. (34.) 30,000 men. 

(35.) 200 miles from Edinburgh. (36.) In 56 hours. 

XXV. 

(1.) a;2-2i» + l, a;2 + 2aa; + a^ a;2-10a;+25, a;H6a; + 9. 

(2.) 4a;H4a5 + l, 9a;2-.6a; + l, 4a;H12a; + 9, 9a;2-12a!H-4. 

(3.) £c*--2aa;2 + a2^ 4xV + 4a;2/ + l, ^x^ -VHax^ ^-^a^, a'x'' 
-8a6x2 + 1662. 

(4.) a;2 + 2/H2^—2iC2/ + 2x2-2^2, ix^ + dy^ + z^ + 12x1/ -4:xz 
-62/z, x^+4ty^+2bz^'-ixy^lOxz-\-20yz, 4a;- + 16y + l-16a7/ 
+4a3— 8y. 

(5.) 4a* + 4a3 + ISa^ + 6a + 9, 9a* - 24a» + 22a^- 8a + 1, 
a*-.4a3-4aH16a + 16. 



(6.) 2401, 9604, 990025. 



XXVI. 



(1.) ar^-l; a2-9;f4-a;2. 

(2.) 4ar»-l; 25a2-4; lex^-a^ 



ANSWERS. 



149 



years. 



) 12 lbs. 
) 13. 



)er cent. 
»er cent. 



.) 12. 
.) 1330. 
) men. 
hours. 



9. 
H-4. 

\xy—4:xz 
1— 16ay 



[• 



(8.) a*-a^; a«-l; a»-a;*. 

(4.) 9a*-462 ; 16a«-4a;< ; 49a8-25a«. 

(5.) 2496 ; 9975 ; 489975. 

xxvn. 

(1.) mP-{-n^; p^—g^. 
(2.) m^+l; 1-2'. 
(3.) a;»+27; a^-QL 
(4.) 8as + l; 64a;»-a3. 

(5.) 8a»+276^; 27x3-1252^^ 
(6.) x^-l; x^^-a^ 

xxvni. 

(1.) 03^—03 + 1; aj*— 03^+03'^— aj+1. 

(2.) x^+x+1; x^+x^+x'^+x+1, 

(3.) x-1; a;3-a!2+a;-l. 

(4.) x + 1; x^+x^+x + l. 

(5.) 2a-3&. 

(6.) 3a;3+2a. 

(7.) W-x\ 

(8.) 4a2— 6a5 + 9&2; a*-2a2+4. 

(9.) 9a2 + 3ab + &2 ; 4,^4 ^ 5^2 j ^ 95a, 

XXIX. 
(1.) (-a)3, (2x)3, (ari/'^)^, {2a*hhy. 
(2.) (2a-l)^ («-& + in (x3-l)2. 
(3.) { (xy ] \ { (-2ay } \ { {^axf \ \ { {^aVf ] 

(4.) { (a-hf ] ^ { (x'-iy ] ^ { {x-'-3x^2f ] \ 

(5.) (a^^)^ U-2x)M^ {{a'^hfW [(x-afW 
{{x'-ax^Vf]K 
(6.) {(-a)M^ {{x'^fW { (403.1)2 }», {(a^-a»)M 



A3 I 2 



XXX. 



(1.) jbS 8< a;», 8ic», 81a;«. 
(2.) aV, a^r^ oVy'. 



\ 



150 



ANSWERS. 



^. 






(3.) aWc\ a«6V, Sa'fe'c". 
(4.) x^Y^z^y a'&'*c". 

(5.) «H2a; + l, 4a;2-12a; + 9, a!*-10a;2+25, a;«-4aV+4a*. 
(6.) a;*+4a;H10a;2 + 12a; + 9, a;*-6a!'+17a;2-24a; + 16, 
4a;«~4a;»+a;*+20cc3-10xH25. 

XXXI. 
(1.) V^, Ja^\ V^^^. Va:2-3a;+4. 

(2.) ^":::^«, ^"3^3, ^^^ir^, ^(a3-3a+4). 

(3.) V^^x, V^^^-l, Vl/'^^l . 

(4.) \/^2, V4^3^^1, A//y2(?^^^^iH3. ^ 
(5.) 4^"^, ^'^, y W, !p^ . 

xxxn. 

(1.) 2a'&, 5a;2/', 9a; V. (2.) 4aj + 5. 

(3.) 6a;-3. (4.) l+3a^. (5.) x^h 

(6.) a;-|. (7.) 2a;-J^. (8.) 2a;-32/. 

(9.) a;2+2a;+l. (10.) jcHaj + l. 

(11.) aj2-2a;i/+2/''. (12.) 2a;»-aj2-3a;+2. 



(13.) 1+|-|'; remainder l"-!^, 



(1.) ^l\ 
(5.) 7aV2/. 
(9.) 4a6. 



XXXIII. 

(2.) 5a&. (3.) ^xy. 

(6.) aftwv. (7.) 4a26. 

(10.) baW. (11.) 26^ 



(4.) 3aaj. 
(8.) 6xV. 
(12.) 4ww;. 



(1.) 05 + 3. 

(4.) x+3. 



XXXIV. 



X- 



(2.) 

(5.) 2ir-5. 



(3.) a;-3. 
(6.) x-3. 



ANSWERS. 



iS« 



(7.) (B2+10a;+25. 
(10.) 05-2. 
(13.) 2a3+5. 
(16.) aj-3y. 
(19.) a;+3y. 



(1.) a!r(2a;+8). 
(4.) a;-l. 



(1.) a(aj— a). 
(4.) aj+l. 



(8.) a;2-5a;+6,. 
(11.) a;-l 
(14) a;-2. 
(17.) a; +2/. 
(20.) £c2-/. 

XXXV. 

(2.) 2(i»+3). 
(5.) 3x-2. 

XXXVI. 

(2.) 2a(aj-8). 
(5.) a; +2. 

XXXVII. 

(2.) I^^xhj. 

(5.) 240a2&2c2«fcVM;2 



(9.) cc2-9. 
(12.) a;2_3. 
(15.) a;2-2a; + l. 
(18.) a;2+y. 



(3.) x(a;-2). 



(3.) a(a;-l). 
(6.) cc-a 



(3.) a^W. 



(1.) 6a&scy. 

(4) 24a26V. 

(6.) (a;2-.7 + 12)(a; + 2) = (a;2-a;-6)(a;-4). 

(7.) (2a;2-5a;-3) (2iK + 1) = (4*2 +4a; + 1) (a;-3). 

(8.) (3a;2-lla;+6)(2x-l) = (2x'-^-7x+3)(3x-2). 

(9.) (x^ - 4aa;2 + ^o?x - 2a») (x" f 2aa; + 2a2) = {x' - 2a2a; 
.4a3)(aj2-2aa; + a2)- (10.) 24(a? + l)(a;-l)2. 

(11.) (!r-l)n«^+l)'. (12.) xhfix'-y^), 

(13.) a3&(«'-&')- (1^-) 12(a;=^-l)(a'Ha;+l). 

(15.) ip^i-q'')(p'-q'Kp'-pq+f)' 
(16.) (i>*-l)(i>*+pHl). (17.) (b^c)(c^a)(a~b), 
(18.) 24a2&2(a2-62). 



XXXVIII. 



a 



<^-) fc 



(2.) - 
cy 



(4.) 



2a& 



a 



:::&■ 



(^•) ^f^ 



362-6a« 



(3.) 
(6.) 



[JB 



3_ 

4^2' 



x::! 



152 



ANSWERS, 



bvw 



I 



(7.) 



5+2* 



(10.) J,^. 



(13.) 
(16.) 



3+aj 
3-a;' 
3(4a;~ l) 
2(3a;!*+l) 



(8.) 
(11.) 
(14.) 






(9.) _^±y,. 



(12.) 
(15.) 



(17 ^ ^'(^+2y) ris^ 



g! + 3 
05 — 5* 

a;-3 

cc+y' 



a>. 6 2a 



XXXIX. 



(3.)?^^, 3?., 4 



(5.) ::?- , 



y 



a;?/^ xyz xyz 
-2a 3 



aajy axy axy 



,«x a(a; + l ) 2a ^^^ a; + 3 2( g; + l) 

^''•' a;2-l ' x^-r ^''^ (a; + l)(a; + 3) ' (x + 1) (x + 3) ' 






(9.) 



a 



(10.) 
(11.) 



X 



JC — 1 * 03 — 1* 

4(a;'^-l) 3a;(g;-l) 

gi'-l g;'^-a; + l 3a; 



g-l 

a—b * a—b' 



nS.^ 2(«^~2) 3(a;+2) 

^ ^ (a;-l)(a;-2)(x+2) ' (a;-l)(£c-2)(a;+2) * 



(13.) 



a;-l 
(a;-l)(a;-2)(cc + 2)* 



a—c 



(fc«c)(c-a)(a-fe) ' (b-rc)ic-d)(a-b) ' 

6— a 
(6— c) (c— a) (a— &) * 



.,.^ bx(x—b) ax(a—x) 

> ^^ abx(a-b)lx-d)(x~b) ' a&a;(a-&)(a;-a)(c»-6) ' 

(q— &)(ag~a)(ag— &) 

qhy(a-^h) (r —a)(.r~ h) 



ANSWERS, 



«S3 



(1.) 
(4.) 
(6.) 

(8.) 
(11.) 

(14.) 
(17.) 
(19.) 
(21.) 



'+6» 



ah 
3a;~-2 

a;2~3a;+6 

2x^ ' 

2a'' +26' 
a»-5a * 

a+h 
a 



V-1) 

2a; + 2y 



(2.) 



XL. 

aa;+2 
2aa • 



(3.) 



4a2a; + 6a+6 



12aa 



(5.) 
(7.) 



8x«+12ag;- 1 
66-17a 



60 



(9.) 
(12.) 



(15.) 



4:db 



~2 



a;(4a;2-l) 



(10.) 
(13.) 



1- 



a 



2a;» 



a;* + a;2 + l" 



a5-2y 



(16.) 



2x + 6 
a;*-l' 



x^-)rxy-^y^ 



X 



-9 



(aj-l)(a;-2)(aJ+2) 



(18.) 
(20.) 



2a;(2a;Hl) 



£»' 



«-.l 



a^ 



(«—•&)(«— c) 



(a+6+c)aj 



(a; — a) (a; — 6) (as — c) 



(22.) 0, 



(23.) 0. 



a.) 
(4.) 

(6.) 

(8.) 

(10.) 

(12.) 



3+5cc 
a+aic— 05 

2xHg;~l 
^^ • 

^-«2+2a;-3 

^^ 
g— •a'-f a& 
a— 6 

4a;g+9a;-4 

(c + S • 



(2.) 



XLI. 



cc— a'^ 



a' 



(5.) 
(7.) 



(3.) 

3a— aa;+a3 
as 

2x2 + 1 



2a— OS 



a 



OS 



2 • 



^n\ l + 3a? 

a+36 
a + 6 ' 

x^—x y 



(11.) 
(13.) 



«S4 



ANSWERS. 






*■ 



S;i: 



(14.) S+|. 
a 

0.7.) 2aj-l+?. 

a; 

3a;+5 



(19.) 14 



a;2-l' 
(21.) 4-j-^^. 

(23.) 8- ^"^ 



aaj^-oj+l' 



(1.) ^. 
(4.) ^y 

(7.) 
(9.) 






(a+6)2 • 
1 

(11.) 2/»+2aj+ajy. 
(13.) ^. 



(17.) 



oj+y' 



(4.) «:::^. 



(15.) a-±. 



a; 



(16.) ^-1. 



(18.) ^-1+2.. 
(20.) 5--i-^. 
(22.) .-3+1=7, 



(24.) a;2- 



XLH. 

(2.) 1. 

a+b 



x^+x—5 
x^+l ' 






(5.) 



6 • 



(6.) 



x^yz" 

^±abVb^ 
a^-ab + b^' 



(10.) ,4^Y2- 
(a;'*— 05 + 1/ 

(12.) aj-a. 
(14.) -^ . 

(16.) 1. 
2 



(a~6) 



(7.)- 



a+a? 



(2.) 
(5.) 
(8.) 



(18.) 

XLUI. 

4ax^y 

a^+ab + b^ 
a^^ab + b^' 

2 
(a^br 



r 



(3.) 
(6.) 
(9.) 



x+r 

a 
2ap ' 



ANSH^EKS. 



m 



(10.) 
(12.) 



ax 



(11.) ^^rf. 



(1.) 10. 

(4.) 6. 

(7.) -^ 
(10.) 20. 

(13.)f|. 
(16.) f . 

(19.) a+&. 
a+6 ' 



(22.) 



(13.) M. 

XLIV. 

(2.) 8. 
(5.) 8. 
(8.) -1. 

(11.) V. 
(14.) -i 

(17.) i^. 
(20.) a. 

XLV. 
(2.) 40. 



(14.) 1^. 



(3.) 12. 
(6.) 2. 

(9.) -f. 
(12.) 15. 

(15.) 3. 

a+6 * 



(18.) 2. 



(^^•) 4- 



(1.) 40. (2.) 40. (3.) 355. 

(4.) 4 bowled, 3 run out. (5.) ^■^. (6.) £5. 

(7.) £9150 of 3 per cents., £5820 of 3i per cents. 

(8.) A £1250, B £1500. (9.) 100. 

(10.) 7 miles an hour. (11.) 45 and 30 miles an hour. 

(12.) C 42 days, B 84 days, A 168 days. 
(13.) 240, 180, 144 days. (14.) 5^^ minutes past 7. 

(15.) 26^ minutes past 2. (16.) 49^^ minutes past 3. 



(17.) 32^^ minutes past 3. 
(19.) 30 gallons. 
(21.) 70 grains. 
(23.) 3 miles an hour. 
(25.) 10 yards in 200. 
(27.) Imile; half-an-hour. 
(29.) ^6 of a mile behind. 



(18.) 10 gallons. 
(20.) 112 oz. 
(22.) 21 oz. 
(24.) 4i miles. 
(26.) 1430 yards. 
(28.) 5f . 



156 



ANSWERS. 



mk 









a.) ^ -6. 

(4.) 4, -4. 
(7.) 0, 3. 
(10.) 0, f. 
(13.) 3, 5. 
(16.) i, f. 

(19.) ^, -1 



(1.) 2, 4. 
(4) 2, h. 

(J.) h -f. 
(10.) I, f . 
(13.) 2, «^. 
(16.) -1, -I-. 
(19.) 1, 2. 
(21.) 6, f . 



XLVL 
(2.) 3, -3. 
(5.) 2, -2. 
(8.) 0, -12. 
(11.) 0, -^\. 
(14.) -5,7. 
(17.) h -h 
(20.) 0, a, 

XLVn. 

(2.) 5, -1. 
(5.) h -2. 
(8.) i |. 
(11.) 2, i. 
(14.) 14, -10. 
(17.) 4, 4. 

(20.) 3, -|. 

(22.) a, 6. 



(3.) 9, -9. 
(6.) V26, -V26. 
(9.) 0, 19. 
(12.) 0, -|. 
(15.) -1,-3. 
(18.) -f, -h 



(3.) 3, -7. 
(6.) h h 
(9.) 6, -4. 

(12.) V, -10. 

(15.) 4, -1. 

(18.) 1, V. 



XLVIII, 

(1.) 60 ft. by 30 ft. (2.) 13 and 7. 

(3.) 2 or 10. (4.) 12 and 3. 

(5.) 4 or 6. (6.) 5, 4, 3 ; or -1, 0, 1. 

(7.) 30 ft. (8.) £20. (9.) £90 or £10. 

(10.) A's rate 4 miles ; B's 2^ nail^s an hour. 
(11.) |. (12.) 10. (13.) 87. 

(14.) 2 ft. (15.) 11. (16.) 23. 

(17.) 12. (18.) 56. (19.) 1296. 



ANSWERS. 



157 



(1.) 2, 3. 
(4) 6, 12. 
(7.) 4, -1. 
(10.) 2, 3. 

(13.) 13, 3. 
(16.) a, f . 



XLIX. 

(2.) 3, 5. 
(5.) 30, 12. 
(8.) 3, 5. 
(11.) 9, 4. 

(11) 4, 1. 

aib--aG 



(3.) 6, 4. 

(6.) 4, 3. 

(9.) 2, 3. 

(12.) 60, 40. 

nx!^\ a+& a—h 
^^^'^ "2"' "2"* 



(17.) 



ah— he 
b—a * a—h 



L. 

(1.) 2, 1, 3. (2.) 1, 3, 2. 

(4.) 3, 4, -3. (5.) 3, 2, 1. 

(7.) 4, 3, 2. (8.) 19, 7, 4. 

nn^ ^+c— a c+a— & a+6--c 
y.^^-) — o — * — o — » o • 



(3.) 5, 4, 3. 
(6.) 4, 5, 6. 
(9.) 2, 4, 7. 



LI. 

(1.) 72. (2.) 378 and 216. 

(3.) 5 and 4. (4.) V (5) 84. 

(6.) A to B 37 miles, B to C 45 miles, C to A 52 miles. 

(7.) 19 five-franc pieces, 11 two-franc pieces. 

(8.) 296 for A, 1488 for B, 198 for C. 

(9.) The stream 3 miles an hour; the boat 8 miles an 
hour. 

(10.) 148 for A, 750 for C and A, 158 for A and B. 

(11.) The gold coins are half-sovereigns, the silver coins 
are crowns. 

(12.) A to B Hi miles, B to C 7 miles, C to D 5| miles. 

(13.) 24 octavos or 32 duodecimos. 

(14.) A thaler = 2s. lid. ; a franc = W. ; a florin = 
Is. lUd. 



158 



ANSWERS. 



ll 



,«.r 



m 



(15.) A Prussian pound = 16^ oz. ; an Austrian pound = 
191 oz. ; a kilogramme = 35i oz. 



(16.) 3i miles. 



(1.: '/a, «/a, ^a\ ^a^ 



(3.) 



Ml. 



(2.) 



L L L. 



^m V»i° tIP' 



(4.) 2^ a 



- 3 



cc- 



./wi^ 



1. 1. 1 



(5.) x^, m^, n 



(7.) 



a. 3L i 
a;2 ass gjt 



(6.) 
(8.) 



x~\ a~\ a'\ a 



-8 



-1 -i. 



X 



-3 



(9.) 2m-i, 3»-2, lOp 



(1.) 6a;"+*; 4a;'»+2; 4a;*». 



a; 
-1 



_a 



(10.) 2«"2, 5a;"3, 7a;~» 



Lin. 



(2.) 2.x* ; 6x6 . 30a; 



:l 



2n+l 



3n 



fin 



(3.) jc 2 ; 6a;=^ ; a;« 

(5.) a6; 2a"'^; o^. 
(7.) 1; 1; 6; wiw. 



(4) 2a; Ba'^; 30a 



-1 



(6.) a^; a^; a\ 



(1.) a"; a2«. 
(4.) a? ; (B^ 

Sn 

(7.) as"; a;2. 



12. «8 



a': a 



(1.) a 

(3.) a . . 

(5.) a; a^; a*. 



LIV. 

(2.) a»; ai 
(6.) a?; aj^. 

LV. 
(!2.) a-= 



(3.) x^; X. 
(6.) (c; (B*. 



a 



-6. „-12 



-«»• a«: ai2. 



a=« ; a"; a 



9. „io 



(4.) 

(6.) a3 ; as ; a^. 



id =