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=b. In like manner, the result being the same, 6 whether the numerator be multiplied by a whole quantity, or the denominator divided by it, the latter method is to be preferred, when the denominator is some multiple of the multiplier : Thus, let ad

ad be the fraction, and c the multiplier; then

bc

Х adc ad ud ad ad and XC=

as before. b be bec

b? Also, when the numerator of one of the fractions to be multiplied, and the denominator of the other, can be divided by some quantity which is common to each of them, the quotients may be used instead

atb of the fractions themselves; thus, Х

a-b^atb a=7; cancelling a+b in the numerator of the one, and denominator of the other.

40 Es. 1. Multiply by 7 3a X 4a=120°= numerator,

... the fraciion re5 X7=35= denominator;

12a? quired is

35

3.0 +2 Ex. 2. Muliply by

4 Here, (3x+2) X 8x=24x2 +16x=numerator,

and 1x7=28= denominator;

24x2 +-16x Therefore,

(dividing the numera28

6x2 +420 tor and denominator by 1) the product re

7 quired.

82

2 -X

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a.

7x2 Ex. 3. Multiply by

За Here, (a--?)X 7x2 =(a+x) x (a-x) x7.co numerator (Art. 106), and 3a x(a-x)= denominator; see Ex. 15, (Art. 79).

(a + x) x(a--) 722 Hence, the product is

3a X (-2)
(dividing the numerator and denominator by amx)
72% (a +*)_7ax2 +723

3a
Ex. 4. Multiply a+ğ by a

3
5a +

3a - 2 , =

and 5 5

3 3 Then, (5a + x) X(30—2)=15a® – 2ax— 22 = new numerator, and 5 X3=15= denominator : There

15a2 - 2ax -22 2ax +32 fore,

is the product

15 required.

3a

8 2018

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a

=a2

15

157. But, when mixed quantities are to be mul-
tiplied together, it is sometimes more convenient
to proceed, as in the multiplication of integral
quantities, without reducing them to improper frac-
tions.
* Ex. 5. Multiply - 2+3 by 1.+2.

x2 - 1 x + 2
*+2

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3x2 -5%

7a Ex. 6. Multiply

by
14
3x3 - 3x

Sax - 5a
Ans.

402-6
3х2 15.1-30
Ex. 7. Multiply by

5x --10

2.0

9x Ans.

2 2a - 2x

3ax Ex. 8. Multiply

3ab 5а — 5x

by

20

5

ax

Ans.

56 Es. 9.' It is required to find the continual pro« 3a

2002 atb duct of

and 3

2ax + 2ab Ans.

5 Ex. 10. It is required to find the continued pro

at - 4 aty duct of

and a–ya' a2 +2?

Ans. atc. Ex. 11. It is required to find the continued pro

02 – 2 a262 duct of

and
ata
ax—

aa--ab
Ans.

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a

a+b;

Ex. 12. Multiply 2-x+1 by x2-x.

Ans. A-03 +42-X.

1

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To divide one fractional quantity by another:

RULE.

158. Multiply the dividend by the reciprocal of the divisor, or which is the same, invert the divisor, and proceed, in every respect, as in multiplication of algebraic fractions; and the product thus found will be the quotient required.

When a fraction is to be divided by an integral quantity; the process is the reverse of that in multiplication ; or, which is the same, multiply the denominator by the integral, (Art. 120), or divide the numerator by it. The latter mode is to be preferred, when the numerator is a multiple of the divisor.

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by The divisor inverted, becomes , hence

6

50

X

a 6 5cx

is the fraction required. abe

3a-3x 501-5% Ex. 2. Divide

by a+b

atb

50-50 The divisor

inverted, becomes

atb a+b 30-32

За-3

ath
;
hence

х ja-5X

atob 50-50 5Q ----50 3(2-x)_3

is the quotient required. 5(0-2) 5

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a2-62 Ex. 3. Divide by a+b.

X

a+b;

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is the quotient

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a

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1 The reciprocal of the divisor is

;

hence Q2-62

1 x

_(a+b)(ab)_a

atb * +(a+b) required. a2 --62

6 Or =a-b; hence is the fraction

a+6 required.

x2 +a?

22 Ex. 4. Divide

by art ato

x02 -a? atxo-a? Here, at

; then, the -a2 fraction

22 -a2 becomes

х ato

ato ax2

-a3

= the quotient required. axa tcx

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a

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a

a

a

x2

divided by

a

a.

2

159. But it is, however, frequently more simple in practice to divide mixed quantities by one another, without reducing them to improper fractions, as in division of integral quantities, especially when the division would terminate. Ex. 5. Divider --*3+x?- by ?-X. 22 - x)*4 -23 +x2 x(x2 --*: +1

24 - 23

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