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b

b. In like manner, the result being the same, 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

be the fraction, and c the multiplier; then

ad

bcx

bc

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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 of the fractions themselves; thus,

a+b

X

x

a-bˆa+b

=

XC ; cancelling a+b in the numerator of the one,

a

-b

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Here, (3x+2) X8x=24x2+16x numerator, and 4X7 28 denominator;

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Here, (a2x2)×7x2=(a+x) × (a−x)×7x2 = numerator (Art. 106), and 3a × (a-x)= denominator; see Ex. 15, (Art. 79).

Hence, the product is

(a+x)x(α-x)×7x2

3α X(a-x)

(dividing the numerator and denominator by a-x) 7x2(a+x) 7ax2+7x3

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Then, (5a+x)× (3α—x)=15a2-2ax-x2= new

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157. But, when mixed quantities are to be multiplied together, it is sometimes more convenient to proceed, as in the multiplication of integral quantities, without reducing them to improper fractions.

Ex. 5. Multiply x2 −1x+? by }x+2.

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Ex. 9. It is required to find the continual pro

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a+b

and

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

duct of

a1. -304 a+y

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α x

Ans. a+x.

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

duct of

a2 - x2
a+b;

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Ex. 12. Multiply x2-3x+1 by x2.

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8

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.

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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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C

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The divisor - inverted, becomes hence
C

is the fraction required.

b

5cx

ab

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5x

a

The divisor (5a-5x

a+b

5a-5x

; hence

3(a-x)_3

3(a-x) 5

a+b

inverted, becomes

3a-3x a+b 3a-3x

X

a+b 5a-5x 5α-5x

is the quotient required.

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The reciprocal of the divisor is

a2-b2 1 (a+b)(a—b) ___a—b

X

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x a+b x +(a+b) 3C

required.
a2-b2

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Ex. 4. Divide

·by a+

a+c

a

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- 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.

2

Ex. 5. Divide x4x3x2-1x by x2-x.

8

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