bc bc х ab =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. 3а 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 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) 3a 3 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 a =a2 15 157. But, when mixed quantities are to be mul- x2 - 1 x + 2 3x2 -5% 7a Ex. 6. Multiply by Sax - 5a 402-6 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 aa--ab a a+b; Ex. 12. Multiply 2-x+1 by x2-x. Ans. A-03 +42-X. 1 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. 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 х ja-5X atob 50-50 5Q ----50 3(2-x)_3 is the quotient required. 5(0-2) 5 a2-62 Ex. 3. Divide by a+b. X a+b; is the quotient a 1 The reciprocal of the divisor is ; hence Q2-62 1 x _(a+b)(a−b)_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 a 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 |