532 A L G E B K A 32. PEOB. VI. To Add or Subtract Fractions. [FRACTIONS. Rule. Reduce the fractions to a common denominator, and add or subtract their numerators ; and the sum or difference placed over the common denominator is the sum or remainder required. In practice, however, it is generally better to separate the process into two or more parts analogous to the addi tion or subtraction of sums of money, where the pounds are added to the pounds, the shillings to the shillings, &c., and the result afterwards combined. Ex. 1. Add together - and ; a-b b-a The latter fraction is a-b a .: the sum required is , ; = ab ab Similarly, a-b a b -1. 1-a* 1 ar n 1 a" 1 1 a" Ex. 2. Collect into a single fraction 1 1 2a = 1 Since 1 a-b Ex. 3. Collect -f - i b a+b c 1 1 a b a+b 1 2a a + b a 2 b 2 2& a 2 -6 2 2(6 + q)
- a 2 -6 2 :
7 2 a b 4x-8 3x-6 24-12x We observe that x 2 is common to all the denominators : the question may therefore be written, Ex. 4. Collect 1 1 7 4 3 12 x-2 x-2 x-2 1 i 1 x-4y 25x+4y x 2 -16?/ Here we commence by adding the 1st and 3d together, and the 2d and 4th together ; which results in 24x p. 9x 2 -47/ 2 x 2 -16?/ 2 - 1 - . _ ... 4 _ = p. r V9x 2 -4;/ 2 x 2 -16i/V (9x 2 -35x 2 210a> 3 (9x a -42/ 2 )(x 2 -16i/ 2 ) Ex. 5. Find the sum of + -X + X 2 -X 3 1 + X + X 2 + X 3 The numerator will consist of the sum of two products, the one containing + x, exactly in the same way that the other contains - x. If, then, we write down one of these products, and double the even powers of x in it, omitting the odd powers, we shall obtain the required result. The product of the denominators again may be readily obtained by regarding it as that of the difference and sum of 1 + x 2 and x + x 3 . As such processes are of constant occurrence, we will indicate the work in full. Numerator, 1+1+1+1 1+1+1+1 Double of 1 + 3# 2 + 3* 4 +1 Denominator, { 1 + a 2 - (x + X s ) } { 1 + x- + (x + x 3 ) } = -f ;r 2 ) 2 - (x + X s } 2 = 1 And the result is 1+X2-X--X 6 Ex. 6. Collect into one fraction 1 1 Multiply numerator and denominator of the first frac tion by x~ m , &c., and the given quantity becomes , = 1 X+X- 1 Ex. 7. If + X~P x 1 l + l+ln I l + m+ml ml l+n+nm 1 l + l+ln l + m+ml l + n + nm = 1, and _ j_ none of the denominators being zero, then l = m = n. Multiply the first quantity by I, and subtract, thcro results I n, which, when substituted in the first l + n quantity, gives m n, whence the proposition. 33. The converse problem to collecting many fractions Partial into one is frequently as important as the direct the pro- fraction blem, namely, of resolving a compound fraction into its components or partial fractions. For a first example, if it be required to find what simple fractions make up the 2x compound fraction -^ -, we commence by observing that X" Q/" the denominator x- - a? is the product of x + a and x - a. 2x Hence. - ; is the sum of the fractions whose denomina- zP-d 2 tors are x + a and x a. 2x A B Let -T- , -- 1 --- , where A and B arc quantities x 2 a 2 x + a x-cr which involve a only, not x, since # 2 does not appear in the numerator of the sum. By addition, 2x A(x-q , 2 x 2 a 2 /. 2x = A(# - a) + B(# + a). To obtain A and B from this equality, we remark that the equality is an identity, as in Art. 20. We may, therefore, deal with it in either of two ways : 1. Make the x s on the left hand side to coincide with the x s on the right, and the a s in like manner. 2. As in Art. 20, write any thing we please in place of x on both sides. We will in this example take the first method, and illustrate the second method by the subsequent examples. We get 2 = A + B, = A-B; .. A = B = 1, and the result is 2x 1 1 x 2 -a 2 x a x a 1 A B Ex. 2. (x-a)(x-6) - + , xa xb Write a for x. then 1 = A(a - Z>) . *. A = - a b Write Jfor ar,then 1 = B(6 - a) = - (a - &) /. B = - -^ ; a b hence ~~ ^~~ ( ^~^~~ *"- James500 (talk) 22:07, 13 January 2020 (UTC) )
- (x a)(x 6) a b x a x bj
The reader will observe that we have treated - as if a b it were not itself a fraction. In fact, in the application of the subject before us, the letters a and b stand for arith metical quantities, and the fraction = is simply an arith- CL ~~ metical fraction, as contradistinguished from an algebraical
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