(19x - 2)/((5 - x)(1 + 6x)) can be expressed as A/(5-x) + B/(1+6x) where A and B are integers. Find A and B
First we can equate (19x - 2)/((5 - x)(1 + 6x)) to A/(5-x) + B/(1+6x) which means:
(19x - 2)/((5 - x)(1 + 6x)) = A/(5-x) + B/(1+6x). Then we will turn the RHS into a single fraction:
(19x - 2)/((5 - x)(1 + 6x)) = (A(1+6x) + B(5-x))/((5 - x)(1 + 6x)). Since the denominator on RHS = denominator on LHS, the numerators on both sides must be equal to:
19x - 2 = A(1+6x) + B(5-x). Now we can use the method of undetermined coefficients which means to match the coefficients of the powers of x and use the information to solve for A and B. 19x = (6A - B)x and -2 = A +5B. From this we can gather: A = -5B - 2 and 19 = 6A - B
substitue in for A: 19 = 6*(-5B -2) -B. simlifying gives: 31 = -31B therefore B = -1 and A = -5*(-1) -2, hence A is 3.
A = 3 , B = -1
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