(x + 7) / (x2 - x - 2) = A / (x - 2) + B / (x + 1)
Since x2 - x - 2 = (x - 2)(x + 1)
---> (x + 7) / (x2 - x - 2) = (x + 7) / [ (x - 2)(x + 1) ] =
(x + 7) / [ (x - 2)(x + 1) ] = A / (x - 2) + B / (x + 1)
Writing with the common denominator of (x - 2)(x + 1)
(x + 7) / [ (x - 2)(x + 1) ] = [ A(x + 1) ] / [ (x - 2)(x + 1) ] + [B(x - 2) ] / [ (x - 2)(x + 1) ]
Since the denominators are equal, they can be ignored:
x + 7 = A(x + 1) + B(x - 2)
x + 7 = Ax + A + Bx - 2B
x + 7 = Ax + Bx + A - 2B
Set the x-term on the left side equal to the x-terms on the right side: x = Ax + Bx
---> x = (A + B)x ---> 1x = (A + B)x ---> 1 = A + B
Set the number term on the left side equal to the number terms on the right side: 7 = A - 2B
Solving: A + B = 1 ---> x -1 ---> -A - B = -1
A - 2B = 7 ---> A - 2B = 7
Adding down the columns: -3B = 6
B = -2
Since B = -2: A + B = 1 ---> A + -2 = 1 ---> A = 3
(-2, 3)
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