(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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