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There are numbers A and B for which
A/(x - 1) + B/(x + 1) = (x + 8)/(x^2 - 1)
for every number $x\neq\pm1$. Find A-B

 Jun 15, 2022
 #1
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Multiply by (x^2 - 1) on both sides gives:

 

\(A(x + 1) + B(x - 1) = x + 8\\ (A + B)x + (A - B) = x + 8\)

 

Note that this holds true for any real number x.

Therefore, we can compare the coefficients of x on both sides.

 

\(\begin{cases}A + B &=& 1\\A - B &=& 8\end{cases}\)

 

Can you take it from here?

 Jun 15, 2022

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