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Find constants $A$ and $B$ such that \[\frac{x - 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\]for all $x$ such that $x\neq -1$ and  $x\neq 2$. Give your answer as the ordered pair $(A,B)$.

 Aug 27, 2021
 #1
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By method of partial fractions, A - B = 1 and 4A + B = 1.  Solving, we get (A,B) = (-3/5,2/5).

 Aug 28, 2021
 #2
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On the right-hand side make the denominator (x-2)(x+1) 

 

Then the bottoms will be the same and you can just equate the coefficients on the top.

 Aug 28, 2021
edited by Melody  Aug 28, 2021
 #3
avatar+114536 
+1

You want more hints ... ok

 

\(\frac{x - 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\\ \frac{x - 7}{x^2 - x - 2} = \frac{A(x+1)}{(x - 2)(x+1)} + \frac{B(x-2)}{(x + 1)(x-2)} \)

 Aug 29, 2021
 #4
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+1

thank youuuu

Guest Aug 29, 2021

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