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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 $(A,B)$.

Guest Aug 27, 2017
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\(\[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\]\)

 

\( $x\neq -1$\)

 

and

 

\($x\neq 2$\)

 

 

This is a partial fractions problem.....we want to solve this

 

x + 7  =  A(x + 1) + B(x - 2)     simplify

 

x + 7  = (A + B)x + (A - 2B)     equating coefficients, we have the following system

 

A + B  = 1    →   B  = 1 - A     (1)

 

A - 2B  = 7      (2)

 

Sub (1) into (2)  and we have

 

A - 2(1 - A)  = 7

3A - 2  = 7

3A  = 9

A  = 3

 

And B  = 1 - 3  =  -2

 

So  (A, B)  =  (3 , -2)

 

 

 

cool cool cool

CPhill  Aug 28, 2017

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