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avatar+1195 

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 \((A,B)\).

 Sep 14, 2019
 #1
avatar+106536 
+2

We can solve this by the method of partial fractions

 

Note that the first denominator can be factored as  ( x -2) (x+ 1)

 

Multiply  everything through by   (x-2) (x + 1)   and we get that

 

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

 

x + 7  =  (A + B)x  + (A - 2B)        equating terms, we have that

 

A + B  =1      (1)  

A - 2B  =  7    →   -A + 2B  = -7     (2)

 

Add (1) and (2)  and we have that

 

3B = -6

B = -2

 

So   A -2  = 1

A = 3

 

(A,B)  =  ( 3, -2)

 

 

cool cool cool

 Sep 14, 2019
 #2
avatar+1195 
+1

Thank you CPhill!

 Sep 15, 2019

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