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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).

 Jun 16, 2020
 #1
avatar+781 
+3

Notice that multiplying the denominators x-2 and x+1 give us \(x^2-x-2\).

 

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

Now look here, the x terms Ax+Bx must result in x(with a coefficient of 1) so A+B=1 and A-2B must result in 7.

 

So we have a system of equations.

 

\(A+B=1\\ A-2B=7\\ \text{then:}\\ 3B=-6\\ B=-2, A=3\)

 

Therefore the pair is \((3, -2)\)

 Jun 17, 2020
 #2
avatar+794 
0

Thanks!

AnimalMaster  Jun 17, 2020

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