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Find $A$ and $B$ such that \( \frac{4x}{x^2-8x+15} = \frac{A}{x-3} + \frac{B}{x-5}\)for all x besides 3 and 5. Express your answer as an ordered pair in the form $(A, B).$

 Jun 20, 2020
 #1
avatar+781 
+1

\(\frac{4x}{x^2-8x+15}=\frac{A}{x-3}+\frac{B}{x-5}\\ \frac{4x}{x^2-8x+15}=\frac{A(x-5)}{x^2-8x+15}+\frac{B(x-3)}{x^2-8x+15}\\ 4x=A(x-5)+B(x-3)\\ 4x=Ax-5A+Bx-3B\\\)

The x terms, Ax and Bx must add up to 4x so A+B=4. The constant terms, -5a and -3B must add up to 0 so -5A-3B=0.

We have a system of equations:

 

A+B=4

-5A-3B=0

 

Solving, we get A= -6 and B=10.

 Jun 20, 2020
 #2
avatar+794 
-1

thanks!

AnimalMaster  Jun 20, 2020

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