+865 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 $(A, B).$
+781 \(\frac{4x}{x^2-8x+15}=\frac{A}{x-3}+\frac{B}{x-5}\\ \frac{4x}{x^2-8x+15}=\frac{A(x-5)}{x^2-8x+15}+\frac{B(x-3)}{x^2-8x+15}\\ 4x=A(x-5)+B(x-3)\\ 4x=Ax-5A+Bx-3B\\\)
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.
