Question:
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 $(A, B).$

Answer:
Factoring the denominator on the left side gives \[ \frac{4x}{(x-5)(x-3)}=\frac{A}{x-3}+\frac{B}{x-5}. \]Then, we multiply both sides of the equation by $(x - 3)(x - 5)$ to get \[ 4x = A(x-5) + B(x-3). \]If the linear expression $4x$ agrees with the linear expression $A(x-5) + B(x-3)$ at all values of $x$ besides 3 and 5, then the two expressions must agree for $x=3$ and $x=5$ as well.  Substituting $x = 3$, we get $12 = -2A$, so $A = -6$.  Likewise, we plug in $x = 5$ to solve for $B$. Substituting $x = 5$, we get $20 = 2B$, so $B = 10$.  Therefore, $(A, B) = \boxed{(-6, 10)}.$