help Algebra

$\frac{Bx-11}{x^2-7x+10}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\ =\ \frac{A}{x-2}+\frac{8}{x-5}\\~\\ \frac{Bx-11}{(x-2)(x-5)}\cdot(x-2)(x-5)\ =\ \Big(\frac{A}{x-2}+\frac{8}{x-5}\Big)\cdot(x-2)(x-5)\\~\\ Bx-11=\frac{A(x-2)(x-5)}{x-2}+\frac{8(x-2)(x-5)}{x-5}\\~\\ Bx-11\ =\ A(x-5)\ +\ 8(x-2)$

Assuming the previous equation is true for all values of  x ,  it must be true when  x = 0 , so we can say:

$B(0)-11\ =\ A(0-5)+8(0-2)\\~\\ -11\ =\ A(-5)-16\\~\\ 5\ =\ A(-5)\\~\\ A\ =\ -1$

And it must also be true when  x = 5  so we can say:

$B(5)-11\ =\ A(5-5) + 8(5 - 2)\\~\\ 5B-11\ =\ 24\\~\\ 5B\ =\ 35\\~\\ B\ =\ 7$

So   A + B   =   -1 + 7   =   6

Check:  https://www.wolframalpha.com/input/?i=%287x-11%29%2F%28x%5E2-7x%2B10%29%3D%28-1%29%2F%28x-2%29%2B8%2F%28x-5%29

_{(BTW even if the 8 and the B are swapped you can still apply this same method to solve it!)}