How to get the value of $A + B ?$ I have this statement:

If $\frac{x+6}{x^2-x-6} = \frac{A}{x-3} + \frac{B}{x+2}$, what is the value of $A+B$ ?

My attempt was:
$\frac{x+6}{(x-3)(x+2)} = \frac{A(x+2) + B(x-3)}{(x-3)(x+2)}$
$x+6=(x+2)A + B(x-3)$: But from here, I don't know how to get $A + B$, any hint is appreciated.

By your work the coefficient before $x$ it's $1$, which ends.
Also,  $$\frac{x+6}{x^2-x-6}=\frac{x+6}{(x-3)(x+2)}=\frac{x-3+9}{(x-3)(x+2)}=$$
$$=\frac{1}{x+2}+\frac{9}{5}\left(\frac{1}{x-3}-\frac{1}{x+2}\right)=\frac{\frac{9}{5}}{x+3}+\frac{-\frac{4}{5}}{x+2},$$
which gives $A+B=1.$
