How to get the value of $A + B ?$

Question

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.

Answer 1

If $x+6=(x+2)A+B(x-3)$ then $$x+6=x(A+B)+(2A-3B)\implies$$ $$A+B=1\text{ and } 2A-3B=6$$

Answer 2

By equating coefficients, you have the system of equations $$A+B = 1,$$ $$2A-3B = 6.$$

Answer 3

Put $x=-2$ and $x=3$ to get $B=-\frac{4}{5}$ and $A=\frac{9}{5}$ respectively.

Answer 4

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 $A+B=1.$