Answer
$x=4$
Work Step by Step
$\frac{1}{x+3} +\frac{6}{x^2-9}=1$
$\frac{1}{x+3} +\frac{6}{(x-3)(x+3)}=1$
$\frac{1(x-3)}{(x+3)(x-3)} +\frac{6}{(x-3)(x+3)}=\frac{(x-3)(x+3)}{(x-3)(x+3)}$
$x-3+6=(x-3)(x+3)$
$x+3=x^2-9$
$x+12=x^2$
$x^2-x-12=0$
$(x-4)(x+3)=0$
$x-4=0$
$x=4$
$x+3=0$
$x=-3$ (this isn't an answer since the denominators would be zero)
$x=4$
$\frac{1}{x+3} +\frac{6}{x^2-9}=1$
$\frac{1}{4+3} +\frac{6}{4^2-9}=1$
$1/7 + 6/(16-9) = 1$
$1/7 + 6/7 = 1$
