Answer
$(-\infty,-3)\cup(-3,2)\cup(2,+\infty)$
Work Step by Step
$f8x)+g(x)=\frac{8x}{x-2}+\frac{6}{x+3}$ simplify
$\frac{8x(x+3)+6(x-2)}{(x-2)(x+3)}=\frac{8x^2+24x+6x-12}{(x-2)(x+3)}=\frac{8x^2+30-12}{(x-2)(x+3)}$ since the denominator can/t be equal to 0 $x\ne-3$ and $x\ne2$
