Answer
$(-\infty,-8)\cup (-8,4)\cup(4,+\infty)$.
Work Step by Step
The given functions are
$f(x)=\frac{9x}{x-4}$ and $g(x)=\frac{7}{x+8}$
$\Rightarrow f(x)+g(x)=\frac{9x}{x-4}+\frac{7}{x+8}$
LCM of both the denominators $=(x-4)(x+8)$
$\Rightarrow f(x)+g(x)=\frac{(x+8)}{(x+8)}\cdot \frac{9x}{x-4}+\frac{(x-4)}{(x-4)}\cdot\frac{7}{x+8}$
Simplify.
$\Rightarrow f(x)+g(x)=\frac{9x(x+8)}{(x+8)(x-4)}+\frac{7(x-4)}{(x-4)(x+8)}$
Add both numerators because both denominators are equal.
$\Rightarrow f(x)+g(x)=\frac{9x(x+8)+7(x-4)}{(x+8)(x-4)}$
Simplify.
$\Rightarrow f(x)+g(x)=\frac{9x^2+72x+7x-28}{(x+8)(x-4)}$
$\Rightarrow f(x)+g(x)=\frac{9x^2+79x-28}{(x+8)(x-4)}$
In the denominator the value of $x$ must not be $-8$ and $4$.
Hence, the domain is $(-\infty,-8)\cup (-8,4)\cup(4,+\infty)$.
