Answer
x=4
Work Step by Step
$\frac{4}{x-2}-\frac{x}{x+2}=\frac{16}{x^{2}-4}$
$\frac{4(x+2)}{(x-2)(x+2)}-\frac{x(x-2)}{(x-2)(x+2)}=\frac{16}{(x-2)(x+2)}$
$4x+8-x^{2}+2x=16$
$x^{2}-6x+8=0$
$(x-4)(x+2)=0$
x=4 or x=-2
Substitute x=-2 to the original formula and find that $\frac{x}{(-2)+2}$, $\frac{16}{(-2)^2-4}$, the denominator is 0, which is incorrect.
So the answer is x=4
