Answer
{$\frac{4}{3}$}
Work Step by Step
First, we find the LCM of the denominators of the two fractions on the left-hand side of the equation in order to develop a single fraction:
$\frac{x}{x-4}-\frac{2x}{x+4}=-1$
$\frac{x(x+4)-2x(x-4)}{(x+4)(x-4)}=-1$
$\frac{x^{2}+4x-2x^{2}+8x}{(x+4)(x-4)}=-1$
$\frac{-x^{2}+12x}{(x+4)(x-4)}=-1$
Now, we cross multiply:
$\frac{-x^{2}+12x}{(x+4)(x-4)}=-1$
$-x^{2}+12x=-1(x-4)(x+4)$
$-x^{2}+12x=-1(x^{2}-4^{2})$
$-x^{2}+12x=-1(x^{2}-16)$
$-x^{2}+12x=-x^{2}+16$
$12x=16$
$x=\frac{16}{12}$
$x=\frac{4}{3}$
Therefore, the solution set is {$\frac{4}{3}$}.
