Answer
{$6$}
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:
$\frac{2x}{x+2}+\frac{x}{x-2}=3$
$\frac{2x(x-2)+x(x+2)}{(x+2)(x-2)}=3$
$\frac{2x^{2}-4x+x^{2}+2x}{(x+2)(x-2)}=3$
$\frac{3x^{2}-2x}{(x+2)(x-2)}=3$
Now, we cross multiply:
$\frac{3x^{2}-2x}{(x+2)(x-2)}=3$
$3x^{2}-2x=3(x-2)(x+2)$
$3x^{2}-2x=3(x^{2}-2^{2})$
$3x^{2}-2x=3(x^{2}-4)$
$3x^{2}-2x=3x^{2}-12$
$-2x=-12$
$x=\frac{-12}{-2}$
$x=6$
Therefore, the solution set is {$6$}.
