#### Answer

{$-\frac{1}{5}$}

#### 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:
$\frac{2x}{x+1}-\frac{3}{x-1}=2$
$\frac{2x(x-1)-3(x+1)}{(x+1)(x-1)}=2$
$\frac{2x^{2}-2x-3x-3}{(x+1)(x-1)}=2$
$\frac{2x^{2}-5x-3}{(x+1)(x-1)}=2$
Now, we cross multiply:
$\frac{2x^{2}-5x-3}{(x+1)(x-1)}=2$
$2x^{2}-5x-3=2(x-1)(x+1)$
$2x^{2}-5x-3=2(x^{2}-1^{2})$
$2x^{2}-5x-3=2(x^{2}-1)$
$2x^{2}-5x-3=2x^{2}-2$
$-5x-3=-2$
$-5x=-2+3$
$-5x=1$
$x=-\frac{1}{5}$
Therefore, the solution set is {$-\frac{1}{5}$}.