## Elementary Algebra

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