## College Algebra (6th Edition)

$x=-3$
$\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{x^{2}-1}$ Factor the fraction on the right side completely: $\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{(x-1)(x+1)}$ Multipy the whole equation by $(x-1)(x+1)$: $(x-1)(x+1)\Big[\dfrac{2}{x+1}-\dfrac{1}{x-1}=\dfrac{2x}{(x-1)(x+1)}\Big]$ $2(x-1)-(x+1)=2x$ $2x-2-x-1=2x$ Take all terms with $x$ to the left side and all terms without $x$ to the right side. Then, solve for $x$: $2x-2x-x=1+2$ $-x=3$ $x=-3$