## Precalculus (6th Edition)

$\dfrac{x}{x-1}-\dfrac{1}{x+1}=\dfrac{2}{x^{2}-1}$ Factor the denominator of the fraction on the right side of the equation: $\dfrac{x}{x-1}-\dfrac{1}{x+1}=\dfrac{2}{(x-1)(x+1)}$ Multiply the whole equation by $(x-1)(x+1)$: $(x-1)(x+1)\Big[\dfrac{x}{x-1}-\dfrac{1}{x+1}=\dfrac{2}{(x-1)(x+1)}\Big]$ $x(x+1)-(x-1)=2$ Evaluate the indicated operations: $x^{2}+x-x+1=2$ Take $2$ to the left side and simplify: $x^{2}+x-x+1-2=0$ $x^{2}-1=0$ Solve by factoring: $(x-1)(x+1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-1=0$ $x=1$ $x+1=0$ $x=-1$ The original equation is undefined for $x=1$ and $x=-1$, so it has no solution.