## Precalculus: Mathematics for Calculus, 7th Edition

$x=-5$
$\dfrac{x}{x-2}+\dfrac{1}{x+2}=\dfrac{8}{x^{2}-4}$ Factor the denominator of the fraction on the right side: $\dfrac{x}{x-2}+\dfrac{1}{x+2}=\dfrac{8}{(x-2)(x+2)}$ Multiply the whole equation by $(x-2)(x+2)$: $(x-2)(x+2)\Big[\dfrac{x}{x-2}+\dfrac{1}{x+2}=\dfrac{8}{(x-2)(x+2)}\Big]$ $x(x+2)+(x-2)=8$ $x^{2}+2x+x-2=8$ Take $8$ to the left side: $x^{2}+2x+x-2-8=0$ $x^{2}+3x-10=0$ Solve by factoring: $(x+5)(x-2)=0$ Set both factor equal to $0$ and solve each individual equation for $x$: $x+5=0$ $x=-5$ $x-2=0$ $x=2$ The original equation is undefined for $x=2$, so the only solution for this equation is $x=-5$