## College Algebra 7th Edition

$x=-5$
$\frac{x}{x-2}+\frac{1}{x+2}=\frac{8}{x^{2}-4}$ We multiply through by $(x-2)(x+2)$: $x(x+2)+(x-2)=8$ Next we distribute: $x^{2}+2x+x-2=8$ $x^{2}+3x-10=0$ And factor: $(x-2)(x+5)=0$ $x-2=0$ or $x+5=0$ $x=2$ or $x=-5$ However, we have to eliminate $x=2$ because it creates division by 0 in the original equation. So the only solution is $x=-5$.