## Algebra 1: Common Core (15th Edition)

The solutions are $n=0$ or $n=\frac{1}{2}$.
$\frac{n}{n-2}+\frac{n}{n+2}=\frac{n}{n^2-4}$ $\frac{n(n+2)}{(n-2)(n+2)}+\frac{n(n-2)}{(n-2)(n+2)}=\frac{n}{(n-2)(n+2)}$ $\frac{2n^2}{(n-2)(n+2)}=\frac{n}{(n-2)(n+2)}$ $2n^2=n$ $2n^2-n=0$ $n=0$ or $n=\frac{1}{2}$ Check: $\frac{0}{0-2}+\frac{0}{0+2}=\frac{0}{0^2-4}$ $0=0$ $\frac{\frac{1}{2}}{\frac{1}{2}-2}+\frac{\frac{1}{2}}{\frac{1}{2}+2}=\frac{\frac{1}{2}}{(\frac{1}{2})^2-4}$ $\frac{-2}{15}=\frac{-2}{15}$ The solutions are $n=0$ or $n=\frac{1}{2}$.