Answer
The solutions are $n=0$ or $n=\frac{1}{2}$.
Work Step by Step
$\frac{n}{n-2}+\frac{n}{n+2}=\frac{n}{n^2-4}$
$\frac{n(n+2)}{(n-2)(n+2)}+\frac{n(n-2)}{(n-2)(n+2)}=\frac{n}{(n-2)(n+2)}$
$\frac{2n^2}{(n-2)(n+2)}=\frac{n}{(n-2)(n+2)}$
$2n^2=n$
$2n^2-n=0$
$n=0$ 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 $n=\frac{1}{2}$.
