## Algebra 1

$n=0,\frac{1}{2}$
Given: $\dfrac{n}{n-2}+\dfrac{n}{n+2}=\frac{n}{n^2-4}$ $\dfrac{n}{n-2}+\dfrac{n}{n+2}=\frac{n}{(n-2)(n+2)}$ $\dfrac{n(n+2)}{(n-2)(n+2)}+\dfrac{n(n-2)}{(n-2)(n+2)}=\dfrac{n}{(n-2)(n+2)}$ $n(n+2)+n(n-2)=n$ $n^2+2n+n^2-2n=n$ $n^2+2n+n^2-2n-n=0$ $2n^2-n=0$ $n(2n-1)=0$ Hence, $n=0,\frac{1}{2}$