Solution
[tex]\begin{aligned}\frac{1}{h-5}+\frac{2}{h+5}&=\frac{16}{h^2-25}\\\frac{(h+5)+2(h-5)}{(h-5)(h+5)}&=\frac{16}{h^2-25}\\\frac{h+2h+5-10}{h^2-25}&=\frac{16}{h^2-25}\\\frac{3h-5}{\cancel{h^2-25}}&=\frac{16}{\cancel{h^2-25}}\\3h-5&=16\\3h&=16+5\\3h&=21\\h&=\frac{21}{3}\\h&=7\end{aligned}[/tex]
