What is the solution to the equation \( \frac{1}{h-5}+\frac{2}{h+5}=\frac{16}{n^{2}-25} ? \)

To solve the equation \( \frac{1}{h-5}+\frac{2}{h+5}=\frac{16}{n^{2}-25} \), we start by recognizing that \( n^2 - 25 \) can be factored as \( (n-5)(n+5) \). To combine the left-hand side, we need a common denominator. The common denominator of \( h - 5 \) and \( h + 5 \) is \( (h - 5)(h + 5) \). Therefore, we can rewrite the left-hand side: \[ \frac{1}{h-5} + \frac{2}{h+5} = \frac{(h + 5) + 2(h - 5)}{(h - 5)(h + 5)} = \frac{h + 5 + 2h - 10}{(h - 5)(h + 5)} = \frac{3h - 5}{(h - 5)(h + 5)} \] Now, we equate both sides: \[ \frac{3h - 5}{(h - 5)(h + 5)} = \frac{16}{(n - 5)(n + 5)} \] Now we cross-multiply: \[ (3h - 5)(n - 5)(n + 5) = 16(h - 5)(h + 5) \] Calculate \( (h - 5)(h + 5) = h^2 - 25 \): \[ (3h - 5)(n^2 - 25) = 16(h^2 - 25) \] Now expand both sides. The left-hand side becomes: \[ 3h n^2 - 75h - 5n^2 + 125 = 16h^2 - 400 \] Rearranging gives: \[ 3hn^2 - 5n^2 - 16h^2 + 75h + 125 - 400 = 0 \] Combine like terms: \[ 3hn^2 - 5n^2 - 16h^2 + 75h - 275 = 0 \] At this point, we can analyze or substitute values for either \( h \) or \( n \) depending on what solution we're interested in finding. The equation isn't easily factored, so numerical methods or substitutions might be needed for specific values of \( n \) or \( h \) if required. For specific solutions, we might need to provide additional constraints or values for either variable.