1. **State the problem:** Solve the equation $$\frac{2}{x+5} + \frac{1}{x-5} = \frac{16}{x^2 - 25}$$ where $x^2 - 25$ can be factored as $(x+5)(x-5)$. 2. **Identify the common denominator:** The denominators are $x+5$, $x-5$, and $x^2 - 25 = (x+5)(x-5)$. The common denominator is $(x+5)(x-5)$. 3. **Rewrite each term with the common denominator:** $$\frac{2}{x+5} = \frac{2(x-5)}{(x+5)(x-5)}$$ $$\frac{1}{x-5} = \frac{1(x+5)}{(x+5)(x-5)}$$ The right side is already over the common denominator. 4. **Rewrite the equation:** $$\frac{2(x-5)}{(x+5)(x-5)} + \frac{1(x+5)}{(x+5)(x-5)} = \frac{16}{(x+5)(x-5)}$$ 5. **Combine the left side over the common denominator:** $$\frac{2(x-5) + 1(x+5)}{(x+5)(x-5)} = \frac{16}{(x+5)(x-5)}$$ 6. **Since denominators are equal and nonzero, set numerators equal:** $$2(x-5) + (x+5) = 16$$ 7. **Expand and simplify:** $$2x - 10 + x + 5 = 16$$ $$3x - 5 = 16$$ 8. **Add 5 to both sides:** $$3x - \cancel{5} + \cancel{5} = 16 + 5$$ $$3x = 21$$ 9. **Divide both sides by 3:** $$\frac{3x}{\cancel{3}} = \frac{21}{\cancel{3}}$$ $$x = 7$$ 10. **Check for restrictions:** The denominators $x+5$ and $x-5$ cannot be zero, so $x \neq -5$ and $x \neq 5$. Since $x=7$ is allowed, it is the solution. **Final answer:** $$x = 7$$