1. **State the problem:** Solve the equation $$\frac{11}{x^2 - 25} + \frac{2}{5 - x} = -1$$ for $x$. 2. **Rewrite the denominators:** Note that $$x^2 - 25 = (x - 5)(x + 5)$$ and $$5 - x = -(x - 5)$$. 3. **Find a common denominator:** The common denominator is $(x - 5)(x + 5)$. 4. **Rewrite each term with the common denominator:** $$\frac{11}{(x - 5)(x + 5)} + \frac{2}{5 - x} = \frac{11}{(x - 5)(x + 5)} + \frac{2}{-(x - 5)} = \frac{11}{(x - 5)(x + 5)} - \frac{2}{x - 5}$$ 5. **Express the second fraction with the common denominator:** $$- \frac{2}{x - 5} = - \frac{2(x + 5)}{(x - 5)(x + 5)} = - \frac{2x + 10}{(x - 5)(x + 5)}$$ 6. **Combine the fractions:** $$\frac{11 - (2x + 10)}{(x - 5)(x + 5)} = \frac{11 - 2x - 10}{(x - 5)(x + 5)} = \frac{1 - 2x}{(x - 5)(x + 5)}$$ 7. **Set the equation:** $$\frac{1 - 2x}{(x - 5)(x + 5)} = -1$$ 8. **Multiply both sides by the denominator:** $$1 - 2x = -1 \times (x - 5)(x + 5)$$ 9. **Expand the right side:** $$(x - 5)(x + 5) = x^2 - 25$$ So, $$1 - 2x = - (x^2 - 25) = -x^2 + 25$$ 10. **Bring all terms to one side:** $$1 - 2x + x^2 - 25 = 0$$ 11. **Simplify:** $$x^2 - 2x - 24 = 0$$ 12. **Factor the quadratic:** $$x^2 - 2x - 24 = (x - 6)(x + 4) = 0$$ 13. **Solve for $x$:** $$x - 6 = 0 \Rightarrow x = 6$$ $$x + 4 = 0 \Rightarrow x = -4$$ 14. **Check for restrictions:** Denominators cannot be zero, so $x \neq 5$ and $x \neq -5$. Both $6$ and $-4$ are valid. **Final answer:** $$x = 6 \text{ or } x = -4$$