1. **State the problem:** Solve the rational equation $$1 + \frac{2}{x - 8} = \frac{x}{x^2 - 12x + 32}$$ for all values of $x$. 2. **Factor the denominator on the right side:** Note that $$x^2 - 12x + 32 = (x - 8)(x - 4)$$. 3. **Rewrite the equation using the factorization:** $$1 + \frac{2}{x - 8} = \frac{x}{(x - 8)(x - 4)}$$ 4. **Find the least common denominator (LCD):** The LCD is $(x - 8)(x - 4)$. 5. **Multiply both sides of the equation by the LCD to clear denominators:** $$\left(1 + \frac{2}{x - 8}\right)(x - 8)(x - 4) = \frac{x}{(x - 8)(x - 4)} \times (x - 8)(x - 4)$$ 6. **Simplify both sides:** Left side: $$1 \times (x - 8)(x - 4) + \frac{2}{x - 8} \times (x - 8)(x - 4) = (x - 8)(x - 4) + 2(x - 4)$$ Right side: $$x$$ 7. **Expand the terms on the left:** $$(x - 8)(x - 4) = x^2 - 4x - 8x + 32 = x^2 - 12x + 32$$ So left side becomes: $$x^2 - 12x + 32 + 2x - 8 = x^2 - 10x + 24$$ 8. **Set the equation:** $$x^2 - 10x + 24 = x$$ 9. **Bring all terms to one side:** $$x^2 - 10x + 24 - x = 0$$ $$x^2 - 11x + 24 = 0$$ 10. **Factor the quadratic:** $$x^2 - 11x + 24 = (x - 8)(x - 3) = 0$$ 11. **Solve for $x$:** $$x - 8 = 0 \Rightarrow x = 8$$ $$x - 3 = 0 \Rightarrow x = 3$$ 12. **Check for restrictions:** The original denominators cannot be zero. Denominators are $x - 8$ and $(x - 8)(x - 4)$. So $x \neq 8$ and $x \neq 4$. 13. **Exclude $x = 8$ because it makes denominator zero.** 14. **Final solution:** $$\boxed{x = 3}$$