1. **State the problem:** Solve the rational equation $$-\frac{1}{x^2 + 8x} + \frac{9x}{x + 8} = \frac{8}{x + 8}$$ for all values of $x$. 2. **Identify the domain restrictions:** The denominators cannot be zero, so: - $x^2 + 8x = x(x+8) \neq 0 \implies x \neq 0, x \neq -8$ - $x + 8 \neq 0 \implies x \neq -8$ 3. **Rewrite the equation:** $$-\frac{1}{x(x+8)} + \frac{9x}{x+8} = \frac{8}{x+8}$$ 4. **Find a common denominator:** The common denominator is $x(x+8)$. 5. **Express all terms with the common denominator:** - First term is already $-\frac{1}{x(x+8)}$ - Second term: $$\frac{9x}{x+8} = \frac{9x \cdot x}{x(x+8)} = \frac{9x^2}{x(x+8)}$$ - Third term: $$\frac{8}{x+8} = \frac{8 \cdot x}{x(x+8)} = \frac{8x}{x(x+8)}$$ 6. **Combine all terms:** $$-\frac{1}{x(x+8)} + \frac{9x^2}{x(x+8)} = \frac{8x}{x(x+8)}$$ 7. **Multiply both sides by $x(x+8)$ to clear denominators:** $$-1 + 9x^2 = 8x$$ 8. **Rearrange to standard quadratic form:** $$9x^2 - 8x - 1 = 0$$ 9. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=9$, $b=-8$, $c=-1$. Calculate the discriminant: $$\Delta = (-8)^2 - 4 \cdot 9 \cdot (-1) = 64 + 36 = 100$$ 10. **Find the roots:** $$x = \frac{8 \pm \sqrt{100}}{18} = \frac{8 \pm 10}{18}$$ - For $+$: $$x = \frac{8 + 10}{18} = \frac{18}{18} = 1$$ - For $-$: $$x = \frac{8 - 10}{18} = \frac{-2}{18} = -\frac{1}{9}$$ 11. **Check domain restrictions:** - $x=1$ is allowed (not 0 or -8) - $x=-\frac{1}{9}$ is allowed **Final answer:** $$x = 1 \text{ or } x = -\frac{1}{9}$$