1. **State the problem:** Solve the equation $$\frac{6x}{x^2 - 9} = \frac{18}{x + 3}$$ for $x$. 2. **Recall the formula and rules:** The denominators are $x^2 - 9$ and $x + 3$. Note that $x^2 - 9$ can be factored as a difference of squares: $$x^2 - 9 = (x - 3)(x + 3)$$ 3. **Rewrite the equation using the factorization:** $$\frac{6x}{(x - 3)(x + 3)} = \frac{18}{x + 3}$$ 4. **Multiply both sides by the common denominator $(x - 3)(x + 3)$ to clear fractions:** $$6x = 18 \times \cancel{\frac{(x - 3)(x + 3)}{x + 3}} = 18(x - 3)$$ 5. **Simplify the right side by canceling $x + 3$:** $$6x = 18(x - 3)$$ 6. **Expand the right side:** $$6x = 18x - 54$$ 7. **Bring all terms to one side:** $$6x - 18x = -54$$ 8. **Simplify:** $$-12x = -54$$ 9. **Divide both sides by $-12$:** $$x = \frac{-54}{-12} = \frac{54}{12}$$ 10. **Simplify the fraction by dividing numerator and denominator by 6:** $$x = \frac{\cancel{54}^9}{\cancel{12}^2} = \frac{9}{2}$$ 11. **Check for restrictions:** - The original denominators cannot be zero. - $x^2 - 9 = 0 \Rightarrow x = \pm 3$ are excluded. - $x + 3 = 0 \Rightarrow x = -3$ is excluded. Since $x = \frac{9}{2} = 4.5$ is not excluded, it is a valid solution. **Final answer:** $$x = \frac{9}{2}$$