1. **State the problem:** Solve the equation $$\frac{x}{x+3} + \frac{12}{x^2 - 9} = \frac{2}{x-3}$$ for $x$. 2. **Identify the denominators and restrictions:** - Note that $x^2 - 9 = (x-3)(x+3)$. - The denominators $x+3$, $x-3$, and $x^2-9$ cannot be zero, so $x \neq 3$ and $x \neq -3$. 3. **Rewrite the equation with factored denominators:** $$\frac{x}{x+3} + \frac{12}{(x-3)(x+3)} = \frac{2}{x-3}$$ 4. **Find the least common denominator (LCD):** The LCD is $(x-3)(x+3)$. 5. **Multiply both sides of the equation by the LCD to clear denominators:** $$ (x-3)(x+3) \times \left( \frac{x}{x+3} + \frac{12}{(x-3)(x+3)} \right) = (x-3)(x+3) \times \frac{2}{x-3} $$ 6. **Distribute and simplify each term:** $$ (x-3) \cancel{(x+3)} \times \frac{x}{\cancel{x+3}} + \cancel{(x-3)(x+3)} \times \frac{12}{\cancel{(x-3)(x+3)}} = \cancel{(x-3)} (x+3) \times \frac{2}{\cancel{x-3}} $$ This simplifies to: $$ x(x-3) + 12 = 2(x+3) $$ 7. **Expand and simplify:** $$ x^2 - 3x + 12 = 2x + 6 $$ 8. **Bring all terms to one side:** $$ x^2 - 3x + 12 - 2x - 6 = 0 $$ $$ x^2 - 5x + 6 = 0 $$ 9. **Factor the quadratic:** $$ (x - 2)(x - 3) = 0 $$ 10. **Solve for $x$:** $$ x = 2 \quad \text{or} \quad x = 3 $$ 11. **Check for restrictions:** Recall $x \neq 3$ because it makes denominators zero. 12. **Final solution:** $$ \boxed{x = 2} $$