1. **State the problem:** Simplify the expression $$\frac{x+3}{x^2-9} - \frac{2}{x-3}$$. 2. **Recall the formula and rules:** The denominator $x^2-9$ is a difference of squares and can be factored as $$(x-3)(x+3)$$. 3. **Rewrite the expression with factored denominators:** $$\frac{x+3}{(x-3)(x+3)} - \frac{2}{x-3}$$ 4. **Simplify the first fraction by canceling common factors:** $$\frac{\cancel{x+3}}{(x-3)\cancel{(x+3)}} = \frac{1}{x-3}$$ 5. **Now the expression is:** $$\frac{1}{x-3} - \frac{2}{x-3}$$ 6. **Since denominators are the same, subtract numerators:** $$\frac{1-2}{x-3} = \frac{-1}{x-3}$$ 7. **Final simplified expression:** $$-\frac{1}{x-3}$$ This means the original expression simplifies to $$-\frac{1}{x-3}$$, valid for all $x \neq 3$ and $x \neq -3$ (to avoid division by zero).