1. **State the problem:** Simplify the expression $$\frac{x-2}{x+2} + \frac{x+10}{x^2+6x+8}$$. 2. **Factor the denominator:** Note that $$x^2+6x+8$$ factors as $$(x+2)(x+4)$$. 3. **Rewrite the expression:** $$\frac{x-2}{x+2} + \frac{x+10}{(x+2)(x+4)}$$ 4. **Find a common denominator:** The common denominator is $$(x+2)(x+4)$$. 5. **Rewrite the first fraction with the common denominator:** $$\frac{x-2}{x+2} = \frac{(x-2)(x+4)}{(x+2)(x+4)}$$ 6. **Add the fractions:** $$\frac{(x-2)(x+4)}{(x+2)(x+4)} + \frac{x+10}{(x+2)(x+4)} = \frac{(x-2)(x+4) + (x+10)}{(x+2)(x+4)}$$ 7. **Expand the numerator:** $$(x-2)(x+4) = x^2 + 4x - 2x - 8 = x^2 + 2x - 8$$ 8. **Combine terms in the numerator:** $$x^2 + 2x - 8 + x + 10 = x^2 + 3x + 2$$ 9. **Factor the numerator:** $$x^2 + 3x + 2 = (x+1)(x+2)$$ 10. **Simplify the fraction:** $$\frac{(x+1)(x+2)}{(x+2)(x+4)}$$ 11. **Cancel the common factor $x+2$:** $$\frac{\cancel{(x+1)}\cancel{(x+2)}}{\cancel{(x+2)}(x+4)} = \frac{x+1}{x+4}$$ **Final answer:** $$\frac{x+1}{x+4}$$