1. **State the problem:** Solve the rational equation $$\frac{2x}{x+5} + \frac{4}{x+3} = \frac{8}{x^2 + 8x + 15}$$. 2. **Factor the denominator on the right:** Note that $$x^2 + 8x + 15 = (x+5)(x+3)$$. 3. **Rewrite the equation:** $$\frac{2x}{x+5} + \frac{4}{x+3} = \frac{8}{(x+5)(x+3)}$$. 4. **Multiply both sides by the common denominator** $(x+5)(x+3)$ to clear fractions: $$\cancel{(x+5)}\cancel{(x+3)} \left( \frac{2x}{\cancel{x+5}} + \frac{4}{\cancel{x+3}} \right) = 8 \cancel{(x+5)} \cancel{(x+3)}$$ which simplifies to $$2x(x+3) + 4(x+5) = 8$$. 5. **Expand both terms:** $$2x^2 + 6x + 4x + 20 = 8$$ which simplifies to $$2x^2 + 10x + 20 = 8$$. 6. **Bring all terms to one side:** $$2x^2 + 10x + 20 - 8 = 0$$ $$2x^2 + 10x + 12 = 0$$. 7. **Divide entire equation by 2 to simplify:** $$\cancel{2}x^2 + \cancel{2}5x + \cancel{2}6 = 0$$ which is $$x^2 + 5x + 6 = 0$$. 8. **Factor the quadratic:** $$x^2 + 5x + 6 = (x+2)(x+3) = 0$$. 9. **Solve for x:** $$x+2=0 \Rightarrow x=-2$$ $$x+3=0 \Rightarrow x=-3$$. 10. **Check for extraneous solutions:** The original denominators are $x+5$, $x+3$, and $(x+5)(x+3)$. Values $x=-3$ and $x=-5$ make denominators zero, so $x=-3$ is not valid. 11. **Final solution:** $$\boxed{x = -2}$$