1. **State the problem:** Simplify and solve the equation $$\frac{3}{x+5} - \frac{1}{x^2 + 2x - 15} = \frac{1}{x+5}$$. 2. **Factor the quadratic denominator:** Note that $$x^2 + 2x - 15 = (x+5)(x-3)$$. 3. **Rewrite the equation with factored form:** $$\frac{3}{x+5} - \frac{1}{(x+5)(x-3)} = \frac{1}{x+5}$$ 4. **Find a common denominator:** The common denominator is $$(x+5)(x-3)$$. 5. **Rewrite each term with the common denominator:** $$\frac{3(x-3)}{(x+5)(x-3)} - \frac{1}{(x+5)(x-3)} = \frac{1(x-3)}{(x+5)(x-3)}$$ 6. **Combine the left side:** $$\frac{3(x-3) - 1}{(x+5)(x-3)} = \frac{x-3}{(x+5)(x-3)}$$ 7. **Simplify numerator on the left:** $$3(x-3) - 1 = 3x - 9 - 1 = 3x - 10$$ 8. **Set the equation:** $$\frac{3x - 10}{(x+5)(x-3)} = \frac{x-3}{(x+5)(x-3)}$$ 9. **Since denominators are equal and nonzero, set numerators equal:** $$3x - 10 = x - 3$$ 10. **Solve for x:** $$3x - x = -3 + 10$$ $$2x = 7$$ $$x = \frac{7}{2}$$ 11. **Check for restrictions:** Denominators cannot be zero, so $$x \neq -5$$ and $$x \neq 3$$. Since $$\frac{7}{2} = 3.5$$ is allowed, this is the solution. **Final answer:** $$x = \frac{7}{2}$$