1. **State the problem:** Simplify the expression $$\frac{x^2-25}{x^2+5x} \div \frac{xy+6x-5y-30}{5x-15}$$. 2. **Rewrite division as multiplication by reciprocal:** $$\frac{x^2-25}{x^2+5x} \times \frac{5x-15}{xy+6x-5y-30}$$ 3. **Factor all polynomials:** - Numerator of first fraction: $$x^2-25 = (x-5)(x+5)$$ (difference of squares) - Denominator of first fraction: $$x^2+5x = x(x+5)$$ (factor out $x$) - Numerator of second fraction: $$5x-15 = 5(x-3)$$ (factor out 5) - Denominator of second fraction: $$xy+6x-5y-30$$ group terms: $$ (xy - 5y) + (6x - 30) = y(x-5) + 6(x-5) = (x-5)(y+6)$$ 4. **Rewrite expression with factors:** $$\frac{(x-5)(x+5)}{x(x+5)} \times \frac{5(x-3)}{(x-5)(y+6)}$$ 5. **Cancel common factors:** - $(x+5)$ cancels - $(x-5)$ cancels 6. **Simplify remaining expression:** $$\frac{1}{x} \times \frac{5(x-3)}{y+6} = \frac{5(x-3)}{x(y+6)}$$ **Final answer:** $$\boxed{\frac{5(x-3)}{x(y+6)}}$$