1. **State the problem:** Simplify the expression $$\frac{x^2-25}{x^2+5x} \div \frac{xy+6x-5y-30}{5x-15}$$. 2. **Recall the division of fractions rule:** Dividing by a fraction is the same as multiplying by its reciprocal. So, $$\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$$. 3. **Apply the rule:** $$\frac{x^2-25}{x^2+5x} \times \frac{5x-15}{xy+6x-5y-30}$$. 4. **Factor all polynomials:** - $x^2-25 = (x-5)(x+5)$ (difference of squares) - $x^2+5x = x(x+5)$ - $5x-15 = 5(x-3)$ - $xy+6x-5y-30$ group terms: $$xy+6x-5y-30 = x(y+6) -5(y+6) = (x-5)(y+6)$$ 5. **Rewrite the expression with factors:** $$\frac{(x-5)(x+5)}{x(x+5)} \times \frac{5(x-3)}{(x-5)(y+6)}$$ 6. **Cancel common factors:** - $(x+5)$ cancels - $(x-5)$ cancels 7. **Simplify the remaining expression:** $$\frac{1}{x} \times \frac{5(x-3)}{y+6} = \frac{5(x-3)}{x(y+6)}$$ **Final answer:** $$\boxed{\frac{5(x-3)}{x(y+6)}}$$