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:** Dividing by a fraction is the same as multiplying by its reciprocal. $$\frac{x^2-25}{x^2+5x} \times \frac{5x-15}{xy+6x-5y-30}$$ 3. **Factor all polynomials:** - $x^2-25 = (x-5)(x+5)$ (difference of squares) - $x^2+5x = x(x+5)$ - $5x-15 = 5(x-3)$ - Factor $xy+6x-5y-30$ by grouping: $xy+6x-5y-30 = x(y+6) - 5(y+6) = (x-5)(y+6)$ 4. **Substitute factored forms:** $$\frac{(x-5)(x+5)}{x(x+5)} \times \frac{5(x-3)}{(x-5)(y+6)}$$ 5. **Cancel common factors:** - Cancel $(x+5)$ from numerator and denominator - Cancel $(x-5)$ from numerator and denominator Intermediate step showing cancellation: $$\frac{\cancel{(x-5)}\cancel{(x+5)}}{x\cancel{(x+5)}} \times \frac{5(x-3)}{\cancel{(x-5)}(y+6)} = \frac{1}{x} \times \frac{5(x-3)}{y+6}$$ 6. **Multiply remaining factors:** $$\frac{1}{x} \times \frac{5(x-3)}{y+6} = \frac{5(x-3)}{x(y+6)}$$ 7. **Final answer:** $$\boxed{\frac{5(x-3)}{x(y+6)}}$$