Question: top-center: $\frac{x^2-25}{x^2+5x} \div \frac{xy+6x-5y-30}{5x-15}$
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:**
- $x^2-25 = (x-5)(x+5)$ (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 $(x+5)$ in numerator and denominator
- Cancel $(x-5)$ in 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 simplified expression:**
$$\boxed{\frac{5(x-3)}{x(y+6)}}$$