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. **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. **Rewrite the expression using multiplication:**
$$\frac{x^2-25}{x^2+5x} \times \frac{5x-15}{xy+6x-5y-30}$$
4. **Factor all polynomials where possible:**
- $x^2-25$ is a difference of squares:
$$x^2-25 = (x-5)(x+5)$$
- $x^2+5x$ can be factored by taking out $x$:
$$x^2+5x = x(x+5)$$
- $5x-15$ can be factored by taking out $5$:
$$5x-15 = 5(x-3)$$
- $xy+6x-5y-30$ can be factored by grouping:
$$xy+6x-5y-30 = x(y+6) -5(y+6) = (x-5)(y+6)$$
5. **Substitute the factored forms back into the expression:**
$$\frac{(x-5)(x+5)}{x(x+5)} \times \frac{5(x-3)}{(x-5)(y+6)}$$
6. **Cancel common factors:**
- $(x+5)$ appears in numerator and denominator.
- $(x-5)$ appears in numerator and denominator.
Intermediate step showing cancellation:
$$\frac{\cancel{(x-5)}\cancel{(x+5)}}{x\cancel{(x+5)}} \times \frac{5(x-3)}{\cancel{(x-5)}(y+6)} = \frac{1}{x} \times \frac{5(x-3)}{y+6}$$
7. **Multiply the remaining factors:**
$$\frac{1}{x} \times \frac{5(x-3)}{y+6} = \frac{5(x-3)}{x(y+6)}$$
**Final answer:**
$$\boxed{\frac{5(x-3)}{x(y+6)}}$$