Subjects algebra

Fraction Division Bbb852

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1. **State the problem:** Simplify the expression $$\frac{x^2-25}{x^2+5x} \div \frac{xy+6x-5y-30}{5x-15}$$. 2. **Rewrite the division as multiplication by the reciprocal:** $$\frac{x^2-25}{x^2+5x} \times \frac{5x-15}{xy+6x-5y-30}$$ 3. **Factor all polynomials:** - Numerator of first fraction: $$x^2-25 = (x-5)(x+5)$$ - Denominator of first fraction: $$x^2+5x = x(x+5)$$ - Numerator of second fraction: $$5x-15 = 5(x-3)$$ - Denominator of second fraction: $$xy+6x-5y-30$$ 4. **Factor the denominator of the second fraction by grouping:** $$xy+6x-5y-30 = x(y+6) - 5(y+6) = (x-5)(y+6)$$ 5. **Substitute the factored forms:** $$\frac{(x-5)(x+5)}{x(x+5)} \times \frac{5(x-3)}{(x-5)(y+6)}$$ 6. **Cancel common factors:** - Cancel $(x+5)$ from numerator and denominator. - Cancel $(x-5)$ from numerator and denominator. Intermediate step showing cancellation: $$\frac{\cancel{(x-5)}\cancel{(x+5)}}{x\cancel{(x+5)}} \times \frac{5(x-3)}{\cancel{(x-5)}(y+6)} = \frac{1}{x} \times \frac{5(x-3)}{y+6}$$ 7. **Multiply the remaining factors:** $$\frac{1}{x} \times \frac{5(x-3)}{y+6} = \frac{5(x-3)}{x(y+6)}$$ **Final answer:** $$\boxed{\frac{5(x-3)}{x(y+6)}}$$