1. **State the problem:** Simplify the complex fraction $$\frac{\frac{x^2 - 1}{x^2 + x}}{\frac{x - 1}{x + 1}}$$ and find which expression it is equivalent to. 2. **Recall the rule for dividing fractions:** Dividing by a fraction is the same as multiplying by its reciprocal. So, $$\frac{\frac{x^2 - 1}{x^2 + x}}{\frac{x - 1}{x + 1}} = \frac{x^2 - 1}{x^2 + x} \times \frac{x + 1}{x - 1}$$ 3. **Factor all polynomials where possible:** - $x^2 - 1$ is a difference of squares: $$x^2 - 1 = (x - 1)(x + 1)$$ - $x^2 + x$ can be factored by taking out $x$: $$x^2 + x = x(x + 1)$$ 4. **Rewrite the expression with factored forms:** $$\frac{(x - 1)(x + 1)}{x(x + 1)} \times \frac{x + 1}{x - 1}$$ 5. **Multiply the fractions:** $$\frac{(x - 1)(x + 1)}{x(x + 1)} \times \frac{x + 1}{x - 1} = \frac{(x - 1)(x + 1)(x + 1)}{x(x + 1)(x - 1)}$$ 6. **Cancel common factors:** - Cancel $(x - 1)$ in numerator and denominator: $$\frac{\cancel{(x - 1)}(x + 1)(x + 1)}{x(x + 1)\cancel{(x - 1)}}$$ - Cancel one $(x + 1)$ in numerator and denominator: $$\frac{(x + 1)\cancel{(x + 1)}}{x\cancel{(x + 1)}} = \frac{x + 1}{x}$$ 7. **Final simplified expression:** $$\frac{x + 1}{x}$$ 8. **Compare with given options:** None of the options directly match $$\frac{x + 1}{x}$$, so let's check if any option simplifies to this. - Option A: $$-3\frac{x - 1}{x + 3}$$ no. - Option B: $$3(x - 1)$$ no. - Option C: $$3x(x - 1)$$ no. - Option D: $$-\frac{x + 3}{x - 1}$$ no. Therefore, none of the options match the simplified expression $$\frac{x + 1}{x}$$. **Note:** The problem asks which expression is equivalent to the quotient, but none of the options match the simplified form. The simplified form is $$\frac{x + 1}{x}$$.