1. **State the problem:** Solve the equation $$\frac{3}{x+1} - \frac{2}{x^2-1} = \frac{1}{x^2 - x}$$ for $x$. 2. **Identify the denominators and factor where possible:** - $x+1$ is linear. - $x^2 - 1 = (x+1)(x-1)$ (difference of squares). - $x^2 - x = x(x-1)$. 3. **Find the least common denominator (LCD):** $$\text{LCD} = x(x+1)(x-1)$$ 4. **Multiply both sides of the equation by the LCD to clear denominators:** $$3 \cdot x(x-1) - 2 \cdot x = 1 \cdot (x+1)$$ 5. **Write the equation after clearing denominators:** $$3x(x-1) - 2x = x + 1$$ 6. **Expand and simplify:** $$3x^2 - 3x - 2x = x + 1$$ 7. **Bring all terms to one side:** $$3x^2 - 3x - 2x - x - 1 = 0$$ 8. **Combine like terms:** $$3x^2 - 6x - 1 = 0$$ 9. **Solve the quadratic equation:** Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=-6$, $c=-1$. 10. **Calculate the discriminant:** $$\Delta = (-6)^2 - 4 \cdot 3 \cdot (-1) = 36 + 12 = 48$$ 11. **Find the roots:** $$x = \frac{6 \pm \sqrt{48}}{6} = \frac{6 \pm 4\sqrt{3}}{6} = 1 \pm \frac{2\sqrt{3}}{3}$$ 12. **Final answers:** $$x_1 = 1 + \frac{2\sqrt{3}}{3}$$ $$x_2 = 1 - \frac{2\sqrt{3}}{3}$$ **Note:** Check for restrictions where denominators are zero: $x \neq -1$, $x \neq 0$, $x \neq 1$. Both solutions do not violate these restrictions.