1. **State the problem:** Solve the rational equation $$\frac{1}{x} - \frac{x}{x-1} = -\frac{3x}{x-1}$$ for all values of $x$. 2. **Identify the common denominator:** The denominators are $x$ and $x-1$. The common denominator is $$x(x-1).$$ 3. **Multiply both sides by the common denominator to clear fractions:** $$x(x-1) \left( \frac{1}{x} - \frac{x}{x-1} \right) = x(x-1) \left(-\frac{3x}{x-1} \right)$$ 4. **Simplify each term:** $$\cancel{x}(x-1) \cdot \frac{1}{\cancel{x}} - x \cancel{(x-1)} \cdot \frac{x}{\cancel{(x-1)}} = -3x \cancel{x} \cdot \frac{x-1}{\cancel{x-1}}$$ which simplifies to $$(x-1) - x^2 = -3x^2$$ 5. **Rewrite the equation:** $$(x-1) - x^2 = -3x^2$$ 6. **Bring all terms to one side:** $$(x-1) - x^2 + 3x^2 = 0$$ 7. **Combine like terms:** $$x - 1 + 2x^2 = 0$$ 8. **Rewrite in standard quadratic form:** $$2x^2 + x - 1 = 0$$ 9. **Use the quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=1$, and $c=-1$. 10. **Calculate the discriminant:** $$\Delta = 1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9$$ 11. **Find the roots:** $$x = \frac{-1 \pm \sqrt{9}}{2 \cdot 2} = \frac{-1 \pm 3}{4}$$ 12. **Calculate each solution:** - For $+$ sign: $$x = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2}$$ - For $-$ sign: $$x = \frac{-1 - 3}{4} = \frac{-4}{4} = -1$$ 13. **Check for restrictions:** Denominators cannot be zero, so $x \neq 0$ and $x \neq 1$. Both solutions $x=\frac{1}{2}$ and $x=-1$ are valid. **Final answer:** $$x = \frac{1}{2} \text{ or } x = -1$$