1. **State the problem:** Solve the equation $$\frac{x-3}{x-1} - \left(4 + \frac{x}{x+2}\right) = 0$$ for $x$. 2. **Rewrite the equation:** $$\frac{x-3}{x-1} - 4 - \frac{x}{x+2} = 0$$ 3. **Isolate the fractions:** $$\frac{x-3}{x-1} - \frac{x}{x+2} = 4$$ 4. **Find a common denominator:** The common denominator is $(x-1)(x+2)$. 5. **Rewrite each fraction with the common denominator:** $$\frac{(x-3)(x+2)}{(x-1)(x+2)} - \frac{x(x-1)}{(x+2)(x-1)} = 4$$ 6. **Combine the fractions:** $$\frac{(x-3)(x+2) - x(x-1)}{(x-1)(x+2)} = 4$$ 7. **Multiply both sides by the denominator to clear fractions:** $$\cancel{(x-1)(x+2)} \cdot \frac{(x-3)(x+2) - x(x-1)}{\cancel{(x-1)(x+2)}} = 4 \cdot (x-1)(x+2)$$ 8. **Simplify the numerator:** $$(x-3)(x+2) = x^2 + 2x - 3x - 6 = x^2 - x - 6$$ $$x(x-1) = x^2 - x$$ 9. **Substitute back:** $$x^2 - x - 6 - (x^2 - x) = 4(x-1)(x+2)$$ 10. **Simplify the left side:** $$x^2 - x - 6 - x^2 + x = -6$$ 11. **Expand the right side:** $$(x-1)(x+2) = x^2 + 2x - x - 2 = x^2 + x - 2$$ $$4(x^2 + x - 2) = 4x^2 + 4x - 8$$ 12. **Set up the equation:** $$-6 = 4x^2 + 4x - 8$$ 13. **Bring all terms to one side:** $$0 = 4x^2 + 4x - 8 + 6$$ $$0 = 4x^2 + 4x - 2$$ 14. **Divide entire equation by 2 to simplify:** $$\cancel{2} \cdot 0 = \cancel{2} \cdot (2x^2 + 2x - 1)$$ $$0 = 2x^2 + 2x - 1$$ 15. **Use quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=2$, $b=2$, $c=-1$. 16. **Calculate discriminant:** $$b^2 - 4ac = 2^2 - 4 \cdot 2 \cdot (-1) = 4 + 8 = 12$$ 17. **Calculate roots:** $$x = \frac{-2 \pm \sqrt{12}}{2 \cdot 2} = \frac{-2 \pm 2\sqrt{3}}{4} = \frac{-2}{4} \pm \frac{2\sqrt{3}}{4} = -\frac{1}{2} \pm \frac{\sqrt{3}}{2}$$ 18. **Final solutions:** $$x = -\frac{1}{2} + \frac{\sqrt{3}}{2} \quad \text{or} \quad x = -\frac{1}{2} - \frac{\sqrt{3}}{2}$$ 19. **Check for restrictions:** Denominators cannot be zero, so $x \neq 1$ and $x \neq -2$. Both solutions are valid. **Answer:** $$x = -\frac{1}{2} + \frac{\sqrt{3}}{2} \quad \text{or} \quad x = -\frac{1}{2} - \frac{\sqrt{3}}{2}$$