1. **State the problem:** Solve the equation $$\frac{2x - 3}{x^2 - 4} + \frac{x + 1}{x^2 - x - 2} = \frac{4}{x - 2}$$ for $x$. 2. **Factor denominators:** - $x^2 - 4 = (x - 2)(x + 2)$ - $x^2 - x - 2 = (x - 2)(x + 1)$ 3. **Rewrite the equation with factored denominators:** $$\frac{2x - 3}{(x - 2)(x + 2)} + \frac{x + 1}{(x - 2)(x + 1)} = \frac{4}{x - 2}$$ 4. **Find the least common denominator (LCD):** The LCD is $(x - 2)(x + 2)(x + 1)$. 5. **Multiply both sides by the LCD to clear denominators:** $$\left(\frac{2x - 3}{(x - 2)(x + 2)} + \frac{x + 1}{(x - 2)(x + 1)}\right) \times (x - 2)(x + 2)(x + 1) = \frac{4}{x - 2} \times (x - 2)(x + 2)(x + 1)$$ 6. **Simplify each term:** - First term: $(2x - 3)(x + 1)$ - Second term: $(x + 1)(x + 2)$ - Right side: $4(x + 2)(x + 1)$ 7. **Write the simplified equation:** $$(2x - 3)(x + 1) + (x + 1)(x + 2) = 4(x + 2)(x + 1)$$ 8. **Factor out $(x + 1)$ on the left side:** $$(x + 1)((2x - 3) + (x + 2)) = 4(x + 2)(x + 1)$$ 9. **Simplify inside the parentheses:** $$(x + 1)(3x - 1) = 4(x + 2)(x + 1)$$ 10. **Divide both sides by $(x + 1)$, noting $x \neq -1$ to avoid division by zero:** $$3x - 1 = 4(x + 2)$$ 11. **Expand the right side:** $$3x - 1 = 4x + 8$$ 12. **Bring all terms to one side:** $$3x - 1 - 4x - 8 = 0$$ $$-x - 9 = 0$$ 13. **Solve for $x$:** $$-x = 9$$ $$x = -9$$ 14. **Check for restrictions:** - $x \neq 2$ (denominators zero) - $x \neq -2$ (denominators zero) - $x \neq -1$ (denominators zero) Since $x = -9$ is not restricted, it is a valid solution. **Final answer:** $$x = -9$$