1. **State the problem:** Solve the equation $$\frac{x+3}{x-2} = \frac{x-2}{x+3}$$. 2. **Cross-multiply to eliminate the fractions:** $$ (x+3)(x+3) = (x-2)(x-2) $$ 3. **Expand both sides:** $$ (x+3)^2 = (x-2)^2 $$ $$ x^2 + 6x + 9 = x^2 - 4x + 4 $$ 4. **Subtract $x^2$ from both sides:** $$ \cancel{x^2} + 6x + 9 = \cancel{x^2} - 4x + 4 $$ $$ 6x + 9 = -4x + 4 $$ 5. **Add $4x$ to both sides:** $$ 6x + 4x + 9 = -4x + 4x + 4 $$ $$ 10x + 9 = 4 $$ 6. **Subtract 9 from both sides:** $$ 10x + \cancel{9} - 9 = 4 - 9 $$ $$ 10x = -5 $$ 7. **Divide both sides by 10:** $$ x = \frac{-5}{10} = -\frac{1}{2} $$ 8. **Check for restrictions:** The denominators $x-2$ and $x+3$ cannot be zero, so $x \neq 2$ and $x \neq -3$. Our solution $x = -\frac{1}{2}$ is valid. **Final answer:** $x = -\frac{1}{2}$ (option a).