1. **State the problem:** Solve the equation $$\frac{1}{2}x + 1 = \frac{1}{3}x - 2$$ for $x$. 2. **Formula and rules:** To solve linear equations, we isolate $x$ by moving all terms involving $x$ to one side and constants to the other. 3. **Step 1: Move $x$ terms to one side:** Subtract $\frac{1}{3}x$ from both sides: $$\frac{1}{2}x - \frac{1}{3}x + 1 = -2$$ 4. **Step 2: Simplify the $x$ terms:** Find common denominator 6: $$\frac{3}{6}x - \frac{2}{6}x + 1 = -2$$ $$\left(\frac{3}{6} - \frac{2}{6}\right)x + 1 = -2$$ $$\frac{1}{6}x + 1 = -2$$ 5. **Step 3: Move constant term:** Subtract 1 from both sides: $$\frac{1}{6}x + 1 - 1 = -2 - 1$$ $$\frac{1}{6}x = -3$$ 6. **Step 4: Solve for $x$:** Multiply both sides by 6: $$6 \times \frac{1}{6}x = 6 \times (-3)$$ $$\cancel{6} \times \frac{1}{\cancel{6}} x = -18$$ $$x = -18$$ **Final answer:** $$x = -18$$