1. Stating the problem: Solve the equation $$\frac{x + 1}{2} - 3 = \frac{2}{3}x - 1$$ for $x$. 2. Formula and rules: To solve linear equations, we isolate $x$ by performing inverse operations and maintaining equality. 3. Start by simplifying both sides: $$\frac{x + 1}{2} - 3 = \frac{2}{3}x - 1$$ 4. Add 3 to both sides: $$\frac{x + 1}{2} = \frac{2}{3}x - 1 + 3$$ $$\frac{x + 1}{2} = \frac{2}{3}x + 2$$ 5. Multiply both sides by 6 (the least common multiple of denominators 2 and 3) to clear fractions: $$6 \times \frac{x + 1}{2} = 6 \times \left(\frac{2}{3}x + 2\right)$$ $$3(x + 1) = 4x + 12$$ 6. Expand the left side: $$3x + 3 = 4x + 12$$ 7. Subtract $3x$ from both sides: $$\cancel{3x} + 3 = 4x - \cancel{3x} + 12$$ $$3 = x + 12$$ 8. Subtract 12 from both sides: $$3 - 12 = x + \cancel{12} - \cancel{12}$$ $$-9 = x$$ 9. Final answer: $$x = -9$$