1. The problem is to solve the equation $$\frac{1}{2} \div \frac{3}{2} + 2 \frac{1}{2} = 3$$. 2. First, recall that dividing by a fraction is the same as multiplying by its reciprocal: $$a \div b = a \times \frac{1}{b}$$. 3. Calculate $$\frac{1}{2} \div \frac{3}{2}$$: $$\frac{1}{2} \times \frac{2}{3} = \frac{1 \times 2}{2 \times 3} = \frac{2}{6}$$. 4. Simplify $$\frac{2}{6}$$ by canceling common factors: $$\frac{\cancel{2}^1}{\cancel{6}^3} = \frac{1}{3}$$. 5. Convert the mixed number $$2 \frac{1}{2}$$ to an improper fraction: $$2 \frac{1}{2} = \frac{2 \times 2 + 1}{2} = \frac{5}{2}$$. 6. Now add the fractions: $$\frac{1}{3} + \frac{5}{2}$$. 7. Find a common denominator, which is 6: $$\frac{1}{3} = \frac{2}{6}, \quad \frac{5}{2} = \frac{15}{6}$$. 8. Add the fractions: $$\frac{2}{6} + \frac{15}{6} = \frac{17}{6}$$. 9. The left side simplifies to $$\frac{17}{6}$$, but the right side is 3, which is $$\frac{18}{6}$$. 10. Since $$\frac{17}{6} \neq \frac{18}{6}$$, the equation as stated is not true. Final answer: $$\frac{1}{2} \div \frac{3}{2} + 2 \frac{1}{2} = \frac{17}{6} \neq 3$$.