1. State the problem: Solve the inequality $$\frac{2}{3}x + \frac{1}{3} \leq 2 \frac{2}{3}$$. 2. Convert the mixed number to an improper fraction: $$2 \frac{2}{3} = \frac{8}{3}$$. 3. Subtract $$\frac{1}{3}$$ from both sides to isolate the term with $$x$$: $$\frac{2}{3}x + \frac{1}{3} - \frac{1}{3} \leq \frac{8}{3} - \frac{1}{3}$$ $$\Rightarrow \frac{2}{3}x \leq \frac{7}{3}$$ 4. Multiply both sides by the reciprocal of $$\frac{2}{3}$$, which is $$\frac{3}{2}$$, to solve for $$x$$: $$\frac{3}{2} \cdot \frac{2}{3}x \leq \frac{7}{3} \cdot \frac{3}{2}$$ 5. Show cancellation of common factors: $$\cancel{\frac{3}{2}} \cdot \cancel{\frac{2}{3}} x \leq \frac{7}{3} \cdot \frac{3}{2}$$ 6. Simplify: $$x \leq \frac{7}{2}$$ 7. Convert $$\frac{7}{2}$$ to a mixed number if desired: $$x \leq 3 \frac{1}{2}$$ Final answer: $$x \leq \frac{7}{2}$$ or $$x \leq 3 \frac{1}{2}$$.