1. **State the problem:** Solve the inequality $$|2x + 1| \leq 3$$ and graph the solution set. 2. **Recall the rule for absolute value inequalities:** For $$|A| \leq B$$ where $$B \geq 0$$, the inequality is equivalent to $$-B \leq A \leq B$$. 3. **Apply the rule:** Here, $$A = 2x + 1$$ and $$B = 3$$, so $$-3 \leq 2x + 1 \leq 3$$. 4. **Solve the compound inequality:** First, subtract 1 from all parts: $$-3 - 1 \leq 2x + 1 - 1 \leq 3 - 1$$ $$-4 \leq 2x \leq 2$$ Next, divide all parts by 2: $$\frac{-4}{2} \leq \frac{2x}{2} \leq \frac{2}{2}$$ Show cancelation: $$\frac{\cancel{-4}}{\cancel{2}} \leq x \leq \frac{\cancel{2}}{\cancel{2}}$$ Simplify: $$-2 \leq x \leq 1$$ 5. **Interpret the solution:** The solution set includes all $$x$$ values between $$-2$$ and $$1$$, inclusive. 6. **Final answer:** $$\boxed{-2 \leq x \leq 1}$$