1. **State the problem:** Solve the inequality $$|x + 6| \leq 8$$ algebraically. 2. **Recall the rule for absolute value inequalities:** For any real number $A$ and positive number $B$, $$|A| \leq B$$ means $$-B \leq A \leq B$$. 3. **Apply the rule:** Here, $A = x + 6$ and $B = 8$, so $$-8 \leq x + 6 \leq 8$$ 4. **Solve the compound inequality:** Subtract 6 from all parts: $$-8 - 6 \leq x + 6 - 6 \leq 8 - 6$$ 5. **Show cancellation step:** $$\cancel{-8} - 6 \leq x + \cancel{6} - 6 \leq 8 - 6$$ 6. **Simplify:** $$-14 \leq x \leq 2$$ 7. **Final answer:** The solution set is all $x$ such that $$-14 \leq x \leq 2$$. This means $x$ can be any number between $-14$ and $2$, inclusive.