1. **State the problem:** Solve the compound inequality $$-4 \leq \frac{1}{2}(8x + 8) < 12$$ and find the solution set. 2. **Recall the rule:** When solving inequalities, you can multiply or divide both sides by the same positive number without changing the inequality signs. 3. **Multiply both sides of the inequality by 2 to eliminate the fraction:** $$-4 \leq \frac{1}{2}(8x + 8) < 12$$ Multiply all parts by 2: $$2 \times -4 \leq 2 \times \frac{1}{2}(8x + 8) < 2 \times 12$$ $$-8 \leq \cancel{2} \times \frac{1}{2}(8x + 8) < 24$$ $$-8 \leq 8x + 8 < 24$$ 4. **Subtract 8 from all parts to isolate the term with x:** $$-8 - 8 \leq 8x + 8 - 8 < 24 - 8$$ $$-16 \leq 8x < 16$$ 5. **Divide all parts by 8 to solve for x:** $$\frac{-16}{8} \leq \frac{8x}{8} < \frac{16}{8}$$ $$-2 \leq \cancel{8}x \div \cancel{8} < 2$$ $$-2 \leq x < 2$$ 6. **Final solution:** The solution set is all real numbers $$x$$ such that $$-2 \leq x < 2$$. This means $$x$$ can be any number from $$-2$$ up to but not including $$2$$.