Question: Find the solution set for the inequality.
$\frac{1}{3} (3x - 9) \geq 6 - 2(-2x)$
Enter the solution set as a simplified inequality.
1. **State the problem:** Solve the inequality $$\frac{1}{3} (3x - 9) \geq 6 - 2(-2x)$$ for $x$.
2. **Apply distributive property:** Multiply inside the parentheses on the left and simplify the right side.
$$\frac{1}{3} \times 3x - \frac{1}{3} \times 9 \geq 6 - 2(-2x)$$
which simplifies to
$$x - 3 \geq 6 + 4x$$
3. **Isolate variable terms on one side:** Subtract $x$ from both sides.
$$x - 3 - x \geq 6 + 4x - x$$
$$\cancel{x} - 3 - \cancel{x} \geq 6 + 3x$$
which simplifies to
$$-3 \geq 6 + 3x$$
4. **Isolate $x$:** Subtract $6$ from both sides.
$$-3 - 6 \geq 6 + 3x - 6$$
$$-9 \geq 3x$$
5. **Divide both sides by 3:**
$$\frac{-9}{3} \geq \frac{3x}{3}$$
$$\cancel{3} \times -3 \geq x$$
which simplifies to
$$-3 \geq x$$
6. **Rewrite the inequality:**
$$x \leq -3$$
**Final answer:** The solution set is $$x \leq -3$$.