1. The problem is to solve the inequality $4|x-3| > 12$. 2. We start by isolating the absolute value expression. Divide both sides by 4: $$\frac{4|x-3|}{\cancel{4}} > \frac{12}{\cancel{4}}$$ which simplifies to $$|x-3| > 3$$ 3. Recall the rule for absolute value inequalities: For $|A| > B$ where $B > 0$, the solution is $A < -B$ or $A > B$. 4. Applying this rule to $|x-3| > 3$, we get two inequalities: $$x-3 < -3 \quad \text{or} \quad x-3 > 3$$ 5. Solve each inequality separately: - For $x-3 < -3$: $$x < 0$$ - For $x-3 > 3$: $$x > 6$$ 6. Therefore, the solution set is: $$x < 0 \quad \text{or} \quad x > 6$$ This means any $x$ less than 0 or greater than 6 satisfies the original inequality.