## Precalculus (6th Edition) Blitzer

solution set: $\mathbb{R}$
Rewrite as $|x-\displaystyle \frac{11}{3}|+\frac{7}{3} \gt 1\qquad$... $/\times 3$ $3|x-\displaystyle \frac{11}{3}|+7 \gt 3\qquad$... $/-7$ $|3x-11| \gt -4$ Whatever x is, the absolute value is always going to be greater than a negative number, so all real x values are solutions. Solution set: $(-\infty,\infty)$