Answer
solution set: $\mathbb{R}$
Work Step by Step
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: $(-\infty,\infty)$
