Answer
$(-\infty,-6)\cup(0,\infty)$
Work Step by Step
Step 1. From $|\frac{2x+6}{3}|\gt2$, we have$|x+3|\gt3$ which gives $x+3\gt3$ or $x+3\lt-3$; thus the solutions are $x\lt-6\cup x\gt0$ and in interval notation $(-\infty,-6)\cup(0,\infty)$
Step 2. See graph for the interval on a number line.
