## Precalculus (6th Edition) Blitzer

The solution set is $\left( -\infty ,-\frac{5}{3} \right]\cup \left[ \frac{1}{3},\infty \right)$.
Considered the inequality, $\left| 3x+2 \right|\ge 3$ , Now, apply the absolute value inequality: $3x+2\le -3$ Or $3x+2\ge 3$ , Now, subtract 2 on both sides, \begin{align} & 3x+2-2\le -3-2 \\ & 3x\le -5 \end{align} Or \begin{align} & 3x+2-2\ge 3-2 \\ & 3x\ge 1 \end{align} , Divide 3 on both sides, \begin{align} & \frac{3x}{3}\le -\frac{5}{3} \\ & x\le -\frac{5}{3} \end{align} Or \begin{align} & \frac{3x}{3}\ge \frac{1}{3} \\ & x\ge \frac{1}{3} \end{align} So, the solution set is $\left( -\infty ,-\frac{5}{3} \right]\cup \left[ \frac{1}{3},\infty \right)$. Graph: The solution set consists of all real numbers greater than or equal to $-\frac{5}{3}$ and greater than or equal to $\frac{13}{2}$. The graph on the number line is shown below.