[tex]\begin{gathered} 4|x+3|\ge8\text{ First, divide both sides by 4} \\ \frac{4}{4}|x+3|\ge\frac{8}{4} \\ |x+3|\ge2\text{ Then, transform into the following equations} \\ x+3\ge2\text{ and } \\ x+3\leq-2\text{ } \\ \text{Then, solve each equation.} \end{gathered}[/tex][tex]\begin{gathered} x+3\ge2\text{ Solving first equation} \\ x\ge2-3\text{ Isolating x} \\ x\ge-1 \\ \end{gathered}[/tex][tex]\begin{gathered} x+3\leq-2\text{ Solving second equation} \\ x\leq-2-3\text{ } \\ x\leq-5 \end{gathered}[/tex]
Based on the solutions of the equations we find that the answer is the union of the following intervals.
(-∞ , -5]∪[-1, ∞+)
