## Precalculus: Mathematics for Calculus, 7th Edition

$\left( -\infty, -3 \right] \cup \left[ -1,\infty \right)$
Using the properties of inequality, the given expression, $3-\left| 2x+4 \right|\le1 ,$ is equivalent to \begin{array}{l}\require{cancel} -\left| 2x+4 \right|\le1-3 \\\\ -\left| 2x+4 \right|\le-2 \\\\ \left| 2x+4 \right|\ge2 .\end{array} For any $a\gt0$, $|x|\gt a$ implies $x\gt a \text{ OR } x\lt -a$. (The symbol $\gt$ may be replaced with $\ge$.) Using the concept above, the solutions to the given inequality, $\left| 2x+4 \right|\ge2 ,$ is \begin{array}{l}\require{cancel} 2x+4\ge2 \\\\ 2x\ge2-4 \\\\ 2x\ge-2 \\\\ x\ge-\dfrac{2}{2} \\\\ x\ge-1 ,\\\\\text{ OR }\\\\ 2x+4\le-2 \\\\ 2x\le-2-4 \\\\ 2x\le-6 \\\\ x\le-\dfrac{6}{2} \\\\ x\le-3 .\end{array} In interval notation, the solution set is $\left( -\infty, -3 \right] \cup \left[ -1,\infty \right) .$