Answer
$\left( -\infty, -3 \right] \cup \left[ -1,\infty \right)$
Work Step by Step
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)
.$
