College Algebra, Chapter 1, 1.7, Section 1.7, Problem 40

Solve the inequality $3-|2x+4| \leq 1$. Express the answer using interval notation.

$
\begin{equation}
\begin{aligned}
3-|2x+4| &\leq 1\\
\\
-|2x+4| &\leq -2 && \text{Subtract 3}
\end{aligned}
\end{equation}
$



We have,


$
\begin{equation}
\begin{aligned}
-(2x+4) &\leq -2 && \text{and}& -(-(2x+4)) &\leq -2 && \text{Divide each side by -1}\\
\\
2x +4 &\geq 2 && \text{and}& -(2x+4) &\geq 2 && \text{Divide by -1}\\
\\
2x + 4 &\geq 2 && \text{and}& 2x+4 &\leq -2 && \text{Subtract 4}\\
\\
2x &\geq -2 && \text{and}& 2x &\leq -6 && \text{Divide by 2}\\
\\
x &\geq -1 && \text{and}& x &\leq -3
\end{aligned}
\end{equation}
$


The solution set is $(-\infty,-3] \bigcup [-1,\infty)$