College Algebra, Chapter 1, 1.7, Section 1.7, Problem 44

Solve the inequality $\displaystyle 2\left| \frac{1}{2}x + 3 \right| +3 \leq 51$. Express the answer using interval notation.

$
\begin{equation}
\begin{aligned}
2\left| \frac{1}{2}x + 3 \right| +3 &\leq 51\\
\\
2\left| \frac{1}{2}x + 3 \right| &\leq 48 && \text{Subtract 3}\\
\\
\left| \frac{1}{2}x + 3 \right| &\leq 24 && \text{Divide by 2}
\end{aligned}
\end{equation}
$



We have,


$
\begin{equation}
\begin{aligned}
\frac{1}{2}x + 3 &\leq 24 && \text{and}& -\left( \frac{1}{2}x + 3 \right) &\leq 24 && \text{Divide each side by -1}\\
\\
\frac{1}{2}x + 3 &\leq 24 && \text{and}& \frac{1}{2}x + 3 &\geq -24 && \text{Subtract 3}\\
\\
\frac{1}{2}x &\leq 21 && \text{and}& \frac{1}{2}x &\geq - 27 && \text{Multiply by 2}\\
\\
x &\leq 42 && \text{and}& x &\geq -54
\end{aligned}
\end{equation}
$


The solution set is $[-54,42]$