College Algebra, Chapter 1, 1.7, Section 1.7, Problem 42

Solve the inequality $7|x+2| + 5 > 4$. Express the answer using interval notation.

$
\begin{equation}
\begin{aligned}
7|x+2| + 5 &> 4\\
\\
7|x+5| &> -1 && \text{Subtract 5}\\
\\
|x+5| &> - \frac{1}{7} && \text{Divide by 7}
\end{aligned}
\end{equation}
$



We have,


$
\begin{equation}
\begin{aligned}
x+5 &> -\frac{1}{7} && \text{and}& -(x+5) &> - \frac{1}{7} && \text{Divide each side by -1}\\
\\
x+5 &> -\frac{1}{7} && \text{and}& x+5 &< \frac{1}{7} && \text{Subtract 5}\\
\\
x &> \frac{34}{7} && \text{and}& x &< -\frac{34}{7}
\end{aligned}
\end{equation}
$


The solution set is $\displaystyle \left( -\infty, -\frac{34}{7}\right] \bigcup \left[ \frac{34}{7}, \infty \right)$