Answer
$\left( -\infty,-\dfrac{11}{7}
\right)\cup\left( -\dfrac{5}{7},\infty \right)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
|7x+8|-6\gt-3
,$ use the properties of inequality to isolate the absolute value expression. Then use the definition of absolute value inequalities.
$\bf{\text{Solution Details:}}$
Using the properties of inequality, the given expression is equivalent to
\begin{array}{l}\require{cancel}
|7x+8|\gt-3+6
\\\\
|7x+8|\gt3
.\end{array}
Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
7x+8\gt3
\\\\\text{OR}\\\\
7x+8\lt-3
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
7x+8\gt3
\\\\
7x\gt3-8
\\\\
7x\gt-5
\\\\
x\gt\dfrac{-5}{7}
\\\\
x\gt-\dfrac{5}{7}
\\\\\text{OR}\\\\
7x+8\lt-3
\\\\
7x\lt-3-8
\\\\
7x\lt-11
\\\\
x\lt-\dfrac{11}{7}
.\end{array}
Hence, the solution set is the interval $
\left( -\infty,-\dfrac{11}{7}
\right)\cup\left( -\dfrac{5}{7},\infty \right)
.$
