Answer
$x \lt -\dfrac{43}{24} \text{ or } x \gt \dfrac{9}{8}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given inqeuality, $
\left| \dfrac{1+3x}{5} \right| \gt \dfrac{7}{8}
,$ use the definition of a greater than (greater than or equal to) absolute value inequality and solve each resulting inequality. Finally, graph the solution set.
In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$
$\bf{\text{Solution Details:}}$
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}
\dfrac{1+3x}{5} \gt \dfrac{7}{8}
\\\\\text{OR}\\\\
\dfrac{1+3x}{5} \lt -\dfrac{7}{8}
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
\dfrac{1+3x}{5} \gt \dfrac{7}{8}
\\\\
40\cdot\dfrac{1+3x}{5} \gt 40\cdot\dfrac{7}{8}
\\\\
8\cdot(1+3x) \gt 5\cdot7
\\\\
8+24x \gt 35
\\\\
24x \gt 35-8
\\\\
24x \gt 27
\\\\
x \gt \dfrac{27}{24}
\\\\
x \gt \dfrac{9}{8}
\\\\\text{OR}\\\\
\dfrac{1+3x}{5} \lt -\dfrac{7}{8}
\\\\
40\cdot\dfrac{1+3x}{5} \lt 40\cdot\left( -\dfrac{7}{8} \right)
\\\\
8\cdot(1+3x) \lt 5\cdot\left( -7 \right)
\\\\
8+24x \lt -35
\\\\
24x \lt -35-8
\\\\
24x \lt -43
\\\\
x \lt -\dfrac{43}{24}
.\end{array}
Hence, the solution set is $
x \lt -\dfrac{43}{24} \text{ or } x \gt \dfrac{9}{8}
.$
