#### Answer

$\left( -\infty,-\dfrac{2}{3} \right)\cup \left( 4,\infty \right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
\left|5-3x \right|\gt7
,$ remove the absolute value sign using the properties of absolute value inequality. Then use the properties of inequality to isolate the variable.
$\bf{\text{Solution Details:}}$
For any $a\gt0,$ $|x|\ge a$ implies $x\ge a$ OR $x\le -a$ (Note that the symbols $\ge$ may be replaced with $\gt$ and the symbol $\le$ with $\lt.$) Hence, the inequality above is equivalent to \begin{array}{l}\require{cancel} 5-3x\gt7 \\\\\text{OR}\\\\ 5-3x\lt-7 .\end{array}
Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 5-3x\gt7 \\\\ -3x\gt7-5 \\\\ -3x\gt2
\\\\\text{OR}\\\\ 5-3x\lt-7 \\\\ -3x\lt-7-5 \\\\ -3x\lt-12
.\end{array}
Dividing by a negative number and consequently reversing the sign results to
\begin{array}{l}\require{cancel}
x\lt\dfrac{2}{-3}
\\\\
x\lt-\dfrac{2}{3}
\\\\\text{OR}\\\\
x\gt\dfrac{-12}{-3}
\\\\
x\gt4
.\end{array}
Hence, the solution is the interval $
\left( -\infty,-\dfrac{2}{3} \right)\cup \left( 4,\infty \right)
.$