#### Answer

$\left( -\infty,1 \right)\cup \left( \dfrac{11}{3},\infty \right)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $ \left|7-3x \right|\gt4 ,$ 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} 7-3x\gt4 \\\\\text{OR}\\\\ 7-3x\lt-4 .\end{array}
Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} 7-3x\gt4 \\\\ -3x\gt4-7 \\\\
-3x\gt-3
\\\\\text{OR}\\\\ 7-3x\lt-4 \\\\ -3x\lt-4-7 \\\\ -3x\lt-11
.\end{array}
Dividing by a negative number and consequently reversing the sign results to
\begin{array}{l}\require{cancel}
x\lt\dfrac{-3}{-3}
\\\\
x\lt1
\\\\\text{OR}\\\\
-3x\lt-11
\\\\
x\gt\dfrac{-11}{-3}
\\\\
x\gt\dfrac{11}{3}
.\end{array}
Hence, the solution is the interval $
\left( -\infty,1 \right)\cup \left( \dfrac{11}{3},\infty \right)
.$