#### Answer

$\left[ 1,\dfrac{11}{3} \right]$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
\left|7-3x \right|\le4
,$ 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 $a\gt0,$ $|x|\le a$ implies $-a\le x\le a.$ (Note that the symbol $\le$ may be replaced with $\lt.$) Hence, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-4\le 7-3x \le4
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
-4-7\le 7-7-3x \le4-7
\\\\
-11\le -3x \le-3
.\end{array}
Dividing both sides by $
-3
$ and reversing the inequality symbols, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{-11}{-3}\ge \dfrac{-3x}{{-3}} \ge\dfrac{-3}{-3}
\\\\
\dfrac{11}{3}\ge x \ge1
\\\\
1\le x \le\dfrac{11}{3}
.\end{array}
Hence, the solution is the interval $
\left[ 1,\dfrac{11}{3} \right]
.$