Answer
$\left( -\infty,\dfrac{26}{9} \right)\cup \left( \dfrac{34}{9},\infty \right)
$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given inequality, $
\left|\dfrac{5}{3}-\dfrac{1}{2}x \right|\gt\dfrac{2}{9}
,$ 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}
\dfrac{5}{3}-\dfrac{1}{2}x\gt\dfrac{2}{9}
\\\\\text{OR}\\\\
\dfrac{5}{3}-\dfrac{1}{2}x\lt-\dfrac{2}{9}
.\end{array}
Using the properties of inequality, the inequality above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{5}{3}-\dfrac{1}{2}x\gt\dfrac{2}{9}
\\\\
18\left(\dfrac{5}{3}-\dfrac{1}{2}x\right)\gt\left(\dfrac{2}{9}\right)18
\\\\
30-9x\gt4
\\\\
-9x\gt4-30
\\\\
-9x\gt-26
\\\\
x\lt\dfrac{-26}{-9}
\\\\
x\lt\dfrac{26}{9}
\\\\\text{OR}\\\\
\dfrac{5}{3}-\dfrac{1}{2}x\lt-\dfrac{2}{9}
\\\\
18\left(\dfrac{5}{3}-\dfrac{1}{2}x\right)\lt\left(-\dfrac{2}{9}\right)18
\\\\
30-9x\lt-4
\\\\
-9x\lt-4-30
\\\\
-9x\lt-34
\\\\
x\gt\dfrac{-34}{-9}
\\\\
x\gt\dfrac{34}{9}
.\end{array}
Note that division by a negative number entails reversing the inequality symbol.
Hence, the solution is the interval $
\left( -\infty,\dfrac{26}{9} \right)\cup \left( \dfrac{34}{9},\infty \right)
.$
