#### Answer

$\color{blue}{\left\{\frac{11}{27}, \frac{25}{27}\right\}}$

#### Work Step by Step

Add $9$ to both sides of the equation to obtain:
$\left|\dfrac{7}{2-3x}\right|-9+9=0+9
\\\left|\dfrac{7}{2-3x}\right|=9$
RECALL:
$|x|=a \longrightarrow x=a \text{ or } x = -a$
Use the rule above to have:
$\left|\dfrac{7}{2-3x}\right|= 9 \longrightarrow \dfrac{7}{2-3x}=9 \text{ or } \dfrac{7}{2-3x} = -9$
Solve each equation to obtain:
$\dfrac{7}{2-3x}=9
\\7=9(2-3x)
\\7=9(2)-9(3x)
\\7=18-27x
\\7-18=-27x
\\-11=-27x
\\-\frac{11}{-27}=-\frac{27x}{-27}
\\\frac{11}{27}=x$
or
$\dfrac{7}{2-3x}=-9
\\7=-9(2-3x)
\\7=-9(2)-(-9)(3x)
\\7=-18+27x
\\7+18=27x
\\25=27x
\\\frac{25}{27}=\frac{27x}{27}
\\\frac{25}{27}=x$
Thus, the solution set is $\color{blue}{\left\{\frac{11}{27}, \frac{25}{27}\right\}}$.