Answer
$\text{the interval }
\left( \dfrac{7}{8},\dfrac{27}{20} \right]
$
Work Step by Step
Using compound inequalities, the solution to the given inequality, $
\dfrac{1}{6} \lt \dfrac{4x-3}{3} \le \dfrac{4}{5}
,$ is
\begin{array}{l}\require{cancel}
\dfrac{1}{6}\cdot3 \lt \dfrac{4x-3}{3}\cdot3 \le \dfrac{4}{5}\cdot3
\\\\
\dfrac{1}{2} \lt 4x-3 \le \dfrac{12}{5}
\\\\
\dfrac{1}{2}+3 \lt 4x-3+3 \le \dfrac{12}{5}+3
\\\\
\dfrac{7}{2} \lt 4x \le \dfrac{27}{5}
\\\\
\dfrac{\dfrac{7}{2}}{4} \lt \dfrac{4}{4}x \le \dfrac{\dfrac{27}{5}}{4}
\\\\
\dfrac{7}{8} \lt x \le \dfrac{27}{20}
.\end{array}
In interval notation, the solution is $
\text{the interval }
\left( \dfrac{7}{8},\dfrac{27}{20} \right]
.$
