Answer
$\text{the interval }
\left( -\dfrac{1}{2},2 \right)
$
Work Step by Step
Since, for any positive number $c$, $|x|\lt c$ implies $-c \lt x \lt c$, then the solution of the given inequality, $
\left| \dfrac{4x-3}{5} \right|\lt 1
,$ is
\begin{array}{l}\require{cancel}
-1 \lt \dfrac{4x-3}{5} \lt 1
\\\\
-1\cdot5 \lt \dfrac{4x-3}{5}\cdot5 \lt 1\cdot5
\\\\
-5 \lt 4x-3 \lt 5
\\\\
-5+3 \lt 4x-3+3 \lt 5+3
\\\\
-2 \lt 4x \lt 8
\\\\
-\dfrac{2}{4} \lt \dfrac{4}{4}x \lt \dfrac{8}{4}
\\\\
-\dfrac{1}{2} \lt x \lt 2
.\end{array}
In interval notation, the solution is $
\text{the interval }
\left( -\dfrac{1}{2},2 \right)
.$