keeping in mind that, an absolute value expression is in effect a piece-wise, then
[tex]\bf |2z-3|=4z-1\implies
\begin{cases}
+(2z-3)=4z-1\\
\qquad 2z-3=4z-1\\
\qquad -2=2z\\
\qquad \frac{-2}{2}=z\\
\qquad -1=z\\
----------\\
-(2z-3)=4z-1\\
\qquad -2z+3=4z-1\\
\qquad 4=6z\\
\qquad \frac{4}{6}=z\\\\
\qquad \frac{2}{3}=z
\end{cases}[/tex]
