Answer
$z=\dfrac{2}{3}$
Work Step by Step
The given equation, $
|2z-3|=4z-1
,$ is equivalent to
\begin{array}{l}\require{cancel}
2z-3=4z-1
\\\\\text{OR}\\\\
2z-3=-(4z-1)
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
2z-3=4z-1
\\
2z-4z=-1+3
\\
-2z=2
\\
\dfrac{-2z}{-2}=\dfrac{2}{-2}
\\
z=-1
\\\\\text{OR}\\\\
2z-3=-(4z-1)
\\
2z-3=-4z+1
\\
2z+4z=1+3
\\
6z=4
\\
\dfrac{6z}{6}=\dfrac{4}{6}
\\
z=\dfrac{2}{3}
.\end{array}
Since the right side of the given equation is not a constant, then checking of solution/s is required.
Substituting $
z=-1
$ in the original equation results to
\begin{array}{l}\require{cancel}
|2z-3|=4z-1
\\
|2(-1)-3|=4(-1)-1
\\
|-2-3|=-4-1
\\
|-5|=-5
\\
5=-5
\text{ (FALSE)}
.\end{array}
Since the substitution above resulted in a FALSE statement, then $
z=-1
,$ is not a solution (an extraneous solution).
Substituting $
z=\dfrac{2}{3}
$ in the original equation results to
\begin{array}{l}\require{cancel}
|2z-3|=4z-1
\\
\left|2\left(\dfrac{2}{3}\right)-3\right|=4\left(\dfrac{2}{3}\right)-1
\\
\left|\dfrac{4}{3}-3\right|=\dfrac{8}{3}-1
\\
\left|\dfrac{4}{3}-\dfrac{9}{3}\right|=\dfrac{8}{3}-\dfrac{3}{3}
\\
\left|-\dfrac{5}{3}\right|=\dfrac{5}{3}
\\
\dfrac{5}{3}=\dfrac{5}{3}
\text{ (TRUE)}
.\end{array}
Hence, the solutions $
z=\dfrac{2}{3}
.$
