Answer
$c=-\dfrac{1}{3}$
Work Step by Step
Using the properties of equality, the given equation, $
\dfrac{1}{2}|3c+5|=6c+4
,$ is equivalent to
\begin{array}{l}\require{cancel}
2\left( \dfrac{1}{2}|3c+5| \right)=(6c+4)2
\\\\
|3c+5|=12c+8
.\end{array}
Removing the absolute value sign, the expression above is equivalent to
\begin{array}{l}\require{cancel}
3c+5=12c+8
\\\\\text{ OR }\\\\
3c+5=-(12c+8)
.\end{array}
Solving each equation results to
\begin{array}{l}\require{cancel}
3c+5=12c+8
\\
3c-12c=8-5
\\
-9c=3
\\
\dfrac{-9c}{-9}=\dfrac{3}{-9}
\\
c=-\dfrac{1}{3}
\\\\\text{ OR }\\\\
3c+5=-(12c+8)
\\
3c+5=-12c-8
\\
3c+12c=-8-5
\\
15c=-13
\\
\dfrac{15c}{15}=-\dfrac{13}{15}
\\
c=-\dfrac{13}{15}
.\end{array}
Since the right side of the original equation is not a constant, then checking of solutions is required.
Substituting $
c=-\dfrac{1}{3}
$ in the original equation results to
\begin{array}{l}\require{cancel}
\dfrac{1}{2}|3c+5|=6c+4
\\\\
\dfrac{1}{2}\left| 3\left( -\dfrac{1}{3} \right)+5 \right|=6\left( -\dfrac{1}{3} \right)+4
\\\\
\dfrac{1}{2}\left| -1+5 \right|=-2+4
\\\\
\dfrac{1}{2}\left| 4 \right|=2
\\\\
\dfrac{1}{2}(4)=2
\\\\
2=2
\text{ (TRUE)}
.\end{array}
Substituting $
c=-\dfrac{13}{15}
$ in the original equation results to
\begin{array}{l}\require{cancel}
\dfrac{1}{2}|3c+5|=6c+4
\\\\
\dfrac{1}{2}\left| 3\left( -\dfrac{13}{15} \right)+5 \right|=6\left( -\dfrac{13}{15} \right)+4
\\\\
\dfrac{1}{2}\left| -\dfrac{13}{5}+5 \right|=2\left( -\dfrac{13}{5} \right)+4
\\\\
\dfrac{1}{2}\left| -\dfrac{13}{5}+\dfrac{25}{5} \right|=-\dfrac{26}{5}+4
\\\\
\dfrac{1}{2}\left| \dfrac{12}{5}\right|=-\dfrac{26}{5}+\dfrac{20}{4}
\\\\
\dfrac{1}{2}\left( \dfrac{12}{5}\right)=-\dfrac{6}{5}
\\\\
\dfrac{6}{5}=-\dfrac{6}{5}
\text{ (FALSE)}
.\end{array}
Since the substitution above ended with a FALSE statement, then $
c=-\dfrac{13}{15}
$ is not a solution.
Hence, the solution is $
c=-\dfrac{1}{3}
.$
