Answer
$\left\{-\dfrac{1}{3},\dfrac{1}{6}\right\}$
Work Step by Step
Let $z=
\dfrac{1}{3x-1}
$. Then the given equation, $
2+\dfrac{5}{3x-1}=\dfrac{-2}{(3x-1)^2}
,$ is equivalent to
\begin{align*}
2+5\cdot\dfrac{1}{3x-1}&=-2\cdot\dfrac{1}{(3x-1)^2}
\\\\
2+5z&=-2z^2
\\
2z^2+5z+2&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(z+2)(2z+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
z+2=0 & 2z+1=0
\\
z=-2 & 2z=-1
\\\\
& z=-\dfrac{1}{2}
.\end{array}
Since $z=
\dfrac{1}{3x-1}
,$ by back substitution, then
\begin{array}{l|r}
\dfrac{1}{3x-1}=-2 & \dfrac{1}{3x-1}=-\dfrac{1}{2}
.\end{array}
By cross-multiplication and the properties of equality, the equations above are equivalent to
\begin{array}{l|r}
1=-2(3x-1) & 1(2)=(3x-1)(-1)
\\
1=-6x+2 & 2=-3x+1
\\
6x=2-1 & 3x=1-2
\\
6x=1 & 3x=-1
\\\\
x=\dfrac{1}{6} & x=-\dfrac{1}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=\dfrac{1}{6}: & \text{If }x=-\dfrac{1}{3}
\\\\
2+\dfrac{5}{3\left(\frac{1}{6}\right)-1}\overset{?}=\dfrac{-2}{\left(3\left(\frac{1}{6}\right)-1\right)^2} &
2+\dfrac{5}{3\left(-\frac{1}{3}\right)-1}\overset{?}=\dfrac{-2}{\left(3\left(-\frac{1}{3}\right)-1\right)^2}
\\\\
2+\dfrac{5}{\frac{1}{2}-1}\overset{?}=\dfrac{-2}{\left(\frac{1}{2}-1\right)^2} &
2+\dfrac{5}{-1-1}\overset{?}=\dfrac{-2}{\left(-1-1\right)^2}
\\\\
2+\dfrac{5}{-\frac{1}{2}}\overset{?}=\dfrac{-2}{\left(-\frac{1}{2}\right)^2} &
2+\dfrac{5}{-2}\overset{?}=\dfrac{-2}{\left(-2\right)^2}
\\\\
2-10\overset{?}=\dfrac{-2}{\frac{1}{4}} &
\dfrac{4}{2}-\dfrac{5}{2}\overset{?}=\dfrac{-2}{4}
\\\\
-8\overset{\checkmark}=-8 &
-\dfrac{1}{2}\overset{\checkmark}=-\dfrac{1}{2}
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
2+\dfrac{5}{3x-1}=\dfrac{-2}{(3x-1)^2}
$ is $\left\{-\dfrac{1}{3},\dfrac{1}{6}\right\}$.
