Answer
$\left\{-\dfrac{1}{2},3\right\}$
Work Step by Step
Let $a=
(z-1)^{-1}
$. Then the given equation, $
2-6(z-1)^{-2}=(z-1)^{-1}
,$ is equivalent to
\begin{align*}\require{cancel}
2-6a^2&=a
\\
0&=6a^2+a-2
\\
6a^2+a-2=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(2a-1)(3a+2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
2a-1=0 & 3a+2=0
\\
2a=1 & 3a=-2
\\\\
a=\dfrac{1}{2} & a=-\dfrac{2}{3}
.\end{array}
Since $a=
(z-1)^{-1}=\dfrac{1}{z-1}
,$ by back substitution, then
\begin{array}{l|r}
\dfrac{1}{z-1}=\dfrac{1}{2} & \dfrac{1}{z-1}=-\dfrac{2}{3}
.\end{array}
By cross-multiplication and the properties of equality, the equations above are equivalent to
\begin{array}{l|r}
1(2)=1(z-1) & 1(3)=-2(z-1)
\\
2=z-1 & 3=-2z+2
\\
2+1=z & 2z=2-3
\\
3=z & 2z=-1
\\\\
z=3 & z=-\dfrac{1}{2}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }z=3: & \text{If }z=-\dfrac{1}{2}
\\\\
2-6(3-1)^{-2}\overset{?}=(3-1)^{-1} &
2-6\left(-\frac{1}{2}-1\right)^{-2}\overset{?}=\left(-\frac{1}{2}-1\right)^{-1}
\\\\
2-6(2)^{-2}\overset{?}=(2)^{-1} &
2-6\left(-\frac{3}{2}\right)^{-2}\overset{?}=\left(-\frac{3}{2}\right)^{-1}
\\\\
2-\dfrac{6}{2^2}\overset{?}=\dfrac{1}{2^{1}} &
2-6\left(-\frac{2}{3}\right)^{2}\overset{?}=\left(-\frac{2}{3}\right)^{1}
\\\\
2-\dfrac{6}{4}\overset{?}=\dfrac{1}{2} &
2-6\left(\frac{4}{9}\right)\overset{?}=-\frac{2}{3}
\\\\
2-\dfrac{\cancelto36}{\cancelto24}\overset{?}=\dfrac{1}{2} &
2-\cancelto26\left(\frac{4}{\cancelto39}\right)\overset{?}=-\frac{2}{3}
\\\\
\dfrac{4}{2}-\dfrac{3}{2}\overset{?}=\dfrac{1}{2} &
2-2\left(\frac{4}{3}\right)\overset{?}=-\frac{2}{3}
\\\\
\dfrac{1}{2}\overset{\checkmark}=\dfrac{1}{2} &
\dfrac{6}{3}-\dfrac{8}{3}\overset{?}=-\frac{2}{3}
\\\\
&
-\dfrac{2}{3}\overset{\checkmark}=-\frac{2}{3}
.\end{array}
Since all solutions satisfy the original equation, then the solution set of the equation $
2-6(z-1)^{-2}=(z-1)^{-1}
$ is $\left\{-\dfrac{1}{2},3\right\}$.
