Answer
$\left\{-\dfrac{2}{3},1\right\}$
Work Step by Step
Multiplying by the $LCD=
t^2
,$ the given equation $
3-\dfrac{1}{t}=\dfrac{2}{t^2}
,$ is equivalent to
\begin{align*}\require{cancel}
t^2\left(3-\dfrac{1}{t}\right)&=\left(\dfrac{2}{t^2}\right)t^2
\\\\
t^2(3)+t(-1)&=2(1)
\\\\
3t^2-t&=2
\\
3t^2-t-2&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}
(3t+2)(t-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
3t+2=0 & t-1=0
\\
3t=-2 & t=1
\\\\
t=-\dfrac{2}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }t=-\dfrac{2}{3}: & \text{If }t=1:
\\\\
3-\dfrac{1}{-2/3}\overset{?}=\dfrac{2}{(-2/3)^2} &
3-\dfrac{1}{1}\overset{?}=\dfrac{2}{1^2}
\\\\
3+\dfrac{3}{2}\overset{?}=\dfrac{2}{4/9} &
3-1\overset{?}=\dfrac{2}{1}
\\\\
\dfrac{6}{2}+\dfrac{3}{2}\overset{?}=\dfrac{9}{2} &
2\overset{\checkmark}=2
\\\\
\dfrac{9}{2}\overset{\checkmark}=\dfrac{9}{2}
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
3-\dfrac{1}{t}=\dfrac{2}{t^2}
$ is $\left\{-\dfrac{2}{3},1\right\}$.
