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