Answer
$\left\{-\dfrac{7}{3},\dfrac{1}{3}\right\}$
Work Step by Step
Multiplying by the $LCD=
(3z+2)^2
,$ the given equation $
1+\dfrac{2}{3z+2}=\dfrac{15}{(3z+2)^2}
,$ is equivalent to
\begin{align*}\require{cancel}
(3z+2)^2\left(1+\dfrac{2}{3z+2}\right)&=\left(\dfrac{15}{(3z+2)^2}\right)(3z+2)^2
\\\\
(3z+2)^2(1)+(3z+2)(2)&=15(1)
\\
(9z^2+12z+4)+(3z+2)(2)&=15
&(\text{use }(a+b)^2=a^2+2ab+b^2)
\\
9z^2+12z+4+(6z+4)&=15
&(\text{use Distributive Property})
\\
9z^2+(12z+6z)+(4+4-15)&=0
\\
9z^2+18z-7&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(3z+7)(3z-1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
3z+7=0 & 3z-1=0
\\
3z=-7 & 3z=1
\\\\
z=-\dfrac{7}{3} & z=\dfrac{1}{3}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }z=-\dfrac{7}{3}: & \text{If }z=\dfrac{1}{3}:
\\\\
1+\dfrac{2}{3\left(-\frac{7}{3}\right)+2}\overset{?}=\dfrac{15}{(3\left(-\frac{7}{3}\right)+2)^2} &
1+\dfrac{2}{3\left(\frac{1}{3}\right)+2}\overset{?}=\dfrac{15}{(3\left(\frac{1}{3}\right)+2)^2}
\\\\
1+\dfrac{2}{-7+2}\overset{?}=\dfrac{15}{(-7+2)^2} &
1+\dfrac{2}{1+2}\overset{?}=\dfrac{15}{(1+2)^2}
\\\\
1+\dfrac{2}{-5}\overset{?}=\dfrac{15}{(-5)^2} &
1+\dfrac{2}{3}\overset{?}=\dfrac{15}{(3)^2}
\\\\
1-\dfrac{2}{5}\overset{?}=\dfrac{15}{25} &
1+\dfrac{2}{3}\overset{?}=\dfrac{15}{9}
\\\\
\dfrac{5}{5}-\dfrac{2}{5}\overset{?}=\dfrac{\cancelto3{15}}{\cancelto5{25}} &
\dfrac{3}{3}+\dfrac{2}{3}\overset{?}=\dfrac{\cancelto5{15}}{\cancelto39}
\\\\
\dfrac{3}{5}\overset{\checkmark}=\dfrac{3}{5} &
\dfrac{5}{3}\overset{\checkmark}=\dfrac{5}{3}
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
1+\dfrac{2}{3z+2}=\dfrac{15}{(3z+2)^2}
$ is $\left\{-\dfrac{7}{3},\dfrac{1}{3}\right\}$.
