Answer
$\left\{-\dfrac{1}{2},1\right\}$
Work Step by Step
Multiplying both sides by the $LCD=
n(n+1)
,$ the given equation, $
\dfrac{1}{n}+\dfrac{2}{n+1}=2
$, is equivalent to
\begin{align*}
n(n+1)\left(\dfrac{1}{n}+\dfrac{2}{n+1}\right)&=2(n)(n+1)
\\\\
(n+1)(1)+n(2)&=2n^2+2n
\\
n+1+2n&=2n^2+2n
\\
0&=2n^2+(2n-n-2n)-1
\\
0&=2n^2-n-1
\\
2n^2-n-1&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(n-1)(2n+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
n-1=0 & 2n+1=0
\\
n=1 & 2n=-1
\\
& n=-\dfrac{1}{2}
.\end{array}
Hence, the solution set of the equation $
\dfrac{1}{n}+\dfrac{2}{n+1}=2
$ is $
\left\{-\dfrac{1}{2},1\right\}
$.
