Answer
$\left\{-\dfrac{14}{17},5\right\}$
Work Step by Step
Multiplying by the $LCD=
35x(x+2)
,$ the given equation $
\dfrac{1}{x}+\dfrac{2}{x+2}=\dfrac{17}{35}
,$ is equivalent to
\begin{align*}\require{cancel}
35x(x+2)\left(\dfrac{1}{x}+\dfrac{2}{x+2}\right)&=\left(\dfrac{17}{35}\right)35x(x+2)
\\\\
35(x+2)(1)+35x(2)&=17(x)(x+2)
\\
35x+70+70x&=17x^2+34x
\\
105x+70&=17x^2+34x
\\
0&=17x^2+(34x-105x)-70
\\
0&=17x^2-71x-70
\\
17x^2-71x-70&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}
(17x+14)(x-5)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
17x+14=0 & x-5=0
\\
17x=-14 & x=5
\\\\
x=-\dfrac{14}{17}
.\end{array}
Checking the solutions by substitution in the given equation results to
\begin{array}{l|r}
\text{If }x=-\dfrac{14}{17}: & \text{If }x=5:
\\\\
\dfrac{1}{-\frac{14}{17}}+\dfrac{2}{-\frac{14}{17}+2}\overset{?}=\dfrac{17}{35} &
\dfrac{1}{5}+\dfrac{2}{5+2}\overset{?}=\dfrac{17}{35}
\\\\
\dfrac{-17}{14}+\dfrac{2}{-\frac{14}{17}+\frac{34}{17}}\overset{?}=\dfrac{17}{35} &
\dfrac{1}{5}+\dfrac{2}{7}\overset{?}=\dfrac{17}{35}
\\\\
\dfrac{-17}{14}+\dfrac{2}{\frac{20}{17}}\overset{?}=\dfrac{17}{35} &
\dfrac{7}{35}+\dfrac{10}{35}\overset{?}=\dfrac{17}{35}
\\\\
\dfrac{-17}{14}+\dfrac{17}{10}\overset{?}=\dfrac{17}{35} &
\dfrac{17}{35}\overset{\checkmark}=\dfrac{17}{35}
\\\\
\dfrac{-85}{70}+\dfrac{119}{70}\overset{?}=\dfrac{17}{35}
\\\\
\dfrac{34}{70}\overset{?}=\dfrac{17}{35}
\\\\
\dfrac{17}{35}\overset{\checkmark}=\dfrac{17}{35}
.\end{array}
Since both solutions satisfy the given equation, then the solution set of the equation $
\dfrac{1}{x}+\dfrac{2}{x+2}=\dfrac{17}{35}
$ is $\left\{-\dfrac{14}{17},5\right\}$.
