Answer
$R_2=\dfrac{RR_1}{R_1-R}$
Work Step by Step
By multiplying both sides by the $LCD=
RR_1R_2
$, the given equation, $
\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}
,$ is equivalent to
\begin{array}{l}\require{cancel}
R_1R_2(1)=RR_2(1)+RR_1(1)
\\\\=
R_1R_2=RR_2+RR_1
.\end{array}
Using the properties of equality, then, in terms of $
R_2
,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
R_1R_2-RR_2=RR_1
\\\\
R_2(R_1-R)=RR_1
\\\\
R_2=\dfrac{RR_1}{R_1-R}
.\end{array}
