Answer
$R=\dfrac{E-Ir}{I}$
Work Step by Step
By multiplying both sides by the $LCD=
R+r
$, the given equation, $
I=\dfrac{E}{R+r}
,$ is equivalent to
\begin{array}{l}\require{cancel}
I(R+r)=E
.\end{array}
Using the properties of equality, then, in terms of $
R
,$ the equation above is equivalent to
\begin{array}{l}\require{cancel}
IR+Ir=E
\\\\
IR=E-Ir
\\\\
R=\dfrac{E-Ir}{I}
.\end{array}
