First, build a matrix representation of $S$ (Definition MR). We are free to choose whatever bases we wish, so we should choose ones that are easy to work with, such as
\begin{align*}
B&=\set{1,\,x}\\
C&=\set{\begin{bmatrix}1 & 0\end{bmatrix},\,\begin{bmatrix}0 & 1\end{bmatrix}}
\end{align*}
The resulting matrix representation is then
\begin{equation*}
\matrixrep{T}{B}{C}=
\begin{bmatrix}
3 & 1\\
2 & 1
\end{bmatrix}
\end{equation*}
This matrix is invertible since it has a nonzero determinant, so by Theorem IMR the linear transformation $S$ is invertible. We can use the matrix inverse and Theorem IMR to find a formula for the inverse linear transformation,
\begin{align*}
\lt{\ltinverse{S}}{\begin{bmatrix}a&b\end{bmatrix}}
&=\vectrepinv{B}{\matrixrep{\ltinverse{S}}{C}{B}\vectrep{C}{\begin{bmatrix}a&b\end{bmatrix}}}&&\knowl{./knowls/theorem.FTMR.knowl}{\text{Theorem FTMR}}\\
&=\vectrepinv{B}{\inverse{\left(\matrixrep{S}{B}{C}\right)}\vectrep{C}{\begin{bmatrix}a&b\end{bmatrix}}}&&\knowl{./knowls/theorem.IMR.knowl}{\text{Theorem IMR}}\\
&=\vectrepinv{B}{\inverse{\left(\matrixrep{S}{B}{C}\right)}\colvector{a\\b}}&&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}}\\
&=\vectrepinv{B}{
\inverse{\left(
\begin{bmatrix}
3 & 1\\
2 & 1
\end{bmatrix}
\right)}
\colvector{a\\b}}\\
&=\vectrepinv{B}{
\begin{bmatrix}
1 & -1\\
-2 & 3
\end{bmatrix}
\colvector{a\\b}}&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\
&=\vectrepinv{B}{
\colvector{
a-b\\
-2a+3b
}}&&\knowl{./knowls/definition.MVP.knowl}{\text{Definition MVP}}\\
&=(a-b)+(-2a+3b)x&&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}}
\end{align*}