We choose $\vect{x}$ to be any vector we like. A particularly cocky choice would be to choose $\vect{x}=\zerovector$, but we will instead choose
\begin{equation*}
\vect{x}= \begin{bmatrix} 2 & -1 \\ -1 & 4 \end{bmatrix}
\end{equation*}
Then $\lt{R}{\vect{x}}=9+9x$. Now compute the kernel of $R$, which by Theorem KILT we expect to be nontrivial. Setting $\lt{R}{\begin{bmatrix}a&b\\b&c\end{bmatrix}}$ equal to the zero vector, $\zerovector=0+0x$, and equating coefficients leads to a homogeneous system of equations. Row-reducing the coefficient matrix of this system will allow us to determine the values of $a$, $b$ and $c$ that create elements of the null space of $R$,
\begin{equation*}
\begin{bmatrix}
2 & -1 & 1 \\
1 & 1 & 2
\end{bmatrix}
\rref
\begin{bmatrix}
\leading{1} & 0 & 1 \\
0 & \leading{1} & 1
\end{bmatrix}
\end{equation*}
We only need a single element of the null space of this coefficient matrix, so we will not compute a precise description of the whole null space. Instead, choose the free variable $c=2$. Then
\begin{equation*}
\vect{z}=\begin{bmatrix} -2 & -2 \\ -2 & 2\end{bmatrix}
\end{equation*}
is the corresponding element of the kernel. We compute the desired $\vect{y}$ as
\begin{equation*}
\vect{y}=\vect{x}+\vect{z}=
\begin{bmatrix} 2 & -1 \\ -1 & 4 \end{bmatrix}
+
\begin{bmatrix} -2 & -2 \\ -2 & 2\end{bmatrix}
=
\begin{bmatrix} 0 & -3 \\ -3 & 6 \end{bmatrix}
\end{equation*}
Then check that $\lt{R}{\vect{y}}=9+9x$.