For example, let's say we have $-2x+y+z=5$, $2x+2y=4$, and $3y+4z=2$
We want to solve
$$ \begin{bmatrix} -2 & 1 & 1\\ 2 & 2 & 0\\ 0 & 3 & 4\\ \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} =\begin{bmatrix} 5 \\ 4\\ 2 \end{bmatrix} $$
To find $x, y$ and $z$ we'd put the following augmented matrix into reduced row echelon form:
\begin{bmatrix} -2 & 1 & 1 &5 \\ 2 & 2 & 0 &4\\ 0 & 3 & 4&2 \end{bmatrix}
And then read off the values of $x, y$ and $z$
But what if instead we left the $x, y, z$ vector in the augmented matrix (instead of essentially invisibly multiplying it with matrix $A$)? i.e. what if we put this matrix into RREF:
\begin{equation*} A = \begin{bmatrix} -2 & 1 & 1 &x \\ 2 & 2 & 0 &y\\ 0 & 3 & 4&z \end{bmatrix} \end{equation*}
Doing so results in the matrix
\begin{equation*} A = \begin{bmatrix} 1 & 0 & 0 & \frac y2 - \frac{x+y}3 + \frac{z-x-y}9 \\ 0 & 1 & 0 & \frac{x+y}3 - \frac{z-x-y}9 \\ 0 & 0 & 1& \frac{z-x-y}3 \end{bmatrix} \end{equation*}
Do those expressions in the $4$th column involving $x, y$ and $z$ have any significance?
Well, this just means you are letting $\begin{bmatrix} x \\ y \\ z\end{bmatrix}$ play the role of $\begin{bmatrix}5 \\ 4 \\ 2\end{bmatrix}$. So, you are solving for solutions to $$\begin{bmatrix} -2 & 1 & 1\\ 2 & 2 & 0\\ 0 & 3 & 4\\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \end{bmatrix} =\begin{bmatrix} x \\ y\\ z \end{bmatrix}$$ where you are thinking of $x,y,$ and $z$ as constants and $a,b,$ and $c$ as the variables to solve for. The solution you found is just the general expression for $a,b,$ and $c$ in terms of $x,y,$ and $z$.
(This is essentially equivalent to inverting the matrix $B=\begin{bmatrix} -2 & 1 & 1\\ 2 & 2 & 0\\ 0 & 3 & 4\\ \end{bmatrix} $, the solution to $B\begin{bmatrix} a \\ b \\ c\end{bmatrix}=\begin{bmatrix} x \\ y \\ z\end{bmatrix}$ is just $\begin{bmatrix} a \\ b \\ c\end{bmatrix}=B^{-1}\begin{bmatrix} x \\ y \\ z\end{bmatrix}$, by multiplying both sides by $B^{-1}$ on the left.)