Give the linear systems:
(i) $x_1+2x_2=2\\3x_1+7x_2=8$
(ii) $x_1+2x_2=1\\3x_1+7x_2=7$
Solve both systems by incorporating the right-hand sides into a $2\times2$ matrix $B$ and computing the reduced row echelon form of
$$[A|B]=\begin{bmatrix}1&2&\Big|&2&1\\3&7&\Big|&8&7\end{bmatrix}$$
Both two sides of this form is $2\times2$, so I have no idea how to reduce the row echelon form of this one.
Since the coefficient matrix for both systems is $A=\begin{bmatrix}1&2\\3&7\end{bmatrix}$, we can solve them simultaneously. This is because the row transformations required to be done on $A$ to convert it into its row echelon form will be the same for both systems.
$\begin{bmatrix}1&2&\Big|&2&1\\3&7&\Big|&8&7\end{bmatrix}\xrightarrow{R_2\to R_2-3R_1}\begin{bmatrix}1&2&\Big|&2&1\\0&1&\Big|&2&4\end{bmatrix}\xrightarrow{R_1\to R_1-2R_2}\begin{bmatrix}1&0&\Big|&-2&-7\\0&1&\Big|&2&4\end{bmatrix}$
Now the solution for the first system is given $$\begin{bmatrix}1&0&\Big|&-2\\0&1&\Big|&2\end{bmatrix}$$ and the solution of the second system is given by $$\begin{bmatrix}1&0&\Big|&-7\\0&1&\Big|&4\end{bmatrix}$$