#### Answer

The system is consistent.

#### Work Step by Step

The procedure is shown with the matrix notation for simpler understanding.
$x_{1} + 3x_{3} = 2 $
$x_{2} - 3x_{4} = 3$
$-2x_{2} + 3x_{3} + 2x_{4} = 1$
$3x_{1} + 7x_{4} = -5$
This can be depicted in the augmented matrix notation as follows :
$\begin{bmatrix}
1 & 0 & 3 & 0 & 2\\
0 & 1 & 0 & -3 & 3\\
0 & -2 & 3 & 2 & 1\\
3 & 0 & 0 & 7 & -5
\end{bmatrix}$
To get an $x_{1}$ term in the second row, interchange row 2 and row 4:
$\begin{bmatrix}
1 & 0 & 3 & 0 & 2\\
3 & 0 & 0 & 7 & -5\\
0 & 1 & 0 & -3 & 3\\
0 & -2 & 3 & 2 & 1
\end{bmatrix}$
To eliminate the $3x_{1}$ term in the second row, add $-3$ times row 1 to row 2:
$\begin{bmatrix}
1 & 0 & 3 & 0 & 2\\
0 & 0 & -9 & 7 & -11\\
0 & 1 & 0 & -3 & 3\\
0 & -2 & 3 & 2 & 1
\end{bmatrix}$
To get an $x_{2}$ term in the second row, interchange row 2 and row 3:
$\begin{bmatrix}
1 & 0 & 3 & 0 & 2\\
0 & 1 & 0 & -3 & 3\\
0 & 0 & -9 & 7 & -11\\
0 & -2 & 3 & 2 & 1
\end{bmatrix}$
To get an $x_{2}$ term in the third row, interchange row 3 and row 4:
$\begin{bmatrix}
1 & 0 & 3 & 0 & 2\\
0 & 1 & 0 & -3 & 3\\
0 & -2 & 3 & 2 & 1\\
0 & 0 & -9 & 7 & -11
\end{bmatrix}$
To eliminate the $-2x_{2}$ term in the third row, add $2$ times row 2 to row 3:
$\begin{bmatrix}
1 & 0 & 3 & 0 & 2\\
0 & 1 & 0 & -3 & 3\\
0 & 0 & 3 & -4 & 7\\
0 & 0 & -9 & 7 & -11
\end{bmatrix}$
To eliminate the $-9x_{3}$ term in the fourth row, add $3$ times row 3 to row 4:
$\begin{bmatrix}
1 & 0 & 3 & 0 & 2\\
0 & 1 & 0 & -3 & 3\\
0 & 0 & 3 & -4 & 7\\
0 & 0 & 0 & -5 & 10
\end{bmatrix}$
In the equation form,
$x_{1} + 3x_{3} = 2 $
$x_{2} - 3x_{4} = 3$
$3x_{3} - 4x_{4} = 7$
$-5x_{4} = 10$
Now, we know the value of $x_{4}$. Back substituting this value into the other equations, we can subsequently find the values of $x_{3}$, $x_{2}$ and $x_{1}$.
Hence, a solution exists. Thus, the system is consistent.