#### Answer

The system is consistent. Infinitely many solutions will exist.

#### Work Step by Step

The procedure is shown with the matrix notation for simpler understanding.
$x_{1} - 2x_{4} = -3$
$2x_{2} + 2x_{3} = 0$
$x_{3} + 3x_{4} = 1$
$-2x_{1} + 3x_{2} + 2x_{3} + x_{4} = 5$
This can be depicted in the augmented matrix notation as follows :
$\begin{bmatrix}
1 & 0 & 0 & -2 & -3\\
0 & 2 & 2 & 0 & 0\\
0 & 0 & 1& 3 & 1\\
-2 & 3 & 2 & 1 & 5
\end{bmatrix}$
To eliminate the $-2x_{1}$ term in the fourth row, add $2$ times row 1 to row 4:
$\begin{bmatrix}
1 & 0 & 0 & -2 & -3\\
0 & 2 & 2 & 0 & 0\\
0 & 0 & 1& 3 & 1\\
0 & 3 & 2 & -3 & -1
\end{bmatrix}$
To eliminate the $3x_{2}$ term in the fourth row, add $-3/2$ times row 2 to row 4:
$\begin{bmatrix}
1 & 0 & 0 & -2 & -3\\
0 & 2 & 2 & 0 & 0\\
0 & 0 & 1& 3 & 1\\
0 & 0 & -1 & -3 & -1
\end{bmatrix}$
To eliminate the $-x_{3}$ term in the fourth row, add $1$ times row 3 to row 4:
$\begin{bmatrix}
1 & 0 & 0 & -2 & -3\\
0 & 2 & 2 & 0 & 0\\
0 & 0 & 1& 3 & 1\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}$
$x_{1} - 2x_{4} = -3 $
$2x_{2} + 2x_{3} = 0$
$x_{3} + 3x_{4} = 1$
$0x_{4} = 0$
Now, we know that the last equation is true for infinite values of $x_{4}$. Back substituting these values into the other equations, we can find that there will be infinite subsequent values of $x_{3}$, $x_{2}$ and $x_{1}$ for each value of $x_{4}$.
Hence, infinitely many solutions exist. Thus, the system is consistent.