#### Answer

The given system of equations has no solutions. It is inconsistent.

#### Work Step by Step

The procedure is shown with the matrix notation for simpler understanding.
$x_{1} - 3x_{2} + 4x_{3} = -4 $
$3x_{1} - 7x_{2} + 7x_{3} = -8$
$-4x_{1} + 6x_{2} - x_{3} = 7$
This can be depicted in the augmented matrix notation as follows :
$\begin{bmatrix}
1 & -3 & 4 & -4 \\
3 & -7 & 7 & -8 \\
-4 & 6 & -1 & 7
\end{bmatrix}$
To eliminate the $3x_{1}$ term in the second equation, add $-3$ times row 1 to row 2:
$\begin{bmatrix}
1 &-3 & 4 & -4 \\
0 & 2 & -5 & 4 \\
-4 & 6 & -1 &7
\end{bmatrix}$
Next we use the $x_{1}$ term in the first equation to eliminate the $-4x_{1}$ term from the third equation. Add $4$ times row 1 to row 3:
$\begin{bmatrix}
1 & -3 & 4 & -4 \\
0 & 2 & -5 & 4 \\
0 & -6 & 15 & -9
\end{bmatrix}$
Next we use the $2x_{2}$ term in the second equation to eliminate the $6x_{2}$ term from the third equation. Add $3$ times row 2 to row 3:
$\begin{bmatrix}
1 & -3 & 4 & -4 \\
0 & 2 & -5 & 4 \\
0 & 0 & 0 & 3
\end{bmatrix}$
Now the augmented matrix is in a triangular form. For interpreting it, we go back to the equation notation:
$x_{1} + -3x_{2} + 4x_{3} = -4$
$2x_{2} - 5x_{3} = 4 $
$0 = 3$
The equation $0 = 3$ is a short form of $0x_{1} + 0x_{2} + 0x_{3} = 3$.
This is a contradiction, and no values of $ x_{1}, x_{2} $, and $x_{3} $ can satisfy this equation.
Since the final and original equations have the same solution set, we conclude that the given system of equations
$x_{1} - 3x_{2} + 4x_{3} = -4 $
$3x_{1} - 7x_{2} + 7x_{3} = -8$
$-4x_{1} + 6x_{2} - x_{3} = 7$
has no solutions. It is inconsistent.