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The system of linear equations provided is:

\[
\begin{aligned}
1. \quad &x_1 - 3x_2 + 4x_3 = -4 \\
2. \quad &3x_1 - 7x_2 + 7x_3 = -8 \\
3. \quad &-4x_1 + 6x_2 - x_3 = 7
\end{aligned}
\]

We can solve this system using methods such as substitution, elimination, or matrix techniques like Gaussian elimination. I'll proceed with Gaussian elimination.

### Step 1: Write the augmented matrix
\[
\begin{pmatrix}
1 & -3 & 4 & | & -4 \\
3 & -7 & 7 & | & -8 \\
-4 & 6 & -1 & | & 7
\end{pmatrix}
\]

### Step 2: Perform row operations to reach row-echelon form
- Let's make the pivot of the first column 1, which is already the case in row 1.

- Now, eliminate the values below this pivot in rows 2 and 3 by subtracting suitable multiples of row 1.

1. \( R_2 \leftarrow R_2 - 3R_1 \)
\[
R_2 = \begin{pmatrix} 3 & -7 & 7 & | & -8 \end{pmatrix} - 3 \times \begin{pmatrix} 1 & -3 & 4 & | & -4 \end{pmatrix}
= \begin{pmatrix} 0 & 2 & -5 & | & 4 \end{pmatrix}
\]

2. \( R_3 \leftarrow R_3 + 4R_1 \)
\[
R_3 = \begin{pmatrix} -4 & 6 & -1 & | & 7 \end{pmatrix} + 4 \times \begin{pmatrix} 1 & -3 & 4 & | & -4 \end{pmatrix}
= \begin{pmatrix} 0 & -6 & 15 & | & -9 \end{pmatrix}
\]

The matrix now becomes:
\[
\begin{pmatrix}
1 & -3 & 4 & | & -4 \\
0 & 2 & -5 & | & 4 \\
0 & -6 & 15 & | & -9
\end{pmatrix}
\]

### Step 3: Eliminate below the second pivot
We want to eliminate the value below the second pivot, which is the -6 in row 3. First, let's make the second pivot equal to 1 by dividing row 2 by 2.

\[
R_2 \leftarrow \frac{1}{2} R_2 = \begin{pmatrix} 0 & 1 & -\frac{5}{2} & | & 2 \end{pmatrix}
\]

Now eliminate the -6 in row 3:
\[
R_3 \leftarrow R_3 + 6R_2 = \begin{pmatrix} 0 & -6 & 15 & | & -9 \end{pmatrix} + 6 \times \begin{pmatrix} 0 & 1 & -\frac{5}{2} & | & 2 \end{pmatrix}
= \begin{pmatrix} 0 & 0 & 0 & | & 3 \end{pmatrix}
\]

The matrix becomes:
\[
\begin{pmatrix}
1 & -3 & 4 & | & -4 \\
0 & 1 & -\frac{5}{2} & | & 2 \\
0 & 0 & 0 & | & 3
\end{pmatrix}
\]

### Step 4: Interpretation
The third row becomes:
\[
0 = 3
\]
This is a contradiction, meaning the system of equations has **no solution**.

Would you like further details on this process, or do you have any questions?

Here are 5 related questions you might consider:
1. What are the different methods to solve a system of linear equations?
2. How does Gaussian elimination work for systems with unique, infinite, or no solutions?
3. What is the difference between row-echelon and reduced row-echelon form?
4. Can matrix inverses be used to solve this system?
5. What are consistent and inconsistent systems in linear algebra?

**Tip:** Always check for contradictions when solving systems of equations, as they indicate no solutions or inconsistencies.

Solve the system of linear equations: x₁ - 3x₂ + 4x₃ = -4, 3x₁ - 7x₂ + 7x₃ = -8, -4x₁ + 6x₂ - x₃ = 7.

This problem involves solving a system of linear equations using Gaussian elimination. Learn the step-by-step process of converting the system into row-echelon form and identifying if a solution exists. This problem shows how inconsistencies can indicate no solutions in linear systems. Suitable for high school students in Grades 10-12.

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