Find the solution of the following system of equations using Gaussian elimination method and show the steps. 
x-y+z = 4 
5x-y+z = 6 
3x-y+5z= 5 

To solve the system of equations using the Gaussian elimination method, we follow these steps:

### System of Equations:
\[
\begin{align*}
1. & \quad x - y + z = 4 \quad \text{(Equation 1)} \\
2. & \quad 5x - y + z = 6 \quad \text{(Equation 2)} \\
3. & \quad 3x - y + 5z = 5 \quad \text{(Equation 3)}
\end{align*}
\]

### Step 1: Write the Augmented Matrix
First, we convert the system of equations into an augmented matrix.

\[
\begin{pmatrix}
1 & -1 & 1 & \vert & 4 \\
5 & -1 & 1 & \vert & 6 \\
3 & -1 & 5 & \vert & 5 \\
\end{pmatrix}
\]

### Step 2: Apply Row Operations to Get Upper Triangular Matrix

#### 1. Make the first column below the pivot a zero.

We'll use \( R_1 \) to eliminate the \( x \)-terms in the other rows.

- **Operation for \( R_2 \):** \( R_2 \rightarrow R_2 - 5R_1 \)
\[
R_2 = (5 \ -1 \ 1 \ \vert \ 6) - 5(1 \ -1 \ 1 \ \vert \ 4)
\]
\[
R_2 = (5 - 5, \ -1 + 5, \ 1 - 5, \ \vert \ 6 - 20) = (0 \ 4 \ -4 \ \vert \ -14)
\]

- **Operation for \( R_3 \):** \( R_3 \rightarrow R_3 - 3R_1 \)
\[
R_3 = (3 \ -1 \ 5 \ \vert \ 5) - 3(1 \ -1 \ 1 \ \vert \ 4)
\]
\[
R_3 = (3 - 3, \ -1 + 3, \ 5 - 3, \ \vert \ 5 - 12) = (0 \ 2 \ 2 \ \vert \ -7)
\]

The augmented matrix after these operations becomes:
\[
\begin{pmatrix}
1 & -1 & 1 & \vert & 4 \\
0 & 4 & -4 & \vert & -14 \\
0 & 2 & 2 & \vert & -7 \\
\end{pmatrix}
\]

#### 2. Make the second column below the pivot a zero.

We'll use \( R_2 \) to eliminate the \( y \)-term in \( R_3 \).

- **Operation for \( R_3 \):** \( R_3 \rightarrow R_3 - \frac{1}{2}R_2 \)
\[
R_3 = (0 \ 2 \ 2 \ \vert \ -7) - \frac{1}{2}(0 \ 4 \ -4 \ \vert \ -14)
\]
\[
R_3 = (0 \ 2 - 2, \ 2 + 2, \ \vert \ -7 + 7) = (0 \ 0 \ 4 \ \vert \ 0)
\]

The augmented matrix now is:
\[
\begin{pmatrix}
1 & -1 & 1 & \vert & 4 \\
0 & 4 & -4 & \vert & -14 \\
0 & 0 & 4 & \vert & 0 \\
\end{pmatrix}
\]

### Step 3: Back Substitution

Now that we have an upper triangular matrix, we can solve the system using back substitution.

1. **Start with the third row:**
\[
4z = 0 \quad \Rightarrow \quad z = 0
\]

2. **Substitute \( z = 0 \) into the second row:**
\[
4y - 4(0) = -14 \quad \Rightarrow \quad 4y = -14 \quad \Rightarrow \quad y = -\frac{14}{4} = -\frac{7}{2}
\]

3. **Substitute \( y = -\frac{7}{2} \) and \( z = 0 \) into the first row:**
\[
x - \left(-\frac{7}{2}\right) + 0 = 4 \quad \Rightarrow \quad x + \frac{7}{2} = 4 \quad \Rightarrow \quad x = 4 - \frac{7}{2} = \frac{8}{2} - \frac{7}{2} = \frac{1}{2}
\]

### Final Solution:
\[
\boxed{x = \frac{1}{2}, \quad y = -\frac{7}{2}, \quad z = 0}
\]

Would you like any more details or have any questions? 

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Here are some related questions you might find helpful:

1. How does Gaussian elimination differ from other methods for solving systems of equations?
2. Can Gaussian elimination be used for non-square matrices?
3. What happens if a row in the augmented matrix becomes all zeros?
4. How can we use Gaussian elimination to determine if a system has no solutions or infinitely many solutions?
5. How is Gaussian elimination related to matrix inversion?
6. What are some applications of solving systems of linear equations in real life?
7. How does partial pivoting improve the stability of Gaussian elimination?
8. Can Gaussian elimination be performed on complex numbers?

**Tip:** Always check for any row with all zeros except for the last entry in the augmented matrix, which would indicate the system has no solutions.

Learn how to solve a system of equations using the Gaussian elimination method with step-by-step instructions. This problem involves linear algebra concepts and is suitable for high school students studying systems of equations.

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