solve the following system of equations using matrices and elementary row operations. x+y=6, 2x+4y+z=5, 2x+3y+z=6

To solve the system of equations using matrices and elementary row operations, follow these steps:

### System of Equations:
1. \( x + y = 6 \)
2. \( 2x + 4y + z = 5 \)
3. \( 2x + 3y + z = 6 \)

### Step 1: Write the augmented matrix
The augmented matrix for this system is:

\[
\begin{bmatrix}
1 & 1 & 0 & 6 \\
2 & 4 & 1 & 5 \\
2 & 3 & 1 & 6 \\
\end{bmatrix}
\]

### Step 2: Perform Elementary Row Operations
1. **Make the first pivot 1**: The pivot in the first row is already 1.

2. **Eliminate the first column below the pivot**:
   - Subtract \( 2 \times \text{Row 1} \) from Row 2:
     \[
     R2 = R2 - 2R1
     \]
   - Subtract \( 2 \times \text{Row 1} \) from Row 3:
     \[
     R3 = R3 - 2R1
     \]

   Resulting matrix:
   \[
   \begin{bmatrix}
   1 & 1 & 0 & 6 \\
   0 & 2 & 1 & -7 \\
   0 & 1 & 1 & -6 \\
   \end{bmatrix}
   \]

3. **Make the second pivot 1**:
   Divide Row 2 by 2:
   \[
   R2 = \frac{R2}{2}
   \]
   Resulting matrix:
   \[
   \begin{bmatrix}
   1 & 1 & 0 & 6 \\
   0 & 1 & \frac{1}{2} & -\frac{7}{2} \\
   0 & 1 & 1 & -6 \\
   \end{bmatrix}
   \]

4. **Eliminate the second column below and above the pivot**:
   - Subtract Row 2 from Row 3:
     \[
     R3 = R3 - R2
     \]
   - Subtract \( 1 \times \text{Row 2} \) from Row 1:
     \[
     R1 = R1 - R2
     \]

   Resulting matrix:
   \[
   \begin{bmatrix}
   1 & 0 & -\frac{1}{2} & \frac{5}{2} \\
   0 & 1 & \frac{1}{2} & -\frac{7}{2} \\
   0 & 0 & \frac{1}{2} & \frac{1}{2} \\
   \end{bmatrix}
   \]

5. **Make the third pivot 1**:
   Multiply Row 3 by 2:
   \[
   R3 = 2R3
   \]
   Resulting matrix:
   \[
   \begin{bmatrix}
   1 & 0 & -\frac{1}{2} & \frac{5}{2} \\
   0 & 1 & \frac{1}{2} & -\frac{7}{2} \\
   0 & 0 & 1 & 1 \\
   \end{bmatrix}
   \]

6. **Eliminate the third column above the pivot**:
   - Add \( \frac{1}{2} \times \text{Row 3} \) to Row 1:
     \[
     R1 = R1 + \frac{1}{2}R3
     \]
   - Subtract \( \frac{1}{2} \times \text{Row 3} \) from Row 2:
     \[
     R2 = R2 - \frac{1}{2}R3
     \]

   Resulting matrix:
   \[
   \begin{bmatrix}
   1 & 0 & 0 & 3 \\
   0 & 1 & 0 & -4 \\
   0 & 0 & 1 & 1 \\
   \end{bmatrix}
   \]

### Step 3: Interpret the Solution
The final augmented matrix corresponds to the equations:
\[
x = 3, \; y = -4, \; z = 1
\]

### Final Solution:
\[
\boxed{x = 3, \; y = -4, \; z = 1}
\]

Would you like more details on the process or further explanations?

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**Related Questions:**
1. How do you verify solutions for a system of equations?
2. Can systems of equations have no solutions or infinite solutions?
3. How does row reduction differ from other methods like substitution?
4. What are real-world applications of solving systems of equations?
5. How does matrix rank relate to solving systems of equations?

**Tip:** When solving systems, always double-check your augmented matrix setup to avoid mistakes in the row operations.

Explore how to solve the system of linear equations x + y = 6, 2x + 4y + z = 5, and 2x + 3y + z = 6 using matrix representation and Gaussian elimination. This step-by-step guide demonstrates row reduction to find the solution: x = 3, y = -4, z = 1. Suitable for high school students learning linear algebra.

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