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Let's solve the given system of linear equations using **Gaussian elimination**. The augmented matrix is:

\[
\begin{pmatrix}
3 & -2 & 4 & | & 3 \\
4 & 3 & 0 & | & 9 \\
2 & 4 & 1 & | & 0
\end{pmatrix}
\]

The steps involved in Gaussian elimination are:

### Step 1: Convert to row-echelon form.
We need to perform row operations to make the matrix upper triangular.

1. **Row 1 stays the same**: \(\text{R1}: (3, -2, 4, 3)\).

2. **Row 2 (R2) update**: Subtract \( \frac{4}{3} \times \text{R1} \) from R2 to eliminate the first element of R2.
   
   \[
   R2 = R2 - \frac{4}{3}R1
   \]
   \[
   R2 = (4, 3, 0, 9) - \frac{4}{3}(3, -2, 4, 3)
   \]
   \[
   R2 = \left( 4 - 4, 3 - \left(-\frac{8}{3}\right), 0 - \frac{16}{3}, 9 - 4 \right)
   \]
   \[
   R2 = (0, \frac{17}{3}, -\frac{16}{3}, 5)
   \]

3. **Row 3 (R3) update**: Subtract \( \frac{2}{3} \times \text{R1} \) from R3 to eliminate the first element of R3.

   \[
   R3 = R3 - \frac{2}{3}R1
   \]
   \[
   R3 = (2, 4, 1, 0) - \frac{2}{3}(3, -2, 4, 3)
   \]
   \[
   R3 = \left( 2 - 2, 4 - \left(-\frac{4}{3}\right), 1 - \frac{8}{3}, 0 - 2 \right)
   \]
   \[
   R3 = \left( 0, \frac{16}{3}, -\frac{5}{3}, -2 \right)
   \]

The new matrix is:

\[
\begin{pmatrix}
3 & -2 & 4 & | & 3 \\
0 & \frac{17}{3} & -\frac{16}{3} & | & 5 \\
0 & \frac{16}{3} & -\frac{5}{3} & | & -2
\end{pmatrix}
\]

### Step 2: Eliminate the second element in the third row.

To do this, subtract \( \frac{16}{17} \times \text{R2} \) from R3:

\[
R3 = R3 - \frac{16}{17}R2
\]
\[
R3 = \left( 0, \frac{16}{3}, -\frac{5}{3}, -2 \right) - \frac{16}{17} \left( 0, \frac{17}{3}, -\frac{16}{3}, 5 \right)
\]
\[
R3 = \left( 0, 0, \frac{27}{51}, -\frac{67}{51} \right)
\]

Now the matrix is:

\[
\begin{pmatrix}
3 & -2 & 4 & | & 3 \\
0 & \frac{17}{3} & -\frac{16}{3} & | & 5 \\
0 & 0 & 1 & | & -\frac{5}{3}
\end{pmatrix}
\]

### Step 3: Back-substitution.

We can now solve the system by back-substitution.

1. From row 3: \( x_3 = -\frac{5}{3} \).

2. Substitute \( x_3 = -\frac{5}{3} \) into row 2:
   \[
   \frac{17}{3} x_2 - \frac{16}{3} \left( -\frac{5}{3} \right) = 5
   \]
   \[
   \frac{17}{3} x_2 + \frac{80}{9} = 5
   \]
   \[
   x_2 = \frac{5 \cdot 9 - 80}{17} = \frac{-35}{17} = -\frac{35}{17}
   \]

3. Substitute \( x_2 \) and \( x_3 \) into row 1:
   \[
   3 x_1 - 2 \left( -\frac{35}{17} \right) + 4 \left( -\frac{5}{3} \right) = 3
   \]
   Solving this gives \( x_1 \).

Let me calculate \( x_1 \).The solution for \( x_1 \) is approximately \( x_1 = 1.85 \).

Thus, the solutions to the system of equations are:

\[
x_1 \approx 1.85, \quad x_2 \approx -\frac{35}{17}, \quad x_3 = -\frac{5}{3}
\]

Would you like further details or clarifications on any of the steps?

Here are 5 related questions you could explore:
1. What is the difference between Gaussian elimination and Gauss-Jordan elimination?
2. How can Gaussian elimination be applied to systems with more variables?
3. What are some common pivoting strategies to avoid division by zero in Gaussian elimination?
4. How can you perform Gaussian elimination when dealing with non-square matrices?
5. Can Gaussian elimination be used for matrices with complex numbers?

**Tip:** When applying Gaussian elimination, always ensure the pivot element is non-zero. If it is zero, row swapping is needed.

Resuelva el sistema de ecuaciones mostrado utilizando el método de Gauss.

Learn how to solve a system of linear equations using Gaussian elimination. This problem involves converting a matrix to row-echelon form and performing back-substitution to find the values of x1, x2, and x3. The solution provides step-by-step calculations and is suitable for students in high school or early college studying linear algebra.

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