https://file.aiquickdraw.com/mathbot/file/1729362119013095jz6b2.jpg

The system of linear equations given in the image is:

\[
\begin{aligned}
1) &\ 2x_1 + 3x_2 - x_3 = 5 \\
2) &\ 4x_1 + x_2 + 2x_3 = 11 \\
3) &\ -2x_1 + 5x_2 + 3x_3 = 0 \\
\end{aligned}
\]

The problem asks to solve this system using Gaussian elimination, and three steps (Part 1, Part 2, Part 3) are listed:

### Part 1: \( R_2 \leftarrow R_2 - 2 \times R_1 \)
This step performs an elementary row operation to eliminate the \(x_1\) term in the second row. Let's calculate this:

- Multiply row 1 by 2: \( 2(2x_1 + 3x_2 - x_3 = 5) \) gives \( 4x_1 + 6x_2 - 2x_3 = 10 \).
- Subtract this from row 2: 
  \[
  (4x_1 + x_2 + 2x_3 = 11) - (4x_1 + 6x_2 - 2x_3 = 10)
  \]
  This results in:
  \[
  0x_1 - 5x_2 + 4x_3 = 1
  \]
  So the new system is:
  \[
  \begin{aligned}
  2x_1 + 3x_2 - x_3 &= 5 \\
  -5x_2 + 4x_3 &= 1 \\
  -2x_1 + 5x_2 + 3x_3 &= 0 \\
  \end{aligned}
  \]

### Part 2: \( R_3 \leftarrow R_3 + R_1 \)
This step eliminates \(x_1\) from the third row by adding row 1 to row 3.

- Add row 1 to row 3: 
  \[
  (-2x_1 + 5x_2 + 3x_3 = 0) + (2x_1 + 3x_2 - x_3 = 5)
  \]
  This gives:
  \[
  0x_1 + 8x_2 + 2x_3 = 5
  \]
  So the system is now:
  \[
  \begin{aligned}
  2x_1 + 3x_2 - x_3 &= 5 \\
  -5x_2 + 4x_3 &= 1 \\
  8x_2 + 2x_3 &= 5 \\
  \end{aligned}
  \]

### Part 3: \( R_3 \leftarrow R_3 - \frac{8}{5} \times R_2 \)
This step eliminates the \(x_2\) term from the third row by scaling row 2 and subtracting it from row 3.

- Multiply row 2 by \( \frac{8}{5} \): 
  \[
  \frac{8}{5}(-5x_2 + 4x_3 = 1) \Rightarrow -8x_2 + \frac{32}{5}x_3 = \frac{8}{5}
  \]
- Subtract this from row 3:
  \[
  (8x_2 + 2x_3 = 5) - (-8x_2 + \frac{32}{5}x_3 = \frac{8}{5})
  \]
  This results in:
  \[
  0x_2 + \left(2 - \frac{32}{5}\right)x_3 = 5 - \frac{8}{5}
  \]
  Simplifying:
  \[
  0x_2 - \frac{22}{5}x_3 = \frac{17}{5}
  \]
  Multiply by \( -\frac{5}{22} \) to solve for \(x_3\):
  \[
  x_3 = \frac{17}{22}
  \]

### Back Substitution:
Now, substitute \(x_3 = \frac{17}{22}\) into the second equation:
\[
-5x_2 + 4 \times \frac{17}{22} = 1
\]
This simplifies to:
\[
-5x_2 + \frac{68}{22} = 1 \Rightarrow -5x_2 = 1 - \frac{68}{22} = \frac{22}{22} - \frac{68}{22} = -\frac{46}{22}
\]
Thus:
\[
x_2 = \frac{46}{110} = \frac{23}{55}
\]

Finally, substitute \(x_2 = \frac{23}{55}\) and \(x_3 = \frac{17}{22}\) into the first equation:
\[
2x_1 + 3 \times \frac{23}{55} - \frac{17}{22} = 5
\]
Simplifying:
\[
2x_1 + \frac{69}{55} - \frac{17}{22} = 5
\]
Getting a common denominator:
\[
2x_1 + \frac{276}{220} - \frac{170}{220} = 5 \Rightarrow 2x_1 + \frac{106}{220

Solve the following system of linear equations using Gaussian elimination:
2x1 + 3x2 - x3 = 5
4x1 + x2 + 2x3 = 11
-2x1 + 5x2 + 3x3 = 0

Learn how to solve a system of linear equations using the Gaussian elimination method. This problem involves performing elementary row operations and back substitution to find the solutions of the equations. Suitable for college-level linear algebra students.

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