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To solve this system of linear equations, we can use methods such as substitution, elimination, or matrix techniques (like Gaussian elimination). The equations are:

1. \( x_1 - 2x_2 + x_3 = 0 \)
2. \( 2x_2 - 8x_3 = 8 \)
3. \( -4x_1 + 5x_2 + 9x_3 = -9 \)

Let's go through the steps to find the values of \( x_1 \), \( x_2 \), and \( x_3 \).

### Step 1: Simplify Equation (2)
Equation (2) can be simplified by dividing through by 2:
\[
x_2 - 4x_3 = 4
\]
This gives us:
\[
x_2 = 4 + 4x_3
\]

### Step 2: Substitute \( x_2 \) in Equation (1)
Now we substitute \( x_2 = 4 + 4x_3 \) into Equation (1):
\[
x_1 - 2(4 + 4x_3) + x_3 = 0
\]
Expanding and simplifying:
\[
x_1 - 8 - 8x_3 + x_3 = 0
\]
\[
x_1 - 8 - 7x_3 = 0
\]
\[
x_1 = 8 + 7x_3
\]

### Step 3: Substitute \( x_1 = 8 + 7x_3 \) and \( x_2 = 4 + 4x_3 \) into Equation (3)
Now we use Equation (3):
\[
-4x_1 + 5x_2 + 9x_3 = -9
\]
Substituting \( x_1 = 8 + 7x_3 \) and \( x_2 = 4 + 4x_3 \):
\[
-4(8 + 7x_3) + 5(4 + 4x_3) + 9x_3 = -9
\]
Expanding each term:
\[
-32 - 28x_3 + 20 + 20x_3 + 9x_3 = -9
\]
Combine like terms:
\[
-12 + x_3 = -9
\]
\[
x_3 = 3
\]

### Step 4: Substitute \( x_3 = 3 \) back to find \( x_2 \) and \( x_1 \)
Now that we know \( x_3 = 3 \), we can find \( x_2 \) and \( x_1 \):

1. Substitute \( x_3 = 3 \) into \( x_2 = 4 + 4x_3 \):
   \[
   x_2 = 4 + 4(3) = 4 + 12 = 16
   \]

2. Substitute \( x_3 = 3 \) into \( x_1 = 8 + 7x_3 \):
   \[
   x_1 = 8 + 7(3) = 8 + 21 = 29
   \]

### Solution
The solution to the system is:
\[
x_1 = 29, \quad x_2 = 16, \quad x_3 = 3
\]

Would you like more details on any of these steps?

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Here are some related questions that build upon this solution:
1. How would the solution change if the constants in the equations were different?
2. What other methods could be used to solve this system, such as matrix inversion?
3. How can you verify the solution of a system of equations?
4. How would you handle a system where there is no solution or infinitely many solutions?
5. How can Gaussian elimination be applied to solve this system?

**Tip:** When solving systems of equations, it’s helpful to first simplify each equation as much as possible, to make substitution and elimination easier.

Solve the given system of equations: x1 - 2x2 + x3 = 0, 2x2 - 8x3 = 8, -4x1 + 5x2 + 9x3 = -9.

This example demonstrates the solution of a 3x3 system of linear equations: x1 - 2x2 + x3 = 0, 2x2 - 8x3 = 8, and -4x1 + 5x2 + 9x3 = -9. Follow a step-by-step solution using substitution and Gaussian elimination techniques, ideal for students in Grades 11-12.

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