To solve the given system of equations:

\[
\begin{aligned}
1. & \quad x + y - 2z = 5 \\
2. & \quad -x + 2y + z = 2 \\
3. & \quad 2x + 3y - z = 9
\end{aligned}
\]

### Step 1: Choose a method to solve the system

We'll use the **substitution** and **elimination** method to find the values of \(x\), \(y\), and \(z\).

### Step 2: Solve for \(x\) from Equation (1)
The first equation is:

\[
x + y - 2z = 5
\]

Rearrange this equation to solve for \(x\):

\[
x = 5 - y + 2z
\]

### Step 3: Substitute \(x = 5 - y + 2z\) into Equations (2) and (3)
Now, substitute this expression for \(x\) into the second and third equations.

For the second equation:
\[
-x + 2y + z = 2
\]
Substitute \(x = 5 - y + 2z\):
\[
-(5 - y + 2z) + 2y + z = 2
\]
Simplify:
\[
-5 + y - 2z + 2y + z = 2
\]
Combine like terms:
\[
3y - z = 7
\]
This is our new equation (4).

For the third equation:
\[
2x + 3y - z = 9
\]
Substitute \(x = 5 - y + 2z\):
\[
2(5 - y + 2z) + 3y - z = 9
\]
Simplify:
\[
10 - 2y + 4z + 3y - z = 9
\]
Combine like terms:
\[
y + 3z = -1
\]
This is our new equation (5).

### Step 4: Solve the system of equations from (4) and (5)

Now, solve the system of two equations:

\[
\begin{aligned}
3y - z & = 7 \quad \text{(Equation 4)} \\
y + 3z & = -1 \quad \text{(Equation 5)}
\end{aligned}
\]

#### Step 4a: Solve for \(z\) from Equation (5)
From Equation (5), solve for \(y\):

\[
y = -1 - 3z
\]

#### Step 4b: Substitute \(y = -1 - 3z\) into Equation (4)
Substitute this into Equation (4):

\[
3(-1 - 3z) - z = 7
\]
Simplify:
\[
-3 - 9z - z = 7
\]
Combine like terms:
\[
-10z = 10
\]
Solve for \(z\):
\[
z = -1
\]

### Step 5: Substitute \(z = -1\) back into the equation for \(y\)
Substitute \(z = -1\) into \(y = -1 - 3z\):

\[
y = -1 - 3(-1) = -1 + 3 = 2
\]

### Step 6: Substitute \(y = 2\) and \(z = -1\) back into the equation for \(x\)
Substitute \(y = 2\) and \(z = -1\) into \(x = 5 - y + 2z\):

\[
x = 5 - 2 + 2(-1) = 5 - 2 - 2 = 1
\]

### Final Solution
The solution to the system of equations is:

\[
x = 1, \quad y = 2, \quad z = -1
\]

This matches our earlier result.

If you have any questions or need further clarification, feel free to ask!

Here are five related questions for further exploration:
1. How would the steps change if you started solving for \(y\) instead of \(x\)?
2. What happens when the system is inconsistent (i.e., has no solution)?
3. Can Gaussian elimination be used to solve this system? How?
4. How does the determinant of the coefficient matrix relate to the existence of a solution?
5. What if one equation was a multiple of another—how would that affect the solution?

**Tip:** Always keep track of signs when performing substitution and elimination steps to avoid mistakes.

Learn how to solve the system of linear equations x + y - 2z = 5, -x + 2y + z = 2, 2x + 3y - z = 9 using substitution and elimination methods. This problem involves understanding systems of linear equations and applying algebraic methods to find the values of x, y, and z. Suitable for high school students.

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