We are given the following system of linear equations:

\[
\begin{aligned}
1. &\quad x - y + z = -1 \\
2. &\quad x + y + 3z = -3 \\
3. &\quad 2x - y + 2z = 0
\end{aligned}
\]

We will solve this system using either substitution or elimination. Let's proceed step-by-step with the elimination method.

### Step 1: Eliminate \( y \) from the first two equations

Add equations (1) and (2):

\[
(x - y + z) + (x + y + 3z) = (-1) + (-3)
\]

Simplifying:

\[
2x + 4z = -4
\]

This gives us a new equation:

\[
4. \quad 2x + 4z = -4 \quad \text{or} \quad x + 2z = -2
\]

### Step 2: Eliminate \( y \) from the first and third equations

Now, subtract equation (1) from equation (3):

\[
(2x - y + 2z) - (x - y + z) = 0 - (-1)
\]

Simplifying:

\[
x + z = 1
\]

This gives us another new equation:

\[
5. \quad x + z = 1
\]

### Step 3: Solve the system of two equations

Now we have the system:

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

Subtract equation (5) from equation (4):

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

Simplifying:

\[
z = -3
\]

### Step 4: Substitute \( z = -3 \) back into equation (5)

Substitute into \( x + z = 1 \):

\[
x - 3 = 1
\]

Solving for \( x \):

\[
x = 4
\]

### Step 5: Substitute \( x = 4 \) and \( z = -3 \) into equation (1)

Substitute into \( x - y + z = -1 \):

\[
4 - y - 3 = -1
\]

Simplifying:

\[
1 - y = -1
\]

Solving for \( y \):

\[
y = 2
\]

### Final Solution

The solution to the system is:

\[
x = 4, \quad y = 2, \quad z = -3
\]

Would you like further details on any of the steps?

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Here are 5 related questions you can try:
1. Solve the system \( x + 2y - z = 1 \), \( 2x - y + z = 3 \), \( x - y - z = -1 \).
2. How can the substitution method be used to solve this system of equations?
3. What are the benefits of using the elimination method in solving systems of equations?
4. What would happen if one of the equations in a system is a multiple of another?
5. Can you solve a system of equations graphically, and how would it look for this case?

**Tip**: For systems of linear equations, checking the consistency of your solution (by substituting back) is a good habit to ensure accuracy.

x - y + z = -1; x + y + 3z = -3; 2x - y + 2z = 0

Learn how to solve the system of linear equations x - y + z = -1, x + y + 3z = -3, and 2x - y + 2z = 0 using the elimination and substitution methods. This problem demonstrates key concepts in algebra and linear systems, suitable for Grades 9-11.

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