x − y + z = 1 Equation 1
2x − y − 2z = 3 Equation 2
x + 3y − 3z = −1 solve using elimination

We are given the system of equations:

\[
\text{Equation 1: } x - y + z = 1
\]
\[
\text{Equation 2: } 2x - y - 2z = 3
\]
\[
\text{Equation 3: } x + 3y - 3z = -1
\]

### Step 1: Eliminate one variable (let's eliminate \( y \))

First, subtract **Equation 1** from **Equation 2** to eliminate \( y \):

\[
(2x - y - 2z) - (x - y + z) = 3 - 1
\]
Simplifying:
\[
2x - y - 2z - x + y - z = 2
\]
\[
x - 3z = 2 \quad \text{(Equation 4)}
\]

Next, subtract **Equation 1** from **Equation 3** to eliminate \( y \):

\[
(x + 3y - 3z) - (x - y + z) = -1 - 1
\]
Simplifying:
\[
x + 3y - 3z - x + y - z = -2
\]
\[
4y - 4z = -2
\]
Dividing by 2:
\[
2y - 2z = -1 \quad \text{(Equation 5)}
\]

### Step 2: Solve the system of two equations

Now we have the reduced system:

1. \( x - 3z = 2 \) (Equation 4)
2. \( 2y - 2z = -1 \) (Equation 5)

#### Solve for \( x \) in terms of \( z \) from **Equation 4**:
\[
x = 3z + 2
\]

#### Solve for \( y \) in terms of \( z \) from **Equation 5**:
\[
2y = 2z - 1
\]
\[
y = z - \frac{1}{2}
\]

### Step 3: Substitute into the original equation

Now that we have \( x = 3z + 2 \) and \( y = z - \frac{1}{2} \), substitute these into **Equation 1** to find \( z \):

\[
x - y + z = 1
\]
\[
(3z + 2) - \left(z - \frac{1}{2}\right) + z = 1
\]
Simplifying:
\[
3z + 2 - z + \frac{1}{2} + z = 1
\]
\[
3z + \frac{5}{2} = 1
\]
\[
3z = 1 - \frac{5}{2}
\]
\[
3z = -\frac{3}{2}
\]
\[
z = -\frac{1}{2}
\]

### Step 4: Solve for \( x \) and \( y \)

Now that we know \( z = -\frac{1}{2} \), we can substitute this value into the expressions for \( x \) and \( y \):

\[
x = 3z + 2 = 3\left(-\frac{1}{2}\right) + 2 = -\frac{3}{2} + 2 = \frac{1}{2}
\]
\[
y = z - \frac{1}{2} = -\frac{1}{2} - \frac{1}{2} = -1
\]

### Final Solution:

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

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How can substitution be used to solve this system?
2. What happens if you choose to eliminate \( z \) instead of \( y \)?
3. How does elimination compare to solving systems graphically?
4. Can this method be applied to larger systems of equations?
5. What real-world problems can be modeled by systems of equations?

**Tip**: Always double-check each step of elimination to avoid sign errors!

Explore how to solve a system of linear equations using the elimination method with the equations x − y + z = 1, 2x − y − 2z = 3, and x + 3y − 3z = −1. This guide provides a step-by-step breakdown, focusing on algebraic concepts suitable for students in Grades 9-11.

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