You have a system of three linear equations:

\[
\begin{aligned}
1. & \quad 2x + y + z = 18 \\
2. & \quad x + 2y + z = 18 \\
3. & \quad 3x + y + 2z = 23
\end{aligned}
\]

To solve this system, we'll use the method of substitution or elimination. Here's a step-by-step solution using the elimination method.

### Step 1: Subtract the second equation from the first
Subtract equation 2 from equation 1:

\[
(2x + y + z) - (x + 2y + z) = 18 - 18
\]

This simplifies to:

\[
x - y = 0 \quad \text{or} \quad x = y
\]

### Step 2: Subtract the first equation from the third
Subtract equation 1 from equation 3:

\[
(3x + y + 2z) - (2x + y + z) = 23 - 18
\]

This simplifies to:

\[
x + z = 5
\]

### Step 3: Substitute \( x = y \) into the second equation
Using \( x = y \) in equation 2:

\[
x + 2x + z = 18 \quad \text{(since \( y = x \))}
\]

Simplifying:

\[
3x + z = 18
\]

### Step 4: Solve for \( x \), \( y \), and \( z \)
Now, we have the following simplified system:

\[
\begin{aligned}
x &= y \\
x + z &= 5 \\
3x + z &= 18
\end{aligned}
\]

Subtract the second equation from the third:

\[
(3x + z) - (x + z) = 18 - 5
\]

This simplifies to:

\[
2x = 13 \quad \Rightarrow \quad x = \frac{13}{2} = 6.5
\]

So:

\[
y = 6.5
\]

Now substitute \( x = 6.5 \) into \( x + z = 5 \):

\[
6.5 + z = 5 \quad \Rightarrow \quad z = 5 - 6.5 = -1.5
\]

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

\[
x = 6.5, \quad y = 6.5, \quad z = -1.5
\]

Would you like more details, or do you have any questions?

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Here are 5 related questions:
1. How would the solution change if the first equation was \(2x + y + z = 20\)?
2. Can you solve a system with 4 variables and 4 equations using the elimination method?
3. What does it mean if a system of linear equations has no solution?
4. How do you determine if a system of equations is dependent?
5. What happens if two of the equations in a system are identical?

**Tip:** For systems with more equations or variables, matrix methods like Gaussian elimination can be very effective!

Learn how to solve a system of three linear equations: 2x + y + z = 18, x + 2y + z = 18, and 3x + y + 2z = 23 using the elimination and substitution methods. This solution provides a step-by-step approach suitable for students in Grades 10-12.

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