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The system of equations provided is:

\[
\begin{aligned}
x + 2y + z &= 9 \\
x - y + 3z &= 13 \\
2z &= 10
\end{aligned}
\]

### Step 1: Solve for \( z \)
From the third equation:
\[
2z = 10 \implies z = \frac{10}{2} = 5
\]

### Step 2: Substitute \( z = 5 \) into the first and second equations
Substitute \( z = 5 \) into the first equation:
\[
x + 2y + 5 = 9 \implies x + 2y = 9 - 5 = 4 \quad \text{(Equation 1)}
\]

Substitute \( z = 5 \) into the second equation:
\[
x - y + 3(5) = 13 \implies x - y + 15 = 13 \implies x - y = 13 - 15 = -2 \quad \text{(Equation 2)}
\]

### Step 3: Solve the system of two equations in \( x \) and \( y \)
We now have:
\[
\begin{aligned}
x + 2y &= 4 \quad \text{(Equation 1)} \\
x - y &= -2 \quad \text{(Equation 2)}
\end{aligned}
\]

To eliminate \( x \), subtract Equation 2 from Equation 1:
\[
(x + 2y) - (x - y) = 4 - (-2)
\]
This simplifies to:
\[
x + 2y - x + y = 6 \implies 3y = 6 \implies y = \frac{6}{3} = 2
\]

### Step 4: Substitute \( y = 2 \) into Equation 2
Substitute \( y = 2 \) into the equation \( x - y = -2 \):
\[
x - 2 = -2 \implies x = 0
\]

### Final Solution:
The solution to the system is:
\[
x = 0, \quad y = 2, \quad z = 5
\]

This corresponds to the third option: \( \boxed{x = 0, y = 2, z = 5} \).

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Would you like further details on this solution or have any other questions?

Here are five related questions to explore:
1. How do you solve a system of linear equations using substitution?
2. What is the elimination method, and how does it work for solving systems of equations?
3. How can you solve a system of equations using matrices?
4. How do you determine if a system of equations has no solution or infinitely many solutions?
5. What are the conditions for a unique solution in a system of linear equations?

**Tip:** When solving a system of equations, always start by simplifying one variable if possible, like we did with \( z \) in this example. This can make the system easier to manage.

What is the solution to the following system? x + 2y + z = 9, x - y + 3z = 13, 2z = 10

Learn how to solve a system of three linear equations: x + 2y + z = 9, x - y + 3z = 13, and 2z = 10. This detailed step-by-step solution involves substitution and elimination methods. Suitable for students in Grades 9-12.

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