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The problem provided shows a triangular system of linear equations. The method suggested is **back-substitution**. Let's break it down:

We are given the system:

\[
\begin{aligned}
x + 5y + z &= 6 \\
-y + 3z &= 3 \\
2z &= 2
\end{aligned}
\]

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

### Step 2: Substitute \( z = 1 \) into the second equation to find \( y \)
The second equation is:
\[
-y + 3z = 3
\]
Substituting \( z = 1 \):
\[
-y + 3(1) = 3 \implies -y + 3 = 3
\]
\[
-y = 3 - 3 = 0 \implies y = 0
\]

### Step 3: Substitute \( y = 0 \) and \( z = 1 \) into the first equation to find \( x \)
The first equation is:
\[
x + 5y + z = 6
\]
Substituting \( y = 0 \) and \( z = 1 \):
\[
x + 5(0) + 1 = 6 \implies x + 1 = 6
\]
\[
x = 6 - 1 = 5
\]

### Final Solution:
\[
(x, y, z) = (5, 0, 1)
\]

Would you like more details on this or have any questions? 

Here are some related questions to explore:

1. How does back-substitution work for triangular systems in general?
2. What would happen if we had a non-triangular system?
3. Can Gaussian elimination be used to solve this system?
4. How can we determine if a system of equations has no solution or infinite solutions?
5. How does the substitution method compare to matrix methods like row reduction?

**Tip:** Always check your final solution by plugging the values back into the original equations!

Use back-substitution to solve the triangular system: x + 5y + z = 6, -y + 3z = 3, 2z = 2

Solve the triangular system of linear equations x + 5y + z = 6, -y + 3z = 3, and 2z = 2 using the back-substitution method. This example demonstrates step-by-step algebraic reasoning, suitable for high school students.

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