https://file.aiquickdraw.com/mathbot/file/1726371707567iw058zxy.jpg

The system of equations visible in the image is:

\[
\begin{aligned}
1. \quad x + y + z &= 6 \\
2. \quad x + 2y + 3z &= 14 \\
3. \quad -x + y - z &= -2
\end{aligned}
\]

### Solving the system of equations:

We'll use the substitution or elimination method to solve this system.

#### Step 1: Eliminate one variable
Let's eliminate \(x\) by adding the first and third equations.

From the first and third equations:
\[
(x + y + z) + (-x + y - z) = 6 + (-2)
\]
Simplifies to:
\[
2y = 4 \quad \Rightarrow \quad y = 2
\]

#### Step 2: Substitute \(y = 2\) into the first and second equations

1. Substitute \(y = 2\) into the first equation:
\[
x + 2 + z = 6 \quad \Rightarrow \quad x + z = 4 \quad \Rightarrow \quad x = 4 - z \tag{4}
\]

2. Substitute \(y = 2\) into the second equation:
\[
x + 2(2) + 3z = 14 \quad \Rightarrow \quad x + 4 + 3z = 14 \quad \Rightarrow \quad x + 3z = 10 \tag{5}
\]

#### Step 3: Substitute \(x = 4 - z\) from (4) into (5)
Substitute \(x = 4 - z\) into \(x + 3z = 10\):
\[
(4 - z) + 3z = 10 \quad \Rightarrow \quad 4 + 2z = 10 \quad \Rightarrow \quad 2z = 6 \quad \Rightarrow \quad z = 3
\]

#### Step 4: Solve for \(x\)
Now substitute \(z = 3\) into equation (4):
\[
x = 4 - 3 = 1
\]

#### Final solution:
The solution to the system of equations is:
\[
x = 1, \quad y = 2, \quad z = 3
\]

Let me know if you'd like a detailed breakdown of any step, or if you have further questions!

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### Related questions:
1. What happens if we try solving this system using matrices?
2. Can the system of equations be solved using substitution for a different variable?
3. How would the solution differ if one of the equations was non-linear?
4. How does Gaussian elimination work for this system?
5. What is the significance of having a unique solution in this case?

**Tip:** When solving systems of linear equations, it's useful to try both substitution and elimination methods to see which simplifies the process more efficiently.

Solve the system of equations: x + y + z = 6; x + 2y + 3z = 14; -x + y - z = -2.

Learn how to solve the system of linear equations: x + y + z = 6, x + 2y + 3z = 14, -x + y - z = -2. This step-by-step solution demonstrates the substitution and elimination methods. Suitable for students in Grades 9-12.

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