To solve the system of linear equations:

\[
2x - y + 3z = 8
\]
\[
x - 2y - z = -4
\]
\[
3x + y - 4z = 0
\]

we can use either substitution, elimination, or matrix methods like Gaussian elimination. Let's proceed with elimination.

### Step 1: Rewrite the equations
The system is:
1. \(2x - y + 3z = 8\)
2. \(x - 2y - z = -4\)
3. \(3x + y - 4z = 0\)

### Step 2: Eliminate one variable
We'll eliminate \(x\) by manipulating the equations.

1. **Multiply Equation (2) by 2** to align with the coefficient of \(x\) in Equation (1):
   \[
   2(x - 2y - z) = 2(-4)
   \]
   \[
   2x - 4y - 2z = -8
   \]

2. **Subtract the modified Equation (2) from Equation (1):**
   \[
   (2x - y + 3z) - (2x - 4y - 2z) = 8 - (-8)
   \]
   \[
   3y + 5z = 16
   \]

So, we have a new equation:
\[
3y + 5z = 16 \quad \text{(Equation 4)}
\]

3. Now, **eliminate \(x\) using Equations (2) and (3):**

   **Multiply Equation (2) by 3** to match the \(x\) coefficient in Equation (3):
   \[
   3(x - 2y - z) = 3(-4)
   \]
   \[
   3x - 6y - 3z = -12
   \]

4. **Subtract the modified Equation (2) from Equation (3):**
   \[
   (3x + y - 4z) - (3x - 6y - 3z) = 0 - (-12)
   \]
   \[
   7y - z = 12
   \]

So, we have another equation:
\[
7y - z = 12 \quad \text{(Equation 5)}
\]

### Step 3: Solve for \(y\) and \(z\)
Now we have a system of two equations with two variables (\(y\) and \(z\)):

1. \(3y + 5z = 16\)
2. \(7y - z = 12\)

Let's solve these by elimination or substitution.

1. **Multiply Equation (5) by 5** to align with the coefficient of \(z\) in Equation (4):
   \[
   5(7y - z) = 5(12)
   \]
   \[
   35y - 5z = 60
   \]

2. **Add Equation (4) and the modified Equation (5):**
   \[
   (3y + 5z) + (35y - 5z) = 16 + 60
   \]
   \[
   38y = 76
   \]
   \[
   y = 2
   \]

### Step 4: Substitute \(y = 2\) back to find \(z\)

Using Equation (4):
\[
3(2) + 5z = 16
\]
\[
6 + 5z = 16
\]
\[
5z = 10
\]
\[
z = 2
\]

### Step 5: Substitute \(y = 2\) and \(z = 2\) back to find \(x\)

Using Equation (1):
\[
2x - y + 3z = 8
\]
\[
2x - 2 + 3(2) = 8
\]
\[
2x - 2 + 6 = 8
\]
\[
2x = 4
\]
\[
x = 2
\]

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

Would you like further details on any steps? 

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Here are some related questions that build on this solution:

1. How would you solve this system using matrix row operations?
2. What is the significance of unique solutions in systems of equations?
3. How can substitution be used to solve systems of equations?
4. What conditions create infinitely many or no solutions in linear systems?
5. How would you verify if a system has an inconsistent solution?

**Tip:** When solving systems of equations, ensure you’re consistent with signs and coefficients—errors often arise from small mistakes.

2x - y + 3z = 8, x - 2y - z = -4, 3x + y - 4z = 0

Explore how to solve the system of linear equations 2x - y + 3z = 8, x - 2y - z = -4, 3x + y - 4z = 0 using elimination and substitution methods. Follow detailed steps to find x = 2, y = 2, z = 2, covering key linear algebra concepts. Suitable for Grades 10-12.

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