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

Answer

To solve the system of linear equations given by:

1. \( x + 2y + z = 4 \) (Equation 1)
2. \( -x + y + z = 0 \) (Equation 2)
3. \( x - 3y + z = 4 \) (Equation 3)

we will follow these steps:

### Step 1: Eliminate one variable
We can start by eliminating one of the variables. Let's eliminate \( z \) by manipulating the equations.

**Add Equation 1 and Equation 2:**
\[ (x + 2y + z) + (-x + y + z) = 4 + 0 \]
This simplifies to:
\[ 3y + 2z = 4 \quad \text{(Equation 4)} \]

**Add Equation 2 and Equation 3:**
\[ (-x + y + z) + (x - 3y + z) = 0 + 4 \]
This simplifies to:
\[ -2y + 2z = 4 \quad \text{(Equation 5)} \]

### Step 2: Solve the new system of equations
Now we have a new system of equations (Equations 4 and 5):
1. \( 3y + 2z = 4 \) (Equation 4)
2. \( -2y + 2z = 4 \) (Equation 5)

**Now, let's eliminate \( z \) by subtracting Equation 5 from Equation 4:**
\[ (3y + 2z) - (-2y + 2z) = 4 - 4 \]
This simplifies to:
\[ 5y = 0 \]
Thus, we find:
\[ y = 0 \]

### Step 3: Substitute \( y \) back into one of the equations
Now that we have \( y = 0 \), we can substitute this value back into Equation 4 to find \( z \):
\[ 3(0) + 2z = 4 \]
This simplifies to:
\[ 2z = 4 \quad \Rightarrow \quad z = 2 \]

### Step 4: Substitute \( y \) and \( z \) back into one of the original equations to find \( x \)
Now, we can substitute \( y = 0 \) and \( z = 2 \) back into Equation 1:
\[ x + 2(0) + 2 = 4 \]
This simplifies to:
\[ x + 2 = 4 \quad \Rightarrow \quad x = 2 \]

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