x+y-z=1
3x+y-2z=3
x-y-z=-1

Answer

To solve the system of linear equations given by:

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

we will follow these steps:

### Step 1: Eliminate one variable
We can start by eliminating \( y \) from Equations 2 and 3.

From Equation 3, we can express \( y \) in terms of \( x \) and \( z \):
\[ y = x - z + 1 \]

### Step 2: Substitute \( y \) in Equation 2
Now, substitute \( y \) in Equation 2:
\[ 3x + (x - z + 1) - 2z = 3 \]
This simplifies to:
\[ 3x + x - z + 1 - 2z = 3 \]
Combining like terms gives:
\[ 4x - 3z + 1 = 3 \]
Subtracting 1 from both sides:
\[ 4x - 3z = 2 \quad (Equation 4) \]

### Step 3: Eliminate \( y \) using Equations 1 and 3
Now, we will eliminate \( y \) using Equations 1 and 3. Adding Equation 1 and Equation 3:
\[ (x + y - z) + (x - y - z) = 1 - 1 \]
This simplifies to:
\[ 2x - 2z = 0 \]
Dividing by 2 gives:
\[ x - z = 0 \quad (Equation 5) \]
So, we can express \( z \) in terms of \( x \):
\[ z = x \]

### Step 4: Substitute \( z \) in Equation 4
Now substitute \( z = x \) into Equation 4:
\[ 4x - 3(x) = 2 \]
This simplifies to:
\[ 4x - 3x = 2 \]
Thus:
\[ x = 2 \]

### Step 5: Find \( z \)
Now that we have \( x \), we can find \( z \):
\[ z = x = 2 \]

### Step 6: Substitute \( x \) and \( z \) back to find \( y \)
Now substitute \( x \) and \( z \) back into Equation 1 to find \( y \):
\[ x + y - z = 1 \]
Substituting the values:
\[ 2 + y - 2 = 1 \]
This simplifies to:
\[ y = 1 \]

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