x+2y+z=7
x+3z=11
2x-3y=1

Answer

To solve the system of linear equations given by:

1. \( x + 2y + z = 7 \) (Equation 1)
2. \( x + 3z = 11 \) (Equation 2)
3. \( 2x - 3y = 1 \) (Equation 3)

we will follow these steps:

### Step 1: Multiply Equation 1 by 3
We will multiply the first equation by 3 to eliminate \( z \) later when we combine it with Equation 2.

\[ 3(x + 2y + z) = 3(7) \]
This gives us:
\[ 3x + 6y + 3z = 21 \quad \text{(Equation 4)} \]

### Step 2: Write Equation 2
We will keep Equation 2 as it is for now:
\[ x + 3z = 11 \quad \text{(Equation 2)} \]

### Step 3: Subtract Equation 2 from Equation 4
Now we will subtract Equation 2 from Equation 4 to eliminate \( z \).

\[ (3x + 6y + 3z) - (x + 3z) = 21 - 11 \]

This simplifies to:
\[ 3x - x + 6y + 3z - 3z = 10 \]
\[ 2x + 6y = 10 \quad \text{(Equation 5)} \]

### Step 4: Simplify Equation 5
We can simplify Equation 5 by dividing everything by 2:
\[ x + 3y = 5 \quad \text{(Equation 6)} \]

### Step 5: Use Equation 3
Now we will use Equation 3:
\[ 2x - 3y = 1 \quad \text{(Equation 3)} \]

### Step 6: Solve Equations 6 and 3
Now we have a new system of equations:
1. \( x + 3y = 5 \) (Equation 6)
2. \( 2x - 3y = 1 \) (Equation 3)

We can add these two equations to eliminate \( y \):
\[ (x + 3y) + (2x - 3y) = 5 + 1 \]
This simplifies to:
\[ 3x = 6 \]
Thus, we find:
\[ x = 2 \]

### Step 7: Substitute \( x \) back into Equation 6
Now we will substitute \( x = 2 \) back into Equation 6 to find \( y \):
\[ 2 + 3y = 5 \]
Subtracting 2 from both sides gives:
\[ 3y = 3 \]
Thus, we find:
\[ y = 1 \]

### Step 8: Substitute \( x \) into Equation 2 to find \( z \)
Now we will substitute \( x = 2 \) into Equation 2 to find \( z \):
\[ 2 + 3z = 11 \]
Subtracting 2 from both sides gives:
\[ 3z = 9 \]
Thus, we find:
\[ z = 3 \]

### Final Solution
The values of \( x \), \( y \), and \( z \) are:
\[ x = 2, \quad y = 1, \quad z = 3 \]