x+y+z=1
x-2y+3z=2
5x-3y+z=3

Answer

To solve the system of linear equations given by:

1. \( x + y + z = 1 \) (Equation 1)
2. \( x - 2y + 3z = 2 \) (Equation 2)
3. \( 5x - 3y + z = 3 \) (Equation 3)

we will follow these steps:

### Step 1: Eliminate one variable
Let's eliminate \( x \) from Equations 1 and 2. We can do this by manipulating the equations.

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

Now, substitute this expression for \( x \) into Equation 2:
\[ (1 - y - z) - 2y + 3z = 2 \]
Simplifying this gives:
\[ 1 - y - z - 2y + 3z = 2 \]
\[ 1 - 3y + 2z = 2 \]
Now, rearranging this gives:
\[ -3y + 2z = 1 \quad \text{(Equation 4)} \]

### Step 2: Eliminate \( x \) from Equations 1 and 3
Now, let's eliminate \( x \) from Equations 1 and 3. Again, substitute \( x = 1 - y - z \) into Equation 3:
\[ 5(1 - y - z) - 3y + z = 3 \]
Simplifying this gives:
\[ 5 - 5y - 5z - 3y + z = 3 \]
\[ 5 - 8y - 4z = 3 \]
Rearranging this gives:
\[ -8y - 4z = -2 \]
Dividing through by -2 gives:
\[ 4y + 2z = 1 \quad \text{(Equation 5)} \]

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

We can eliminate \( z \) by subtracting Equation 4 from Equation 5:
\[ (4y + 2z) - (-3y + 2z) = 1 - 1 \]
This simplifies to:
\[ 4y + 2z + 3y - 2z = 0 \]
\[ 7y = 0 \]
Thus, we find:
\[ y = 0 \]

### Step 4: Substitute back to find \( z \)
Now substitute \( y = 0 \) back into Equation 4:
\[ -3(0) + 2z = 1 \]
This simplifies to:
\[ 2z = 1 \implies z = \frac{1}{2} \]

### Step 5: Substitute back to find \( x \)
Finally, substitute \( y = 0 \) and \( z = \frac{1}{2} \) back into Equation 1 to find \( x \):
\[ x + 0 + \frac{1}{2} = 1 \]
This simplifies to:
\[ x + \frac{1}{2} = 1 \implies x = 1 - \frac{1}{2} = \frac{1}{2} \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = \frac{1}{2}, \quad y = 0, \quad z = \frac{1}{2} \]