`{:(3x - y - 2 - 0),(2x + y - 8 = 0):}`

Answer

To solve the system of equations given by:

1. \( 3x - y - 2 = 0 \) (Equation 1)
2. \( 2x + y - 8 = 0 \) (Equation 2)

we will follow these steps:

### Step 1: Rearranging the equations
We can rearrange both equations to express \( y \) in terms of \( x \).

From Equation 1:
\[ 3x - y - 2 = 0 \implies y = 3x - 2 \]

From Equation 2:
\[ 2x + y - 8 = 0 \implies y = 8 - 2x \]

### Step 2: Setting the equations equal to each other
Now, we can set the two expressions for \( y \) equal to each other:
\[ 3x - 2 = 8 - 2x \]

### Step 3: Solving for \( x \)
Now, we will solve for \( x \):
\[ 3x + 2x = 8 + 2 \]
\[ 5x = 10 \]
\[ x = \frac{10}{5} = 2 \]

### Step 4: Substituting \( x \) back to find \( y \)
Now that we have \( x = 2 \), we can substitute this value back into one of the original equations to find \( y \). We'll use Equation 1:
\[ y = 3(2) - 2 \]
\[ y = 6 - 2 = 4 \]

### Final Solution
Thus, the solution to the system of equations is:
\[ x = 2, \quad y = 4 \]

---