`x - 2y + 2 = 0, 2x + y - 6= 0 `

Answer

To solve the system of linear equations given by:

1. \( x - 2y + 2 = 0 \)
2. \( 2x + y - 6 = 0 \)

we will use the elimination method. Here are the steps to find the values of \( x \) and \( y \):

### Step 1: Rewrite the equations
First, we can rewrite the equations in a more standard form:

1. \( x - 2y = -2 \) (Equation 1)
2. \( 2x + y = 6 \) (Equation 2)

### Step 2: Make the coefficients of \( x \) equal
To eliminate \( x \), we can multiply Equation 1 by 2 so that the coefficients of \( x \) in both equations will be the same:

\[ 2(x - 2y) = 2(-2) \]
This gives us:

\[ 2x - 4y = -4 \quad \text{(Equation 3)} \]

Now we have:

1. \( 2x - 4y = -4 \) (Equation 3)
2. \( 2x + y = 6 \) (Equation 2)

### Step 3: Subtract the equations
Next, we will subtract Equation 2 from Equation 3 to eliminate \( x \):

\[ (2x - 4y) - (2x + y) = -4 - 6 \]

This simplifies to:

\[ -4y - y = -10 \]

Combining like terms gives us:

\[ -5y = -10 \]

### Step 4: Solve for \( y \)
Now, we can solve for \( y \):

\[ y = \frac{-10}{-5} = 2 \]

### Step 5: Substitute \( y \) back to find \( x \)
Now that we have \( y = 2 \), we can substitute this value back into either of the original equations to find \( x \). We will use Equation 1:

\[ x - 2(2) + 2 = 0 \]

This simplifies to:

\[ x - 4 + 2 = 0 \]
\[ x - 2 = 0 \]

Thus, we find:

\[ x = 2 \]

### Final Solution
The solution to the system of equations is:

\[ x = 2, \quad y = 2 \]