` 2x - 5y + 4= 0 `,
` 2x + y - 8 = 0 `

Answer

To solve the system of equations given by:

1. \( 2x - 5y + 4 = 0 \) (Equation 1)
2. \( 2x + y - 8 = 0 \) (Equation 2)

we will use the elimination method.

### Step 1: Rearranging the equations
First, let's rearrange both equations into the standard form \( Ax + By + C = 0 \).

From Equation 1:
\[ 2x - 5y + 4 = 0 \implies 2x - 5y = -4 \]

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

### Step 2: Subtract Equation 2 from Equation 1
Now, we will subtract Equation 2 from Equation 1 to eliminate \(2x\):
\[ (2x - 5y) - (2x + y) = -4 - 8 \]
This simplifies to:
\[ -5y - y = -12 \]
\[ -6y = -12 \]

### Step 3: Solve for \(y\)
Now, divide both sides by -6:
\[ y = \frac{-12}{-6} = 2 \]

### Step 4: Substitute \(y\) back into one of the original equations
Now that we have \(y = 2\), we can substitute this value back into Equation 2 to find \(x\):
\[ 2x + 2 - 8 = 0 \]
This simplifies to:
\[ 2x - 6 = 0 \]
\[ 2x = 6 \]

### Step 5: Solve for \(x\)
Now, divide both sides by 2:
\[ x = \frac{6}{2} = 3 \]

### Final Solution
The solution to the system of equations is:
\[ x = 3, \quad y = 2 \]