Let's solve the system of equations given:
- \( x + 2y = 4 \) (Equation 1)
- \( 2x - 2y = 5 \) (Equation 2)
First, let's simplify Equation 2. We can divide the entire equation by 2:
\[ x - y = \frac{5}{2} \quad \text{(Equation 2 simplified)} \]
Now we can express \( x \) from Equation 1:
From Equation 1:
\[ x = 4 - 2y \]
Now, substitute \( x \) from Equation 1 into the simplified Equation 2:
\[ (4 - 2y) - y = \frac{5}{2} \]
This simplifies to:
\[ 4 - 3y = \frac{5}{2} \]
To eliminate the fraction, we can multiply the entire equation by 2:
\[ 8 - 6y = 5 \]
Now solve for \( y \):
\[ -6y = 5 - 8 \] \[ -6y = -3 \] \[ y = \frac{-3}{-6} = \frac{1}{2} \]
Now that we have \( y = \frac{1}{2} \), we can substitute it back into the equation for \( x \):
\[ x = 4 - 2\left(\frac{1}{2}\right) = 4 - 1 = 3 \]
Thus, the solution to the system of equations is:
\[ (x, y) = (3, \frac{1}{2}) \]
Looking at the answer options given:
- (3, -5.5)
- (3, 0.5)
- no solution
- infinitely many solutions
The correct solution is:
\[ \boxed{(3, \frac{1}{2})} \]