What is the solution to this system of equations?

Question

x + 2 y = 4. 2 x minus 2 y = 5.
(3, Negative 5 and one-half)
(3, one-half)
no solution
infinitely many solutions

Answer 1

Let's solve the system of equations given:

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

1. (3, -5.5)
2. (3, 0.5)
3. no solution
4. infinitely many solutions

The correct solution is:

\[ \boxed{(3, \frac{1}{2})} \]