1 Solve for \( x \) and \( y \) in the fol a \( 3 y+x=2 \) \( y^{2}+x=x y+y \)

To solve this system of equations, we start with the first equation: 1. From \( 3y + x = 2 \), we can express \( x \) in terms of \( y \): \[ x = 2 - 3y \] Now, we substitute \( x \) in the second equation: 2. The second equation is \( y^2 + x = xy + y \). Replacing \( x \) with \( 2 - 3y \): \[ y^2 + (2 - 3y) = (2 - 3y)y + y \] 3. Simplifying both sides, we have: \[ y^2 + 2 - 3y = 2y - 3y^2 + y \] \[ y^2 + 2 - 3y = -3y^2 + 3y \] 4. Now, moving all terms to one side: \[ y^2 + 3y^2 - 3y - 3y + 2 = 0 \] \[ 4y^2 - 6y + 2 = 0 \] 5. Dividing everything by 2: \[ 2y^2 - 3y + 1 = 0 \] 6. Now, we can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \] \[ y = \frac{3 \pm \sqrt{9 - 8}}{4} \] \[ y = \frac{3 \pm 1}{4} \] Thus, \( y = 1 \) or \( y = \frac{1}{2} \). 7. Now substituting these values back to find \( x \): - For \( y = 1 \): \[ x = 2 - 3(1) = -1 \] So one solution is \( (x, y) = (-1, 1) \). - For \( y = \frac{1}{2} \): \[ x = 2 - 3\left(\frac{1}{2}\right) = 2 - \frac{3}{2} = \frac{1}{2} \] Hence, another solution is \( (x, y) = \left(\frac{1}{2}, \frac{1}{2}\right) \). The final solutions are: 1. \( (x, y) = (-1, 1) \) 2. \( (x, y) = \left(\frac{1}{2}, \frac{1}{2}\right) \)