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To find the solution for the equations \( 3y + x = 2 \) and \( y^2 + x = xy + y \), let's tackle this systematically. From the first equation, we can express \( x \) in terms of \( y \): \[ x = 2 - 3y \] Now, we can substitute this expression for \( x \) into the second equation: \[ y^2 + (2 - 3y) = (2 - 3y)y + y \] Simplifying both sides will help us find the values of \( y \): 1. Left-hand side: \[ y^2 + 2 - 3y \] 2. Right-hand side: \[ (2y - 3y^2) + y = 2y - 3y^2 + y = 3y - 3y^2 \] Now, equating both sides: \[ y^2 + 2 - 3y = 3y - 3y^2 \] This simplifies to: \[ y^2 + 2 - 3y - 3y + 3y^2 = 0 \] \[ 4y^2 - 6y + 2 = 0 \] We can use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = -6, c = 2 \): \[ y = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} \] \[ y = \frac{6 \pm \sqrt{36 - 32}}{8} \] \[ y = \frac{6 \pm 2}{8} \] So we have: 1. \( y = 1 \) 2. \( y = \frac{1}{2} \) Now substituting back to find \( x \): For \( y = 1 \): \[ x = 2 - 3(1) = -1 \] For \( y = \frac{1}{2} \): \[ x = 2 - 3\left(\frac{1}{2}\right) = 2 - \frac{3}{2} = \frac{1}{2} \] Thus, the solutions to the system are \( (x, y) = (-1, 1) \) and \( \left(\frac{1}{2}, \frac{1}{2}\right) \).
