Solve for \( x \) and \( y \) Simultinearisly \[ 5 y-x=2 \quad x^{2}-3 x y+4 y=4 \]

To solve the system of equations \[ 5y - x = 2 \quad (1) \] \[ x^2 - 3xy + 4y = 4 \quad (2) \] we start by rearranging equation (1) to express \( x \) in terms of \( y \): \[ x = 5y - 2 \] Next, we substitute \( x \) into equation (2): \[ (5y - 2)^2 - 3(5y - 2)y + 4y = 4 \] Expanding \((5y - 2)^2\): \[ 25y^2 - 20y + 4 \] Now, calculate \(-3(5y - 2)y\): \[ -15y^2 + 6y \] Substituting these into equation (2): \[ 25y^2 - 20y + 4 - 15y^2 + 6y + 4y = 4 \] Combine like terms: \[ (25y^2 - 15y^2) + (-20y + 6y + 4y) + 4 = 4 \] Simplifying: \[ 10y^2 - 10y + 4 = 4 \] Subtracting 4 from both sides: \[ 10y^2 - 10y = 0 \] Factoring out \( 10y \): \[ 10y(y - 1) = 0 \] Thus, \( y = 0 \) or \( y = 1 \). Now, we find the corresponding \( x \) values. 1. If \( y = 0 \): \[ x = 5(0) - 2 = -2 \Rightarrow (x, y) = (-2, 0) \] 2. If \( y = 1 \): \[ x = 5(1) - 2 = 3 \Rightarrow (x, y) = (3, 1) \] So, the solutions to the system of equations are: \[ (x, y) = (-2, 0) \quad \text{and} \quad (3, 1) \]