1.2 Solve for \( x \) and \( y \) simultaneously if: \( x+4=2 y \) and \( y^{2}-x y+21=0 \)

To solve the equations \( x + 4 = 2y \) and \( y^2 - xy + 21 = 0 \) simultaneously, we can start by expressing \( x \) in terms of \( y \) from the first equation: 1. Rearranging gives us: \[ x = 2y - 4 \] 2. Now substitute \( x \) in the second equation: \[ y^2 - (2y - 4)y + 21 = 0 \] This simplifies to: \[ y^2 - (2y^2 - 4y) + 21 = 0 \] \[ y^2 - 2y^2 + 4y + 21 = 0 \] \[ -y^2 + 4y + 21 = 0 \] 3. Multiplying through by -1 gives: \[ y^2 - 4y - 21 = 0 \] 4. Now, we can factor this quadratic equation: \[ (y - 7)(y + 3) = 0 \] Thus, \( y = 7 \) or \( y = -3 \). 5. Now we can find the corresponding values of \( x \): - For \( y = 7 \): \[ x = 2(7) - 4 = 14 - 4 = 10 \] - For \( y = -3 \): \[ x = 2(-3) - 4 = -6 - 4 = -10 \] Thus, the solutions are: \[ (x, y) = (10, 7) \quad \text{and} \quad (x, y) = (-10, -3) \]