Solve for \( x \) and \( y \) : \[ x=-2 y \] \[ \quad \text { and } \] \[ x^{2}-3 x y=40 \] INSTRUCTIONS: - You should get two coordinate pairs in the form \( \left(x_{1} ; y_{1}\right) \) and \( \left(x_{0} \cdot y_{0}\right) \)

To solve the system of equations given by 1. \( x = -2y \) 2. \( x^2 - 3xy = 40 \) we start by substituting the expression for \( x \) from the first equation into the second equation: Substituting \( x = -2y \) into \( x^2 - 3xy = 40 \): \[ (-2y)^2 - 3(-2y)y = 40 \] This simplifies to: \[ 4y^2 + 6y^2 = 40 \] Combining like terms results in: \[ 10y^2 = 40 \] Next, we solve for \( y^2 \): \[ y^2 = 4 \] Taking the square root of both sides yields: \[ y = 2 \quad \text{or} \quad y = -2 \] Now we can find the corresponding \( x \) values using \( x = -2y \): 1. If \( y = 2 \): \[ x = -2(2) = -4 \quad \Rightarrow \quad (x_1, y_1) = (-4, 2) \] 2. If \( y = -2 \): \[ x = -2(-2) = 4 \quad \Rightarrow \quad (x_0, y_0) = (4, -2) \] Therefore, the solutions in coordinate pair form are: \[ \left(-4; 2\right) \quad \text{and} \quad \left(4; -2\right) \]