\( x+2 y=0 \) and \( y-x y=\frac{1}{2} x^{2}-2 \)

To solve the system of equations, start by rewriting the first equation \( x + 2y = 0 \) in terms of \( y \): \( y = -\frac{1}{2} x \). Next, substitute this expression for \( y \) into the second equation \( y - xy = \frac{1}{2} x^2 - 2 \): \[ -\frac{1}{2} x - x\left(-\frac{1}{2} x\right) = \frac{1}{2} x^2 - 2 \] Simplifying gives: \[ -\frac{1}{2} x + \frac{1}{2} x^2 = \frac{1}{2} x^2 - 2 \] This means we can eliminate \( \frac{1}{2} x^2 \) from both sides, leading to: \[ -\frac{1}{2} x = -2 \] Multiplying both sides by -2 results in \( x = 4 \). Now substitute \( x = 4 \) back into \( y = -\frac{1}{2} x \): \[ y = -\frac{1}{2}(4) = -2 \] Thus, the solution to the system of equations is \( (x, y) = (4, -2) \).