Question

Solve the following system of equations:

Answer 1

$xy = 4 \\ 2y = x - 6 \\ \\ xy = 4 \\ x = 6 + 2y$

$(6 + 2y)y = - 4 \\ {y}^{2} + 3y + 2 = 0 \\ d = {3}^{2} - 4 \times 2 = 9 - 8 = 1 = {1}^{2}$

$y = \dfrac{ - 3 + 1}{2} = - 1 \\ \\ y = \dfrac{ - 3 - 1}{2} = - 2$

$- 2x = - 4 \\ x = 2 \\ \\ - x = - 4 \\ x = 4$

Answer: B) (2; -2), (4; -1)