Помогите решить эту задачу

$2 {x}^{2} + {y}^{2} + 3xy = 12 \\ 2 {x}^{2} + 2 {y}^{2} + 4xy - {y}^{2} = 14 \\ \\ 2 {x}^{2} + {y}^{2} + 3xy = 12 \\ 2 {x}^{2} + {y}^{2} + 4xy = 14 \\ \\ xy = 2 \\ 2 {x}^{2} + {y}^{2} = 6 \\ \\ x = \frac{2}{y} \\ 2 \times \frac{4}{ {y}^{2} } + {y}^{2} = 6 \\ \\ x = \frac{2}{y} \\ \frac{8 + {y}^{4} - 6 {y}^{2} }{ {y}^{2} } = 0 \\ \\ {y}^{2} = t \\ {t}^{2} - 6t + 8 = 0 \\ t1 = 4 \\ t2 = 2 \\ 1) {y}^{2} = 4 \\ y1 = 2 \\ x1 = 1 \\ y2 = - 2 \\ x2 = - 1\\ \\ 2) {y}^{2} = 2 \\ y3 = \sqrt{2} \\ x3 = \sqrt{2} \\ y4 = - \sqrt{2} \\ x4 = - \sqrt{2}$
Получили решения:
(2; 1), (-2; -1),
(sqrt(2), sqrt(2)), (-sqrt2; -sqrt2)

x1*y2 + x2*y1 = 2*(-1) + (-2)*1 = -2 - 2 = -4