Question

помогите срочно пожалуйста ​

Answer 1

$\displaystyle\bf\\\left \{ {{x + 2y = 6} \atop { {x}^{2} + xy + {y}^{2} = 12 }} \right. \\ \displaystyle\bf\\\left \{ {{x = 6 - 2y} \atop {(6 - 2y) {}^{2} +y (6 - 2y) + {y}^{2} = 12 }} \right. \\ \\ 36 - 24y + 4 {y}^{2} + 6y - 2 {y}^{2} + {y}^{2} - 12 = 0 \\ 3 {y}^{2} - 18y + 24 = 0 \\ {y}^{2} - 6y + 8 = 0$

По теореме Виета:

${x}^{2} + bx + c = 0 \\ x_{1} + x_{2} = - b \\ x_{1} x_{2} = c$

${y}^{2} - 6y + 8 = 0 \\ y_{1} + y_{2} =6 \\ y_{1} y_{2} =8 \\ y_{1} = 2 \\ y_{2} = 4 \\ \\ x_{1} = 6 - 2 \times 2 = 6 - 4 = 2\\ x_{2} = 6 - 2 \times 4 = 6 - 8 = - 2$

Ответ: ( 2 ; 2 ) и ( - 2 ; 4 )