[tex]\displaystyle\bf\\\left \{ {{2x - y = -1 } \atop { - 3 {x}^{2} + {y}^{2} = 6 }} \right. \\ \displaystyle\bf\\\left \{ {{y = 2x + 1} \atop { - 3 {x}^{2} + (2x + 1) {}^{2} = 6 }} \right. \\ \\ - 3 {x}^{2} + 4 {x}^{2} + 4x + 1 - 6 = 0 \\ x {}^{2} + 4x - 5 = 0 \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ x_{1}  +  x_{2} = - 4  \\ x_{1} x_{2} =  - 5 \\ x_{1} =  - 5\\ x_{2} = 1 \\ \\ y_{1} = 2 \times( - 5) + 1 = - 10 + 1 = - 9 \\ y_{2} = 2 \times 1 + 1 = 2 + 1 = 3[/tex]

Ответ: ( - 5 ; - 9 ) и ( 1 ; 3 )