Помогите решить систему: 3x+y=-1 x-xy=8

$\left \{ {3x+y=-1} \atop {x-xy=8}} \right. \\ \\ \left \{ {y=-1-3x} \atop {x(1-y)=8}} \right. \\ \\ \left \{ {y=-1-3x} \atop {x(1+1+3x)=8}} \right. \\ \\\left \{ {y=-1-3x} \atop {x(2+3x)=8}} \right. \\ \\ 3 x^{2}+2x -8=0 \\ \\ D=4+4*3*8=100 \\ \\ x_{1}= \frac{-2+10}{2*3}= \frac{4}{3} ~~ y_1=-1-3* \frac{4}{3}= -5 \\ \\ x_2= \frac{-2-10}{6}=-2 ~~y_2=-1-3*(-2)=5 \\ \\$
Ответ (4/3; -5), (-2; 5)