Решите под Б пожалуйста

$\left \{ {{x^2y^2-xy=12} \atop {x+y=2}} \right. \\ \left \{ {{x^2y^2-xy-12=0} \atop {x+y=2}} \right. \\\ \left \{ {{(xy-4)(xy+3)=0} \atop {x+y=2}} \right.$

$\left \{ {{xy-4=0} \atop {x+y=2}} \right. \\\ x=2-y \\\ (2-y)y-4=0 \\\ y^2-2y+4=0 \\\ D_1=1-4<0$

$\left \{ {{xy+3=0} \atop {x+y=2}} \right. \\\ x=2-y \\\ (2-y)y+3=0 \\\ y^2-2y-3=0 \\\ D_1=1+3=4 \\\ y_1=1+2=3 \\\ y_2=1-2=-1 \\\ x_1=2-3=-1 \\\ x_2=2+1=3$

Ответ: (-1; 3); (3; -1)