Решить систему неравенств (x-y)^2-x+y=0 x^2y^2-xy-2=0

Решите задачу:

$\left \{ {{(x-y)^2-x+y=0,} \atop {x^2y^2-xy-2=0,}} \right. \left \{ {{(x-y)^2-(x-y)=0,} \atop {(xy)^2-xy-2=0,}} \right. \\ x-y=a, xy=b; \\ a^2-a=0, \\ a(a-1)=0, \\ a_1=0, \\ a-1=0, a_2=1; \\ b^2-b-2=0, \\ b_1=-1, b_2=2; \\ \left \{ {{ \left [ {{x-y=0,} \atop {x-y=1;}} \right. } \atop { \left [ {{xy=-1,} \atop {xy=-2;}} \right. }} \right. \\ 1) \left \{ {{x=y} \atop {xy=-1;}} \right. \left \{ {{x=y} \atop {y^2=-1;}} \right. \\ y^2=-1<0, \\ y\in\varnothing;$
$2) \left \{ {{x=y,} \atop {xy=2;}} \right. \left \{ {{x=y,} \atop {y^2=2;}} \right. \\ y^2=2, \\ y_1=- \sqrt{2} , y_2= \sqrt{2} , \\ x_1=- \sqrt{2} , x_2= \sqrt{2};\\ 3) \left \{ {{x-y=1,} \atop {xy=-1;}} \right. \left \{ {{y=x-1,} \atop {x(x-1)=-1;}} \right. \\ x^2-x+1=0, \\ D=-3<0, \\ x\in\varnothing; \\ 4) \left \{ {{x-y=1,} \atop {xy=2;}} \right. \left \{ {{y=x-1,} \atop {x(x-1)=2;}} \right. \\ x^2-x-2=0, \\ x_3=-1, x_4=2, \\ y_3=-2, y_4=1.$