Question

{x^2y^2-xy=12;  {x+y=2.

Answer 1

$begin{cases}x^2*y^2-xy=12\x+y=2=>x=2-yend{cases}\\xy(xy-1)-12=0\(2-y)*y*((2-y)*y-1)-12=0\((2-y)*y)^2-(2-y)*y-12=0\(2-y)*y=a\a^2-a-12=0\a_1=4 ;a_2=-3\(2-y)*y=4 ; (2-y)*y=-3\y^2-2y+4=0 ; y^2-2y-3=0\y_{1,2}=1^+_-sqrt{1-4} ; y_{3,4}=1^+_-2\error y_3=3 y_4=-1\. x_3=-1 x_4=3$
$OTBET:(-1;3);(3;-1)$