Question

Система 3x+y^2=6 x-y=2

Answer 1

$\left \{ {{3x+y^2=6} \atop {x-y=2}} \right.\\ \left \{ {{3(2+y)+y^2=6} \atop {x=2+y}} \right.\\ \left \{ {{6+3y+y^2-6=0} \atop {x=2+y}} \right.\\\left \{ {{y(3+y)=0} \atop {x=2+y}} \right.\\ \left \{ {{y=0} \atop {x=2}} \right.\ \ our\ \ \left \{ {{y=-3} \atop {x=-1}} \right. \\\boxed{OTBET:(0;2);(-3;-1).}}$

Answer 2

$\left \{ {{3x+y^2=6} \atop {x-y=2}} \right. \\x=2+y \\3(2+y)+y^2-6=0 \\6+3y+y^2-6=0 \\y^2+3y=0 \\y(y+3)=0 \\y_1=0 \\y_2=-3 \\x_1=2 \\x_2=2-3=-1$
Ответ: (2;0), (-1;-3)