Question

система уравнени x²-y+xy=x; 2x²+y²=12

Answer 1

$left { {{x^2-y+xy=x} atop {2x^2+y^2=12}} right. ; left { {{x(x+y)-(x+y)=0} atop {2x^2+y^2=12}} right. ; left { {{(x+y)(x-1)=0} atop {2x^2+y^2=12}} right. \\1); left { {{x+y=0,to y=-x} atop {2x^2+(-x)^2=12}} right. ; left { {{y=-x} atop {3x^2=12}} right. ; left { {{y=-x} atop {x^2=4}} right. ; left { {{y_1=-2,; y_2=2} atop {x_1=2,; x_2=-2}} right. ; Rightarrow ; (2,-2),; (-2,2)\\2); left { {{x=1} atop {2+y^2=12}} right.$

$left { {{x=1} atop {y^2=10}} right. ; left { {{x=1} atop {y=pm sqrt{10}}} right. ; Rightarrow ; (1,sqrt{10}),; (1.-sqrt{10}).$