Question

Система уравнении x+y-xy=1; x²+y²-xy=3

Answer 1

$\left \{ {{x+y=1+xy} \atop {(x^2+y^2+2xy)-2xy-xy=3}} \right. \; \left \{ {{x+y=1+xy} \atop {(x+y)^2=3xy+3}} \right. \; \left \{ {{x+y=1+xy} \atop {(x+y)^2=3(xy+1)}} \right. \; \\\\\left \{ {{x+y=1+xy} \atop {(x+y)^2=3(x+y)}} \right. \; \to \; (x+y)^2-3(x+y)=0,\; (x+y)(x+y-3)=0\\\\1)\; \left \{ {{x+y=0} \atop {x+y=1+xy}} \right. \; \left \{ {{x=-y} \atop {-y+y=1-y^2}} \right. \; \left \{ {{x=-y} \atop {0=1-y^2}} \right.\; \left \{ {{x=-y} \atop {y^2=1}} \right.$

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