Решите системы уравнений : X+y=3 { X^2+3xy+y^2-x-y=2

Решите задачу:

$\left \{ {{x+y=3} \atop {x^2+3xy+y^2-x-y=2}} \right. \; \left \{ {{x+y=3} \atop {x^2+3xy+y^2-(x+y)=2}} \right. \; \left \{ {{(x+y)^2=9} \atop {(x^2+2xy+y^2)+xy-3=2}} \right. \\\\ \left \{ {{x+y=3} \atop {(x+y)^2+xy=5}} \right. \; \left \{ {{x+y=3} \atop {9+xy=5}} \right. \; \left \{ {{x+y=3} \atop {xy=-4}} \right. \; \left \{ {{x+y=3} \atop {y=-\frac{4}{x}}} \right. \\\\x-\frac{4}{x}-3=0\; \; \to \; \; \; x^2-3x-4=0\; \; \to \; \; x_1=-1,\; x_2=4\\\\y_1=-\frac{4}{-1}=4,\; y_2=-\frac{4}{4}=-1$

$Otvet:\; \; (-1,4)\; ,\; (4,-1)\; .$