Решите систему уравнений x+xy+3y=3 2x-xy-y=2

Решите задачу:

$\left \{ {{x+xy+3y=3} \atop {2x-xy-y=2}} \right. \; \left \{ {{3x+2y=5} \atop {x+xy+3y=3}} \right. \; \left \{ {{x=\frac{5-2y}{3}=\frac{5}{3}-\frac{2}{3}y} \atop {(\frac{5}{3}-\frac{2}{3}y)+(\frac{5}{3}y-\frac{2}{3}y^2)+3y=3}} \right. \\\\-\frac{2}{3}y^2+4y-\frac{4}{3}=0\, |\cdot (-3)\\\\2y^2-12y+4=0\, |:2\\\\y^2-6y+2=0\\\\(y-3)^2-7=0\\\\y-3=\pm \sqrt7,\\\\y_1=3-\sqrt7,\; y_2=3+\sqrt7\\\\x_1=\frac{5}{3}-\frac{2}{3}(3-\sqrt7)=\frac{2\sqrt7-1}{3}\\\\x_2=\frac{5}{3}-\frac{2}{3}(3+\sqrt7)=$

$x_2=\frac{-2\sqrt7-1}{3}\\\\Otvet;\; \; (\frac{2\sqrt7-1}{3};\, 3-\sqrt7),\; (-\frac{2\sqrt7+1}{3};\, 3+\sqrt7)$