Решите систему уравнений \left \{ {{x-xy=-3} \atop {y+xy=8}} \right. [/tex]

Решите задачу:

$x-xy=-3, \ y+xy=8\\ x(1-y)=-3, \ xy=8-y\\ x=- \frac{3}{1-y} = \frac{3}{y-1} , \ x= \frac{8-y}{y} \\ \frac{3}{y-1} = \frac{8-y}{y} \\ (8-y)(y-1)=3y\\ 8y-8-y^2+y-3y=0\\ -y^2+6y-8=0\\ y^2-6y+8=0\\ D=6^2-4*1*8=36-32=4\\ y_1= \frac{6+2}{2} =4\\ y_2= \frac{6-2}{2} =2\\ x_1= \frac{8-4}{4} =1\\ x_2= \frac{8-2}{2} =3\\ Answer: \ (1;4), \ (3;2)$