Question

Решить системы уравнений
1){х+ху-3у=-3
{2х-ху+у=2
2){y-xy+3x=-3
{2у+ху-х=2

Answer 1

$1)\; \left \{ {{x+xy-3y=-3} \atop {2x-xy+y=2}} \right. \; \left \{ {{x+xy-3y=-3} \atop {3x-y=-1}} \right. \; \left \{ {{x+x(3x+1)-3(3x+1)=-3} \atop {y=3x+1}} \right.\\\\x+3x^2+x-9x-3=-3\\\\3x^2-7x =0\\\\x(3x-7)=0\\\\x_1=0,\; y_1=3\cdot 0+1=1\\\\x_2=\frac{7}{3},\; y_2=3\cdot \frac{7}{3}+1=8\\\\Otvet:\; (0,1),\; (\frac{7}{3},8).$

$2)\; \left \{ {{y-xy+3x=-3} \atop {2y+xy-x=2}} \right. \; \left \{ {{y-xy+3x=-3} \atop {3y+2x=-1}} \right. \; \left \{ {{y-(-\frac{3}{2}y-\frac{1}{2})y+\frac{3}{2}\cdot (-3y-1)=-3} \atop {x=\frac{-3y-1}{2}}} \right. \\\\y+\frac{3}{2}y^2+\frac{1}{2}y-\frac{9}{2}y-\frac{3}{2}=-3\\\\\frac{3}{2}y^2-3y+\frac{3}{2}=0\, |\cdot 2\\\\3y^2-6y+3=0\, |:3\\\\y^2-2y+1=0\\\\(y-1)^2=0\\\\y=1\\\\x=\frac{-3-1}{2}=-2\\\\Otvet:\; (-2,1)$