Помогите решить систему неравенств 2x+3y=3xy+1 x-2y=xy-2

$\left \{ {{2x+3y=3xy+1} \atop {x-2y=xy-2}} \right.$
$\left \{ {{2x+3y=3xy+1} \atop {xy= x-2y+2}} \right.$
$\left \{ {{2x+3y=3(x-2y+2)+1} \atop {xy= x-2y+2}}$
$\left \{ {{2x+3y=3x-6y+7} \atop {xy= x-2y+2}}$
$\left \{ {{-x+9y=7} \atop {xy= x-2y+2}}$
$\left \{ {{x=9y-7} \atop {xy= x-2y+2}} \right.$
$\left \{ {{x=9y-7} \atop {(9y-7)y= 9y-7-2y+2}} \right.$
$\left \{ {{x=9y-7} \atop {9y^2-7y= 7y-5}} \right.$
$\left \{ {{x=9y-7} \atop {9y^2-14y+5=0}} \right.$
выходим из системы и работаем со вторым уравнением:
$9y^2-14y+5=0$
$D =196-180 = 16 = 4^2$
$y_{1}= \frac{14+4}{18} =1$
$y_{2}= \frac{14-4}{18} = \frac{5}{9}$
$\{ {{x_{1}=9y-7} \atop {y_{1}=1}}$
$\{ {{x_{1}=9-7} \atop {y_{1}=1}}$
$\{ {{x_{1}=2} \atop {y_{1}=1}}$
$\{ {{x_{2}=9* \frac{5}{9} -7} \atop {y_{2}= \frac{5}{9} }}$
$\{ {{x_{2}=5 -7} \atop {y_{2}= \frac{5}{9} }}$
$\{ {{x_{2}=-2} \atop {y_{2}= \frac{5}{9} }}$
ответ: $\{ {{x_{1}=2} \atop {y_{1}=1}}$
$\{ {{x_{2}=-2} \atop {y_{2}= \frac{5}{9} }}$