Решите систему уравнений: 2ху+у^2=28 х+2у=10

$\left \{ {{2xy+y^2=28} \atop {x+2y=10}} \right. \\ \left \{ {{2(10-2y)y+y^2=28} \atop {x=10-2y}} \right. \\ \left \{ {{20y-4y^2+y^2=28} \atop {x=10-2y}} \right. \\ \left \{ {{3y^2-20y+28=0} \atop {x=10-2y}} \right.$
$3y^2-20y+28=0 \\ D=20^2-4*3*28=400-336=64 \\ y_1= \frac{20+8}{6} = \frac{14}{3}=4 \frac{2}{3} \\ y_1= \frac{20-8}{6} = \frac{12}{6}=2$
при $y_1= \frac{14}{3}$ $x_1=10-2* \frac{14}{3} = \frac{30-28}{3} = \frac{2}{3}$
при $y_2=2$ $x_2=10-2*2=10-4=6$
Ответ: $(6; 2)$ и $( \frac{2}{3}; 4\frac{2}{3} )$