Помогите пожалуйста! Решите систему уравнений {у+2ху=7; х+2ху=9

$\left \{ {{y+2xy=7} \atop {x+2xy=9}} \right. \left \{ {{2xy=7-y} \atop {x(2y+1)=9}} \right. \left \{ {{x=\frac{7-y}{2y}} \atop {\frac{7-y}{2y}(2y+1)=9}} \right. \\\frac{7-y}{2y}(2y+1)=9|\cdot2y\\(7-y)(2y+1)=18y\\14y+7-2y^2-y-18y=0|\cdot(-1)\\2y^2+5y-7=0\\D=25+56=81=9^2\\y_{1,2}=\frac{-5\pm9}{4}=\left |{ {{1} \atop {-3,5}} \right.$

$\left[\begin{gathered}\left\{\begin{gathered}x+2xy=9 \hfill\\y=1\hfill\\\end{gathered}\right.\hfill\\\left\{\begin{gathered}x+2xy=9\hfill\\y=-3,5\hfill\\\end{gathered}\right.\hfill\\\end{gathered}\right.$ $\left[\begin{gathered}\left\{\begin{gathered}x=3 \hfill\\y=1\hfill\\\end{gathered}\right.\hfill\\\left\{\begin{gathered}x=1,5\hfill\\y=-3,5\hfill\\\end{gathered}\right.\hfill\\\end{gathered}\right.$

Ответ: $(3;1);(1,5;-3,5)$