Решить систему уравнений x+3y=5x^2+2y^2-4xy=-2

[latex]\begin{cases}x+3y=5=\ \textgreater \ x=5-3y\\x^2+2y^2-4xy=-2\end{cases}\\(5-3y)^2+2y^2-4y(5-3y)=-2\\25-30y+9y^2+2y^2-20y+12y^2=-2\\23y^2-50y+27=0\\y_{1,2}=\frac{25^+_-\sqrt{625-621}}{23}=\frac{25^+_-2}{23}\\y_1=\frac{27}{23}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y_2=1\\x_1=5-3\frac{12}{23}=1\frac{11}{23}\ \ x_2=2\\\\OTBET:(1\frac{11}{23};\frac{27}{23});(2;1)[/latex]