Помогите решить. (Задание решается ч-з дискриминант 1 уравнения?)

$\left \{ {{x^2y^2+2xy=3} \atop {(x+y)^2-(x+y)=0}} \right. \\ \left \{ {{x^2y^2+2xy=3} \atop {(x+y)(x+y-1)=0}} \right. \\(xy)^2+2xy=3 \\xy=a \\a^2+2a-3=0 \\D=4+12=16=4^2 \\a_1= \frac{-2+4}{2} =1 \\a_2= \frac{-2-4}{2} =-3 \\1) \left \{ {{xy=1} \atop {x+y=0}} \right. \\x=-y \\-y^2=1 \\y^2=-1 \\y\in \varnothing \\2) \left \{ {{xy=1} \atop {x+y-1=0}} \right. \\x=1-y \\(1-y)*y=1 \\-y^2+y=1 \\-y^2+y-1=0 \\y^2-y+1=0 \\D=1-4\ \textless \ 0\Rightarrow y \in \varnothing$
$3) \left \{ {{xy=-3} \atop {x+y=0}} \right. \\x=-y \\-y^2=-3 \\y^2=3 \\y_1=\sqrt{3} \\y_2=-\sqrt{3} \\x_1=-\sqrt{3} \\x_2=\sqrt{3} \\4) \left \{ {{xy=-3} \atop {x+y-1=0}} \right. \\x=1-y \\(1-y)*y=-3 \\-y^2+y+3=0 \\y^2-y-3=0 \\D=1+12=13 \\y_3= \frac{1+\sqrt{13}}{2} \\y_4=\frac{1-\sqrt{13}}{2} \\x_3=1- \frac{1+\sqrt{13}}{2} = \frac{2-1-\sqrt{13}}{2} = \frac{1-\sqrt{13}}{2} \\x_4=1- \frac{1-\sqrt{13}}{2} = \frac{1+\sqrt{13}}{2}$
Ответ: $(-\sqrt{3};\sqrt{3}),\ (\sqrt{3};-\sqrt{3}),\ ( \frac{1-\sqrt{13}}{2} ; \frac{1+\sqrt{13}}{2}),\ ( \frac{1+\sqrt{13}}{2};\frac{1-\sqrt{13}}{2})$