Question

помогите!!!!!пожалуйста ​

Answer 1

$\displaystyle \left \{ {5x^2+3y^2=18} \atop {5x^2-3y^2=12}} \right.\\\\ \left \{ {{5x^2+3y^2=18} \atop {5x^2=12+3y^2}} \right. \\\\ 12+3y^2+3y^2=18\\y=-1,y=1\\5x^2=12+3*(-1)^2;5x^2=12+3*1^2\\x=-\sqrt{3},x=\sqrt{3},x=-\sqrt{3},x=\sqrt{3}\\(x_{1},y_{1})=(-\sqrt{3},-1);(x_{2},y_2)=(\sqrt{3},-1);(x_{3},y_3)=(-\sqrt{3},1);(x_4,y_4)=(\sqrt{3},1)$