Решите систему уравнений х^2+у^2=10 ху=3

$\left \{ {{x^2+y^2=10} \atop {xy=3}} \right. \\ y= \frac{3}{x} \\ x^2 + ( \frac{3}{x} )^2=10 \\ x^2 + ( \frac{3^2}{x^2} )=10 \\ t=x^2 \\ \\ t + \frac{9}{t} =10 \\ t^2+9=10t \\ t^2-10t+9=0 \\ D/4=25-9=16 \\ t=5б4 \\ t_1=1; t_2=9 \\ \\ x^2=t \\ x_{1,2} ^2=1;x_1=1; x_2=-1 \\ x_{3,4} ^2=9;x_3=3; x_4=-3 \\$

$y_{1,2} =б \frac{3}{1} =б3 \\ y_{3,4} =б \frac{3}{3} =б1 \\$

Ответ: (-1; -3); (1; 3); (-3; -1); (3; 1).