Роз"яжіть систему рівнянь х²+у²=13 ху=6

Решите задачу:

$\left \{ {{x^2+y^2=13} \atop {xy=6}} \right. \; \left \{ {{x^2+y^2+2xy=13+12} \atop {xy=6}} \right. \; \left \{ {{(x+y)^2=25} \atop {xy=6}} \right. \; \left \{ {{x+y=\pm 5} \atop {xy=6}} \right. \\\\a)\quad \left \{ {{x+y=5} \atop {xy=6}} \right. \; \left \{ {{y=5-x} \atop {x(5-x)=6}} \right. \; \left \{ {{y=5-x} \atop {-x^2+5x-6=0}} \right. \; \left \{ {{y=5-x} \atop {x^2-5x+6=0}} \right. \; \left \{ {{y_1=3\; ,\; y_2=2} \atop {x_1=2\; ,\; x_2=3}} \right.$

$b)\quad \left \{ {{x+y=-5} \atop {xy=6}} \right. \; \left \{ {{y=-x-5} \atop {x(-x-5)=6}} \right. \; \left \{ {{y=-x-5} \atop {-x^2-5x-6=0}} \right. \; \left \{ {{y=-x-5} \atop {x^2+5x+6=0}} \right. \; \left \{ {{y_1=-3\; ,\; y_2=-2} \atop {x_1=-2\; ,\; x_2=-3}} \right. \\\\Otvet;\; \; (2,3),\; (3,2),\; (-2,-3),\; (-3,-2) \; .$