решите систему Х+У=3 Х в квадрате+У в квадрате= 25+2ХУ

Решите задачу:

$\left \{ {{{x+y=3} \atop {x^2+y^2=25+2xy}} \right. \\ \left \{ {{{x+y=3} \atop {x^2+y^2-2xy=25}} \right. \\ \left \{ {{{x+y=3} \atop {(x-y)^2=25}} \right. \\ \left \{ {{{x=3-y} \atop {(3-y-y)^2=25}} \right. \\ \left \{ {{{x=3-y} \atop {(3-2y)^2=25}} \right. \\ \left \{ {{{x=3-y} \atop {3-2y=5}} \right. \ \left \{ {{{x=3-y} \atop {3-2y=-5}} \right. \\ \left \{ {{2y=-2} \atop {x=3-y}} \right. \ \left \{ {{{2y=8} \atop {{x=3-y}} \right. \\ \left \{ {{y=-1} \atop {x=3-y}} \right. \ \left \{ {{{y=4} \atop {{x=3-y}} \right. \\ \left \{ {{y=-1} \atop {x=4}} \right. \ \left \{ {{{y=4} \atop {{x=-1}} \right. \\$