Решите систему уравнения , буду благодарна!!!

$\left \{ {{x^2+xy+y^2=13} \atop {x+y=4}} \right. \left \{ {{(x+y)^2-xy=13} \atop {x+y=4}} \right. \left \{ {{16-xy=13} \atop {x+y=4}} \right. \left \{ {{xy=3} \atop {x+y=4}} \right. \left \{ {{(4-y)y=3} \atop {x=4-y}} \right. \\(4-y)y=3\\4y-y^2-3=0\\y^2-4y+3=0\\ \left \{ {{y_1y_2=3} \atop {y_1+y_2=4}} \right. \left |{ {{y=3} \atop {y=1}} \right. \\ \left [{ {{\left \{ {{y=3} \atop {x=1}} \right.} \atop {\left \{ {{y=1} \atop {x=3} \right. }} \right.$

Ответ: $(1;3);(3;1)$