Решите систему уравнений :

$\displaystyle \left \{ {{x+y-7=0} \atop {x^2+xy+y^2=43}} \right. \Rightarrow \left \{ {{x+y=7} \atop {(x+y)^2-xy=43}} \right. \Rightarrow \left \{ {{x+y=7} \atop {7^2-xy=43}} \right. \\ \\ \\ \left \{ {{x+y=7} \atop {xy=6}} \right. \Rightarrow \left \{ {{x=7-y} \atop {(7-y)\cdot y=6}} \right. \\ \\ -y^2+7y=6\\ y^2-7y+6=0$
По т. Виета:
$y_1=1\\ y_2=6$
Тогда
$x_1=7-y_1=7-1=6\\ x_2=7-y^2=7-6=1$

Ответ: $(6;1),\,\, (1;6).$