Помогите со вторым заданием

Решите задачу:

$\left \{ {{x+y=10} \atop {x^2-y^2=40}} \right. \; \left \{ {{x+y=10} \atop {(x-y)(x+y)=40}} \right. \; \left \{ {{x+y=10} \atop {10(x-y)=40}} \right. \; \left \{ {{x+y=10} \atop {x-y=4}} \right. \; \left \{ {{2x=14} \atop {2y=6}} \right. \; \left \{ {{x=7} \atop {y=3}} \right. \\\\7\in (2,10)\\\\3\in (2,10)\\\\Otvet:\; \; A.$