Решить систему уравнений

$\left \{ {{x+y=4} \atop {x^2-y^2=32}} \right. \\\\ \left \{ {{x+y=4} \atop {(x-y)(x+y)=32}} \right. \\\\ \left \{ {{x+y=4} \atop {(x-y)*4=8*4}} \right. \\\\ \left \{ {{x+y=4} \atop {x-y=8}} \right. \\\\ \left \{ {{x+y=4} \atop {x+y+x-y=4+8}} \right. \\\\ \left \{ {{y=4-x} \atop {2x=12}} \right. \\\\\ \left \{ {{y=4-6=-2} \atop {x=6}} \right. \\\\ \left \{ {{x=6} \atop {y=-2}} \right.$

Ответ: $(6;\ -2)$