Решение систем уравнений второй степени

$\left \{ {{x^2-y^2=6} \atop {x-y=6(x+y)}} \right. \left \{ {{(x-y)(x+y)=6} \atop {x-y=6(x+y)}} \right. \left \{ {{6(x+y)(x+y)=6} \atop {x-y=6(x+y)}} \right. \left \{ {(x+y)^2=1 \atop {x-y=6(x+y)}} \right.\\ \left \{ {{ \left {[ {{x+y=1} \atop {x+y=-1}} \right. } \atop {x-y=6(x+y)}} \right. \\(1) \left \{ {{x+y=1} \atop {x-y=6(x+y)}} \right. \left \{ {{x+y=1} \atop {x-y=6}} \right. |+ \left \{ {{2x=7} \atop {x-y=6}} \right. \left \{ {{x=3.5} \atop {y=-2.5}} \right.$
$(2) \left \{ {{x+y=-1} \atop {x-y=6(x+y)}} \right. \left \{ {{x+y=-1} \atop {x-y=-6}} \right. |+ \left \{ {{2x=-7} \atop {x+y=-1}} \right. \left \{ {{x=-3.5} \atop {y=2.5}} \right.$
$\left[{ {{ \left \{ {{x=3.5} \atop {y=-2.5}} \right. } \atop { \left \{ {{x=-3.5} \atop {x=2.5}} \right. }} \right.$
Ответ: $(3.5;-2.5);(-3.5;2.5)$