Вопрос в картинках...

Решите задачу:

$\left \{ {{(x-2)(y-2)=4} \atop {x^2y+xy^2=72}} \right. \; \left \{ {{xy-2x-2y+4=4} \atop {xy(x+y)=72}} \right. \; \left \{ {{xy=2(x+y)} \atop {2(x+y)(x+y)=72}} \right. \; \left \{ {{xy=2(x+y)} \atop {(x+y)^2=36}} \right. \\\\\left \{ {{xy=2(x+y)} \atop {x+y=\pm 6}} \right. \\\\a)\; \; \left \{ {{xy=2(x+y)} \atop {x+y=6}} \right. \left \{ {{xy=12} \atop {y=6-x}} \right. \; \left \{ {{x(6-x)=12} \atop {y=6-x}} \right. \; \left \{ {{x^2-6x+12=0} \atop {y=6-x}} \right. \\\\x^2-6x+12=0\; ,\; D/4=3^2-12=-3<0\; ,\; x\in \varnothing$

$b)\; \; \left \{ {{xy=2(x+y)} \atop {x+y=-6}} \right. \left \{ {{xy=-12} \atop {y=-x-6}} \right. \; \left \{ {{x(-x-6)=-12} \atop {y=-x-6}} \right. \; \left \{ {{x^2+6x-12=0} \atop {y=-x-6}} \right. \\\\x^2+6x-12=0\; ,\; \; D/4=3^2+12=21\; ,\; \; x_{1/2}=-3\pm \sqrt{21}\\\\\left \{ {{x=-3-\sqrt{21}} \atop {y=3+\sqrt{21}-6}} \right. \; \left \{ {{x=-3-\sqrt{21}} \atop {y=-3+\sqrt{21}}} \right. \\\\\left \{ {{x=-3+\sqrt{21}} \atop {y=3-\sqrt{21}-6}} \right. \; \left \{ {{x=-3+\sqrt{21}} \atop {y=-3-\sqrt{21}}} \right. \\\\Otvet:\; \; (-3-\sqrt{21}\; ,\; -3+\sqrt{21})\; ,\; (-3+\sqrt{21}\; ;\; -3-\sqrt{21})\; .$