|x-y|=2 |x|+|y| = 4 Решите систему

$\left \{ {{|x-y|=2} \atop {|x|+|y|=4}} \right. \to \left \{ {{ \left[\begin{array}{ccc}x-y=2\\-x+y=2\end{array}\right} \atop {|x|+|y|=4}} \right. \\ \\ \left \{ {{x-y=2} \atop {|x|+|y|=4}} \right. \to \left \{ {{x=y+2} \atop {|y+2|+|y|=4}} \right. \\ \\ \left \{ {{y \geq 0} \atop {y+2 \geq 0}}\atop {y+2+y=4} \right. \\ 2y=2 \\ y=1 \\ x=3$

$\left \{ {{y<0} \atop {y+2 \geq 0}} \right. \\ y+2-y=4 \\ 0=2 \\ x=\O \\ \\ \left \{ {{y<0} \atop {y+2<0}} \right. \\ -y-2-y=4 \\ -2y=2 \\ y=-1 \\ x=-3$

Теперь второе

$\left \{ {{-x+y=2} \atop {|x|+|y|=4}} \right. \to \left \{ {{x=y-2} \atop {|y-2|+|y|=4}} \right. \\ \left \{ {{y \geq 0} \atop {y -2\geq 0}} \right. \\ y-2+y=4 \\ 2y=6 \\ y=3; \\ x=1$

$\left \{ {{y<0} \atop {y-2<0}} \right. \\ -y+2-y=4 \\ -2y=2 \\ y=-1 \\ x=-3 \\ \\ \\ \\ OTBET:(3;1),(-1;-3),(1;3),(-3;-1).$