|x-y|=2 |x|+|y| = 4 Решите систему

[latex] \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[/latex] [latex] \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[/latex] Теперь второе [latex] \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[/latex] [latex] \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).[/latex]

/x-y/=2 /x/+/y/=4 x-y=2 или х-у=-2 х=2+у  или х=y-2 1)х=у+2 /2+y/+/y/=4 a)y<-2 -2-y-y=4 -2y=6 y=-3⇒x=2-3=-1 б)-2≤y≤0 2+y-y=4 Нет решения в)y>0 2+y+y=4 2y=2 y=1⇒x=1+2=3 2)х=у-2 /у-2/+/у/=4 а)у<0 2-y-y=4 2y=-2 y=-1⇒x=-1-2=-3 б)0≤y≤2 2-y+y=4 Нет решения в)y>2 y-2+y=4 2y=6 y=3⇒x=3-2=1 Ответ (-1;-3),(1;3),(-3;-1),(1;3)