Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Here is my solution for this problem, it's idea is to consider all cases without much effort and find the solutions Firstly we should say that /x+y/<4 (1) beacuse /x+y+1/+/1-y/=4 Now consideration, 1st case, let's say x+y>=0 then if y<=1 then 4-y=/1-x/ if x<=1 then x=y-3, x+y=2*y-3<0 that contradicts our assumption. if x>1 then x=5-y, x+y=5 contradicts (1); if y>1 then y+2=/1-x/ if x<=1 then y+2=1-x, x+y=-1<0 taht contradicts our assumption if x>1 then x=y+3, checking we get y=1/3, x=10/3, but y<1 and again contradiction 2nd case, let's say -1<x+y<0, and opening absolute values we obtain /1-x/+/1-y/=9 if y<=1 then 8+y=/1-x/ if x<=1 then x+y=-7 contradicts out assumption if x>1 then x=9+y, checking we get y=-7, x=2 but /y+x/=5>4 contradicts (1) as we see the expression is symetric that means none of (x,y) is less then or equal to -1 so (x>1 y>1) but then x+y=11 and contradicts the assumption 3rd case, x+y<=-1 then 1+/1-x/=/1-y/ if y<=1 then y+/1-x/=0 if x<=1 then x=y+1 and checking we get one solution for the given system y=-7/3; x=-4/3; if x>1 then x+y=1 and contradiction if y>1 then y-2=/1-x/ if x>=1 x+y>2 and contradiction again as well as if x<1 x+y=3 So we considered all possible cases and found the only solution for the given system (x=-4/3, y=-7/3); ie//

I graphed both relations and got two parallelograms which intersected at two points (-14/3,13/3) and (-2,-1).