cazul I
x+1≤2
x≤2-1
x≤1
cazul II
-(x+1)≤2
-x-1≤2
-x≤2+1
-x≤3 inmultim -1
x≥-3
deci -3≤x≤1
A={-3,-2,-1,0,1}
sau
x+1=0
x=-1
x -α -1 +α
x+1 - - - - - - - - - - - - - - - - 0+++++++++++++++++++++
1)daca x∈(-α, -1]
-x-1≤2
-x≤2+1
-x≤3 inmultim -1
.x≥-3
x∈[-3,1] prima solutie (1)
2) daca x∈(-1,+α)
x+1≤2
x≤2-1
x≤1
x∈[-1,1] a doua solutie (2)
din (1)U(2)⇒x∈[-3,1]
A={-3,-2,-1,0,1}