A={x∈ℝ / |2x+1| ≤5}
|2x+1| ≤5
=> -5 ≤2x+1 ≤ 5 /-1
-6≤ 2x≤ 4/:2
-3 ≤x ≤2
x∈ℝ => A=[-3;2]
B={x∈ℝ / -3<(x+1)/2 ≤ 1}
-3<(x+1)/2 ≤ 1 /×2
-6<x+1≤2 /-1
-7<x ≤1
x∈ℝ => B=(-7;1]
A∩B=[-3;-1]
A\B=(1;2]
B\A=(-7;-3)
