Solve for x, the inequalities in |x – 1| ≤ 5,

|x – 1|≤ 5

There are two cases,

1:-

x – 1 ≤ 5

Adding 1 to LHS and RHS

⇒ x ≤ 6

2:-

⇒ -(x – 1) ≤ 5

⇒ -x + 1 ≤ 5

Subtracting 1 from LHS and RHS,

⇒ -x ≤ 4

⇒ x ≥ -4

From cases 1 and 2, we have

⇒ -4 ≤ x ≤ 6 …[i]

Also,

|x| ≥ 2

⇒ x ≥ 2 and

⇒ -x ≥ 2

⇒ x ≤ -2

⇒ x ∈ (∞, -2] ∪ [2, ∞) …[ii]

Combining equation [i] and [ii], we get

x ∈ [-4, -2] ∪ [2, 6]