If |x – 1| > 5, then

Given,

|x – 1| > 5

There are two cases,

(x – 1) > 5

⇒ x > 6

⇒ x ∈ (6, ∞) [1]

And

-(x – 1) > 5

⇒ -x + 1 > 5

⇒ -x > 4

⇒ x < -4 [Multiplication by negative inverts the inequality sign]

⇒ x ∈ (-∞, -4) [2]

From [1] and [2], we get

⇒ x ∈ (-∞, -4) ∪ (6, ∞)