Part I : 5x + 2 < 3x + 8

⇒ 2x < 6

⇒ x < 3

Part II :

Case I : -3x + 6 < 0 and x – 1 > 0

⇒ x > 2 and x > 1

By taking intersection x ∈ (2, ∞)

Case II : -3x + 6 > 0 and x – 1 < 0

⇒ x < 2 and x < 1

By takin intersection x ∈ (-∞, 1)

Taking union of case I and case II, x ∈ (-∞, 1) (2, ∞)

We have to take intersection of part I and part II, we have

final answer i.e., x ∈ (-∞, 1) (2, 3)