|2x – 3| < 1

Square both sides

⇒ (2x – 3)² < 1²

⇒ 4x² – 12x + 9 < 1

⇒ 4x² – 12x + 8 < 0

Divide throughout by 4

⇒ x² – 3x + 2 < 0

⇒ x² – 2x – x + 2 < 0

⇒ x(x – 2) – 1(x – 2) < 0

⇒ (x – 1)(x – 2) < 0

Observe that when x is greater than 2 (x – 1)(x – 2) is positive

And for each root the sign changes hence

We want less than 0 that is negative part

Hence x should be between 1 and 2 for (x – 1)(x – 2) to be negative

Hence x ∈ (1, 2)

Hence the solution set of |2x – 3| < 1 is (1, 2)