Given

For this inequation to be true, there are two possible cases.

i. 2x – 3 ≥ 0 and x – 1 > 0

⇒ 2x – 3 + 3 ≥ 0 + 3 and x – 1 + 1 > 0 + 1

⇒ 2x ≥ 3 and x > 1

However,

Hence,

ii. 2x – 3 ≤ 0 and x – 1 < 0

⇒ 2x – 3 + 3 ≤ 0 + 3 and x – 1 + 1 < 0 + 1

⇒ 2x ≤ 3 and x < 1

However,

Hence, x ∈ (–∞, 1)

Thus, the solution of the given inequation is.