The inequality is

(2x−1)

3

≥

(3x−2)

4

-

2−x

5

Multiplying each term by L.C.M. of 3, 4, 5, i.e., by 60

(2x−1)

3

× 60 ≥

(3x−2)

4

× 60 -

2−x

5

× 60
or, 20(2x – 1) ≥ (3x – 2) × 15 – (2 – x) × 12 or, 40x – 20 ≥ 45x – 30 – 24 + 12x
or, 40x – 20 ≥ 57x – 54
Transposing 57x to L.H.S. and –20 to R.H.S., we get
40x – 57x ≥ –54 + 20 or –17x ≥ –34
Dividing both sides by –17, we get x ≤ 2
∴ The solution is (– ∞, 2].