The inequality is $\frac{2x-1}{3}$ ≥ $\frac{3x-2}{4}$ - $\frac{2-x}{5}$ Multiplying each term by L.C.M. of 3, 4, 5, i.e., by 60 $\frac{2x-1}{3}$ × 60 ≥ $\frac{3x-2}{4}$ × 60 - $\frac{2-x}{5}$ × 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].