The inequality is $\frac{1}{2}\left(\frac{3x}{5}+4\right)$ ≥ $\frac{1}{3}(x-6)$ or , $\frac{1}{2}\left(\frac{3x+20}{5}\right)$ ≥ $\frac{1}{3}(x-6)$ Multiplying both sides by 30, 3(3x + 20) ≥ 10(x – 6) or, 9x + 60 ≥ 10x – 60 Transposing 10x to L.H.S. and 60 to R.H.S., we get ∴ 9x – 10x ≥ – 60 – 60 or –x ≥ –120 Multiplying both sides by –1, we get x ≤ 120 ∴ The solution is (–∞, 120].