The solution set of the inequation $\frac{1}{2}(\frac{3x}{5}+4)≥\frac{1}{3}(x-6)$ is
Answer & explanation
Correct answer: option 2
We have,
$\frac{1}{2}(\frac{3x}{5}+4)≥\frac{1}{3}(x-6)$
$⇒\frac{1}{2}(\frac{3x+20}{5})≥\frac{1}{3}(x-6)$
$⇒\frac{3x+20}{10}≥\frac{x-6}{3}$
$⇒ 3(3x+20) ≥10 (x-6)$ [Multiplying both sides by 30 i.e. the LCM of 10 and 3]
$⇒ 9x+60 ≥10x-60 $
$⇒ 9x-10x≥60-60 $
$⇒ -x≥120 $
$⇒ x ≤120 $
$⇒ x ∈ (-∞, 120]$
Hence, the solution set of the given inequation is (-∞, 120].