Practicing Success
The solution set of the inequation $\frac{1}{2}(\frac{3x}{5}+4)≥\frac{1}{3}(x-6)$ is |
[120, ∞) (-∞, 120] [0, 120] [-120, 0] |
(-∞, 120] |
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]. |