Solve the given inequality for real x: $$\dfrac{1}{2}(\dfrac{3x}{5}+4)≥\dfrac{1}{3}(x-6)$$

Asked by Abhisek | 1 year ago |  64

##### Solution :-

$$\dfrac{1}{2}(\dfrac{3x}{5}+4)≥\dfrac{1}{3}(x-6)$$

$$3(\dfrac{3x}{5}+4)≥2(x-6)$$

$$\dfrac{9x}{5}+12≥2x-12$$

$$12+12≥2x- \dfrac{9x}{5}$$

$$24≥\dfrac{10x-9x}{5}$$

$$24≥\dfrac{x}{5}$$

$$120≥x$$

Thus, all real numbers x, which are less than or equal to 120, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (-∞, 120]

Answered by Pragya Singh | 1 year ago

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