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

Asked by Pragya Singh | 1 year ago |  87

Solution :-

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

$$\dfrac{(2x-1)}{3}≥\dfrac{5(3x-2)-4(2-x)}{20}$$

$$\dfrac{(2x-1)}{3}≥\dfrac{15x-10-8+4x}{20}$$

$$\dfrac{(2x-1)}{3}≥\dfrac{19x-18}{20}$$

$$20(2x-1)≥3(19x-18)$$

$$40x-20≥57x-54$$

$$-20+54≥57x-40x$$

$$34≥17x$$

$$2≥x$$

Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (-∞, 2]

Answered by Abhisek | 1 year ago

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