Solve the given inequality for real x: $$\dfrac{x}{3}$$ > $$\dfrac{x}{2}$$+ 1

Asked by Abhisek | 1 year ago |  65

##### Solution :-

=  $$\dfrac{x}{3}$$ > $$\dfrac{x}{2}$$+ 1

$$\dfrac{x}{3}- \dfrac{x}{2}>1$$

$$\dfrac{2x-3x}{6}>1$$

$$- \dfrac{x}{6}>1$$

$$-x>6$$

= x < -6

Thus, all real numbers x, which are less than -6, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (-∞, -6)

Answered by Pragya Singh | 1 year ago

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