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Mathematics Question on inequalities

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

Answer

(2x−1)3≥(3x−2)4−(2−x)5\frac{ (2x -1)}{3} ≥ \frac{(3x -2)}{4} - \frac{(2-x)}{5}
⇒(2x−1)3≥5(3x−2)−4(2−x)20⇒ \frac{(2x -1)}{3} ≥ \frac{5(3x -2) - 4(2-x)}{20}
⇒(2x−1)3≥15x−10−8+4x20⇒ \frac{(2x -1)}{3} ≥ \frac{15x -10 - 8 + 4x}{20}
⇒(2x−1)3≥19x−1820⇒\frac{ (2x -1)}{3} ≥ \frac{19x - 18}{20}
⇒40x−20≥57x−54⇒ 40x - 20 ≥ 57x - 54
⇒−20+54≥57x−40x⇒ - 20 + 54 ≥ 57x - 40x
⇒34≥17x⇒ 34 ≥ 17x
⇒2≥x⇒ 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].