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Question

Mathematics Question on inequalities

Solve the given inequality for realx:12(3x5+4)≥13(x−6)x: \frac{1}{2}(\frac{3x}{5}+4) ≥ \frac{1}{3}(x-6).

Answer

12(3x5+4)≥13(x−6)\frac{1}{2}(\frac{3x}{5}+4) ≥ \frac{1}{3}(x-6)
⇒3(3x5+4)≥2(x−6)⇒ 3(\frac{3x}{5} + 4) ≥ 2(x - 6)
⇒9x5+12≥2x−12⇒ \frac{9x}{5} + 12 ≥ 2x - 12
⇒12+12≥2x−9x5⇒ 12 + 12 ≥ 2x - \frac{9x}{5}
⇒24≥10x−9x5⇒ 24 ≥ \frac{10x - 9x}{5}
⇒24≥x5⇒ 24 ≥ \frac{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].