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Solve the following inequalities for real x. \(\frac{2x-3}{4}\) − 3< \(\frac{x-4}{3}\) −2, x ∈ R. |
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Answer» We have, \(\frac{2x−3}{4}\)−3 < \(\frac{x−4}{3}\)−2, ⟹ \(\frac{2x−9}{4}\) < \(\frac{x−10}{3}\) ⟹ \(\frac{x}{2}\) − \(\frac{x}{3}\) < \(\frac{−10}{3}\) + \(\frac{9}{4}\) [Transposing \(\frac{x}{2}\)to LHS and \(\frac{9}{4}\) to RHS] ⟹ \(\frac{x}{6}\) < \(\frac{13}{12}\) ⟹ x < \(\frac{13}{2}\) ∴ x ∈ (−∞, \(\frac{13}{2}\)) |
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