Solve:
2x+33≥3x−14\dfrac{2x+3}{3} \ge \dfrac{3x−1}{4}32x+3≥43x−1 where x is positive even integer.
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Given,
2x+33≥3x−14⇒2x3+33≥3x4−14⇒2x3+1≥3x4−14⇒2x3−3x4≥−14−1⇒8x−9x12≥−54⇒−x12≥−54⇒x12≤54⇒x≤54×12⇒x≤15\dfrac{2x+3}{3} \ge \dfrac{3x−1}{4}\\[0.5em] \Rightarrow \dfrac{2x}{3} + \dfrac{3}{3} \ge \dfrac{3x}{4}-\dfrac{1}{4}\\[0.5em] \Rightarrow \dfrac{2x}{3} + 1 \ge \dfrac{3x}{4} - \dfrac{1}{4}\\[0.5em] \Rightarrow \dfrac{2x}{3} - \dfrac{3x}{4} \ge -\dfrac{1}{4}-1\\[0.5em] \Rightarrow \dfrac{8x-9x}{12} \ge - \dfrac{5}{4}\\[0.5em] \Rightarrow -\dfrac{x}{12} \ge -\dfrac{5}{4}\\[0.5em] \Rightarrow \dfrac{x}{12} \le \dfrac{5}{4}\\[0.5em] \Rightarrow x \le \dfrac{5}{4} \times 12\\[0.5em] \Rightarrow x \le 1532x+3≥43x−1⇒32x+33≥43x−41⇒32x+1≥43x−41⇒32x−43x≥−41−1⇒128x−9x≥−45⇒−12x≥−45⇒12x≤45⇒x≤45×12⇒x≤15
Since, x is a positive even integer.x = {2, 4, 6, 8, 10, 12, 14}.
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