Solve $\dfrac{x + 2}{3} = \dfrac{2x - 1}{4}$.

Solve $\dfrac{x + 2}{3} = \dfrac{2x - 1}{4}$.
  1. $\dfrac{11}{2}$
  2. $5$
  3. $2$
  4. $3$

Solution

$4(x+2) = 3(2x-1) \Rightarrow 4x + 8 = 6x - 3 \Rightarrow 11 = 2x$.

Asked in: IMO

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