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

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

Solution

LCM 20: $4(x+3) - 5(x-1) = 20 \Rightarrow 4x + 12 - 5x + 5 = 20 \Rightarrow -x + 17 = 20 \Rightarrow x = -3$.

Asked in: IMO

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