.Find the value of x such that \(\frac{1}{x}\) +\(\frac{4}{3x}\) - \(\frac{5}{6x}\) + 1 = 0

  • A  \(\frac{1}{6}\)
  • B  \(\frac{1}{4}\)
  • C  \(\frac{-3}{2}\)
  • D  \(\frac{-7}{6}\)
waec 2022mathematics

The correct answer is C.  \(\frac{-3}{2}\)

\(\frac{1}{x}\) +\(\frac{4}{3x}\) - \(\frac{5}{6x}\) + 1 = 0

using 6x as lcm

→ \(\frac{6+8-5+6x}{6x}\) 

→ \(\frac{9+6x}{6x}\) = 0

9+6x = 0

6x = -9

x = \(\frac{-9}{6}\)  or \(\frac{-3}{2}\)

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