Find the value of x such that the expression \(\frac{1}{x}+\frac{4}{3x}-\frac{5}{6x}+1\) equals zero

  • A \(\frac{1}{6}\)
  • B \(\frac{1}{4}\)
  • C \(\frac{-3}{2}\)
  • D \(\frac{-7}{6}\)
waec 2000mathematics

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

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

\(\frac{6 + 8 - 5 + 6x}{6x} = 0\)

\(\frac{9 + 6x}{6x} = 0 \implies 9 + 6x = 0\)

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

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