WAEC - Mathematics (2022 - No. 17)
.Find the value of x such that \(\frac{1}{x}\) +\(\frac{4}{3x}\) - \(\frac{5}{6x}\) + 1 = 0
\(\frac{1}{6}\)
\(\frac{1}{4}\)
\(\frac{-3}{2}\)
\(\frac{-7}{6}\)
Explanation
\(\frac{1}{x}\) +\(\frac{4}{3x}\) - \(\frac{5}{6x}\) + 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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