Question on: WAEC Mathematics - 2000
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
\(\frac{-7}{6}\)
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Correct Option: C
\(\frac{1}{x} + \frac{4}{3x} - \frac{5}{6x} + 1 = 0\)
\(\frac{6 + 8 - 5 + 6x}{6x} = 0\)
\(\frac{9 + 6x}{6x} = 0 \implies 9 + 6x = 0\)
\(6x = -9 \implies x = \frac{-3}{2}\)
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