Find the value of t such that the expression \(\frac{1}{t}\) + \(\frac{4}{3}\) – \(\frac{5}{6t}\) + 1 is equal to zero
- \(\frac{1}{6}\)
-
\(\frac{1}{-14}\)
- - \(\frac{3}{2}\)
- \(\frac{7}{6}\)
- \(\frac{2}{5}\)
The correct answer is: B
Explanation
\(\frac{1}{t}\) + \(\frac{4}{3}\) - \(\frac{5}{6t}\) + 1 = 0
collect like terms
\(\frac{4}{3}\) + 1 = \(\frac{5}{6t}\) - \(\frac{1}{t}\)
\(\frac{4 + 3}{3}\) = \(\frac{-1}{6t}\)
cross multiply
42t = - 3
t = \(\frac{-3}{42t}\)
t = \(\frac{-1}{14}\)