Find the value of t such that the expression \(\frac{1}{t}\) + \(\frac{4}{3}\) - \(\frac{5}{6t}\) + 1 is equal to zero

A.

\(\frac{1}{6}\)

B.

\(\frac{1}{-14}\)

C.

- \(\frac{3}{2}\)

D.

\(\frac{7}{6}\)

E.

\(\frac{2}{5}\)

Correct answer is B

\(\frac{1}{t}\) + \(\frac{4}{3}\) - \(\frac{5}{6t}\) + 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 =  \(\frac{-1}{14}\)