Question

\[\frac{6x - 5}{4x + 1} < 0\]

Answer 1

\[\frac{6x - 5}{4x + 1} < 0\]
\[\text{ Equating } 6x - 5 \text{ and } 4x + 1 \text{ to zero, we obtain } x = \frac{5}{6} \text{ and } - \frac{1}{4} \text{ as the critical points } . \]

\[\therefore x \in \left( \frac{- 1}{4}, \frac{5}{6} \right)\]
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