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Question
\[\frac{2x - 3}{3x - 7} > 0\]
Solution
\[\frac{2x - 3}{3x - 7} > 0\]
\[\text{ Equating } 2x - 3 \text{ and } 3x - 7 \text{ to zero, we obtain } x = \frac{3}{2}, \frac{7}{3} \text{ as the critical points } . \]
\[\therefore x \in \left( - \infty , \frac{3}{2} \right) \cup \left( \frac{7}{3}, \infty \right)\]
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