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