Solving Rational Inequality

Rearranging the inequality: $\frac{1}{x+5} - \frac{1}{2x-3} \le 0$ $\frac{(2x-3) - (x+5)}{(x+5)(2x-3)} \le 0 \implies \frac{x - 8}{(x+5)(2x-3)} \le 0$

Critical points are $x = -5, \frac{3}{2}, 8$. Testing intervals:

• $x < -5$: $\frac{(-)}{(-)(-)} = - \le 0$ (Satisfied)
• $-5 < x < \frac{3}{2}$: $\frac{(-)}{(+)(-)} = + > 0$ (Not satisfied)
• $\frac{3}{2} < x \le 8$: $\frac{(-)}{(+)(+)} = - \le 0$ (Satisfied)
• $x > 8$: $\frac{(+)}{(+)(+)} = + > 0$ (Not satisfied)

Note that $x \ne -5$ and $x \ne \frac{3}{2}$ due to division by zero. Hence, $x < -5$ or $\frac{3}{2} < x \le 8$.