Answer
$(-\infty,-\frac{2}{3})U(0,\frac{3}{2})$
Work Step by Step
Step 1. Rewrite the inequality as $6x-5-\frac{6}{x}\lt0$ or $\frac{(2x-3)(3x+2)}{x}\lt0$. Identify boundary points (zeros and asymptotes) as $x=-\frac{2}{3},0, \frac{3}{2}$ which divide the x-axis into four intervals $(-\infty,-\frac{2}{3}),(-\frac{2}{3},0),(0,\frac{3}{2}),(\frac{3}{2},\infty)$
Step 2. Choose test values in each interval $x=-1,-\frac{1}{3},1,2$ and test the inequality to get $True,\ False,\ True,\ False$
Step 3. Thus the solution intervals are $(-\infty,-\frac{2}{3})U(0,\frac{3}{2})$
