Solve for $(5x-1)(x-3)<0$ The inequality $(5x-1)(x-3)<0$ is true when 
$(5x-1)<0$ and $(x-3)>0$ 
or 
$(5x-1)>0$ and $(x-3)<0$.
If I solve for $x $ in the first scenario, $x < \frac{1}{5}$ and $x > 3$ which is wrong. But if I solve for $x$ in the second scenario, $x > \frac{1}{5}$ 
and $x < 3$ which is correct. 
Why is such kind of contradiction occurring in the first scenario?

As $x$ increases, the two linear factors are negative, then positive, and each changes sign once, at a root. So one changes sign before the other, and the combinations $--,+-,++$ are possible, but not $-+$.
