How to solve: $\frac{3·x-5}{8·x-2}<6$ I'm trying to solve $\frac{3x-5}{8x-2}<6$ ?
I'm not sure which first step to take. I mean if I multiply both sides by $8x-2$ then I'm not sure if the sign would switch, as this could be positive or negative depending on $x$.
 A: Hint
You can't multiply by $8x-2$ without discuss on its sign. The best way to answer the question is:
$$\frac{3x-5}{8x-2}<6\iff\frac{3x-5}{8x-2}-6=\frac{-45x+7}{8x-2}<0$$
and now draw a signs table for this quotient.
Edit The sign table is

so the answer is $$\left(-\infty,\frac7{45}\right)\cup \left(\frac14,+\infty\right)$$
A: Multiply both sides of $\frac{3x-5}{8x-2}<6$ by $(8x-2)^2$ will make you free from the worry (if the sign would switch). 
A: Assume $8x-2>0$ (i.e. $x>\frac14$) and multiply both sides: $3x-5<48x-12$, or $x>\frac7{45}$.
Symmetrically, when $x<\frac14$ you get $x<\frac7{45}$.
Combined, $x>\frac14\lor x<\frac7{45}$.
A: Check if the denominator $=0$
Otherwise, 
$$\frac{3x-5}{8x-2}<6\iff \frac{3x-5}{8x-2}-6<0\iff\frac{45x-7}{8x-2}>0\iff (45x-7)(8x-2)>0$$
as for real $x,(8x-2)^2>0$
$$\implies\left(x-\frac7{45}\right)\left(x-\frac28\right)>0$$
If $\displaystyle(x-a)(x-b)>0,$ where $a<b$ can you prove that either $x<a$ or $x>b$ for real $a,b$
A: $$\frac{3x-5}{8x-2}<6\iff\frac{3x-5}{8x-2}-6<0\iff\frac{3x-5-48x+12}{8x-2}<0\iff$$ 
$$\iff\frac{7-45x}{2(4x-1)}<0\iff\frac{7-45x}{4x-1}<0\iff$$
$$\left(7-45x>0\text{and}4x-1<0\iff x<\frac{7}{45}\text{and} x<\frac{1}{4}\iff x<\frac{1}{4}\right)$$
or
$$\left(7-45x<0\text{and}4x-1>0\iff x>\frac{7}{45}\text{and} x>\frac{1}{4}\iff x>\frac{7}{45}\right)\iff$$
$$\iff x\in\mathbb R-\ [1/4,7/45\ ]$$
