Simple inequality $\left|\frac{3x+1}{x-2}\right|<1$ 
$$\left|\frac{3x+1}{x-2}\right|<1$$

$$-1<\frac{3x+1}{x-2}<1$$
$$-1-\frac{1}{x-2}<\frac{3x}{x-2}<1-\frac{1}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}$$
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2} \text{ , }x \neq 2$$
$${-x+1}<{3x}<{x-3} \text{ , }x \neq 2$$
$${-x+1}<{3x} \text{  and } 3x<{x-3} \text{ , }x \neq 2$$
$${1}<{4x} \text{  and } 2x<{-3} \text{ , }x \neq 2$$
$${\frac{1}{4}}<{x} \text{  and } x<{\frac{-3}{2}} \text{ , }x \neq 2$$
While the answer is 
$${\frac{1}{4}}>{x} \text{  and } x>{\frac{-3}{2}}$$

Hint:
For $x\ne2$,
$$|3x+1|<|x-2|.$$
Then we need to distiguish three cases:


*

*$x\le-\frac13\to-3x-1<-x+2$, or $x>-\frac32$,

*$-\frac13\le x\le 2\to3x+1<-x+2$, or $x<\frac14$,

*$2\le x\to 3x+1<x-2$ or $x<-\frac32$, which is impossible.
