<s>[INST] 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 
$${\frac{1}{4}}>{x} \text{  and } x>{\frac{-3}{2}}$$
 [/INST] Multiplying by $x-2$ changes sign of the inequality when $x-2<0$ though you could multiply by something which is always $>0$ for example multiplying by $(x-2)^2$(assuming $x\neq2$) we get
$$\frac{-x+1}{x-2}<\frac{3x}{x-2}<\frac{x-3}{x-2}\\(1-x)(x-2)<3x(x-2)<(x-2)(x-3)$$
Which is equivalent with
$$(4x-1)(x-2)>0\land (x-2)(2x+3)<0$$
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