<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] You can solve it more easily by splitting the modulus and squaring both sides:
Assuming $x\neq2$,
$$|3x+1|<|x-2|\implies(3x+1)^2-(x-2)^2<0$$
$$\implies(3x+1+x-2)(3x+1-x+2)<0$$
So $$(4x-1)(2x+3)<0\implies-\frac 32<x<\frac 14$$
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