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 
$${\frac{1}{4}}>{x} \text{  and } x>{\frac{-3}{2}}$$
 A: We can not multiply by $x-2$ even if it is $\ne0$
In fact the inequality sign remains valid only if $x-2>0$ 
$$-1<\dfrac{3x+1}{x-2}\iff0<\dfrac{3x+1-(x-2)}{x-2}=\dfrac{x+3}{x-2}$$
So, we need $(x-2)(x+3)>0$
Now if $(x-a)(x-b)>0;a\le b$ we can prove either $x>b$  or $x<a$
A: Square to remove the absolute value: if $x\neq 2$,
$$\left|\frac{3x+1}{x-2}\right|<1\iff (3x+1)^2<(x-2)^2\iff8x^2+10x-3<0.$$
Now this quadratic polynomial has reduced discriminant $\Delta'=5^2+24=49$, whence the roots $-3/4$ and $1/4$, and it is negative between the roots, so the solutions are
$$-\frac32<x<\frac14.$$
A: 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 $$(4x-1)(2x+3)<0\implies-\frac 32<x<\frac 14$$
A: When you multiply by $x-2$, you have to worry about whether it's negative, in which case the direction of the inequalities would be reversed.  So break things into two cases: $x>2$ and $x<2$.  
If $x>2$, you can do what you did or be a little more efficient:
$$-(x-2) < 3x+1 < x-2$$
$$-x +1 < 3x < x-3$$
$$1<4x \mbox{ and } 2x<-3$$
$$\frac{1}{4}<x \mbox{ and } x<-\frac{3}{2}$$
This is the empty set because no $x$ satisfies both inequalities.
Repeat for $x<2$ but reverse the inequalities when you multiply by it,
to get 
$$\frac{1}{4}>x \mbox{ and } x>-\frac{3}{2}$$
A: $$-1<\frac{3x+1}{x-2}<1$$
Multiplying expression by $x-2$,
$-x+2<3x+1<x-2$
$-x+2<3x+1$ and $3x+1<x-2$
$1<4x$ and $2x<-3$
$\frac 14<x$ and $x<\frac{-3}{2}$
We have either $x<2$ or $x>2$.
A: 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.
A: 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
$$(4x-1)(x-2)>0\land (x-2)(2x+3)<0$$
