How to solve this absolute value inequality? I am new to absolute value inequalities.I was looking trough a book and I found this inequality,
I know a little bit about absolute value inequalities.
The inequality is given below:
$$
\left| \frac{n+1}{2n+3} - \frac{1}{2} \right| > \frac{1}{12}, \qquad n \in \mathbb{Z}
$$
 A: We have that
$$\left|\frac{n+1}{2n+3}-\frac12\right|=\left|\frac{-1}{2(2n+3)}\right| =\frac{1}{|2(2n+3)|}>\frac1{12}\iff |2n+3|<6 $$
$$\iff -6<2n+3<6 \iff -9<2n<3 \iff -\frac92<n<\frac32$$
that is $n\in\{-4,-3,-2,-1,0,1\}$.
A: First simplify the expression inside the absolute value:
$$\left|\frac{n+1}{2n+3}-\frac12\right|=\left|\frac{-1}{4n+6}\right|=\left|\frac{1}{4n+6}\right|>\frac1{12}$$
Now take the reciprocal of both sides:
$$|4n+6|<12$$
and this can be interpreted as:
$$-12 < 4n + 6 < 12$$
then finish the rest.
A: As $2n+3\ne0$,
$$\left|\frac{n+1}{2n+3}-\frac12\right|>\frac1{12}$$
can be written (multiplying by $12|2n+3|$)
$$6>|2n+3|.$$
The odd numbers below $6$ are $1,3,5$ (in absolute value), from which we draw
$$n=-4,-3,-2,-1,0,1.$$

As the number of solutions is small, I preferred to exhaust them rather than work out the inequalities.
A: As it is an absolute value
Case 1)
$$\frac{n+1}{2n+3} - \frac{1}{2} > \frac{1}{12} $$
Case 2)
$$\frac{n+1}{2n+3} - \frac{1}{2} < \frac{-1}{12}$$
Now I can solve it a little for you , as in Case 1
$$\frac{n+1}{2n+3} - \frac{1}{2} - \frac{1}{12}>0 $$
$$\frac{n+1}{2n+3} - \frac{7}{12} >  0 $$
A: For case 1:
$ \frac{n+1}{2n+3} - \frac{1}{2} > \frac{1}{12} $
$ \frac{n+1}{2n+3} - \frac{2n+3}{2(2n+3)} > \frac{1}{12} $
$ \frac{-1}{2(2n+3)} > \frac{1}{12} $
$ \frac{-6}{2n+3} > 1 $
$ \Rightarrow 2n+3 < 0 $ and $ 2n+3 > -6 $
$ \frac{-9}{2} > n < \frac{-3}{2} $
