Question

How do you solve `1<abs(s-2)<3`?

Answer 1

It's complicated.

Explanation:

First let's split this into two parts.

`1 < |s-2|` and `|s-2|<3`

Here's where things get confusing. Absolute value means that the number could either be positive or negative. If it's negative, the sign is flipped.

`1 < s-2` or `-1 > s-2` and `s-2<3` or `s-2> -3`

`3 < s` or `1>s` and `s < 5` or `s> -1`

Wait, look at that 2nd part. If s can be less than 5 (a bigger number), or more than -1 (a smaller number), then s can be anything (at least in one conditional). So this can be rephrased as:

`3< s` or `1>s` and `s =`all real numbers

The "and" means that 's' can be anything where these 2 conditionals overlap. It's pointless to ask if something overlaps with all real numbers, so we can get rid of that, leaving us with the final answer:

`3< s` or `1>s`

Answer 2

There is no one sentence "How".
The solution set is: `(-1,1) uu (3,5)`

Explanation:

Algebraic solution

`abs (s-2) = s-2 " or " (-s-2)`. So solve the two inequalities:

`1 < s-2 < 3` which gives us `3 < s < 5`

`1 < -s+2 < 3` which yields `-1 < -s < 1`. Multiply through by `-1` (change the direction of the inequalities) to get `1 > s >-1`

Geometric solution

`abs (a-b)` is the distance between `a` and `b`.

So `1 < abs (s-2) <3` means that `s` is more than `1` and less than `3` away from `2`.

On the number line, find `2`.
Look on the right for the points (numbers) more than `1`, but less that `3` away from 2. These are the numbers between 3 and 5 (not including 3 and 5.)
Look on the left for the points (numbers) more than `1`, but less that `3` away from 2. These are the numbers between 1 and -1 (not including 1 and -1.)

So we need s to be between `-1` and `1` OR between `3` and `5`.