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

Jun 17, 2015

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 $1 > s$

Jun 17, 2015

There is no one sentence "How".
The solution set is: $\left(- 1 , 1\right) \cup \left(3 , 5\right)$

#### Explanation:

Algebraic solution

$\left\mid s - 2 \right\mid = s - 2 \text{ or } \left(- s - 2\right)$. 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 $1 > s \succ 1$

Geometric solution

$\left\mid a - b \right\mid$ is the distance between $a$ and $b$.

So $1 < \left\mid s - 2 \right\mid < 3$ 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 $5$.