# Absolute Value

#### Petrus

##### Well-known member
Hello MHB,
solve $$\displaystyle ||x|-2|+|x+1|=3$$
and we find that $$\displaystyle x=0,-2,2,-1$$
I got problem to find the 'case',

Regards,
$$\displaystyle |\rangle$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Hello MHB,
solve $$\displaystyle ||x|-2|+|x+1|=3$$
and we find that $$\displaystyle x=0,-2,2,-1$$
I got problem to find the 'case',

Regards,
$$\displaystyle |\rangle$$
That doesn't look like the right solution...

Talking about cases, what if x<-2? Or -2<x<-1? And what about -1<x<0? Or perhaps 0<x<2? And x>2?

#### Petrus

##### Well-known member
That doesn't look like the right solution...

Talking about cases, what if x<-2? Or -2<x<-1? And what about -1<x<0? Or perhaps 0<x<2? And x>2?
I don't mean those are the answer, but those are the point we should check $$\displaystyle \geq$$ or $$\displaystyle \leq$$, I hope you did understand.
Can I also check this one insted of -1<x<0
-2<x<0?

Regards,
$$\displaystyle |\rangle$$

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I don't mean those are the answer, but those are the point we should check $$\displaystyle \geq$$ or $$\displaystyle \leq$$, I hope you did understand.
Can I also check this one insted of -1<x<0
-2<x<0?
Sure you can. It's just that |x+1| does something funny at x=-1.

#### Petrus

##### Well-known member
Sure you can. It's just that |x+1| does something funny at x=-1.
Thanks, got the correct answer now Regards,
$$\displaystyle |\rangle$$