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- #1

#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

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,

\(\displaystyle |\rangle\)

- Thread starter Petrus
- Start date

- Thread starter
- #1

- Feb 21, 2013

- 739

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,

\(\displaystyle |\rangle\)

- Admin
- #2

- Mar 5, 2012

- 9,314

That doesn't look like the right solution...

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,

\(\displaystyle |\rangle\)

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?

- Thread starter
- #3

- Feb 21, 2013

- 739

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.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?

Can I also check this one insted of -1<x<0

-2<x<0?

Regards,

\(\displaystyle |\rangle\)

- Admin
- #4

- Mar 5, 2012

- 9,314

Sure you can. It's just that |x+1| does something funny at x=-1.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?

- Thread starter
- #5

- Feb 21, 2013

- 739

Thanks, got the correct answer nowSure you can. It's just that |x+1| does something funny at x=-1.

Regards,

\(\displaystyle |\rangle\)