# Absolute Value Inequalities

• Oct 21st 2011, 03:27 PM
benny92000
Absolute Value Inequalities
Abs(x-2) > -1 (That is, the absolute value of x-2 is greater than -1.) Now I have always been accustomed to setting up two inequalities, one x-2 > -1 and the other x-2 < 1. When I do this, I don't get the write answer (2.412). Is the method I use with abs value wrong? I use this whenever there is an inequality sign or an equals with an abs value on one end like in the problem above. Thank you.
• Oct 21st 2011, 03:51 PM
skeeter
Re: Absolute Value Inequalities
Quote:

Originally Posted by benny92000
Abs(x-2) > -1 (That is, the absolute value of x-2 is greater than -1.) Now I have always been accustomed to setting up two inequalities, one x-2 > -1 and the other x-2 < 1. When I do this, I don't get the write answer (2.412). Is the method I use with abs value wrong? I use this whenever there is an inequality sign or an equals with an abs value on one end like in the problem above. Thank you.

$|x-2| > -1$

x can be any value ... $|anything| \ge 0 > -1$

sure you copied the problem correctly?
• Oct 22nd 2011, 06:34 AM
benny92000
Re: Absolute Value Inequalities
I copied the answer wrongly. My book has the symbol "Z" in the back. Is my method wrong? How did you go about solving it?.
• Oct 22nd 2011, 06:41 AM
skeeter
Re: Absolute Value Inequalities
Quote:

Originally Posted by benny92000
I copied the answer wrongly. My book has the symbol "Z" in the back. Is my method wrong? How did you go about solving it?.

no "method" required ... only common sense. the absolute value of anything is greater than any negative value. kapish?
• Oct 22nd 2011, 06:44 AM
benny92000
Re: Absolute Value Inequalities
I understand that. But does that mean the method I am accustomed to using is incorrect?
• Oct 22nd 2011, 06:55 AM
skeeter
Re: Absolute Value Inequalities
Quote:

Originally Posted by benny92000
I understand that. But does that mean the method I am accustomed to using is incorrect?

your method is correct, you just interpreted the results incorrectly.

$|x-2| > -1$

this means that ...

$x-2> -1$ or $-(x-2) > -1$

$x > 1$ or $x < 3$

the union of these two sets is the set of all real numbers.