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.
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.
$\displaystyle |x-2| > -1$
x can be any value ... $\displaystyle |anything| \ge 0 > -1$
sure you copied the problem correctly?
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?.
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?
Re: Absolute Value Inequalities
I understand that. But does that mean the method I am accustomed to using is incorrect?
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.
$\displaystyle |x-2| > -1$
this means that ...
$\displaystyle x-2> -1$ or $\displaystyle -(x-2) > -1$
$\displaystyle x > 1$ or $\displaystyle x < 3$
the union of these two sets is the set of all real numbers.