Absolute Value Equations

Date: 10/16/98 at 22:24:55
From: Calvin DeRoo
Subject: Absolute value in algebra

The solution set is -3 < x < 11, and I'm supposed to make an equation 
out of it involving absolute value. I just don't know how to set this 
problem up right. Please help. I made a number line but I still don't 
know. Thank you.

Date: 10/16/98 at 22:52:54
From: Doctor Pat
Subject: Re: Absolute value in algebra

Calvin,

The number line is a good first step. 

For absolute value problems it is easiest to think of center and 
distance. I'll explain. Look at the set of numbers on the number line 
you have described by -3 < x < 11. In English, we want to express all 
the numbers "between -3 and ll". To do that with absolute value, first 
find the number in the center. The midpoint of the segment of the 
number line from -3 to 11 is at 4.  

Now look at four, and the numbers in your set. Some of them are very 
close to 4, some are a little farther away, but how far are the most 
distant points?  Well, 11 is seven units away, and -3 is also seven 
units away. So another way to describe these is by saying they are the 
numbers that are LESS THAN 7 units away from 4. Absolute value and 
distance are related because the distance of a number, x, on the number 
line away from the number 4 is given by |x-4|. The absolute value is 
needed because 3 and 5 are both 1 unit away, and without the absolute 
value we would get a distance of -1, which doesn't make sense in 
geometric distance (although we could use it as a VECTOR to tell us 
direction).  

From all this I hope you can see that to express the numbers centered 
at four and less than 7 units away, we could write |x-4| < 7  

Can you see that:

    |x-6| < 2  would be the numbers between 4 and 8   
   |x-15| < 5  would be between 10 and 20 
    |x+5| < 3  would be between -8 and -2 because |x-(-5)| < 3

To show points farther than a given distance we would change the sign.
The set x > 10 or x < 6 would be |x-8| > 2 since they are all MORE 
than 2 away from the center, 8.  

Hope that helps. Good luck.

- Doctor Pat, The Math Forum
  http://mathforum.org/dr.math/