View Full Version : Interval Notation
MsTaylor23
Jun 7, 2007, 12:36 PM
|5x-1|<11
Above is the problem. This is my answer.
5x+1<11
5x+1-1<11-1
5x<10
x<10
ebaines
Jun 7, 2007, 12:55 PM
Not quite. You can always check your answer by substituting it for x back in the original equation. If you did that here you'd get:
|5*10-1|<11, or
49<11, which is clearly not right.
It looks like you may have a typo in the original formula: is it |5x+1|<11, or is it |5x-1|<11?
In general, I find it easiest to start by converting the single inequality using absolute value signs to a dual inequality without the absolute values. For example, here the first step would be:
|5x-1| < 11
-11 < 5x-1 <11
Now you can use normal algebra rules to get to an expression like this: a < x < b, which tells you the range for x is from a to b. Now, can you determine what the range is for this problem?
MsTaylor23
Jun 7, 2007, 01:48 PM
It is |5x-1|<11
ebaines
Jun 7, 2007, 01:54 PM
OK, so from what I told you before can you determine the interval for x?
MsTaylor23
Jun 7, 2007, 01:56 PM
I am stomped!! I was thinking that you would add the 1 on both sides of the 11 and that leaves 5x by itself
ebaines
Jun 7, 2007, 01:58 PM
Yep - which gets you to -10<5x<12. Then divide everything by 5. That leaves x by itself, in the form a<x<b.