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Mathematics 21 Online
OpenStudy (anonymous):

How do I convert an inequality to an inequality with absolute symbols?

OpenStudy (anonymous):

Depends, is it descrete?

OpenStudy (anonymous):

here's an example....-3 < x < 3

OpenStudy (anonymous):

Oh, you mean absolute value.. \[|a| < b \iff -b < a < b\]

OpenStudy (anonymous):

The arrows mean that each side (left or right) means the same thing. Anytime you have one, you can write the other instead.

OpenStudy (anonymous):

\[\iff\] Those arrows I mean.

OpenStudy (anonymous):

How would I write this equation as an absolute value inequality?

OpenStudy (anonymous):

So look at my example.. What do you have that is in the same spot as the 'a' ?

OpenStudy (anonymous):

I just gave my example in general terms. you can use any value for a you want, and any value for b you want (as long as \(b \ge 0\))

OpenStudy (anonymous):

-b < a < b In your problem, what is b, and what is a?

OpenStudy (anonymous):

- b is less than a...b is great than a..

OpenStudy (anonymous):

No, I mean you have a problem in front of you that looks the same, but has numbers instead of a and b. What is the number that corresponds to the b and what is the number that corresponds to the a?

OpenStudy (anonymous):

b and a are both 3?

OpenStudy (anonymous):

No, b is 3, a is x.

OpenStudy (anonymous):

-b < a < b -3 < x < 3 See the similarity?

OpenStudy (anonymous):

So if: -b < a < b is the same as this: |a| < b Then what would -3 < x < 3 be the same as?

OpenStudy (anonymous):

IxI < 3?

OpenStudy (anonymous):

Yep! Nicely done. And if you had -4 < f < 4 ?

OpenStudy (anonymous):

IxI < 4?

OpenStudy (anonymous):

The x would be an f. But yeah.

OpenStudy (anonymous):

Here's a little trickier one.. -5 < a+b < 5

OpenStudy (anonymous):

Ia + bI < 5?

OpenStudy (anonymous):

Yes! awsome. Also remember you can go the other way.. |b| < g can be written as -g < b < g too.

OpenStudy (anonymous):

They mean the same thing.

OpenStudy (anonymous):

Can you explain another one? :)

OpenStudy (anonymous):

Sure

OpenStudy (anonymous):

So how would you simplify |3x| > 27, again?

OpenStudy (anonymous):

Ok, so here we are gonna go the other way.

OpenStudy (anonymous):

Rewrite it as an inequality without the absolute values

OpenStudy (anonymous):

-27 < 3x < 27?

OpenStudy (anonymous):

Yes. Now we want to get the x by itself, so what do we do?

OpenStudy (anonymous):

Sorry, when I say 'by itself' I mean without the 3 in front of it.

OpenStudy (anonymous):

we divide the negative 27 and 27 by 3.

OpenStudy (anonymous):

Yes. We divide everything by 3.

OpenStudy (anonymous):

the -27, the 3x, and the 27 are all divided by 3 what do we have now?

OpenStudy (anonymous):

Are each divided by 3 rather.

OpenStudy (anonymous):

leave it as is?

OpenStudy (anonymous):

Yeah, we have: \[-\frac{27}{3} < \frac{3}{3}x < \frac{27}{3}\] Which simplifies to \[-9 < x < 9\] Then we can rewrite in absolute value form again...

OpenStudy (anonymous):

actually I just needed to get it into standard inequality form..

OpenStudy (anonymous):

Oh, ok well there ya go.

OpenStudy (anonymous):

thanks. hopefully this will all stick soon...:)

OpenStudy (anonymous):

The only tricky thing to remember is that when you multiply or divide everything by a negative you have to change the direction of the inequality.

OpenStudy (anonymous):

I tend to remember that, for some reason..but anyway,thanks again..got to get back to my math. :)

OpenStudy (anonymous):

good luck =)

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