|5-8x|<=1

- anonymous

|5-8x|<=1

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- schrodinger

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- DebbieG

What does absolute value mean?

- anonymous

The value inside the notations will always be positive.

- DebbieG

Well, kind of... :) I want you to start thinking of absolute value a little bit differently.
Absolute value is DISTANCE FROM 0. Got that?

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## More answers

- anonymous

Right

- DebbieG

So, |5|=5 since 5 is 5 units from 0. and |-5| = 5 since it is also 5 units from 0, just on the other side.

- DebbieG

So if:
|something|<=1 then what does that MEAN? what does it tell you about the "something"?

- anonymous

The number will be less than or equal to 1 from zero

- anonymous

It has to be somewhere between 1 and 0 or even be those themselves

- DebbieG

OK, the number... or variable... or expression.. whatever the "stuff" inside the abs value sign is, it lives WITHIN 1 unit of 0.
So if I have |z|<=1, then that means that z is WITHIN 1 unit of 0, so:
-1<=z<=1
If I have |t+3|<=1, then that means that (t+3) is WITHING 1 unit of 0, so:
-1<=t+3<=1
So do you see how an understanding of what that abs value expression really MEANS allows me to re-write it as a plain "between inequality" that doesn't involve abs value?

- DebbieG

So what does that mean for your problem: |5-8x|<=1
How can you rewrite that, as an equivalent mathematical statement that does not involve the abs value signs?

- anonymous

Oh, that makes it much clearer. -1 <= 5-8x <= 1

- DebbieG

Be careful, when you say "It has to be somewhere between 1 and 0 or even be those themselves" I worry that you are forgetting what abs value means. It is DISTANCE FROM 0. So if |z|<=1 then it could be that z=-1, or z=-0.4 or z=1... anything BETWEEN -1 and 1 will satisfy the abs value inequality.

- DebbieG

Yes, exactly! Now do you know how to solve that inequality?

- anonymous

Nope

- DebbieG

You solve an inequality just like an equation, except that IF you multiply or divide by a negative number, then you have to flip the direction of the ineq. symbol.
when you have a "compound inequality" like this one which is called "between notation", you have a couple of options.
You can either solve it just like it's a "3 part equation", so anything that you do to one part you do to the other 2.
e.g. if I had:
-3<1-x<3 I would subtract 1 from all "3 parts"
-4<-x<2 and then I would multiply each part by -1
4>x>-2 Notice I had to flip the inequalities!
Then I would probably re-write it with the order reversed, just because I think it makes more sense to look at it that way:
-2

- anonymous

So then it would be 1/2 <= x <= 3/4

- DebbieG

Option 2 is to break it apart into 2 separate inequalities, with an "and" between them, and solve them separately, following all the same steps:
-3<1-x<3
-3<1-x AND 1-x<3
-4<-x AND -x<2
4>x AND x>-2
Notice when I'm finished and I re-combine into between notation, I have:
-2

- DebbieG

oh wait, I didn't even see your post there... lol

- anonymous

This makes so much more sense. Thank you so much

- DebbieG

Yup, that's what I got - looks right! And you're welcome, happy to help. :)

- anonymous

What do I do when the absolute value is less than the integer on the right?

- anonymous

such as (2x-3)/2 >= 8

- anonymous

Sorry, I mean greater than

- DebbieG

So I'm assuming that's an abs value in the ( )?
Is it:
\[\Large \frac{ \left| 2x-3 \right| }{ 2 }\ge8\] ??
Or is the den'r with the 2 also in the abs value?

- anonymous

The two is also in the abs value

- DebbieG

Very good intuition that you need to something a little bit different. :) But again, remember that abs value is the DISTANCE FROM 0.
so if |z|>3, then there are TWO ways that can happen.... either z is more than 3 units TO THE LEFT OF 0 or it is more than 3 units TO THE RIGHT of 0. Either way makes it true... so I can rewrite:
|z|>3 as ---> z<-3 OR z>3
Notice that it's again a "compound inequality", except this time I have an OR in the middle, not an AND. That's important. And notice that it makes sense, since z certainly can't be BOTH z<-3 and z>3, but it certainly could be one or the other!

- DebbieG

So it doesn't matter what the "something" is... if:
|something|>a where a is some positive number, then the "something" is AT LEAST a units from 0, so:
something<-a OR something>a

- DebbieG

So you have:\[\Large \left| \dfrac{2x-3}{2} \right| \ge8\]

- DebbieG

So that stuff in the abs values is the "something".... so your compound inequality is?

- anonymous

Oh, I get it now. Thanks!! It would be x>= 19/2 or x<=-13/2?

- DebbieG

Yes, that's it! Notice that it's a "2 part" solution set, unlike the first one that is a "one part" set. :) If you are doing interval notation and set unions & intersections, you would write this one as a UNION of 2 sets. But if not, the inequality notation for the solution set should be fine too. :)

- anonymous

Alright I'll be able to finish the rest of this smoothly now. Thanks again!

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