Question

|5-8x|<=1

Answer 1

What does absolute value mean?

Answer 2

The value inside the notations will always be positive.

Answer 3

Well, kind of... :) I want you to start thinking of absolute value a little bit differently.

Absolute value is DISTANCE FROM 0. Got that?

Answer 4

Right

Answer 5

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.

Answer 6

So if:

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

Answer 7

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

Answer 8

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

Answer 9

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?

Answer 10

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?

Answer 11

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

Answer 12

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.

Answer 13

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

Answer 14

Nope

Answer 15

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<x<4

Answer 16

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

Answer 17

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<x<4 same as above.

So either way works, just kind of depends on preference.

Is that example enough for you to try your problem now? Does it make sense?

Answer 18

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

Answer 19

This makes so much more sense. Thank you so much

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

Sorry, I mean greater than

Answer 24

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?

Answer 25

The two is also in the abs value

Answer 26

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!

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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. :)

Answer 32

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