anonymous
  • anonymous
|5-8x|<=1
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
DebbieG
  • DebbieG
What does absolute value mean?
anonymous
  • anonymous
The value inside the notations will always be positive.
DebbieG
  • 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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Right
DebbieG
  • 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
  • DebbieG
So if: |something|<=1 then what does that MEAN? what does it tell you about the "something"?
anonymous
  • anonymous
The number will be less than or equal to 1 from zero
anonymous
  • anonymous
It has to be somewhere between 1 and 0 or even be those themselves
DebbieG
  • 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
  • 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
  • anonymous
Oh, that makes it much clearer. -1 <= 5-8x <= 1
DebbieG
  • 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
  • DebbieG
Yes, exactly! Now do you know how to solve that inequality?
anonymous
  • anonymous
Nope
DebbieG
  • 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
  • anonymous
So then it would be 1/2 <= x <= 3/4
DebbieG
  • 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
  • DebbieG
oh wait, I didn't even see your post there... lol
anonymous
  • anonymous
This makes so much more sense. Thank you so much
DebbieG
  • DebbieG
Yup, that's what I got - looks right! And you're welcome, happy to help. :)
anonymous
  • anonymous
What do I do when the absolute value is less than the integer on the right?
anonymous
  • anonymous
such as (2x-3)/2 >= 8
anonymous
  • anonymous
Sorry, I mean greater than
DebbieG
  • 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
  • anonymous
The two is also in the abs value
DebbieG
  • 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
  • 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
  • DebbieG
So you have:\[\Large \left| \dfrac{2x-3}{2} \right| \ge8\]
DebbieG
  • DebbieG
So that stuff in the abs values is the "something".... so your compound inequality is?
anonymous
  • anonymous
Oh, I get it now. Thanks!! It would be x>= 19/2 or x<=-13/2?
DebbieG
  • 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
  • anonymous
Alright I'll be able to finish the rest of this smoothly now. Thanks again!

Looking for something else?

Not the answer you are looking for? Search for more explanations.