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Absolute value inequalities word problem

Sal solves a word problem about a carpenter by writing an appropriate absolute value inequality and solving it. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • marcimus pink style avatar for user Sammy Rea (>")> <("<) <3
    couldnt you just add 2.50 to 150 and subtract 2.50 from 150 and get the anwser?
    (13 votes)
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  • blobby green style avatar for user chang.nou012
    How would you solve an equation like this: 8 | x - 3 / 2 | > 40 ? *This is only an example.
    (3 votes)
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    • male robot donald style avatar for user Jeremy
      Chang,

      I would solve it like this:
      First, divide both sides of the equation by the 8, leaving:
      |x-3/2|>5

      Now, you have to deal with the absolute value symbol. The easiest way to do this is to turn the one equation into two equations. Absolute values have the effect of changing what is inside the symbols into a positive number if they weren't already, so we have to consider the case where they were already positive, and the case where what's inside the symbol was negative to begin with. This will give you two answers.

      The first case: Everything was already positive:
      |x-3/2|> 5 becomes x-3/2>5. So you add 3/2 to both sides and one of the two answers is x>6 1/2.

      Second case: (x-3/2) was a negative number, whose absolute value was greater than 5. That means that (x-3/2) was a number that was less than negative 5:

      |x-3/2|>5 becomes (x-3/2) < -5 (NOTICE THE DIRECTION OF THE INEQUALITY SWITCHED).

      Now add 3/2 to both sides and x < -6 1/2.

      So the two answers are x>6 1/2 and x<-6 1/2.
      (10 votes)
  • male robot hal style avatar for user Ger
    From the initial question we are told that there is a margin of error of 2.5mm, so my question is why was the answer w<=152.5 and w>=147.5 which would give an overall difference of 5mm not 2.5 right?
    (5 votes)
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  • aqualine ultimate style avatar for user haynerom000
    How do you write inequalities?

    For example: "Passengers have to be at least 12 years of age."
    (5 votes)
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    • aqualine ultimate style avatar for user Robby Olivam
      An inequality is written on a keyboard as:
      Less than: <
      Greater than: >
      Less than or equal to: <=
      Greater than or equal to: >=

      When converting sentences to inequalities, it's helpful to plug in a couple numbers to see if they fit. If they need to be at least 12, then that means they can be 12. That's a flag telling you that the equal sign will be included in the answer. An 11 year old would not be able to ride the bus, but a thirteen year old would. Essentially, when a person's age is greater than 12, the equation also works.
      Therefore: age >= 12
      (6 votes)
  • starky sapling style avatar for user Elsa
    At , how do we know that it is a less than or equal to sign? How can you tell which sign it is going to be?
    (5 votes)
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  • blobby green style avatar for user Mrs Dennis
    I can't seem to find practice problems for Absolute value inequalities? Only lessons?
    (4 votes)
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  • blobby green style avatar for user carle
    At , the term of "absolute value" is supplemented by the term "absolute error". Neither of these terms is explained in detail. I can assume I understand what "absolute value" means but I would appreciate a little elaboration, as well as the use of the two lines on either side of the figure. I may have missed the lesson on this but it is the first time I have ever seen these symbols. Please explain their usage with a few examples outside of the present one, for comparison.
    (0 votes)
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  • female robot amelia style avatar for user Walking Dead Lover
    Is there a video for proving inequlities using substitution
    (2 votes)
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  • piceratops seed style avatar for user Gabriel R.
    |w-150| means it doesn't matter if will be up to 2.5 milimeters missing or there will be up to 2.5 milimeters left or leaving(I don't know, i'm foreign 7;^(] ) and the second part just says that it must be someway about 2.5 milimeters.
    (2 votes)
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  • blobby green style avatar for user Brigitte Kock
    How to compute for example: ∣∣8x−1∣∣ ≤ ∣∣4x−6∣∣? (The double stripe are absolute lines)
    And what about: 5/(x−1) < 3/(x−8)?
    Thanks for the rest through! ;)
    (2 votes)
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Video transcript

A carpenter is using a lathe to shape the final leg of a hand-crafted table. A lathe is this carpentry tool that spins things around, and so it can be used to make things that are, I guess you could say, almost cylindrical in shape, like a leg for a table or something like that. In order for the leg to fit, it needs to be 150 millimeters wide, allowing for a margin of error of 2.5 millimeters. So in an ideal world, it'd be exactly 150 millimeters wide, but when you manufacture something, you're not going to get that exact number, so this is saying that we can be 2 and 1/2 millimeters above or below that 150 millimeters. Now, they want us to write an absolute value inequality that models this relationship, and then find the range of widths that the table leg can be. So the way to think about this, let's let w be the width of the table leg. So if we were to take the difference between w and 150, what is this? This is essentially how much of an error did we make, right? If w is going to be larger than 150, let's say it's 151, then this difference is going to be 1 millimeter, we were over by 1 millimeter. If w is less than 150, it's going to be a negative number. If, say, w was 149, 149 minus 150 is going to be negative 1. But we just care about the absolute margin. We don't care if we're above or below, the margin of error says we can be 2 and 1/2 above or below. So we just really care about the absolute value of the difference between w and 150. This tells us, how much of an error did we make? And all we care is that error, that absolute error, has to be a less than 2.5 millimeters. And I'm assuming less than-- they're saying a margin of error of 2.5 millimeters-- I guess it could be less than or equal to. We could be exactly 2 and 1/2 millimeters off. So this is the first part. We have written an absolute value inequality that models this relationship. And I really want you to understand this. All we're saying is look, this right here is the difference between the actual width of our leg and 150. Now we don't care if it's above or below, we just care about the absolute distance from 150, or the absolute value of that difference, so we took the absolute value. And that thing, the difference between w a 150, that absolute distance, has to be less than 2 and 1/2. Now, we've seen examples of solving this before. This means that this thing has to be either, or it has to be both, less than 2 and 1/2 and greater than negative 2 and 1/2. So let me write this down. So this means that w minus 150 has to be less than 2.5 and w minus 150 has to be greater than or equal to negative 2.5. If the absolute value of something is less than 2 and 1/2, that means its distance from 0 is less than 2 and 1/2. For something's distance from 0 to be less than 2 and 1/2, in the positive direction it has to be less than 2 and 1/2. But it also cannot be any more negative than negative 2 and 1/2, and we saw that in the last few videos. So let's solve each of these. If we add 150 to both sides of these equations, if you add 150-- and we can actually do both of them simultaneously-- let's add 150 on this side, too, what do we get? What do we get? The left-hand side of this equation just becomes a w-- these cancel out-- is less than or equal to 150 plus 2.5 is 152.5, and then we still have our and. And on this side of the equation-- this cancels out-- we just have a w is greater than or equal to negative 2.5 plus 150, that is 147.5. So the width of our leg has to be greater than 147.5 millimeters and less than 152.5 millimeters. We can write it like this. The width has to be less than or equal to 152.5 millimeters. Or it has to be greater than or equal to, or we could say 147.5 millimeters is less than the width. And that's the range. And this makes complete sense because we can only be 2 and 1/2 away from 150. This is saying that the distance between w and 150 can only at most be 2 and 1/2. And you see, this is 2 and 1/2 less than 150, and this is 2 and 1/2 more than 150.