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## Algebra (all content)

### Course: Algebra (all content)>Unit 8

Lesson 3: Solving absolute value inequalities

# 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.

## 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.