Formulating absolute value linear equations

0:00- [Instructor] We're told that Antonella is a quality control inspector at a factory that manufactures ball bearings. Each bearing has a target diameter of 40 millimeters, but some variation in size is possible due to the manufacturing process. Based on industry standards, the factory will only ship bearings that are within 4%, within 4% of the target diameter.

0:24Antonella needs to determine the minimum and maximum acceptable diameters for these ball bearings, right? An absolute value equation that appropriately models the situation. So have a go at that and then we will do it together. Okay, so if we were to say that X is going to be the measurement of the diameter of any one of the bearings, we wanna figure out the distance between X, and in this case 40 millimeters.

0:51This, we can sometimes refer to as our central value. So we wanna take the difference between that and 40 millimeters. Now, sometimes X might be larger than 40 millimeters, where this is gonna be a positive number, but sometimes it's gonna be less than, but all we care about is the distance.

1:06We don't care if it's positive or negative. So we're going to take the absolute value of that. So this would tell you the distance between any one of those bearings and the ideal, the target diameter. Now, we know that we want to be within 4% of the target diameter.

1:23So what is 4% of the target diameter? That would be, this would be 0.04 times 40 millimeters. So we want, this right over here is 4% of target. Now, if we just wanted to say, what are all of the ball bearings that would satisfy that? We would say something like this needs to be less than or equal to for something to be acceptable.

1:52But they're not asking us for what are all of the ones that are going to be acceptable, they wanna know what are the minimum and maximum acceptable values. And so for the minimum and maximum acceptable values, we actually want to do equal here because this is a situation where the distance is maximally far away.

2:11And it'll be the minimum in the cases where X is less than 40, and it'll be the maximum in the cases where X is greater than 40. And so this right over here is our equation. If we wanted to write it out a little bit differently, we could write the absolute value of X minus.

2:28I'll stick to those same colors. X minus 40 is equal to, and 0.04 times 40, we can think of it this way, zero point. Well, let me write, well, zero point is. Let me do it this way. 4 times 40 is 160, but we have two places behind the decimal. So two places behind the decimal, so it's 1.60.

2:58And then we can actually solve this. If the absolute value of something is equal to 1.60, or we could even say 1.6, that means that what's ever in the absolute value and that's X minus 40, that means that X minus 40 is either going to be equal to positive 1.6, is gonna be equal to positive 1.6 or X minus 40, whatever was in the absolute value, is equal to negative 1.6.

3:31Is equal to negative, no, I'm running outta space, negative 1.6. And we know that because whether it's, if it's positive 1.6, you take the absolute value, you're gonna get 1.6. And if it's negative 1.6, you take the absolute value and you're going to get to 1.6.

3:48And now we can solve each of these by subtracting 40 from both sides. And I could, or lemme say, add 40 actually, to both sides. And on this left-hand one, we are left with X is equal to 41.6, or X is equal to, on the right hand side, we have to add 40 everywhere.

4:15So plus 40 and plus 40 there as well. Make sure I do that, all right. We're doing algebra, we have to do things to both sides. Or in this case we have X is equal to negative 1.6 plus 40. 40 minus 1.6 is 38.4. And so you can see this is the minimum value that a ball bearing has to meet and this is the maximum value.