Common Core Algebra I.Unit 4.Lesson 1.Proportional Relationships
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Learning the Common Core Algebra I.Unit 4.Lesson 1.Proportional Relationships by eMathInstruction
Hello and welcome to another common core algebra one lesson. By E math instruction. My name is Kirk Weiler, and today we're going to be doing unit four lesson one on proportional relationships. Before we get into the lesson, let me remind you that you can find a worksheet. And a homework that go along with this video. By clicking on the video's description, or by visiting our website at WWW dot E math instruction dot com. Also, don't forget about the QR codes that are on the upper right hand corner of all of our worksheets. Those codes will allow you to scan with your smartphone or a tablet. And come right to these videos. Great. Let's get into it.
The basis of all linear relationships relationships that when graphed look like lines are what are known as proportional relationships. Proportional relationships of things that you've been working with since 6th grade. So hopefully, this will seem pretty familiar to each other. But let's take a look at the technical definition. Two variables have a proportional relationship if their respective values are always in the same ratio. They have the same size relative to one another. An equation form, why divided by X is a constant. Now, what this really is getting at is the idea that when two variables are in a proportional relationship. One of them is always two times the other, or one half the other, or 7 fifths of the other, right? In other words, you can always produce one by multiplying the other by that constant. That constant is known as the constant of proportionality, the constant of proportionality. Okay? So today, what we're going to be doing is playing around with proportionality problems that you probably could have done in previous courses. But there really is the lead in to getting into more deep, linear functions. Clear out that little blue line.
Let's jump into a great proportional relationship problem. All right. Let's talk about apples. At a local farm stand, 6 apples can be bought for $4. So 6 apples, half a dozen, for $4. Determine how much it would cost to buy the following amounts of apples. Round to the nearest scent, when necessary. All right. So I want to know how much it's going to cost for a dozen apples. Now, that's probably not going to be too tough. If 6 apples cost $4, then a dozen apples is going to cost $8. But let's set this up as a proportion, right? We're looking for the amount of money, right? And what we know is that we'll spend $4 for 6 apples, and we can set up what's known as proportion, $4 is to 6 apples as X dollars is two 12 apples. We can then use what's known as the method of cross multiplying. I can multiply these two together. I'm going to leave off the units now to get 6 times X and I can multiply these two together to get 48. All right, dividing both sides by 6 to solve my resulting equation. Gives me X equals 8 or 8 dollars.
Now I get it. That was easy enough, right? 6 apples cost $4 doesn't apples. It's going to cost $8. We knew that. But it gets a little bit trickier when we have something like 20 apples. There, setting up that proportion is going to help a lot, right? Because now, if we say, well, $4 is to 6 apples. X dollars is to 20 apples. Make sure if apples is in the denominator on one side, it's in the denominator on the other side. I can now cross multiply giving me 6 times X is four times 20 or 80. I can now divide both sides by 6, and I'll find out that X is a little bit ugly. It doesn't turn out to be that nice. Is 1333. Obviously that extends forever it's 13.3333. But that's not a reasonable way to leave our answer because we're talking about money here, right? And certainly the cashier would round to the nearest scent. So we have $13 and 33 cents. But that's it. That's a proportional relationship. Two variables, the ratio of two variables always stays the same in a proportional relationship. And the cost of items based on how many you buy is a great example of a proportional relationship.
All right, I'm going to clear this out. It's a copy down anything you need to. All right, here we go. Let's keep moving on with this one. We're going to keep with this one exercise, see what else we can do with it. All right. We've still got this 6 apples for $4. Now let's take it a little bit further. Letter C says if C is the total cost. Okay, always remember what or always note what your variables mean. C is the total cost. And N is the number of apples bought, we want to write a proportional relationship between CNN. Solve this equation for the variable C all right, so we're going to do exactly what we did before. On one side of the equation, we're going to have $4. Is to 6 apples. But on the other side of the proportion, we're going to have C dollars total cost and N apples. Now, I'm going to get rid of these units. I'm just going to have four, 6 equals C divided by N we could certainly cross multiply. There'd be nothing wrong with that. But I'm going to use a property of equality. And just multiply both sides of this equation by N those two cancel and what I get is I get C is equal to four 6th times N that is a perfectly good way to leave our answer. Or we could reduce that four 6, that looks like the letter E now.
It's supposed to be the letter C, and it was, or we can reduce that four 6 down to two thirds. And get this relationship. Both are completely good. Okay? And this is great now. Because now if I've got a total number of apples, I can just simply substitute it into this equation. Multiply by four sixths or by two thirds, and I can get a final answer. All right? Now, letter D asks us to graph the relationship below. This is kind of cool. Because we could totally create a table of values or something like that. But watch how easy this is going to be. We can actually just generate this in our head, right? At zero apples, we're paying zero total cost. When we hit 6 apples, we're paying $4, which is right here. 12 apples, we're paying $8, which is right here. And what would we have then? One more up, 18 apples. We would have $12. When does that make sense? So 6 apples gives us $4. 12 apples gives us $8. And 18 apples. Gives us $12. We can now graph this beautiful relationship, got it already by just doing this.
I think it's appropriate to have an error on this, given that we could keep buying apples, maybe not forever. There's got to be a finite amount, but that's a good relationship right there. Now, letter E, a little function notation. According to the graph, C of 15 equals ten. Illustrate this on your graph. All right, let's do it. Let's do it in red. Everything else is in blue. Remember, this is our input. So if we go up to 15, what we see is our output is ten. Now how do we interpret this? What would we write down? What we would say is 15 apples. Cost. $10. That's our interpretation. The input was the number of apples, and the output was the number of dollars, so when the input was 15 apples, the output was $10. All right, little proportional relationships there. Okay? I'm going to clear this text out. So write down what you need to. All right, here comes the scrub. It's all gone. Okay. Let's move on to another problem involving proportional relationships. Exercise two. If Jenny can run 5 meters in two seconds, which of the following gives the distance D, she can run over a span of T seconds going at the same constant rate. Show the work that leads to your answer.
All right, so watch. What we're really trying to do is establish a relationship between distance traveled and time run. So what do we know? We know that she can run 5 meters every two seconds. And that must be in proportion to D meters every T seconds. Now I'm going to get rid of those units really quick, otherwise they're going to be kind of bulky. Notice that every one of these things is solved for D remember the letters everywhere lesson if you were with me for it. I'm going to solve this for D so what I'm going to do because D is being divided by T so I'm going to multiply by T on both sides. And what I'm going to find is that D equals 5 halves times T and there it is. Choice four. Okay? It's pretty simple. Pretty easy. All right, I'm going to clear this out. And let's keep going. Now, one of the most important sort of proportional relationships that we come upon is this relationship between the distance we've traveled and the time that we've traveled. All right? So we're going to keep exploring that in exercise number three. And apparently we're going to just keep clearing the page. So it says Erika is driving at a constant rate.
All right, she travels 120 miles in the span of two hours. If Erica travels at the same rate, how far will she travel in three hours? Why don't you set up a proportion or some other type of thinking and figure this out, okay? I think that you can do it. But hit pause on the video now and take some time to figure out what array. All right, let's do it. If you got a 180 miles, you're right. Now, this can be done in a lot of different ways, with a lot of different types of thinking. The basic kind of cut and dry way of doing it is saying, well, all right. I travel 120 miles in two hours, right? How far will I travel? I'll call it big D, because I'm going to do that later on in the problem. In three hours. So your classic cross multiplying, we would get two times D is three times one 20. That's 360. I would then solve that by dividing both sides by two, and I would get 180 miles. Now, on the other hand, if you just said, look, I travel a 120 miles in two hours, therefore I travel 60 miles in one hour, and I'm traveling for three hours, so 60 times three gives me a 180. Wonderful. That is the best type of proportional thinking you can do. So nothing wrong with that.
In fact, letter B says write a proportional relationship between the distance D that Eric will drive over the time T that she travels. Assuming she continues at the same rate. So this is always based on the idea that Erica is not speeding up. She's not slowing down. She's always going at the same rate. This is kind of what we did before. We have a 120 miles divided by two hours equals D miles divided by T hours. I'm going to multiply both sides of this equation by T to solve for D what I'm going to find is that D is 120 divided by two times T but it would be better if I do 120 divided by two. And I just get this. The distance she travels is always 60 times T all right? Now let her see asks, what is the value of the proportionality constant? Okay? That's this guy right here. All right. And the value of the proportionality constant is 60. What are its units, this is cool, right? We can track them, they're right here, miles per hour. In other words, that's her speed. That's her speech. She's traveling at 60 mph. Finally, how much time will it take for Erika to travel 150 miles?
Well, we know that she, the relationship between the distance she's traveled and the time she's traveled is here. D equals 60 T what they're telling me with the 150 miles is this value. So if I set that equal to one 50 and then divide both sides by 60 and we'll find where is it? It takes Erica 2.5 hours. To travel 150 miles. Sorry. And again, you can probably do that one another other ways, right? You know it has to be between two hours, three hours, because at two hours she's traveled one 20, and at three hours, she's traveled one 80. But the straightforward equation solving method works quite well quite nicely as well. Tongue tied today. All right, I'm going to clear out the text, so pause the video if you need to. All right, here we go. Okay, continuous. Let's graph D as a function of T on the axis to the right. And I wrote down our equation again. It's kind of helpful. D equals 60 T again, we could create kind of a little table of values here, and that might help, right? It should be straightforward. If I haven't been driving at all, we haven't gone any miles.
After one hour, we've gone 60 miles. After two hours, we've gone a 120 miles. And after three hours, we've gone a 180 miles. So simple enough, zero. Oh, what do these go by one, two, three, four, 5? Go by 20s. Watch out. It'd be easy to think that they go by 25s. They go by 20, right? All right, so that kind of makes it nice because then that means after one hour, we're at 60. After two hours, we're at one 20 and after three hours we're at one 80. So. There's my graph. Notice that proportional relationships are always linear. Now I was follow along a line, and they always go through the origin. In other words, when one variable is equal to zero, the other variable is equal to zero, so they pass through the point zero zero always. Okay? Now, what we talked about a little bit already was what that constant of proportionality represents about the graph. All right? Well, we didn't talk about what it represents about the graph. What we did, and I don't know why letter C is hanging out here, random letter C letter C? But what we did is we talked about the fact that it represented her speed. But what does it represent about this graph? This goes back to some things that you talked about in previous courses.
So pause the video for a moment. What does the 60 represent about this graph? All right, let's talk about it. It represents. The lines slope. Or its average rate of change. We saw that in the last unit. But I prefer you almost to say that it just represents the line slope. I mean, look, what was slope for a line, right? Slope was always rise divided by run. I'm trying to do the most basic thing I can here. Rise divided by run. So if I took these two points, right? And I went over by a run of one, then my rise would be equal to 60, right? So that slope of 60 divided by one, or just 60. Represents the rise divided by the run. It also represents the average rate of change because it's telling us how fast the distance is changing compared to how fast the time is changing, right? 60 miles for every hour that goes by. All right. So let's get rid of this.
Okay. And let's finish up. All right? Oh. Actually, the C needed to be there for the sentence to be correct. How silly. Okay. So today, we reviewed basic ideas about proportional relationships, how they form a line that goes through the origin. How the constant of proportionality is the slope of that line. And sometimes has physical significance, like the speed at which a driver is driving. Right? We can solve these proportions, hopefully, from information we knew from previous courses by setting up a proportion, cross multiplying, if needed, all right? And we're going to use these as the basis of all other linear functions. So it's good to have a grip on that. For now though, let me just thank you for joining me for another common core algebra one lesson by E math instruction. My name is Kirk Weiler. And until next time, keep taking and keep solving problems.