# Welcome to Algebra I: Determining if an Inverse Variation Exists

25 minutes

Hey guys. Welcome to Algebra I. Today's lesson focuses on determining if an inverse variation exists. Your knowledge of independent and dependent variables and patterns in the coordinate plane will help you get through this lesson. You ready to get started? Let's go. Okay, so before we start talking about inverse variation, I want to do a right variation so you understand the difference between the two. Okay? "Megan earns \$10 per hour working at an electronic store. How much money will she earn after working 4 hours?" We could represent this situation a couple of ways to get the answer. We could derive a table of values and look for a pattern, or you may be able to see an equation that represents the situation. If we generated a table of values to figure out how much Megan will earn after 4 hours-- need a little space here-- I could say, after 1 hour, she will have earned \$10. After 2 hours, it would be \$40. I need more room for our work. After 3 hours--oop-- back this up a little bit. After 2 hours, she would have only earned \$20 because she is making \$10 an hour. I tried to give her a raise. After 2 hours, \$20. After 3 hours, \$30-- extend that. After 4 hours, \$40. If she's making \$10 per hour, we could tell here that after 4 hours, she'll make \$40. If you didn't attack it by generating a table, maybe you understood that I could represent the situation by the equation y equals 10x. If I let y represent the amount of money she's earning after her work shift, and x represents the number of hours she works, and I want to determine after 4 hours how much money she will have earned, then I could say, 10 times 4, that's 40. After 4 hours, Megan earns \$40. This is a direct variation situation. We've seen these before. In a direct variation situation, as x increases, y increases at some constant rate, right? Or as x decreases, y decreases at some constant rate, but they are changing in the same direction. Now, how inverse variation is different is, as x increases, y decreases at a rate proportional, so that the product of x and y is the same. You really need to see that so it makes sense. Let me first show you this.

(female describer) Inverse Variation. y equals k over x or k equals xy; k is the constant of variation.

(teacher) Get that out of the way. This is our equation for an inverse variation, so it differs from that of a direct variation. We have, y equals k divided by x, or we can say that k equals x times y. This is what I meant when I said that y changes in a rate proportional to x so the product of x and y is always the same. That second equation represents what I was saying right there. Let's get an example so you can see what I mean. "In which of these tables does y vary inversely as x?" That's what we're figuring out here. To figure out which one of these is an inverse variation, we find the product of x and y. For each ordered pair, we multiply x times y, and we look for the table that gives us the same answer every time. The table where the product of x and y is constant will be the table that represents an inverse variation. Let's see what we've got here. I put the work underneath because it is our first time doing it. This time we're multiplying, so 1 times 2-- Let's get our pen back. So 1 times 2, that's 2. Now, 5 times 3/5. So we throw back a little bit to remember how to work with those fractions, but we got it. 5 times 3/5; I can make that 5. I can represent it as a fraction by putting it over 1, right? Let's line that up better so it looks more in sync here. Get that out of the way. Bring that up. There we go. I have 5 over 1 times 3/5. We'll multiply straight across to get to our answer. So 5 times 3, that's 15. 1 times 5, that's 5. And 15 divided by 5, that's 3. I can tell now, k is not constant. I don't have a constant product for each ordered pair. I don't really need to bother to check that last one because I can tell the products are different. So I know that table A is not the table in which y varies inversely as x, okay? Let's check out B here. Let's see if we get the same answer for each product of x and y, okay? Let's get a little space underneath here, okay. For the first pair, negative 10 times 3-- that's negative 30. 1 times 5, that's 5. So, I can tell right there, k isn't constant. I'm not getting the same thing for every product, my products of "x"s and "y"s. I can tell right now, B is not the answer. So process of elimination-- it should be C, but let's run through these and see that we get the same product each time, right? Move that out of our way so we can work right under there. For the first pair, negative 1 times negative 5-- get that pen back. So, negative 1 times negative 5, that's a positive 5, okay. For the second pair, 5 times 1, that's a positive 5. Looking good so far. For the last pair, we have 10 times 1/2. Like we saw in the first table, I put that 10 over 1 to represent it as a fraction, and multiply straight across. So, 10 times 1, that's 10. Going to need some more room. 1 times 2, that's 2. I am going-- Let's scoot this over a bit so I could fit my answer right beside that. There we go. Get that pen back here. Now I'm ready to do 10 divided by 2, and that is 5. So, I do get that constant variation every single time. The product in the third table, the product of each x and y, each ordered pair, was indeed 5. So, C is the table that represents an inverse variation. Okay? All right. Now it is time for you to try one of these, okay? I'll move this so you can see that table there. Press pause and take a few minutes. Work your way through these problems. Look for the table that is actually an inverse variation. To find that out, check those products. When you're ready to compare your answers, press play.

(female describer) There are three tables, each with a column x and column y. In Table A, the ordered pairs are negative 5, negative 10; 2, 4; and 6, 12. In Table B, the ordered pairs are negative 1/2, negative 4; 6, 1/3; and 8, 1/4. In Table C, the ordered pairs are 2, 3; 3, 2; and 4, 1. Let's see how you did here. Let's scroll down here; we can get room to work. Let's check this first table. Now, remember, this time we're checking the products. So, negative 5 times negative 10. I write that right here. Make sure I've got the pen. That's 50. 2 times 4, that's 8. So, I can tell, it can't be A. Let's check out B. We've got some fractions. So, let me work underneath here to handle that. Negative 1/2 times negative 4. So, negative 1/2 times negative 4-- I put that negative 4 over 1 so I can visualize it as a fraction. Negative 1 times negative 4 is positive 4; 2 times 1 is 2; 4 divided by 2 is 2. For the first product, I got 2. I've got another fraction. I need more room to work so I can do that multiplication. Put these in a calculator if you want. It's completely up to you. This time I have 6 times 1/3, so 6 times 1/3. I'm going to put that 6 over 1. 6 times 1 is 6, 1 times 3 is 3, 6 divided by 3 is 2. So, B is looking good now. Let's see what happens at the last product. So 8 times 1/4-- going to scroll down here so I can do that work-- 8 times 1/4. Going to need a little more space here, okay. I put that 8 over 1, so I can visualize it as a fraction. 8 times 1 is 8, 1 times 4 is 4, 8 divided by 4 is 2. I see I did have a constant variation in this table. It was the same for each ordered pair. It was 2, so that tells me, it has to be B. But I have to show you C because the math teacher in me has to check to make sure that it's not C. Let's get those products quickly for C. So 2 times 3, that's 6. I'll squeeze it there and circle it so you know that was the answer for that product. 3 times 2, that's 6. But 4 times 1, that was 4. So, it couldn't have been C because you didn't get that same product every time. So we were right; the answer to this one was B. Let's keep going with these here. Got a word problem for you. "Determine if this situation represents "an inverse variation: "3 people painted the room in 6 hours, and 9 people painted the room in 2 hours," okay? I want to discuss inverse variation a little bit. Earlier, I said that in an inverse variation, as x increases, y decreases. See what I mean by that in this problem. Make sure it's a constant change, but you do see how 3 people painted the room--6 hours. If I increase the number of people to 9, it took 2 hours-- a fewer number of hours. As the more people were working, the less time it took to paint. We have to make sure that time was actually constant, that we saw a constant variation with that. For this situation, we want to use our inverse variation equation that is k equals y times x because we're doing the same thing as in that table. We're checking the products. They presented the values in a different way here because we have a word problem. We start after we have the equation written down. We identify our independent and our dependent variables. We can control the number of people working. Right? So x, that's our number of people. And that is our independent variable. The time it takes to paint the rom depends on the number of people, so that's our y-- the number of hours. And that's our dependent variable here. We make sure that we have the same product for each situation here to know if we have the inverse variation situation. Let's get more work space. For the first part, 3 people painted the room in 6 hours. So, k, I already have y times x-- You can multiply in any order; it doesn't matter. I'll follow the y and x to be consistent. If y is the number of hours-- for the first situation, it was 6-- and x is the number of people, so that's 3. So, I'd have 6 times 3; that's 18. If I also get 18 when I find that product, then I do indeed have an inverse variation situation. Okay? Let's see here. Get a little more space. We have 9 people painted the room in 2 hours. Okay? This time, y is our number of hours. So, that's 2. And x is the number of people, so that's 9. 2 times 9 is 18. So, we do have a constant variation. It's the same thing-- the same constant for each part. So, we do indeed have an inverse variation situation here. Okay? All right. Good job on that one. Let's do another one together. We want to see, does this represent an inverse variation? We've got 6 landscapers that earn \$40 each and then 3 landscapers that earned \$120 each. Let's see what's going on. We're testing out k to see if it's the same thing-- k equals y times x. So, here we can control the number of landscapers. And how much money each earns? It depends on how many landscapers we have. If you have more people, they divide that money. It's a smaller amount that each person gets. Because it is divided amongst more people there. Here, because we can control the number of landscapers, that's our dependent variable-- I'm sorry-- that's our independent variable, the number of landscapers. This is independent because that's what we can control. The money they earn, that's our dependent variable. We'll say the money earned, that's our dependent variable because the money earned depends on the number of landscapers working, okay? Let's see here; let's test out these products. For the first part, we have 6 landscapers, \$40 each. So, k equals--y is the money earned, so \$40. x is the number of landscapers, so that's 6. Either mental math or your calculator to get this answer-- 40 times 6, that's 240. So, we've got the first value for k. Let's figure out what we get for the second part. Now we have 3 landscapers that earned \$120 each, okay? Let's get a little more space. Here y is the amount of money earned, so this time it was 120. x is the number of landscapers, so this time, that's 3. Either the calculator or mental math-- 120 times 3, that's 360. You see that it's not constant. The first time I got 240; the second time I got 360. This does not represent an inverse variation situation. I didn't get that same product for each part of this problem. All right. Okay, let's keep going here. It is your turn to try one on your own. Press pause and take a few minutes. Work your way through these problems. When you're ready, press play.

(female describer) Determine if the situation represents an inverse variation-- 4 campers assembled a tent in 15 minutes, and 5 campers assembled a tent in 12 minutes.

(teacher) Let's see how you did here. We want to determine if this is an inverse variation situation, so the equation we're working with is k equals y times x. We see we have 4 campers assembled a tent in 15 minutes; 5 campers assembled a tent in 12 minutes. We can control the number of campers. And the time it takes to assemble this tent depends on the number of campers we've got working on it. Let's get that part down-- x is what we control, and we said we can control the number of campers, so that is our independent variable. y, that's going to be the time, as the time it takes the campers to assemble the tent depends on the number of campers working, right? So, that's our dependent variable. Let's test these values and see if we get the same product each time. Let's get a little more work space here. For the first part, 4 campers assembled the tent in 15 minutes. So, k equals-- y is the time, so 15 minutes. x is the number of campers, so that's 4. In your calculator or mental math-- 15 times 4, that's 60. Okay, let's test that second part and see if we get 60. This time we have 5 campers that assembled a tent in 12 minutes. So, k equals--y is the time, which, this time, is 12 minutes. And x is the number of campers, which this time is 5. In a calculator or mental math, 12 times 5 is 60. So, we did get that same product. We have the same k for each part of this problem, so the answer is yes. This is an inverse variation situation here. Okay, good job on that one. Let's move on and discuss the graphs of inverse variations. Earlier, we said that y equals k divided by x is also an inverse variation equation. It's just solve for y. We use that one a lot when we're dealing with another type of word problem and the graphs of inverse variations. Let's get that out of our way; there you go. This is actually the graph of y equals-- to show an example-- of y equals 6 divided by x. And you see for inverse variation, you don't have a line. You actually have two curves for a graph of an inverse variation. So, if you see this for your graph, you know that you have an inverse situation-- or an inverse variation situation. A little tongue-tied there. There're two curves, and they're reflections of each other. Remembering that a reflection is a flip, if you envision yourself folding this graph diagonally, one graph would lay right on top of the other. Two curves are reflections of each other is what you have for a graph of an inverse variation. Let's do a little practice with the graphs. "Select the graph in which y varies inversely as x." Which one of these graphs represents an inverse variation? We're looking for two curves that are reflections of each other. We've got two graphs up top, and two we're going to look at underneath. If you look at A, I have curves, but don't have 2 curves that are reflections of each other. This graph doesn't have that same style or that same set-up of the example we saw earlier. So, it's not A, okay? For B, I have a V-shaped graph. I don't have curves at all for this one, so I can immediately tell here, it's not B. Let's scroll down; look at the two underneath here. For C, we have a line that passes through the origin. This is actually a direct variation situation, not an inverse variation situation. So, it's not C. Let's take a look at D. You've got two curves. They are reflections of each other. So, this is the graph that represents an inverse variation. D is the one that you're looking for, all right? Keep that in mind, and I want you to look at this one. Press pause and work your way through this problem. See if you can identify which graph represents an inverse variation. When you're ready to compare your answer, press play.

(female describer) Select the graph in which y varies inversely as x. Graph A has a straight line above the x axis, and it crosses the y axis. Graph B has two curves opposite each other in the lower-left quadrant and the upper-right quadrant. They reflect each other. Graph C has a straight line which passes diagonally through the origin. All right, let's see how you did. For A, we have a horizontal line. For inverse variation, I'm looking for two curves that are reflections of each other, so it's not A. Does appear to be B. I've got my two curves reflecting each other. So, the inverse variation here was B. To check, if you take a look at C, you have a line that passes through the origin. That's actually y varying directly as x, not inversely as x, so it's not C. Our right answer here was B, okay? Good job on that. You've reached the end of your lesson on inverse variation. I hope you saw how knowledge of the coordinate plane and independent and dependent variables and even direct variation knowledge helped you through this lesson. Hope to see you back here for more Algebra I. Bye.

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In this program, students learn to determine if an inverse variation exists for the given data. Inverse relationships exist if one variable increases while the other decreases. Part of the "Welcome to Algebra I" series.

## Media Details

Runtime: 25 minutes

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