Algebra 1 – Module 11, Topic 2 – Determining if an Inverse Variation Exists

Hey guys, welcome to Algebra I.  Today's lesson is going to focus on Determining if an Inverse Variation Exists, so your knowledge of independent and dependent variables and patterns in the coordinate plane is really going to help you get through this lesson. You ready to get started? Let's go. 
Before, we start talking about inverse variation I want to go back a second to do a right variation because I you want to understand the difference between the 2. Here we've got Megan earns $10 per hour working at a local at electronic store. How much money will she earn after working 4 hours? We could represent this situation a couple of different ways in order to get to the answer. We could derive a table of values and the 4 patterns or you may intuitively be able to see an equation that can represent 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 would have earned $10. After 2 hours, it would be $40. I'm going to need a little more room for our work. After 3 hours, ... Back this up a little bit. After 2 hours, she would have only earned $20, right, because she is only making $10 an hour. I tried to give her a little 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.
Now, if you didn't attack it this way by generating a table, maybe you understood that I could represent the situation by the equation Y=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 10x4 that's 40. After 4 hours, Megan earns $40. This is a direct variation situation. We've kind of seen these before. In a direct variation situation, as X increases Y increases at some constant rate 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. Now that, I know you really need to see, so that it makes sense, but let me first show you this. Get that out of the way. This is our equation for an inverse variation, so you can see that it differs from that of a direct variation. We have that Y=K÷X or we can say that K=X times Y. This is what I meant by when I said that Y changes in a rate proportional to X so that 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 going here, so you can really see what I mean. In which of these tables, does Y vary inversely as X that. That's what we're trying to figure out here. In order to figure that out, which one of these is an inverse variation, we're going to find the product of X and Y. For each ordered pair, we're going to multiply X times Y and we're looking for the table that gives us the same answer every time. The table in which the product of X and Y is constant will be the table that represents an inverse variation.

All right, so let's see what we've got going here. I'm going to put the work right underneath because it is our first time doing it. This time we are multiplying, so 1×2, that's 2. Now, 5×3/5, so we're going to have to throwback a little bit to remember how to work with those fractions, but we got it, we got it. 5×3/5, I can make that 5 ... I can represent it as a fraction by putting it over 1. Let's line that up a little better, so it looks a little more in sync here. Let's get that out of the way. Bring that up a little bit. There we go. Now, I have 5 over 1x3/5 and we'll just need to multiply straight across to get to our answer, so 5×3 that's 15, 1×5 that's 5, and 15÷5 that's 3. I can tell right now, K is not constant. I don't have a constant product for each of these ordered pairs. I don't really need to bother to check that last one because I can already tell the products are different, so I know that table A is not the table in which Y varies inversely as X.

Let's check out B here. Let's see if we get the same answer for each product of X and Y. Let's get a little space underneath here. For the first pair, -10×3 that's -30, 1×5 that's 5, so I can tell right there K isn't constant. I'm not getting the same thing for every product in my table, my products of X and Y's. I can tell right now B is not the answer. All right, so process of elimination it should be C, but let's actually run through these and see that we do get the same product each time.

Let's move that out of our way, so we can work right under there. For the first pair, we have -1x-5, get that pen back, so -1x-5 that's a +5. For the second pair, 5×1 that's +5, looking good so far. For the last pair, we have 10×1/2. Again, like we saw in the first table, I can put that 10 over 1 to represent it as a fraction and then just multiply straight across, so 10×1 that's 10. Going to need some more room here. 1×2 that's 2. I am going to ... Let's scoot this over just 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÷2 and that is 5, so I see 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.

Now, it is time for you to try one of these. Let me move this out of your way, so you can see that table there. Press pause and take a few minutes and work your way through these problems. You're looking for the table that is actually an inverse variation. Remember to find that out, you just need to check those products. When you're ready to compare your answers against mine, go ahead and press play.

Let's see how you did here. Let's scroll down here, we can get a little room to work. Let's check this first table. Now remember, this time we're checking the products. -5x-10, I'm just going to write that right here to the side. Make I've got the 10. That's 50, 2x4 that's 8, so I can already tell it can't be A. Let's check out B, we've got some fractions, so let me do a little work underneath here to handle that. -1/2x-4, I'm going to put that -4/1, so I can visualize it as fraction. -1x-4 is +4, 2x1 is 2, 4÷2 is 2. For the first product, I got 2. I've got another fraction here, so I'm going to need a little more room to work, so that I can do that multiplication. You could also put these in a calculator, if you want, it's completely up to you.
This time I have 6×1/3, so 6×1/3, I'm going to put that 6/1. 6x1 is 6, 1×3 is 3, 6÷3 is 2, so B is looking pretty good right now. Let's see what happens at the last product. 8×1/4, going to scroll down here, so I can do that work. 8×1/4, going to need a little more space here. I'm going to put that 8/1, so I can visualize it as fraction. 8×1 is 8, 1×4 is 4, 8÷4 is indeed 2. I see I did have a constant variation in this table and it was the same for each ordered pair. It was 2, so that tells me it has to be B, but you know that I'm going to have to show you C because the math teacher in me has to still check to make sure that it's not C. Let us get those products quickly for C. 2×3 that's 6. I think I might be able to squeeze it right there, but I'll circle it so you know that was the answer for that product. 3x2 that's 6, but 4×1 that was 4, so it couldn't have been C because you didn't get that same product every time. We were right; the answer to this one was C.
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. Now, I want to discuss inverse variation a little bit. Earlier, I said that an inverse variation as X increase Y decreases. You can really see what I mean by that in this problem. We just have to make sure it's a constant change, but you do see how 3 people painted the room 6 hours, but if I increase the number of people to now 9 people painted the room, I take 2 hours, so it took a fewer number of hours. As the more people were working, the less time it took to paint the room. We just have to make sure that that time was actually constant, that would be so I constant variation with that. For this situation, we want to use our inverse variation equation that is K=Y times X because we're basically doing the same thing we were doing in that table. We're checking the products. It's just that they presented it to us, the values, in a different way here because of have a word problem. 

We do want to start out after we have the equation written down, you want to identify our independent and our dependent variables. We can control the number of people working. X, that's our number of people, and that is our independent variable. Now, the time it takes to paint the room depends on the number of people, so that's our Y. The number of hours and that's our dependent variable here. We need to make sure that we have the same product for each situation in here to know if we actually have the inverse variation situation. Let's get a little more work space. For the first part, 3 people painted the room in 6 hours, so K, I already have Y times X ... You know you can multiply in any order you want, it doesn't matter. I'm just going to follow the Y and X, just to keep it consistent up here. 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 6×3 that's 18. In the second part of this, if I also get 18 when I find that product, then I do indeed have an inverse variation situation.

Let's see here. Get a little more space. Now, we have 9 people painted the room in 2 hours. This time Y is our number of hours, so that's 2. X is the number of people, so that's 9. 2×9 is 18, so we do have a constant variation. It's the same thing. The same constant for each part of this problem, so we do indeed have an inverse variation situation here. Good job on that one. Let's do another one together.

Now, 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 here. We know we're testing out K to see if it's the same thing. K=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're going to have to divide that money up. It's going to be 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 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 amount money earned depends on the number of landscapers working. Let's see here, let's test out these products.

For the first part we have 6 landscapers, $40 each, so K=, 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, 40×6 That's 240, so we've got the first value for K. Let's figure out what we get for the second part of this. Now, we have 3 landscapers that earned $120 each. 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 for this, 120×3 that's 360, so you see that it's not constant. The first time I got 240, the second time I got 360, so 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. I think we have reached the point that it is your turn to try one on your own, so go-ahead and press pause and take a few minutes and work your way through these problems. When you're ready to compare your answers with me, press play.

All right, let's see how you did here. We wanted to determine if this is an inverse variation situation, so the equation we're going to work with his K=Y times X. Here 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 go ahead and get that part down. X is what we can 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, because the time it takes the campers to assemble the tent depends on the number of campers working, so that's our dependent variable. Okay, so let's test out 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, we've got 4 campers that assembled the tent in 15 minutes. K=, Y is the time, so 15 minutes. X is the number of campers. so that's 4. Either in your calculator or mental math here, 15x4 that's 60. Let's test out that second part and see if we also get 60. This time we have 5 campers that assembled a tent in 12 minutes. K=, Y is the time, which this time is 12 minutes. X is the number of campers, which this time is 5. Either in a calculator or mental math here, 12×5 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.
Now, let's move on and discuss the graphs of inverse variations a little bit. Earlier, we said that Y=K÷X is also an inverse variation equation. It's just solve for Y there. We use that one a lot when we're dealing with another type a word problem and the graphs of inverse variations. Let's get that out of our way there. There you go. Now, this is actually the graph of Y=, just to show you an example, of Y=6÷X and you see for inverse variation you don't have a line at all. You actually have 2 curves for a graph of an inverse variation, so if you ever see this set up for your graph, you know that you have an inverse variation situation. A little tongue-tied there. They're 2 curves and they're a reflections of each other, remembering that a reflection is a flip. If you envision yourself folding this graph diagonally, that one graph would lay right on top of the other. It's 2 curves that 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, so in other words, which one of these graphs represents an inverse variation. We know we are looking for 2 curves that are reflections of each other. We've got 2 graphs up top here, then we have 2 that we're going to scroll down and look at underneath here. If you start out by looking at A, I do have curves, but I don't have 2 curves that are reflections of each other. This graph doesn't have that same style, that same set up, of the example we saw earlier, so it's not A.

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 2 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 a D. We've got 2 curves and they are reflections of each other, so this is the graph that represents an inverse variation. D is the one that you are looking for. Keep that in mind and I want you to take a look at this one. Press pause and take a few minutes and work your way through this problem. See if you can identify which of these graphs represents an inverse variation. When you're ready to compare your answer against mine, go ahead and press play.
All right, let's see how you did. For A, we have a horizontal line and I know for inverse variation I'm looking for 2 curves that are reflections of each other, so it's not A. Does appear to be B. I've got my 2 curves and they are reflections of each other, so the inverse variation here was B. Just to double 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. Good job on that.

All right guys, you've reached the end of your lesson on inverse variation. I hope that you saw how your knowledge of the coordinate plane and independent and dependent variables and even a little bit of direct variation knowledge helped you get through this lesson. Hope to see you back here for more Algebra I. Bye.