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Welcome to Algebra I: Determining if a Direct Variation Exists

25 minutes

Hey guys, welcome to Algebra 1. Today's lesson's going to focus on determining if a direct variation exists. Now, your knowledge of independent and dependent variables and patterns and the coordinate plane are really gonna help you get through this lesson. You ready to get going? Let's start.

(Describer) She stands at a screen.

To warm you up a little bit about direct variation, I want to first start with this particular word problem. Let's take a look. "Meagan earns $10 per hour "working at a local electronics store. How much will she earn after working four hours?" There's a couple of different ways you could attack this and get to the answer. Maybe you would want to use a pattern and just develop a few situations until you got to the point where Meagan had worked four hours. This is what I mean. After the first hour,

(Describer) She writes.

Meagan will have earned $10 if she's making $10 for every hour she works. At 2 hours, that would be $20 that she's earned. After 3 hours, that would be $30. Let's scroll down a little bit. After 4 hours, she will have earned $40. Maybe you could generate, which really is a table of values, and be able to tell that after 4 hours, Meagan will have learned $40. That's one strategy. Something else you may have done in order to figure out how much Meagan earned after four hours is maybe intuitively you figured out an equation to represent this situation. If I wrote an equation, y=10x, let's say, and I want to say that y is going to be the amount of money Meagan earns, and x is going to be the number of hours she works. If she's working for 4 hours, that would be 10x4, and I'd see there that y is 40. I could get to my answer there and be able to tell that after 4 hours working, Meagan will have earned $40. What you did right here without even knowing is you were working with a direct variation situation. When you have a relationship that's a direct variation, all that means is y changes directly as x changes. It varies directly. As x increases, y increases at some constant rate. That's all a direct variation is. Let's keep going with this. I want to show you the equation formally because there's actually a couple of different equations you'll work with for direct variation. Let's get that out of our way. I could write a direct variation equation as y=kx, where k is that constant of variation. Like in our previous example-- let's get that up here just to remember-- we said that we could work with the equation y=10x. We would call ten the constant of variation. It's that constant rate at which you're seeing that change. When you want to figure out what your constant rate is, say you've already been given a table of values, and you want to know what is that constant change, you could figure out what k is by using the equation k=yx. All that is is that we've taken y=kx and we've solved for k, like a literal equation. Let me show you what I mean just right underneath here. If I have y=kx and I want to solve for k, I could divide both sides by x from what we know about solving literal equations. The xs would cancel out, and I can see right here if I just flipped it around, law of symmetry, that k=y/x. You're going to be working with these two different equations as we go through these direct variation problems. Let's keep going, let's get some practice with this now.

(Describer) The next example has three tables. Each has an x column and a y column, and each column has three numbers.

(female describer) The next example has three tables. Each has an x column and a y column, and each column has three numbers. Let's take a look at this. "Which of the following tables shows that y varies directly as x?" What we want to do in order to solve this problem is we want to figure out what is k. Which one of these tables has a constant of variation, where my y divided by x is the same for every ordered pair. We're just gonna run through these three tables and see which one actually gives us what we're looking for. Let's see here. Working with table A, I'm just gonna put my work right underneath this, because I know this is the first time we've tackled this style. If we see four divided by one. We want to know y divided by x, what is that? We're trying to figure out what k is. In this case, it's four. Let's take that second ordered pair in that table, I'm going to scroll down a little bit. Now I'm going to do six divided by two. Let's see what that is. Six divided by two, that's three. I really don't need to go any further because I can tell right now that I don't have a constant of variation. In my first ordered pair, y was obtained by multiplying x by 4 because 1x4 would give me 4. In that second pair, my constant was three. I got y by multiplying 2x3. I'm not getting a constant rate of change for each of these ordered pairs. I can tell right now, A is not gonna be the table that I'm looking for. Let's cross that one out. Now let's tackle B. Again, what we're going to do is just divide. Y divided by x. We're going to do that for every ordered pair. We want to see which is the table that gives us the same value every time we do y divided by x. We want that constant. Let's try this one out. The first ordered pair we have three divided by one. Let's close that up a bit, there we go. That's three. If this next ordered pair is three when we do y divided by x, then we're on a good roll here. As I'm looking at it, eight divided by two, that's four. Again, I don't have a constant rate of change for each set of ordered pairs. B does not represent a direct variation. Process of elimination, it has to be C, but you know the math teacher in me has to work through these to verify that. Let me show you. For that first set, five divided by one, that's five. Now, ten divided by two is also five. Let's scroll, get a little more work space up here. Scroll up a bit, catch that table. The last set of ordered pairs is 15 divided by 3 which is 5. You see, we got five every single time. We found that quotient, that y divided by x. Table C, that is the table that's representing a direct variation. That's the one you want. Let's try another one here. Actually, I do believe it is your turn. So go ahead and press pause. Take a few minutes and I want you to figure out which one of these tables represents a direct variation. Remember from our previous example, to check that, just y divided by x for every set of ordered pairs in your table. You're looking for that constant. After you've gotten your answer, go ahead and press play so I can see what you got and compare our answers.

(Describer) Title: Which of the following is a direct variation? Each of the three tables has three numbers in an x-column on the left and a y-column on the right. In Table A, the ordered pairs are 1 and 6, 2 and 10, and 3 and 9. In Table B, the ordered pairs are 1 and 8, 2 and 16, and 3 and 24. In Table C, the ordered pairs are 1 and 10, 2 and 24 and 3 and 18.

(describer) Title: Which of the following is a direct variation? Each of the three tables has three numbers in an x column on the left and a y column on the right. In Table A, the ordered pairs are one and six, two and ten, and three and nine. In Table B, the ordered pairs are one and eight, two and 16, and three and 24. In Table C, the ordered pairs are one and ten, two and 24, and three and 18. Let's see how you did. Gonna work through these, and let's figure out which one of these tables represents a direct variation. For table A, we know we're doing y divided by x. This time, I'm just gonna write my answers just right off to the side here since I know that you know that we're dividing y divided by x. Six divided by one, that's six. Ten divided by two, that's five. If you did keep going, nine divided by three, that's three. It can't be A because I didn't have a constant. It wasn't the same answer every time. Let's try B. Eight divided by one, that's eight. Sixteen divided by two, that's eight. Looking good so far. Twenty-four divided by three, also eight. The correct answer for this one, it was B. I do want to run through C just to make sure that you didn't make any mistakes there and maybe picked C for the answer. If we find these quotients, ten divided by one, that's ten. Twenty-four divided by two, that's 12. Eighteen divided by three, that's six. We can see, C wasn't the answer. Our correct answer for this one was indeed B. Eight was our k. Eight was that constant of variation. Let's keep going with these. We're gonna do a few word problems here. I'm gonna read through it and then I'm gonna show you how you attack these. "Determine if the situation represents a direct variation. "Yesterday, Juliet ran two miles in ten minutes. Today she ran three miles in 15 minutes." Basically, what we want to figure out here is, is Juliet running at a constant rate? What I do want to start out by doing is I want to write out those direct variation equations. I want to show you which one you'll use here to figure out what's going on. We know we have y=kx, and then we also know that we can use k=y/x. Now, because we're trying to determine if there's a constant rate of change, what is that constant of variation, we're going to use this equation here to figure out if this does indeed represent a direct variation.

(Describer) Y over X.

Now what we have to figure out is which one of these values is y and which one is x. We've got to throw back to your pre-algebra skills here. If you remember, y is your dependent variable. Let me write it out this time. I don't want to abbreviate it the first time because I want to make sure that you know what I mean by that. Let me just scoot it in, I'll write it in the middle. Y is our dependent variable and x is our independent variable. What that means is, we can control x. We can control the independent variable. The value of y depends on what the value of x was. For example, in this situation, Juliet ran two miles in ten minutes. The number of miles she runs is going to depend on how many minutes she spends running. In this situation, I could say that y is the miles, the distance that she's running, and x is the minutes, or the amount of time she's spending running. She can control how long she runs. That's her independent variable. The distance she runs, the number of miles, is the dependent variable. Now that we've identified that, what we want to do is figure out for the first situation, the yesterday, what's our k there, what's our constant of variation, and we want to determine for the second part, for the part she ran today, what's our k there, what's our constant of variation. If we get the same thing each time, then we know that this is a direct variation situation. Let's do that math, let's see what we get. If our independent variable is the minutes, our dependent variable is the miles. We're trying to figure out what k is. K=y/x. For the first situation, let's put a little line here to separate this. Yesterday, she ran two miles in ten minutes. The miles, that's my y. The minutes is my x. I'll say, okay, two divided by ten. Let's see what that fraction simplifies to. If you throw back to your fraction skills, I know I can simplify this fraction if I just divide the numerator and the denominator by two. Two divided by two, that would be one. Ten divided by two is five. For that first situation, k is 1/5. When I figure out what's going on for the today part, for what she ran today, the three miles and the 15 minutes, if I also get 1/5, then I have a direct variation situation here. Let's see what we get there. I think I can fit that right beside. Let's get it right underneath, actually. Just get it a little spaced out for us. For today, she ran three miles. We know the miles is our y. In 15 minutes, so that's our x. Again, we end up with a fraction and we can simplify this. If we divide our numerator and our denominator by three, three divided by three is one, and 15 divided by three is five. We see, we did have a constant of variation here. We did get 1/5 for both times. Does the situation represent a direct variation? Yes it does. We did have a constant rate of change. Let's try another one together. Then I'm gonna let you try one on your own and see how you do. Here we go. "James prepared six sandwiches in three minutes and then prepared 12 sandwiches in four minutes." We're assuming here James must work at some fast food restaurant or something like that because he's working pretty fast here. First, let's write our direct variation equations. What we saw from the last one, this is actually the one that we need here. We need the k=y/x. We'll go ahead and we'll start by writing that one down. Now we need to figure out, we know that y is our dependent variable. This time I'll abbreviate it because I know that you know what I mean. X is our independent variable. In this situation, the number of sandwiches that James prepares depends on how much time he's spending working. He can control the time. The number of sandwiches depends on the length of time. For this situation, we'll say our dependent variable, that's our sandwiches, and our independent variable, that's the minutes, that's the time. We've got that part sorted out, so now we can go ahead and figure out what exactly is the answer to this question. Is it a direct variation situation or not. Let's see here. James prepared six sandwiches in three minutes. That's the first situation. Six sandwiches, so that's the y. In three minutes, so that's the x. We have six divided by three, that's two. If, when I figure out the second part of this, if I also get two, then I know that this is a direct variation situation. Let's test that out. This time, then he prepared 12 sandwiches. That's the y. In four minutes, that's the x. Twelve divided by four, that's three. Because I did not get the same k, the same constant, for both situations, this is a no. This does not represent a direct variation situation. My constant is not constant. I'm not getting the same thing every time here. Good job with that. Now it's time for you to try one. Go ahead and press pause, take a few minutes, work your way through this problem, and when you're ready to check your answer against mine, press play.

(describer) Titles:

(Describer) Titles: Determine if the situation represents a direct variation. Yesterday, Alan washed five cars in 20 minutes. Today, he washed 8 cars in 32 minutes.

Determine if the situation represents a direct variation. Yesterday, Alan washed five cars in 20 minutes. Today, he washed eight cars in 32 minutes. Let's work this one out. Let's see how you did. "Yesterday Alan washed five cars in 20 minutes. Today he washed eight cars in 32 minutes." You know we've been starting out by first writing the direct variation equation that we're going to be using, then we go ahead and we identify what's our independent variable, what's our dependent variable. We know that y is the dependent variable. We know that x is the independent variable, the one we can control. In this situation, the number of cars that Alan washes depends on how much time he's washing the cars. How much time he's working. Our dependent variable, that's the number of cars he gets washed. Our independent variable, that's the minutes, that's the time. I abbreviated minutes there. What we need to do is, we need to see, for yesterday and then for today, do we have a constant of variation here. Let me scroll a little bit and let's check this out. For the first situation, y is the cars. X is the minutes. Five cars in twenty minutes. Five divided by twenty. I can simplify this fraction if I divide the numerator and the denominator by five. Five divided by five, that's one. Twenty divided by five, that's four. We're good there. Now let's check out the second part of this. Today he washed eight cars in 32 minutes. The number of cars is y, the minutes is x. Eight divided by thirty-two. Let's move that out of our way there. I think it can simplify this by dividing the numerator and the denominator by eight. Eight divided by eight is one. Thirty-two divided by eight is four. What do you know; we do have a constant of variation here. We got 1/4 for each part of this problem. Does this represent a direct variation situation? Yes, it does. Good job on that one. Let's do a little more with these direct variations here. This time, we're gonna discuss the graph of a direct variation. To understand this, we actually use the other direct variation equation, the y=kx. It probably looks a little familiar to you because it's not much unlike slope intercept form.

(Describer) Y equals mx plus b.

In the case of direct variation, your y intercept, or your b, is always zero. K is kind of like your slope. You're thinking about it in the same kind of way. Thinking back to our first example, when Meagan was working at an electronics store earning $10 per hour, that could be represented by the equation y=10x. If we take a peek at the graph of that equation, let's get this out of our way here,

(Describer) She uncovers it.

then you can see that that actually does go through the origin. Let's erase that point. Don't think it's part of the graph. We see that the graph of y=10x does pass through the origin. That's true for every single direct variation equation. Regardless of what your k is, what your constant is, the graph of a direct variation will always pass through the origin. It's always a line that passes right through the origin. It may have a positive slope or a positive rate of change, it may have a negative one, but it will always go through the origin. Let's look at some graphs. "Select the graph in which y varies directly as x." We have two up top, then I'm gonna scroll down, then we have two right underneath. Let's analyze these, keeping in mind that a direct variation, the graph of that will always be a line that passes through the origin. If we take a look at A, you don't have a line there, you've got some curves. You can tell that A is not what we're looking for here. This is not the direct variation. B, you see I have a V-shaped graph. That's actually the graph of an absolute value. You'll get more into that in Algebra 2. I know that what I'm looking for when y varies directly as x, when I have a direct variation, I'm looking for a line that passes through the origin. I don't have a line here. It's not B. Let's look at these two underneath here. For C, I'm getting better here because I do have a line, but this line doesn't pass through the origin. It's not C. Let's check out D. I've got a line and my line passes through the origin. D is actually the graph of a direct variation. You see how we analyze all those graphs to be able to determine which one was a line that passed through the origin. That's the graph that's going to be your direct variation. Let's move on to the next one. It is your turn, so press pause. Take some time and analyze these graphs. See if you can determine which one represents a direct variation. When you're ready to check your answer, press play.

(Describer) Title: Select the graph in which y varies directly as x. Three graphs are displayed. In Graph A, a straight line crosses the y-axis, It is above the x-axis and is parallel to it. In Graph B, a straight diagonal line sloping downward crosses the positive sections of the x-axis and the y-axis. In Graph C, a straight diagonal line sloping downward goes through the origin, where the axes cross.

(describer) Title: Select the graph in which y varies directly as x. Three graphs are displayed. In Graph A, a straight line crosses the y axis. It is above the x axis and is parallel to it. In Graph B, a straight diagonal line sloping downward crosses the positive sections of the x axis and the y axis. In Graph C, a straight diagonal line sloping downward goes through the origin where the axes cross. Let's see how you did here. For A, I did have a line, but it didn't pass through the origin. It wasn't A. For B, I did have a line, but it didn't pass through the origin. It's not B. Now, for C, I've got a line and that line passes through the origin. C is the graph that represents a direct variation. Great job, guys, working with those direct variation problems. I hope you saw how your knowledge of independent and dependent variables and the coordinate plane and patterns helped you get through this lesson. Hope to see you soon for more Algebra 1, bye.

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Direct variation problems are solved using the equation y=kx, and this equation describes a particular form of dependence of one variable on another. In this program, students learn to determine if a direct variation exists for the given data. Part of the "Welcome to Algebra I" series.

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Runtime: 25 minutes

Welcome to Algebra I
Episode 1
31 minutes
Grade Level: 7 - 12
Welcome to Algebra I
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25 minutes
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17 minutes
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Episode 10
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