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Welcome to Algebra I: Representing an Inverse Variation Algebraically

25 minutes

Hey, guys. Welcome to Algebra I. Today's lessons are going to focus on representing inverse variations algebraically. Your prior knowledge about what the equation of an inverse variation looks like will help you get through this lesson. Ready? Let's go.

(Describer) She gets a stylus.

Just to jog your memory, let's look at this example where you're asked to determine: Is this relation an inverse variation?

(female describer) On a screen, a table has two columns,

(Describer) On a screen, a table has two columns, one labeled x, one labeled y. Vertically in the x-column are minus 1, 5 and 10. Vertically in the y-column are minus 5, 1 and one-half.

one labeled x, one labeled y. Vertically, in the x column, are -1, 5, and 10. Vertically, in the y column, are -5, 1, and 1/2. If you remember, for an inverse variation, k was a product of y and x, right? That was that constant you were looking for every time. If you went through your table and saw that each product for each x and y pair was the same, you knew you had an inverse variation situation. So, we tested this one out: -1 times -5, that's 5.

(Describer) She writes.

5 times 1, that's 5. 10 times 1/2-- Let's step off to the side. 10 times 1/2-- I'll make that a 10 over 1 to see it as a fraction. 10 times 1 is 10, 1 times 2 is 2, 10 divided by 2 is 5. I see that I do have the same product for each ordered pair in this table. So, I could say, yes, that this is an inverse variation situation. Okay? Keep that in mind. Jog your memory a little bit more here. Okay. If you remember-- Let me get that out of your way.

(Describer) She changes the screen.

You could represent an inverse variation by two different equations: y equals k over x, or k equals x times y, or y times x, because you could multiply in any order that you want. Feel free to swap those variables, the x and y. These are the formats of your inverse variation equations, and you know that k is your constant of variation. How we're gonna step the problems up a notch is we're actually gonna write that inverse variation equation. Keep this in mind, and let's look back at this problem. This is the same table we had a second ago, when we determined that the constant of variation-- let's get the pen back-- was indeed 5. We found that each of these products was 5. We knew this was an inverse variation. What that means is that k equals 5.

(Describer) She writes.

And if we want to write an equation to represent this inverse variation, then you just need to know that y equals k over x, or k divided by x. And if I know now that k is 5, then I can say that y equals 5 over x. And I have written an inverse variation equation for this situation. Okay? Keep that in mind, and let's do that one more time, but this time, we'll look at a set of ordered pairs.

(Describer) The ordered pairs are inside braces. Each pair is in parentheses and has a comma between them. They're 2 and 36, 6 and 12, and 8 and 9.

(describer) The ordered pairs are inside braces. Each pair is in parenthesis and has a comma between them. They're 2 and 36, 6 and 12, and 8 and 9.

(instructor) Again, we're asked to write an equation to represent this inverse variation, Because they've already told us that this is an inverse variation, I don't have to test out each of these products. It's guaranteed to me that this is an inverse variation situation. All I need to do is pick one of these ordered pairs and use them to figure out what my k is, because I know k is constant for this. I'm gonna pick the smallest numbers of ordered pairs so I can use mental math to figure it out.

(Describer) Eight and nine.

I know that k is x times y,

(Describer) She writes.

or y times x, however you like to write it. So, 8 is x, 9 is y. So, k equals 8 times 9, which is 72. If I'm writing an inverse variation equation, I can write, if I fill into the y equals k over x style-- scroll down a little bit-- then I could write y equals 72 over x, and I'd be all done. If you didn't want to write your equation in this style, and you wanted to write it as k equals x times y-- I'm gonna rewrite this so you get a good visual of what I mean. Because you know k is 72, 72 equals x times y. That's a valid equation also. That is also how you could represent this inverse variation. Generally you'll see it in this style where it's solved for y, but if you wrote it in this way, you're not wrong. Just be familiar with both of these, but commonly you'll see the equation where it's already solved for y. Let's keep moving. And it is time for you to try a couple. You're given a set of ordered pairs and a table, and you're being asked to write the-- actually not the direct variation equation, you are being asked to write the inverse. Let's get that changed. The inverse variation represented in the set of ordered pairs and in the table. You got to figure out what your k is, and then fill in to that y equals k over x style. When you're ready to compare your answers with mine, press play.

(Describer) Relation A is a set of ordered pairs: 1 and 12, 3 and 4 and 6 and 2. Relation B is a table. Under the x and y columns respectively are 2 and 9, 3 and 6 and 6 and 3.

(describer) Relation A is a set of ordered pairs, 1 and 12, 3 and 4, and 6 and 2. Relation B is a table. Under the x and y columns respectively are 2 and 9, 3 and 6, and 6 and 3. Let's see how you did here. For A-- let's get this moved out of the way-- you should have gotten y equals 12 over x, or y equals 12 divided by x. And for B, y equals 18 divided by x. If you want to see how I got that, I'll show you.

(Describer) She changes the screen to just A.

For the first one, remember, the first thing you have to do is to figure out what is k. Because you were told that these were inverse variation situations,

(Describer) She writes.

you know that k is gonna equal x times y. Pick any pair you want for x and y, and then determine what your k is. I'm picking the first pair. It's the easiest one since I'm multiplying by 1. K would equal 1 times 12, which is 12. Right? If I'm filling into the y equals k over x format, and now I know that k is 12, then I can say y equals 12 over x. And I'm done. I've gotten my equation to represent this inverse variation. Okay? Let me show you how I got B. Had a table here. So again, same process. First get k. K is a product of x and y. Pick any pair because you've already been told this is an inverse variation situation. I'll pick the last pair this time, 6 and 3. So, k would equal 6 times 3, which is 18, right? And if I'm gonna fill into the y equals k over x format, then I would have y equals 18 over x. And I'm done. Okay? It was as simple as that. Figure out your k, and then drop it into that y equals k over x format. You know I couldn't let you out of here without looking at a few word problems. Let's look at this. I'll get my highlighter.

(Describer) She reads.

"The amount of time it takes to paint a room "varies inversely with the number of painters working. "If 5 painters "finished painting a room in 64 minutes, "how long would it have taken 8 painters to complete the same room?" When you're presented with a lengthy word problem, there are just three steps that you're gonna want to follow in order to solve it. Here are the steps. First, determine your x, your y, and your k. Get those important values. Then write the equation. Write that inverse variation equation, and then go ahead and answer the question. If you follow this process, you'll be able to work through these word problems easily. Determine x, y, and k, write the equation, use it to answer the question. All right? Let me show you what I mean. Jump back to example three.

(Describer) The word problem.

I've got my highlighter ready. Let me start by figuring out the key information that'll help me determine what's x, what's y, what's k. "The amount of time it takes to paint the room varies inversely." The amount of time varies inversely with the number of painters.

(Describer) She highlights.

Okay. "If 5 painters finished painting a room in 64 minutes." So, 5 painters, 64 minutes. "How long would it have taken 8 painters--" So, how long would it have taken 8 painters to complete the same room? All right. Let's begin working our process here. Scroll a little bit. I'm gonna label this the first step, which is to determine what's x, what's y, what's k. If we look back at what we highlighted here at the top, it tells us that the amount of time varies inversely with the number of painters. So, the time depends on the number of painters. All right? So, y is the dependent variable,

(Describer) She writes.

in this case, and it's the time, and x is the independent variable, because we can control the number of painters. Lock in that order here. Let's erase that. And let's write that y is the independent variable. And that's the number of painters. That's what we can control here, how many people are actually painting the room. We know y is the time, x is the number of painters, so we can figure out what k is for this situation. They told us that it took 5 painters 64 minutes. I'm gonna use that information to figure out k. Get some space here. So, k equals x times y, and it took 5 painters 64 minutes. So, 64 minutes is my y, and 5 is my x. I noticed a little something up here too.

(Describer) She erases a misspelling of "independent".

Let's get that fixed to "independent." There we go. Make sure that's spelled correctly. All right. So, 5 painters, 64 minutes. So, 5 painters, that's my x. 64, that's my y, and I just need to do this multiplication. Because these numbers are large, I'm gonna switch to the calculator to make the multiplication easier. So, we need 5 times 64. Let's clear my memory here. Just like to do that whenever I first start working on the calculator. 5 times 64. We have 320. Okay? So, back to our problem. Let's scroll down here. K equals 320. We've gotten step one taken care of. We know the values of x and y, we know what they represent, and we know the value of k. We can go ahead and write our inverse variation equation now that we know what k is. Step two--because we're gonna fill into y equals k over x, and we know that k is 320-- then we'll have y equals 320 over x. Okay? And we've got step two handled. We've written our inverse variation equation. Now it's time to scroll up so we can use that equation to actually answer the question. How long will it have taken 8 painters to paint the same room? We know that the number of painters-- that's our x, and this time x is 8. All right? For step three, y equals 320 divided by 8. All right? I think I can handle this division here. 320 divided by 8, that's 40. What that means is that for 8 painters in that same room, it would have only taken them 40 minutes to paint that same room. All right? And you have solved your first inverse variation word problem. I'm gonna scroll back through it to bring it all together for you. After reading through it, we highlighted that key information. Then our first step was to identify what x and y represented and then used the information they gave us to solve for k. We figured out that k was 320. Then we wrote our inverse variation equation. We used the y equals k over x format. Y equals 320 divided by x, and then they asked us to determine how long would it have taken 8 painters to paint that room. 8 was our value for x, and 320 divided by 8 is 40. It would have taken 8 painters 40 minutes to paint that same room. Keep that process in mind, and let's try another one together. Let me do a little moving around here. Let's move that over there and get the highlighter ready. We know what's coming, right?

(Describer) She reads.

"The length of a guitar string varies inversely "with the frequency of its vibrations. "If an 8-inch guitar string "vibrates at a frequency of 628 cycles per second, "find the length of a guitar string that vibrates at a frequency of 314 cycles per second." Okay. We read through it. Now let's highlight that key information. "The length of a guitar string--" "The length of a guitar string varies inversely with the frequency of its vibrations."

(Describer) She highlights.

"If an 8-inch guitar string "vibrates at a frequency of 628 cycles per second, "find the length of a guitar string that vibrates at a frequency of 314 cycles per second." So, we've got the key information highlighted, so now let's start working our process, and step one is to determine x, y, and k. All right. If we look back at our problem, we see that the length of the guitar string varies inversely with the frequency of the vibrations. Because we know, in an inverse variation situation, we're always told that y varies inversely as x. So, I can tell here the length of the string, that's y.

(Describer) She writes, "y: length of string".

And the frequency, that's x.

(Describer) She writes, "x: frequency."

Now that we know that y is the length of the string, x is the frequency, we can determine the value for k. We know that for an inverse variation situation that k equals x times y. Right? So, we're told here that an 8-inch guitar string vibrates at 628 cycles per second. So, 8-inch, 628, okay? Those are our values for x and for y. So, k would equal-- the frequency for that situation was 628 for the 8-inch string. It's gonna be pretty large numbers. I'm going to the calculator to do this multiplication. 628 times 8. All right. So, let's go here. 628 times 8. And we have 5,024. I knew it was gonna be big. All right. So, let's go back here. All right. So, k equals 5,024. All right. So, now step two. Because we know that for our inverse variation situation y equals k over x, I can represent this situation by y equals 5,024 divided by x. Right? Now we're ready for step three because we've gotten our equation down. Let's see what we've been asked to find here. "Find the length of a guitar string that vibrates at a frequency of 314 cycles per second." Our frequency here, or our x, is 314. Let's scroll down. Okay. And then step three... would be y equals 5,024 divided by 314. Make sure I got that number right. That was actually, yep, 314. Once I do this division, I'll be able to determine exactly how long this guitar string is, right? That's gonna be some interesting division, so I'm going to the calculator. Those numbers are kind of large. So, 5,024 divided by 314. 5,024 divided by 314. And we have 16. All right. Let's get back to our problem. Let's scroll down so we can write the answer. 16 is what we get for y. What that means is a guitar string that's vibrating at 314 cycles per second is 16 inches long. All right? And you're all done. Good job on that one. Now it's time for you to try one on your own. Press pause and take a few minutes, work your process, and work your way through this problem. Determine x, y, and k, write your equation, and answer the question. When you're ready to compare your answer against mine, press play.

(Describer) Titles: The amount of time it takes to tile a floor varies inversely with the number of people working. If six contractors completed a room in 80 minutes, how long would it have taken 12 contractors to complete the same room?

(describer) The amount of time it takes to tile a floor varies inversely with the number of people working. If 6 contractors completed a room in 80 minutes, how long would it have taken 12 contractors to complete the same room? Let's see how you did here. "The amount of time it takes to tile a floor "varies inversely with the number of people working. "If 6 contractors completed a room in 80 minutes, "how long would it have taken 12 contractors to complete the same room?" Let's highlight the key information here. The time it takes, the amount of time, "varies inversely with the number of people working." "If 6 contractors completed a room in 80 minutes, "how long would it have taken 12 contractors to complete the same room?" So, we've got the key information highlighted. Our first step is to determine what's x, what's y, what's k. All right. Here we have the time "varies inversely with the number of people working." Because we know, with an inverse variation, we're given the information as y varies inversely as x. That means that our amount of time is y. And x is the number of people working. That's what we can control. Okay. Let's keep going here. And to get k, what we're gonna do is use this first part of information we were given: 6 contractors, 80 minutes. Okay. And because we know for inverse variation that k equals x times y, and I know that, for this situation, k will equal-- Gonna need some more room. I had 6 contractors, and they took 80 minutes. So, 6 times 80-- I think I can handle this multiplication here-- should be 480. Okay. Your value for k in this situation is 480. Let me scroll down a bit more, get some space. Now I'm ready for step two, and I can write that inverse variation equation. I'm gonna write it in the y equals k over x format. Now that I know that k is 480-- Let's get more space. Okay. So, y equals 480 divided by x. So, I've got that equation. I've finished step two. Let's scroll up so we can answer the question. "How long would it have taken 12 contractors to complete the same room?" So, 12 contractors has to be a value for x. That's my number of people. Let's scroll down here. So, y equals 480 divided by 12. Right? And I think I can handle that...division. That should be 40. Okay? Let me double-check myself just to be sure here. Because I've been doing a lot of mental math. 480 divided by 12 is indeed 40. All right. Had that one right. What that means is, if we bring this back to the problem so we can get the meaning of this answer for step three. Okay. We were asked to determine how long it would have taken 12 contractors to complete that same room. So, if 12 contractors had been working--scroll down-- it would have only taken them 40 minutes to complete that same room. Okay? Good job on that one. You have reached the end of your lesson on representing inverse variations algebraically. Hope to see you back here soon for more Algebra I. Bye.

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Inverse variations describe a relationship between two variables when one variable increases the other decreases in proportion so that the product is unchanged. In this program, students learn about representing inverse variations algebraically. Part of the "Welcome to Algebra I" series.

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Runtime: 25 minutes

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