### Video Transcript

In this video, we’ll be looking at direct variation. There’s a bit of terminology and notation to cover as well as the basic concepts, so we’ll talk about what direct variation is and how to express it and then we’ll look at some typical questions.

First, before we talk about what it is, let’s talk about what people call it. 𝑦 is directly proportional to 𝑥 and that little fish symbol there means “is directly proportional to”. 𝑦 varies directly as 𝑥 or 𝑦 varies directly with 𝑥 or 𝑦 is directly proportional to 𝑥 or even 𝑦 equals 𝑘 times 𝑥. 𝑥 and 𝑦 vary directly, 𝑥 and 𝑦 are directly proportional, or 𝑦 is a simple multiple of 𝑥. And this last one here really sums up what’s at the heart of direct proportionality; one variable is just a simple multiple of the other variable. And obviously that links back to this one here: 𝑦 is a simple multiple; 𝑘 is a constant, okay then? It’s called the constant of proportionality or the constant of variation, and typically people use the letter 𝑘 but you might use a different letter depending on where you live. So there are lots of different ways that you might encounter this concept in a question. And really this one here, as we said 𝑦 is a simple multiple of 𝑥 or 𝑦 equals 𝑘 times 𝑥, this will be the best way of describing it. But sadly that’s the way that you’re probably least likely to encounter it in a question. So you’re just gonna have to learn to recognise all of the other forms.

This graph represents one example of direct proportionality. So it’s a graph of time taken to travel and certain number of miles by train travelling at a constant speed. The more miles you travel, the more time it takes; if I travel one more mile, it takes the same amount more time whether I’ve already travelled zero miles or a hundred miles. So the gradient of that line or the slope of that line is always constant and with direct proportionality, the line will always pass through the origin. So when I’ve travelled no miles, that takes no time. Now remember, the slope of the line is the change in 𝑦-coordinates when I increase my 𝑥-coordinate by one. So in this case if I increase my 𝑥-coordinate by one the 𝑦-coordinate goes up by 𝑘. So the slope is 𝑘, and the equation of that line is 𝑦 is equal to 𝑘 times 𝑥. And because it cuts the 𝑦 axis at zero, it goes through the origin, I’m adding zero onto the end, which — In fact from adding zero, I don’t actually need to write that down. So the equation, the general equation, of one of these directly proportional relationships is 𝑦 is equal to 𝑘 times 𝑥. Now depending how fast the train is travelling, that will affect the value of 𝑘. And just a bit more terminology, that 𝑘 value is sometimes called the constant of proportionality or the constant of variation. Now the first skill that you have to master is recognising when two variables are in direct variation. So you might get a question like this: the graph shows the relationship between variables 𝑥 and 𝑦. Does 𝑦 vary directly with 𝑥? Well it’s a straight line relationship, and it passes through the origin. So those two facts together tell us, “Yes, 𝑦 does vary directly with 𝑥.” So it’s pretty easy to spot; the only thing that you need to look out for is the different forms of wording. So they might have said does 𝑦 vary directly as 𝑥 or are 𝑥 and 𝑦 directly proportional or do 𝑥 and 𝑦 vary directly even.

Another type of question that asks you to recognise whether or not to variables are in direct variation; it’d be something like this: the coordinates of some points on a line are given in the table of values below. Does it show that 𝑦 is directly proportional to 𝑥? And then in our table, we’ve got 𝑥 is one when 𝑦 is two, 𝑥 is two when 𝑦 is four, 𝑥 is three when 𝑦 is six, 𝑥 is four when 𝑦 is eight, and 𝑥 is five when 𝑦 is ten. Now with each of these pairs of coordinates, 𝑥 and 𝑦, we can see that 𝑦 is always twice the value of 𝑥, plus nothing. So the question says it’s a line, so the equation of the line is 𝑦 equals two times 𝑥. So 𝑦 is always a simple multiple of 𝑥. And when 𝑥 is equal to zero, then 𝑦 will be two times zero. That would also be zero, so it goes through the origin. And another way to check that is that like every time the 𝑥-coordinate goes down by one, then the 𝑦-coordinate goes down by two. So if I decrease the 𝑥-coordinate from one to zero, then the corresponding 𝑦-coordinate would go down from two to zero. So either way, we’ve got two different ways of checking whether it passes through the origin. Overall it meets our two criteria: so 𝑦 is a simple multiple of 𝑥, and it do- and the line does go through the origin. So the answer is yes, it shows that 𝑦 is directly proportional to 𝑥.

Another example is do the coordinates in the table of values below show that the variables 𝑥 and 𝑦 are in direct variation. And then we’ve got the coordinate pairs two and six, four and sixteen, and six and twenty-four. So the first criterion is if these are going to be in direct proportion or direct variation, then 𝑦 is gonna be a simple multiple of 𝑥. So let’s work out the multiple for each coordinate pair. When 𝑥 is two, 𝑦 is six, so the 𝑦-coordinate is three times the 𝑥-coordinate When 𝑥 is four, 𝑦 is sixteen, so 𝑦 is equal to four times the 𝑥-coordinate. And when 𝑥 is six, 𝑦 is twenty-four, so 𝑦 is equal to four times the 𝑥-coordinate there. Now we’ve got different values here. Sometimes 𝑦 is equal to three times the 𝑥-coordinate; sometimes 𝑦 is equal four times the 𝑥-coordinate. So that is not just a simple constant multiple of 𝑥, so this is not direct variation. And because the data broke our first rule, we didn’t even have to go on and check the second rule to see whether we thought it would go through zero zero, through the origin.

Here’s another one: the coordinates of some points on a straight line are given in the table of values below. Does it show that 𝑦 varies directly as 𝑥? And then we’ve got the co-ordinate pairs three, eleven, six, seventeen, and nine, twenty-three. So just looking at the numbers there, every time I increase the 𝑥-coordinate by three, the corresponding 𝑦-coordinate goes up by six. So these numbers do support the idea that what it says in the question is true, that the points are on a straight line. But let’s just check to see whether or not they go through the origin. Well I’ve got the point three, eleven. So if I take away three from the 𝑥-coordinate, I need to take away six from the 𝑦-coordinate. And that gives me the point zero, five, so it doesn’t pass through the origin. So it’s a straight line, but it doesn’t pass through the origin. So no, it doesn’t show that 𝑦 varies directly with 𝑥. In fact looking at the numbers, the analysis of the differences between 𝑥 and 𝑦 tells us that the slope was two. And this tells us that it cuts the 𝑦 axis at five. So the equation of that line is 𝑦 equals two 𝑥 plus five. So all directly varying relationships would have a plus zero rather than plus something nonzero on the end.

Okay, here’s three questions for you to try. What I want you to do is pause the video and then say whether or not these three situations represent directly proportional relationships. Right, the first one is a straight line relationship, and it passes through the origin. So yes, 𝑦 is directly proportional to 𝑥. The second one is a straight line relationship, but it doesn’t pass through the origin. So no, 𝑦 is not directly varying with 𝑥. And the third one, the cost in dollars of fares for a cab company is one point five times the number of miles, so we’ve got this linear relationship. But because we always add three to the result, that means that the 𝑦-coordinate if you like, if we drew the graph of that, wouldn’t always be a simple multiple of the 𝑥-coordinate because we’ll be shifting the whole line up by three. More importantly, if we did zero miles, one point five times zero would be zero. But then add three to that, the 𝑦-coordinate, the fare would be three dollars. So it doesn’t pass through the origin, which means that that’s not a directly proportional relationship.

Okay now we can recognise directly varying relationships. Let’s explore them in a bit more detail. Now we need to be able to work with equations which represent directly varying relationships. So this question here says given that 𝑦 varies directly with 𝑥, write an equation for 𝑦 in terms of 𝑥 if 𝑘 is the constant of proportionality. So when it says 𝑦 varies directly with 𝑥, this means that 𝑦 is directly proportional to 𝑥. So we can write that like this. And that means that 𝑦 is always gonna be equal to some number, some constant number, times 𝑥. And the question tells us that 𝑘 in this case is the constant of proportionality, so the value we can put in here is 𝑘. So our answer is 𝑦 equals 𝑘 times 𝑥. Now in some questions, you’re told to write an equation for 𝑦 in terms of 𝑥 and then you’re given some basic information that will help you to do that. So we’re told that 𝑦 and 𝑥 are directly proportional and that when 𝑥 is equal to twelve, then 𝑦 is equal to thirty-six. We’ll start off by saying if 𝑦 is directly proportional to 𝑥, then that means that 𝑦 is equal to 𝑘 times 𝑥, some number times 𝑥. But in this case, the question told us that when the 𝑥-coordinate is twelve, the 𝑦-coordinate is thirty-six, so we can plug those into our equation. So we get thirty-six is equal to 𝑘 times twelve. So that’s just a specific point on the line. Now dividing both sides by twelve, then we can cancel on the right hand side; twelve divided by twelve is one twelve divided by twelve is one, so we’re just left with 𝑘. And thirty-six divided by twelve on the left-hand side is three so 𝑘 is equal to three. Well now we know that 𝑘 is three, we can write that back in our original equation: 𝑦 is equal to three times 𝑥.

So another example would be: write an equation for 𝑦 in terms of 𝑥 given that 𝑦 varies directly as 𝑥 and 𝑥 equals thirty-five when 𝑦 equals nine. So again, we can say 𝑦 is directly proportional to 𝑥. And that means the 𝑦 is just some constant number, let’s call it 𝑘, times 𝑥 But we were given a specific pairing of 𝑥 and 𝑦 in the question, so we can put those values in for 𝑥 and 𝑦 to work out what 𝑘 is. So 𝑦 is nine and 𝑥 is thirty-five, so nine is equal to 𝑘 times thirty-five. Now I can divide both sides by thirty-five to leave myself with just 𝑘 on the right-hand side, and 𝑘 is equal to nine over thirty-five. Well that won’t cancel down. So sometimes 𝑘 comes out as a not such a nice number. In this case, it’s just a fraction. Sometimes it’s an integer. Sometimes it’s positive. Sometimes it’s negative. So in this case, our equation is 𝑦 is equal to nine thirty-fifths of 𝑥.

Now some questions will ask you to not only write the equation but then to substitute a particular value into it and solve that equation. So we’ve been given 𝑦 is directly proportional to 𝑥. When 𝑥 is seven, then 𝑦 is twenty-one. And we’ve gotta find the value of 𝑦 when 𝑥 equals eleven. So the general approach to this is: first of all, work out the equation, and then substitute and find the value of 𝑦 So 𝑦 is directly proportional to 𝑥 means that 𝑦 is some constant number times 𝑥; let’s call it 𝑘. And we’re told specifically that when 𝑥 is seven, then 𝑦 is twenty-one. So we can put those values in for 𝑥 and 𝑦. And that gives us twenty-one is equal to 𝑘 times seven. So if we now divide both sides by seven to find out the value of 𝑘, that leaves us with 𝑘 on the right-hand side and twenty-one divided by seven is three, so 𝑘 is equal to three. So the equation that governs this relationship is 𝑦 is equal to three times 𝑥. So now we need to go on and find the value of 𝑦 when 𝑥 is equal to eleven. So we can plug that value of 𝑥 into the equation. So we’ve got 𝑦 equals three times eleven, which means that when 𝑥 is eleven, 𝑦 is thirty-three.

Now they can make that sort of question slightly harder by asking you to find the value of 𝑥 given a particular value of 𝑦. So we’ve got 𝑦 is directly proportional to 𝑥, and when 𝑥 equals five 𝑦 equals three. Find the value of 𝑥 when 𝑦 equals twenty. So we start again in the same place: 𝑦 is directly proportional to 𝑥 means that 𝑦 is equal to some constant number times 𝑥, which we’re calling 𝑘 in this case. Then we’ll substitute the particular pair of 𝑥𝑦-values we were given into our equation to find the value of 𝑘. And that means that three is equal to 𝑘 times five. Now dividing both sides by five to find the value of 𝑘 leaves us with 𝑘 on the right-hand side. And that tells us that 𝑘 is equal to three-fifths. Now 𝑘 is equal to three-fifths and 𝑦 is equal to 𝑘 times 𝑥. That means our general formula is 𝑦 is equal to three-fifths times 𝑥. And we can use that formula to find the value of 𝑥 when 𝑦 is equal to twenty So we’ve got twenty is equal to three-fifths times 𝑥. Well if I multiply both sides of that by five, five times twenty is a hundred and three over five times five, the five’s will cancel leaving us just with three. So the right-hand side just becomes three 𝑥. And then if I divide both side of that by three, I’ve got 𝑥 is equal to a hundred divided by three. So that’s thirty-three and a third.

So this question was pretty much like the last question, but the problem was we had the unknown here and we had some things that we had to come unpack that equation to work out the value of 𝑥. Now we could’ve organised our equation in a quite different way. So let’s- let’s kind of turn this round. Instead of saying 𝑦 is directly proportional to 𝑥 means that 𝑦 is equal to 𝑘 times 𝑥, let’s say that that means that 𝑥 is equal to 𝑘 times 𝑦. And of course that’s true. It’ll be a different value of 𝑘 for that equation, so I’m actually gonna use a different letter to represent it just to make that point. But let’s run through that calculation again. So starting from the beginning, 𝑦 is directly proportional to 𝑥 means that 𝑥 is equal to some number times 𝑦. So we’ll call that 𝑐 in this case. And we’re told that when 𝑥 is five, 𝑦 is three. So this means that five is equal to 𝑐 times three, so we divide both sides of that by three. We’ve got 𝑐 is equal to five-thirds. That means 𝑥 is equal to five-thirds of 𝑦.

And this is the case with all directly proportional relationships though actually we could come up with two different equations to describe that relationship. We’ve been coming up with 𝑦 equals three-fifths of 𝑥 here, but also 𝑥 equals five-thirds of 𝑦. These two constants here are the reciprocal of each other. And you know in this particular case, if we use the second one and we were given a value of 𝑦, we would we could easily put 𝑦 equals twenty in there and then just get straight to our answer for 𝑥, making that part of the calculation a bit easier. So sometimes it’s easy to do it one way. Sometimes it’s easier to do it the other.

One last question then: the variables 𝑥 and 𝑦 are directly proportional. Some values are shown in the table of contents below. Find the constant of variation and the value of 𝑥 when 𝑦 equals nineteen. So we know that 𝑦 is directly proportional to 𝑥, and that tells us that 𝑦 is equal to some constant times 𝑥. And the question gave us three pairs of values: when 𝑥 is three, 𝑦 is four point five; when 𝑥 is six, 𝑦 is nine; and when x is ten, 𝑦 is fifteen. So we can pick any of those that we like in order to work out this constant. So I’m just gonna use when 𝑥 is six, 𝑦 is nine cause they’re like relatively easy numbers. And then that gives me nine is equal to 𝑘 times six. So dividing both sides by six, I’ve got 𝑘 is equal to nine over six, which they’re both divisible by three, so that becomes three over two. So our constant of variation is three over two.

Now it’s worth noting that although I taught you little trick of doing this the other way round to make the maths easier in the previous question, you’ve gotta be careful; in most of these questions, they’re expecting you to do the 𝑦 is directly proportional to 𝑥 and 𝑦 is equal to 𝑘 times 𝑥. If you do the calculation the other way round, 𝑥 is directly proportional to 𝑦, and find out this constant variation, you’ll get a different value, the reciprocal of the value you’re looking for. Anyway that gives us an equation of 𝑦 equals three over two 𝑥, which we can rearrange to 𝑥 equals two over three 𝑦. And substituting in 𝑦 equals nineteen gives us 𝑥 is twelve and two-thirds.