### Video Transcript

In this video, weβll be looking at
direct variation. Thereβs a bit of terminology and
notation to cover as well as the basic concept. So weβll talk about what direct
variation is and how to express it. And then weβll look at some typical
questions.

First, before we talk about what it
is, letβs talk about what people call it. π¦ is directly proportional to
π₯. And that little fish symbol there
means βis directly proportional to.β π¦ varies directly as π₯ or π¦
varies directly with π₯. Or π¦ is directly proportional to
π₯. Or even π¦ equals π times π₯. π₯ and π¦ vary directly. π₯ and π¦ are directly
proportional. Or π¦ is a simple multiple of
π₯. And this last one here really sums
up whatβs 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. π¦ is a simple multiple. π is a constant. Okay, then itβs called the constant
of proportionality or the constant of variation. And typically, people use the
letter π. 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 π¦ is a simple multiple of π₯ or π¦ equals π times π₯, this will be the best
way of describing it. But sadly, thatβs the way that
youβre probably least likely to encounter it in a question. So youβre just gonna have to learn
to recognise all of the other forms.

This graph represents one example
of direct proportionality. So itβs 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 Iβve already travelled zero miles or 100
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 Iβve travelled no miles,
that takes no time. Now remember, the slope of the line
is the change in π¦-coordinates when I increase my π₯-coordinate by one. So in this case, if I increase my
π₯-coordinate by one, the π¦-coordinate goes up by π. So the slope is π, and the
equation of that line is π¦ is equal to π times π₯.

And because it cuts the π¦-axis at
zero, it goes through the origin, Iβm adding zero on to the end, which in fact from
adding zero I donβt actually need to write that down. So the equation, the general
equation, of one of these directly proportional relationships is π¦ is equal to π
times π₯.

Now depending how fast the train is
travelling, that will affect the value of π. And just a bit more terminology,
that π 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 π₯ and π¦. Does π¦ vary directly with π₯? Well, itβs a straight-line
relationship. And it passes through the
origin. So those two facts together tell
us, βYes, π¦ does vary directly with π₯.β

So itβs 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 π¦
vary directly as π₯ or are π₯ and π¦ directly proportional or do π₯ and π¦ vary
directly even.

Another type of question that asks
you to recognise whether or not two variables are in direct variation itβd be
something like this.

The coordinates of some points on a
line are given in the table of values below. Does it show that π¦ is directly
proportional to π₯? And then in our table, weβve got π₯
is one when π¦ is two. π₯ is two when π¦ is four. π₯ is three when π¦ is six. π₯ is four when π¦ is eight. And π₯ is five when π¦ is 10.

Now with each of these pairs of
coordinates, π₯ and π¦, we can see that π¦ is always twice the value of π₯, plus
nothing. So the question says itβs a
line. So the equation of the line is π¦
equals two times π₯. So π¦ is always a simple multiple
of π₯. And when π₯ is equal to zero, then
π¦ 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 π₯-coordinate goes down by one, then the π¦-coordinate goes
down by two. So if I decrease the π₯-coordinate
from one to zero, then the corresponding π¦-coordinate would go down from two to
zero. So either way, weβve got two
different ways of checking whether it passes through the origin. And overall, it meets our two
criteria. So π¦ is a simple multiple of π₯,
and it do- and the line does go through the origin. So the answer is yes, it shows that
π¦ is directly proportional to π₯.

Another example is, do the
coordinates in the table of values below show that the variables π₯ and π¦ are in
direct variation? And then weβve got the coordinate
pairs two and six, four and 16, and six and 24. So the first criterion is if these
are gonna be in direct proportion or direct variation, then π¦ is gonna be a simple
multiple of π₯. So letβs work out the multiple for
each coordinate pair.

When π₯ is two, π¦ is six. So the π¦-coordinate is three times
the π₯-coordinate. When π₯ is four, π¦ is 16. So π¦ is equal to four times the
π₯-coordinate. And when π₯ is six, π¦ is 24. So π¦ is equal to four times the
π₯-coordinate there. Now weβve got different values
here. Sometimes π¦ is equal to three
times the π₯-coordinate. Sometimes π¦ is equal four times
the π₯-coordinate. So that is not just a simple
constant multiple of π₯. So this is not direct
variation.

And because the data broke our
first rule, we didnβt even have to go on and check the second rule to see whether we
thought it would go through zero, zero, through the origin.

Hereβs another one.

The coordinates of some points on a
straight line are given in the table of values below. Does it show that π¦ varies
directly as π₯? And then weβve got the coordinate
pairs three, 11; six, 17; and nine, 23.

So just looking at the numbers
there, every time I increase the π₯-coordinate by three, the corresponding
π¦-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 letβs just check to see whether
or not they go through the origin.

Well, Iβve got the point three,
11. So if I take away three from the
π₯-coordinate, I need to take away six from the π¦-coordinate. And that gives me the point zero,
five. So it doesnβt pass through the
origin. So itβs a straight line, but it
doesnβt pass through the origin. So no, it doesnβt show that π¦
varies directly with π₯.

In fact, looking at the numbers,
the analysis of the differences between π₯ and π¦ tells us that the slope was
two. And this tells us that it cuts the
π¦-axis at five. So the equation of that line is π¦
equals two π₯ plus five. So all directly varying
relationships would have a plus zero rather than plus something nonzero on the
end.

Okay, hereβs 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, π¦ is directly proportional
to π₯. The second one is a straight line
relationship. But it doesnβt pass through the
origin. So no, π¦ is not directly varying
with π₯. And the third one, the cost in
dollars of fares for a cab company is 1.5 times the number of miles. So weβve got this linear
relationship. But because we always add three to
the result, that means that the π¦-coordinate if you like, if we drew the graph of
that, wouldnβt always be a simple multiple of the π₯-coordinate because weβd be
shifting the whole line up by three.

More importantly, if we did zero
miles, 1.5 times zero would be zero. But then add three to that, the
π¦-coordinate, the fare, would be three dollars. So it doesnβt pass through the
origin, which means that thatβs not a directly proportional relationship.

Okay, now we can recognise directly
varying relationships. Letβs 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, βGiven
that π¦ varies directly with π₯, write an equation for π¦ in terms of π₯ if π is
the constant of proportionality.β

So when it says π¦ varies directly
with π₯, this means that π¦ is directly proportional to π₯. So we can write that like this. And that means that π¦ is always
gonna be equal to some number, some constant number, times π₯. And the question tells us that π
in this case is the constant of proportionality. So the value we can put in here is
π. So our answer is π¦ equals π times
π₯.

Now in some questions, youβre told
to write an equation for π¦ in terms of π₯. And then youβre given some basic
information that will help you to do that.

So weβre told that π¦ and π₯ are
directly proportional and that when π₯ is equal to 12, then π¦ is equal to 36. So we will start off by saying if
π¦ is directly proportional to π₯, then that means that π¦ is equal to π times π₯,
some number times π₯. But in this case, the question told
us that when the π₯-coordinate is 12, the π¦-coordinate is 36. So we can plug those into our
equation. So we get 36 is equal to π times
12. So thatβs just a specific point on
the line.

Now dividing both sides by 12, then
we can cancel on the right-hand side. 12 divided by 12 is one. 12 divided by 12 is one, so weβre
just left with π. And 36 divided by 12 on the
left-hand side is three, so π is equal to three. Well, now we know that π is three,
we can write that back in our original equation. π¦ is equal to three times π₯.

So another example would be, write
an equation for π¦ in terms of π₯ given that π¦ varies directly as π₯ and π₯ equals
35 when π¦ equals nine. So again, we can say π¦ is directly
proportional to π₯. And that means that π¦ is just some
constant number β letβs call it π β times π₯. But we were given a specific
pairing of π₯ and π¦ in the question. So we can put those values in for
π₯ and π¦ to work out what π is. So π¦ is nine and π₯ is 35, so nine
is equal to π times 35.

Now I can divide both sides by 35
to leave myself with just π on the right-hand side. And π is equal to nine over
35. Well, that wonβt cancel down. So sometimes π comes out as a not
such a nice number. In this case, itβs just a
fraction. Sometimes itβs an integer. Sometimes itβs positive. Sometimes itβs negative. So in this case, our equation is π¦
is equal to nine thirty-fifths of π₯.

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 weβve been given π¦ is directly
proportional to π₯. When π₯ is seven, then π¦ is
21. And weβve gotta find the value of
π¦ when π₯ equals 11. So the general approach to this is,
first of all, work out the equation and then substitute and find the value of
π¦. So π¦ is directly proportional to
π₯ means that π¦ is some constant number times π₯ β letβs call it π. And weβre told specifically that
when π₯ is seven, then π¦ is 21. So we can put those values in for
π₯ and π¦. And that gives us 21 is equal to π
times seven.

So if we now divide both sides by
seven to find out the value of π, that leaves us with π on the right-hand
side. And 21 divided by seven is three,
so π is equal to three. So the equation that governs this
relationship is π¦ is equal to three times π₯.

So now we need to go on and find
the value of π¦ when π₯ is equal to 11. So we can plug that value of π₯
into the equation. So weβve got π¦ equals three times
11, which means that when π₯ is 11, π¦ is 33.

Now they can make that sort of
question slightly harder by asking you to find the value of π₯ given a particular
value of π¦.

So weβve got π¦ is directly
proportional to π₯. And when π₯ equals five, π¦ equals
three. Find the value of π₯ when π¦ equals
20. So we start again in the same
place. π¦ is directly proportional to π₯
means that π¦ is equal to some constant number times π₯, which weβre calling π in
this case. Then weβll substitute the
particular pair of π₯π¦-values we were given into our equation to find the value of
π. And that means that three is equal
to π times five.

Now dividing both sides by five to
find the value of π leaves us with π on the right-hand side. And that tells us that π is equal
to three-fifths. Now π is equal to three-fifths and
π¦ is equal to π times π₯. That means our general formula is
π¦ is equal to three-fifths times π₯.

And we can use that formula to find
the value of π₯ when π¦ is equal to 20. So weβve got 20 is equal to
three-fifths times π₯. Well, if I multiply both sides of
that by five, five times 20 is 100. And three over five times five, the
fives will cancel, leaving us just with three. So the right-hand side just becomes
three π₯. And then if I divide both of the
sides of that by three, Iβve got π₯ is equal to 100 divided by three. So thatβs 33 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 kinda unpack that equation to
work out the value of π₯. Now we couldβve organised our
equation in a quite different way. So letβs- letβs kind of turn this
round. Instead of saying π¦ is directly
proportional to π₯ means that π¦ is equal to π times π₯, letβs say that that means
that π₯ is equal to π times π¦. And of course thatβs true. Itβll be a different value of π
for that equation. So Iβm actually gonna use a
different letter to represent it just to make that point. But letβs run through that
calculation again.

So starting from the beginning, π¦
is directly proportional to π₯ means that π₯ is equal to some number times π¦. So weβll call that π in this
case. And weβre told that when π₯ is
five, π¦ is three. So this means that five is equal to
π times three. So we divide both sides of that by
three. Weβve got π is equal to
five-thirds. And that means π₯ is equal to
five-thirds of π¦. And this is the case with all
directly proportional relationships. And actually we could come up with
two different equations to describe that relationship. Weβve been coming up with π¦ equals
three-fifths of π₯ here, but also π₯ equals five-thirds of π¦. 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 π¦, we would- we could
easily put π¦ equals 20 in there and then just get straight to our answer for π₯,
making that part of the calculation a bit easier. So sometimes itβs easy to do it one
way. Sometimes itβs easier to do it the
other.

One last question then.

The variables π₯ and π¦ are
directly proportional. Some values are shown in the table
of contents below. Find the constant of variation and
the value of π₯ when π¦ equals 19. So we know that π¦ is directly
proportional to π₯. And that tells us that π¦ is equal
to some constant times π₯. And the question gave us three
pairs of values. When π₯ is three, π¦ is 4.5. When π₯ is six, π¦ is nine. And when π₯ is 10, π¦ is 15. So we can pick any of those that we
like in order to work out this constant.

So Iβm just gonna use when π₯ is
six, π¦ is nine, cause theyβre like relatively easy numbers. And then that gives me nine is
equal to π times six. So dividing both sides by six, Iβve
got π is equal to nine over six, which theyβre both divisible by three. So that becomes three over two. So our constant of variation is
three over two.

Now itβs worth noting that although
I told you this little trick of doing this the other way round to make the maths
easier in the previous question, youβve gotta be careful. In most of these questions, theyβre
expecting you to do the π¦ is directly proportional to π₯ and π¦ is equal to π
times π₯. If you do the calculation the other
way round, π₯ is directly proportional to π¦, and find out this constant of
variation, youβll get a different value, the reciprocal of the value youβre looking
for. Anyway, that gives us an equation
of π¦ equals three over two π₯, which we can rearrange to π₯ equals two over three
π¦. And substituting in π¦ equals 19
gives us π₯ is 12 and two-thirds.