Lesson video

<v Instructor>Hello, Mr. Robson here.

Welcome to Maths, what a super place to be.

Today, we're changing the subject to suit the context.

Did you know that expressions and formula aren't restricted to the maths classroom? You'll see them all over the world and we're going to see some lovely examples today.

So, if you are as excited as I am, what are we waiting for? Let's get on with the learning.

Our learning outcome, so, we'll be able to evaluate different problems to ascertain when rearranging a formula is beneficial.

Keywords, a few that you'll hear me use throughout the lesson.

The subject of an equation or a formula is a variable that is expressed in terms of other variables.

It should have an exponent of one and a coefficient of one.

For example, A is the subject of this formula.

A equals the square root of C squared minus B squared.

A is the subject because it has an exponent of one and a coefficient of one.

However, A squared is not the subject of the second formula.

A squared equals C squared, minus B squared.

A squared, that's not an exponent of one so A is not the subject of that formula.

Two parts to today's lesson, we're gonna begin by rearranging additive and multiplicative steps.

Multiplicative relationships can be rearranged into a family of facts.

For example, you know four lots of five makes 20, say, five lots of four makes 20 and we can do the inverse, 20 divided by five must be four and 20 divided by four must be five.

We can rearrange multiplicative formula in the same way.

This is a multiplicative relationship between speed, time, and distance.

Speed multiplied by time equals distance.

You may well have seen this elsewhere in your learning and you'd have seen it described as S multiplied by T equals D.

We can rearrange that multiplication in the same way we can rearrange four times five equals 20.

So we could arrange it as T multiplied by S equals D, and we can do the inverse.

D divided by T must leave us with S.

D divided by S must leave us with T.

Just be aware that in algebra you unlikely to see it written with the division sign, you are more likely to see it written with a fraction bar.

So instead of D divided by T equals S, you'll see D over T equals S, and D over S equals T.

The variable we wish to calculate determines which rearrangement we use.

Here's the four rearrangements of that formula.

If we have this context, a car travels at 50 miles per hour for three hours.

How far has it travelled? What we want to know is distance, so we want the rearrangement whereby distance is the subject.

Can you spot which rearrangement that might be? Well done, we can take this one, S multiplied by T equals D.

We could have taken the second one, T multiplied by S equals D, either one would work.

If I start with that arrangement, whereby D is the subject, I can substitute in unknown speed and unknown time and very quickly calculate how far it is we've travelled.

In this case, 50 times three makes 150, the distance is 150 miles.

What if we change the context a little bit? A plane has to fly 6,000 miles and will travel at 400 miles per hour.

I'm interested in this maths because 6,000 miles from me, there's some very beautiful parts of planet Earth.

The question here is how long will the journey take? We want to know time, so this time we want a rearrangement whereby time is the subject.

Can you spot which one it is? Well done, D over S equals T, that's the one where time is the subject.

So we just need to substitute in the known distance and the known speed.

You've got a distance of 6,000 miles and a speed of 400 miles per hour.

6,000 over 400 makes 15, our time is 15 hours.

By making time the subject, you are very quickly able to find how long the journey would take.

In physics, you may encounter a formula for calculating the acceleration of a particle.

The formula is that acceleration equals final velocity minus initial velocity over time.

You'll see those abbreviated to algebraic variables.

Acceleration will be referred to as A, final velocity, V, initial velocity U, and time T.

That enables us to write the formula like this A equals V minus U over T.

In this format we have acceleration as the subject.

This is good for calculating acceleration.

If you know the final velocity V, the initial velocity U and the time T, you can substitute those in, tap it into your calculator and you'll immediately have acceleration.

But what if you knew A, U, and T and you wanted to know V, the final velocity? You wouldn't be able to type that one into your calculator so easily, but what we can do is we can make V the subject.

We rearrange the formula so that V is expressed in terms of the other variables.

You might think of this as isolating the V.

Let's have a look at how we might do that.

We could start by multiplying both sides of the formula by T.

Then we can add U to both sides of the formula and we've done it, V is now the subject.

We could substitute in the other variables to quickly find the value of V.

Quick check, you've got what I've said so far, which of these rearrangements would be most useful if we knew A, V, and T and we wanted to know U, the initial velocity? Pause, have a think about those three options.

Which one is it? See you in a moment for the answer.

Welcome back, I wonder what you thought.

I hope you didn't go for option A.

If we want to know U, we want U to be the subject, in this case, V was the subject.

I hope you did go for option B, I'll explain why in a moment but not option C.

Option C wasn't a valid rearrangement, so why was option B the right one? Well, we can start with our formula, multiply both sides by T, and if we add U to both sides, we get a positive U term.

We'll subtract AT from both sides and there it is, U equals V minus AT, U being the subject to the formula, that's the most efficient way to have that formula arranged if we wanted to calculate U.

It's not just in maths and physics that we see formulae.

Break-even is a crucial formula in business, especially for new entrepreneurs.

When you are older you might choose to set up your own business, so you'll definitely want to know this formula.

Break-even quantity equals fixed costs, over selling price minus variable cost per unit.

Well, what's this going to tell us? It's gonna tell us something really important.

The break-even point is the quantity of units a business has to sell in order to start making a profit and you are going to want to make profit.

For example, if you were to start a coffee shop with 16,000 pounds of setup costs.

Once up and running, you can sell each coffee for three pounds and they cost you one pound each to produce.

When do you break-even? Or rather, when do you start making money? You're welcome to pause and have a play with that information now.

See if you can figure it out or stick with me and I'll show you how the formula is applied.

Break-even quantity equals fixed costs over selling price minus variable cost per unit.

The breakeven quantity in this case will be a fixed cost of 16,000 pounds over three pound minus one pound.

What's the three pound minus one pound mean? Well, it gives us two pounds i.

e.

, for every coffee we sell we make two pounds, we turn a pound into three pounds, each coffee we sell is two pound of profit.

But we need to make that 16,000 pound of setup cost back.

That's why we do 16,000 divided by two and we get 8,000.

That's 8,000 units to break-even.

Now you need to ask yourself, can you sell 8,000 coffees and how long will it take you? You might have promised somebody that you'll pay them that 16,000 pounds back within a year, could you do it? 8,000 units sounds like a lot, doesn't it? It's only about 20 a day, I think you can do it.

But what if you knew the quantity of units that you could sell? You'd studied your market, you knew how much people would buy, if you knew the break-even quantity or the quantity of units you could sell, and you knew your fixed costs and you knew your variable cost per unit, then you could calculate the selling price you'd have to charge in order to make a profit.

The problem is selling price is in a really awkward position in that formula but we could rearrange.

If we can make selling price the subject of this formula, it'll be far easier to calculate.

We'll start by abbreviating those terms, break-even quantity Q, fixed costs F, selling price S, variable cost per unit V.

Now we can make selling price the subject of the formula.

We're gonna multiply both sides of the formula by S minus V.

Notice I'm keeping them grouped when I multiply three.

From there, we'll divide both sides by Q and then we'll add V to both sides and look, S, the selling price, is now the subject of the formula.

Selling price will equal fixed costs over breakeven quantity minus variable cost per unit.

We can now calculate what price we need to sell each unit for in order to start making a profit.

Once again, when you've got your own business in the future, you're going to want to make a profit.

Quick check, you've got that.

Cost control is a really important element of business.

Rearrange our break-even formula to make variable cost per unit, V, the subject.

There's the formula, get rearranging, make V the subject.

You'll need a minute to do that, so pause and I'll see you again in a moment.

Welcome back, making V the subject of this formula.

Let's start by multiplying through by S minus V.

Then we can divide through by Q, but we've still got the V term as a negative term, let's add V to both sides, we've made V a positive term.

Then we can subtract F over Q and there it is, V variable cost is the subject of the formula.

Practise time now.

Question one, I'd like you to rearrange the formula for circumference of a circle so that'd be more efficient for finding the radius if you knew the circumference.

Circumference equals two pi R, but what form would we want that formula in if we wanted to find the radius? Part B, it's about the area of a triangle.

I'd like you to rearrange that formula so that'd be more efficient for finding the height if you knew the area and the base.

Pause and do some rearranging now.

For part C of question one, I'd like you to rearrange the formula for surface area of a cylinder so that it'd be more efficient for finding the height if you knew the surface area and radius.

For question two, this lovely formula is used for converting Celsius temperatures into Fahrenheit temperatures.

I'd like you to show me two different ways that you can rearrange the formula so that it converts Fahrenheit into Celsius.

Currently, it's converting Celsius into Fahrenheit.

Can you see why? So what change do you need to make so that it's converting Fahrenheit into Celsius? Pause, have a think about that and a good go at that now.

Welcome back, feedback time.

Rearrange the formula for circumference of a circle so that'd be more efficient for finding the radius if you knew the circumference.

Well, that sounds like we want radius to be the subject of the formula.

We can do that by dividing through by two pi.

In that form, R, the unknown we want to find is now the subject.

We could really quickly find the radius if you were to tell me the circumference.

For part B, rearranging the formula for area of a triangle so that'd be more efficient for finding the height if you knew the area and the base.

That sounds like we wanna make height the subject of the formula.

A useful first step there, is to multiply by the reciprocal of a half, multiply both sides by two and the formula looks immediately more simple, two A equals B multiplied by H.

Divide both sides by B and we've isolated our H.

We've made H the subject.

If that's the unknown we want to find it's so much easier when the formula is in that form.

Question one part C, we were rearranging this beautiful formula for surface area of a cylinder to make it more efficient for finding the height if you knew the surface area and the radius.

Looks like a really awkward formula, it's not, it's just an additive relationship.

If two pi R squared plus two pi RH gives me the surface area, I can rearrange that to area minus two pi R squared, will give me two pi RH.

From there, I want to isolate the H.

I'm gonna divide through by two pi R.

I leave H on its own on the right hand side and leave A minus two pi R squared, over two pi R on the left hand side.

We've made H the unknown that we want to find, the subject.

That'll make it much quicker to find when that formula is in that arrangement.

For question three, this lovely formula converts Celsius into Fahrenheit.

You give me any Celsius temperature, I just plug it in where it says C and I can quickly turn that into a Fahrenheit temperature for you.

But it can be rearranged so that it goes the other way, turns Fahrenheit into Celsius.

I asked you to tell me two different ways that you could rearrange it.

So option one, would be to make subtract 32 from both sides your first step.

If we do that we get Fahrenheit minus 32 on the left hand side and nine lots of the Celsius over five on the right hand side.

Our next step, be to multiply both sides by five.

And then we're looking to make Celsius, C, the subject, so we'll divide both sides by nine.

Oh look, that will now turn Fahrenheit into Celsius because we've made Celsius the subject.

A second way you might have done that would be to multiply all terms by five.

When we do that, we get five F equals nine C plus 160.

From there, again we're looking to isolate the C, make C the subject.

I'll subtract 160 from both sides, divide through by nine, so this time the Celsius as the subject, we've got five F minus 160 over nine.

Onto the second part of the lesson now, where we'll be rearranging equations to find the gradient.

A different context but a beautiful one and a most useful one.

If we have a linear equation in the form Y equals MX plus C, there are key features we can quickly spot.

For example, the linear equation Y equals two X minus three, it's represented on that graph there.

We can very quickly see a gradient of two, i.

e.

, for every increase of one in X we have an increase of two in Y and that gradient is repeated everywhere along that line.

Oh look, a gradient of two, the coefficient of X was two.

A Y intercept of zero negative three, that's that coordinate there where the line intersects the Y axis.

Oh look, a Y intercept of zero negative three and our C value was negative three.

In the form Y equals MX plus C, M tells us our gradient, the coefficient of X, in this case two, told us we had a gradient of two.

The C value tells us our Y intercept, in this case is negative three.

Crucially, in this form Y is the subject.

What if we change the form in which we see that linear equation? Sometimes we see linear equations in the form, AXE plus BY equals C, Y is not the subject.

In this form, the gradient is not so easy to spot.

Aisha, Izzy and Alex are discussed in the equation, four X plus five Y equals 40.

Aisha suggests, "I think the gradient will be four because that's the coefficient of X." Izzy thinks something's different here, "I think the gradient might be five, the coefficient of Y this time." Whereas Alex suggests, "I think it will be either 40 divided by five or 40 divided by four." What do you think? Who do you agree with? Pause, have a conversation with the person next to you or a good think to yourself.

I'll see you in a moment.

Welcome back, did you find one of our three pupils you could agree with or did you disagree with all of them? I wonder.

The equation was four X plus five Y equals 40 and Aisha suggests if we make Y the subject, we'll know what the gradient is.

We're looking for the form Y equals MX plus C, so we wanna make Y the subject.

So start with the equation four X plus five Y equals 40, subtract four X from both sides and divide through by five or multiply by a fifth, however you choose to think about that.

If you think of it as multiplying by a fifth on both sides, we're left with Y on the left hand side and then crucially you can see on the right hand side, one 1/5 of 40, that's eight and 1/5 of negative four X, that's negative four over five X.

Izzy says with Y now the subject, we can see the gradient is negative 4/5.

So none of them were right, but they put some really good suggestions forward and then really importantly tested their ideas.

You can see that gradient of negative 4/5 when I show you the line, that's a change of negative four in Y for every change of positive five in X.

You can also see the Y intercept zero, eight.

Let's check you've got that.

I'm gonna do a question and then ask you to repeat a very similar skill.

My question, I'm asked to work out the gradient and the Y intercept of the equation below by rearranging to make Y the subject.

My equation is three X plus seven Y equals 28.

I'm reminded that I need to make Y the subject, so I'm gonna start to isolate it.

I'm gonna subtract three X from both sides of that equation.

Currently, Y is being multiplied by seven so I'm gonna multiply both sides by the reciprocal of seven i.

e.

, 1/7.

By thinking of it this way, I'm reminded that I need to take 1/7 of both those terms on the right hand side.

The left hand side simplifies to Y, on the right hand side I get 1/7 of 28, which is positive four, and then 1/7 of negative three X, which is negative three over seven X.

Now that I've got Y as the subject, I'm in the form Y equals MX plus C.

M, the coefficient of X is gonna be my gradient, i.

e.

, negative 3/7, and the C value will tell me my Y intercept.

I've got C value of four, so that Y intercept is the coordinate zero, four.

Can you repeat that skill for this question? I'd like you to make Y the subject and from there, can you read the gradient and the Y intercept? Pause and give this a go now.

Welcome back, et's see how you did.

We're looking to make Y the subject.

Let's start by subtracting two X from both sides and then we'll multiply through by a fifth.

That will isolate Y on the left hand side, on the right hand side we'll get 1/5 of 30, that's positive six and 1/5 of negative two X.

That's negative two of the five X.

From there, your gradient must be negative two over five and your Y intercept the coordinate zero, six.

Practise time now, I'd like you to work out the gradient and the Y intercept of the lines given by these equations by rearranging to make Y the subject.

Pause and give these problems a go now.

For question two, I'd like to rearrange this general form of a linear equation to make Y the subject and find an expression for the gradient and Y intercept when in this form.

This format we've been looking at, where Y is not the subject, linear equations in the form, AXE plus BY equals C.

Can you write a generalised expression for the gradient and a generalised expression for the Y intercept? Tricky problem, but you've got this.

Pause and give it a go now.

Feedback time, rearranging to make Y the subject and reading the gradient and Y intercept.

For part A, I'm gonna add negative X to both sides and then multiply through by a quarter.

So in Y as the subject, we have the form Y equals three minus a quarter X.

That'll give us a gradient of negative quarter and a Y intercept of zero, three.

For part B, we'll start by adding negative two X to both sides.

So on the right hand side, negative two X minus 18 now.

We'll divide through by six on both sides or multiply both sides by a sixth.

That leave Y is a subject on the left hand side and on the right hand side we've got 1/6 of negative two X, which would give you negative two over six X, which would cancel to negative one over three X and 1/6 of negative 18, which is negative three.

The gradient therefore, negative a third, the Y intercept, zero, negative three.

For part C, eight X minus two Y equals 15.

Something different here, let's start by adding two Y to both sides.

We're gonna want our Y term to be positive, so adding two Y to both sides is a good start.

Let's subtract 15 from both sides and multiply both sides by half.

Half of eight X is four X, half of negative 15 is negative 15 over two, Y is now the subject.

We can see the gradient must be four and the Y intercept must be zero, negative 15 over two.

For part two, this is interesting.

In the form AXE plus BY equals C, can you immediately see the gradient in that form? Can you immediately see the Y intercept in that form? You absolutely can if you can rearrange and generalise from here.

We want to make Y the subject, so let's subtract AXE from both sides.

Y is currently being multiplied by B, so let's multiply by the reciprocal of B, which should be one over B.

That'll leave Y as the subject on the left hand side and we multiply out the bracket on the right hand side, we get C over B, minus A over BX.

In that form, our gradient is gonna be negative A over B and our Y intercept is gonna be zero, C over B.

Sadly, that's the end of the lesson now, but in summary, rearranging can be beneficial depending upon which unknown in a formula you are trying to find.

This will change depending on the context.

Changing the subject requires similar skills to solving an equation.

I hope you've enjoyed this lesson as much as I have, and I hope you're looking forward to using formula like these in your future learning and future life.

I will see you again soon for more wonderful mathematics.