Lesson video

Thank you for choosing to learn using in this video.

My name is Ms. Davies, and I'm gonna be helping you as you work your way through this lesson.

We're looking at a lot of our algebra skills, we're also looking at our fraction skills.

You are definitely gonna want to write things down as we go through.

You learn best in mathematics when you give things a go, so make sure that you've got some paper and a pen to hand so that you can try things out.

Of course, I'm gonna help you as we go through, so if there's bits that you're not sure about what the next step would be, have a think yourself, but then watch the video, and I will show you how I would do these questions.

As with a lot of things in algebra, there are different ways of doing things.

Just because your way is not the same as my way doesn't mean that you are doing it wrong, so bear that in mind as you're exploring some of these concepts.

Let's get started then.

Welcome to this lesson, where we're gonna be checking, and securing, your understanding of changing the subject with simple algebraic fractions.

First thing we need to be confident with today is the idea of a subject.

So 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.

What that means is in the formula a equals the square root of c squared minus b squared, a is the subject, it's being expressed in terms of other variables, and it has a coefficient of one and exponent of one.

a squared is not the subject of the formula a squared equals c squared minus b squared, it can only be the subject if it has an exponent of one and a coefficient of one.

So, we're gonna start by reminding ourselves of how to rearrange multiplicative relationships.

So, what is the subject of this equation? Hopefully, you said a, it is expressed in terms of the other variables, it has an exponent of one, and a coefficient of one.

Okay, what about this equation? Pause the video, see if you can explain to someone what you think.

Well done, if you said that it doesn't have a subject.

One over x could be written as x to the negative one, remember, so it doesn't have an exponent of one, it has an exponent of negative one, so x is not the subject.

Equally, y cannot be the subject, because it has a coefficient of two, so we'd have to rearrange that if we wanted to have either x the subject or y the subject.

So, we can rearrange equations using inverse operations.

So let's think about what we could do to make b the subject of this equation.

Well, where the equation contains fractions, it's often easiest to start by multiplying to remove the fractional terms. So let's think about that right-hand side, b over c is the same as b times one over c, so we can multiply by the reciprocal of one over c, which is c.

If we multiply both sides by c, we get ac for the left-hand side, and just b for the right-hand side.

We've now made b the subject So, now we've got two forms of this equation, what do you think we could do to make c the subject? Pause the video, what would you do? Now, using that second equation is gonna be easiest, we've already done the first step, to make c the subject, we need to divide both sides by a, or you can think about that as multiplying both sides by one over a, which is the reciprocal.

So, ac times one over a, equals b times one over a, that just leaves us with c on the left-hand side, and b over a on the right-hand side.

If you like to think of it as divide in both sides by a, that's absolutely fine.

So, we've now made c the subject.

Izzy says, "We use multiplicative relationships loads in geometry." "Yeah, like trigonometry," but Lucas sometimes gets confused with which values he is dividing.

"If you know one relationship, you can rearrange to get the others." Lucas knows that sine theta is the opposite side divided by the hypotonus.

So let's think about how Lucas could rearrange this to get an equation where the hypotenuse is the subject.

So, if you've looked at any right angle trigonometry at any point, you will recognise this equation, if not, don't worry, we can rearrange it just like any equation.

So let's write this as sine theta equals o over h.

To make the hypotenuse the subject, we need to multiply both sides by h first to remove any fractional terms, and then we can divide both sides by sine theta so you get h equals o over sine theta.

So, sometimes we need to use multiple operations, so let's rearrange this one to make c the subject.

Just like before, it's gonna be easiest to remove fractional terms by multiplying by the denominator, so multiply by five, then we can add three c to both sides, or you might prefer to subtract b from both sides, it doesn't matter.

Then we can subtract five a, and then we can divide by three.

We can write that as b minus five a over three equals c.

And, again, we have c as the subject.

Izzy is trying to make a the subject of this equation.

Ooh, Izzy normally likes to start by multiplying by the denominator to remove fractional terms. Can you see what the problem might be with this idea? So, the fraction has been square rooted, so if she wants to multiply by the denominator, she's gonna have to rewrite this.

Now ,it could be written as s equals the square root of a over the square root of r.

That'd be absolutely fine, and then she could use this method.

Once she's got to this point, she could square both sides to get what a is.

However, that does require us to deal with square roots.

What other option does Izzy have for a first step which might be a bit better? Can you see it? The easier option is probably to square both sides of the equation first.

That gives us s squared equals a over r.

Now we can multiply both sides by r, and we have a as the subject.

Yeah, Izzy's reflected a little bit here, and said, "I need to look at each question and decide the sensible first step." There's not always gonna be one method that always works nicely, you're gonna have to look at what the equation looks like, and make a sensible decision.

So, where there are multiple turns in the denominator, we can do exactly the same, we just have to make sure we're maintaining equality.

So we're gonna make x the subject of this equation.

Let's start by multiplying both sides by the denominator.

We need to multiply by the whole expression, x plus four.

So that means five lots of x plus four, and I'm gonna use brackets, equals y.

What could we do now? Lucas says we could expand the brackets.

Jacob says we could divide both sides by five.

I wonder which one you went with? Both options are gonna be valid here.

However, expanding the brackets is unnecessary, and sometimes, by expanding the brackets, we cause ourselves to do trickier calculations than we need to.

Here, however, expanding the brackets isn't too bad.

We get 5x plus 20 equals y, 5x equals y minus 20, and then x equals y minus 20 over 5.

If we just divide it by five first, we get x plus four equals y over five, so x equals y over five, minus four.

Now, those two things are equivalent, pause the video and just check if you're not sure why.

All right, your turn.

Which of these is the most efficient first step to make r the subject of this equation, what do you think? Well done if you said multiply both sides by two p.

that's gonna remove any fractional terms and make it easier to manipulate.

Okay, which of these is the correct way to multiply both sides by two p? Give it a go.

Well, I dunno if you spotted, there's two answers.

We could use brackets, and write it as two p lots of t plus five on the right-hand side, or we can expand those brackets and write it as two pt plus 10 p.

Remember that left-hand side, if we multiply it by two p, it just gives us r plus four.

Right, time to have a practise then.

So, for one, you are making o the subject of both those equations, you might recognise them as the trigonometry ratios.

For two, you need to make a the subject.

Then for three, you need to make h the subject.

You might recognise b as the formula for the volume of a cuboid.

And for c, it's actually the formula for the area of a triangle.

Can you make each the subject of all three of those? Well done.

This time, you're gonna have a go at making x the subject.

Think carefully about your priority of operations, and the order in which you're doing things.

For question five, you need to make r the subject.

Give those a go.

Well done.

For 6a, I would like you to make a the subject.

For 6b, you've got the same equation, but I'd like b to be the subject.

Give those a go.

Well done, if you've got this far, there was quite a lot to do in those previous ones.

There's a little challenge for you to stretch yourself at the end here.

In each row of the table, can you identify an equation which is not equivalent to the others? Have a bit of a play around, see if you can find it.

Off you go.

Well done.

So, for a, o equals h sine theta, or h times sine theta.

For b, o equals a tan theta, or a times tan theta.

For two, a equals h cos theta.

For b, a equals o over tan theta, you've got two steps for that one.

For three, you've got h equals a over cos theta.

For b, you've got h equals v divided by lw.

It's easiest if you group that lw together and divide by the whole thing.

For c, it might be easiest to double both sides first, you get 2a equals bh, so h equals 2a over b.

Right, for question four, I'd like you to pause the video and just check your answers.

Think about where I've used brackets, and making sure that you've got equivalent equations to mine.

The same for five.

Again, you need to think carefully about the order that you've done these in, so pause the video and check through my working, do they align with yours? For question six, we'll do this one together.

So I added one to both sides, then I multiply both sides by the square root of a.

Then I divided both sides by eight c plus one, and then I squared both sides.

There's no reason to write that in another form, it's absolutely fine as it is.

For b, you didn't have to do any work.

Hopefully, you spotted that you'd already got to that part by making a the subject.

And finally, this is a little bit of a challenge, I'd like you to pause the video and see if you got the same odd ones out.

If you didn't quite get to that final one, don't panic, there was lots of skills involved in the previous questions, we're just playing around with 'em a little bit more in this final puzzle.

Well done, though, if you did give that last one a go.

Right, so now we're gonna have a look at rearranging to suit the context.

So, using those skills we did in the first part of the lesson, and thinking about some problems in context.

So, here is one way to write the formula for the area of a trapezium.

What, at the moment, is the subject of this formula? Right, at the moment, capital A, which is representing the area, is the subject of this formula.

If we knew the length of the two parallel sides, so a and b, and we knew the height, we could substitute to calculate the area.

Right, how could we rearrange this to make a the subject? If you'd like to try this yourself, pause the video now.

Let's have a look then.

So, we can multiply both sides by two, we get two A equals h multiplied by a plus b.

Then we can divide both sides by h, and subtract b.

We've now made a the subject.

Right, when might this be useful? Pause the video, can you think of a time where having a as the subject is gonna be helpful to us? Right, you might have said, when we know the area, the height, and one of the parallel sides, but want to calculate the other.

So when we want to calculate A, but we have all the other pieces of information, having a as the subject could be helpful.

Let's have an example.

So here, we've got the area, we've got the length of b, and we've got the height.

So now all we need to do is substitute the required values.

So, two lots of 90 over 9 minus 13 equals a.

And then we can either use a calculator, or we can use our mental arithmetic, and we get a is seven.

Right, let's look at another formula that you might have seen before.

So, the area of a sector which has a radius, r, and an angle, theta, is given by this formula.

So, there's a sector on the right-hand side, it's got an angle of theta, it's got a radius in this case of 20.

Lucas says, "I want to calculate theta, so I need to make that the subject." Let's give that a go.

So, if we group pi r squared together, we can just divide by the whole of pi r squared, so we get A over pi r squared equals theta over 360, and then we can multiply both sides by 360.

So we get 360A over pi r squared equals theta.

"Now I can substitute the values I know," says Lucas.

So, let's give it a go.

Definitely gonna need a calculator for this one.

So, I can type that into my calculator as an entire fraction, that's absolutely fine.

And if I'm rounding it to the nearest degree, it's 155.

Ooh, Izzy says you could have substituted first without changing the subject.

Is she correct, what do you reckon? Yeah, of course you can.

You can substitute the values that you know, and then solve the equation for those values.

Let's have a look.

So, at this stage, we could have substituted 541 for A, and 20 for r.

And then we could actually do 20 squared, 'cause that's 400.

Then we can divide both sides by 400 pi, I'm gonna need my calculator for that.

And then times the answer by 360.

So, essentially we've done the same thing, we just substituted to get numbers first, and then rearranged it, rather than rearranged with the algebra first, and then substitute the numbers.

You might find that one is sometimes easier than the other.

So, sometimes it can be more efficient to change the subject first, especially if you are using the same formula multiple times.

So we've now got formula for any angle, given the radius and the area, so we could use that over and over again, without having to keep rearranging it each time.

However, sometimes it can be easier to substitute values first, particularly if the formula involves lots of different variables, and lots of different algebraic fractions, for example.

So, just be aware that you do have two different methods that you can make use of.

Let's see if we can try one ourselves then.

The formula for calculating the surface area of a cylinder is given below.

If you've got A equals two pi r squared plus two pi rh, where r is the radius of the cross section of the cylinder, and h is the height, and you can see it on the diagram there.

Why might you want to make h the subject? Can you put this into words? So, you might have said something like, "When you want to find the height, and you know the surface area and the radius." It looks like quite a complicated formula, but the only variables we've got are A, which is the surface area, r, which is the radius, and h, which is the height.

We know that pi is actually just a constant.

So, Jacob wants to work out the height of a cylinder.

It has a surface area of 408 centimetres squared, and a radius of 5.

He says, "I'm gonna substitute the values into the original equation." What might be the advantage to this method? Right, lots of different things you could have talked about.

You might have said, "Well, that means he doesn't have to make h the subject first." We said that it looks like quite a complicated formula, even though it isn't, so making h the subject might have required a bit of rearranging that he doesn't want to do.

This formula does have a lot of terms, so if he's gonna make h the subject first and rearrange, it's possible that he might make a mistake.

Once he substitutes the numbers in, this actually just becomes a linear equation, so you might find that a bit easier.

Right, he's given it a go then, I'd like you to fill in the blanks in his working.

Off you go.

Right, well done, if you spotted it, that's gonna be 50 pi plus 10 pi h, we've just substituted in the radius as 5.

The next step, you could have written as 408, subtract 50 pi, or if you evaluated that to 250.

92 approximately.

The final height then is going to be eight centimetres to the nearest whole number.

Right, time for you to have a go.

This formula is for changing degrees celsius into degrees fahrenheit.

So, C is representing degrees Celsius, and F is temperature in degrees Fahrenheit.

I'd like you to make C the subject, and then answer those three questions.

When you've got your answers to those, come back for the next bit.

Well done.

This time, we've got a formula for the volume of a cylinder.

V is the volume, r is the radius, and h is the height.

What I'd like you to do is calculate the radius of the cross section of the cylinder, and it's marked on the diagram for you, given that the volume is 503 centimetres cubed, and the height is 10 centimetres.

Now, you've got a choice here as to whether you want to rearrange your formula to change the subject, or whether you want to substitute in first and then rearrange.

It's entirely up to you.

And question three, this is a formula for motion with constant acceleration.

You may not have seen this before, that's absolutely fine, I'm gonna talk you through it now.

So, v is the final velocity, u is the starting velocity, a is acceleration, and t is the time it takes.

So what this means is, if an object is accelerating, so that could be a vehicle, or it could be a runner, it could be an animal, if you take the final velocity, and subtract the starting velocity, so that's the change in velocity, and then you divide it by the acceleration rate, that'll tell you the time that has passed between that first velocity and that final velocity.

For this question, you don't need to understand that in great detail at the moment, because all I'd like you to do to start with is to use that formula to tell me what t would be when v is 10, u is 2, and a is 4.

Then I would like you to rearrange to make a the subject, and then think about how you could tell me what a is when v is 23, u is 12 and t is 10.

As you're doing this question, think about why it was useful to rearrange to make a the subject first.

What did that allow us to do? Give that one a go.

And finally, we've got a different formula for motion this time.

This time, our formula links s, which is displacement, u, v, and t, that we saw before.

So, have a look at our formula.

Can you make v the subject? Be very careful, it's easy to make your u'S and your v's look the same, so make sure your v's and your u's look distinct.

Once you've made v the subject, try making u the subject, and then t the subject, and then think about which of those equations is gonna help you work out d.

When you're happy with your answer, we'll look at it together.

Fantastic.

So, making C the subject, I've subtracted 32 from both sides, multiplied by five and then divided by nine.

You could also do that as multiplying by five ninths, which is the reciprocal of nine over five.

Now, there's three different ways of writing that, given below, so check whether yours aligns with any of those.

Then all we need to do is substitute in the value in degrees Fahrenheit to get the value in degrees Celsius.

And because we're gonna apply this to three different questions, it's really helpful that we've rearranged our formula to make c the subject.

So I can just substitute the three different values for F to find out the three different values for C.

So, to start with, I'm going to do 5 ninths of 50 minus 32, and that gives me 10.

For c, I'm gonna do five ninths of 104 minus 32, and that gives me 40 degrees C, so 104 degrees Fahrenheit is 40 degrees Celsius.

I thought this one was a little bit interesting, negative 40 degrees Fahrenheit is the same as negative 40 degrees Celsius.

So here, you had a choice of methods, if you rearranged first, you would've got r equals a square root of v over pi h, and then you could just substitute your values of v and h.

If you substituted first, you'd get 503 equals pi r squared times 10, which gives you 50.

3 equals pi r squared.

R squared is 16.

01, and then either way, you get R as four centimetres, 4.

0, to one decimal place.

I think, for this example, either method was absolutely fine.

You're essentially doing exactly the same thing.

You are rearranging first, you're just rearranging with variables, and then substituting in.

If you're substituting first, you're doing exactly the same steps, you are just doing them with numbers so that you can evaluate at each stage and get an actual number rather than an expression.

But you do get exactly the same answer.

So, have a think about which way you preferred and why.

So, this formula for motion, then.

So all you need to do is substitute in the correct values, and you get t is two.

To make a the subject, you need to multiply both sides by a, and then divide both sides by t so you get the v minus u over t.

Just out of interest, what that is saying is, if you do the final velocity, subtract your initial velocity, and divide by the time it took for you to get from that first velocity to that final velocity, that will be the acceleration.

And that is what acceleration is, it's the change in velocity over time.

Not something you need to know at this exact moment in time, but a little bit of interest, if you're wondering where that formula came from.

Then that helps us with the next question, 'cause all we have to do is substitute in those numbers, and we get a is 11 over 10, or 1.

1.

For four, to make v the subject, you can multiply both sides by two, divide both sides by two, and subtract u.

Now, that's actually gonna help us make u the subject, 'cause we can just add y to both sides and subtract v instead.

So you get two s over t, subtract v, equals u.

Making the subject was a little bit trickier, you need to divide both sides by v plus u over two, but of course that's the same as multiplying by the reciprocals, the same as multiplying by two over v plus u.

So, s multiplied by two over v plus u, it's 2s over v plus u.

Well done, If you've got C.

Then the best equation to work out D, 'cause we want to find out what U is, is that second equation.

So, two s over t minus v equals u.

Substitute the values in, and we get U as 10.

Fantastic work today, guys.

We've drawn on our rearranging equation skills, we've looked at in the subject, especially with algebraic fractions, and we played around with some real life context at the end as well.

We looked at those formula for motion, there's quite a few different formula out there for motion, especially when you've got constant acceleration rates, which is not something you have to be confident with at the moment.

Might be something that you want to go and explore in your own time.

Thank you for all your hard work today, I look forward to seeing you implementing those skills in your mathematics in the future, and I hope you'll join us for another lesson again.