Lesson video

Thank you for choosing to learn using this video.

My name is Miss Davis 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're 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 today's lesson on changing the subject where the variable appears in multiple terms. We're gonna be talking a lot about making different variables the subject today, if you need a reminder of what the subject of a formula is, pause the video and have a read through that now.

So we're gonna start by looking at factorising to change the subject.

So here is a rectangle made up of two smaller rectangles.

I've told you that the total area is C.

What equations can we write for the area of the whole rectangle? Pause the video, jot an idea down, or share it with someone near you.

So I think the most obvious one that you may have written is ab + 3a = c.

If you've gone for other forms, that's absolutely fine.

Right, if one length is A, how could we express the other length? Pause and have a go at this one.

Well, if the height is A, then the length of the first rectangle is B, and the second rectangle is three, so we can write the whole of that length as b + 3.

What this means is we can also write the equation, a(b + 3) = c.

I wonder if this is one of the ones you wrote down to start with.

Once we're in this form, how would we rearrange this to make A the subject? Perfect, we're just gonna divide by b + 3, so we get a = c over b + 3.

So when the required subject appears in an equation more than once, factorising can be a useful tool, and that's what we just did there with the area of the rectangles.

Let's look at this example.

I want X to be the subject of xy + 4 = 2x.

Alex says we can just divide both sides by two.

Do you agree? See if you can explain your answer.

If we do what Alex says, we'd have xy + 4 over 2 = x, and no X is not the subject, 'cause there's still other terms containing the variable X in the equation.

Right, he's had another go.

This time I've rearranged so all the X terms are on the same side.

Does that help? Yes it does, 'cause now we can factorise to isolate X.

So we can see that X appears in two terms. We can't additively combine them, but we can factorise.

So we've got 4 = x(2 - y).

And then just like the example we did with the rectangles, we can divide both sides by 2 - y.

So it's just the same as making any variable the subject, but we've got that added step of factorising to isolate X, and it's just a useful tool to be aware that we can use when necessary.

Okay, let's look at one where we might have to rearrange a little bit more.

So what could we do first to make A the subject of this equation? Pause the video and start this off.

Because we've got A in that bracketed expression on the right hand side, we're gonna need to expand the brackets.

We can't get around that.

So we've got ab + ac = 2b + 2a.

What would you do now? So we would rearrange, so all terms containing A are on the same side, and all the other terms are on the other side.

How can we isolate A now? Pause the video and finish off this question if you haven't already.

All right, so because A is in three terms and we can't additively combine them, we can factorise.

So I can write this as a(b + c - 2) = 2b.

Then I can divide by that bracketed expression and I get 2b over b + c - 2.

Lovely, I'm gonna run one on the left hand side and then it'll be your turn have a go.

So we're gonna make P the subject, because it appears in multiple terms, I'm gonna expand the brackets and then have a go at rearranging.

I've expanded my brackets, then I've rearranged, so all terms containing P are on one side of the equation, and the other terms are on the other side.

Now I can factorise out to P and write that as p(q + r - 3) = 8 + 3q.

So I get p = 8 + 3q divided by q + r - 3.

Fantastic.

See if you can do the same with this one on the right hand side, you're looking to make A the subject.

Give it a go.

So we're gonna need to start by expanding our bracket.

And then we need to rearrange to get all terms containing A on one side.

It's entirely up to you which side you rearrange.

I generally like to avoid as many negative terms as possible so I'm gonna subtract 2a from both sides and subtract ac from both sides.

That gives me 2a - ac, then I've added 3b to both sides so I've got 3b + 8.

If you rearrange the other way, that's absolutely fine.

You're just gonna have each of your terms will have the opposite sign.

Now I can factorise.

And make A the subject by dividing by 2 - c.

If you rearrange slightly differently you might have -3b - 8 over c - 2, that's fine as well.

Right, time to have a practise.

For each one, I'd like you to make A the subject of the formula.

When you're happy with your answers, come back for the next bit.

Well done.

This time you're looking to make X the subject of your formula.

Have a play around, remembering that skill of factorising to isolate the subject, and when you're happy with your answer, come back for the next bit.

For question three, we've got a cuboid.

It has dimensions L, W and H, length, width and height.

Can you explain why the surface area can be written as that formula? So think about what you know about surface area, can you explain where that formula has come from? When you've done that, can you rearrange the formula to make H the subject? Give it a go.

And now we've got a sector of a circle.

I wonder if you've had a look at sectors before.

So this diagram shows a sector, it has as a radius, R, and an angle of theta.

I'm telling you that the perimeter of the sector can be written as p = 2r + 2πr times Θ over 360.

If you'd like to pause the video and have a look at your notes on sectors, see if you can remind yourself of where that formula has come from, feel free to do that now.

You don't need to though 'cause what I want you to do is rearrange it to make R the subject, and I want you to show that it can be written in that form.

When you're happy with that, come back and we'll look at the answers.

Let's have a look then.

So the first one, you just need to factorise, then you can divide by 2 + b.

So you've got a = c over 2 + b.

For B, there's an extra step we need to rearrange first.

So I've subtracted A from both sides.

And then I can factorise, and I get a = b over b - 1.

If you rearrange the other way, you might have got -b over 1 - b.

For C again we need to rearrange to get all terms containing A on the same side.

The easiest way is to subtract 3a.

Then I can factorise.

And I get 2c - 4 over bc - 3.

Now for D and E you didn't actually need to factorise, so you just have to have a quick look before you start and see what skills you're gonna need.

A is almost the subject already.

All we need to do is divide both sides by b + c.

So you get 5(b + 2) over b + c.

Obviously if you choose to expand the numerator that's absolutely fine, you'd get 5b + 10 over b + c.

Now E was possibly a little bit trickier to spot, but again you don't actually need to factorise.

You can start by dividing both sides by B, and then you get a + c = 5(b + 2) over b, and then you can subtract C from both sides.

There are other forms that you may have written that in, so just play around and see if it's the same as what I have written.

F however, because the A is in two terms, it's in two separate brackets, I'm going to need to expand.

So I've got ab + bc = 5a + 10, I can rearrange, you can rearrange either way, I've decided to subtract 5a from both sides.

So you get ab - 5a = 10 - bc.

And then I can factorise that left hand side, so I get a = 10 - bc over b - 5.

If you made a different choice on how to rearrange, you may have got bc - 10 over 5 - b.

It's exactly the same thing, we've just divided the numerator and the denominator by -1.

So making X the subject of these.

So we're gonna need to expand and rearrange, and then factorise.

You get x = 4y over 2y + 3z.

For B, there might have been a few more steps.

If you've got x²y and you're square rooting that product, you can square root x², and you can square root Y separately.

So I've written that as X square root of Y.

Once I've expanded everything and rearranged, you get the third line down, and then you can factorise X out, 5w - 6y = x(8 - √y).

Then you can divide.

At this stage I have no problem with there being a third in the denominator.

If you wanted to play around with practising rationalising the denominator, you could absolutely do that, but there's no need in this case.

If I was gonna substitute numbers into that formula, that form is absolutely fine.

So let's look at our cuboid.

So you might have said something along these lines.

Surface area is the total area of all six faces.

There are three different sized faces.

For hw, lw and hl, you multiply the two lengths.

Because there's two of each size face, you can write it as the sum of the area of the three different faces, all multiplied by two.

And that's how I've written it.

To make H the subject, I've decided to expand the brackets again to help me.

You could start by doing a divided by 2, but I think that's a little bit trickier.

So I've expanded the bracket, I've subtracted 2lw from both sides.

Now I'm gonna factorise again, and then divide by 2w + 2l.

There are other forms that that could take, I think that was the easiest way.

And finally perimeter of a sector.

So I've multiplied every term by 360, and then divided every term by two.

Now I've factored out R, and divided by that bracket.

And that's in the format that we want it.

Of course if you wanted to now, you apply that to any question where you know the perimeter and you know the angle and you want to work out the radius.

Perfect, we're gonna extend that now to have a look at questions with algebraic fractions.

So let's try an example like this.

None of the expressions factorise and there's no common factors at the moment.

So the best first step is to multiply by the denominator.

We've got r(m + 1) = m + 2.

We've got M in multiple terms so we're gonna need to expand and rearrange.

And then this looks just like the questions we looked at before.

We can factorise.

And then divide to make M the subject.

We're just making sure that we're not missing out any steps, we're not making any assumptions, we're thinking carefully about our algebraic manipulation.

Jun and Sophia are having a look at this question.

This is what Sophia does.

Ooh, Jun got something different.

He wants to know if he went wrong somewhere.

Have a look at his working.

Can you spot what has happened? Right, their answers are actually equivalent.

We touched on this in the previous task.

If you multiply the numerator and the denominator of Sophia's answer by -1, you will get Jun's answer.

At this point they rearranged in a different order.

So Sophia rearranged to get all the terms containing X on the left hand side, but Jun rearranged to get all the terms containing X on the right hand side, and that meant that all of his terms had the opposite sign to Sophia's.

It's just something to be aware of.

If you've got a different answer to somebody else or you know what the answer is supposed to be but yours looks slightly different, try multiplying the numerator and the denominator by -1 and see if you then get the answer you are looking for.

Let's have a look at this question.

We're gonna make A the subject.

What do you think we could do first? What skills can you bring to this question? Right, we should be getting the hang of this now, factorising is gonna help us see the common factors.

So I factorised everything that will factorise.

And what this shows me is there's no common factors which allow us to simplify a single fraction, but the denominators share a common factor and that's gonna help us add these.

So the lowest common multiple of the denominator is gonna be 12(a + 2).

That means I can multiply the numerator and the denominator at the first fraction by four, and the second fraction by three.

Now I can additively combine them.

What would you do next? You've got two choices, you can either multiply both sides by 12(a + 2) or you can expand the brackets first.

I'm gonna multiply both sides by 12(a + 2).

And I'm gonna need to calculate the product of two binomials on the right hand side.

So I've got A multiplied by B, A multiplied by one, two multiplied by B, and two multiplied by one.

So they're my four partial products.

None of them simplify.

So all I can do now is expand my remaining brackets.

And then do a bit of rearranging.

If you want to pause the video and rearrange that yourself, do that now.

Hopefully you ended up with this equation.

All we need to do now is factorise out an A because it's A that we want to make the subject, and then we can divide both sides by the expression 12b + 4.

Just to be aware, I chose not to fully factorise the right hand side, because it's A that I want to make the subject, it's only a that I need to factorise out.

So I've factorised out an A, and I've got 12b + 4 in the bracket.

I'm well aware that they still have a common factor, but for my purposes that's fine.

And there's my final answer.

Sophia says there was lots of steps to that one.

This is why we have been practising writing clear, working out, it really helps.

So I'm gonna show you one on the left, and then you're going to have a go.

So I want to make X the subject.

I'm gonna start by factorising, see if I can find any common factors.

And I can.

So this is gonna simplify to x over x - 2.

I'd like you to see if you can get to the same stage with your example.

Off you go.

So the numerator is gonna factorise to (x - 3)(x - 5), and the denominator b(x- 3).

That's helpful 'cause we have a common factor.

So you've got x + 1 = x - 5 over b.

Now I want to make X the subject.

I'm gonna need to multiply by that denominator by x - 2, I'm gonna expand and rearrange so all my X terms are on the same side.

Then I can factorise, we're getting really good at that now, and then divide through.

All right, see if you can have a go at doing the same thing with yours.

Think carefully about that first step.

Make sure you're using brackets where necessary.

Give it a go.

So if you multiply both sides by B, then expand our brackets, rearrange, you could have rearranged in a slightly different way, but that I think was probably the simplest, factorise, and divide by 1 - b.

Well done.

Right, your chance to bring all those skills together then.

So for each one I'd like you to make A the subject.

Think about all those things we've practised today.

When you're happy with your answers, come back for the next bit.

Well done.

This time you're looking to make X the subject.

You might want to think about when factorising would be helpful, to see if you can spot any of those common factors.

Try and make your life as easy as possible.

And the final challenge I'd like you to make R the subject of this equation.

Again, you're mathematicians now, you want to be making sure you're doing things as efficiently as possible.

Give that a go.

Let's have a look then.

So I'm gonna multiply both sides by A, rearrange and factorise.

I get a = 4 over b - 1.

The second one I've gotta make sure I'm multiplying both sides by the whole expression a + b.

I've then expanded, rearranged, and factorised.

My final answer is 4 - b² over b - 1.

Just like we've already talked about, if you've got b² - 4 over 1 - b, it's exactly the same thing.

And then for C, just like before, I'm multiplying by A, expanding and rearranging and I get 4 over b + 1.

For B, I'm doing exactly the same thing, I'm multiplying by that denominator.

You've gotta be really careful that you're multiplying the whole expression b + 2, by whole expression a + b.

So that means I've got a(b + 2) + b(b + 2).

Gives me ab + 2a + b² + 2b.

Make sure you've got all four partial products.

Then rearrange to get all the terms containing A on one side.

If we subtract A from both sides, I've got ab + a now 'cause 2a - a is a, then 4 - 2b - b².

Factorise to isolate A, and then divide by b + 1.

Then for 2A, we've got two fractions that we're adding together, so we need a common denominator.

Once I've got my common denominator, I can multiply by that denominator, which is axe, rearrange, and then factorise again.

So I've got X on the outside of my bracket, ay - 3 = 2a - c.

Dividing them by the bracketed expression gives me x = 2a - c over ay - 3.

Check whether yours is equivalent to that.

Right, this one looks fun.

Let's have a look.

Now I'm looking at that first bit and going, hang on a minute, I want some common factors here.

Let's factorise and see if I can find any.

So I've factorised everything that will factorise, and what that has showed me is there's a common factor in the denominators of x - 2.

So my lowest common multiple is gonna be 4(x-2).

So I can leave the first fraction as it is, and multiply the numerator and the denominator of the second fraction, by four.

So that's 12(y + 5) over 4(x - 2).

Adding them together I get x - 1 + 12(y + 5) all over 4(x - 2) = 2 - y.

I've kind of done two steps for the next bit.

I've expanded the left hand side and I've multiplied both sides by 4(x - 2).

And then I'm gonna need to do a little bit of rearranging.

So if I simplify the left hand side, I've got x + 12y + 59.

The right hand side, I've got the product of two binomials, so I've got 2x - 4 - xy + 2y.

Then a little bit of rearranging.

And then I just need to factorise out X like we've been doing for all the questions.

So you get 4y + 75 over 7 - 4y.

Pause the video and have a look over those steps if you need to.

And then the final challenge, again, we're gonna do this step by step.

You might wanna watch a few steps and then give this a go if you didn't get to a final answer yourself.

So I'm gonna factorise everything and see what will help me.

This was a good decision, 'cause I can see that the first fraction simplifies to five over four.

The second fraction simplifies to r + 5 over 3p.

And the right hand side simplifies to 5r over 6.

Now I'm rearranging, so all fractions containing R are on the same side.

Then I need a common denominator, 6p is gonna be the lowest common multiple.

And then I can combine my fractions.

Okay, if we look at the right hand column, all I've done here is I've multiplied both sides of the equation by 6p.

I'm just gonna quickly look at the question again and remind myself what I'm trying to make the subject, it's R that I'm trying to make the subject, that's fine.

So I've then done two steps together.

I've seen that 30p over 4 simplifies to 15p over 2.

And then I've added 10 to both sides.

But 10 is 20 over 2, so I've got 15p over 2, add 20 over 2, which is 15p + 20 over 2.

Now I need to factorise R, so I've got R on the outside, 5p - 2.

And then I can divide by 5p - 2.

I've got 15p + 20 over 2(5p - 2).

Now you've got a choice now, you can either factorise both expressions and that will help us see that there's no common factors left, or you can leave them in expanded form, so you've got either answer at the bottom.

Right, I hoped you enjoyed playing around with that last one.

There's lots of skills that we were bringing together and you've got that real sense of achievement when you manage to get to that final answer.

Even if you don't get it the first time, being able to go back through and make sure you know how to get it also helps with your algebraic manipulation skills.

Right, thank you for joining us today.

We've had a look now at how to change the subject when the variable appears in multiple terms, and that is a skill that can be quite useful.

So look for that option of factorising to isolate a variable, where it might be of use to you.

Thank you for joining us today.

I hope you enjoyed some of the challenge in that lesson, and I look forward to seeing you again.