Lesson video

Well done for taking that first step and choosing to load this video to learn today.

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

Lots of skills that we're gonna be using, so feel free to pause the video and give things a go, give yourself time to think about what you are doing, and then I'll help you out by adding any hints, any tips as we work our way through.

Let's get started then.

Welcome to this lesson on changing the subject with multiple algebraic fractions.

By the end of this lesson, you are gonna be experts at changing the subject of a formula which involves multiple algebraic fractions.

If you need a reminder of what the subject of a formula is, pause the video and read through the key words now.

So we're gonna start by looking how we can manipulate our answers into a specific form.

So when changing the subject of a formula or equation, we sometimes require our answers to be in a specific form.

For example, let's look at how we can make x the subject of this equation.

If you want to pause the video and give it a go first, feel free to do so.

So the quickest way is to multiply both sides of the equation by 2a + 1.

I've put it in brackets to make sure we multiply by the whole expression.

So we've got 8 lots of 2a + 1 = x + 3a.

Now, we want to make x the subject, so I can subtract 3a from both sides.

At this point, we have made x the subject.

However, we can simplify by combining like terms, and it'll depend why you're making something the subject as to whether you need to do this or not.

If you're gonna use it for something in the future, it's gonna be a lot more helpful if it's in its simplest form.

So to simplify, we can expand our brackets and collect like terms. So let's look at another example.

So we've been asked to show that this equation could be written in this form.

When we manipulate algebraic fractions, it's quite possible that we can all end up with very different looking equations.

That's because there's lots of different orders in which we could perform our operations, or we could choose to simplify certain bits or expand certain bits, whereas other people might not.

What that means is that questions can be asked where we have to show that this can be written in a certain format.

So what do you think we could do first? Pause the video.

What would you do? Just like before, the easiest first step is gonna be to multiply by the denominator.

Now we've gotta be careful here 'cause we need to multiply the whole expression 3 + 2b by the whole expression 3 - 2b, so you can see that I've put it in brackets.

Now, what do you notice about the left-hand side of the equation? How else could we write this? Well done if you spotted that the product of these binomials can be written as the difference of two squares.

I'm gonna remind you what that looks like, but you might be able to get the expanded form straight away using that knowledge.

So expanding these brackets, we've got 3 lots of 3 - 2b add 2b lots of 3 - 2b.

That gives us 9 - 6b + 6b - 4b squared.

Because this is the difference of two squares, those middle terms are a zero pair, so we end up with 9 - 4b squared.

Of course, once we'd identified this was the difference of two squares, we could have just squared the 3, squared the 2b and shown it was the difference between the two.

So we've got 9 - 4b squared = 2a - 5.

Now we can add 5 to both sides and divide by two.

And it's now in that format that we were asked to write our answer in.

So we can also use our algebraic fraction skills to write answers as a single fraction.

So let's look at making r the subject of this formula.

So because we want r to be the subject, we can rearrange so all the fractions not containing r are on one side and all the fractions containing r are on the other.

We can now just multiply both sides by 5, but we wanted our answer as a single fraction, so let's manipulate that right hand side, so it's a single fraction.

We're gonna need a common denominator, so 3q.

So the numerator and the denominator of the first fraction have both been multiplied by q, and the numerator and denominator of the second fraction have both been multiplied by 3.

Then we can combine the two fractions and we've got q lots of p + 4 - 6 all over 3q.

Right, we haven't made r the subject yet.

What would you do now? So you might have said multiply both sides by 5, or you might have said that you would expand the brackets first.

Either is fine.

So I've multiplied both sides by 5, and then because I want it as a single fraction, I've written that as 5q, p + 4 - 30.

Remember I have to multiply the whole of the numerator 5.

If you wanted to expand the brackets, you absolutely could do, and you get 5pq + 20q - 30 all over 3q.

It's always useful to look out for common factors to help us simplify our answers.

So let's try making x the subject of this equation.

Just like before, we can multiply by the denominator, and we get b lots of a + 1 over 5a + 5.

Now, we have made x the subject, but we can simplify our answer.

Andeep says, "Now we can expand the numerator." Sam says, "I think we should factorise the denominator." Who do you agree with? Well, if you said factorise the denominator, doing that will show us a common factor.

We can write this as b lots of a + 1 over 5 lots of a + 1.

That simplifies then to b over 5.

Sam's now having a go at this one.

Have a read through Sam's working.

What is the problem with their statement? That's right, a has not been written in terms of other variables.

The expression which a is equal to still contains a, so we have not made a the subject.

So what do you think Sam should do instead? Pause the video.

What would be your first step? So often the easiest step is to combine the fractions containing the required subject.

So let's rearrange our equations, so we've got the two fractions containing a on one side.

Now we need a common denominator, and then we can add them.

Just expanding and simplifying gives us this equation.

What do you think we could do now to make a the subject? Pause the video.

So now, we can multiply by that denominator 'cause it's a reciprocal of 1/4b, so if we multiply both sides by 4b.

The left hand side becomes 11a + 35.

The right hand side is 16b squared.

Now we can subtract 35 from both sides and divide by 11.

That's it, Sam, a is the subject now, as it does not appear anywhere else in the equation.

Time for a check then.

I'd like you to select the most efficient first step to make b the subject of that equation.

Well done if you said add the fractions.

You might have thought that factorising would be a good first step 'cause that's gonna help us find the common denominator.

But out of those choices, it's combining the fractions by adding.

Okay, so what is the lowest common multiple of 4a - 20 and 3a - 15? What do you think? Did you notice that I'd given you a bit of a clue earlier? You need to factorise first, now we can see that the lowest common multiple is 12a - 5.

So we're gonna use that then to have a go at making b the subject.

So can you fill in the missing steps in this working? Off you go.

So, we're multiplying the numerator and the denominator of the first fraction by 3 and the second fraction by 4.

Now we can combine them by adding, and then I've multiplied by the denominator, so that's going to be 12c multiplied by a - 5, and then I can divide by 10.

If you wanted to expand the brackets, you could obviously do that as well.

Well done, you have all the skills you need.

So I'd like you to have a go at making a the subject of each equation.

Try writing your answers in its simplest form where possible.

Off you go.

Well done.

This time you're making p the subject, but I want your answers as a single fraction and all expressions in expanded form.

Give those a go.

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

So Andeep and Sam are making x the subject of this equation.

Andeep's method is shown below.

He started by subtracting 2y from both sides and then multiplying by the denominator.

Sam has started by combining the fractions by making a common denominator.

What I would like you to do is show that their answers are equivalent.

Because this is a show that question, you need to make sure you are clearly showing every step of your working and how you move from one line of working to the next line of working.

Give that a go.

Come back when you're ready for the answers.

Let's have a look then.

So for a, you need a common denominator of 10, so that you can combine the fractions containing a, and then you get 30b + 10c over 9.

For b, your common denominator is going to be 3b.

Once you've got a common denominator, you can just equate the numerators, and then solve to get b squared + 10b over 9.

Pause the video if you need to look at those in more detail.

Let's have a look at 2a.

There's quite a few steps here.

So first, we factorised to help us find a common denominator, and we've gone with 10 lots of q + 1.

Make sure that you're subtracting the entire fraction, so we've got 5p subtract 2p, and we've got 20 subtract 12, so we've got 3p + 8 over 10 lots of q + 1 = q + 8.

Now we can multiply by the denominator, and we've got the product of two binomials, so we can expand those and simplify.

Your final answer should be 10q squared + 90q + 72 all over 3.

Pause the video if you need to check through any of those steps.

And for 2b, you'll see that I factorised all the expressions that it's possible to factorise, and that's helped me out a lot because I can see that the first fraction simplifies just the 2/5.

Then I've rearranged to get all the fractions not containing p on one side, and helpfully they've already got a common denominator of 5, so I can write that right hand side as 3r - 7 all over 5.

If we look at the second column of working, I've now multiplied both sides of the equation by 2q + 1, then I'm gonna need to expand the product of those two binomials, being careful that I'm getting the 4 partial products correct.

Then I can add 4 to both sides, and I've gotta be careful doing that.

Because my fraction is all over 5, if I'm adding 4, remember 4 is the same as 20 over 5, if you want a common denominator.

<v ->7 + 20 is 13,</v> and that's where that 13 comes from.

And finally I can divide by 3.

Well, if I've got an expression divided by 5 and I'm also dividing it by 3, that's the same as dividing by 15.

Again, if you need to spend some time, just pause the video and read through those steps.

Then Andeep and Sam.

So to show that Andeep and Sam's answers are the same, you need to expand the product of two binomials for Andeep's answer.

So make sure you've got the 4 correct partial products, and you'll see that I've simplified 2y over y to 2.

Sam's answer takes a little bit more work.

If you expand the two single brackets and then collect like terms. And Sam's is currently all over y, so if you divide each individual term by y, you get the same as the expression we had before for Andeep's.

Well done.

Now we're gonna have a go at rearranging when the subject is in the denominator.

So Laura's manipulating this equation.

She says, "If I divide both sides by c, I get one 1/a = b/c." Is she correct? Yes, she is.

c/a is c times 1 over a, so she's maintained equality by dividing both sides by c.

Has she made a the subject? Of course not.

1/a is a to the negative 1, and for something to be the subject, it needs to have an exponent of 1.

Aisha says, "Can we just find the reciprocals of both fractions?" Well, let's have a look.

So we've got to this stage, 1/a = b/c.

We know we can maintain the quality by multiplying both sides by the same value, so let's go with a, because it's the reciprocal of 1/a.

So that means I have 1 on the left hand side and ab/c on the right hand side.

Now, if I multiply both sides by c, I get c = ab.

We wanted a to be the subject, so we've got a = c/b.

So let's compare that with what we started with.

The reciprocal of 1/a is a, the reciprocal of b/c is c/b.

And we've just shown then that if two fractions are equal, then their reciprocals will be equal.

This is gonna be a big help to us.

So let's try this example.

Laura says, "We can start by multiplying both sides by a." Let's have a look.

"Then multiply both sides by c." What would you do next? So we're gonna divide both sides by 5m + 5.

Can you see a way that we can simplify our answer? What do you think? Yeah, of course, if we factorise, we can see a common factor.

So our final answer is a = c/5.

So what Laura did is she multiplied by one denominator then by the other denominator to remove the fractional terms. Aisha says, "I think it would've been quicker to find the reciprocals of both fractions first." Let's have a look.

So if 5m + 5 over c = m + 1 over a, their reciprocals will be equal to each other.

Now I can just multiply both sides of the equation by m + 1, and just like before, factorising helps me simplify.

So Aisha's method works well when both sides of the equations are a single fraction, and we've shown algebraically why this works.

All right, let's look at a slightly different example.

We're gonna have a go at making u the subject of this equation.

If you'd like to pause a video and have a go, feel free to do that, and then we'll talk through it together.

Right, so just like before, let's move all the fractions not containing u to one side of the equation.

Laura says, "We could now multiply by u." And then we can divide by that entire expression in a bracket.

So u is gonna be equal to 1 divided by 1/f - 1/w.

However, this is quite a complicated fraction to simplify.

It's not a form that we're gonna want to use, and it's gonna take a bit of manipulating to get it into a fraction not containing fractions in the denominator.

So what we could have done to make this easier is to actually combine the two fractions earlier on.

So let's come back to that step.

Once we got to this point, if we made a common denominator, we can actually write that left hand side as w - f over fw.

Pause the video if you just want to check through that.

Right, how does this form now help us? What can we do? Right, well, both sides of the equations are now a single fraction, so we can find the reciprocals really easily.

So if we stick with our original method of multiplying both sides by u, and dividing by that whole expression in the brackets, one divided by that fraction is the same as one times its reciprocal, and then we've got our final answer.

You may be saying, "Well, hang on a minute, we don't need to do that.

Once we get to the step where we've got what 1/u equals, we can just find the reciprocals." That's fine as well.

So time for a check.

If a/b = c/d, which of these are also true? What do you think? Yeah, their reciprocals are also going to be equal.

Okay, if 1/a = 1/b + 1/c, which of these is also true? Hopefully you didn't fall into the trap here.

They need to be combined into a single fraction before we can find the reciprocal.

So if we get a common denominator of bc, we've got c + b over bc, and then we can find the reciprocal, a is bc over c + b.

Time for a practise then.

I would like you to make a the subject of each equation.

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

Well done.

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

I want your answer as a single fraction in simplest form.

Give it a go.

And for question three, I've asked you to write that equation in a specific form.

So think about the order in which you are doing your steps again.

You're making x the subject, but I want it in that form.

Give it a go.

And your final challenge, I would like you to show that this equation when you make x the subject can be written in that form.

You might wanna think about how you're going to get a simplified answer.

You might want to factorise some things to help you.

When you've played around with that one and you're happy with your answer, we'll look at them together.

Well done.

So for the first one, you can just find the reciprocals of both expressions.

So a = 1 over b + 4.

For b, you can do exactly the same thing.

So you can find the reciprocals of both fractions.

Well done if you also spotted that the fraction on the right hand side simplifies.

If you factorise the numerator, you'll see that they have a common factor of b + 4, so that just simplifies to 2.

For c, we're gonna need to combine the right hand side before we can find the reciprocals.

So if you write 2 as 2c/c, you can add the fractions and then find the reciprocal.

For question 2a, same as before, but I'm gonna want to start by rearranging, so I've got the fractions containing x on one side and the ones not containing x on the other.

I've got a common denominator of yz, so I've got 4/3x = 2y - 2z over yz.

I could find the reciprocals of those fractions, and then multiply both sides by 4.

At that stage, I've spotted that there's a common factor of two, so I've simplified and then divided by 3.

I wanted simplest form, so I've decided to expand all my brackets.

So I get x = 2yz over 3y - 3z.

If you did write the denominator as 3 brackets y - z, that's absolutely fine.

And for 2b, there was lots to do here, but factorising is going to help us.

So I've rearranged the start, so I've got 2/x on the left and everything else on the right.

Factorising the denominator shows me that my lowest common multiple will be two lots of y + 3, y + 4.

I can then combine my fractions and find the reciprocals.

You might wanna pause and check you've got that left hand side sorted.

On the right hand side then, I've multiplied both sides of the equation by two, and then I have expanded and simplified.

You could leave your final answer in factorised form or expanded form, they're both as simple as each other.

And then we had our show that question.

Because it's a show that question, we need to make sure we're presenting our working really neatly and we're not skipping out any steps.

So I've started by rearranging, then I found a common denominator.

Then I've combined my fractions, making sure I'm subtracting all the terms. Then I found the reciprocal of my fractions.

I've multiplied both sides of my equation by two, and that gets me to that right hand column.

From that stage, I've subtracted 1 from both sides, and so that I can write this as a single fraction, I've written 1 as y squared - 3y - 6 over y squared - 3y - 6.

That allows me to combine my fractions and simplify as required.

Pause the video if you need to read through that again.

And finally, well done if you got this far.

We're gonna look through this step by step, so you might want to follow along as we go through.

If you didn't get a chance to try this one, you might want to watch your first few steps and then try it yourself.

So to start with, I factorised everything.

Then I've rearranged, so all the equations not containing x are on one side of the equation.

That's the second line.

And then I need a common denominator.

Because I factorised, it's really easy to see that the lowest common multiple of those three denominators is 4 lots of 2w - 1.

So that means I can multiply the numerator and the denominator of the first fraction by 2 lots of 2w - 1.

The second fraction, the numerator and the denominator can be multiplied by 2.

And the third fraction I don't need to change.

Now I can combine them, making sure that I'm adding and subtracting the terms as required, and that gets you down to that bottom line on the left hand side.

Then I found the reciprocals, and then I've decided to make my life a little bit easier, I'm gonna divide both sides by two, so that's how I get x/5.

Then I need to multiply both sides by 5, and then I've just expanded my numerator, so it's in the required form.

Again, feel free to pause the video and have another look through if you're not sure about all those steps.

Fantastic effort today.

There were some really lovely questions in that last part.

Couple of things that we have learned today then.

We've talked about how making the variable the subject means that there should be no other terms containing that variable in the equation.

We've talked about how combining any fractions can make it easier when rearranging the formula.

And this really important bit that if two fractions are equal, then their reciprocals are equal, and that's gonna help us solve some of the more complicated problems. Thank you for joining me today, and I really look forward to seeing you again.