Lesson video

Hello.

Mr. Robson here, and welcome to maths.

It's a lovely place to be, especially because we're changing the subject of complex formula today.

And you know me, I love a lesson with the word complex in the title.

It means the maths is gonna be pretty exciting.

So what are we waiting for? Let's get learning.

Our learning outcome is that we'll be able to apply an understanding of inverse operations to a complex formula in order to make a specific variable the subject.

Keywords that you'll be hearing throughout the lesson: subject, 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 root of C squared minus B squared.

But A squared is not the subject of this formula.

A squared equals C squared minus B squared.

That's not an exponent to one, so we don't call A squared the subject.

Two parts to today's lesson, and we're gonna begin by changing the subject with two steps.

Rearranging is a key skill in mathematics.

It helps us in a lot of situations.

For example, how we solve this equation.

Five X plus 12 equals 42.

We're trying to isolate the X.

If we perform this operation to both sides at negative 12, we'll maintain equality.

When we simplify there, we've rearranged five X plus 12 equals 42 to read five X equals 42 minus 12.

The right hand side simplified is 30, 42 minus 12, but the X is not yet isolated.

We need to divide by five or multiply by one fifth, however you choose to think of that, and then we'll isolate the X, X equals six.

We've isolated it.

Another example of how we might solve that equation isn't the thought that we've performed two inverse operations of minus 12 and then multiply by a fifth in order to isolate X.

If we minus 12, we get to that stage.

That's a rearrangement of five X plus 12 equals 42.

And then if we multiply both sides by a fifth, we're left with just X on the left hand side, and we've got got four two minus 12 over five on the right hand side.

This is a rearrangement of five X equals 42 minus 12, whereby X is now the subject.

Rather than thinking of it as isolating X, we could think of it as making X the subject of the equation.

When rearranging, the subject can often be thought of as the unknown you're trying to find the value of.

We're not always seeking to find a given value like a solution to an equation when rearranging.

For example, the linear function, two X plus Y equals 12.

This form enables us to efficiently graph the line.

For example, if X equals zero, Y must be 12.

And if Y equals zero, X must be six.

If we plus those two coordinates and connected a line through them, we'll have that linear function, so that's a useful form.

For any additive relationships like this, though, we can quickly see four arrangements.

If two X plus Y equals 12, then we know that Y plus two X equals 12, and we know that Y equals 12 minus two X and two X equals 12 minus Y.

You might have seen an additive relationship rearranged like that and described as a fact family or a family of rearrangements.

With Y as the subject, it's the most useful rearrangement because we can immediately see the gradient.

This is closest to the form Y equal MX plus C, so the negative two, the coefficient of X tells us that's the gradient.

That's a useful rearrangement of the equation.

If we look at this form, we can manipulate that, multiply both sides by a half, and isolate the X, or make X the subject.

If we have it in the form X equals 12 minus Y over two with X as the subject, we can efficiently find the X coordinate if we know the Y coordinate.

If I said to you the Y coordinate was four, you could substitute it in there, do 12 minus four, divide that by two, and you'd find the X coordinate.

There'll be different moments when we want different rearrangements of such equations and formula.

Quick check you've got what I've said so far.

I'd like you for the linear function three X plus Y equals 15, to make Y the subject.

Pause.

Write that down now.

See if you can make Y the subject.

See you in a moment for the the answer.

Welcome back.

Let's see how we got along.

It's a lovely bit of maths, this, because there's multiple ways to think about it.

Option one is thinking of it in terms of inverse operations.

If we start with three X plus Y equals 15, if we want to isolate that Y, we need to do the inverse of ad three X, so we'll add negative three X to both sides.

That'll leave Y alone on the left hand side and a 15 minus three X on the right hand side.

We've made Y the subject.

Another way to think of it is it's an additive relationship, so you can immediately tell me four arrangements.

If three X plus Y equals 15, Y plus three X equals 15, three X equals 15 minus Y, and Y equals 15 minus three X.

Oh, look.

Y equals 15 minus three X.

By knowing that family of rearrangements you can make Y the subject.

Another check that you've got that now.

For the linear function three X plus Y equals 15 use inverse operations to make X the subject.

I'm gonna make X the subject of this equation and then ask you to do something similar.

My equation is three X plus Y equals 15.

I'm gonna do the inverse of add Y.

I'm gonna subtract Y from both sides.

That'll lead me with the arrangement three X equals 15 minus Y.

Am I finished? Absolutely not.

That subject has to have a coefficient of one, and it doesn't.

X has got a coefficient of three.

I need to undo that.

What's the inverse of multiplying by three? We could think of it as dividing by three or multiplying by a third.

If I multiply both sides by a third, I get X equals 15 minus Y over three.

You might see that written as 15 over three minus Y over three, the 15 over three simplifying to five.

So you might see it written as five minus Y over three, or you might see it written as 15 minus Y over three.

Both are perfectly acceptable.

Your turn now.

I'd like you, for the linear function seven X minus Y equals 28, to use inverse operations to make X the subject.

Pause.

Give this a go now.

I'll see you in a moment for the solution.

Welcome back.

Let's see how you got on.

Did you start by adding positive Y to both sides? That takes you to seven X equals 28 plus Y, and you know that X is not yet the subject.

We need to multiply by the reciprocal of seven, we're gonna multiply both sides by one seventh.

That'll leave X isolated on the left hand side and have Y plus 28 over seven on the right hand side.

You may have written that as Y over seven plus four.

Alex and Aisha are writing formula.

Formula one is D plus five then divided by two makes E.

Formula two is D divided by two then plus five makes E.

Alex says, "They both say, 'plus five and divide by two,' "so I think we'll get the same formula." That's logical, isn't it? They're both plus five and divide by two.

They're the same thing, aren't they? Hmm.

Aisha says, "I think we'll get two very different formulae." Who do you agree with? Pause.

Have a conversation with a person next to you or a good think to yourself, and I'll reveal the answer in a couple of seconds.

Welcome back.

Who did you agree with? It was Aisha who was correct.

D plus five then divided by two makes E.

We'll give us that formula.

Whereas D divided by two and then plus five equals E will give us that formula, and they are two very different formula.

Alex says, "Thanks, Aisha.

"I can see that order matters now." The order mattered when we generated these formula so we can be certain that the order will matter when we rearrange them.

If we wanted to make D the subject, the inverse operations for both are to subtract five and to multiply by two, but the order in which we apply them matters.

Formula one was D plus five over two equals E.

What's gonna happen to this formula if we do minus five first? It's a little bit awkward because you can't just remove that five in the numerator.

It's not really a five, it's five over two.

So if we go to add negative five to both sides, we end up with this.

Then we have to do some fraction work on the left hand side to create a common denominator.

From there we can simplify that left hand side and then we've got D minus five over two.

I can undo the over two or divide two by doing the inverse and multiply both sides by two and we'll end up with D minus five equals two E minus 10.

Then we'll add five to both sides, D equals two E minus five.

But wow, that was a a lot of work.

This was really inefficient because that's not really a five.

Like I said, it's five over two.

So could we consider it like this? We could write that as D over two plus five over two and then add negative five over two to both sides and then multiply through by two on both sides and get to D equals two E minus five.

That was slightly more efficient, but it was still a lot of work.

So would it be better if we performed the inverse operation multiplied by two first? I hope you are shouting at the screen, "Absolutely, Mr. Robson.

"That would be far more efficient." If we multiply both sides of this equation by two, the right hand side simplifies beautifully to two E and the left hand side simplifies beautifully to simply a numerator, D plus five.

From there we have the add negative five step to isolate the D, D equals two E minus five.

That was so much more efficient.

When rearranging equations and formula involving fractions, it's sensible to consider which order to proceed with.

We mathematicians like to make our lives as easy as possible.

How does this all look for rearranging our second formula? What happens if we add negative five first? We'll end up with D over two equals E minus five.

Next we'll multiply both sides of the equation by two.

That'll leave us with D on the left hand side and two lots of bracket E minus five on the right hand side.

You might see that left in that form or you might see the brackets expanded.

Two lots of E two, lots of negative five.

You might see the right hand side written as two E minus 10.

How does that contrast with if our first step had been to multiply both sides by two? We multiply both sides by two, it's really important to remember to multiply all the terms by two.

That's why I write two outside of the bracket on the left hand side of the equation.

That will give us D plus 10 equals two E, and from there we can subtract the 10.

And look, D equals two E minus 10, the exact same result just in a different order.

There's not much difference in efficiency here between the two methods.

I think the one on the right edge is it for me, but that wasn't a great deal of difference.

Quick check you've got that.

Which is the most efficient first step to take when rearranging this formula: Y plus seven over three equals X? Which will you go for? Subtract seven from both sides? Minus seven thirds from both sides? Or multiply three on both sides? Pause.

Have a conversation with the person next to you, a think to yourself, and I'll see you in a moment for the solution.

Welcome back.

I hope we didn't go for option A.

It's really rather ugly if we add negative seven to both sides.

It's going to become incredibly inefficient like the example I showed you a few moments ago.

So let's not do that.

Option B, it's an option.

I asked you for the most efficient one.

It'll work but it's not the most efficient.

Option C was the winner.

Multiply three on both sides, and that will transform Y plus seven over three equals X to Y plus seven equals three X.

That is highly efficient as a first step.

Practise time now.

Question one, I'd like you to rearrange each formula to make C the subject.

Pause.

Give these a go, see you shortly for the answers.

For question two, Laura has been rearranging these equations to make Y the subject.

"I've checked my answers and they are all wrong.

"Can you help me spot where I went wrong?" says Laura.

In each case can you circle Laura's errors and write a sentence to explain what she should do differently? Pause and do that now.

Feedback time, now.

Question one, I asked you to rearrange each formula to make C the subject.

For part A, we'll start by adding negative seven to both sides of the formula.

From there we can divide through by six, giving a C equals 19 minus seven over six.

For part B we'll start adding negative eight to each side.

From there we can divide through by three, giving a C equals D minus eight over three.

For part C, we'll add C to both sides of this equation.

It's nice to have the C term positive.

So on the right hand side we've now got 19 plus C.

We'll add negative 19 to both sides giving us three D minus 19 equals C.

There you go, C is now the subject.

For D, we've again got a negative coefficient of C, D minus three C.

I'm gonna add positive three C to both sides.

Then we can subtract 19 from both sides, multiply by a third, D minus 19 over three equals C.

For part E, order's about to become really important in terms of efficiency.

Quick and simple start here at negative three to both sides, then multiply three by nine and we've made C the subject of the formula.

You might see those brackets expanded on the right hand side written as nine D minus 27.

Either one of those is perfectly acceptable.

For F, this is now C plus three then being divided by nine, so let's undo the divide by nine first.

Let's multiply both sides by nine, and then we can subtract three from both sides.

C equals nine D minus three, C is now the subject.

For question two.

Laura was rearranging these equations to make Y the subject and she made some errors.

This is absolutely fine, and we should reassure Laura, we won't get everything right first time in maths.

And when we don't, we should be studying our work to try and spot the errors so we can understand them further and not make them again in the future.

So how can we help Laura here? I hope for part A you said two Y is where the error is and wrote something along the lines of for Y to be the subject, the coefficient needs to be one.

You'd need to divide both sides by two from this stage.

For part B, the error was in the negative five in the second step of working.

Why is that an error? Our feedback would be something along the lines of when multiplying both sides by two, be sure to multiply all terms. She's multiplied Y over two by two to leave Y, but she hasn't multiplied the negative five by two.

That line should have read Y minus 10 equals two X.

For part C, the error was the positive three.

Really common error this one.

Laura isn't the only one to make it.

I'm certain of that.

What we should have said to Laura was you meant to add negative three to both sides.

We can't just jump the positive three across to the right hand side.

We think of it as adding negative three to both sides, so that should read seven X minus three on the right hand side.

Onto the second part of the lesson now.

We're going to be rearranging more complex formulae.

Some of the formula we'll need to rearrange will involve exponents.

For example, when we're asked to make X the subject.

Can you see the difference for this one? Y equals X squared plus five.

X no longer has an exponent of one, it's got an exponent of two, that's X squared plus five.

We'll treat this case no differently as in we'll still use inverse operations.

Inverse operations, the first one's nice and simple.

Add negative five to both sides, and then the next thing we need to do is the inverse of squaring.

Do you remember what that is? I can hear you shouting at the screen, "Mr. Robson, it's the square root." Well done.

We need to take the square root of both sides.

On the right hand side, we've now got X being squared and then having the square root taken, that just simplifies to X.

The left hand side now reads the square root of Y minus five and we group everything under the square root sign.

X is now the subject.

Laura is rearranging this equation to make X the subject, Y equals X squared minus nine.

That's the right first step Laura, well done.

And then we need to take the square root.

So Laura writes square root of Y plus a square root of nine plus a square root of X squared.

Laura says, "The inverse of squaring is to square root "so I can do this?" question mark.

She's not certain.

Aisha comes up with a brilliant idea.

"I don't know, but we could test it "with a numerical example." There is a simple equation, 16 plus nine equals 25.

Can you spot why I chose those numbers to make an equation? Well done, they're perfect squares.

This is useful because we can now test whether we can do this.

Take the square root of the terms individually.

Is that an equation? 16 plus nine equals 25 is an equation, that's a mathematical truth.

Is the square root of 16 plus a square root of nine equal to the square root of 25? Pause and just confirm whether it is a truth or not.

Welcome back.

I hope you identified that is not a truth.

The square root of 16 is four, the square root of nine is three, and the square root 25 is five, and four plus three is not five.

So that's not an equation.

So it must mean that this is not an equation.

Laura says, "So I can't take the square root of each term.

"I have to group them." Aisha says, "So Y plus nine equals X squared "becomes the square root of Y plus nine "grouped on the left hand side equals X." Laura corrects her work and there says, "Wonderful.

"Thanks, Aisha." Just wanna check you've got that now.

I'm gonna do two questions and then give you two pretty similar ones to have a go at yourself.

I'm being asked to make X the subject Y equals X squared plus seven.

Well, I'll subtract seven from both sides of the equation and then I'll take the square root, remembering to group all the terms on the left hand side.

I've got the square root of Y minus seven being equal to X, X being the subject.

The next one's a little trickier.

I'm being asked to make X the subject of Y equals X cubed minus seven.

That's slightly different, but is it any more difficult? No, I'll just add seven to both sides.

Next I'm gonna take the cube root of both sides.

The cube root is the inverse cubing.

I needed to undo X being cubed, so I took the cube root.

I'm left with the cube root Y plus seven equals X.

Your turn now.

For both of those cases, I'd like you to make X the subject your working out will look a lot like mine.

Pause, give that a go.

See you in a moment with the answers.

Welcome back.

Let's find out how you did.

Making X the subject Y equals X squared plus Z.

Well let's add negative Z to both sides and then let's undo that squaring by taking the square root of both sides.

You should have X as a subject in the equation square root of Y minus Z equals X.

In the next case, Y equals X to the power of four minus one, let's add one to both sides of the equation and then let's take the fourth root.

You should have root four of Y plus one equals X.

When the term of the exponent that is not one has a coefficient that is not one, then order really matters.

For example, we're asked to make X the subject and the equation is Y equals three X squared.

Is this going to rearrange as the square root of Y equals three X and then divide three by three giving us root Y over three equals X? Or would it be Y over three equals X squared, therefore the root of Y over three equals X? Which of those twos correct? Pause.

Have a conversation with a person next to you, a really good think to yourself and I've see in a few seconds for the answer.

Welcome back.

Which one did we go for? I'm hoping you said it's the one on the right hand side.

It's the root of Y over three equals X.

Understanding the expression three X squared is useful in helping us understand why this rearrangement is correct.

Three X squared.

Is this X squared then multiplied by three, or is it X multiplied by three and then squared? Which is it? Hmm? Well done.

It's the top one.

It's X squared then being multiplied by three.

Once you've acknowledged that, we can understand where the expression came from.

X being squared and then multiplied by three to give us three X squared.

So if we wanted to undo all that, start at three X squared, do the inverse of times three, the inverse of squaring, and we'll get back to X.

Can you see how that applies to our rearrangement now? The first thing we undid was the multiply by three.

We divided by three and then we took the square root.

That's why that one was the correct arrangement.

Just wanna check you've got that now.

Which is the correct rearrangement of this formula to make B the subject? A equals five B squared.

Three to choose from.

One of them is right.

Which one is it? Pause.

Have a good think.

See you in a moment.

Hello again.

How did we get on? I hope you said it's not A.

Divide by five is the correct first step, but the exponent of B must be one if we're going to call it the subject.

Did you identify that it was in fact B? If we start with A equals five B squared and then divide both sides by five and then take the square root, we need the square root of all of A over five.

That's why B was correct.

C was not.

Making X the subject is different for this formula.

Y equals brackets three X close brackets squared.

Let's think about the order here.

What's happening to X? Because the presence of those brackets, this is X being multiplied by three before being squared.

That's what gives us the bracket three X close bracket squared.

So if we wanted to undo this, the first thing we need to undo is the squaring.

We'll undo that with a square root and then we'll undo the multiply by three.

The inverse of multiply by three, of course being to divide by three.

That would take us back to X.

Same, the first thing to undo is the squaring.

We'll end up with root Y equals three X and then we'll divide by three.

So if we wanted to make X subject for that one, we end up with root Y over three equals X.

Let's check you've got that.

Which is the correct rearrangement of this formula to make B the subject? A equals brackets five B close brackets squared.

Three to choose from.

Pause, have a think, see you in a moment.

Hello again.

How did we do? I hope you said it's not A.

Why is it not A? Well done.

Square root is the correct first step, but the exponent of B must be one if we're gonna call B the subject.

It was not part B either.

It was option C.

Why? Because we need to take the square root of both sides first and then divide through by five, giving us root A over five equals B.

Practise time now.

Question one, I'd like you to rearrange each formula to make X the subject.

You are going to have to pay particular attention to the right order in which to perform your inverse operations.

Some of these are tricky.

You might make some mistakes.

That's perfectly fine.

I'd just like to see you having a good go at all of them.

Good luck.

See you shortly for the solutions and to correct any errors that we might make.

Feedback time.

Rearranging to make X the subject.

The first step for part A is to divide both sides by four and then take the square root, giving us the root of Y over four equals X.

For part B, the first step was to take the square root of both sides and then divide by four, giving us root Y over four equals X.

For part C, we want to add negative one to both sides, then divide three by four and then take the root of both sides.

Square root of all the terms on the left hand side Y minus one over four being equal to X.

For part D, the first thing we wanna do is divide through by four, leaving X square plus one on the right hand side.

We get rid of the plus one by adding negative one to both sides, and then we'll take the square root.

Absolutely, we will group all those terms on the left hand side under our square root side.

For part E, we need to undo the divide four on the right hand side.

We'll multiply both sides by four and then take the square root.

For part F, tricky one, this, the inverse of divide by X squared will be to multiply by X squared.

So let's do that.

And now we've got X squared being multiplied by Y.

The inverse of multiply by Y is divide by Y.

We'll do that to both sides.

From there we can take the square root, so as the subject X equals root of four Y.

Part G, three X equals the square root of Y.

We want X as the subject.

What's happening to X is being multiplied by three.

Let's undo that.

Let's divide both sides by three.

Quite simple, that one really.

I don't think H is gonna be quite as simple.

The square root of three X equals Y.

Okay, we need to isolate that X.

I'm gonna undo the square root.

I'm gonna square both sides and then divide by three, so X equals Y squared over three.

I, hmm, that's the root of X being multiplied by three.

Let's do the inverse of multiplied by three.

Let's divide both sides by three.

What's coming next? Well done, square both sides.

X equals bracket Y over three close bracket squared.

You might see that written as Y squared over three squared, which is Y squared over nine, or you could leave it in bracketed four.

Sadly, that's the end of the lesson now.

But in summary, we can apply an understanding of inverse operations to a complex formula in order to make a specific variable the subject.

For example, to make E the subject of this formula, we'll subtract eight from both sides, then divide both sides by five.

Hope you've enjoyed this lesson as much as I have, and I should look forward to seeing you again soon for more mathematics.

Goodbye for now.