Lesson video

Hello, Mr. Robson here.

Super choice to join me for maths today! We're evaluating expressions with and without changing the subject.

That sounds interesting.

Shall we find out what it's all about? Our learning outcome is I'll be able to evaluate an expression where the subject has been changed and where it has not.

Some key words you'll hear throughout the lesson.

"Subject" is an important one.

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

It should have an exponent of 1 and a coefficient of 1." For example, A is the subject of this formula.

A equals the root of C squared minus B squared.

A is the subject because there's an exponent of 1 and a coefficient of 1.

But A squared is not the subject of this formula, A squared equals C squared minus B squared.

A squared, that's not an exponent of 1, so A squared is not what we call the "subject." Look out for that word throughout the lesson.

Two parts to our learning today, and we're going to begin by rearranging in order to solve.

I'd like to start with a problem with which you'll be very familiar.

Andeep and Sam are studying geometry.

Andeep says, "We can use this formula to work out the area." And there's the formula for area of a triangle: area equals half of base times height.

They start to write down the problem.

Area equals base times height.

Substitute in the known base and the known height.

Simplify the three lots of 6, and then take the half of 18.

The area is 9 centimetres squared.

I hope that you are very familiar with that problem.

But what if the problem was working backwards, so to speak? Working backwards is where pupils frequently make mistakes with formula.

The area is 30 centimetres squared.

The height is 6 centimetres.

Find the base.

Can you see how this problem is different? Andeep says, "I think we start with 30 divided by 6, which is 5." Then Sam says, "Yes.

Then it's something to do with halving, so I think the base is 2.

5 centimetres, half of 5." Do you agree with Andeep and Sam? Pause, have a conversation with the person next to you, or a good think to yourself.

Do you agree or not? See you in a moment.

Welcome back, I wonder what you said.

Let's see what Andeep and Sam have realised.

Andeep says, "I'll test our answer." What a useful and smart thing to do.

We can test it by substituting it back into the formula for area of a triangle.

Area equals half times 2.

5 times 6.

That gives us an area of half of 15, which is 7.

5 centimetres squared.

Do you see the problem? 2.

5 cannot be right.

It didn't give us our original area of 30 centimetres squared.

This is a frequent error that we see in mathematics.

The reason that area was so easy to find in the first problem that I showed you was because area was the subject of the formula.

When you see the area of a triangle expressed as A equals half of B times H, area's the subject.

We just substitute in B and H and we can find it.

If the unknown we're trying to work out is the subject of the formula, we can work more efficiently.

Making B the subject of the formula will make Andeep and Sam's problem more straightforward.

How do we do that? We'll start with the area of a triangle formula.

We'll multiply both sides by 2, because that's the reciprocal of a half.

We'll end up with 2A equals B multiplied by H.

B multiplied by H, what's the inverse of multiply by H? We'll divide both sides by H.

Now, 2A over H equals B.

B is the subject.

So let's look at Andeep and Sam's problem again.

It's the same problem.

The area's 30 centimetres squared, the height is 6 centimetres, find the base.

Andeep was paying attention to the rearranging I just did.

He says, "We can use this arrangement of the formula to work out the base." And there it is, my rearrangement where base is the subject.

They write that arrangement of the formula, substitute in the area of 30, the height of 6, to get 60 over 6, and Sam says, "The base is 10 centimetres.

That was much easier after rearranging the formula!" Quick check you've got that now.

The area is 10 centimetres squared, the base is 8 centimetres, find the height.

I'd like you to rearrange the formula to make H the subject and then evaluate to find the height.

I'd like you to pause, give that a go now.

See you in a moment.

Welcome back, let's see how we did.

Tricky little problem this, so if we've made mistakes that's perfectly allowed, we just pay attention and see if we can correct them as we go.

So to start by rearranging to make H the subject, I'm gonna multiply by 2, the reciprocal of a half.

I'll turn the formula into 2A equals B times H.

I want H as the subject, so I'll do the inverse of multiply by B.

That'll be to divide both sides by B.

We end up with 2A over B equals H.

There, H is the subject.

Now we can substitute in the area of 10 and the base of 8 into that rearrangement.

We'll end up with two lots of 10 over 8.

That's 20 over 8.

20 divided by 8 is 2.

5.

The height is 2.

5 centimetres.

Practise time now.

Question one.

The formula for volume of a cuboid can be written as V equals LWH.

That's length multiplied by width multiplied by height.

I'd like you to rearrange the formula to make W the subject.

Then evaluate to find the width of a cuboid with length 1.

6 centimetres, height 2.

25 centimetres, and volume 1.

44 centimetre cubed.

You will want to use a calculator to do the substitution.

It's okay to use a calculator.

In fact, it'll make this problem far more efficient.

Pause, rearrange, evaluate.

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

For question two, the formula for the area of a circle is A equals pi R squared.

I'd like to rearrange the formula to make R the subject.

Then evaluate to find the radius of a circle with area 80 centimetres squared.

Once you've done that, I'd like you to give your answer to two decimal places.

Once again, you may use a calculator to do the substitution.

In fact, if you can do this one without a calculator, I really will be impressed.

I will be using my calculator.

Anyway, pause, get rearranging and evaluating.

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

Welcome back, feedback time.

Question one, formula for volume of a cuboid.

I asked you to rearrange the formula to make W the subject and find the width of a cuboid with length 1.

6 centimetres, height 2.

25 centimetres, and volume 1.

44 centimetres cubed.

Let's start with that rearrangement.

We want W to be the subject.

To do that, we'll isolate the W.

We'll multiply through by one over LH.

That'll give us V over LH equals W.

W is the subject.

We then substitute in our known volume of 1.

44 in the V position, and the known length and height of 1.

6 and 2.

25 into that multiplication, L times H, in the denominator position.

Press equals, and my calculator tells me it's 0.

4.

Because W is the subject, that means W equals 0.

4 centimetres.

How efficient was that, after we'd rearranged? For question two, we were rearranging the formula for area of a circle to make R the subject, and evaluating to find the radius of a circle with area 80 centimetres squared.

Let's start with the rearrangement.

Slightly trickier rearrangement this, but we can do it.

I'm gonna isolate the R term by dividing through by pi.

I've got R squared on the right-hand side.

What's the inverse of squaring? Well done, the square root.

I need to take the square root of everything on both sides.

Notice on the left-hand side, it's the square root of A over pi, and you'll want that grouped when you type it into your calculator.

When I put this into my calculator, I substitute in the area, 80, in the numerator position, put the pi symbol in the denominator position, and I need the square root of that entire fraction.

When I press equals, I get 5.

04626, et cetera.

We want to round that to two decimal places, so we can say that the radius is 5.

05 centimetres to two decimal places.

On to the second half of the lesson now.

Evaluating without changing the subject.

Andeep and Sam are now studying trapeziums. The area of a trapezium is 70 centimetres squared.

The lower base length, that's the A position, is 9 centimetres, and the upper base length, that's that B position on the diagram, is 5 centimetres.

Calculate the height.

Andeep says, "We can rearrange this formula for area to make H the subject." You've seen this formula before, area of a trapezium.

The area of a trapezium is half, brackets A plus B, close brackets, multiplied by H.

Slightly trickier formula this one, but we can still manipulate it and rearrange it.

Andeep wants to rearrange it to make height the subject.

What do you think the first step's going to be? Well done, we're gonna multiply by the reciprocal of a half.

If we multiply both sides by 2, we get 2A on the left-hand side, and we're left with, brackets A plus B, close brackets, multiplied by H on the right-hand side.

When making H the subject, what do you think the next step's going to be? Well done, we want to isolate it, so we're gonna divide through by that grouped term, A plus B.

Divide both sides by A plus B, and H is now the subject.

Now we can substitute in that known area of 70, the known lower base of 9, and the known upper base of 5.

Once simplified, that'll give us 140 over 14.

We get a height of 10 centimetres.

Sam says, and this might sound a little controversial, "I'm not sure that method was any more efficient on this problem!" Well, that's interesting, let's explore that idea.

Sam says, "We could evaluate without first rearranging." Let's have a look at what that looks like.

We'll start with the same formula, and we'll substitute in the things that we know.

We know the area 70, we know the lower base length is 9, and the upper base length is 5, and we substitute them in.

We get 70 equals a half, bracket 5 plus 9, multiplied by H.

You can simplify from here.

What's 5 plus 9? 14.

What's half of 14? 7.

So the right-hand side simplifies to 7H.

Oh look, a really simple linear equation.

If 7H equals 70, what's H? Well done, H equals 10 centimetres.

Andeep says, "Well done Sam! Your method was much more efficient this time!" We can evaluate without changing the subject, and on occasions this may be more efficient.

This is one of those lovely parts of mathematics where there's multiple ways to find the same answer.

What we mathematicians like to do is find the most efficient way to solve a problem.

Let's check you've got that.

A cuboid has a width of 4.

5 centimetres, a height of 20 centimetres, and a volume of 630 centimetres cubed.

Substitute into the formula V equals L by W by H, and evaluate without rearranging, that's without rearranging, to find the length L.

Your work will look a little bit like Sam's did a moment ago.

Pause, have a go at this now.

Welcome back, let's see how we did.

We should start with writing down the formula V equals LWH and then substituting in the things that we know, in this case, a width of 4.

5, a height of 20, and a volume of 630.

We'll simplify that right-hand side to 90L.

Now we've got a linear equation, 90L equals 630.

Divide both sides by 90, and we find L equals 7 centimetres.

This was very efficient.

Practise time now.

Question one, missing dimensions in a trapezium.

Part A, a trapezium has area 95 centimetres squared, a lower base length of 23, an upper base length of 15.

I'd like you to substitute without rearranging to find the height.

For part B, a trapezium has area 88 centimetres squared, a lower base length of 14 centimetres, and a height of 8 centimetres, substitute without rearranging To find the upper base B.

For part C, I'd like you to think back on what you've done in part A and part B and then think which was the harder problem, A or B? I then want you to write a sentence explaining why it was harder and suggest an alternative method for solving.

I'd like you to pause and have a go at that now.

Question two, a right triangle has hypotenuse 13 centimetres and one shorter length of 12 centimetres.

For part A, I'd like you to substitute and evaluate without rearranging to find the remaining shorter length.

For part B, I'd like you to rearrange first to make A the subject, and then substitute to find the remaining shorter length.

Once you've done parts A and B, I'd like you to compare the approaches.

Look at your working alongside each other and see how they were similar and how they were different.

Pause and give this a go now.

Feedback time.

Question one, part A, substitute without rearranging to find the height H.

We start by writing out the formula for area of a trapezium, and then we can substitute in the known area, the known lower base, the known upper base lengths.

We can evaluate from there, simplify that right-hand side.

19H equals 95, there's a linear equation to be solved.

Divide through both sides by 19.

H equals 5 centimetres.

For part B, we were given the area of the lower base length and the height and we were asked to find the upper base without rearranging.

So start with the formula for area of a trapezium, substitute in that which we know, and then try to find B without rearranging.

So I can start by simplifying that right-hand side, half of 8 is 4, so I've got that bracket being multiplied by 4.

I can then expand that bracket.

I need four lots of 14, and four lots of B.

From there we need to do 88 minus 56.

4B has a value of 32, which means B is 8 centimetres.

For question C, I said compare the two and explain why one was more difficult than the other.

So part B was the harder problem.

It had a lot more steps and involved a lot more mathematical skills, such as expanding a bracket, therefore there were more opportunities to make an error.

And we see a lot of errors from pupils in this kind of problem.

So what was an alternative method? You always have that alternative method of rearranging the formula to make B the subject.

If you rearrange your formula for area of a trapezium, you can make B the subject and have it in the form, 2A over H minus A equals B.

When you substitute in the known values there, in one swift calculation, you found that B equals 8 centimetres.

You might have chosen to do that method as an alternative.

Another really smart thing to do would be use both methods.

If you get the same answer of 8 centimetres both times, you know you're right.

Question two, we had a right triangle with a hypotenuse of 13 centimetres and one shorter length of 12 centimetres.

For part A, I asked you to substitute and evaluate without rearranging to find the remaining shorter length.

We can substitute in the known length to A squared plus B squared equals C squared, and then subtract 12 squared from both sides.

13 squared is 169, 12 squared is 144.

We could simplify the right-hand side because they're like terms. If A squared equals 25, then A must be 5 centimetres in length.

For part B, I asked you to rearrange to make A the subject, then substitute to find the remaining shorter length.

If we wanna make A the subject, we'll subtract B squared from both sides.

And then we need to take the square root of both sides.

Notice, the square root groups the C squared minus B squared on the right-hand side.

When we substitute in, we end up with A equals the square root of 13 squared minus 12 squared.

When you type that into your calculator, make sure the 13 squared minus 12 squared is grouped underneath your square root sign.

It should look just like my calculator display.

When you press equals you once again find that A equals 5 centimetres.

And no surprise that with the different method we get the exact same result.

For part C, I asked you to compare the two methods.

If I put my working out for part A alongside my working out for part B, you can see there's a lot of similarity between the two methods, and they give us the same result with a similar level of effort in this problem.

That's the end of the lesson now.

In summary, when finding unknowns in equations and formula, we can first change the subject, or evaluate an expression where the subject has not been changed.

Both methods will give us the same result.

It's down to us to decide when each method is most appropriate.

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

Goodbye for now!.