Lesson video

Hello, Mr. Robson here.

Maths.

You're in the right place.

And we're problem solving with linear equations today.

And I love problem solving.

So let's get started.

Our learning outcome is I'll be able to use our knowledge of solving linear equations to solve problems. A nice variety of problems coming your way today.

Some keywords that we're gonna need, equation and solution.

Two parts to today's lesson, and two very different styles of problems that we're going to solve.

We'll start by solving geometric problems. We'll encounter many examples where we use our knowledge of algebra to solve geometric problems. For example, x equals 3, find the perimeter of this rectangle.

I imagine you can do this, so get yourself warmed up by doing so.

x equals 3, find the perimeter.

Pause and work that out now.

Lots of ways that you might have done this.

I like to start with an expression for the perimeter, and then gathering those like terms to make that expression a lot more simple, and then substituting in my x equals 3 to find that that perimeter is 32.

But what if we had to work backwards in such a problem? For example, the perimeter of this rectangle is 100.

Find the value of x.

Can you see how this is now a very different problem? First, we need to form an equation that represents this situation.

There's my equation.

I've got 2 lots of the 4x minus 7 length, 2 lots of the 5x plus 3 length, and we're told that the perimeter is 100.

So those lengths are adding up to 100.

That's my equation.

From there, lots of routes we could have taken to solve this.

This time I've gone for expanding out those brackets, 2 lots of 4x, 2 lots of negative 7, 2 lots of 5x two, 2 lots of positive 3.

The next step I can take is to simplify that left hand side.

Now we've got a nice two-step equation to solve.

Add positive 8 to both sides, divide 3 by 18, x equals 6.

We were asked to find the value of x, and we found it.

Quick check you've got that now.

Which equation will help us to find the value of x if we know that the perimeter of this rectangle is 80? Is it option A, option B, or option C? Which of those 3 is going to help us to find x? Pause and have a good think about that now.

Welcome back.

I hope you said it's not option A.

It's not option B.

It's option C.

Whilst B is indeed an accurate expression for the perimeter, it's not an equation.

We can't solve anything from B.

We need that equals 80 to make it an equation that we can solve.

Lucas is helping Sam solve this problem.

Find the perimeter of this square.

Sam says, "I know that an expression for the perimeter would be 2 lots of bracket 4x plus 7 plus 2 lots of bracket 5x minus 3 but I don't know where to go from there.

Lucas says, "What else do you know about the properties of a square?" What do you think Lucas's hinting at? Pause this video, and have a little think, perhaps a conversation with the person next to you.

What is Lucas hinting at? Sam's got Lucas's hint.

Equal lengths! That means 4x plus 7 is equal to 5x minus 3.

Once you identified that, the square's got equal lengths we're told the lengths, or rather we're told expressions for two of the lengths.

If those expressions are equal, we like that.

That's an equation, two expressions being equal.

From there, we can rearrange the equation to find that x equals 10.

If we know that, substitute 10 into one of those lengths 4 lots of 10 plus 7 is 47.

We could have substituted it into the other expression.

5 lots of 10 minus 3 would also have been 47.

Either way, we just need to know one length, because to find the perimeter we just need 4 lots of that length, 4 lots of 47 is 188.

The perimeter equals 188.

Well done Sam, well done Lucas, you got there together.

Quick check you've got that.

To find the perimeter of this square, which is a useful starting point? Is it A, B, or C? Pause, and have a good think.

I hope you said it's not A.

It is B, and it's not C.

If you look at B, that's an equation.

We knew from the problem, from the word "square," in the problem that the lengths were equal, so 2x plus 1 is equal to 4x minus 23.

From there we've got an equation which we can solve, which will help us to find the perimeter.

Practise time now.

Question 1, the perimeter of this rectangle is 7.

Work out the value of x.

You'll need to form an equation, and solve it yourself.

For question 2, slightly different, similar looking, but there's something different about this one.

Find the perimeter of this equilateral triangle.

Pause, and have a go at those two problems now.

Question 3, two identical rectangles make this compound shape with perimeter 49.

Work out the value of x.

My key hints to you, two identical rectangles.

You need to use that knowledge.

It's the same rectangle twice.

And use a visual representation.

If I had a picture of this in front of me, I'd be scribbling ideas all over it to help me solve this problem.

Pause now, and have a good go at that.

For question 4, the rectangle, and the square have equal perimeters.

Work out this perimeter.

Really tough looking problem this one.

The key to unlocking it, both shapes have equal perimeters.

Those words might help you to form an equation to solve.

Pause, and give this one a go.

Feedback time now.

Question 1.

We were asked to work out the value of x, if we know the perimeter is 7.

That's an equation which will help us to solve this problem.

2 lots of our 6x minus 1 length, 2 lots of our 8x plus 1 length.

They make up the perimeter, and we're told that the perimeter is 7.

That's why we form that equation.

Lots of ways we can solve this.

One way, expand out those brackets, and then simplify the left hand side.

Once you know that 28x equals 7, we can divide both sides through by 28.

x equals 7 over 28, and that fraction would cancel to 1/4.

For question 2, find the perimeter of this equilateral triangle.

What do we know about equilateral triangles? An equilateral triangle has equal lengths.

Therefore we can form an equation.

5x plus 20 equals 8x minus 7 because they're equal lengths.

Lots of ways we can solve from here.

I've added negative 5x to both sides.

I'll add positive 7 to both sides, and then I'll divide 3 by 3.

x equals 9.

Finished right? Well done.

You're telling me when not finished.

We found the value of the unknown, however we were asked to find the perimeter.

If we want to find the perimeter, we can substitute x equals 9 back into either of those lengths.

5 lots of 9 plus 20 tells us that each length is 65.

3 lots of 65 tells us the perimeter of this triangle is 195.

Question 3.

I like to think of the simplest possible starting point being this little visual representation here.

If I can find that length, I can unlock this problem.

I need to take the shorter length away from the longer length to find that missing length, so 6x minus 2 take away x plus 4.

A common error here would be to not expand that second bracket correctly.

I need to take away both the x, and the 4, so I get negative x and negative 4 when I expand that second bracket.

I can simplify that expression to 5x minus 6, and then label it on my diagram.

The shorter length of the rectangle is 5x minus 6.

They're identical rectangles, so the shorter length on the other rectangle is 5x minus 6, and the same opposite.

This feels like we're getting somewhere now.

What are we missing? We want to know the perimeter, aha! We need to label the long lengths of the upper rectangle.

The long length of the rectangle we were told from the very start.

It's 6x minus 2, and we can label that both sides.

From here we can write an expression for the perimeter of the whole thing.

Around the outside of that compound shape, I see 3 lengths of 6x minus 2, 3 lengths of 5x minus 6, and one length of x plus 4.

We were told that the perimeter is 48, so I can say that expression is equal to 48.

Expand each of those brackets, simplify the left hand side, and we've got a two-step equation to solve.

Add positive 20 to both sides, divide through by 34, x equals 2.

Question 4, the rectangle, and the square have equal perimeters.

Work out this perimeter.

We need to start with that equation.

What does that equation mean? Well, the left hand side is an expression for the perimeter of that rectangle.

2 lots of the length, 8 minus x, and 2 lots of length, 5x plus 2.

The right hand side is the square.

The perimeter of the square is 4 lots of the length, 6x minus 5.

Expand those brackets.

Simplify the left hand side.

Lots of ways we can solve from here.

If I add negative 8x to both sides, and add positive 20 to both sides, I'll find 16x equals 40.

Divide through both sides by 16.

x equals 2.

5, but we're not finished yet.

We need to take that unknown, substitute it into, well, I've used the square for substitution here.

6x minus 5 if we know x to be 2.

5.

6 lots of 2.

5 minus 5 equals 10.

Each length on that square is 10, 4 lots of 10 is 40.

The perimeter of both those shapes is 40.

Onto the second part of the lesson now.

We're gonna form, and solve equations from worded problems. An example, a plumber charges £100 call out fee, and then £65 for every hour of labour.

How much would a 5 hour job cost? I expect that you'd be able to work this out, so pause, and work it out.

What you were doing was substituting in.

This is an expression for that cost.

£100 for the call out fee plus 5 lots of £65.

We can calculate that, 100 plus £225 is £425.

But what if the problem were this? Working backwards from there, if you will? Another plumber has a different call out fee, and charges £60 for every hour of labour.

A 4 hour job costs £370.

How much is the call out fee? The problem is different, because the call out fee is now the unknown.

By forming and solving an equation, we can find its value.

I'll start with wording this one out.

The call out fee plus £60 per hour equals the total cost.

We know that it was a 4 hour job, so instead of writing £60 per hour, I'll write 4 lots of £60.

We also know that the total cost of the job was £370, so I can substitute that in on the right hand side.

Finally, I'm gonna call that unknown call out fee an unknown, I'll use x.

We've got an equation to solve now.

x plus 4 lots of £60 equals £370.

We know x is £130.

Quick check you've got that.

A taxi driver has a fixed charge, and then charges £2 per mile.

A 3 mile journey is £11.

Which equations would help work out the fixed charge? A, B, or C? Which of those equations would help? Pause, and have a think about that now.

I hope you said option A, not option B, and option C.

Option A is our unknown plus 3 lots of £2 making the total cost.

If we could find x from there, we'll have the fixed charge.

For option C, you'll notice it's the exact same as A, in the exception I've called the unknown F instead of x.

The unknown doesn't have to be referred to as x.

In contextual problems you'll see lots of other more relevant letters used.

In this example F, for fixed charge.

Andeep and Jun are making up number problems. Andeep says, "I have a number whereby if you multiply it by 3 and add 13, it's the same as if you multiply it by 5 and minus 11." Jun says, "I can use trial and improvement to find your number.

My starting guess is that your number is 5." Will Jun's method work? Pause, and have a conversation with the person next to you.

It will work.

We can try it.

If Jun's right, and Andeep's number is 5, if you multiply it by 3 and add 13, you get 28.

And that would be the same as if we take that 5, multiply by 5 and minus 11.

But 5 is not Andeep's number, so we'll try something different.

Let's try 10.

10 multiplied by 3, add 13.

10 multiplied by 5 minus 11.

Ooh, have you noticed something? We're closer.

The gap between the value of the two expressions is less.

We must be getting closer.

So let's try something else.

What about 11, multiply it by 3, add 13.

Multiply it by 5 minus 11.

We're incredibly close.

Not there yet, but that gap is narrowing.

12, multiply by 3, add 13, multiply it by 5 minus 11.

Oh hurrah, we're there.

Andeep's number must be 12.

We got there.

Jun's method works but it was very inefficient.

That was a lot of work.

One of the many reasons we use algebra in problem solving is because it can be really efficient.

The same number problem, but let's think about it algebraically.

Andeep says, "I have a number." But we don't know Andeep's number.

We could say it's unknown.

Let's use x.

If you multiply by 3 and add 13, we can express that.

It's the same.

It's the same, that means equals, as if you multiply it by 5 and minus 11.

That very same number multiplied by 5 minus 11.

We formed an equation.

My starting point is to simplify x multiplied by 3.

We'd like to write that more concisely as 3x.

The same for x multiplied by 5, more concisely as 5x.

There's an equation unknown both sides.

I'm gonna add 11 to both sides, and then add negative 3x to both sides, and then divide through by 2, x equals 12.

Andeep's number is 12, and algebra enabled us to solve the problem far more efficiently.

Quick check you've got that.

There is a number whereby if you multiply it by 7, and minus 9, it's the same as if you multiply it by 5 and add 8.

Which equation represents this problem? Pause, have a good read.

Is it A, B, or C? I hope you said A, and B, but not C.

A is a correct equation.

It's just not yet simplified.

The difference between A and B is we've simplified to 7x and 5x rather than x multiplied by 7, and x multiplied by 5.

Another one to check.

There is a number whereby if you minus 9, and multiply by 7, it's the same as if you multiply it by 5 and add 8.

Which equation represents this problem? Is it A, B, or C? Pause, have a good read.

Have a good think.

I hope you said it's not A, it's not B, and it is C.

What was different this time? On the left hand side, we've now got a bracket.

If you read the order of the words in that sentence, if you minus 9 and multiply it by 7.

If we're gonna take 9 away from this number before multiplying by 7, we need to use a bracket to express that.

Accurate use of brackets is vital in modelling problems algebraically.

Laura, Sam, and Alex are buying presents for their teacher, oh.

Sam spends £15 more than Laura, and Alex spends twice as much as Laura.

In total they spend £95.

How much does Sam spend? Pardon? I can't possibly work that one out.

There's too much going on.

We can work it out, and there is a lot going on, so we use all the tools at our disposal to make it as simple as possible.

Sometimes official representation can support us in writing our equations.

So Laura, Sam and Alex.

We don't know what they spend, but we are told Sam spends £15 more than Laura.

I can represent that using those visual representations.

Laura's unknown amount, I'll represent as x, and that same unknown amount, with another 15, represents what Sam is spending.

What do you think I'm going to draw next? Alex spends twice as much as Laura.

What's Alex's going to look like? That's right, twice as much as Laura.

2 lots of that unknown.

In total they spend £95.

How much does Sam spend? The model I've drawn here is gonna enable us to write an equation which we can solve.

4 lots of this unknown amount plus 15 is equal to 95.

I can solve that.

Add negative 15 to both sides.

Divide through by 4, x equals 20.

So Laura spends 20, Alex spends 40, Sam spends 20 plus 15.

Sam spends £35.

The visual representation made it so much easier to write the equation on this occasion.

Sofia, Jacob and Izzy are buying presents for their teacher, ah.

Jacob spends £10 more than Sofia, and Izzy spends 3 times as much as Jacob.

In total they spend £130.

How much does Izzy spend? To check that you can do this, I'm gonna ask you to draw a model to represent this situation.

Write the accompanying equation and solve it.

Some of the model I'll draw for you.

Pause, and do that now.

How did we get on? Let's find out.

Sofia has an unknown spend.

But we do know that Jacob spends £10 more.

Next, Izzy spends 3 times as much as Jacob.

So Izzy spends once, twice, 3 times as much as Jacob.

In total they spend £130.

Can you see how my visual representation is a model of those sentences? From there I can turn it into an algebraic equation.

5 lots of this unknown x plus 4 lots of 10.

Oh, that's 5x plus 40, and we know that in total it's £130.

From there, add negative 40 to both sides.

Divide through by 5.

x equals 18.

So what was our original question? How much does Izzy spend? Izzy spends 18 plus 10 plus 18 plus 10 plus 18 plus 10.

Well that's three 18s plus 30, that's £84.

Again, the visual representation really supporting our use of algebra there.

Practise time now.

Question 1, part A.

An electrician has a call out fee, and then charges £45 per hour.

For an 8 hour job, they charge £385.

Form an equation and calculate their call out fee.

For part B, another electrician has a call out fee of £50, and for an 8 hour job they charge £370.

Form an equation, and calculate their hourly rate.

I'd like you to pause, and have a go at those problems now.

Question 2, Andeep has made two number problems. Write an equation and find the number in each case.

For problem A, "I have a number whereby if you multiply it by 3 and add 60, it's the same as if you multiply it by 7 and minus 8." Again, you need to form an equation, and then solve it to find the number.

No trial and error thank you.

For part B, Andeep says, "I have another number whereby if you multiply it by 5 and minus 7, it's the same as if you multiply it by 2 and minus 5." Pause and have a go at those two problems now.

For question 3, part A, Jun, Aisha, and Jacob are buying a present for a classmate, ah.

Aisha spends £2 more than Jun, and Jacob spends 3 times as much as Jun.

In total they spend £39.

50.

How much does Jacob spend? Good luck.

My advice would be you might choose to use a visual representation of this problem to help you write your equation.

Pause, and try it now.

Question 3, part B, Jun, Aisha, and Jacob are buying a present for a second classmate, ah.

Aisha spends £5 more than Jun, and Jacob spends four times as much as Aisha.

Jacob spends £33 more than Aisha.

How much does Jun spend? Good luck with this one.

You might choose to use a visual representation of this problem to help you write your equation.

Pause, and try this one now.

Feedback time.

An the electrician with a call out fee, and then a charge of £45 per hour.

An 8 hour job charging £385 forming an equation.

We have that call out fee, and 8 lots of the £45 an hour, coming to £385.

That call out fee plus £360 must be £385.

That call out fee is just £25.

For part B, another electrician has a call out fee of 50.

We know the call out fee this time, and for an 8 hour job they charge £370.

We don't know their hourly rate.

So a £50 call out fee plus 8 hours, at an unknown rate coming to £370.

Those 8 hours must cost £320.

Each hour must cost £40.

Their hourly rate is £40.

For question 2, part A, an unknown number multiplied by 3 and add 60, that's my expression on the left hand side, is the same as if you multiply by 7 and minus 8.

That's my expression on the right hand side.

Simplify those.

3x plus 60 equals 7x minus 8.

Lots of ways you can solve it from here.

I quite like adding positive 8 to both sides, because all the terms are now positive.

Add negative 3x to both sides.

Divide 3 by 4.

Andeep's number is 17.

We could always check that by substituting it back in.

17 multiplied by 3, add 60, that's 111.

17 multiplied by 7 minus 8, that's 111.

Super! For part B, a number multiplied by 5 minus 7, that's the left hand side, is the same as if you multiply it by 2 minus 5.

That's the right hand side.

Simplify there.

Lots of ways to solve from here.

I will add negative 2x to both sides.

Add positive 7 to both sides.

Divide through by 3.

x equals 2/3.

Question 3, part A.

Aisha spends £2 more than Jun.

Jacob spends 3 times as much as Jun.

Let's try and model that.

I need to visually represent this one.

There's Jun's unknown spend.

There's Aisha spending £2 more than Jun's unknown spend.

Jacob spends 3 times as much as Jun.

Okay, so that's 3 lots of the unknown in Jacob's spend.

In total it's £39.

50.

All right, I feel confident now I can write an equation from what I see.

That's 5x plus 2 being £39.

50.

5x must be £37.

50.

The unknown x is £7.

50.

How much does Jacob spend? Jacob spends 3 lots of x, so 3 lots of £7.

50.

Jacob has spent £22.

50.

If you missed anything from your model, from your equation, from your calculations, and manipulations, you might wanna pause, just copy it down now.

Question 3, part B.

This one read hideously.

It feels like a horrible problem.

The visual representation just eases that pain so much.

Aisha spends £5 more than Jun, and Jacob spends 4 times as much as Aisha.

Let's model that.

As Jun's unknown spend, Aisha's spending 5 times more than that.

Jacob's spending 4 times as much as Aisha.

Can you see how my model represents that? Next, Jacob spends £33 more than Aisha.

That's that length there representing how much more Jacob spends than Aisha.

And the sentence tells us it's got a value of 33.

So now we've got a pretty simple equation.

I can see 3 lots of x and 5 being equal to 33.

Divide both sides through by 3, add negative 5, to both sides, x equals 6.

That was a lot less painful than I thought it would be.

How much does Jun spend? Well, Jun spends one of those unknown x's.

Jun spent £6.

I quite like that problem actually.

Sadly, that's the end of our lesson.

In summary, we can apply our knowledge of solving linear equations to efficiently solve a variety of problems from geometric to number and monetary ones.

I hope you've enjoyed problem solving today as much as I have, and I'd like to see you again soon for more mathematics.

Good bye for now.