Lesson video

Well done for deciding to learn using this video today.

My name is Ms. Davies, and I'm gonna be working with you as you work your way through these really exciting algebra topics.

Now that you've got the basics, we are gonna start bringing together all the things you know about algebra to tackle some really interesting problems. So I hope you're looking forward to it and I hope that you're spending the time to make sure you really understand what is going on.

The more you understand with mathematics, the easier it is to replicate things in the future, even if situations are slightly different to ones you've seen before.

With that in mind, make sure you pause the video when you need to, check your answers, and you can use calculators for bits and pieces, especially to check your answers as well.

Make sure then that you've got everything you need, and let's get started.

Welcome to our problem solving lesson with algebraic manipulation.

This is the most exciting bit about mathematics.

We are now gonna be able to use our skills to solve problems. So some key words that you may have come across before, just take a moment to read through those 'cause I'll be using them in today's lesson.

So we're gonna split this lesson into two parts.

We're gonna start by looking at using algebra to solve number puzzles, and then we're gonna bring that together with our geometry skills and use our algebra skills to solve geometric problems. So we're gonna look at some number puzzles to start.

Whenever we have an unknown value, whether it's in a problem or a scenario, we can use a letter to represent this.

So Aisha says, "I'm thinking of a number." So that's an unknown value.

We don't know what Aisha's value is.

You can use whatever letter you like to stand for the number that Aisha is thinking of.

Jacob says, "If you double your number and subtract 17, what do you get?" "I should get -32." So let's think about how we can write this using our algebraic notation.

Well, we said that we can use any letter to represent Aisha's number.

I'm going to use n.

So let n be the number that Aisha is thinking of, and I'm making it really clear what n means.

It's the number that Aisha is thinking of.

She doubles her number and then subtract 17.

I can write that as 2n.

So that's Aisha's number doubled, subtract 17.

2n - 17.

Then we found out that she gets the value of -32.

So I can now form an equation, 2n - 17 = -32.

We have formed an equation for Aisha's number.

This equation can now be solved.

It's a linear equation, so we can solve it by doing inverse operations.

And we can use that to find that value of Aisha's number.

Quick check then.

Alex's grandfather is three times as old as Alex's sister, who is two years younger than Alex.

Which of the following is a correct expression for Alex's grandfather's age? We're gonna use a to represent Alex's age.

There's lots of information there, so you might want to write some things down so that you get the correct expression.

Read through that again and then which of those is the correct expression? Okay, so I would've done this in two steps.

First, I would've thought about Alex's sister.

Now if Alex is a years old, Alex's sister would be a - 2.

Alex's grandfather is three times as old as Alex's sister.

So that's 3 lots of a - 2.

The important thing here is that the a - 2 would be in brackets, so it'd be 3 bracket a - 2.

Now that isn't one of our options.

And that's because that's the same as 3a - 6.

Okay, a little bit of reasoning now.

True or false, two consecutive even integers could be expressed algebraically as n and n + 1.

Is that true or false? And think about your justification.

Good thinking, especially if you realise that that's going to be false.

n and n + 1 are two consecutive integers.

However, we want both to be even.

If n is even, then n + 1 will not be even.

If n is even, n + 2 would be the next even number.

If you think about a number line, even numbers are two apart on a number line.

Time for you to put those skills into practise.

For each question, think about whether you can form an equation which you can then solve.

A reminder that if your equation is quadratic, if it has a exponent of two, so an x squared or an n squared, that you are going to need it equal to zero in order to solve.

We know that when equations are equal to zero, you can factorised and then solve.

So a little bit of a hint that if you have a quadratic equation, you want it to be equal to zero and use your factorising method to solve.

Give those two problems a go and then we'll have a look at the next one.

Well done.

So the same again, we're gonna need to form an equation.

So Sofia's older sister recently had a birthday.

Sofia pointed out that, in 15 years time, her sister's age will be the square of her age 15 years ago.

Think about how you can write that as an equation.

Then you're gonna need to manipulate your equation.

So it's in a format to solve.

Little bit of a hint at the bottom.

Remember that Sofia is in secondary school, so you might have to think about your answers and whether they make sense for the fact that Sofia is in secondary school and we are talking about her older sister.

Lots of clues there.

Give that one a go, then we'll look at all our answers.

Well done.

There was a range of problems in that one that required you to use a range of skills, really hoping that you enjoyed playing around with the numbers and finding values that worked.

So let's think about Aisha.

Aisha's thinking of a number.

If she subtracts 6, you can write that as n subtract 6, but then that needs to be multiplied by her original number.

Well, we called that n, didn't we? So we've got n subtract 6 multiplied by n.

And she's told us that that gives us -5.

So that's not in a format to solve.

It's a quadratic equation.

It's not linear, therefore we need it equal to zero so we can factorised.

So in order to make that equal to zero, we're gonna have to start by expanding our brackets.

So we get n squared - 6n, and then we need to rearrange equal to zero.

So n squared - 6n + 5 = 0.

Fantastic if you got that equation yourself.

If you didn't, you might wanna pause the video and then see what you do from this stage before I finish off the answers.

Okay, so now we've got an equation equal to zero and it's a quadratic equation.

So we need to factorised to work out the solutions.

Now lucky for us, this factorises really nicely.

In order to have a constant of positive 5, we know that both values must be negative.

So both constants in the binomial must be negative and they must multiply to 5.

Now there's not many options here 'cause 5 is a prime number.

So it must be n - 5 and n - 1 as our binomials.

Now that I know those two things multiply to zero, that means I can solve, 'cause that means either n - 5 = 0 or n - 1 = 0.

Well, if n - 5 = 0, n would be 5.

And if n - 1 = 0, n would be 1.

So those are our two options.

Now be a really sensible idea here to check that they work.

So let's say Aisha was thinking of 5.

She subtracts 6, that's -1 and then multiplies by her original number, which was 5.

<v ->1 x 5 is -5,</v> and that's what she said she was going to get.

Let's try the same with one.

So subtract 6.

1 subtract 6 is -5, multiplied by her original number, -5 times 1 is -5.

So both those values work.

And doesn't it feel fantastic when you find the values that work and you have a way of checking that you must be correct.

Well done.

Let's think about these two consecutive even integers.

So again, the first task is gonna be to form an equation.

Now we know that if the first even integer that we are using is n, then the next even integer must be n + 2.

We talked about that earlier.

We know the product is 624.

Product is the value of these two things multiplied together? So we've got n multiplied by n + 2 = 624.

Now just like before, if we are gonna solve this, we need it to be equal to zero, because it is not a linear equation.

So we can't use inverse operations.

We're gonna have to use our factorising to solve method.

So if we expand our bracket, we get n squared + 2n, and then we need to equal to zero.

So we need to rearrange by subtracting 624.

Again, if you haven't got that far, I suggest you write that down now and you think about what your next step would be.

Okay, so from here, we want to factorised.

Now you might be thinking, "But I don't know what multiplies to 624." That's okay.

There's some clues here that we can use.

Now firstly, we know that one value is going to be positive and one value is going to be negative in our binomials.

We also know that when we add them, we're gonna get 2n.

That means the absolute value, so if you ignore whether they're positive or negative, the absolute values of those constants are gonna have a difference of two.

That means they're not far apart from each other on our number line.

Now one way you can think about that is you can think about the square root of 624.

So if you square root 624, it'll tell you two numbers that are exactly the same that multiply to 624.

Now the square root of 624 is not an integer, but the square root of 625 is 25.

You use a calculator if you want to check.

So that means our numbers must be roughly around 25, but we know that they've got to be two apart.

So let's try 24 and 26, or 24 times 26 is 624.

Remember, we needed -624.

So it's gonna be positive 26 and negative 24.

It's a really useful skill, especially if you've got access to the calculator to find those factors.

Now it's straightforward, we can see that our solutions are gonna be -26 or 24.

Now we're gonna have to go back to the question and remind ourselves what we're looking for.

And n was the first even integer.

So if n is -26, then the next even integer would be -24.

So one option is -26 and -24.

The other option is 24 and 26.

Well done.

Let's have a look at this final one.

So Sofia's older sister recently had a birthday.

Let's represent Sofia's sister's age as x.

So that's her current age.

In 15 years time, we could write our age as x + 15.

15 years ago her age would be x - 15.

So let's check what we're saying.

In 15 years time, her sister's age will be the square of her age 15 years ago.

So if we subtract 15 and square it, that'll give us her age in 15 years time.

Well done if you formed an equation for that problem, there was quite a lot to wrap your head around.

Now we just need to manipulate this algebraically.

So x - 15 all squared, means x - 15 multiplied by x - 15.

So that's x squared - 30x + 225, and then we have to rearrange to equal zero.

So x squared - 31x + 210.

Now on the face of it, it looks like a tricky one to factorised, but actually 210 is 21 x 10.

And if we have a look at our x term, we need it to be -31.

So it must be -21 and -10.

Our solutions then are x = 21 or x = 10.

Now remember x was Sofia's older sister's age.

And if we think about it, Sofia is in secondary school and her sister is older.

So it's unlikely that her sister is going to be 10, isn't it? Therefore, her sister must be 21 years old.

If you think about it, 15 years ago, if Sofia's sister was only 10, she wouldn't have been born yet.

So this problem doesn't make sense.

So the only valid solution is that Sofia's sister is 21.

Well done.

Lots of hard work there.

Doesn't matter if you didn't get the final answer every single time.

The important thing is that you are thinking about how you can set up an equation and what skills you can use to tackle these problems. The more you practise, the better you get as with everything.

We're now gonna apply this to geometric problems. So we can use expressions to represent the length of different shapes.

So here is a triangle.

The fundamental properties of these shapes have not changed.

So anything you know about triangles applies in this case even though the side length are expressions rather than numerical values.

So let's think about how we can express the area of this triangle and the perimeter.

Pause the video if you've got an idea.

So the area of a triangle is half of the base times the height or the base times the height divided by two.

So in fact, we don't need the 5x - 6 at the moment for the area.

We can write this as a 1/2 multiplied by 2x + 3 multiplied by 7.

Now of course, we could then write that different ways 'cause we could do 2x + 3 multiplied by 7 to get 14x + 21, and then it would be a half of that.

Or we could write that as 7x + 10.

5.

There's different ways to write that expression depending on what you then want to do with it.

The perimeter actually is even easier.

Because it's an isosceles triangle, we know that we've got the base is 2x + 3 and then 2 lots of 5x - 6.

If we wanted to, we could write that as 2x + 3, add 10x - 12, or 12x - 9.

True or false then.

To calculate the volume of a prism, you can always multiply the height, the width and the length.

So little bit of geometry knowledge there.

To calculate the volume of a prism, you always multiply the height, the width, and the length.

Is that true or false? And think about your justification.

Well done if you said that is false.

If you said that was true, you are probably thinking about a cuboid, so a rectangular prism.

That's not gonna be the case with other types of prisms. Well done if you then said to find the volume of a triangular prism, you calculate the area of the cross section, so the area of the triangle, and then multiply by the length of the prism.

Read that through again, if you want to remind yourself how to find the volume of a triangular prism, 'cause you're gonna use that in your task.

Perfect, this is one of my favourite problems. We're gonna use all our algebraic manipulation skills to calculate some missing lengths in some puzzles.

So here, we have two rectangles and we know some of the side lengths and we know the areas but they've been expressed as quadratics, but that's fine.

We've got our quadratic skills that we can use.

So I want you to think, how would you calculate the value for a? You don't need a final answer, just what do you need to know? What information would help you find out what a is? Have a think, and then we're gonna work through it together.

Okay, so we're gonna start by looking at that dashed line, that green dash line, which is the bottom of that right hand rectangle.

How can we calculate an expression for the length of the dashed line? What do we need to do? Have a look at your options.

Which one's correct? Well done.

If you notice, we're gonna need to factorised.

We know that the area of that rectangle is x squared + 2x - 8.

So to find a side length, we need to find values that multiply to x squared + 2x - 8.

And that's what we're doing when we factorised.

We're looking for things that multiply to give us that product.

Let's do that together then.

Let's factorised x squared + 2x - 8, and that gives us x - 2, x + 4.

Now we've got a little bit of a a hint, haven't we? We know that one of the side length is x - 2, so that hopefully helps us find the other side length.

That means the dash line must be x + 4 'cause those two multiply to give us the area.

Okay, let's think about our next step then.

So think about how you are gonna find the length for the dashed line this time, the bottom of that left hand rectangle.

What are you gonna need to do to find the length of that dashed line? Can you write an expression? Off you go.

If you needed a hint, the top length is 2x + 2, and we know that the right hand rectangle is x + 4.

So subtracting those will give us the missing length.

Let's have a look at it then.

So 2x + 2 subtract x + 4, and we need to subtract the whole of that expression.

So put it in brackets.

That would be the same as 2x + 2 subtract x and subtract 4, which gives us x - 2.

We getting there now.

We've solved quite a lot of this problem.

We're just trying to work out how to get a.

What we can do is we can use the area of the left hand rectangle to find the height.

Now if the area is x squared - 2x, then if we factorised that, we get x lots of x - 2, 'cause x is a common factor.

We can see that the bottom length is x - 2.

So the left hand length must be x.

Right, we've got all the information we need now.

So in order to find a, what are we going to do and then which of these are valid ways to express a? Off you go.

So hopefully, you saw that if we do x and then subtract x - 2, we'll find a.

x subtract x - 2, we need to subtract the whole of expression.

So C is a correct way of writing that expression.

However, A is also correct Because if I do x subtract x, I get zero.

If I subtract negative 2, I get positive 2.

So this would be the same as saying 0 + 2 or 2.

If you use a little bit of reasoning skills, if the right hand length is x - 2 and the left hand length is x, we can see that the difference between them is going to be 2.

Well done.

There's loads of steps there, but we clearly thought about what we were gonna do, what information we needed and we showed our working as we went through.

Time for you to put all that into practise.

So for question one, the area of this rectangle is 50 centimetre squared.

I'd like you to form an equation for x.

All you need to do to start is form an equation.

Then I'd like you to solve to find the length and the width.

So you're gonna need to solve it to find x.

Remembering that if you're solving by factorising, it needs to be equal to zero.

Once you know what x is, you can find the length and the width.

There's a nice way to check that your final answer works as well.

Have fun with that one.

Come back when you're ready for the next one.

And this time, we've got the volume of this prism.

It's a triangular prism.

We talked earlier how you find the volume of a triangular prism.

I'd like you to form an equation for x.

Think about which of those values you need to find the volume and how can you then write an equation if the volume equals 88? In part B, you need to calculate the surface area.

Now you can't do that without finding out x first.

So solve for x and then use to work out the surface area.

Give that one a go.

Come back for the last one.

And finally, we've got one of these puzzles to have a go at.

I'd like you to work out the length of the side marked a.

Start by thinking out what you know, what you can work out, and then use that to get a.

You might have to do it in a few steps.

Draw over the diagram.

That will definitely help you.

Give that one a go.

Come back when you're ready for the answers.

Well done.

Let's have a look at this area one.

So an equation for x would be x + 3 multiplied by x - 2 = 50.

We've got the product of two binomials.

You don't have to do any more than that.

You have formed an equation for x.

If you chose to find the product of those binomials and rearrange, that's fine.

We are gonna do that in our next step.

So in order to find the length and the width, we need x.

Our factorised form is not equal to zero at the moment.

So we're gonna need to expand our brackets and rearrange our equation.

So we have x squared + 3x - 2x - 6 = 50, or x squared + x - 6 = 50.

We need this equal to zero.

If we think about factorising this.

We need values that multiply to 56.

The absolute values are gonna be quite close to each other because they're gonna have a difference of one to get our 1x.

So 8 and 7 sound reasonable.

In order to get 1x, that must be positive 8 and negative 7.

So our binomials are x + 8, x - 7.

If they multiply to give you zero, then our solutions were x = -8 and x = 7.

Now we're gonna need a bit of logic.

If x was -8, that means the height of our rectangle is -10 and the width is -5.

That doesn't make sense in the context of this problem, so that's not going to work.

So instead, x must be 7.

Okay, so if x is 7, 7 subtract 2 is 5, 7 + 3 is 10.

So our dimensions are 5 and 10.

And I said didn't I that we could check that works 'cause the area is supposed to be 50, is 5 x 10, 50? Yeah, of course, it is.

Okay, so we know that we must have found the correct answers.

Well done, guys.

And the volume.

So the volume is the area of the triangle at the end.

So that's x - 6 times 2 times a half because the area of a triangles a half, the base times height, and then we're gonna to multiply it by the length.

So we've got 1/2 times 2 times x - 6 times x + 12, and we know that equals 88.

There's other ways of writing that which is absolutely fine, but that's the easiest way at the moment.

Now if I've got 1/2 multiplied by 2, that's just 1.

So that means we can simplify this as 1 lot of x - 6, x + 12 = 88.

Expand our brackets like we did before, which gives us x squared + 6x - 72 = 88.

And rearranged to equals zero, x squared + 6x - 160 = 0.

Again, looks like it's gonna be quite complicated, but 160 is 16 times 10, and that's gonna work, isn't it? If we have positive 16 and negative 10, they multiply to -160 and add to 6.

We've got two solutions then.

We've got 1x is -16 or 1x is 10.

Just like our previous problem, x being -16 is not gonna work 'cause -16 subtract 6 is a negative length, therefore x must be 10.

Now I know we've worked really hard so far, but we haven't quite got to our answer.

Remember we wanted the surface area of the prism.

So since x is 10, we can substitute in to work out the base and the length.

So the base is gonna be 4 and the length is 22.

If you wanted to, you could check now that that gave you the volume that you wanted.

Is 2 times 4 times 22 divided by 2, giving you 88? Yes, it does.

But what we wanted was the surface area.

So remember, we need to find the area of all the faces and sum them together.

So the triangular faces add the two slanted rectangular faces, add the base.

So we have 2 times 1/2 times 2 times 4, those are the two triangles.

Plus 2 times 22 times 5 are the two slanted rectangles.

Plus 4 times 22 for the base.

All of those together gives us 316 centimetres squared.

There was loads of steps to that problem, but as long as you work through it step by step, showing you are working, you've got the skills to get to that final answer.

And finally, think about what it is that we know and what it is that we can work out.

So we can work out that if you factorised x squared + 3x - 18, you get x + 6, x - 3.

We know that x + 6 is the height of that rectangle.

So x - 3 is the length.

Equally, if we look at that top rectangle, if we have x + 2 is one length and the area is x squared - 4.

That's the difference of two squares.

So the height is gonna be x - 2 because x squared - 4 can be written as x + 2, x - 2.

Now we can work out the height of the other rectangle because if the height of the whole shape is x + 6, and the small rectangle is x - 2, the difference is gonna give us our height.

So x + 6 subtract x - 2, gives us 8.

Now we know that that height is 8.

We can work out the length of the rectangle.

You factorised 8x + 40, you get 8 lots of x + 5, and then we can use that final bit to work out what a is.

If the bottom of that rectangle is x + 5, and the rectangle sitting on top of it has a length of x + 2, then that value of a must be 3.

I know that there was loads of steps to that problem, so you should be feeling super proud if you managed to follow through with that answer.

We've looked at today then that using a letter to represent a variable or unknown helps us solve these problems. It is important to clearly define what the letter is representing.

If you think about that problem we had with Sofia's sister, x was Sofia's sister's current age, and then we use that to find an expression or an equation for her age in 15 years time.

We know then that when a possible solution is found, we need to check that it works and then see if it's valid, 'cause sometimes it might not be a valid solution 'cause it might be too big or too small or negative where it has to be positive.

So we can check that those values give us what we want.

Having a way to check that our answer works is really helpful, gives us that confidence that we've got to that right answer.

If you don't think your answer is right, then because you set your working out really well, you'll be able to go back through your working and see where you might have gone wrong.

Well done, guys, and I hope you find some other problems and things that you can solve using your algebra skills.