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Thank you for joining us for today's algebra lesson.

My name is Ms. Davis, and I'm really excited to get stuck into what I think is the best part of mathematics.

Simultaneous equations, once you get used to pronouncing those words, is one of the topics in mathematics that's really fun, can help us solve lots of problems and there's loads of ways to check that our answers are right so that we can be really confident in what we are doing.

So I hope you're looking forward to it as much as I am.

Let's get started.

Welcome to today's lesson on solving simultaneous equations from a context by elimination.

Because today's lesson is all about real-life contexts, you're gonna be able to use a bit of your logic skills as well as things you've come across before when solving problems, to help you with this.

By the end of the lesson, you are going to be experts at solving linear simultaneous equations in a context.

To do that, we need to be clear with what a simultaneous equation is.

So equations which represent different relationships between the same variables are called simultaneous equations.

So what you generally have is 2 equations and you're solving them at the same time to find values that satisfy both equations.

Elimination is a technique to help solve simultaneous equations.

It's where one of the variables in a problem is removed.

So we're gonna start by solving problems with 2 unknowns.

So we can solve problems with 2 unknowns as long as we have enough information.

So Aisha and Jun are playing a game where they collect two different types of counters.

The counters are worth a different number of points.

So Aisha has finished the game with 3 squares and 2 circles, and that's 21 points.

Can you work out what each counter is worth? Pause the video and play around with some numbers.

So you might have come up with some values that work.

For example, if a square was 1 point, a circle would be 9 points and that would add up to 21.

Or if a square was 2 points, a circle would need to be 7.

5 points to add up to 21.

You may have come up with even more ideas.

This problem, though, has two unknown values, the value of the square and the value of the circle.

What that means is we cannot find the solution to this problem with just one equation.

In fact, there's infinite pairs of values that work.

What we need is another piece of information.

So Jun has finished the game with 3 squares and 3 circles, and he's got 24 points.

Now we have two pieces of information about the values of the square and the circle.

Aisha says, "Squares could be 3 and circles could be 6 points." Is she correct? Pause the video and have a think.

You may have spotted that this works for her counters 'cause 3 lots of 3 plus 2 lots of 6 gets 21, but it does not work for Jun.

3 threes are 9, so his squares would be 9, but 3 sixes are 18, so in total he'd have 27 points, but he only has 24.

Because we have two equations linking the squares and circles, we need the values to work for both sets of counters.

Jun says, "We can just keep trying some values, then." Let's try a square as 1 and a circle as 9.

So Aisha would have 21 and Jun would have 3 times 1 add 3 times 9, oh, but that's 30 and he only has 24.

Okay, let's try something else.

If a square was 2, the circles would have to be 7.

5 to work for Aisha, then check it with Jun's.

Oh, but that gets him 28.

5 points and he only had 24 points.

Okay, well if a square was 3, a circle would have to be 6 for it to work for Aisha's counters.

Check it with Jun's, 3 lots of 3 plus 3 lots of 6, that's 27.

That's the one we tried before.

So at the moment we've not found a pair of values that works for Aisha and for Jun, but we are getting closer.

That method is called trial and error.

We tried something, it didn't work, we changed the numbers, we tried something else.

We realised that we were getting closer.

So by making the square worth more and the circle worth less, we might be get even closer and that can be a really useful tool for finding solutions to problems. However, it can be time consuming.

Especially if the answers are negative or non-integer, it's often very difficult to spot answers through trial and error.

Instead, what we can do is explore the structure of the problem and find a method that we can use so we don't have to keep guessing until we get it right.

Jun thinks lining up this way will help.

What he's done is he's lined up the squares and he's lined up the circles, and that allows us to compare the two scores.

You can see that they both have the same number of square counters, but Jun has an extra circular counter.

Pause the video.

What does this mean? Can you figure this out? So a little bit of logic, then.

That must mean that the extra counter that Jun has must be worth 3 points.

Because he had 3 more points, the only extra thing he has is one more circle.

Well, that means a circle must be 3.

So all the circles must be 3.

See if you can now work out the value of the square.

Okay, so a little bit of logic and a little bit of arithmetic.

If we look at Jun's scores, his 3 circles make 9, 24 subtract 9 is 15.

So that must mean his 3 squares must be worth 15.

Each square then must be worth 5.

15 divided by 3 is 5.

What we can now do, and this is why simultaneous equations are great, is we can check that our answer works using Aisha's counters.

So we've got 5, 5, 5, 3 and 3.

Does that equal 21 points? Yes it does.

So we know we must have the right values 'cause it works for Jun's and it works for Aisha's.

Okay, your turn.

Andeep and Sofia are playing a different game with counters.

Which of these values for the counters would work for both pupils? So Andeep has 3 squares and 3 triangles, and that's 30 points.

Sofia has 3 squares and 1 triangle, and that's 22 points.

So you've got four options.

Which of those work for both pupils? Off you go.

I wonder how you approached this one.

What you can do is just try it out.

So the top one, if a square is 4, then for Andeep, that would be 12 for the squares and 30 for the triangles, so that would give you 42, which would definitely not work.

For B, for Andeep, if the squares were 5, that would be 15 for the squares and 15 for the triangle.

So that would work for Andeep.

Now let's check it for Sofia.

If the squares were 5, that would be 15 plus 5 for the triangle.

Oh, that would only be 20 and she had 22.

So that second one doesn't work.

The third one, the squares were 6, that'd be 18 for Andeep and the triangles were 4, that would be 12.

That gets you 30.

Let's try it for Sofia.

3 sixes are 18 and 4 is 22.

So it's that third one.

Let's just check that D doesn't work.

It's quite possible for a triangle to be worth negative points, negative 3 points in this game.

So let's check if it works.

3 squares, so 3 sevens at 21, subtract 9.

Well that doesn't work for Andeep's and it also doesn't work for Sofia's.

However, you might have had a better approach rather than just trying them out to find that value.

Keep that in your heads.

We're gonna look at it now.

So which of these statements are true for those points? Read the four statements, select any true ones.

Okay, so the difference in their scores is 8.

That is true.

What that means is 2 triangles must be worth 8.

So those are your correct statements.

Andeep has 2 extra triangles, they must be worth 8 points.

That means one triangle is worth 4 points, and then you can work out the squares.

Okay, time for a practise, then.

I'd like you to use the clues to work out the amount of points for each type of counter.

Okay, second set, exactly the same idea.

I've maybe thrown in something with a little bit more challenge at the end.

It does require you to think a little bit more logically, see if you can work it out.

And finally, Izzy has used trial and error to try and find the values of these counters.

For each statement, is Izzy correct? And explain your answer.

So have a read of the problems and her statements and see if you can explain what she's thinking and whether she's correct.

Off you go.

So I'm expecting you to have got all of these right because you should be able to check your answers really well.

It doesn't matter if you haven't the first time, but you should be able to look at those and go, "Oh that doesn't work." And then adapt your method.

So for A, a square is 4 points and a circle is 2 points.

For B, a triangle is 7 points and a rhombus is 1 point.

I might have caught you out with C, 'cause a rectangle is 5 points and a pentagon is negative 2 points.

What you need to notice to start with is that the bottom row had one extra pentagon, but was less points.

So that extra pentagon must be negative 2.

That was a little bit trickier, so well done if you got that one first time.

For the second set, the trapezium is 8 points.

A right-angle triangle is 9 points.

For E, a parallelogram is 10 points, and then the cross shape was negative 3 points.

What you should have noticed with that one is that there's 3 extra parallelograms in the top row, which is 30 extra points.

Once you know that each parallelogram is 10 points, if you look at the bottom row, 2 parallelograms is 20, you've only got 17 so that cross must be negative 3.

Now this bottom one had a little extra element to it.

Well done if you got this.

What you might have seen is because one hexagon and one arrow is 15, that means 2 hexagons and 2 arrows must be 30.

So if you double the top one, 2 hexagons and 2 arrows are 30, which means the extra arrow in the bottom one must be the 3 points, and then you can check that works.

Okay, let's look at what Izzy is doing then.

So yes, she is correct for the first one.

You need some way of showing that she's correct.

The easiest way to do that is with substitution.

So 2 lots of 3 plus 4 lots of eight is 38.

It works for the top equation.

You've also got to check it works for the bottom one.

So 3 lots of 3 plus 4 lots of 8 is 41, so it works for the bottom set as well.

For B, she's incorrect.

2 lots of 4 plus 2 lots of 9 is 26, which does work for that top equation.

It doesn't work for the second set.

4 lots of 4 plus 2 lots of 9 is 34, not the 44 points.

You might have noticed she's got them the wrong way around.

It's the circle that's 9 and the rhombus that's 4.

This can sometimes happen when you're looking at more complicated questions, is that you can get the right answers but then forget which way around it went.

So just make sure you're laying your working out really clearly and that you're checking your answers at the end to make sure they do work.

And finally, that is incorrect.

It is not impossible.

I reckon she thinks it's impossible because the top one has more counters but less points.

But we've already seen questions like this.

That means some of the counters must be negative points.

It is possible, but using trial and error would be very difficult here.

In fact, the trapezium is worth negative points, negative 1 point, and the triangle is worth 7.

5 points.

She's probably not gonna get that by just trial and error.

Instead, she needs to use that method we talked about before, spotting the difference between the equations and using that to work out the values.

So you've got the general idea.

We're now gonna bring in a little bit of algebra and have a look at some more problems. So Laura and her friends go to a school fundraising event.

In the cafe they order 3 cups of tea and 2 glasses of juice.

Their order came to £9.

90.

Is it possible to work out the price of a cup of tea from this information? In fact, it's not.

We have two unknown values, but only one statement of a quality.

We know that 3 teas plus 2 juices cost £9.

90, but we don't know the individual prices.

It could be anything at this stage.

Lucas is gonna help us.

"We ordered 2 teas and 2 glasses of juice and it cost us £7.

60." Now we can figure this out.

What piece of information can we figure out from this? Can you spot it? Well done if you saw that we can work out the price of a cup of tea from this information.

If you can't see it yet, don't worry, we're gonna draw a diagram on the next page.

So there's Laura's 3 teas and 2 glasses of juice, and Lucas's 2 teas and 2 glasses of juice.

I assume they're sharing them with other people.

We can see now that Laura bought an extra cup of tea, but otherwise, their orders were the same.

Then we can look at the difference in prices.

It cost Laura an extra £2.

30.

That means the extra cup of tea she bought must cost £2.

30.

We can put that on our diagram.

All the cups of tea then must cost £2.

30, and we can use that to work out the juice.

So 2 cups of tea would cost £4.

60, so those juices that Lucas bought must cost £3.

00.

A little bit of our logic skills.

If 2 glasses of juice costs £3.

00, each one must be £1.

50.

So a cup of tea is £2.

30, and a glass of juice is £1.

50.

Of course what we can now do is we can check it works.

If you wanna to in parts of this lesson, you can use a calculator to check some of your answers.

I suggest that you do challenge yourself to have a go at your decimal skills without a calculator as well, and use that calculator just to back you up and check your answers.

So check it works for Laura.

3 lots of £2.

30 plus 2 lots of £1.

50 does give us £9.

90.

And then for Lucas's, 2 lots of £2.

30 plus 2 lots of £1.

50 is £7.

60.

What we're gonna do now is we're gonna look at this structure to help us solve some trickier problems. We're gonna do that by using some equations.

We're not gonna want to be drawing cups of tea and glasses of juice all the time.

Let's say t is the cost of a tea, and I've gone with s to represent the cost of a juice.

Doesn't matter what letter I use.

So that means 3 t's plus 2 s's is 9.

I'm doing everything in pounds.

For Lucas's, 2 teas plus 2 juices is £7.

60.

So 2 t plus 2 s is 7.

I'm gonna label these equations so that I can talk about them and refer to them in my solution.

We can solve these simultaneously.

What we're gonna do is we're gonna spot what's different, a bit of a game of spot the difference.

So the number of juices is the same, but the number of teas is different.

What we then did is we compared this with the difference in price.

To do this algebraically, we can subtract one equation from the other.

Difference in mathematics means subtract.

So if we do equation one, subtract equation two, we get t plus zero s, 'cause 2 s subtract 2 s is 0s, equals 2.

What we found is that the extra tea is 2.

Of course in context, that was the price of £2.

30.

Once we worked out the teas, we substituted it back in to work out the prices of a glass of juice.

You can use either equation, you can use Lucas's or you can use Laura's.

I'm gonna use Lucas's.

So 2 lots of the teas.

So 2 lots of £2.

30 plus his 2 juices is 7.

60.

So 2 lots of 2.

3 plus 2s is 7.

6 plus 2s is 7.

Subtract 4.

6 from both sides gives us 2s equals 3, which gives us s equals 1.

That might have looked like it took us ages, but that's 'cause we are really explaining every step.

It's exactly what we did with the pictures of the cups of tea and the glasses of juice, just with letters instead.

Time to have a go, then.

It's entirely up to you whether you want to draw out the pictures or use the algebra for each.

You might wanna try both.

So 3 apples and 2 oranges cost £2.

50.

3 apples and four oranges cost £3.

20.

Work out the individual cost of an apple and an orange.

I'm gonna do this one with pictures.

I'm gonna compare my equations and see that 2 extra oranges must be worth 0.

7, 'cause 3.

2 subtract 2.

5 is 0.

2 oranges cost 70p, so one costs 35p.

If you're changing between pounds and pence, just be a little bit careful that you know which one you are working with.

I know that 2 oranges cost 70p, I've got that written down.

So those 2 oranges are 0.

So if I subtract them for the cost of that top row, that gets me 1.

8 or £1.

80.

So the 3 apples must be £1.

80.

So each apple costs 60p.

What I need to do now is substitute those back in and check they work.

Again, it is your choice how much you want to practise your decimal calculation skills or whether you want to use your calculator to support you.

Right, I'd like you to try this one on the right-hand side.

Off you go.

Let's have a look then.

So we've got an extra 3 pears in the bottom row.

9 subtract 1.

4 is 1.

So those 3 extra pears cost the extra £1.

50.

So each pear costs 50p.

Then we can put that back into our top row if we like, or all back into the bottom row, it doesn't matter.

If we put it into the top row, we know that pear is 50p.

£1.

40 subtract 50 is 90p, or 0.

9 if you're working in pounds.

So the 3 lemons costs 90p.

So one lemon cost 30p.

And again, you're going to want to substitute that back in and just check it works for both rows.

Okay, same idea.

For a play, 2 adult tickets and 2 child tickets cost £25.

40.

4 adult tickets and 2 child tickets cost £39.

80.

I'm gonna use algebra to help me with this one.

So my equation's, 2a plus 2c equals 25.

4, where a is the price of an adult ticket and c is the price of a child ticket.

Also, 4a plus 2c equals 39.

Now I'm gonna subtract the equations.

It makes more sense for me to do equation two subtract equation one.

So you either rewrite them out underneath, or just remember that you're subtracting that way around.

So 4a subtract 2a is 2a, 2c subtract 2c is nothing.

39.

8 subtract 25.

4 is 14.

What I've shown then is the 2 extra adult tickets costs £14.

40.

So each one costs 7.

2, or £7.

20.

Now I know that, I can substitute that back into either equation.

I'm gonna go with the top one.

So 2 lots of £7.

20, that's the 2 adult tickets, plus 2c equals £25.

40.

So 14.

4 plus 2c is 25.

So 2c is 11.

So each child ticket is £5.

50.

And again, you are gonna want to substitute them back in to check that works for both equations.

Right, your turn.

Have a go at answering this question and then make sure you check your answer as well.

Let's have a look at our answers then.

So our equation's 1a plus 5c equals 52, and 1a plus 2c equals 28, where a is the price of an adult ticket, and c is the price for a child ticket.

That means the 3 extra child tickets must be £24.

00.

I'm doing equation one, subtract equation two.

a subtract a is zero, 5c subtract 2c is 3c, 52 subtract 28 is 24.

That means each child ticket is worth 8, 24 divided by 3, that's £8.

00 and I can substitute that back in.

I'm gonna use the second equation.

So that means a plus 2 lots of 8 equals 28, so a plus 16 is 28.

So the adult ticket must be 12.

And absolutely check that works for both equations.

You'll find that it does.

Well done.

Lots to have go at practising now.

For each question, you need to work out the cost of each item.

Even if you start by drawing out the pictures, I suggest that as you work through, you try using letters to represent the variables and forming some equations.

It's gonna save you time and help when you're looking at questions that aren't in context.

Absolutely check your answers work for both pieces of information, and that way you'll know you've definitely got the right answers.

Give those two a go and then there's more to come.

Lovely.

Two more to have a look at this time.

And then another two.

And finally, Jun says, "My family bought 2 adult tickets and 1 child ticket to my school play and it costs them £14.

50." Laura said, "I'm in the same play and 2 child tickets and 1 adult ticket cost £12.

50." I'd like you to write an equation for each statement and tell me what your variables represent.

Then have a go at answering B, C, D, and E, and then we'll go through it together.

Off you go.

Well done.

Right, I'd like you to look really carefully at my working out.

You'll see that I've labelled my equations, and then I've told you what I've done underneath.

For example, equation two subtract equation one.

You should have drinks as 50p, and crisps as 40p, juice as 80p and chocolate as 35p.

For C, you should have apples as 10p and pears as 15p.

For D, milkshakes as 80p and cheese sticks as 12p.

Marshmallows as 8p and lemonade as 45p.

And finally, you had to pay particular attention to this one 'cause I changed the order in which I wrote the items. It was also quite difficult to get your iced tea and your ice cream so that you didn't get them confused.

What I suggest you do is you write down like I've done, iced tea is a and ice cream is b, and that helps you get everything correct.

Your iced tea is £1.

35 and your ice creams are £2.

95.

So equations, we should have 2a plus c equals 14.

5, and 2c plus a equals 12.

Obviously, you could write those terms either way round.

So for the second equation you might have written a plus 2c equals 12.

And then you need to tell me that a is the price of an adult ticket in pounds, and c is the price of a child ticket in pounds.

Why will subtracting these equations not eliminate a variable? Well, that's because the coefficients of a are not the same in both equations.

The coefficients of c are also not the same in both equations.

So this one we can't solve in the same way we've been doing, just spotting the difference between the two scenarios is not gonna give us the price of a ticket.

So Izzy says, "A child ticket is £2.

00." Is she correct? So no, she's not correct.

She probably said £2.

00 because the difference between the two prices is £2.

00.

But if we try it out, if a child ticket was £2.

00, then for Jun, that means an adult ticket would have to be £6.

25.

That then doesn't work for Laura's statement because 2 lots of 2 plus 1 lot of £6.

25 doesn't add up to £12.

50.

Well done if you articulated that.

Now, this is an interesting point.

Aisha says, "4 child tickets and 2 adult tickets cost £25.

00." I wonder if you saw that that's double the amount that Laura's family bought? If she bought twice the amount of tickets, it should cost twice the price.

So if they bought the same amount again, it would cost them another £12.

50, and in total they'd have four child and 2 adult tickets and have spent £25.

00.

We can actually use that to find the cost of each type of ticket.

So Aisha's statement was 2 adults plus 4 children is £25.

00, and Jun's was 2 adults plus 1 child ticket is £14.

50.

Now we can find the difference.

It's 3 extra child tickets for Aisha's statement, which is an extra £10.

50.

So each child ticket is £3.

50.

If you substitute that back in, that means 2 adult tickets are 11 and each one is 5.

50, and you can check that works for Jun's statement, Laura's statement, and of course, Aisha's statement as well.

Amazing if you got that last one.

If you didn't, you might wanna spend some time just having a look through it.

But do remember this is something that you can practise in the future.

Well done.

As I promised you, you are now experts at solving simultaneous equations in context.

I'm hoping you found that really sat well with your logic skills and your numeracy skills, and that's something you'll be able to use going forward.

Thank you for joining us today and I look forward to seeing you again.

I've finished the video