Lesson video

Thank you for joining us in this lesson.

My name is Miss.

Davis and I'm gonna help you as you work your way through this lesson.

There's some really exciting algebra coming up so make sure that you've got everything you need and that you're really looking forward to getting stuck in.

Let's get started then.

Welcome to today's lesson on solving more complex, simultaneous equations by elimination.

This is where things start to get really fun/ Even though the simultaneous equations we're gonna be solving are gonna be a little bit trickier than ones that you might have seen before.

The great thing is we're still gonna start from our understanding of general logic and how things work.

There's also gonna be loads of opportunities to check our answers.

By the end of the lesson, you'll be able to solve two complex linear simultaneous equations using elimination.

If you're not sure about what simultaneous equations are, or this idea of elimination, pause the video and read through the keywords now.

we're gonna start by looking at how we can multiply one equation before eliminating.

So we're gonna start with a real world example.

The cost of three hot chocolates and two juices is 16 pounds.

The cost of three hot chocolates and one juice is 14 pounds.

We can use elimination to find the cost of each item.

You can see here that the first set has one extra glass of juice.

So if we subtract those equations, we can see that one glass of juice must be worth two pounds.

By substituting, we can then work out the cost of a hot chocolate.

So three hot chocolates must be 12 pounds, so each one is four pound.

But what would happen if the number of hot chocolates were not the same in both questions? So in a different shop, two hot chocolates and three juices cost 13 pound.

One hot chocolate and two juices cost seven pounds.

Let's see what this looks like in a diagram.

So this time if we were to subtract, we would find that one hot chocolate and one glass of juice is equal to six pounds.

There's two things that's different in our two scenarios.

There's the number of hot chocolates is different and the number of juices are different.

So all we can tell by subtracting those is that one hot chocolate and one juice equals six pounds.

This new piece of information is true, however, it doesn't get us closer to finding the cost of each item.

There's still lots of choices for two numbers that add up to six.

So what can we do? So we can use that second piece of information to create another piece of information that's gonna be of more use to us.

So if one hot chocolate and two juices cost seven pounds, what happens if we double that order? Well, that would double the price.

So that means two hot chocolates and four juices must be 14 pound.

Now when we subtract, there's only one thing that's different.

So one juice must be worth one pound.

Then we can substitute as normal to find the hot chocolate.

So let's see what this looks like with algebra.

So let's go with one adult and two child tickets to a museum cost 21 pounds, three adults and five child tickets cost 57 pounds.

Work out the ticket prices.

So we're gonna use some letters to represent our variables and form some equations.

We've got a + 2c = 21 and 3a + 5c = 57.

Neither of the like terms have the same coefficient or coefficients which are a zero pair.

So at the moment we can't eliminate by adding or subtracting, but we could multiply one of the equations so that there is a common coefficient.

Just like in the previous question, we doubled the order so that we had two hot chocolates in both orders.

We can do the same here.

We could double or triple or multiply by any value so that there is a common coefficient.

So I'm gonna multiply that first equation by three.

Pay attention to how I'm using my notation to show people that are following my work exactly what I'm doing.

So equation one multiplied by three gives me 3a + 6c = 63.

Making sure I'm multiplying every term to maintain equality.

Now I can write the second equation.

It doesn't need to change because I've now got three a's in both equations.

So I can write the second equation underneath and then we can subtract.

3a subtract 3a is zero.

6c subtract 5c is c, 63 subtract 57 is 6.

So we've got c = 6.

Of course, just like normal we can substitute that back in to find our value for a.

So a must be 9.

As always, we can substitute both answers into the original equations and check they work.

So let's look at these ones.

What do we need to do first before we can solve these equations simultaneously? Pause the video, have a read.

What would you do? By far, the quickest way will be to multiply the second equation so that the coefficients of b are a zero pair.

If you multiply that second equation by four, you'd have 8b's and then -8 and 8 are the additive inverses of each other.

So let's try that.

Multiplying equation two by four gives me 8a + 8b = 96, making sure I multiply every term.

And then I can write the other equation underneath because the coefficients of b are a zero pair, I can eliminate by adding.

So 8a + 3a = 11a, and 96 plus 14 is 110.

A therefore must be 10 and we can substitute that back into get b, which we're already really confident with.

Remembering that we can check our answers by substituting.

So Jacob's having a go at these two.

Jacob says this doesn't seem to work.

Have a look at the equations, see if you can work out why he says this.

He's absolutely correct.

We cannot solve these equations simultaneously.

Let's look at what happens.

So let's say we were gonna multiply equation one by three so that the coefficients of a were the same.

We'd get 3a + 6b = 30 and the other one will be 3a + 6b = 33.

Well, that doesn't make sense.

How can the same two things added together be two different answers? If you subtract, you end up with a statement zero equals negative three, which of course does not work.

The reason this doesn't work is that both equations have the same variables but are equal to different values.

The original equations, if you scaled one of them up, is the same as the other equation just with slightly different constants.

This means that there are no solutions which satisfy both equations simultaneously.

Most pairs of equations you can solve simultaneously.

Any equations that you can write, you can find values for a and b that satisfy them.

However, there are some exceptions.

Some pairs of equations would not have any solutions and that'll be in situations like this where if you multiply both of the variables by the same thing, it gets the variables in the other equation.

Sam's had a go at solving these simultaneous equations.

I'd like you to use substitution to check Sam's answers.

Off you go.

Let's have a look.

So three lots of negative 10, subtract negative 44 is 14, let's check that.

So -30 + 44 = 14 and that is true.

So they work for the first equation.

Let's also check the other equation.

Five lots of negative 10, subtract two lots of negative 44 is 24.

So -50 + 88 = 24.

Ah, that does not work.

We get 38 = 24 and that's not correct.

So these solutions could not be correct.

They don't work for both equations.

So here's Sam's working.

I'd like you to read through what mistake have they made.

Okay, this is a really common mistake.

They have not remembered to multiply the right hand side of the equation as well.

They've multiplied both variables.

So 3a x 2 and -b x 2, but they haven't remembered to multiply the constant as well.

Just something to double check that you are doing each time.

And if you're using that substitution method to check if your answers are correct and you find that they're not correct, it's a good place to look for a mistake.

Have you made sure that you've multiplied all your terms by the same value? Time to have a go then.

So I'd like you to find the cost of each item.

It'll definitely help you to write these as equations and solve them simultaneously.

And then don't forget to substitute your answers back in to check their work.

Give it a go.

And a couple of more for you to have a go at.

Give it your best shot, come back when you're ready for the next bit.

Fantastic.

So we've removed the context now, but we're gonna do exactly the same thing.

So I'd like you to solve these simultaneous equations and I'd like you to think really carefully about how you are showing you are working out and making sure you're checking your answers.

Because the previous questions were all in context, it didn't make sense for there to be negative values.

This time you want to look out for where there might be zero pairs and also expect that some of your answers may well be negative.

Three more to have a look at.

You'll need to rearrange most of these first in order to then use the elimination method.

Once you're happy with your answers, come back for the last bit.

And finally, Andeep asked a computer programme to generate some equations to solve simultaneously.

I don't think all of these work.

So I'd like you to have a look, which pairs of equations can Andeep solve and which ones can he not solve simultaneously? Even better if you can explain why they will or will not work.

Off you go.

Let's have a look then.

So we should have soda as 60p and popcorn as 50p.

If you left your answer as 0.

5 pounds, that's fine, just make sure you've put your units in.

For b, limes are 20p and kiwis are 15p.

For c, cherryade is one pound and five pence.

Chocolate is 55 pence.

And for d, flapjacks a one pound 10 and cheese is 60p.

Right, I'd like you to pause the video and have a look at the methods for each one.

Gonna draw your attention to question c.

You're gonna have choices of methods for c, you could have doubled the first equation so that you've got 4x in both and then subtract.

Or you could have multiplied the second equation by three so that the Y terms are a zero pair and then added.

Either way you'll get the same answer.

Once you're happy with your methods, we'll have a look at the next set.

So for the first one you're going to want to rearrange and then you're gonna multiply the second equation by five.

So there's 5x in both.

Then you can subtract the equations to get a value for y and substitute back in for x.

You get x is 10 and y is 20.

For e, there's a lot of choices here.

You want to expand the brackets on equation one first and then rearrange.

Now there's different ways to rearrange these equations.

So you might not have ended up with exactly the same equations as me.

If you did, you will have got choices again as to how to eliminate.

We could multiply equation one by two, then we've got 2x in both and we can subtract or we can multiply equation two by two.

Then we've got 2y and -2y, which are a zero pair and we could add them.

And finally for f, again, it's gonna depend how you rearrange these as to what choices you have.

I have ended up multiplying the second equation by two so that there is -6x in both and then I can subtract.

You had to be really careful there with subtracting your negative values.

Your x is -3 and y is 1.

2.

And finally, hopefully you spotted that A has no solutions.

The left hand side of both equations are multiples of each other.

So if you double the top equation, you get 2x + 2y and it's 2x + 2y = 24.

Well, you can't have 2x + 2y = 24 and 2x + 2y = 30 That's not going to work.

So we can't solve those simultaneously.

B is absolutely solvable.

If you multiply the first equation by five, you get 5x - 5y = 25.

That's fine because we can add those to eliminate the y terms, which would give us 10x equals 30 and then we can solve.

So it's okay for there to be variables that are zero pairs.

We can't have exactly the same variables.

And finally, C is actually gonna have infinite solutions.

Well I dunno if you spotted this one.

That's because both equations are actually equivalent.

If you multiply the top one by five, you get 5x - 10y = 5.

which is the same as the bottom equation.

What that means is we've essentially actually only got one equation.

So again, there's infinite solutions for what x and y could be.

Well done.

We're gonna build on those skills now and we're gonna have a look at some slightly trickier ones where we might have to do a little bit more work.

So Sofia is trying to solve these equations.

She says this looks a little trickier than the ones I've solved before.

Have a look.

What do you notice about the equations? So what you might have noticed is that the coefficients of r are not multiples of each other and the coefficients of t are not multiples of each other.

So we're not gonna be able to multiply one equation by an integer and then have equivalent coefficients.

What we could do, we've got a couple of options.

We could multiply by a non integer value instead.

To get from 2r to 3r, you can multiply by 1.

5.

So you could multiply the top equation by 1.

5 However, that might start getting quite tricky.

So there is another method we could use instead.

What we could do is we can multiply both equations by different values so that they do have the same coefficient of r or t.

So what could we multiply these equations by so that we have the same value of r in both? Pause the video, think about what we could do.

Right, I wonder if you thought about a common multiple.

Two and three have a common multiple of six.

So if we were to multiply the top equation by three, but the bottom equation by two, they'd both have 6r.

Let's give that a go.

Making sure we multiply the whole of the top equation by three and then making sure that when we multiply the second equation, we're multiplying it by two.

Now we have the same coefficient of r in both, so we can subtract.

I'm doing the second equation, subtract the first equation.

So 16t - 9t, and 26 - 12 gives me 7t = 14.

So t is 2.

And just as before, we can substitute into one of our equations.

In fact, we've got four equations now that we could substitute into.

The first equations have the smallest coefficients, so they're likely to be the easier ones to substitute into.

So we've got 2r plus three lots of two equals four.

So r must be negative one.

Jacob, Sam, and Andeep are solving this pair of simultaneous equations.

Jacob says, I'm gonna do equation one multiplied by three and equation two multiplied by two.

Sam's gonna do equation one multiplied by five and equation two multiplied by three and Andeep's gonna do equation one multiplied by six and equation two multiplied by four.

Have a read through.

Which method do you think is going to be best? Let's look at Jacob's.

Jacob is gonna use the lowest common multiple of four and six, which is going to be 12.

So he multiplies the top equation by three, he'll have 12x and multiplies the bottom equation by two, he'll have 12x.

And 12 is the lowest common multiple of four and six.

That's gonna be a fairly easy method to use.

Because he is used the lowest common multiple, he's multiplying by the smallest values possible and that does make future calculations a bit easier.

Let's look at Sam's.

Sam is using the lowest common multiple of three and five.

So equation one will have a term of 15y, equation two will have negative 15y.

They can then eliminate by adding.

They're multiplying by slightly bigger values than in Jacob's method.

However, they're still gonna be keeping their values reasonably easy to use.

And the good thing about this method is you're then adding to eliminate and adding can sometimes be a little bit easier than subtracting, especially if you're dealing with negative values.

And Andeep's method.

Andeep's method's absolutely gonna work.

He's found a common multiple of four and six by doing four times six.

Both equations will then have a term of 24x.

The only problem with Andeep's method is he's not using the lowest common multiple.

So some of his values might get quite big and a little bit trickier to manipulate.

It's still a absolutely fine method to use as long as you are aware that there's sometimes easier values to multiply to make your life a little bit easier.

Let's have a look at Jacob and Sam's method stem.

Being careful here that we do 9y - -10y and 81 - -14.

If 19y is 95, then y must be five.

Then we can substitute back in and get our final answers.

Let's have a look at Sam's method.

We're adding this time.

Notice that it gives a slightly larger coefficient of x for us to deal with, but less of that subtracting negatives that can sometimes get us in a muddle.

Substituting back in and we get the same values either way.

Okay, have a look at Andeep's working so far.

Can you spot his mistake? What do you think? Well I don't know if you spotted he's not multiplied the whole of the second equation by the same value.

This is quite common.

If you've multiplied the whole of the top equation by six, it's quite easy to forget that you are multiplying the second equation by something different.

So he's multiplied the whole of the top equation by six correctly.

And then the bottom equation, he's multiplied the first term by four, but then he's multiplied the other two by six.

And he says, I think I'm gonna look for the lowest column multiples next time.

Some of these values got very big.

It is a useful thing to keep your eye out for.

Time for you to have a go then.

So there's a mixture of questions here.

There's somewhere you might need to multiply one equation to eliminate.

There's somewhere you might need to multiply both equations to eliminate.

You might even find some where you don't need to multiply in order to eliminate.

There's a couple that need rearranging as well.

So take your time, make sure you've got clear working out, and then use that to find the value of each variable.

When you've done that, there is a code at the bottom for you to crack.

So what you need to do is find the value of each letter.

So for example, if b equaled seven, everywhere in the table that you could see a seven, you'd write b underneath it.

You might notice that there are no sevens in our code, so b is probably not going to equal seven.

Another way that might help you check that your answers are correct.

If you do that correctly, you'll end up with a phrase.

Give that a go and come back when you've got your answers.

Okay, let's see if we can use our skills now to solve this problem.

Jacob has some 10 pence coins and some 20 pence coins.

He has 38 coins in total with a value of six pound 10.

Can you work out how many of each coin he has? You might wanna form some equations and use our skills to find those values.

Off you go.

Well done.

We've got another problem to solve this time.

I'd like you to use this information to work out the cost to stay one night and have one breakfast.

So it's the total cost of one night and one breakfast, which is gonna be your final answer.

Give that a go, come back when you think you've got it.

Fantastic.

So hopefully with these ones, you are able to check your answers back.

Pause the video and check those now.

And then if you managed to crack the code, you should see it spells the phrase check by substituting.

I think we might have already mentioned that that's a useful thing to do when solving simultaneous equations.

Let's have a look at Jacob's coin problem.

So we can set up the equations.

10a + 20b = 610.

'Cause it's 10 times the number of 10 pence coins plus 20 times the number of 20 pence coins.

And that'll give us our value in pence.

So we need to make sure that says 610.

But also we know he has 38 coins in total.

So the number of 10 pence plus the number of 20 pence must be 38.

Now we can solve those.

We multiply the second equation by 10.

We've got 10a in both, and then we can subtract.

Okay, so 10b is 230, so b is 23.

So we have 23 20 pence coins.

Now we can find our 10 pence coins.

If I use that second equation, so a + b = 38.

We've already worked out that b is 23, so a must be 15.

So 15 10p coins and 23 20p coins.

And finally we set up our equations for our hotel.

We could multiply the first one by three and the second one by two so that we have 6bs in both.

Subtracting gives us a = 100 If we substitute in, we get b is 15.

So the cost to stay for one night and have one breakfast is 115 pounds.

Now I'm gonna draw your attention to something.

Really well done if you spotted this yourself.

But all we actually wanted was the price of one night stay and one breakfast.

We didn't actually need the individual costs, we just needed the combined costs of the two.

So actually if you just subtracted the original equations, 4a - 3a is 1a, 3b - 2b is 1b, we would've got a + b is 115.

Now that's no use to us if we want to just find out what a is or just find out what b is.

But that's perfectly okay for this question.

'cause all we wanted to know was the cost of 1a + 1b.

So the combined cost is 115 pounds.

We can do that without working out the individual prices.

Just something to keep your eye out for that might save you a little bit of time.

Fantastic effort today.

We've looked at loads of problems now that we can solve with simultaneous equations.

We've looked at some real life context and we've had a look at some trickier ones where we might need to expand some brackets, do some rearranging, and bring all our algebra skills together.

Have a read over what we've looked at today, remembering that simultaneous equations are amazing because we can always check our answers.

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