Lesson video

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

My name is Ms. Davis.

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, it 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're doing.

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

Let's get started.

Welcome to today's lesson on solving simultaneous linear equations by substituting.

So we're gonna look at a method called substitution to help us solve linear simultaneous equations.

We are gonna compare this to the elimination method.

So if you haven't already looked at how to solve simultaneous equations by adding or subtracting to eliminate a variable, you might want to do that first so that you can then compare that method to this new one.

By the end of the lesson, you'll be able to solve two linear simultaneous equations using this substitution method.

Even better, we're gonna have options.

We're gonna think about when's gonna be the right time to use each method.

So a couple of terms to be aware of: Elimination is a technique to help solve equations simultaneously, as well where of the variables in a problem is removed, normally by adding the equations or subtracting the equations.

Then we're gonna talk about substituting.

Now you might have seen substituting before.

Substituting means to put in place of another.

In algebra, substitution can be used to replace variables with values, but also with terms or expressions, which might be a little bit new to you.

We're also gonna talk about making variables the subject.

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

So for example, in y equals 5x plus 2, y is the subject.

The subject should have an exponent of one and a coefficient of one.

So one y equals something or one x equals something.

So let's have a look then at solving by substitution.

When we have two simultaneous equations with variables x and y, we could rearrange them into the format axe + by = c and solve by elimination.

So this is the other method of solving.

So for example, if I have x = 3y + 6 and x = 5y + 2, we could rearrange those into x - 3y = 6 and x - 5y = 2 and then solve by subtracting, because the coefficients of x are the same when we subtract, they'll be eliminated.

Negative 3y subtract negative 5y is 2y, six subtract two is four, and then we can solve for y.

Of course we can then substitute that in to get our value for x.

So quick reminder on the screen if you've forgotten how to do that, and then we end up with our solutions, x equals 12 and y equals two.

Ah, but Izzy spotted something, hang on.

If both those expressions are equal to x, then both have to equal the same value.

What do you think Izzy means by this? So because the right hand side of both equations are equal to x, that means the right hand side of both equations must be equal to each other.

That means that 3y + 6 must have the same value of 5y + 2, because they both have the same value as x.

We can write a new equation then 5y + 2 = 3y + 6.

Those two things must be equal.

What we've essentially done is we've substituted 5y + 2 into the first equation, because x = 5y + 2, where we see an x in the other equation, we can write 5y + 2 instead.

So here I'm substituting equation two into equation one.

So I'm gonna put brackets around the value I'm substituting in.

So instead of writing x, I can now write 5y + 2 = 3y + 6.

And then we can solve it like any other linear equation.

We can add negative 3y to both sides, excuse me, 2y + 2 = 6.

So 2y is 4, and y is 2.

which is exactly the same as when we did it by eliminating.

Just like before, we still need to substitute back into an equation to find the value of the other variable.

Don't forget when you've done all that hard work, you still need to find the other variable.

Substituting, we get x is 12.

Just like with the elimination method, we can check our values by substituting into the other equation.

So we've always substituted into one.

Let's have a look at equation two.

So if x = 12, we've got 12 = 5 lots of 2 + 2, and that is correct, 12 = 10 + 2.

Okay, I'm gonna show you how to do this one on the left, and then you're gonna have a go at one yourself.

Listen carefully to my explanation, and then think about how you're gonna set out your working.

So in both these equations, I've got y as the subject.

I'm gonna substitute equation one into equation two.

So instead of writing y in equation two, I can write 5x - 7.

So 5x - 7 = 14 - 2x.

Then I can add 2x to both sides, which gives me 7x - 7 = 14.

The 7x = 21 and x = 3.

As always, I can pick either equation to substitute into.

I've picked equation one, so y is five lots of three minus seven, so 15 minus seven, which is eight.

Okay, time for you to have a go.

Can you solve those equations simultaneously? Think about using this new method of substitution off you go.

Right because y is the subject of both.

It doesn't matter which one we choose to substitute into which one.

So I'm gonna substitute equation two into equation one.

Instead of writing y, I can write 12 plus 4x.

So I've got 12 plus 4x equals one minus 7X.

Adding 7x to both sides gives me 11x plus 12 equals one.

So 11x equals negative 11, so x is negative one.

Then pick either equation to substitute into.

I went for two, so y equals 12 plus four lots of negative one, so y equals eight.

And again we could check our final answers.

Well done.

That's the basics then of this solving by substituting.

So we can use the same method when one variable is the subject of an equation but not the other.

So in the examples we've just looked at, y was the subject in both equations.

In this example, y is the subject of one equation but not the other.

We can still substitute in exactly the same way.

So because y is the subject of equation two, we can substitute 4x minus 25 for y in the other equation.

This is what that looks like.

So we're gonna have 2x plus three lots of y, but instead of y we're gonna substitute 4x minus 25 equals two.

Now it's really important that we use the brackets here.

Now we can manipulate and solve so we can expand our brackets.

So 2x plus 12x minus 75 equals two.

Collect like terms 14x equals 77.

So there I've collected like terms and I've added 75 to both sides.

Now I'm doing 77 divided by 14, which I have decided to do in two steps to help me.

I've divided both sides by seven to get 2x equals 11.

And then I can see that x is 5.

5.

As always substitute back in, and we get y is negative three.

Pause the video and just check those two solutions work And they do.

So let's have a look at one together and then you are gonna give this a go.

So I'm noticing that a is the subject of the second equation.

So wherever I see an a in the first equation, I can actually substitute that for 16 minus 2b.

So substituting equation two into equation one, negative five equals eight lots of 16 minus 2b subtract 3b, expanding my brackets, negative five equals 128 minus 16b minus 3b, collecting like terms, I've got negative five equals 128 minus 19b.

So I'm adding 19b to both sides and I'm adding five to both sides.

19b is 133, that means b must be seven.

I can use a couple of things to help me there.

I know that 140 divided by 20 is seven, so I'm looking at that and thinking, hmm 133 divided by 19 is gonna be close.

So I could just do 19 times seven to check to see if it works.

Little tricks like that can help you find these trickier divisions.

Substituting that back in.

So I've chosen the second equation because a is the subject, it's the easiest one to substitute into.

So 16 minus two, lots of seven, so a is two.

Okay, your turn, see if you can solve these equations simultaneously and check your answers.

Off you go.

So b is the subject of the top equation.

So we can substitute b into the bottom equation.

So substitute equation one into equation two.

We've got 3a minus two, lots of five minus 2a equals 18.

Expand the bracket 3a minus 10 plus 4a equals 18.

be careful there where you're expanding the bracket.

Negative two multiplied by negative 2a is positive 4a.

Collecting like terms 7a minus 10 is 18, 7a is 28 and A is four.

Substituting backing again equation one has b as the subject.

That's gonna be the easiest one to substitute into.

So five subtract two lots of A or five, subtract two lots of four, which is five subtract eight which is negative three.

And again we can check those work.

Lovely and time to have a practise.

For this first set, I'd like you to spot the mistake for each set of working, you're gonna need to read every step carefully and see if you can work out what's gone wrong.

Give that a go.

Well done, time for you to have a go at solving these yourself then.

Think carefully about some of the common mistakes that were made on the previous question.

Make sure you avoid those when doing your answers.

Also remember that you could check your answers when you are done.

you could use a calculator if you wanted to support you.

Make sure you've got that final answer correct, give it a go.

Three more to have a look at here when you are happy with your answers, come back.

Okay, we've got a problem in context this time.

So five highlighters and three notebooks cost £14.

90.

A notebook cost £1.

50 more than a highlighter.

Work out the cost of each item.

If you want to think about setting up some equations and then solving, off you go.

Let's have a look at our answers then.

The mistake on the first one was they substituted for the wrong variable.

It was x that equaled 3y minus four, so it should be x that was substituting into the top equation.

Instead, this person substituted in for y and that meant they then got in a muddle later on 'cause they ended up having y as our next terms still.

So then they tried to add the coefficient of unlike terms. So just be aware that when you're substituting in, you're substituting into the right variable.

For the second one, the expanding the brackets were the problem here.

They did not expand the brackets properly.

We needed to do negative two multiplied by negative 0.

4, so we should have positive 0.

8.

And finally the mistake was in the rearrangement.

The substituting was absolutely fine, but then when this person tried to rearrange the equation, they subtracted four instead of added four.

So we should actually have 2x equals 10 and x is five and then that will change our y value as well.

Fantastic.

So I'd like you to pause the video and have a read through these workings.

For the first two, one of the variables was the subject of both equations, so you could have just equated the right hand sides.

For the third one, you needed to substitute eight minus 4y in for x in that second equation.

Have a read, once you're happy we'll look at the next set.

And for 2D, now with 2D, you might have decided to expand the brackets first in that second equation, which would've you 6p minus 21 equals q And then you've got a choice as to whether to substitute q into the top equation or p into the bottom equation.

I did it slightly differently.

I decided to substitute p into the bottom equation straight away and then deal with expanding my bracket.

So all depends how confident you are with expanding brackets, especially where you've got a bracket expression inside a set of brackets.

It's entirely up to you but have a read through of my working to help you.

For E, again we're substituting 9.

7 minus 3x in for y in the top equation and then we're just being careful with some of our decimal skills.

You get x is 3.

5 and y is negative 0.

8.

And finally same again just testing some of our fractional skills.

So we can substitute a half x plus five over three in for y in the second equation and then we get x as two over three and y as two.

Pause and check those through again if you need to.

And then we'll have a look at that final question.

And finally we can set up some equations.

So we've got 5h plus 3n equals 14.

9 where h is the cost of a highlighter and n is the cost of a notebook, but also we need a notebook is £1.

50 more than a highlighter so it can write n equals h plus 1.

5.

Then I can substitute that into n in the first equation and then solve, pause the video if you need to read through my steps.

We get h is 1.

3 or £1.

30 in context and then substitute back into either equation.

Equation two is the easiest.

If h is £1.

30 and a notebook is £1.

50 more, that must be £2.

80.

Don't forget to write your final answer out properly.

Highlighters cost £1.

30 notebooks cost £2.

80.

Perfect.

you've got the skills now of substitution.

We're just gonna talk about when that is the most efficient method and what things to look out for.

So sometimes neither equation has the variable as the subject.

So it's up to you to decide which method would be most efficient.

So if we look at this example, a plus 3b equals 14, 2b equals 10a minus 12.

To solve by substitution.

What we could do is make one of the variables the subject.

You've got an option, you could rearrange the top equation to make a the subject so a equals 14 minus 3b.

Or you could divide the bottom equation by two to make b the subject.

Equally, don't forget, we could solve by elimination so you could rearrange those equations.

None of the variables are the same, so you need to multiply the top equation by 10 and then solve by either adding or subtracting depending on how you've rearranged.

What we are gonna do is we are gonna make b the subject at the bottom equation.

So if we divide everything by two, we get b equals 5a minus six.

Again remembering to maintain a quality we have to divide the whole equation by two.

Then we can substitute that value for b into that top equation.

So a plus three lots of b or a plus three lots of 5a minus six equals 14, expand, rearrange and solve as normal.

Of course, as soon as we've got our value for a, we can find our value for b.

We can substitute back into either of our original equations or the third equation where we divided through by two.

That's actually gonna be the easiest because b's already the subject and we get b is four.

Let's check by substituting into the original equations.

So that is correct.

Okay, have a look at these equations, which method you think will work best? Sophia says we can make y the subject of equation two and substitute into equation one.

Alex says we can make x the subject of equation one and substitute into equation two.

And Izzy says we could double equation one, rearrange and eliminate by subtracting.

Have a read of those ideas which do you think is going to work best.

Again, this is gonna be personal preference as to what things you prefer to work with when you're solving equations.

Now for Sophia, if we make y the subject of two, we get y equals 1.

25x minus 5.

5 or using fractions y equals five over 4x minus 22 over four.

We would then need to substitute into the first equation.

There's nothing wrong with that, but it might not be the most efficient method because we're gonna have to deal with those decimals and fractions.

Alex says we can make x the subject of one a substitute into two.

Well making x the subject of one is gonna give us x equals negative two thirds y.

We're definitely gonna wanna use fractions here because that would be a recurring decimal.

Again, there's absolutely nothing wrong with that.

We are good at fractions, we can use our fraction skills.

It's not a problem, but you might decide that it is not the most efficient way.

Let's have a look at Izzy then.

Izzy said e could double and rearrange and eliminate by subtracting.

I think this is possibly the most efficient here as it avoids fractional or decimal terms. However, there's absolutely nothing wrong with working with fractions and decimals.

When the first equation is doubled, we're gonna have 4y in both so we could subtract to solve.

Let's have a look at that method then.

So let's rearrange.

So we've got our variables on one side and constants on the other.

Multiply the first equation by two.

If I rewrite the second equation underneath and now I can subtract remembering 6x subtract negative 5x is 11x.

And zero subtract negative 22 is positive 22.

As always I can substitute back in to find my y value.

Okay, have a look at this one.

Which method do you think will work best this time? Sophia wants to make x the subject of two and substitute into one.

Alex wants to rearrange and eliminate by adding, what would you go for? So making x the subject will involve dividing five by three.

So we definitely want to use fractions for that.

It also involve dividing negative two by three.

Because the coefficients of x are a zero pair, I think Alex's method is probably gonna be most efficient.

Rearrange the bottom one to be 3x minus 5y equals negative two.

And then add.

Izzy says though I think we should use substitution but not by making x the subject.

Can you see what Izzy has spotted? And are we allowed to do that? Can we use substitution without making one of the variables the subject? What do you think? So what she's spotted is there's a term of 3x in both equations, negative 3x in the top equation at the moment, 3x in the bottom equation.

What we could do is we could substitute that right hand side in for 3x in the top equation.

We don't actually have to make x the subject first and then multiply it by three again.

We could use the fact that we've got 3x equals 5y minus two and just substitute that in.

Let's see how that would work.

So if I substitute equation two into equation one, I'm gonna replace the 3x with 5y minus two, I've got 4y minus 5y minus two equals negative two.

Expand the brackets, remembering that I'm doing negative one lots of 5y and negative one lots of negative two and I get y is four.

Again, substitute into any equation, second one's gonna be the easiest and get our value for x.

Of course Alex said we could rearrange and solve by elimination by adding the two equations that would've worked just as easily.

Right Sofia is solving these simultaneous equations using the substitution method.

Can you spot the mistake she has made? Off you go, Let's have a look then.

She started off really well.

There was a couple of steps to make x the subject to the first equation.

She actually subtract 4y and then divide by two.

But she's done that correctly.

So we've got x as 16 minus 2y.

But then she's gone really wrong with her substitution.

She substituted back into the same equation that she's just rearranged.

If you rearrange one equation, when you substitute, it must go into the other equation.

Remember we need two pieces of information to solve equations with two variables.

So we can't just ignore that second equation.

If she substituted into the other equation, she would've been absolutely fine because she substituted back into the same equation.

She ended up with the statement 32 equals 32, which has got her nowhere.

Okay, I'd like you to carry on from where Sophia was doing really well and find the correct value for y.

Off you go.

Lovely, so now we need to remember to substitute into the second equation.

So we've got three lots of 16 minus 2y plus 3y is 18, expanding our brackets, collecting like terms. We get 3y is 30 and y is 10.

So x is negative four and y is 10.

Sophia has got it now.

It's one of those mistakes that once you've made it once you're not likely to make it again.

Right time to have a go then, I'd like you to solve each pair of simultaneous equations.

I'm gonna ask you to do it twice.

you're gonna use the substitution method and the elimination method.

And then can you write me a sentence explaining which was the most efficient and why, give it a go.

Okay, so Alex has tried to solve these equations simultaneously by making y the subject of both equations.

I'd like you to start by checking his solutions by substituting.

So he is got an answer of x equals one, y equals five, is that correct? And then can you suggest a more efficient method? Would you do it exactly the same way or do you think there's an easier way, off you go.

So this time you can choose any method you wish.

I'd like you to read the questions through, think about what's gonna be the most efficient and solve them simultaneously.

When you're happy with your answers, come back.

Let's have a look then.

So to solve by substituting, the easiest choice is definitely to divide the bottom equation by five.

So x equals 11 at minus y, substitute into the top equation, rearrange and solve for y.

Then we get y is four and x is seven.

So to eliminate, I'm gonna want to rearrange them first.

And then I haven't got any coefficients the same or a zero pair.

So I'm gonna need to multiply both equations probably in this case.

So I'm gonna multiply the top equation by five and I'm also gonna rearrange at the same time to the y terms in the same order.

And the bottom equation by three.

If you want to pause and just check my equations, feel free to do so.

I've got 15y in both, so I'm going to subtract, doesn't matter which way round you subtract, I've done equation two, subtract equation one so that I end up with positive terms. I get x is seven and substituting to get y is four.

I wonder whether you agree with me then, that in this case making x the subject of the second equation was easier than that elimination method.

It was so easy because all the turns were divisible by five, which didn't leave me with any fractions or decimals to deal with.

And the elimination I had to rearrange and I had to multiply both equations before I could eliminate.

Let's have a look at this one.

I'm gonna make y the subject of the bottom equation again by dividing through by five you can use fractions or decimals, substitute into the top equation.

And I get x as negative one.

The trickiest bit with that is sorting out your decimal skills, 13 divided by five being 2.

6.

And then when you're substituting in, having to multiply those decimals by negative three and rearrange and so on.

Substituting in you get y is three.

So we had to do a lot of decimal calculations despite the fact that the answers were actually integers.

For the second one if I rearrange, I can then multiply the bottom equation by four, eliminate by subtracting and substitute in.

Feel free to pause and check that if you need to.

I think this was less obvious as to what was easier.

As we've already said, making x or y the subject did involve fractions or decimals.

The elimination was reasonably efficient, but we did have to rearrange and we did have to multiply one of our equations.

So again, it's entirely up to you which you think is most efficient here.

I would've gone with that second method.

Let's have a look at Alex.

So let's check by substituting 10 lots of five equals three lots of one plus 47, that is true.

Two lots of five plus three lots of one equals 13.

That is true.

So Alex is correct.

If we have a look at his method, he's rearranged both equations to make y the subject.

That means he's had to deal with decimals or fractions in both.

Then he is put them equal to each other and solved.

Absolutely nothing wrong with that.

But I wonder if you have suggested a more efficient method.

So you might have said there's 3x in both equations.

So we could have rearranged one equation to isolate the 3x term.

So let's say make the bottom one 3x equals 13 minus 2y.

And then you could have substituted that whole thing in for 3x in the top equation.

Equally you could rearranged that top equation to 10 y minus 3x equals 47 and solved by adding, eliminated by adding.

'Cause they would then be a zero pair, the 3x and the negative 3x.

And finally, this is entirely up to you, which is your preferred method.

I'll explain my choices to you.

For the first one, y is already the subject of the second equation.

So I definitely think it's easiest just to substitute that into the first.

You should get answers x is two and y is 11, however you've done it.

Four B, I think there's loads of options as to what was gonna be the most efficient here.

I have decided to double the second equation to make it 6y equals 4x minus 20, and then I substituted the whole of that in for 6y in the first equation.

However, you might have decided that rearranging and that eliminating would work best for you there.

I definitely wouldn't bother dividing through by three and dealing with fractions if I could help it.

I find it much easier to double that bottom equation and then substitute it in for 6y.

Either way, you get x is two and y is negative two.

The last one was entirely up to you.

They're set up to be kind of ready to solve by eliminating, but you would need to multiply both equations to get the coefficients the same.

So I probably would've multiplied the top equation by two, the bottom equation by three.

So I could add to eliminate.

However, there's nothing wrong with rearranging the first equation to make x the subject or the second equation to make y the subject or vice versa.

And then solve by substitution.

Make sure you take the time to check that you are happy with those.

The whole idea of using substitution to check our final answers means that if you didn't get it right, you then need to be able to go back through your working and see where you've gone wrong.

So make sure you are presenting your working in a way that you can then go back through it, to see where the problem lies.

Well done, so hopefully now you've got a range of methods that you can use to solve simultaneous equations.

The elimination method can be a really nice method to use, but you wanna look for when substitution is gonna make your life easier, especially if one of the variables is already the subject.

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