In this video, we find out that simultaneous equations are equations that have more than a single unknown and can have an infinite number of solutions.

We break down the 2 methods by which you can calculate simultaneous equations; cancelling and substitution.

The most common method for solving a simultaneous equation is the elimination method which is where one of the unknowns is removed from the question, meaning that the remaining unknown can then be calculated.

As well as break down some AQA GCSE maths exam paper questions and answers.

We are looking at simultaneous equations and we're going to look at two methods that we can use to solve these. Method one is cancelling method two, substitution. And we're then going to look at a practise, start a question at the end to apply these two methods. So what do I mean by cancelling method one? Well, by cancelling, what we want to do is we want to basically get rid of either all of the y's or all of the x's.

And so we end up with just a linear equation with just x's or just y's and then we can then solve for that value, right? If I end up with say, I don't know, five x plus two equals twelve, I can solve that, right, I know. Therefore x has to equal two. If we look at this equation at the moment, there's no way to kind of hone in on what the values of x and Y are. They could be anything.

There's actually infinitely many solutions to this equation because we've got two unknowns with only one equation. And so we want to get down to a stage where we get rid of either the x's or the y's. We can do that by either adding or subtracting multiples of one equation from another equation. And we'll see what I mean by that in a second, right? So I'm going to decide that I want just x's, I want to get rid of all the Y's.

And I can do that by at the moment I've got a plus four Y and I've got a minus two Y. And so what it might make sense to do is I want to basically turn this two Y into a four Y and that will then allow me to cancel the Y's out. So what I'm going to do is I'm going to take my equation two and I'm going to times it by T. What that's going to give me is 16 x minus four Y equals 20. And I'm going to call this equation three.

Well, now we see, right, we've kind of arrived at a nice junction. We've got plus four Y minus four y and it seems as though we should be able to then cancel out these Y's, right? What it turns out we can do is we can basically take equation one and add it onto equation three. So we can literally do equation one plus equation three. We're going to add the left side together and the right side together.

You might wonder how can we add a whole equation to whole equation? We're probably used to thinking of we could add 14 to the right hand side and we have to then add 14 to the left hand side. We're probably used to that kind of math, right? It's standard rearranging math. Well, we know in this case that 14 is the exact same as x plus four y.

We're told it from equation one. And so instead of writing plus 14, I can write plus x plus four y. And because of the same thing that they're interchangeable. So what we end up doing is basically adding the left hand side of equation one and the right hand side of equation two onto this new equation onto equation three, if that makes any sense.

What we end up with x plus four y. What we end up with is we do 16 x plus x. That gives us 17 x. We do minus four y plus four y and that's going to cancel out the Y's. And that's why we do this, right?

That's why we do this whole thing. It's because we end up with zero y, right? Because plus four Y minus four Y just cancels out and we end up with 14 plus 20 and that's going to give us 34. I want to get x on its own. I want to cancel out that multiply by 17, while the opposite of multiplying by 17 is dividing by 17.

So I can divide by 17 on both sides. That's going to leave me with x equals t. And now I've got my value for x. I've got x equals t. I can take that value and whenever I see x, I can substitute in x equals two.

And so let's look at equation one. If I copy equation one down here again, I've got x plus four y equals 14. Well, instead of x I can put in t because I know x equals t. And so what I end up with is t plus four y equals 14. I can get rid of the two.

I can subtract t from both sides. So four y equals twelve, I can divide both sides by four and y equals three. And so we get our answers x equals two and y equals three. So that's how we can use cancelling to solve simultaneous equations. Let's look at the exact same question again.

But we'll do method t, we'll do substitution this time. So substitution is again, it's a way of basically getting rid of all the x's or all of the y's. But rather than kind of adding and subtracting entire equations from one another, it's slightly different, right? So we'll do an example and in this case, we've already got x on its own. We've got just x here.

So it probably makes sense to get rid of this x and substitute in something to do with y in its place. So what we're going to do is I'm going to use equation two to allow me to do that. And so if I use equation two, I want to get x on its own, right? I want to do better maths and I want to end up with x equals something and then I can take that something and I can put it in for x there. So how do we get excellent saying I can add two Y to both sides to take care of that minus two y.

So I get eight x equals y and I can divide both sides by eight. I get x equals ten plus two y divided by eight. And so that's kind of what goes in that box, right? We've kind of made x the subject of that equation. So I can then take that and I can put it into equation one.

So I'm going to copy out equation one down here, x plus four y equals 14. I'm going to substitute this in for x. So what that gives us is it's going to get slightly complicated. We're going to get ten plus two y over eight plus four Y equals 14. I could times everything by eight.

And so I get ten plus two y equals four times 832 and 14 times eight. I'm going to be lazy in the calculator.

And I've now gotten two y plus 32 y. So I'm going to get 34 Y for those two and I can take care of the ten that's going to leave me. If I subtract ten from both sides, I get 34 y equals one two. I can divide both sides by 34. And if I do that, I get y equals three.

Again, I can then substitute that in. I actually have an equation already up here, right, that tells me exactly what x equals if I know what y equals. And so if I just copied that out down here, I know x equals ten plus two y over eight and I can take that y equals three and put it in to that equation. And so what we end up with is x equals ten plus two times three, which is six over eight. And so that's going to give me 16 rates, which gives me T.

So again, we get the exact same results with both methods. You can probably see method one is slightly quicker than method two. We use method two normally when we're dealing with a quadratic equation, when one of our equations are quadratic. So unfortunately, we can't just ignore method two. We kind of get forced to use it in certain circumstances.

Let's do a quick example question. So let's focus on 13 for now. And this is going to be kind of deriving a single settings equations from the text. So we've got a shop selling two coffees and three cakes. And so if we were to write this in an equation where I've got two coffees, which I'm going to call coffees, that can be C, and three cakes.

So I'm going to call cakes, I'm going to call cakes A and that costs 995. And I also know that the same shop sells one coffee and four cakes. So one coffee and four cakes for 1035, that's not 1030 05:00 a.m. £10. So it probably makes sense to use method one for this, right?

Because we've got two C here and C here. Again, if I call this like equation one equation t. What I can do is I can take my equation t and I can times it by t. That's going to give me two C plus eight A equals £20 and 70 pence. And then what we see here, right, we've got a two c term and a two c term.

Well, I know that if I want to get rid of that two c term, what I can do is I can take equation, of course, equation three, I'm going to take equation three. I'm going to subtract equation one from it. So I'm going to do equation three minus equation one. That's going to give me, well, two C minus two C is just nothing, right? We get zero C.

Eight A minus three A gives me five A and 20 pound 70 minus 995 calculator. 20 pound 70 minus 995 is going to give me 1075. And then on A on its own, I want to get just the cakes on their own. So I divide both sides by five. And if I do that, I get two pound 15.

So I know a cake costs £2.15 p 15. And I now want to work out what a coffee costs. Right, I've got the cakes that's all sorted. I need to work out my coffee as well. So there are a few different ways of doing this.

I could pick either equation one, two or three and just substitute in my value for eight. I'm going to pick two. And so if I take equation two and I want to make C the subject, I get c is equal to 1035 minus four A, and I know that A is equal to 215. And so I can take that value, I can take that value for A and I can put it in my equation. So that's going to give me, that's going to give me c is equal to 1035 minus four times 215.

Well, four times 215 is going to give me £8.16. So 1035 is going to be super lazy. To make sure I get this right, 1035 minus eight pound 60 is going to leave me with £1.

That's the answer. So that's how we can take information from text and actually put it into some of those equations and then use those to solve for our unknowns. And that is simultaneous equations.