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This is Ms. Bridgett.

What we're going to be doing this lesson is trying to solve simultaneous equations, algebraically.

You're going to need to pen.

You're going to need some paper.

You're going to need to remove any distractions.

Okay, let's make a start.

On the screen you can see two tables filled with symbols.

Now, in each of those tables, every symbol represents a number.

So in the top table, the acorn represents a number, the aeroplane represents a number and the spider represents a number and the sum of each row is given at the end of it.

So in that top table, those three acorns sum together, add together to make a total of 12.

What I would like you to do is to find the value of the symbols and while you're doing that I want you to think about which of those two tables is easier and which is more difficult and why do you think that is? Pause the video and off you go.

Okay, so I got the following solutions.

The acorn is 4 the spider is 2 and the aeroplane is 3.

And in the second table I got the following.

The acorns is 5 the spider was 2 again and the aeroplane was 1.

Now what I was really interested in here is which one you thought was more difficult.

So see if you have the same idea as me.

Now I thought the first one was easier than the second one.

I thought the second table was more difficult and I think the reason behind it is that very top row in the first table.

What we've got there is just acorns.

We've got an equation with just one unknown.

Now in the second table, there wasn't any equation where we had just one unknown.

Each of those equations, although they were related, although the table was related, all of those equations contains more than one unknown and I think that's what made that one a little bit more difficult.

A few lessons ago, I don't know if you remember but we looked at a situation where Cala and Antoni were buying different amounts of stationary.

And we worked out the cost of a pen and the cost of a ruler.

We're not going to revisit that scenario and the first thing that I would like you to do is to see if you can rewrite that information using algebra.

Now what I'm going to ask you to do is to let p represent the cost of a pen and let q be the cost of a ruler.

Pause the video and have a go at rewriting that information using algebra.

Off you go.

Okay, here's how I wrote the information.

So I know that p is the cost of a pen and q is the cost of the ruler so from Cala's information, we get 7p + 2q = 4.

36.

And from Antoni's information, we get 3p + 2q = 2.

28.

What we're now going to do is see if we can solve this set of equations, algebraically.

Now, if you remember those first two tables that we had in the last task, we said that one of them was easier than the other because it had a situation with only one unknown.

Now these equations at the moment have got two unknowns in them.

So what we're going to try to do is to manipulate them so we are left with only one unknown.

And we're going to do this very much in the same way that we did when we were looking at your original scenario.

Now in the original scenario, we compared the two shopping baskets and we looked at the difference between them.

We're going to do exactly the same thing here.

So I'm going to look at the difference between these two equations.

So I'm going to do 7p subtract 3p, 2q subtract 2q, 4.

36 subtract 2.

28.

And that's going to leave me with 4p is equal to 2.

08.

Four times the cost of a pen is equal to two pounds and eight pence.

That is exactly what we did in the original situation.

We identified that those four pens, the cost of four pens was making the difference in the cost of those baskets.

Now what we've done here by subtracting those two equations, by looking at the difference between them, is that we've created an equation with just one unknown in it.

Think back to that table, that top table was easier because it only had one unknown in it.

So now we know that 4p is equal to 2.

08, £2.

08.

We can solve that.

We can figure out the value of p.

Once we've got the value of p, the cost of a pen, we can now go back and work out the cost of the ruler.

So when we were looking at the original scenario, we revisited Antoni's basket.

And the reason we chose Antoni's basket was because it was easier to deal with than Cala's basket.

There was less in it.

I'm going to do the same thing again.

I'm going to use this equation and I'm going to use that equation because it's slightly easier than the equation above it.

Now we know that 3p + 2q = 2.

28, £2.

28.

I now know the value p.

I know that it's 0.

52.

So let's replace that 3p with what we know it's equal to.

Let's substitute that 3p in.

And have a look at what that does.

What that does is it stops us having an equation with two unknowns in, p and q and leaves us with just one unknown, q.

As soon as we've got one unknown, we can solve this.

We know how to solve linear equations, so we can manipulate that equation to find the value of q Now, what we looked at last lesson was taking our solutions and checking that they were correct.

So we can now put those values in and know beyond any reasonable doubt that we have got these values correct.

Now think about how, what we've just done there, related to the original scenario.

So we wrote them algebraically.

We looked at the difference between them and that enabled us to have an equation with one unknown.

We then revisited one of the equations.

Once we'd found the value and again, that gave us an equation with just one unknown.

Here are four pairs of simultaneous equations.

Have a go at solving them.

Once you've got you solution, substitute them back into the original equations to see if you were correct or not.

Pause the video and off you go.

Okay, you probably don't need me to tell you whether or not you've got them correct because you've already done that but let me just run through the answers, just in case.

So for the first one, I think the value of p is 1 and the value of q is 2.

For the top right, I rewrote this one slightly.

So I kept the first equation as it was 12p + 17q = 65 but I rewrote the second equation.

So rather than 21 =, I had = 21.

And the reason that I did that was it helped me to line up the p's and the q's and it enabled me to find the difference a little bit more easily.

So in this one, I got that p = 4 and q = 1.

In the bottom left, again, I rewrote it.

So I kept the top equation as it was but I rewrote the bottom equation, I changed the order, so that it's 5p + 6q = 39.

Again, this enabled me to line up the p's and the q's and to find the difference that little bit more easily.

So p = 3 and q = 4.

For the final one, again, I use the commutative property of addition to rewrite that second equation to line up my p's and my q's and I found that p = 4 and q = 5.

For our final task today, we've got two incomplete simultaneous equations.

What I would like you to do is to place 1, 2 and 3 into those boxes to create a pair of simultaneous equations.

So, for example, you might choose to do x + y = 1 and 2x + y = 3.

So I've put my 1, 2 and my 3 in those boxes.

Now my questions to you are how many different pairs of equations are there? So using that 1, 2 and 3 how many different pairs of equations can we make? My second question is how many solutions are there? Are all of the equations that you've created possible to solve? And finally, which solutions give you integers and which aren't? If there's anything else that you notice in addition, that's great.

Pause the video and off you go.

So I think there are six sets of equations.

So there's the one that we just looked at.

I tried to work systematically.

Here are my six pairs of equations.

Now these two, I think are particularly interesting.

And the reason I think they're interesting is because they're not solvable.

It is impossible to have two numbers, that at some point add together to make 2, and at a different time add together to make 3.

There are no numbers that will do that.

That's completely impossible, so these two do not have solutions.

The other four, however, we can solve.

So you will already know whether you're correct or not, because you'll have substituted your values back in to check them but I think the solutions are 2 and -1, 1/2 and 1/2, 1/2 and 5/2 and finally -2 and 5.

So two of our equations were unsolvable.

Two of our equations have integer solutions and two of our equations have non-integer solutions.

I wonder what would have happened if we'd use a different set of numbers other than 1, 2 and 3? Unfortunately, we're not going to look at that right now but thank you so much for your time today and next lesson, we're going to go on and we're going to carry on looking at these simultaneous equations.