If p+6=11, what is the value of p?

If p=11 and p+q=20. What is the value of q?

If q=5 and p+2q=10. Find the value of p.

If p+1=6 and p+q+1=7, what is the value of q?

If p+q=11 and 2p+q=1, what is the value of 3p+2q?

Exploring systems of equations

﻿Okay, so in today's lesson, we're going to head back to the stationary shop that we visited in the first lesson. Now, in the stationary shop today, Antoni is buying three pens and two rulers. Antoni's spending two pounds 28. Carla is buying seven pens and two rulers. She's spending a total of four pounds 36. They're the same kind of pen with the same cost, and the same kind of ruler with the same cost. Can you work out the cost of 10 pens and four rulers, and if you've done that, are there any other quantities that you can calculate? Take a moment to think about that, and pause the video now. There are lots of different ways to go about working this out, so I'm going to share with you the way that I did it. If you look at the number of pens that Antoni bought, it's three, and the number of pens that Carla bought is seven, so all together between them, they bought 10 pens. If we go and look at the rulers, between them they bought four rulers. Antoni's bought two, and Carla's bought two, and that just happens to be the exact number that we want. If we add together Antoni's basket and Carla's basket, that is going to give us the cost of 10 pens and four. There are lots of other quantities that you might have calculated as well. Maybe you thought about six pens and four rulers, so doubling Antoni's basket. Maybe you thought about buying 10 lots of Carla's basket, so 70 pens and 20 rulers. What we've got here is a set of two simultaneous equations. We've got two unknowns, the cost of the pen, and the cost of the ruler, and we've got two sets of connected information. The cost of the pen is the same in both of those baskets, and the cost of the ruler is the same in both of those baskets. What we're going to do is take a closer look at them to see if we can figure out a way of getting those unknowns to get to the cost of the pen, and to get to the cost of the ruler. Carla's basket has got seven pens and two rulers, and that is a total of four pounds 36, and Antoni's basket is three pens and two rulers, and that's the cost of two pounds 28. What I would like you to do is to have a think at what is the same, and what is different about their shopping. Ultimately, we're going to use that to find the cost of one pen and the cost of one ruler. Think about what's the same and what's different. Pause the video. Okay, so what's the same about their baskets is that they both bought pens and they both bought rulers. What else is the same is the number of rulers. They both got two rulers in their baskets. What's different is the amount of pens. Carla has bought four more pens. What's also different is the amount of money. Carla has spent more money. Carla's actually spent an extra two pounds and 8p, and that extra two pounds 8p must've been caused by those four extra pens. The only thing that's different is four extra pens. That must be the thing that's cost the extra money. If we know that those four pens cost two pounds and 8p, if we know the cost of four pens, that means we can find out the cost of one pen. Four pens cost two pounds 8p. One pen must be 52p. We can divide the two pounds and 8p by four. If we know the cost of a pen is 52p, how can we figure out the cost of our ruler? Have a think about it, and pause the video. Okay, so I've decided to take another look at Antoni's basket. The reason that I chose Antoni's basket and not Carla's basket is because there's less in it, so I thought it would be simpler and less work for me. However, it doesn't matter if you've chosen Carla's basket to work with instead. It doesn't make a difference. We know that the pen cost 52 pence. We know that each of those pens in Antoni's basket was 52 pence, 52p, 52p, 52p. We know that the total cost of the basket was two pounds 28. Those rulers must be the difference between the total, and the cost of the pens. The difference between that two pounds 28 and those three lots are 52p, the cost of those three pens. If we calculate that, the two pounds 28, subtract the cost of those three pens, we get left with 72 pence. Those rulers must've cost 72 pence. If we know the cost of two rulers is 72p, I think we can get the cost of one ruler. If two rulers was 72p, we know that one ruler must be 36p. I'm not going to stop there. I'm just going to go back and double check that I've got this right. I think that pen cost 52p, and a ruler cost 36p. If I'm correct, three lots of 52p and two lots of the 36 pence, two lots of the cost of the ruler should give me a total of two pounds 28. If I put that into my calculator, three lots of the pen, three lots of the 52 pence, two lots of the ruler, two lots of the 36 pence, I do get in fact a total of two pounds 28, and I can check it against Carla's basket as well, so I know absolutely that I've definitely 100% got this correct. On the screen, we can see four sets of simultaneous equations. What I would like you to do is for each one of those sets of simultaneous equations to figure out the cost of each of the items. The cost of the protractors, the rulers, the pens, the calculators, et cetera. For each set of information, think carefully about what is the same, and what's different, and what is creating that extra cost. Pause the video and have a go. Okay, let's have a look at the answers. Now, in reality, you should know whether you got your answers correct or not, because you'll have been able to substitute them back into the original information, so you will already know whether your answers are right or not. In the first one, let's just have a quick look at it though. What's different is the two extra rulers, and those two extra rulers cost an extra 80p. From there, we can figure out that one ruler was 40p, and from there, we can then go on to figure out that one protractor was also 40p. Let's have a look at the pens and calculators in the bottom left. Now, the same number of calculators was bought. What was different with the number of pens, seven extra pens bought in that top basket, so seven extra pens for an extra four pounds 20. That means we can figure out the cost of a pen with 60p, and from there, we can go on to figure out the cost of the calculator. In the top right, we bought the same number of highlighters in each basket, but in the top basket, one extra pencil was bought, and that cost an extra 30p. We know that one pencil cost 30p, and from there, we can go on to figure out the cost of a highlighter, 30 pence as well, and then in the bottom right, pencil cases and notepads, there were different amounts of notepads that were bought, but the same amount of pencil cases. The seven extra notepads were two pounds 80 is what we can use to work this out. There's seven extra notepads for two pounds 80, means that one notepad was 40p, and the pencil case must've been one pound 50. Now again, I don't need to tell you the answer to these, because you can substitute your answers back in, and check them against the original scenario, and make absolutely certain that you're correct. Your final test for this lesson, I've given you a price list, and on that price list, pencils cost 30 pence, rulers cost 50 pence, and calculators cost a pound. What I would like you to do is to use that price list to come up with three sets of simultaneous equations of your own. Try to come up with one that is quite easy, one that's a little bit trickier, and one that is quite peculiar. Pause the video and off you go. Okay, here are the examples that I came up with. Probably, you've come up with something different, but maybe you've got something the same. My first example from my easier example is one pencil and one ruler is 80 pence, and one pencil and two rulers cost one pound 30. The reason that I thought this one was quite easy, because there's just one extra ruler that has been bought in a second basket, so straight away, we can get the cost of one ruler, and then because there was only one pencil in each basket, there isn't a huge amount of work to do to find the price of the second object. My trickier example is this one. 13 pencils and 17 calculators cost 20 pounds 90, and two pencils and 17 calculators cost 17 pounds 60. The reason that I thought this one was trickier is because there are 11 more pencils that have been bought, so a little bit more work to do to find the cost of the first object, and again, because it's 17 calculators, there's more work to do to find the cost of one calculator. For my unusual example, my peculiar example, I came up with this. A pencil and a calculator cost 80 pence. A ruler and a calculator cost one pound 50, and I think a calculator and a pencil would cost one pound 30. The reason that I thought this one was more unusual is because all the examples we did in the independent practise have two unknowns, and two sets of information, and here we've got three unknowns, and three sets of information. If you're able to, it would be great if you could take a picture of your equations, and share them with your teacher, and if you'd like to, you could ask your parent or carer, and you can send a picture of your work to @OakNational on Twitter so I could see it too. Thank you for your time today, and I will see you next lesson when we're going to be writing worded equations algebraically.

The cost of 1 protractor is 30 pence. If I spent £2.70, how many did I buy?

A compass costs 50 pence. 8 protractors and 3 compasses costs £8.70. What is the cost of a protractor?

1 compass and 1 protractor costs £1.10. 2 compasses and 1 protractor costs £1.60. What is the cost of a compass?

If 10 compasses and 3 protractors cost £4.20. 10 compasses and 6 protractors cost £5.40. Find the cost of 1 protractor.

If 10 compasses and 3 protractors cost £4.20. 10 compasses and 6 protractors cost £5.40. Find the cost of 1 compass.

Finding two unknowns from simultaneous equations.

In this lesson, we will solve a word simultaneous equation. We will then interpret the problem and create two equations from it and model a solution.

Maths

Y9 Maths

Key Stage 3

Year 9

Solving simultaneous equations

Solving linear simultaneous equations algebraically

Algebra

THEME

UNIT

Reviewing solving equations

Equivalent equations

Linear relationships

Combining algebraic relationships

Exploring expressions with two variables

Writing simultaneous equations algebraically

Checking the solution of simultaneous equations

Solving a simultaneous equation algebraically

Eliminating a variable

Manipulating systems of equations

Solving word problems

Using substitution to solve simultaneous equations

Substitution or Elimination

Simultaneous equations by trial and error

Let p be the number of protractors bought. If each protractor costs £2, which of the following represents the total amount spent on protractors (in pounds)?

If x=2 and y=3, what is the value of 2x+3y?

Use substitution to decide which of the following is the solution to x+y=3 and 5y+3x=15.

What number am I thinking of, if half of the sum of my number and 4 is the same as 3 lots of my number subtracted by 2?

Fill in the gaps: The "n" in the nth term represents the ____.

Which sequence is formed as a result of adding "4, 9, 14, 19, ..." and "12, 8, 4, 0, ..." together?

I want to buy 3 rulers and 3 protractors. If they cost 40 pence each, how much will I spend?

4x +8y=116 AND 4x +4y=68. Solve

Is there a solution with x value between 0 and 1?

Look at the picture below and then fill in the gap: If two equal expressions are both __________ by the same quantity, then the two resulting expressions will be equal.

3x+2y=5 AND 3x-2y=7.  If I add these two equations together, which unknown will be eliminated?

Lesson overview:Solving simultaneous equations

Core Content

Solving linear simultaneous equations algebraically: