x+y=6. Multiply this equation by 3.

x+y=3 AND 2x-y=5 This pair of equations has been transformed.  Which of the following are equivalent to the original pair?

x+y=3 AND 2x-y=5 This pair of equations has been transformed. Which of the following are equivalent to the original pair?

2x+5y=9 AND 5x+4y=14. Solve the simultaneous equations.

10x+5y=5 AND 3x-5y=34. Solve the simultaneous equations.

Manipulating systems of equations

﻿Hey year nine, this is Ms Bridget. Today we're going to be looking at solving some worded problems. You're going to need a pen and you're going to need some paper. Take a moment to clear away any distractions to make sure that you are ready to learn. And if you are ready, let's go. Anthony has been given these two numbers, 61 and 52, and he's telling us that the sum of these two numbers is 113. What I would like you to do is to write some other statements about these numbers and how they are connected together. They might be quite straightforward like the one we've got here, or they could be more complicated. Pause the video and off you go. Okay. There are hundreds and hundreds and hundreds of different statements you could have come up with. So here are a few possibilities. So maybe you wrote down that the difference between the two numbers is nine, that the sum of the digits for both of the numbers is seven. the mean of the two numbers is 56.5 and the products of them is 3,172. This time Anthony's presenting the information differently. So rather than giving us the two numbers and then telling us a fact about them, he's telling us a fact about them, and were able to work out the two numbers. So Anthony is telling us sum of his two numbers is what, 113, and the difference between them is 31. Have a go at working out what the two numbers are. Pause the video and have a try. Okay. I'd be really interested to know how you went about solving this because as always, there are lots of different ways to go about this. Now, one method that you could have used is trial and improvement, or trial and an error. So you could have just simply tried out two numbers and two numbers and two numbers and two numbers until you got to the answer. Now, another way of doing it, which will take us directly to the answer is using simultaneous equations. So that's the way that we're going to have a look at. And the advantage of simultaneous equations over trial and improvement is they will take us straight to the answer. So what I'm going to do is I'm going to call Anthony's numbers A and B, I could have chosen any two letters, I've chosen A and B. Now he tells us that the sum of his numbers is 113. So if we write that algebraically A + B = 113 the sum of A and B is 113. He also tells us that the difference between them is 31. So A - B = 31. At this point, we've got two simultaneous equations. Pause the video and have a go at solving them. Okay, now you might have done this by eliminating the As you might have done this by eliminating the Bs, you might have done something different. What I decided to do was to add the two equations together and eliminate the Bs. That gave me 2A = 144. I eliminated one of the unknowns. So I ended up with an equation, just with A and X. From there, I can solve for A, substitute back into one of my regional bits of information and I can solve for B. Now, we're just going to double check that we got this right. 72 plus 41 does in fact give us 113, and 72 and 41, the difference between those two numbers is 31, so we know we've got that correct. So the simultaneous equations will take us directly to the solution of this problem. Here I've given you three sets of information. Read through each set of information and see if you can identify the numbers in each scenario. Can you do this please by using simultaneous equations rather than by using another method, such as trial and improvement. Pause the video and off you go. Okay, now you should know whether or not you've got these, correct because you would have been able to take your solutions and substitute them back into the original information. So you don't need me to tell you whether or not these are correct, but we're just going to do a quick run through them anyway. So for the first set of information, I formulated my two equations very much like we did in the connect task this lesson. So I go A + B = 99, A -B = 27, and when I solved those, I got the solutions of A = 63 and B = 36. Now this next example, the one underneath, it's slightly tricky to formulate. Now I've got the mean of two numbers is 10. Now we know that to calculate the mean, we sum the numbers and in this case would divide by two. So if A and B have a mean of 10, that means they must sum to 20. So A + B = 20. The next bit of information, I've got the difference between them, which is A subtract B. And I know that three times that is equal to 12. So three lots A - B is equal 12, and we can multiply out those brackets. Once we solve that, we get the A = 12 and B = 8. Now my final example, twice one number added to three times another number is equal to 31 and the difference between them is three. We can formulate that as follows. So 2A + 3B = 31. A - B = 3 We can solve and we'll get A = 8 and B = 5. For the final task this lesson, I've given you a set of information about three numbers. You can see it there in the box. Now, what I would like you to do is to see if you can identify those three numbers, but this time I want to see how many different ways you can do it. So don't just try simultaneous equations, or just trial and improvement, see what other ways you can find to solve it. Pause the video and have a go. Okay, as always, there are many different ways to go about solving this. Some very direct, some very indirect. I'm going to have a look at one method that I used, which I think is really nifty for want of a better word. So what I did was I created a table. I call my three numbers, A, B and C, and I've created this table so I can summarise the information that we've been given. Now, I know that the sum of the first two numbers is 21. So A + B = 21. My next two numbers is 23. So I've recorded that in the table. And my final piece of information, the first and the third number sum to 22. Now, if you look really carefully at that table, each of the columns have got something in common. And what they've got in common is that each of those letters appears twice. So in the A column I can see A appearing twice, in the B column I can see B appearing twice. So if we sum all of those columns, we get this. Two As plus two Bs plus two Cs must be 66. Now, if I know that two of everything adds to 66, then that means that one of everything must add up to 33, it must be half that. Now just compare that top row and that bottom row. On the top row, I've got A + B = 21, and on the bottom row I've got A + B + C = 33. Now what's the same and what's different about those two rows. The only thing that's different is the C. So that seems to be what's creating difference in the total. So C must be equal to 12. Soon as we've got C, we can use the second road to get to B. We can use the third row to get to A. So this isn't necessarily the best way of doing this problem, I just think it's a very nice way. And I don't know if you recognise that this is very, very similar to the arithmetical problem that we looked at back in lesson two. Okay, we're going to finish there for today, but thank you as always for all of your time and attention. And next lesson, we're going to move on and we're going to look at different ways that we can deal with these simultaneous equations.

In this lesson, we will use the techniques learned in this unit using simultaneous equations to investigate and model solutions to word problems.

Maths

Y9 Maths

Key Stage 3

Year 9

Solving word problems

Solving linear simultaneous equations algebraically

Algebra

THEME

UNIT

Reviewing solving equations

Equivalent equations

Linear relationships

Combining algebraic relationships

Exploring expressions with two variables

Exploring systems of equations

Solving simultaneous equations

Writing simultaneous equations algebraically

Checking the solution of simultaneous equations

Solving a simultaneous equation algebraically

Eliminating a variable

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?

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

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 word problems

Core Content

Solving linear simultaneous equations algebraically: