﻿Hi everyone This is Ms Bridgett. We are going to today carry more simultaneous equations and we're going to look at a method of solving them which is called solving them by substitution rather than elimination which is what we have been doing so far. So you're going to need a pen you're going to need some paper. You're going to need to take a moment to remove any distractions. So if you're ready, let's go. Have a look at this spider diagram on the screen. Now in the centre of it I've told you that Y is equal to two X so I have defined Y as being equal to two X. Now what we can do is we can use that fact to fill in the other blanks on the spider diagram. So if we do one as an example. If you look just up to the top of the row on the right you can see three Y but because I know what one Y is equal to two X I can work out what three Y is equal to which is six X. Have a go at filling in the blanks the completely empty bubbles if you use to come up with one of your own and create one of your own. So of pause the video now and have a go. Okay let's start in the top right and we have rounds and compare our answers. So we've just worked out three Y and we knew that was six X in this situation. So let's move on to three Y plus six. Now I know what three Y is so three Y plus six is going to be six X plus six, moving down the leg of that spider we're then taking that three Y plus six and dividing it by six which is going to give me X plus one. Take a look at the equations on the screen I've got three X plus Y is equal to 14. Y equals two plus X. Just take a moment to look at them and think about what is the same about them and what is different about them. So pause the video now. Okay now, there are lots of different things as always that you might have come up with but one of the things that strikes me about this pair of equations is that the second one looks slightly different to the first one. They've both got Y and X in them they've both got two unknowns in them there's an indifferent about that second one and it's also quite different to all of the ones that we've been looking at so far. And what's different about it is that this Y is isolated it's the subject of that equation. So in that second equation it says Y is equal to, two plus X whereas neither the X nor the Y are isolated in that top equation. Now what that means is that we can use that for a different way of solving these simultaneous equations. So let's just take a think about that. I know that three X plus Y is equal to 14, but from the second equation we know something about Y we know that Y is equal to two plus X. So what we could do is we could swap that Y straight out for what we know Y is equal to so we can replace the Y with what it's equal to which is the two plus X which leaves us three X plus two plus X equals 14. Now, this is a very exciting breakthrough because what we've done here is we've now created an equation with one unknown in it rather than two. So in that first line the three X plus Y equals 14 we've got two unknowns once we substitute out the Y we get an equation just with one unknown in it. Pause the video and just have a try at solving that equation. Okay. So the linear equation we're left with you might have solved it this way you might have done it slightly differently. So I simplified the left hand side to give four X plus two is equals 14 I then solved that to get the X as equal to three. Now once we've got one unknown as we know, we can go back we can substitute and we can find second unknown And if you look at that second equation Y is equal to two plus X. We can use that to help us to figure out what Y is and then check it against the first equation. And what we've been here is we've solved this simultaneous equation using a different method what we were doing before was we were lining up those unknowns we were lining up the X's and the Y's and we were trying to eliminate one of them. What we've done here is something different we substituted one of them we substituted in this case the Y out and that left us with our equation with just one unknown in it. And that is the way that we're looking at solving simultaneous equations today. Have a go at solving these pairs of simultaneous equations, using substitution. Remember to check your answers at the ends. Pause the video off you go. Okay let's just have a quick run through the way that you might going about solving this using substitution. So in that first pack we've got three X plus two Y is equal to 30 from the first equation we know what Y is equal to. So we can substitute in the X minus five in place at the Y. That leaves us with three X plus two lots of the difference of X and five is equal to 30. So that's the equation with one unknown and we'll go on to solve it and we can find, we find that X is equal to eight and Y is equal to three. In the second pair of equations the second equation we can see that it's been written with X as a subject so we can take that second equation what X is equal to and we can substitute it into that first one. So rather than having three X plus five Y is equal to 10. We have three lots of what we know the X is equal to the two Y plus seven plus five is equal to 10. We can simplify and manipulate that equation to solve for Y once we've got Y we can go back and find X. So I got the X was equals to five and Y was equal to one. Remember we need both of these solutions solving just for X or solving just for Y isn't solving simultaneous equations. For the final set the second equation I've got Y as the subject Y is equal to. So I'm going to take that second equation and substitute into the first one. Before I do that though I just did you don't have to do this, but I thought I'd just play with that first equation a little bit and rearrange it to make it a little bit nicer. So rather than having two X plus three Y divided by five is equal to eight. I rearranged it to get the two X plus three Y equals 40. And that meant it was slightly nicer for when I substituted the Y in. You don't have to do that it's not the only way to go about solving it. However, we go about doing this what we substitute, the Y in we get an equation with just one unknown in just the X which we can solve, and we get to X as equal to five and Y is it equal to 10. So for your final task for today I've given you a magic square. Now the way that magic squares work is each row and each column and each diagonal add to the same total. So whatever that bottom row adds up to, it will be exactly the same total as one of the columns. What I would like you to do is to see if you can figure out the missing numbers. Now there are lots and lots of different ways to go about this. Some might be quite quick some might take quite a long time. So you might be able to see this quite quickly. You might need to spend a little bit longer on it either way is absolutely fine. So pause the video and off you go. Okay. I'm going to reveal the answers to you. So the missing numbers are ten, three, eight, five, seven, four. Now I didn't reveal them in the order that I calculated them in I just revealed them in a nice logical order this might not been the order that you found them in. And as I said, there were lots and lots of different ways to go about solving this I'm sure your math teacher would love to find out how you did go about solving this so if you are able to please do take a photo of your work and send it in to your math teacher. That is it for today. So thank you as always all of your time and attention. Next lesson we're going to move on and we're going to compare elimination and compare substitution and see which one we might use when.

In this lesson, we will introduce solving simultaneous equations using the method of substitution.

Maths

Y9 Maths

Key Stage 3

Year 9

Using substitution to solve 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

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

Manipulating systems of equations

Solving word problems

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:Using substitution to solve simultaneous equations

Core Content

Solving linear simultaneous equations algebraically: