The purpose of this video is to look at the solution of elementary simultaneous equations. Before we do that, let's just have a look at a relatively straightforward single equation. Equation we're going to look at 2X minus Y equals 3. This is a linear equation. It's a linear equation because there are no terms in it that are X squared, Y squared or X times by Y or indeed ex cubes. The only terms we've got a terms in X terms in Y and some numbers. So this represents a linear equation. We can rearrange it so that it says why equal something, so let's just do that. We can add Y to each side so that we get 2X. Equals 3 plus Yi. Did say add why two each side and you might have wondered what happened here. Well, if I've got minus Y and I add why to it, I end up with no wise at all. Here we've got two X equals 3 plus Y, so let's take the three away from each side. 2X minus three equals Y. So there I've got a nice expression for why. If I take any value of X. Let's say I take X equals 1, then why will be equal to two times by 1 - 3, which gives us minus one. So for this value of XI, get that value of Yi can take another value of XX equals 2, Y will equal. 2 * 2 - 3, which is plus one. Another value of XX equals 0 Y equals 2 times by 0 - 3 will 2 * 0 is 0 and that gives me minus three. So for every value of XI can generate a value of Y. I can plot these as point so I can plot this as the .1 - 1 and I can plot this one on a graph as the .21 and this one on a graph as the point nort minus three. So let's just set that up. Pair of axes. Let's mark the values of X that we've been having a look at. So that was X on there and why there? And let's put on the values of why that we got. When X was zero hour value of why was minus three? So that's. There. When X was one hour value was minus one. And when X was two hour value was one. Those three points lie on a straight line. Y equals 2X minus three, and that's another reason for calling this a linear equation. It gives us a straight line. OK. You've got that one. Y equals 2X minus three. Supposing we take a second one 3X plus two Y equals 8, a second linear equation, and supposing we say these two. Are true at the same time. What does that mean? Well, we can plot this as a straight line. Again, it's a linear equation, so it's going to give us a straight line. Now I don't want to have to workout lots of points for this, So what I'm going to do is just sketch it in quickly on the graph. I'm going to say when X is 0. And cover up the Exterm 2 Y is equal to 8, so why must be equal to four which is going to be up there somewhere? And when? Why is 03 X is equal to 8 and so X is 8 / 3 which gives us 2 and 2/3. So somewhere about their two 2/3 and we know it's a straight line, so we can get that by joining up there. This is the equation 3X plus two Y equals 8. So what does it mean for these two to be true at the same time? Well, it must mean it's this point here. Where the two lines cross. So when we solve a pair of simultaneous equations, what we're actually looking for is the intersection of two straight lines. Of course, it could happen that we have one line like that. And apparel line. They would never meet. And one of the examples that we're going to be looking at later will show what happens in terms of the arithmetic when we have this particular case. But for now, let's go back and think about these two. How can we handle these two algebraically so that we don't have to draw graphs? We don't have to rely on sketching, we can calculate which is so much easier in most cases that actually drawing a graph. So let's take these two equations. And we're going to look at two methods of solution, so I'm going to look at method one. Now, let's begin with the original equation that we had two X minus. Y is equal to three and then the one that we put with it 3X plus two Y equals 8. Our first method of solution, well, one of the things to do is to do what we did in the very first case with this and Rearrange one of these equations. It doesn't matter which one, but we'll take this one. So that we get Y equals. And we know what the result for that one is. It's Y equals 2X minus three. So that's equation one. That's equation. Two, so this is now, let's call it equation 3 and we got it by rearranging. 1. What we're going to do with this is if these two have to be true at the same time, then this relationship must be true in this equation, so we can substitute it in, so let's. Substitute 3 until two so we have 3X plus. Two times Y. But why is 2X minus three that's equal to 8? And you can see that what we've done is we've reduced. This. To this equation giving us a single equation in one unknown, which is a simple linear equation and we can solve it. Multiply out the brackets 3X plus 4X minus 6 equals 8. Gather the excess together. 7X minus 6 equals 8. At the six to each side, Seven X equals 14, and so X must be 2. That's only given as one value. We need a value of Y, but up here we've got an expression which says Y equals and if we take the value of X that we've got and substituted in. Therefore, why will be equal to 2 * 2 - 3 gives US1? And so we've got a solution X equals 2, Y equals 1. Are we sure it's right? Well, we used this equation which came from equation one to generate the value of Y. So if we take the values of X&Y and put them back into here, they should work, should give us the right answer. So let's try that. X is 2. 3 times by two is 6 plus, Y is 1 two times by one is 2 six and two gives us eight. Yes, this works. This is a solution of that equation and of that one. So this is our answer to the pair of simultaneous equations. Let's have a look at another one using this particular method. The example we're going to use is going to be said. Open X. +2 Y equals 47 and five X minus four Y equals 1. Now. We need to make a choice. We need to choose one of these two equations. And Rearrange it so that it says Y equals or if we want X equals. The choice is entirely ours and we have to make the choice based upon what we feel will be the simplest and looking at a pair of equations like this often difficult to know which is the simplest. Well, let's pick at random. Let's choose this one and let's Rearrange Equation too. So we'll start by getting X equals this time. So we say 5X is equal to 1 and I'm going to add 4 Y to each side plus 4Y. Now I'm going to divide throughout by the five so that I have. X on its own. Now I've got to divide everything by 5. Everything so I had to put that line there to show that I'm dividing the one and the four Y. So this is a fraction. I'm sure you can tell this is not going to be as easy as the previous question was. In fact, it's going to be quite difficult because I have to take this now and because it came from equation too. I'm going to have to take it and substitute it back into equation one, and this isn't looking very pretty, so let's give it a try sub. 3. Until one. So I have 7X but X is this lump of algebra here 1 + 4 Y all over 5. +2 Y equals 47. I can see this is becoming quite horrific. Multiply throughout by 5 why? Because we're dividing by 5. We want to get rid of the fraction. The way to do that is to multiply everything by 5 and it has to be everything. So if we multiply that by 5 because we're dividing by 5, it's as though we actually do nothing to the 1 + 4 Y. That leaves a 7 * 1 + 4 Y. We need five times that that's ten Y and we have to have five times that remember, an equation is a balance. What you do to one side of the balance you have to do to the other. If you don't, it's unbalanced. So we're multiplying everything by 5. So 5 * 47 five 735, five falls of 22135 altogether. Now we need to multiply out the brackets 7 + 28 Y plus 10 Y equals 235. So we take this equation. Write it down again so that we can see it clearly. Now we can gather these two together gives us 38Y. And we can take Seven away from each side, which will give us 228. Exactly big numbers coming in here 228 / 38 'cause we're looking for the number which when we multiplied by 38 will give us 228 and that's going to be 6. So we've established Y is equal to 6. Having done that, we can take it and we can substitute it back into the equation that we first had for X. So remember that for that we had X was equal to. And what we had for that was 1 + 4 Y all over five. We substitute in the six, so we have 1 + 24 or over 5 and quickly we can see that's 25 / 5. So we have X equals 5. So again we've got our pair of values. Our answer to the pair of simultaneous equations. We haven't checked it though. Now remember that this came from the second equation, so really to check it we've got to go back to the very first equation that we had written down that one. If you remember was Seven X +2, Y is equal to 47. So let's just check 7 * 5. That gives us 35 + 2 * 6. That gives us 12, so we 35 + 12 equals 47. And yes, that is what we wanted, so we now know that this is correct, but I just stop and think about it. We got all those fractions to work with. We got this lump of algebra to carry around with us. Is there not an easier way of doing these? Yes there is. It's useful to have seen the method that we have got simply because we will need it again when we look at the second video of simultaneous equations, but. That is a simple way of handling these, so let's go on now and have a look at method 2. Now this method is sometimes called elimination and we can see why it gets that name and this is the method that you really do need to practice and become accustomed to. So let's start with the same equations that we had last time. And see. How it works and how much easier it actually is? OK method of elimination. What do we do? What we do is we seek to make the coefficients in front of the wise. Or in front of the axes. The same. Once we've gotten the same, then we can either add the two equations together or subtract them according to the signs that are there. By doing that, we will get rid of that particular unknown, the one that we chose. To make the coefficients numerically the same. So. This one what would we do? Well, if we look at this and this. Here we have two Y and here we have minus four Y. So if I were to double that, I'd have four. Why there? And it's plus four Y Ana minus four Y there, and that seems are pretty good thing to do, because then they're both for Y. One of them is plus and one of them is minus. And if I add them together they will disappear. So let me just number the equations one and two. And then I can keep a record of what I'm doing. So I'm going to multiply the first equation by two and that's going to lead Maine to a new equation 3. So let's do that. 2 * 7 X is 14X plus two times, that is 4 Y equals 2 times that, and 2 * 47 is 94. Now equation two. I'm leaving as it is not going to touch it. Now I've got two equations. This is plus four Y and this is minus four Y. So if I add the two equations together, what happens? I get 14X plus 5X. That's 19 X. No whys. 'cause I've plus four Y add it to minus four Y know wise at all equals. 95 and so X is 95 over 19, which gives me 5 which if you remember, is the answer we have to the last question. Now we need to take this and substitute it back. Doesn't matter which equation we choose to substitute it back into. Let's take this one. X is 5, so five times by 5 - 4 Y equals 1. And so we have 25 - 4 Y equals 1. Take the four way over to that side by adding four Y to each side. So that will give us 25 is equal to four Y plus one. Take the one away from its side, 24 is 4 Y and so why is equal to 6 and we've got exactly the same answer as we had before. And let's just look. How much simpler that is? How much quicker that answer came out. One thing to notice. Well, two things in actual fact. First of all, I try to keep the equal signs underneath each other. This is not only makes it look neat, it enables you to see what it is you're doing. Keep the equations together so the setting out of this work actually helps you to be able to check it. Second thing to notice is down this side. I've kept a record of exactly what I've done multiplying the equation by two, adding the two equations together. That's very helpful when you want to check your work. What did I do? How did I actually work this out? By having this record down the side, you don't have to work it out again. You can see exactly what it is that you did. Now let's take a third example and again. Will solve it by means of the method of elimination. Just so we've got a second example of that method to look at three X +7 Y is 27 and 5X. +2 Y is 60. OK, we've got a choice to make. We can make either the coefficients in front of the axe numerically the same, or the coefficients in front of the wise. Well, in order to do that, I'd have to multiply the Y. Certainly have to multiply this equation by two to give me 14 there and this one by 7:00 to give me 14 there. How do I make that choice? Well? Fairly clearly 2 times by 7 is 14, so 1 by 1, one by the other. But I don't really like multiplying by 7 difficult number. I prefer to multiply by three and five, so my choices actually governed by how well I think I can handle the arithmetic. So let's multiply this one by 5 and this one by three will give us 15X an 15X number. The equations one. 2. And I'll take equation one and I will multiply it by 5 and that will give me a new equation 3. So multiplying it by 5:15 X plus 35 Y is equal to 135. And then equation two, I will multiply by three and that will give me a new equation for. Oh, here we go. Multiplying this by three 15X. Plus six Y is equal to 48. These are now both 15X and they're both plus 15X. So if I take this equation away from that equation, I'll have 15X minus 15X no X is at all. I live eliminative, the X, I'll just have the Wise left, so let's do that equation 3 minus equation 4. 15X takeaway 15X no axis 35 Y takeaway six Y that gives us 29 Y. And then 135 takeaway 48? And that's going to give us A7 their 87 altogether. And so why is 87 over 29, which gives us 3? Having got that, I need to know the value of X so I can take Y equals 3 and substituted back let's say into equation one. So I have 3X plus Seven times Y 7 threes are 21 is equal to 27 and so 3X is 6 taking 21 away from each side and access 2. Check this in here 5 twos are ten 2306 ten and six gives me 16 which is what I want so I know that this is my answer. My solution to this pair of simultaneous equations and again look how straightforward that is. Much, much easier than the first method that we saw. Also think about using letters as well. If we've got letters to use instead of coefficients the numbers here. So we might have a X plus BY. Again, this is a much better method to use. Again, notice the setting down keeping it compact, keeping the equal signs under each other and keeping a record of what we've done. So do something comes out wrong, we can check it, see what we are doing. Now all the examples that we've looked at so far of all had whole number coefficients. They might have been, plus they might be minus, but they've been whole number, and everything that we've looked at as being in this sort of form XY number XY number. Well, not all equations come like that, so let's just have a look at a couple of examples that. Don't look like the ones we've just done. First of all, let's have this One X equals 3 Y. And X over 3 minus Y equals 34 pair of simultaneous equations. Linear simultaneous equations again 'cause they both got just X&Y in an numbers, nothing else, no X squared's now ex wise etc. We need to get them into a form that we can use and that would be nice to have XY number. So let's do that with this One X equals 3 Y, so will have X minus three Y equals 0. This one got a fraction in it. Fractions we don't like, can't handle fractions. Let's get rid of the three by multiplying everything in this equation by three. So will do that three times X over 3 just leaves us with X. Three times the Y minus three Y equals 3 times. This going to be 102. Problem. These two bits here are exactly the same. But these two bits are different. What's going to happen? Well, clearly if we subtract these two equations one from the other, there won't be anything left this side when we've done the subtraction X from X, no access. Minus 3 Y takeaway minus three Y know why is left, and yet we're going to have 0 - 102 equals minus 102 at this side. In other words, we're gonna end up with that. Which is a wee bit strange. What's the problem? What's the difficulty? Remember right back at the beginning when we drew a couple of graphs? In the first case we had two lines that actually crossed, but in the second case I drew 2 lines that were parallel. And that's exactly what we've got here. We have got 2 lines that are parallel because they've got this same form. They are parallel lines so they don't meet. And what this is telling us is there in fact is no solution to this pair of equations because they come from 2 parallel lines that do not meet no solution. There isn't one fixed point, so we would write that down. Simply say no solutions. And it's important to keep an eye out for that. Check back, make sure the arithmetic's correct yes, but do remember that can happen. Let's take just one more final example, X over 5. Minus Y over 4 equals 0. 3X plus 1/2 Y equals 70. Now for this one. We've got fractions with dominators five and four, and we need to get rid of those. So we need a common denominator. With which we can multiply everything in the equation and those get rid of the five in the fall. The obvious one to choose is 20, because 20 is 5 times by 4. Let us write that down in falls 20 times X over 5 - 20 times Y over 4 equals 0 be'cause. 20 * 0 is still 0. Little bit of counseling 5 into 20 goes 4. 4 into 20 goes 5. So we have 4X minus five Y equals 0. So that was our first equation that was our second equation. This one is now become our third equation. So equation one has gone to equation 3. Let's look at equation two. Now that we need to deal with it, it's got a half way in it. So if I multiply every. Anything by two. This will become just why? So we have 6X. Plus Y equals 34 and so equation two has become. Now equation for. We want to eliminate one of the variables OK, which one well I'd have to do quite a bit of multiplication by 6:00 AM by 4. If it was, the ex is that I wanted to get rid of look, there's a minus five here and one there, so to speak. So if we multiply this one by five, will get these two the same. So let's do that 4X minus five Y equals 0, and then times in this by 5. 30X plus five Y equals and then we do this by 5, five, 420, not down and two to carry 5 threes are 15 and the two is 17, so that gives us 170 and now we can just add these two together. So equation three state as it was equation for we multiplied by 5. So that's gone to equation 5 and now. Finally, we're going to add together equations three and five, and so we have 34 X equals 170 and wise have been illuminated. 34 X is 170 and so X is 170 / 34 and that gives us 5. We need to go back and substituting to one of our two equations. It's just have a look which one? Doesn't really matter, I think. Actually choose to go for that one. Why? because I can see that five over 5 gives me one, and that's a very simple number. Might make the arithmetic so much easier. So we'll have X over 5 minus Y. Over 4 equals 0. Take the Five and substituted in. 5 over 5. That's just one, and so I have one takeaway Y over 4 equals 0, so one must be equal to Y over 4. If I multiply everything by 4I end up with four equals Y. So there's my pair of answers X equals 5, Y equals 4 and I really should just check by looking at the second equation now, remember. 2nd equation was 3X plus 1/2 Y equals 17. 3X also, half Y equals 17. So let's substitute these in. X is 5, three X is. Therefore AR15, three fives plus 1/2 of Y. But why is 4 so 1/2 of it is 2. That gives me 17, which is what I want. Yes, this is correct. Let's just recap for a moment. Apparel simultaneous equations. They represent two straight lines in effect when we solve them together, we are looking for the point where the two straight lines intersect. The method of elimination is much, much better to use than the first method that we saw. Remember also in the way that we've set this one out. Keep a record of what it is that you do. Set you workout so that the equal signs come under each other and so that at a glance you can look at what you've done. Check your working. Finally, remember the answer that you get can always be checked by substituting the pair of values into the equations that you began with. That means strictly you should never get one of these wrong. However, mistakes do happen.