 Hi, I'm Zor. Welcome to new Zor Education. Today we will solve a couple of systems of quadratic equations. Now, before getting into it, I'd like to assure you that most of you will never in your life have to solve systems of quadratic equations. So, in this particular case, I can say basically exactly the same as I said many times before. All these math problems are needed only for one very, very important purpose, to make you more creative, more logical, to develop your mind towards certain analytical considerations. Because in real life, you will always have these problems where your creativity, your logic, your analytical skills will be needed for completely different purposes, maybe not even related to mathematics, maybe in literature, music, wherever. So, again, consider all these math problem solving exercises. In exactly the same fashion, you consider your physical exercises in the gym, those develop your muscles, your stamina, math problems develop your mind. Okay, with this, let me start. Now, this lecture is part of the course called Maths for Teens, presented on Unisor.com. I suggest you to watch this lecture from the website, from Unisor.com. You go to this side, and then there is a Maths for Teens course. This is part of the algebra, and among algebra topics, you will find this one. Now, the reason for this is that every lecture on the website has, on the side, complete textual notes, basically repeated whatever I am saying here in the video, but sometimes in more details. For example, today, I am not planning to solve all these problems down to the very, very end, with the answer and the checking, etc. These things I actually include into the textual part. During the video, I will just try to basically show the way how to do it, and then you should actually do it yourself, complete it yourself. That's exactly the kind of approach which I would recommend you. As soon as you know how to do it, just do it yourself, whether I did it in front of the board or not. Alright, so let's start. We are talking about quadratic systems of quadratic equations, and the previous lecture basically explained what it is, so I am just basically solving the problems. Okay, the first problem is I have a system of three equations with three unknowns. Okay, this is my system. Well, let's think about how to approach basically solving this system. First, let's try to do it straight, which means basically we can resolve, let's say, the third equation for z and substitute into a previous one, and we'll see what happens. So z would be equal, z, not y, z would be equal 20 over y. Now, if I will substitute it in the previous one, I will have a new system of equation, and x times z would be 20x over y equals minus 15. Now, just a side note, obviously this doesn't make sense if y is equal to zero, but considering our equation, we definitely know that y is equal to zero is not a solution because otherwise we would have here zero and here zero, right? So y is not a zero, we don't lose anything by substituting. Very important, sometimes we can lose. Okay, from this, we can always resolve it for y is equal to minus 2015's, which is 4 third x, and substitute it into this. And finally, we get what? We get x times y, which is minus 4 third x square equals to 12, from which x is x square is equal to minus 9, correct? I mean minus 12, I'm sorry. So it's plus 9. I see something strange, x square is equal to negative. We are solving equations usually among real numbers only, at least right now. Okay, which is x is equal to plus or minus 3. So we got that. So we have two different solutions. The solution would be x equal to 3, y equals minus 4, and z is equal to minus 5. The second solution is x is equal to minus 3, y is equal to 4, and z is equal to 5. So we have two different solutions. 3 minus 4 minus 5, and minus 3, 4, and 5. This is the end of it, right? No, we have to check. Don't forget, always check your solution. And in this particular case, well, let me just check, for instance, this one as an example. The first equation was x, y is equal to minus 12, x, y is equal to minus 12, x, z minus 15, and y, z, 20, correct? The second one is also checking. Okay, that's it for the first problem. That's the easiest part. However, regardless of how easy it is to solve this one, I would like to point out to another method of solving this equation, which in my personal view is, it's maybe a little clever on the side of this, like a trick, but if you see this particular method, it's really very good. And here's what I suggest. Let me just return it again. You see, they're kind of symmetrical. What if I will multiply all of them? I will have two x's, two y's, and two z's. I have x, y, z squared equals to what? 15 times 20 is minus 300 times 12. That's minus and minus 3, 600, which is x, y, z is equal to 36, 6, 0. Now, if I will divide this by this, I will have only z left, right? So z is equal to 6 divided by... By the way, it's plus or minus. So let's consider a plus. So 6 divided by 12 would be minus 5. Divided by minus 15 will have minus 4, right? So x, y, z divided by... So y is equal to minus 4, and 60 divided by y, z will have x left, right? So x is equal to 3. And if I will use minus 60, divided by minus 12 would be 5. Divided by minus 15 would be 4, and divided by 20 would be minus 3. The same two answers, two triplets. But again, in my case, just my personal opinion, this looks more stylish, if you wish. I might actually use some words like beautiful or whatever it is. But anyway, it's definitely a more elegant solution. So I would like to point out that this is, again, in my view, a better solution. Some people might not agree with me, but it's a matter of taste, if you wish. But sometimes this kind of a trick, basically, is the only way you can solve the problem. Okay, now, next problem. x plus 2 times y plus 2 equals minus 10. x, y plus x, y minus 3 is equal to 15. Okay, well, that's complicated. Let's put it this way. At least from the first look at it. However, well, since I don't know how to solve this equation, there is no kind of general approach. There is no such thing as going straight forward, like, find out what's the value of y in terms of x, substitute it here, it would be too much, too complicated. So I have to look for something and that's the beauty of these type of problems. There is no prescription how to solve this type of equation. You have to really think about what's the peculiar about this equation, which will allow to basically use some kind of a trick. So the whole lecture, this lecture is about little tricks which you can use and that's what develop your creativity. So let me think about it. What's kind of a, you know, brings me to the solution that this is true and this is true. So it's kind of symmetrical. You see, if I will replace x and y, it will be exactly the same equation. So I have to look for something which is symmetrical relative to x and y, like x plus y or x multiplied by y. These are very typical symmetrical combinations of variables. So they're plus or they're multiplication. Now, but this is not actually x plus or x times y, right? However, but let me just open the parenthesis in the first one. What do we have? I have xy plus 2y plus 2x plus 4. You see, this is xy, symmetrical, and this is 2 times x plus y, right? Okay, now I see that there are only symmetrical functions here. x plus y, x plus y, x times y, x times y. And what's typical in this case, so let me just replace this equation with this. Now, now what's very important in this particular case and what actually brings us closer to a solution that I can always use with this symmetrical function. I will replace x plus y with, let's say, a and x times y with b. Because this is multiplication, my higher degree equation will have a lower degree if I will substitute it this way. Let's see what happens. So this is a times b minus 3 equals 15. And this is b plus 2a equals minus 14, right? Now, this is better because here I can just use the substitution. I will put b is equal to minus 14 minus 2a, substitute into this, and I will have a quadratic equation for a, which I can solve. Knowing a, I know b, and if I know these two, I can very easily solve this one by less elegant method of basically substituting, finding y in terms of x and substituting it into here. We will have another quadratic equation. Or I remember the theorem about quadratic equation z squared plus pz plus q The roots of this equation are z1 times z2 is equal to q and z1 plus z2 is equal to minus p, right? So knowing a and b, I can basically construct an equation, which is z squared minus az plus b, solve this equation, and its solution would be x and y. Now, in every case on this way, you will have multiple solutions because you have quadratic equation here, quadratic equation here, and you always have to consider both roots. Again, more detailed explanation is in the text for this, but you understand the principle, the road along which you can really go to obtain the solution. You notice the symmetry. You use the symmetrical functions as new arguments, which simplified our system into a plain quadratic equation, basically. And then another quadratic equation, so you will have multiple solutions. You consider these two, and with each of these, you consider two different roots of this equation, but that would give you a complete set of solutions. And don't forget to check all of them. Okay, that's my second problem. Well, I deliberately do not go to all the details of these calculations and solutions for two reasons. Number one, I can mix up myself and make mistakes. And number two, I would like actually you to do it very accurately very scrupulously to get all the solutions, do the checking, and verify it against the textual description for this lecture, which I have. Next. Next is xy squared minus x. Minus 2y equals minus 2 xy plus y. Minus 2. Okay, now. It doesn't look in any way symmetry or anything else, et cetera. It looks actually scary, because in both cases it's kind of difficult to... Well, you can actually get x from here that would be some kind of a really complex formula and then substituted here would be a lot. So we need again some kind of a peculiarity of this system which will allow us to simplify it. And peculiarity of this system is the following. Let's take x out of the first. We will have y squared minus 1. y squared minus 1 I represent as y minus 1 and y plus 1. Okay, so that's x squared minus x, y squared minus x. This gives you x and y squared minus 1 will be y squared, will be y minus 1, y plus 1. Now here I will get everything into this. I will have what? 2 minus 2y. And that's interesting. 2 minus 2y, you can factor out 2 will be 1 minus y. So I already have y minus 1, right? So if I will put minus 2 I will have y minus 1 equals to 0. That's 1, right? And y minus 1, y minus 1. So the first solution is already have y is equal to 1. y is equal to 1, I substitute into this and this. I will have let's say from this equation I will have x plus 1 is equal to 2, which means x is equal to 1. Okay, that's a solution. And you can check it here. This is 1 minus 1 is 0 minus 2 will be minus 2 minus 2 and this will be 1 plus 1 is equal to 2, right? So we have one solution. After which we can divide by this. Again, if we just divide by this without talking about y is equal to 1 that means we will lose a solution. So we have not lost a solution and what's left? Now we have x times y plus 1 minus 2 is equal to 0 from this. And this stays x, y plus y is equal to 2. So let me just simplify. Instead of this I will have x, y plus x equals 2, right? 2 goes here and this is I open. Now, what's peculiar about this? Well, it's basically the same thing except this. So if I will subtract one from another I see x is equal to y, right? y minus x will be equal to 0. Okay, I will substitute into any one of them. Let's say for this one. I will have x square plus x is equal to 2 and I also have a quadratic equation with two solutions and each of them will give me another pair of solutions to original equation. Okay, so what was important in this case? We have noticed that this will contain y square minus 1 if I will factor out x which is divisible by y minus 1 and this also is divisible by y minus 1 and that gave me the first root after which everything was much simpler. Next. This looks even scarier. Number one, which is absolutely obvious x plus y, x plus y, x y plus y, x y plus y and the first very, very obvious substitution is x plus y. I will substitute this letter. Let's say u and x minus y equals v and my system would be already much simpler. 1 over u plus 1 over d is equal to 36 fifths. u square plus v square is equal to 13 eighths. It looks much simpler, right? It's a must, it's obvious. What's next? Well, again, you see symmetry if you will exchange u and v, you will have exactly the same. So we have to really look, if you have a symmetry, you might look for u plus v and times v as new variables. But somehow we have to really go into this. Okay, let's say this is p and this is q. Again, new variables. Will it be easier? Well, if you will get it to a common denominator which is u v, you will have u plus v. Now u plus v is p. u times v will be q. So that's my first equation. Right? What's my second equation? Well, u square plus v square reminds to what? u plus v square. U plus v square is a little bit more than that. There is a 2u v, right? So I can say that u square plus v square is p square. That would be u square plus 2u v which is 2q plus v square. So I will subtract 2q. Subtract 2u v. So p square u plus v square would be u square plus v square plus 2u v. And then I subtract minus 2u v which is minus 2q. 13, 18. Is it easier? Well, absolutely. Because I can get q from here substituted to here and it will be a quadratic equation. So q would be equal to 536 of p. I will substitute it here and I will have a square p square minus something p equals to something. So it's a quadratic equation which can be solved for p. Knowing p will get q, etc. So that's the way how to do it. And again, I don't go through all the details. My purpose was to push you towards one particular direction. Number one, you notice that you can simplify the whole thing. Number two, you realize that there is a symmetry here. And whenever there is a symmetry you can use something like this. That's it. That's it for today. Thank you very much. Check the notes for this lecture. Again, you go to unizord.com Myth for teens course. Choose algebra topic and then you will find this particular lecture. Alright, thank you very much and good luck.