 Hi, I'm Zor. Welcome to a new Zor education. Let's talk about systems of equations, well, in general. I mean, systems of equations which do not really fall into some specific category, like linear or quadratic or something. They are general, so you can expect anything, but you still have to deal with it somehow. Well, there are certain particular types of systems which you still can solve. And there is no general recipe for general systems of equations. It's always a specific approach which you might actually take to solve this or that equation. And to find this approach is actually the act of creativity, ingenuity, guessing, if you wish. But again, this is basically all about this course. I'm trying to give certain tools for you to develop your creativity and ingenuity and even the guessing ability. So here are a few systems of equations which do not fall into any previous category and they are still solvable and the question is basically to find the right approach. Obviously, the more systems like that you will solve, the better your chances are to solve something. In the future, related or even not related to mathematics because it's just a general development of these qualities of being able to find your way in an unknown territory. Alright, so here they are. System number one, p square root of q minus q square root of p equals 6. p square root of p minus q square root of q equals 19. Well, as you see, it's completely different type of system of equations. So what can be done in this particular case? Well, nobody likes radicals. Nobody likes square roots or any kind of roots. So I think the first thing which comes to mind is to have a substitution. What kind of substitution? Well, obviously, square root of p might be x, square root of q might be y in which case p is equal to x square and q is equal to y square. Now, does it make our life easier? Well, in a way, yes. Because now we have p times square root of q is actually x square y minus q which is y square x equals to 6. And the p square root of p is actually x q minus y q equals to 19. So now, is it better? To tell you the truth, it looks better. And why it looks better? Because it's just polynomial. Granted, it's not quadratic, but it's still kind of polynomial. So it's easier to deal with these square roots, etc. Now, what can be done about this? Look at this. It's kind of symmetrical, right? So what I can see right from this is that this can be represented as if I will factor out x y, I will have x minus y equals to 6, right? And this also can be factored out x y and I have x square plus x y plus y square equals 19. If you don't remember this formula, you can actually go back to one of my previous lectures on induction, mathematical induction. One of the problems which I present there is exactly the representation of a to the power of n minus b to the power of n as a minus b times some polynomial of n minus first degree. Alright, now, why is this better? Well, here is why. Now, I can make another substitution. Namely, x minus y can be u and x y can be v. Why is it better? Because now this equation would look like u times v equals to 6, which is easy. And this one would look like u times. Now, let's think about this. If you square x minus y, you will get x square minus 2 x y plus y square. Now, you don't have minus 2 x y, you have plus x y, which means it's x minus y square minus 3 x y, which is u square minus 3 v equals 19, which is actually u v equals 6 and u cube minus 3 u v equals 19. Is this easier? Of course, u v can be substituted here, so it will be 3, which is 18. So u cube would be, oh, I'm sorry. This should be plus, because this is x square minus 2 x y plus y square. And we need x y, so we have to add 3 v, so it's plus. See something is wrong here. So u v is 6, so u cube is equal to 19 minus 18, which is 1. So u cube is equal to 1. I just remember there should be some simple solution. That's why it's structurally 1. You have minus here, it's definitely wrong. All right. From which we conclude that u is equal to 1. All right. If u is equal to 1, then v is equal to 6. So basically we have this. Now we go backwards. From u v, we go back to x y. And for x y, we have this system, which is very easily solvable. x is equal to y plus 1, right? And I'll substitute it to this. I will have y square plus y minus 6 equals to 0. All right. x is equal to 1 plus, so we substitute. Instead of x y plus 1, we get y square plus y. And this goes here. So we have what? y is equal to 2 and y is equal to minus 3, right? From which x is equal to 3 or x is equal to minus 3. So this is one pair, this is another pair. Now, quite frankly, by definition of square root of p, square root of p is always arithmetic value, always positive thing. So we can actually completely ignore these. And just consider these two things. So the solution is to have square root of p is equal to 3 and square root of q equal to 2. From which p is equal to 9 and q is equal to 4. These are solutions. Let's try to check it out. p is 9 times 2, that's 18 minus q is 4 times 312, 18 minus 12, 6. This is 9 times 3 is 27 minus 4 times 2, 8, 27 minus 8, 19. So everything is fine. That's the end of it. I just want to spend maybe some time to these two values. I basically discarded the negative things, negative minus 3 and minus 2. I think it's for a good purpose and we can always consider that this is the right thing. If you consider the negative things, then your p still has to be 9 and q. I mean, it's still the same thing. But that's not really what this particular problem meant. The problem meant that the square root of p is arithmetic value and always positive. Okay, next one. x square plus y equals 56, y square plus x equals 56. Well, that's interesting. What's peculiar about this? Well, obviously symmetry of these things. Is it coincidence or not that they have the same number on the right? Well, if they have the same number and it's number like 56, I think this system is just asking for subtracting one from another to get zero on the right. So let's try to do it. x square plus y minus y square minus x equals to zero. x square minus y square plus, let's do minus, minus x minus y equals to zero. Right? Minus x plus y. Or x minus y x plus y, because that's what x square minus y square is. And xy goes to the right, it will be x plus y, x minus y. Now, that's much better, right? Well, first of all, maybe you can just reduce by x minus y and have x plus y is equal to one, right? Well, yes and no. You can actually do this, but you have to check. Maybe x minus y is equal to zero. So maybe x is equal to y and this is a solution. Well, let's try. What if x is equal to y? Or x minus y is equal to zero. Let's assume that this is the case and let's just check. Well, if we will find the solution for one of them, obviously another will be exactly the same because x is equal to y, right? So let's try. x square and instead of y I will put x minus 56 equals zero. So solutions to this are minus one plus minus one plus 224, right? Four times 50 is 200 plus 24, right? So it's square root of 225, which is 15, which is minus one plus minus 15 over two. And we have the value of x is equal to, if plus it's 16 it's eight and x is equal to, sorry, it's minus eight. Minus eight and if it's plus then it's seven, right? And y is also equal to minus eight and y is equal to seven. So we have one set and we have another set. Well, let's just check just in case. Eight square would be 64 plus y, which is minus eight, which is 56. So it's correct. How about this? Seven square is 49 plus seven is 56. Correct. So we have these two solutions already just by considering that x might be equal to y and this would be satisfied. Now, another thing is if the second one is x plus y is equal to one, because if that is satisfied, then this is also true. I mean, if we assume that x equals to y we have already taken care of and x no longer is equal to y, then we can just reduce this and we will have x plus y equals to one, right? So that's what I have here, which means that y equals to one minus x substituted to this thing. So we will have x square plus y, which is minus x plus one. I switched it. Equals to 55. So actually I can put, instead of plus one and equal to 56, I will put minus 55. Right? Now, what will be in this case? Well, so the solutions to this would be one. Plus minus square root of one plus y is equal to one minus x. So it will be x square minus x plus one minus 55. Yeah, looks like it. Well, which basically means one plus minus square root of 221. The square root is not exactly the nice number. But in any case, from this value of x, we can always find out y, which is one minus x. And then basically check if this particular solution fits our equations. I think it would take a little bit more time for me to check, but you're welcome to do it yourself. If it is a solution, it is. If not, that's true too. At least we found a couple of good solutions. All right. Always don't forget to do this checking. If I don't do it, it's just because to save some time, but you really have to do the checking. So this is the value of x. We have two different values. And with each of them, you will have the corresponding value of y. So you have one plus square root of 221 divided by 2. That's x. And if y is one minus this, it would be one minus square root of 221 divided by 2. This would be another pair. And if x is equal to this, then y is equal to one minus this. It's 2, so it's one plus. That's another. But I didn't check these ones. You really should. Okay, but that's how this particular system is solved. What's the lesson to be kind of learned from this? You see, 56, 56, it just kind of prongs you. Let's subtract it. And then something like this would definitely be very easy. So it's easy to reduce this system into this type of thing. And to two cases when x minus y is equal to zero or x plus one is equal to one. And then both cases are easily solved. Okay, the next one looks more scary. Okay, here is the system. x plus a power of one-third plus y plus v power one-third equals c, x plus y equals d. Well, we do have, as you see, more or less general numbers, a, b, c, and g. So how to approach it? So I will basically talk about how to approach it, not the complete solution, because it's a, b, and c, and it will be really kind of a complicated thing. If a, b, and c, and g are chosen in some specific way, then we might get some integer solutions, etc. I didn't do it. I wanted actually to show the approach how to do it rather than get to the concrete solution. But with any specific combination of a, b, and c, and g, you can definitely go down to the solution and basically do all the things. But again, how to approach it? Well, what I suggest to do is the following. Let's just raise the top to the third, to the power of three. Now, just for you to remember, if you have, let's say, p plus q to a certain degree, it would be p cubed plus 3p squared q plus 3p squared plus q cubed. Now, I remember it, but you don't really have to remember it. You can always multiply p plus q times p plus q squared, which you probably do remember, p squared plus 2q plus q squared, and you will get this formula. So that's easy to basically, to get. Now, using this formula, which again, I don't want you to remember if you don't, you just always, you always can derive it. Now, using this formula, and I'm raising the top equation to the power of three. It will be the cube of the first one. Now, this is x plus a to the power of one-third. If I will raise it to the power of three, it will be multiplication. Three times one-third, powers are multiplied. Remember this, right? p to the power m to the power of n is equal to p to the power of mn. Again, if you don't remember this, go to the previous lectures. So, it will be the cube of the first plus three square of the first, which is x plus a. It was raised to the power of one-third, and then square, which is two-third. Again, powers are multiplied. This remains as it is, plus. Three now, x plus a remains as it is, and y plus b is squared. And finally, y plus b, one-third to the third degree, which will be one. So, this is all equal to c. Now, what's important here is the following. You see, x plus y, x plus y. So, this is basically the constant, d, right? So, we can group together this x and this y, and we will get d. So, let's just move all the constants to the right, and we will have three. x plus a, two-third, y plus b, one-third. Plus three, x plus a, one-third, y plus b, two-third. Equals c minus, minus a, minus b, and minus x plus a, with the x plus y, which is d. So, it's minus a, minus b, minus d, which is a constant, right? It's a known value. Now, how can we simplify this? So, this is kind of obvious, because we can always factor out one-third of both. And what we'll do left would be x plus a. We have one-third, and this is two-third. So, we'll have x plus a to the power of one-third. And from this, we will have y plus b to the power of one-third. That would be in parentheses. Now, remember this. So, this is equal to c, from which we conclude that three times x plus a, one-third, y plus b, one-third equals c minus a, minus b, minus c, divided by c, right? Because this is, because this is a c in the parentheses. So, it's a known value. It's constant. Now, what actually, again, is like asking us to do, what is this equation? It is asking us to, obviously, raise to the power of three to get rid of this one-third. Now, both members, x plus y, x plus a and y plus b, both are in the one-third, in the power of one-third. So, if we will raise it to the power of three, we will get rid of this, get rid of this. It will be x plus one-to-seven, that's three to the power of three, x plus a times y plus b, and that's c minus a, minus b, minus c, over c to the power of three. That's what we have reached, right? If we will raise this to the power of three. Each one of them will be to the power of three and this to the power of three. Well, this is almost there. Now, how to solve this? Well, easy. Now, this is x plus a is equal to p, y plus b is equal to q. So, what I have is x plus y is equal to d, which means p plus q is equal to a plus b plus d. And this is p times q equals to c minus a minus b minus c divided by c cubed. Now, this is a system which is very easily solvable. It's a plane quadratic equation. You just substitute p in terms of q, let's say, as a plus b plus d minus q into this and you will get a quadratic equation for q. So, this is an approach. I didn't really reach the final conclusion because the final conclusion would be a very big formula of a, b, c and g. I don't want to do this, but the problem here is basically, you know, solved as far as the approach is concerned. That's how you approach this formula. And again, how did I guess? Well, my first intention was to get rid of this one-third, obviously. And to get rid of the one-third as a power, you have to raise it to the power of three. And then whatever happens happens. In this case, something was relatively well thought through and the cards fit. Okay. So, basically, I want to end this lecture on this note. If time will allow me, I will try to put more problems of this type and solve them. And some of them will be probably as self-exercise. Well, anyway, I hope it was interesting. And again, you have to really understand that the purpose of the whole course are these particular lectures where I'm trying to approach non-standard, non-trivial examples, which you really don't have the recipe. You have to look at the problem and find a particular solution which fits this particular problem. That's what develops your creativity. Thank you very much. And don't forget that everything is on Unisor.com, free site where you can actually learn a lot. And if you register it, then you will actually have the opportunity to take exams and have the whole course under control. Thank you very much.