 Hi, I'm Zor. Welcome to a new Zor education. Problems. That's the ultimate goal of this course, solving problems. So, today I will solve a few problems which are general systems of equations, general in sense that it's not really like linear or quadratic. There's something, basically, whatever somebody's mind is set to present them. Now, obviously, you have to really try to solve these problems yourself. That's some kind of a requirement. And I will show you my solutions to these problems. Your solutions might be different, better, worse, whatever, but I do encourage you to try to spend some time and solve these problems. All of them, as usually, are on unizord.com. So, this is in a topic which I called problem number two of general systems of equations. Okay, three problems. I'm going to present them and solve them. And then, you can compare your solutions to mine. All right, problem number one. p square root of q minus q square root of p equals 2. p square q plus q square p equals 20. Okay, as you are solving problems, you probably have to develop certain feel about what does make sense, what doesn't make sense to do with equations or systems of equations. Now, in this particular case, I just don't like the radicals. I don't like square roots. I mean, they're always kind of preventing nice solutions to come up. And I do hope the solutions are nice, otherwise I wouldn't probably give this problem. So, my first thought about this particular system would be to raise the first equation to the power of two. Now, why? Because this will be square of the first, which is exactly this one. You see p square q. This is exactly the square of this one. This is exactly the square of this one. So, it might actually reduce certain things by subtracting afterwards. So, first of all, let's just raise to the power of two. It would be p square q minus two pq square root of pq, right? The product of these two. Plus square of this, which is q square p. And this is equal to four. Now, the second one, I will leave as is. Now, obviously, what actually prompts us is to subtract one from another because these will be basically deleted. So, if I subtract from this, I subtract this. I will have 16 on the right. And on the left, I will have basically two pq square root of pq. That's what I will have. Well, is it easier? Well, obviously it is easier. Why? You see, I can actually get the pq from it. Well, this is pq, this is square root of pq. So, basically, if I will square all together, I will get four pq second degree and the first degree cube is equal to square of 16 is 256, right? So, pq is equal to 64, pq cube, right? So, pq is equal to what's the four, right? Four to the cube is 64. So, I have pq is equal to four. All right, that's good. Now, what change do I have now? Well, this second equation can obviously be rewritten as this. And since pq is equal to four, pq is equal to five, right? So, this is four, so this is five. So, now I have a very familiar kind of system of two equations. p plus q is equal to five and pq is equal to four. Now, this is something which is like classical system of two equations with two variables. It's very easily converted into a quadratic equation because this is linear function of, let's say, p in terms of q, you substitute it, you will get quadratic equation. On the other hand, for every quadratic equation, you know that there are two solutions. The product of them is equal to the free member and the sum is coefficient at the first five. At the member of the first degree, so it's minus five x and x. So, this is basically a quadratic equation. Solutions of this are p and q, right? I mean, so it's basically classical theorem about quadratic equations. So, the product of roots is the free member and the sum of roots is coefficient at the first degree of unknown with an opposite sign. Well, and obviously I can guess the roots to this particular equation are kind of obvious. It's four and one, right? Two roots. The sum is indeed equal to four and sum is indeed equal to five. Well, you can use the formula. For the quadratic equation, you can use the formula. Alright, so I have four and one and obviously considering the symmetry, my solutions are either p is equal to four, q is equal to one, or p is equal to one and q is equal to four. These are two solutions to my system of two equations with two unknowns. Now, if you just look at this particular system, well, this is a multivariate polynomial of the third degree, right? Which is, you know, not obvious, and this one contains seven square roots of the unknown. So, it does look a little bit more complicated, but if you do the right thing, you basically come up with something relatively simple. That's basically what we are looking for right now, to develop certain techniques to see how you can simplify the problem from something which looks extremely complicated to more or less known problems and with no methodology to solve them. Alright, next problem. 2x plus y times xy equals minus 24. 2x minus y times xy equals minus 48. Okay. Well, again, we have multivariate polynomial of the third degree, right? This is the first degree, this is the first, and this is the first, so together it's the third degree. Not typical. You can't really use one of the formulas or something like this. So, you have to think about what to do. Well, again, just let's look at this particular system, and well, I do see that xy is here and here. Besides these numbers, 24 and 48, they're kind of look alike, one is twice as much as another. I think, again, it's a good idea to divide one into another by another, and so what we will have? Well, if I divide one into another, now xy is here and xy is here, so it will be xy in the numerator and xy will be in the denominator, so it will be basically reduced, and it's not equal to zero, by the way, so we don't have any problem with losing any solutions. Why is it not equal to zero xy? Because otherwise, these would be zero, right? And on the right, we will have minus 48 divided by minus 24, which is 2. Well, this is much easier, right? Because this is kind of a linear thing. Again, 2x plus y is not supposed to be equal to zero, otherwise, again, this particular equation would not be equal to 24, so I can safely put 2x minus y is equal to 2 times 2x plus y, right from this, which results in 2x minus y equals 4x plus 2y, 2x equals minus 3y, right? So that's what I've got. This is good. It's simple, right? Now what I can do? Obviously, I have to substitute either x or y into this particular equation, one of those. Well, let's say I will choose y. So y is equal to minus 2 third of x, right? From this equation. And now I will substitute it to each one of those. Let's take the first equation, let's say. 2x plus y, which is minus 2 third x times x, and times y again, minus 2 third x equals 24. Well, that's easy. That's basically x to the third power times some coefficient minus 24. So let's just do it. So we have 6 minus 2, so it's 4x times x. Now this minus and this minus, so I will skip the minuses, times 2x divided by 3 times 3, which is 9, equals 24. That's right? So x to the third equals 4 times 2, 8. So I can reduce by 8 both sides. I will have 3 here and 9 here. So 9 times 3, which is 27. So x is equal to 3. x cubed is equal to 27, so x is equal to 3. And y is equal to from here, minus 2. And this is the solution. Now, I didn't do the checking in the previous problem and I'm not going to do it right now, but you have to understand that whenever you do all kinds of manipulation with these complicated systems, you always have to do the checking. All right, so I just trust that these are solutions because I actually knew it before. So just substitute and take the identities. So what was interesting here? You just notice x, y here, x, y here, so you divide it and that makes much simpler relationship. All right. And the last problem I wanted to present today is a square plus b square over a plus b equals 5. 1 over a plus 1 over b equals minus 1 half. That's the system. Well, what I suggest again is something which you really have to notice. These are symmetrical things. And symmetrical things, you remember this particular system of two equations, p plus q equals something and p times q equals something. This is kind of a typical example of a system of two equations which is very easy to solve. And if you can reduce your system into something like this, you will make your life much easier. So that's what I'm kind of aiming. I would like to do something like this. Now, what's pluses and minuses? Minuses obviously does not like this. However, there are some pluses. Consider a square plus b square, for instance. Well, this is a plus b square minus 2ab, right? a plus b square minus 2ab. It's a square plus 2ab and minus 2ab. So let's reduce plus b square. So that's what it is. So I can express this in terms of this. Now, obviously, this is already in terms of this. Now, how about this? Well, if you use the common denominator, which is ab, you will have a plus b on the top, right? b over ab plus a over ab. So it's a plus b over b. And again, we are expressing in these terms. So it looks like my life would be much easier if I will indeed do this. I will do p equals a plus b and q equals a times b. And let me see what happens in this case. On the top, I will have a square plus b square and, as I said, this is p square, which is a square plus 2ab plus b square. So to get rid of 2ab, I have to subtract 2q, right? So that's on the top, right? So let me write it down. a square plus b square equals a plus b square minus 2ab, which is p square minus 2q in these terms. And that's what I can bring. On the bottom, I have plane p, which is a plus b. That's equal to 5. Here, the common denominator is ab, and ab is my q. And on the top, I have b plus a, which is p. Equals minus 1 half. Does this look simpler? Oh, absolutely. It's absolutely simpler. Now, this is basically a very simple thing because q from here minus q equals 2p. So q is equal to minus 2p, right? That's from this equation. Now, we can substitute it here. So 2q minus 2q is actually minus 2q is what? 4p, right? So minus 2 times q, that's 4p. So what I have is p square minus 4p divided by p equals 5. All right. Now, can p be equal to 0? Is this a plausible solution? Well, no. p is a plus b, so it's definitely prohibited here. However, here we have a plus b on the top, which is p. So we would never have minus 1 half if p is equal to 0. So we can definitely say that p equals 0 is not a solution. That's why we can basically reduce it by p because we will have p minus 4 equals 5. Oh, I'm sorry. It's plus 4. Minus times minus. Minus 2 times minus 2 is plus 4. So p plus 4 is equal to 0. Which is p equals 1, so p is equal to 1 and q is equal to minus 2. So that's what we have, right? Now, let's get back to what we have started with, which is this. What are p and q? So p and q are these. So what we have right now is a plus b is equal to 1 times b is equal to minus 2. And this is, again, a very, very much the same system of two equations with two variables. And, obviously, the solutions are x square minus x minus 2 equals to 0. The product of two solutions is the free member and the sum of two solutions to this equation is the coefficient at x to the minus sign. So the roots of this equation are a and b. And, well, I guess I can... So it's 2 and minus 1, right? So I just guessed. 2 and b is equal to minus 1. Or vice versa, a is equal to minus 1 and b is equal to 2. Now, both are solutions to original system of equations. Now, obviously, the symmetry between a and b is the course of this. So that's it. Yes, that's the right solutions. You can do the checking. As you saw in every system of equations which I have presented, there is something peculiar which you just have to see and use to solve the equation. Well, again, as I was saying, this is the purpose of all these exercises. And try to go through these same problems again if you didn't do it before. And good luck for any future problems. Thank you.