 Hi, I'm Zor. Welcome to Unizor Education. I will talk about equations, but not in the details. It's just an introductory lecture to equations. There is a bigger portion in the algebra part, so all the details are there. This is just a very brief introduction to basically what equations are all together. All right, so let's just consider something which everybody is familiar with, very, very simple equation, and then we will think about what it means. Let's just consider this. It's a very simple equation and everybody knows that solution to this is x equal to 30. Now, what does it actually mean? What does it mean that we have an equation? Well, let me go back to a previous lecture where we're talking about operations, especially like in this particular case, unary operations. Basically, what we can say is that there is some set from which we are looking for certain element. Now, what do we know about this element? This is an unknown element, and we know that if we have performed certain operation on this element, which is subtracting five in this particular case, operation, unary operation, we will get another element, point to five in this case, which we do know. So basically, you can say that equation is something like this. You have certain operation of use for whether f, like a function, of some unknown element of certain set. Let's say in this particular case, it's a set of real numbers, and you know that the result of this operation is certain known element. In this case, it's 25. So the function f of x, or operation, unary operation, which has only one particular argument, is subtraction of five. So generally speaking, an equation is basically this. You have a unary operation, you have the result of the unary operation, which is some other element, and you have to know what's the argument of this unary operation, which causes the result of the operation to be the known element a. Okay, now, the purpose of the solution means, what does it mean to solve the equation? Well, it means to find that particular x, which if a particular known operation is applied against, will result in a. Now, you remember that unary operation might have an inverse operation. So first I have defined in the operations an operation of identity, which for any element, results in the same element. And then if there is certain operation f, let's say, and then another operation g, and the combination of these is basically the same element a. So g combination with f is identity operation, as well as f combination with g is identity operation. So if this is the case, so if for operation f exists g, which is an inverse in this particular sense, so the combination in any order that you do an identity operation, then to solve this is really very easy. How? Well, obviously what we have to do is, now, this is an element and this is an element of the same set. So let's apply the inverse operation g to both sides. So I will have g of f of x is equal to g of a. But since we know that the g is an inverse operation, g of f of x is x, it's identity. And this gives us the solution. So now we know which exactly element was taken as an argument to this particular operation to get the element a. Now, we know the operation f. Like in my example on the top, it was subtraction of five. So in theory, we might know the inverse operation. It's a simple case. Well, basically in case of subtraction of five, we know that addition of five will be an inverse operation because if you subtract something and then add exactly the same thing, you will return to the same element. But this is an easy case. So let me repeat again. If there is an inverse operation to the one which actually forms an equation, then everything is simple. You just apply this inverse operation to the result and you will get an argument. Okay, let's try to complicate the issue a little bit. Let's say that you do not have exact expression for inverse operation. Let's just make such an example. Two times x minus five equals 25. Now, this is slightly more complex. Now, in the previous case, when I had only x minus five, it was obvious that x plus five would be an inverse operation. Now, in this case, I mean, I'm sure it's simple, but it's not like obvious what exactly the operation is. However, if you would like to return to x from two x minus five, you might consider doing it in two steps. First, you inverse one operation. So you have function g which is x plus five. Then you have function h which is division by two. And what you can say is that if you first apply g of x to function to x minus five and then you apply h of this. So first you apply g which means plus five, you will have two x. Then you apply h to the result which is division by two. Then you will get x. So what I'm saying is that it's not one particular operation which you can just write, which you have found to be an inverse to f. So this is f. So you don't have really an expression for one particular operation which is an inverse to this, but you do have two different operations which combined together g of h, h of g will give the inverse operation. Now, this is just a little bit more complicated than the first case. But if you know that your inverse function can be actually found by doing this step by step inversion and you know that h of g of f is identity operation, then you basically do exactly the same thing as in the previous case, but instead of one function which is inverse, you apply two functions which in combination give you the inverse. So you do h combination with g of 25, that should give you the solution because h combination with g from the left part since this is inverse to f will give you x. So that's why when you apply it to 25, you will get x equals two. Now, how to apply this combination? Well, this is h of g of 25, which is h of g is addition of five, which is 30, and h is division by two, which is 15. So you've done basically the same thing. You've found an inverse function. Just in this case, it's a combination of two other functions. Well, by the way, if you check 15 times two, it's 30 minus 5.25 hours in this part. Now, what if it's not two? What if I will give a little bit more complex case here something like, I don't know, x minus five times three plus 17, whatever. Now, how to inverse this particular function f? Well, obviously first you subtract 17. So you have g one of x equals to x minus 17. Then you divide it by three. So you have g two of x is equal to x divided by three. So you get x minus five. Then you plus five. And g three of x is equal to x plus five. So the combination of these three, first is g one, then g two, and then g three. Applied against the result. Let's say the result is, let's say it's 19. No, I have to do it by three, so I don't have to think about this much more. So let's say 11. No, let's have it more. We don't want negative numbers. So let's have 28. Let's say, right? 28. No, not even that, sorry. 38. So it will be 21. Yeah, all right. So what if we will apply these three functions against this particular number? Well, first we apply g one from 38. Then we apply g two from the result. And then we apply g three from the result, which is equal to g three of g two of g one of 38. You subtract 17. So you will get 21, right? Now g two, g two of 21 is division by three, which is seven. And g three is addition of five. So it's 12. Well, let's just check. 12 minus five, seven times 3.1 plus 17, 38. That's okay. So in this case, I have come up with an inverse operation to this one as a combination of three more elementary operations. Again, fine. Everything works fine. Now, if you understand this, then basically the next statement would be kind of obvious. Basically you are simplifying, if I can say so, the original operation with each step you're making in analyzing it and finding what kind of components would give you an inverse operation. So first I realized, okay, this is 17. It's basically the final result. So which means that whatever this is, it's adding 17, which means my first step should be subtracting the 17. I am gradually simplifying this complex expression using certain functions, which in combination would reduce it down to x, basically. In some cases it is obvious how to do it. In some other cases it might not be that obvious. And my point is that solving equations is, I mean the real equations, not the one like this, is a real challenge sometimes, because you have absolutely no idea in the beginning how to come up with this inverse operation step by step to completely reverse whatever the initial original operation f is doing. And this is kind of a creativity, which I'm always trying to talk about. Solving problems is a creative process because you don't really know how to do it. It's not a skill which you can learn. It's an art which you should master. Now, how to develop this type of ability to solve different equations? Well, obviously there is only one way. Just solve the equations and the more the merrier and the more different they are, the more unorthodox in some way they are. Now this is a very orthodox kind of equation. I mean everybody knows how to do that. Well, try to do something more complex and I will actually, on the website on Unisor.com, I will present lots of different equations with different degrees of complexity. And yes, you do have to do that. Now, why is it very important? It's not important for mathematics. As I told many times, mathematics has basically some kind of a purpose to teach you to do something in your real life. Now, solving equations as well as solving any mathematical problems is actually preparing you to solve the real life problems because what is the problem? The problem is something which you have absolutely no idea how to do, how to solve. You have to think about it. You have to invent certain ways and there might not be actually the ways which you have already done before. I mean, it's good if you have the same problem again and again solving with just different numbers or something like this. So let's consider you have only linear equations like this and your whole life you're actually doing basically the same thing. It's a routine job and if that's what makes you satisfied, it's okay. But for those people who are not satisfied with this routine style of work who would like to see something new and to get involved in a very, very creative process solving equations as well as, for instance, geometrical construction problems in geometry. I mean, all these are really preparing you for real life problems. Okay, so I was just trying to do this introductory lecture and we talked about equations from the operations standpoint and solving equations as basically coming up with an inverse operation. A little later I will probably talk about some concrete equations like quadratic equations and some other polynomial equations. But this is just an introductory and it's supposed to serve as just warming you up to the idea that solving something which you have not solved before is really very good for you from creativity standpoint from your preparation for your real life. All right, so that's it for today, enough philosophy. And please go to Unisor.com for all other educational material. For parents and supervisors, it's an excellent tool to control the educational process of their students. They can enroll students, they can check the score obtained on exams and they are in control of basically saying that, this particular topic is complete or not complete, etc. That's it for today, thank you very much.