 Hi, I'm Zor. Welcome to Unizor education. Today, we will talk about equations. This is the problem number one, and it's not really a problem. It's a very easy exercise, but I would like to use it as an opportunity to talk about invariant transformations because they are in the heart of any kind of equations solving. The first problem contains actually three equations, which I put down here, and I'll talk about each of them separately. Again, they're very easy and it's not really for solving these equations that I'm talking about. It's just to illustrate that invariant transformations are extremely important, and how to properly apply them. Basically, this is all about it. Equation number one, let me put it here, 2 times x plus 4 minus 10 equals 0. As you know, invariant transformations can be applied to both sides of equation and because these transformations are invariant, they do not change the solutions of the equation. So instead of solving original equation, we can solve transformed equation without any lost or gained solutions. Now, obviously, what kind of invariant transformations can be applied to this? So I will just use this language of invariant transformations to explain how to solve this equation. Okay. The first invariant transformation is transformation T, which converts anything into anything plus 10. Now, why is this transformation invariant because it always has an inverse transformation, and it's one-to-one correspondence between an argument and a function in this particular case? So if I apply this transformation to both sides of the equation, I will get 2 times x plus 4 minus 10 plus 10 equals 0 plus 10, which I can obviously reduce using associative law of addition. Let's just be specific in this particular case, because this is one argument minus another plus another. If we can change the sequence, we can use the result of the latest addition and then apply it to the left one. So basically to put it on a proper rigorous basis, if you wish, I can say that I can change this and I can put parentheses around the latest. That's using the associative law. We have to understand that it's not just from the blue or whatever. The reason associative law of addition, which we can use this equals to, in this case, 0 plus 10 is calculated as 10. Since I put parentheses around it, I can execute this first, which obviously reduces to 0, and I have 2 times x plus 4 equals 10. We use an invariant transformation and associative law of addition to get to this. Next invariant transformation, which converts any number into this number divided by 2. Now, is this transformation invariant? Yes, because there is always the reverse one. That's the multiplication by 2. As you understand, division is always invariant as long as divisor is not equal to 0. So, what do we do in this particular case? Okay, let's divide it by 2. We get 2 times x plus 4 divided by 2 is equal to and divided by 2. Okay, I'll continue here. Now, what do we do? Well, what we can do here is two things. We can use the commutative law of multiplication and transfer these, which will give us x plus 4 times 2 divided by 2 equals to 5. I divide it directly. And now, again, since division is actually a multiplication, as you remember, so basically any multiplication by an inverse number. So, in this particular case, we can use the same type as before. We can use associative law of multiplication and put parenthesis around this. So, 2 divided by 2 gives us 1 and 1 multiplied by x plus 4 will be x plus 4 equals to 5. And obviously, the next invariant transformation will be to subtract 4 from those sets, x plus 4 minus 4 equals to 5 minus 4. Again, using associative law, I put parenthesis around it, so I will get x equals to 1. Now, this is a solution. Okay? Question number one, do I have to check this solution? Well, to tell you the truth, checking is always a good thing to do. In this case, however, it's not really mandatory. It's a good thing to do, but it's not mandatory. Why? Because every transformation which we did, we did three transformations here, all of them are invariant. So, considering these are invariant transformations, we did not lose any solutions and it would not gain any new solutions, which are not really solutions to original equation. That's why in this particular case, it's not mandatory, just a desirable action to do the checking. Well, and if you do the checking, it will put x equals to 1, it will be 5 times to 10 minus 10, zero ever, so it's fine. Okay. So, I went to really very, very small details of how to solve this particular equation as a demonstration of what actually is happening behind very simple rule, like, okay, we can transfer 10 into the right side of the equation with a changed sign. This is just a very fast thing and then we have two times x plus four equals to 10, immediately. Then let's divide both sides by two and let's subtract four. So, basically this is exactly what we have done, but we did it in more details to show that there is some theoretical background for all these seemingly obvious rules. Okay, so let's do the next one, hopefully much faster than this one. And again, all of these are no more than illustrative examples of how to use invariant transformation. It's not like difficult equations which I'm trying to solve. All right, so the next one will be x plus four divided by two minus 10 equals to zero. Okay, I'll do it faster now. First transformation obviously is we transform by adding 10. That's the transformation, okay? What we get is x plus x plus four divided by two equals 10. I immediately put the result obviously. Minus 10 plus 10 will be zero and plus 10 on the right side will be 10. Next transformation is multiplication by two. Invariant transformation because the multiplication by anything not equal to zero is invariant. So, I will have x plus four divided by two multiplied by two with the x plus four and this is equal to 20, correct? I multiply by two, correct. And the next third transformation is I subtract four from both sides and I will get x equals to 16. Not necessary, not mandatory, but still a good thing to do. Checking 16 plus four, 20 divided by two, 10 minus 10, zero, everything is fine. Okay, that's the problem number two and the problem number three. Equating number three, I should really say, just a little bit more writing x plus four divided by two minus 10 multiplied by x plus one equals zero. Okay, now there are many ways to solve this particular equation. Probably the best way is first simplify the left part and simplification is as follows. If we will divide x plus four by two separately, x and four, we are using a distributive law of multiplication and addition. In this case, it's division. So the distributive law says that this is equal to x divided by two plus four divided by two. That's the distributive law. Again, we have to understand that simple rules have their theoretical foundation. Here, the same thing, distributive law of multiplication versus addition. So it's 10 times x minus 10 equals to zero. Next, next we obviously have to use the commutative and associative laws to change things around and put them together. So, what I will have is the following. x divided by two minus 10x. Four divided by two is two minus 10 equals zero. Now, now we can do the transformation, very easy transformation. T, we multiply by two. Y by two just to get rid of this. Now, what we will have is the following. On the left part, if I multiply by two and again use a distributive law of multiplication versus addition and subtraction, I will have x divided by two minus 10x plus. I will put it together, it will be minus eight. I multiply by two equals to zero multiplied by two. Which is, now we're using distributive law again. Two times x divided by two is x. Two times minus 10 will be minus 20x. And two minus times minus eight will be minus 16 equals to zero. Using associative law, I can put parenthesis around this and I will have minus 19x minus 16 equals to zero. I hope I didn't make any arithmetic mistakes. Well, how to solve this thing? Again, using the transformation plus 16, we convert this into minus 19x minus 16 plus 16. Equals 16, now this is zero using associative law. So we have minus 19x equals 16 and transformation is divided by minus 19 to get rid of this. And I will have x equals minus 16, 19. Now, I don't know why I have such a strange number. Maybe I made a mistake. Maybe not. Let me just check. x plus four divided by two minus 10x minus 10 equals to zero, that seems to be correct. Now, x divided by two minus 10x, this is two minus 10, right? Times two. So that would be x divided by two minus 10x minus eight times two. That's zero minus x minus 20 and minus 16 seems to be correct. Okay. Now, you see, I'm kind of hesitating right now because again, the number is a little strange and I have to check it not because I did something incorrect with transformations but because I just made, I might made an arithmetic mistake. All right, well, let's try. Let's try to check it. If not, we will just have to do it from the beginning. All right, so let's put minus 16, 19s minus 16, 19s. Plus four. What is this? It's equal to, it's 19 times four. Will be 38, 76, 76 minus 16, minus 16 plus 76, 19s, which is 60. 19s, right? Now, if I divided by two, that would be 30, 19s. Okay, now x plus one will be minus 16, 19s plus one, which is minus 16 plus 19, 19s, three 19s, times 10. So 10 times minus 16, 19s plus one is equal to 30, 19s. And now this is equal to 30, 19s. This is equal to 30, 19s. And there is a minus here. So it looks like we checked fine. And regardless of such a weird number, which is a true solution, it didn't make any mistakes. So again, I was just trying to illustrate how simple equations, which present absolutely no problem to solve them, well, except some risk calculations, actually are based on certain principles and certain theory of invariant transformations. Now, the next problem will include non-invariant transformations, which are also possible to apply in the case of solving certain equations, but that should be done very carefully. So go to the next problem and we will discuss this particular issue. Thank you.