 Hi, I'm Zor. Welcome to Inizor Education. I would like to continue talking about systems of inequalities. And this lecture is basically a continuation of the previous one. I just want to present a little bit more problems and how it can be solved. Well, considering this is just like a training session, obviously you are encouraged to try to solve these problems just by yourself before listening to this lecture. If you didn't do it yet, just press the pause button on the video and try to do it yourself. And then just compare your results with whatever I'm presenting. All right, so let's start. The problem number one is a system of these inequalities. Okay, let me just repeat something which I was talking about in the previous lecture. This is about geometric representation of the systems of inequality. Result, the solution of this and any other system is presented as an area on the co-ordinate plane. So, let's start with co-ordinate plane. We are talking about Cartesian co-ordinates, rectangle co-ordinates. Now, if you see an inequality like this, what I usually recommend is to start with equation. One, y is equal to x cubed. It's this one. Now, this curve divides the whole plane into two parts. On the curve itself, y is equal to x cubed. Outside of this curve, y is either less or greater than x cubed. And what's important is that one part is completely devoted to one side of the equation, like less than, for instance, and the other is completely filled up with points where y is the opposite, whatever. If this is less, then this is greater than x cubed. I did present the logic behind it. It's very simple actually. If you allow two points on the same part of the plane to be such that in one point, y is, let's say, greater than x cubed and another is less than x cubed, and then we will probably connect them with some curve which is not intersecting this one because these two points are in the same part. So if this is the place where y is greater than x cubed and this is where y is less than x cubed, since our function is smooth in some sense, it should be somewhere, the point in between, where y is equal to x cubed. And all the y is equal to x cubed supposed to be on this curve, so we cannot really have this point here. So that's why we cannot have two points with different relationship between y and x cubed. Here, y is less than x cubed, and here, y is greater than x cubed. And on the line itself, y is equal to x cubed. Okay, now, why I decided that this particular part is where y is less than x cubed and not the other way around? Well, the simplest thing is, just substitute some point. For instance, the point where x is equal to zero and y is equal to minus one. x zero, y is minus one, so minus one is less than zero, which is true. Okay, so that means that this point belongs to the area of y less than x cubed, and therefore the whole area underneath is where y is less than x cubed. Now, in this case, I'm looking for something which is greater or equal than x cubed, so let me just wipe out this piece. And that's where y is greater or equal to x cubed, and that includes the curve itself. Now, the next one, y is less than x four, or equal. Now, x four goes above x cubed, and this is an even function, because this is an even exponent, four. So it's symmetrical. Well, now, let's look at this one. Now, we are interested in y less or equal to x four, and obviously this is the part which is below the curve. It's this one. And finally, x greater or equal than zero, it means it's this part of the plane. So basically what I can do, I can just completely wipe out this piece, and intersection of all these areas is obviously this one. Looks like a form. So that's the answer, including the border lines, because we always have greater or equal, and including the point zero, zero. That's it for this problem. Next one, it's a combination of equality and inequality. Here it is, y minus x times y minus x over two times y minus two x times y plus x times y plus x over two times y plus two x times y minus two x times y plus x times y plus x over two times y plus two x times x square plus y square minus y equals to zero. That's my first component of the system, and this is an equation. Notice that, that's very important. Two others are inequalities, and the last one is less than 36. So again, what we need to do is we have to combine, intersect, if you wish, this, this, and this areas. Well, let's start from the beginning, this area. Now, I would like to refer you to the lecture on graphs, where I was explaining an interesting manipulation with graphs. If you have a graph of one function, let's talk about f of x, y equals to zero. You have a graph of this equation, and then you have a graph of another equation. Doesn't matter what are these graphs, but these are two equations, and each of them has its own graph. Set of points x, y, coordinates x, y, which substituted into this, converts it into zero, substituted into function g, will converts it into zero. How to produce an equation which has a graph, which is a union of these two graphs. Union means put all the points of one and all the points of another together in a graph. Well, and I have suggested to use the following function times g. So this particular equation, when h is equal to zero, or this is equal to zero, means the following. When is the product of two function, two values actually, when is this equals to zero? Obviously when either this one or this one. When this is equal to zero, then the point belongs to this graph. If this is equal to zero, the point belongs to this graph. So since I'm talking about either or, this is a definition of the union of two sets. The union of all the points which transform this into zero, unionized with all the points which transform this into zero. So basically my equation has a graph which is a union of this and this. We will use this feature to decipher our first ecology. So let's graph it. Now, as I was just saying, the graph of this is a union of graph of this is equal to zero, this equals to zero, this, this, this, et cetera, et cetera. So each component of this product when you equate it to zero will result in certain graph. So the graph of the whole thing is a union of all these. So that's just the graph of y is equal to x. It's this one, y is equal to x, right? y is equal to half, x over two, right? So if this is equal to zero, y minus x over two is equal to zero, so y is equal to x minus two. That's something like this. Now, y minus two x is equal to zero. Well, that's this. y plus x is equal to zero, so it's y is equal to minus x, it's this. This y is equal to minus x over two, and this is y is equal to minus two x. The last one, x square plus y square minus nine is equal to zero, that's x square plus y square is equal to nine, which is three square. So as we know, this is a circle. A circle of the radius three, because three square would be nine. Remember that. x square plus y square equals r square, this is a definition of a circle with a radius r around the center of coordinates. All right, let's draw this circle. So this is three, and I have a circle. So the combination of all these, whatever the number of these lines is, is the graph of the top equation. So all of these together, the union of all these graphs is the graph of the equation on the top. Now, by the way, these are not really two-dimensional areas. These are lines or curves, right? Now these are the areas. What is x square plus y square greater than or equal to nine? Well, this is a circle where x square plus y square is equal to nine. Where is it greater than nine? That's where the distance from the point x and y to a center is greater than three, right? So it's all these points outside of this circle. So whatever is red, this is error which we are talking about, including by the way the circle itself. So the inside, we should really wipe out because it does not correspond to our second component in this system. About this one. Well, this one, if you put an equal sign, it's also the equation of the circle with a radius six. So somewhere, we have a point six, and we have another circle. Now this time, we are talking about lesser equal. Now that means inside of this particular circle. So whatever is outside, we should wipe out. It does not belong to our area. So what's left? What's left is something which looks like a wheel in a car. It's two concentric circles. One is radius three and other is radius six. And these pieces of radiuses, which are in between these two circles. So that's the result basically. This is, and again, this is not a two-dimensional area. It's just the combination of these lines and these two circles. That's all. And again, as I was saying, it looks like a wheel or a tire or something, whatever. Okay, that's it with this. So remember that if you have some equation like this one on the top, the best way to graph it is to graph each individual component equals to zero, graph and then unionize them together. So it looks a little complicated, but really it's divided into relatively simple pieces like lines and circles, look at you. Next one. X square plus Y square minus one. Four minus X square minus Y square greater than zero. Okay. Here again, we will use this logic when I'm dealing with this product. When the product of two numbers is positive, when either both are positive or both are negative, right? So let's first draw an area which is represented by this. Then we will draw an area which is represented by this and then we will unionize them because either when this is true or when this is true, this is true, correct? So let's just build this and then let's build that and unionize them. Now this is basically X square plus Y square greater than one, right? And this one is X square plus Y square less than four. Now what is this? If it's greater than one, it means it's outside of this circle, right? So this circle is X square plus Y square equals two to one. Now whatever is inside is like zero zero when it's less than one. Whatever is outside, that's when it's greater than one. So not including the circle itself because this is a strict greater sign. Now this one is related to the circle with a radius four. Sorry, it's related to radius two. My mistake. Four is two square, right? So radius is two, which is this. But in this case, we're talking about less than sign. So points like zero zero do belong to it. So everything inside the bigger circle is the area which we're interested. So finally, their intersection is this ring. So everything inside this ring, outside of a small circle, but inside the bigger circle is the solution to this system of inequalities. Not including the boundaries, by the way. Now let's talk about these guys. Again, let's transform it slightly. It's X square plus Y square less than one. X square plus Y square greater than four. Well, we are talking about the same two circles. But in this case, the first means inside the smaller circle. The second means outside of the bigger circle and intersection of this thing is empty set. There are no points which are simultaneously inside a small circle and outside of the bigger circle. So we are unionizing this ring with an empty set and the result will be exactly what we are started from because unionizing means all of these plus all of those, the random of those, so only these points remain in the final solution. So that's the solution. Okay, the last problem which I wanted to present to you today is this. Log base one half of X square plus Y square less than or equal to minus two. And log base two of the same thing is less than two. Okay, I had a couple of problems in the previous lecture where logarithms are presented. The typical approach which we can say is the following in this particular case. I would prefer to have the same log here and the same log here. And then by comparing these two functions, the logarithmic functions with the same base, considering the logarithmic function is monotonous. Either monotonous, the increasing or monotonous, the decreasing depending on the base, greater than one or less than one. So based on that, I can actually judge what's the relationship between whatever is under the logarithms. And that's exactly what I'm doing. So I would like to put it this way. Okay, what should I put here to get minus two? Well, obviously log four is minus two because minus two raised to the power of, I mean one half raised to the power of minus two would be one half minus means negative. It means it will invert, so it will be two and then power of two will be four. So that's what it is. Similarly, I will do here. And I will use log base two, same as this. Now, again, what should I put here to get two? What is the power I have to raise two? I mean, what's the result if I will raise two to this power? It would be four as well, right? So log four base two is two because two to the power of two is four. Log four base one half is minus two because one half raised to the power of minus two gives four, right? So instead of these two, I will use these two. They are absolutely equivalent. But now I can use the properties of the monotonous functions. Now, the log base one half has a graph like this, monotonously decreasing. This is log base one half. Now, log base two has this graph. It's monotonously increasing, which means that in this particular case, the relationship between two arguments which are under the logarithms is opposite to the relationship between logarithms because this is a decreasing function. So the greater function, the greater function, the smaller argument. The greater argument, the smaller function. So I will revert the sign of the inequality to get x squared plus y squared greater or equal to four in this case. Now, this is monotonously increasing function. So the relationship between function is the same as the relationship between arguments. So in this case, I have x squared plus y squared less than four. That's interesting. Two seemingly opposite inequalities, right? Well, they are really opposite. But the problem is that there is an equal sign here and equal sign there. So this is greater than equal to four. And this is less or equal. And this is less than or equal. So there is a commonality. The commonality is equal. So that means that we have to take graph of the function x squared plus y squared is equal to four, which is a circle of a radius two. So this is y squared plus y squared equals four. Four is two squared. So that's why the radius is true. Now, the first inequality gives basically all points outside of this circle, including the border belong to this area. This is all points inside this circle, including the border belong to this area. So what's the intersection of these two things? Well, obviously it's just the circle itself. So this is a set of points which satisfy this system of inequality. Only points which are on the circle itself, they satisfy both these equations. Well, that's it for today. I do recommend you to go through the same problems again. They are in notes for this lecture on Unizord.com. And well, good luck. Thank you very much.