 Hi, I'm Zor. Welcome to Unisor Education. I continue talking about systems of linear inequalities. In this case, this lecture is dedicated to solving a couple of problems. I do recommend you to go to Unisor.com website and notes for this particular lecture are there. So try to solve these inequalities yourself first. Then you can listen to whatever I'm saying and again, my methodology is not necessarily the best, but I will present my opinion about how to solve these systems. But it will be great if you have your own first. All right. Anyway, I have three different systems, and let me just go straight to solving them first. Minus x minus 2 less than 0, x minus 2 less than 0, minus y minus 3 less than 0, y minus 3 less than 0. This is a system of four linear inequalities with two variables. Well, first of all, there is nothing wrong with having four inequalities and only two variables. That's quite all right. It's only when we're talking about equalities equations actually, the number of variables and number of our equations usually is the same. But in case of inequalities, it's not. So every particular inequality sets out of the coordinate plane where points x, y are represented. A certain half. So half of the plane is defined by the first inequality. Another half of the same plane, but from another line, which will be solutions to the second one, the third, the fourth, and then we have to basically intersect all these four areas. Each one of them is half a plane from some particular line. So the methodology, as I was explaining in the previous lecture about the systems of linear inequality is. First of all, let's just separately find these half planes for each case, and then we will go and try to intersect them. All right. So we will have one coordinate plane, now the first inequality. Well, let's first draw the graph of the function minus x minus 2 equals to 0. Okay. Minus x minus 2 equals to 0, it means x is equal to minus 2. Right? That's the solution. So the graph of the function x is equal to minus 2 is minus 2 is here. So all the points on the coordinate plane, which have x coordinate equal to minus 2 lie on this particular line. Now, we are interested in less than 0. So is it left or right from this line? Well, let's just substitute 0 for instance. If x is equal to 0, this inequality becomes true. Minus 2 is less than 0. So the point 0, 0 really belongs to the area which we are interested in. Right? So this is the area we are interested in. Great. Next. x minus 2 equals to 0, x is equal to 2. What is this particular line? Well, it's this vertical line which intersects the x-axis at point x is equal to 2, and y can be in. Now, left or right half of the plane, they're interested in. Well, again, let's just check it out with the easiest point 0. If I will substitute, I will have true statement minus 2 less than 0, which means 0 belongs, point 0, 0 belongs to the half of the plane which we need. Right? So the plane, the half of the plane which we need is to the left. So this is, okay, next. Minus y minus 3, what if it's equal to 0? Y is equal to minus 3. Now, minus 3 is here. So what's the graph of this function? It's this one. It's horizontal. All y's are supposed to be minus 3, and all x's can be n's. Now, again, the same question. Which area are we interested in? Higher or lower than this particular line? Let's substitute 0, 0. We will have minus 3 less than 0, which is true statement, which means 0 is supposed to be in our half plane. So our half plane is to the top from this particular line. So it's this one. Finally, y minus 3, what if it's equal to 0? Means y is equal to 3. This is a line which is going here. It divides again our coordinate plane into two parts, upper and lower, which one are we interested in? Let's just substitute 0, 0. It will be a true statement. So 0, 0 is here, which means we're interested in lower part. Now, we have to intersect all these four half planes. The one which is to the right of this vertical, the one which is to the left of this vertical, which is higher than this and lower than this. And obviously, this is this particular rectangle. So solutions to this system of four linear in equations with two variables is the area on the coordinate plane, which is inside this rectangle. That's the answer. This is a solution. Granted, it's obtained graphically, if you wish. But in any case, it's as good as anything else. And if you can just word it in a way that, OK, that this is a solution. It's within all the points which are solutions to this system of inequalities lie on the coordinate plane inside this particular rectangle. Now, the rectangle is defined by either these four sides. Or if you wish, you can define the vertices. Now, this is obviously 2, 3. This is obviously minus 2, 3. This is minus 2, minus 3. And this is 2, minus 3. So these are three vertices. So either these lines or these four vertices, they define the rectangle. And all the solutions are inside. Border is not included because these are strict inequalities, strictly less, not less or equal. This is the end of the first problem. So write minus y, minus 3, less than 0. x plus y, minus 3, less than 0. Minus x plus y, minus 3, less than 0. OK, same methodology. Coordinate plane. Well, let's draw the graph. Graph of the first function. Minus y, minus 3, less than 0. So we have to graph the function minus y, minus 3 equals to 0, which is y is equal to minus 3. y is equal to minus 3. It's horizontal line, which intersects here. Now, we are interested in less than 0, which means the point 0, 0 belongs to our half the plane. Because if substituted, you will get a true statement. So we are interested in everything above. Everything above this line, this half plane, is basically the solution to this particular inequality. Next one. If x plus y, minus 3 is equal to 0, then the graph of this function can be expressed as y equals minus x plus 3. And we can build this particular graph by building y is equal to minus x, which is this line with 45 degrees, but this direction, and lift it up by three units. So it will be something like this. Or alternatively, we can assign what if x is equal to 0, then y is equal to 3. So it's supposed to go through the point 0, 3. Or if y is equal to 0, x is equal to 3. It's supposed to go to point 3, 0, and 0, 3. So in both cases, we get this particular line. It's at 45 degrees to the x-axis and to the y-axis. And it's lifted up by three units. Now, again, which half of the plane, this or this, we are interested in? Substitute 0, 0, which is this. You will get a true statement. So this particular half of the line is something which we are interested in. So right now, we have this section of the plane satisfying two inequalities. But we have the third one. So what's the third one? Minus x plus y minus 3 equals to 0. Or, again, y is equal to x minus 3, sorry, plus 3. Which is what? y is x is this one. Plus 3, it would be lifted up by 3. So it will go from 3 to minus 3. And again, you can use whatever I said before. If x is equal to 0, y is equal to 3. So at this point, and if x is equal to minus 3, y is equal to 0, it goes through this point. So which part of this plane we are interested in? Again, 0.00 would turn this into a true statement. So this 0.00 belongs, so this area is ours. So what do we have? We have higher than this towards this and towards that, which is this particular triangle. This triangle is our solution. Again, not including the borders, because these are strict inequalities. So the answer to the system of 3, in this case, inequalities, linear inequalities with two variables, is represented geometrically on the coordinate plane. Are all the points inside this particular triangle? Now it's very easy, actually, to find the vertices of this triangle. This is obviously 0.03. Now what are these? Well, it's an intersection between y is equal to minus 3 and this one, which is the second inequality. So it's y is equal to minus 3 and x plus y minus 3 is equal to 0. The intersection of them, well, if y is equal to minus 3, it's minus 3, minus 3, so it's minus 6. So x is supposed to be equal to 6. So this is 6. And this is 6 minus 2. And similarly, this would be minus 6, minus 3. These are coordinates of the vertices of this triangle. So when it's equal sign, these are equations of the lines. Now these are vertices of the same triangle. And everything inside this triangle is solution. When I'm saying inside, I do mean that the border is not included, otherwise I would say it's inside and on the border. But this is not the case in this case. OK, so that's it with the second one. And the third is y minus 2 less than 0, 1 minus y minus 2 less than 0, x minus y minus 2 less than 0, minus x plus y minus 2 less than 0. OK, what are these? OK, now we have four different straight lines to be drawn. And then we have to choose which half of the plane we are interested relative to each line. And then intersect them, exactly the same thing. Now from this, it follows y equals to 2. From this, y is equal to minus 2. So if this is my coordinate plane, these are two lines, which are the first two lines which characterize this particular inequality. Now 0.00 belongs to the first one and to the second one, which means from the first line it should go down, from the second line it should go up. So basically, right now, I have this area of the coordinate plane already representing solutions to only two of these inequalities. Now let's add two others. Now this means y equals to x minus 2 and y is equal to x plus 2. Now this line dropped down by two units, which is this. And this line means lift it up. This at 45 degrees y is equal to x, lift it up by two units. Now again, 0.00 fits here and 0.00 fits here, which means from this line I should go to the left and from this line to the right. So these are my half planes, which I am interested in. And before I had from these two towards 0.00. So as a result, I have this parallelogram between these two parallel lines and these two parallel lines. So the parallelogram represents the solution. So the answer is these four inequalities, linear inequalities, have a solution represented by this parallelogram on the coordinate plane. So every point inside this parallelogram has coordinates which are basically correspond to each and every of these equations. So that's basically the result. Now you can obviously find the vertices as well. So these are obviously 0,2 and 0, minus 2. And this point is what that can be. I guess this is 4, 4, 2. Yeah, and this is minus 4, minus 2. How did I get this? Well, very simply, this is the intersection of this and this. This is y is equal to 2 and this is y is equal to x minus 2. Is that right? Right, so if 2 is here, then x is supposed to be 4. 4 minus 2 is 2, right? And similarly here, if you're combining this and this, then if this is minus 2, then it's actually supposed to be minus 4, right? Minus 4 plus 2 minus 2. Right, I'm right. OK, these are three problems. I wanted to solve it basically with you. If you have some other way to solve these problems, that's perfectly OK as long as you have the same solution. I didn't really mention it, but probably if I think it's worth mentioning, if instead of less, you have less or equal, then obviously the border of this parallelogram should also be included. Well, that's kind of an obvious observation. And if some of them are with equal sign, some of them without, then those with equal sign will include the border. Let's say you have this and this. What does it mean? It means that not only everything below this, but also the line itself belongs. And here also not only above this line, but also line itself belongs. But these are strict inequalities, which mean these lines do not belong. So the answer in this particular case is the same parallelogram, but with this and this side, including into a set of all the solutions to this system of inequalities. That's it. Thank you very much. Don't forget to go through the Unisor other problems. It would be great if you can register with some supervisor or your parents, and you can basically just draw the whole course for you. Thanks again. Good luck.