 Hi, I'm Zor. Welcome to the user education. I would like to talk today about systems of inequalities, primarily the linear systems of inequalities, or systems of linear inequalities, whatever your choice is. Well, first of all, let's talk about definition. Obviously, we have to define what we are dealing with. So what is a system of linear inequalities? Well, that's basically two or more greater or equal or less than or less than or equal conditions on certain number of linear functions, which depend on the same set of arguments. So let's say you have a set of arguments. It can be one argument, it can be two arguments, it can be 25 arguments, and you have a certain number of linear functions of these arguments, like function of one argument is basically something like a x plus b function of two arguments. I'm talking about linear functions in something like this, etc. So you have certain number of arguments, and you have certain number of linear functions. Now, we are conditioning the values of these linear functions. I will always use less than sign, less than zero primarily because everything else or other comparisons are basically derivable from here. I will put zero here because if there's a constant, we can always use the constant on the left using the invariant transformation of the inequalities. So my basic linear function of certain number of arguments looks like this. In this case, it's two arguments, x and y and the linear function looks like the combination, the result of summation of every argument multiplied by some multiplier, maybe by zero doesn't really matter. What's important is that not all of these coefficients at arguments are simultaneously equal to zero. Now, this is not a linear function conditioned on whatever the condition is. So we are considering only cases which look like these, and the combination of two or more conditions on linear functions which depend on the same set of arguments is called a system of linear inequalities. So that's the definition. Now, I just give you a couple of examples. The first one was conditioned on one particular linear function, the second one was another. Both functions are linear, so basically that's the example. Now, what is a solution to a system of linear inequality? So let's just again assume that you have certain system let's say this one, x plus e, y plus f, let's say this is a system. Now, the system means that x and y are arguments which are participating in two different linear inequalities. So what is a solution to this particular inequality? Well, any pair, let's call it x0 sub 0 and y sub sub 0, so any pair of values of the argument. In this case, I'm using the word pair because I have two arguments. If I have 10 arguments, it would be 10 set. So in this case, it's any pair which if substituted into these linear inequalities, simultaneously transforms them into true statement, then this is called a solution. It's absolutely similar to an equation if you have. If you have an equation, doesn't really matter whether it's linear or not linear, what is a solution to this particular equation? This is an argument if substituted into this function would convert it into identity. Well, in this case, it's a true statement. If you substitute x0, if this is a solution to these both in this case in equalities, they all simultaneously are true. Now, if I have 10 different linear function, with 25 different arguments, what is a solution in this case? Well, this is a combination of 25 values that the first argument has the first value, the second argument has the second value, et cetera, and all these 25 values of the 25 arguments are simultaneously substituted against the variables in all these 10 inequalities, and all of them simultaneously are true then this is a solution. The last one which is more of a philosophical thing, what does it mean to solve a system of let's say linear inequalities in this particular case? Well, it means to find all the pairs or triplets or sets of 25 different values, all of them which are solutions. So to find all solutions for the system is the purpose of solving it. Whatever the methodology will use to solve these equations, basically it's the methodology which is directed towards finding all the different pairs or triplets or whatever the others of arguments, each one of them is a solution. Now, how can we find it? Well, that's a thing which we really have to talk about a little bit. You see, the problem with equations is a little simpler, because in solving the equations, you really set up finite solutions. Let's say you have an equation and 2x plus 3 is equal to 0. Well, there is a one solution actually, x is equal to minus 3 or 2. If you have a system of let's say two different linear equations with two different arguments, you would still have a certain finite number of solutions. Even if it's a polynomial, not just the linear function, it's still finite number of solutions. In case of inequalities, the number of solutions can be actually infinite and here are a couple of examples basically. Right, I don't really need this general format. So, I will present you a couple of cases when there are different number of solutions, maybe infinite, maybe one solution, maybe no solutions at all, et cetera. But in any case, I think the right now, what's important is to understand on a couple of examples, which I have, what exactly I meant about, let's say I have this particular system of inequalities. So, it's a system because there are two linear functions conditioned on being less than zero. Now, there is only one argument. There is nothing wrong with this, by the way. You can have a system of M equations in equalities with M arguments and it's fine. No problem about that. So, in this particular case, I have two conditions, linear conditions, and only one argument. So, let's just think about what are these conditions and we will use quite extensively the graphical representation of these conditions. Since it's one argument, then the graphical representation of all the values of this argument is straight line where all the real numbers can be presented as points. I didn't really mention it before, but we do assume that in 99.9 percent of the cases, we are talking about real values of the variables. So, all the real values are on straight line, and let's just think about these two inequalities. Now, we can definitely simplify them. In the first case, we can add eight to both sides, and I will have two x less than eight, and we can divide by two, which is a positive number, which means that the sign, this equation less than is supposed to be retained. So, I will have x less than four. So, this is my resulting inequality from the first equation. Now, the second one, we will add minus nine to both sides. That's what I will get. Now, we will divide by minus three, both sides. But now, since minus three is a negative number, the sign of the inequality is supposed to be reverted into greater in this particular case, and I did explain why in the lecture about inequalities. So, I will have x greater than minus nine divided by minus three, which is three. This is another condition. So, from these a little bit more complicated conditions, I have derived two relatively simple conditions, and I would actually like to present it in a graphical form. So, if this is my various of x, this is zero. So, this is one, two, three, four, five, six, minus one, minus two, minus three, minus four, et cetera. So, this is my numerical axis. Now, where are all the x less than four numbers here? Well, they're here. All the numbers to the left of the four not including the boundary four, because it's strictly less. It's not less or equal. So, equal four doesn't really fit. But everything to the left from the four fits fine. Now, what is this? Well, it's this. Greater than three. Again, not including the boundary. Now, you remember, I was saying that solution is a value of the variable or variables, which simultaneously transforms all the inequalities in the system into a true statement. So, both x less than four and x greater than three must be true in order for some value to be a solution. So, 3.5 fits both of them. Now, these two are completely equivalent to these two, right? So, that's why I can check on these two. So, 3.5 fits fine. Now, zero fits less than four, but it doesn't fit this one. So, zero is not a solution. So, solution is intersection without the borders, without these endpoints between these two sets. Greater than three and less than four. So, everything which is from three to four, not including three and four is a solution. And this is the end of solving this particular system in equalities. We have found a set of all the variables, all the values for the variables, which if substituted into all inequalities in our system, transform them all into true statements. Now, let me give you an example of another system which has no solution, which also can happen. So, it's this one, minus plus, minus plus. How about this? Well, again, let's do the manipulation and transform these two inequalities into both simpler format. So, in this case, I can add minus eight, negative eight to both sides. I will get minus two x less than minus eight, divide by minus two. And since it's a negative number, this sign is supposed to be changed to the opposite. And I will have x greater than four. Now, this one, I add nine to both. It will be three x less than nine, divide by three positive. So, the sign of inequality is retained x less than three. So, I have to deal with these two. And again, it's supposed to be simultaneously true. Now, again, let's just draw the line. One, two, three, four. So, less than three are these. Greater than four are these numbers. You see, there is no intersection, which means there is no point, there is no number which would satisfy both simultaneously. This simultaneousness is extremely important. So, it means that this particular system has no solution. End of story. Okay, so, let's talk about a general form of linear inequality of two variables. Now, let me finish maybe the previous example. This was example of one variable and two inequalities. Looks simple, right? So, the next is two variables. And again, we can have two or more than two inequalities which combine together certain linear functions of two arguments. Now, why I specifically emphasize one and two, and maybe three in some other lecture cases. One, two, and three arguments. Well, because all these can have certain visual representation. If you have one argument, it can be represented on a straight line on a numerical axis. If you have two arguments, it can be represented as a point. Every pair of two arguments can be representing a point on the coordinate plane. And any three variables analogously can actually represent a point in a three-dimensional space. Now, we don't go any further than that because it's kind of difficult to imagine four-dimensional space. So, we will concentrate on one, two, and three. And two actually would be probably more, well, familiar maybe, because it's easier to present it on the board. Well, 3D is also fine, but two-dimensional problems are a little bit kind of easier to view. All right, so let's talk about two-dimensional case when I have only two variables. And this is a general expression of a linear function of two arguments conditioned on, let's say, less than zero. Now, similarly to whatever I did in a one-dimensional case. If you remember, I just solved this one-dimensional case and basically cut my numerical line into left and right. Whatever was on the left or was on the right doesn't really matter, it was good, which would satisfy one particular inequality. And whatever was on the other side did not satisfy it. So, I basically solved each inequality graphically by itself and then took intersection of areas. So, if I had one-dimensional case, I had something which was there, something which was there, and then I said, okay, this is intersection and that's the solutions. So, it's easy to present it graphically. Now, in case of one-dimensional, I also could do it algebraically, which looks fine, basically. In case of two-dimensions, it's not really easy to express because you have basically two variables and they are interrelated. So, it's not easy to express algebraically, but it's very easy to express graphically the solution to system of linear inequalities of two arguments. And here's how. Again, we will follow exactly the same strategy. If you have more than one linear inequality, let's just find out graphically where are solutions of each one of them and then do an intersection between these areas. So, what is the solution to, let's say this particular inequality? Here is an interesting thing. If you have an equality equals to zero, what is a locus of points on the coordinate plane, the coordinates of which satisfy this particular equation? Well, everybody knows this is a general concept of a straight line on the coordinate plane. I'm sorry, I made a little mistake here. I meant why? Well, to get it into more familiar form, if non-A, no coefficient at the variable is equal to zero, so A is not equal to zero and B is equal to zero, you can always transform this into y is equal to mx plus m, right? That's easy, especially if B is not equal to zero. Now, if B is equal to zero, you have something like basically, well, ax plus c is equal to zero or x is equal minus c over a. A is not equal to zero if B is equal to zero because if both are equal, then it's not a linear function. So one of them is not equal to zero. So if B not equal to zero is y is equal to some function of x, which you know is a linear function, a straight line. And if B is equal to zero, it's this line. What is this? It's a vertical line, which goes through this point. So all x's on this particular line have x, all points in this line have x coordinate equal to this and y is not restricted to total because this is equal to zero, right? So anyway, whatever a and b are, this is a straight line. So I'm kind of assuming that this issue is closed. It's a straight line somehow positioned on the coordinate plane. Maybe it's a vertical line, maybe it's a horizontal line, maybe it's some kind of line at the angle. But anyway, the locus of all the points on the coordinate plane which satisfy this equation is a straight line, one straight line. That's very important, which means that outside of this straight line, this is not equal to zero. Is this logical? Quite logical. It can be less than zero or it can be greater than zero. Now, another important point. This function a x plus b y plus c is smooth. In the sense that if arguments are changing very slightly, the function value is changing very slightly. So if I have two points at one point on the plane, the function is positive. And at another point on the plane, the function is negative. Now, what if I'm moving from this to this? Doesn't really matter whether it's straight line or some curve, doesn't really matter. I'm changing very smoothly my arguments x and y from this point to this point. So my function should change smoothly from some positive value to some negative value, which means that somewhere along the line it must cross the point where the function is equal to zero. If I'm moving from a positive area to a negative area and my movement is smooth, then somewhere I have to cross the value zero. So there is some point in between where the function is equal to zero. What does it mean actually in our case? Well, here it is. What if my graph of the function ax plus by plus c is equal to zero in this straight line? What I'm actually stating right now is that all the positive values of this function ax plus by should be on one side, let's say this one, and all negative should be on another side. Because if positive and negative exist on one side of this line, then I can always go from here to here and I can find point where my function is equal to zero. But that's not possible because all the zero points of this function are on this straight line. So I cannot have on one side of the plane both positive and negative values of this function. Because otherwise I would have more zero points and all zero points are on this line. So my entire half of the plane, one half of the plane should be positive and another half of the plane should be negative. Then obviously when I'm moving from this to this and I am intersecting my line, that's where it's equal to zero, that's no problem. But if I have on both sides positive and negative, this is not possible because then that would exist a point where my function is equal to zero and it does not belong to the straight line which we have agreed is not the case. So all the zero points are there. So my conclusion is that this particular straight line divides an entire plane into two areas. One where this linear function is positive and another negative. If we are interested in negative, let's say, we just have to choose which half to take. How can we choose it? Well, we can just take any point and check what's the value. If the value is positive, it means it's not our half of the plane. We'll just go to another half of the plane. If it's negative, then it satisfies the equation, then this is the half of the plane which we need. So this is an inequality which has a solution represented as half of the coordinate plane and the edge of this half is the graph of this particular function. Now, how can we graph this function? Well, there are many ways, quite frankly. Let's just assume that, let's say, I have a few cases, yeah, I have a few cases. If a is equal to zero, then my function actually is BY plus C is equal to zero or Y is equal to minus C over B which is a horizontal line which goes through point minus C over B on the Y axis. So it's set of all the points on this horizontal line. So the graph of the function in this case is horizontal line which is crossing the Y axis at point Y is equal to minus C over B. Now, if B is equal to zero, then I have an equation AX plus C is equal to zero, X is equal to minus C over A and again, if this is my coordinate plane, let's say my X is, let's say it's here, minus C over A. Then it's a vertical line which goes through this point. This line represents the graph of the function in this case. So it's either, so if A is equal to zero, it's a horizontal line. If B is equal to zero, it's a vertical line but in any case, we know how to construct it. Now, finally, not finally, what if A is not equal to zero and B is not equal to zero? Now we have a choice of C, all right? So what if C also not equal to zero? Then what can we say? Well, again, there are many ways to construct this particular line but I might suggest something like the following. Substitute X is equal to zero and you will get Y equals to minus C over B. So this is one point because this is a solution to this equality, right? Zero means this is zero, minus C over B, it would be minus C plus C is equal to zero. Now we can substitute zero into Y and you can get the value of X which is minus C over A. So this is another point. Now where are these points? Okay, X is equal to zero is here and Y is equal minus C over B is here. Now another point is Y is equal to zero and X is equal so it's something quite here for instance. This is minus C over A, this is minus C over B. So we have two points so the straight line goes through them, right? To build a line we need two points. So this is one of the ways to do it. Where are any, there are others as well. Now, so we know how to construct this particular case and finally how to construct the case when A is not equal to zero, B is not equal to zero but C is equal to zero, then what this is actually it's AX plus BY is equal to zero Y is equal to minus A over BX correct? So again, you know what this is. This is a straight line which goes through the beginning of the coordinate and minus A over B is some kind of an angle depending on what you want. For instance, you can put one as X and minus A over B as Y and you will get another point. So zero zero is one point and one and minus A over B is another point. So you have these two points and you have straight line. So we know how to build the graph. Okay, so first what we do to solve this inequality on the top, we construct this particular graph. Fine, we have constructed it. Now, after we have constructed it, we have to choose which half the plane we really are interested in. And again, I could suggest basically just take any point outside of the graph itself. For instance, if C is not equal to zero, then you can easily put zero zero as coordinate and see where what's the value of this particular linear function is. In this case, it would be C. So if C is let's say positive, then zero zero lies where this linear function is positive. If C is negative, then zero zero point is belongs to basically the solution of this particular inequality. All right, so we know how to solve one particular inequality. It's half the plane which we can construct using graphic of this particular linear function. Now, what if I have two inequalities like this? Well, let's just think about it. You are talking about that you have one half of the plane which represents solution to one inequality. Now, if you have another inequality and this is another line which represents the equal sign, let's say it's DX plus EF plus EY plus F, then you might have, let's say, this particular half of the plane which represents solution to the second one. So what is the solution to a system of two equations inequality, sorry, in this particular case? That's the intersection. It's this area, right? It should be below this line and below this line. So it's this sector which has a vertex where lines intersect. Now, and where are the intersects? The lines, well, that's where all linear functions which are represented are equal to zero, right? If you have two, then you have two lines and they might intersect. Well, is it always like this? Is it always this type of a sector shape which is a solution to two linear inequalities? Well, not necessarily, because you might have a different picture, look at this for example. If you have one line like this and another line like this, so you have these and you have these. Now these are below this line and they represent solution to one particular linear inequality. Everything above this line, these are solutions to a second one. Now intersect these two areas and you will have this area in between two parallel lines, right? And finally, what if intersection is empty? Is it possible? Yes, absolutely. If lines are parallel, then intersection can be empty because for one particular inequality you can have this area and for another you can have that area and they have no intersection. So then you have nothing. So now I think we are ready to basically have a couple of examples. And here is my first example of system of two inequalities, linear inequalities. And I will write them in this way. Now is this a system of two linear inequalities? Well it looks like a solution, right? Well, in a way this is a system because this is a linear function of two arguments X and Y and the coordinates, the coefficient at X is one, coefficient at Y is zero and the free coefficient C in my formula AX plus BY plus C. So A is equal to one, B is equal to zero and C is equal to zero, right? So it's a linear function, same as this one. In this case A is zero, B is one and C is zero. So these are two linear inequalities in their own rights. So let's try to solve them graphically. Okay, so let's take the first one. First we have to resolve X is equal to zero and graph it. Now what is X is equal to zero? This is X, this is Y, zero. X is equal to zero is a vertical line which goes through the beginning of the coordinates, vertical. This is X is equal to zero. Now, where exactly is the area which we are interested in? Well, we have to choose left or right half of the plane from this vertical line. Well, let's take, for instance, to the left. Let's put this point, for instance, which is, let's say, minus five, zero. Let's substitute it into our inequality. Is it true? X is equal minus five. Y is equal to zero, so it doesn't depend on Y anyway. So we have minus five, which is less than zero, which is true. So everything to the left is our area which we really are interested in. Now, next one. Next one is, let's first draw the graph of Y is equal to zero. Well, Y is equal to zero is a straight line which goes through the beginning coordinate and along the X axis, right? This is not right. So all points on this line have coordinate Y is equal to zero and X is not restricted. So which area do we need right now? Which half of the plane to the up or going to the top of this or going to the bottom of this? Well, let's just again take any point. Let's say we'll point this one. Let's say it's one, one. Substitute one, one here. X is equal to one, Y is equal to one. So it depends only on Y, so one less than zero, it's false, right? Which means that this is above the line is not the area we are interested. We're interested in the one which is below this particular horizontal line. So these are where this which we are interested. Finally, where is intersection of these two areas? One is to the left of this line and another is down from this one. Well, intersection is obviously where I have both red and brown which is this one, again, sector. It's a quadrant of the coordinate plane. It's usually called the third one. The first, the second, the third, the first. It's a third quadrant of the coordinate plane. This sector. All the XY pairs which are here satisfy the system of these equations. That's the clean solution in this very, very simple case which obviously you can derive without dragging into the code. So now we can do a little bit more complicated things. All right, the last one. Minus two X minus three Y plus six less than zero and five X minus three Y plus 15 less than zero. So, again, let's do it according to the book. First, we draw the graph of minus two X minus three Y plus six is equal to zero and then we will find which half of the plane fits our less than condition. All right, so minus two X plus, sorry, minus, minus three Y plus six is equal to zero. How to construct the graph of this? Well, again, let's do the way how I suggest it. Not necessarily the best way but something which is easier for me right now. Let's find where is this particular line crosses the X axis. Well, it crosses the X axis where Y is equal to zero, right? Let's find X. So if Y is equal to zero, what's the value of X? Well, obviously it's three, right? Minus two times three would be minus six plus six zero. So if Y is equal to zero, this is three. If X is equal to zero, then the line intersects my Y axis at point two, right? So if this is zero, this is two, it will be minus six plus six zero. So the line is intersecting my two X's at these points. So this is how it looks. Now, we are interested in less than zero. Now, which part of the plane? This one or this one? Well, let's substitute point zero zero. That's the easiest. If you substitute zero zero, you will get six which is wrong. Six less than zero, this is the wrong. This is the false statement. So this is not good for us. In our case, we need this one. So this is the area which we are interested in and that's the solutions to the first inequality. Now the second one. All right. You can get the right one. Okay, if X is equal to zero, so it's somewhere on the vertical line, Y is equal to two, right? So it's, again, this particular point. If Y is equal to zero, oh, I'm sorry. This is the first equation. You see something is wrong. Now I'm talking about the second equation, which is five X minus three Y plus 15 equals to zero. All right. Again, if X is equal to zero, Y is equal to five. So it's somewhere here. This is a five. And if Y is equal to zero, X is equal to minus three. So it's somewhere here. So these are two points. Okay. Now my question is, this or this half of the plane represents solutions to the second inequality. Again, let's just substitute zero zero. And what do we have? We have 15 less than zero, which is false, which means, again, our solutions are on this side of the plane. So what is the total solution? Let me just continue this and continue this. So as you see, the area of the plane where both inequalities are true is here, this sector of the plane. Now, if you wanted to know where is the vertex of this, well, you can actually find the coordinates of this point because it belongs to both lines, which means it's supposed to be a solution to both equations. Well, how can we find a solution? Well, in this particular case, the easiest way is just to subtract from one another. So minus three Y and minus three Y will be reduced. So I will have five minus negative two, which is seven X plus nine equals zero. So X is equal to minus nine over seven. Now, knowing X, I can substitute it into one of these. Let's say the first one. So I will have minus two times nine over seven, but this is minus and this is minus, so it will be plus. So it's 18 sevens, right? So 18 sevens. Now, minus three Y plus six is equal to zero. So Y is equal to eight sevens plus six divided by three. And what will it be? Seven plus six seven times six four to two is 60. 60 over seven and then over three, it will be 20 over seven. Okay. So Y is equal to 20 over seven. So this point has coordinates X is minus 97, something like here. And Y is 20 seconds, almost three, and this is slightly more than one. That's coordinate of the vertex. Again, if you're interested, not necessarily you are. I understand. All right. So basically this completes my explanation about what the linear system of inequality is actually is and how to solve it in case of one and two arguments. Graphically is probably the most important methodology which you would use. And I will probably have a few examples as problems, which I will talk about in the next lecture. To get into the details of the methodology and I do recommend you to follow up with these. That's it. Thank you very much. Don't forget that Unisor.com contains notes for this lecture and for other lectures. It also has exams. So your supervisors or parents can actually control the educational process. They're registered for the course. So your supervisors or parents can actually control the educational process. 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