 Hi, I'm Zor. Welcome to a new Zor education. We will talk about linear inequalities to date, linear inequalities, not equations. Now, why I mentioned equations by mistake is because I was thinking, what can be simpler than linear inequalities, but only linear equations, right? All right. So let's talk about linear inequality. Well, kind of a general form can be something like this, or any other inequality sign like greater or less than equal, greater than equal at second. All right. So what's important? Well, obviously, this is important, a should not equal to zero because otherwise, it's not an inequality, it's some kind of, I don't know what it is. So, and the purpose is to find various of unknown variable x, which are necessary and sufficient condition for linear function a x plus b, to be in this case, less than zero. So that's what it means to solve an inequality. You have some kind of a condition on a function, in this case, linear function, and you have to find the corresponding condition on the argument of this function, which is x in this case, which is necessary and sufficient for the function to satisfy this condition. Well, it's easy, right? So we all know how to solve these type of things, and let me just make very, very brief remarks. Obviously, to solve algebraically inequality, you have to do the same thing, similar things to what you're doing when you're solving equations, which is invariantly, preferably invariantly, if not non-invariantly, but transform the original equality or inequality or equation, whatever you have, into some form, which really provides the information about the argument. Now, in this particular case, the first action which I'm sure you know how to do is add minus b to both sides of inequality. Now, adding any number, positive or negative, or zero, to both sides of inequality, is an invariant transformation, and the sign of inequality is retained. Whatever, if you have two numbers, one is less than another, you add something, doesn't matter what. Positive or negative to both sides of the same, of this inequality, will result in the sum on the left and sum on the right to be in the same relationship. So, if I add minus b to both sides, I will get ax plus b minus b less than zero minus b, which is ax less than minus b. All right, so that's the number one step, which we have done to transform our original inequality into form, which will give us the condition on the argument. Well, naturally, next step, considering a is not equal to zero, otherwise it would not be an inequality, we divide by a. So, on the left, we will have x. On the right, we will have minus b over a. a is not equal to zero, so there's nothing wrong with b over a. Well, question is what sign to put in? In the previous lecture, I was explaining different transformations of inequalities, and I was actually talking about multiplication and division of both sides of the inequality to be an operation which you have to do carefully. If you multiply or divide by positive number, then you retain the sign. If, however, you multiply or divide by negative number, you have to reverse the sign. So, this thing is like this, if a is positive and, sorry, reverse, if a is negative, and this is a solution. So, solution is a condition on the argument x in this case, which is a necessary and sufficient condition for this function to satisfy this condition. And it depends on the sign of the a. Well, basically there is nothing more which really is of any interest in talking about linear equations and their algebraic solution. And as I was thinking about how to make it a little bit more informative, I have decided to spend some time to analyze the graphical approach to this. So, let's talk about graphics. So, we have exactly the same type of inequality, and what I would like to do is basically find graphically the solution of this particular inequality. Well, to find it graphically, we have to construct the graph, right? So, let's construct the graph. Now, first, we start with function y equals x. That's our step number one to construct the graph. And, well, obviously I could construct it, right, you know, immediately, but I would like actually to do it step by step, so, you know, you will understand it better. Now, next step is I will construct the function y equals a times x. So, I have to multiply this graph by a. Now, how to multiply the graph by a? Basically, you have to multiply each y coordinate for each particular argument by the factor a. And here we have a slight difference. If you multiply by something like 2, for instance, then each y coordinate will be basically double in size. So, it will be something like this. If you multiply it by, let's say, 1 half, well, it will be shorter by 2. So, this is y equals 2x, this is y equals 1 half x. Now, what's common in all these graphs? They are all monotonically increasing, and that's very, very important. So, let's remember that whenever you multiply the graph y equals x by any positive number, less than 1 or greater than 1, it will move a little bit up or a little bit down, but it will still be directed upwards. If you go along the x-axis from minus infinity to plus infinity towards increasing of the axis, then the graph will also be increasing. It's monotonically increasing function. That's very important. If, however, you multiply this graph by a negative number, minus 2, minus 1 half, or minus whatever, then instead of being directed in this way, the graph will be directed this way. So, this would be y is equal to minus 2x. It's similar to, it's symmetrical to y equals x relative to the x-axis, right? So, we have to multiply, to multiply by minus 2x, we have to first multiply by 2, which will do this graph, and then multiply by minus 1, which means every positive y will go to a negative sign. So, that's why the direction will be this way. And again, for all negative multipliers, A. And all these graphs with negative multipliers are monotonically decreasing. So, as x is increasing from minus infinity to plus infinity, the y will be decreasing. And this is extremely important for further analysis of this inequality. All right, so we have concluded that multiplication of graph y equals x by positive number retains, although it changes in the scale, but it retains monotonically increasing behavior. And multiplying by negative factor A, we are making our function y equals x into a monotonically decreasing one. All right, so the first step is we multiply by A, and whatever the graph is, I will use the graph with a positive and something like this for negative A. So these are graphs for A positive and for A negative. Next is what we have to do is we have to add B. Now, what is add B? Add B means we have to move vertically, either upwards if B is positive or downwards if the B is negative. So that's how we build the graph AX plus B from the graph AX. So in this particular case, for instance, one more detail. If I'm shifting the graph upwards or downwards, it doesn't really matter. The behavior of the graph of being monotonically increasing or monotonically decreasing does not change. So if I move this particular graph up or down, it will still be monotonically increasing because it's directed upwards. And this one will be directed downwards again, regardless of where I move it up or down. So I will use, for instance, movement up, but it doesn't really matter. So on my next graph, I will put it this way. This would be my Y equals AX plus B with A greater than 0 and B is greater than 0. And I don't want even to draw the graph when B is equal less than 0. Now, in case of A less than 0, I will also move the graph upwards to the same value. This would be Y equals AX plus B where A is less than 0 and B is greater than 0, also greater than 0. So both graphs are moved in this particular up because the B is positive. Now, we can answer the question, what are the axis when this is less than 0? Well, in both cases, we can find the point where it's equal to 0. One graph is equal to 0 here, this one, and another graph is equal to 0 here. But now the question is, this is where this particular graph is equal to 0. It's monotonically increasing. That's important, which means that if X is increasing from this point to the right, graph will grow, it's monotonically increasing, which means it will be greater than 0, which is not good for our purposes. And only in case we move left towards minus infinity from this particular point, only in this particular point, only in this particular case, the graph will be less than 0 because the graph is decreasing if we go left. So what's important is, if you have a solution to AX plus B equals to 0 and solution is minus B over A, then in case A is greater than 0, our solutions are to the left of this point. And in case of A is negative, the graph is decreasing, monotonically decreasing, which means if I increase the X from this point towards plus infinity, the graph will decrease. And this is where it's equal to 0, which means it will do less than 0. So I have to move to the right from this point, which is a solution to this equation. So the answer is X should be greater than minus B over A to satisfy this particular inequality. If I move to the left, the graph will increase because when I'm moving to the left, monotonically decreasing function will increase the value. And the variable will be greater than 0, which is not good for this purpose. So I would like to basically conclude from the graphical representation of my linear function exactly the same thing which I did algebraically. When A is greater than 0, this goes to minus B, and I divide by A. And I will have exactly the same. I retain the sign less and less. When I divide both sides by A algebraically, my sign of inequality is the same, is retained. If, however, A is negative, my algebraic approach said that I have to change the sign. And as you see graphically, it basically gives you exactly the same result. So this is another explanation, if you wish, of the rule which you were talking about many times before, multiplication or division of the inequality by negative number must actually reverse the inequality sign while the positive number doesn't really change anything. OK, that's it for this particular lecture. I do recommend you to review the notes to this lecture on unizord.com. It's basically the first one in the linear equation category. And I will continue talking about linear inequalities in the next lecture. And well, it will be a slightly different approach to solving inequalities with absolute value and linear functions involved. And that's it for today. Thank you very much.