 Hi, I'm Zor. Welcome to InDesire Education. I would like to talk a little bit more about linear inequalities. However, a very simple linear inequality approach was actually explained in the previous lecture. Here I would like to expand this a little bit into linear inequalities which are related to inequalities with absolute values of linear functions. Just as an example of what I mean, you have an inequality, for instance, like this. Absolute value of one linear function minus, in this case, plus whatever, of another linear function is less than zero. It's not exactly linear inequality. However, what's important is it can be analyzed and basically solved as a result of this using an approach which was explained in the previous lecture of linear inequalities. But you have to do something before using these techniques for playing linear inequalities, as I would say. So what are these techniques? Well, let's think about absolute value. What is an absolute value of the number? Well, the absolute value of the number is by definition. So I'm not really asking anybody to prove for anything like that. It's equal to z for z greater or equal to zero. So for all non-negative numbers, z, and I'm talking about real numbers, of course. So for all real numbers which are greater or equal to zero, absolute value of z is exactly the same as z. The absolute value of two is two. Absolute value of 25 is 25. Absolute value of zero is zero. So I don't have to do anything with the number z if it's non-negative to obtain the absolute value of z. I just take it as it is. Now, what if z is negative? Well, if z is minus two, then we know that its absolute value is two. How can I get two from minus two? Well, I have to negate it. So if I negate z, which is negative by itself, then I will get the corresponding positive, right? Minus of minus two is two. Minus of minus 25 is 25. So if I negate my negative number, I will get its absolute value. So this is the definition. I was just trying to explain it. It's not a proof. It's an explanation why this definition is important. So if I want to approach my absolute value of something, I have to first of all figure out where this something is equal to zero. And to the right of this value, I can replace my absolute value with the number itself. To the left of this value, I should really replace it with minus this particular number. Now, if I have a combination of certain number of absolute values of something, the best approach is to have the x-axis and mark where exactly each one of those is equal to zero. Well, obviously this one is equal to zero at minus two, and this one, x minus one, is equal to zero at one. Now, these two points divide my x-axis, which means all the real numbers, into one, two, three intervals. So this is one interval. This is another. And this is the third. And on each of these intervals, I can consider this particular inequality. And on each of these intervals, I can basically convert it into a linear inequality. And solve it. And let's do just this. So we have three different cases. This is area A, this is area B, and this is area C. So let's consider area A. Now, A is x less than minus two, right? Everything to the left of minus two. What do I have? I will free some space for... Okay. So what do I have? x plus two to the left of this, which means this thing is negative, x plus two. So to get the absolute value, I should reverse the sign. So it would be minus. And now I have a regular parenthesis, x plus two minus. What about x minus two? Well, if I am here in this area, it's two to the left of this one. Now, this point is where the expression x minus one is equal to zero. And everything on the left, which means this as well, would necessitate to replace absolute value with a negation of this. So it would be minus minus x minus one. And that should be less than zero. Well, this is a linear inequality. So I replace my x plus two with minus absolute value of x plus two with minus x plus two. I replace my absolute value of x minus one with minus x minus one. And what do I have now? Well, let's open parenthesis. Minus x minus two plus x minus one less than zero minus three less than zero always, which means that if x is less than minus two, I don't have any new restrictions on x, which means all x's which are less than minus two fit my inequality. Which means it's a solution. It's one of the solutions. All right. Next. So this is a solution. It's one of the solutions. Next, let's consider that minus two is less or equal to x less than one. Notice that I'm using less than equal here, not just less than, just in order to include minus two. By the way, in the definition of the absolute value, z is equal to z if z is greater than equal to zero and minus z if z is less than zero. Actually, it doesn't really matter whether we use this or this for z equal to zero, because it would be zero anyway, right? Plus zero or minus zero. So that's why we have to include the points minus two and one into one of the areas. So in this particular case, I always include the point to the area which is to the right of it. So in this case, the area b. That's where minus two will be included. And in this particular case, what I can say is that this particular absolute value should really be like this, because it's positive, right? If x is greater or equal than minus two, I can replace absolute value of x plus two with just x plus two, because it would be zero or positive. Now, up to this point, which is where the b area is ending, this expression will still be negative, which means I should really replace it with its negation. So what's the result of this? x plus two plus x minus one less than zero. Two x plus one less than zero. x less than one than minus one half. Minus one half is somewhere here, I guess. But this is zero minus one, so minus one half is something like this. Now, don't forget that I am supposed to be in this area, so it's not just any x which is less than one one half, but only those x less than minus one half which belong to this area, which means only these guys. Now, in the first one, we got this area completely, everything which is less than minus two. In the second case, we are considering this area, this area, but only those x which are less than minus one half fit the bill. These are not a solution to our equation. And, so let me put it here, minus two less than x less than minus one half. That would be a true solution x less than minus one half, but it's still supposed to be greater. So this should be a really true solution to this. So the first area gives me less than minus two, and the second area gives me starting from minus two including, by the way. So there is no gaps here, the point minus two is included, up to minus one half. Now, the third one, the third area, c is this one. So x is greater or equal to, well, let's put it this way, one less than x. So I prefer to have always less signs. Alright, what's here? Well, we are to the right of this point, where this is equal to zero, and obviously to the right of this point. Which means everything, all absolute values should be just dropped. So it would be x plus two minus x minus one less than zero. Which means x plus two minus x plus one less than zero. x goes out three less than zero, which is not right. Which means none of the values in the c area will be a solution. So there are no more solutions. We found all solutions in the a area, all solutions to the b area, and non-basically no solutions in the c. So only the numbers from minus one half to the left are actually solutions to our equation. So we can combine this and this because they are basically glued together, these two areas. And we can conclude that the whole solution to this particular inequality is x less than minus one half. This is a total solution, which basically we have found out by combining these two. Everything to the left of minus one half. And, well, obviously if x is equal to minus one half, I guess we have to expect it to be equal to zero, right? Let's check it out just in case. So this would be what? One and a half, right? Two minus one half is one and a half. This is minus one half minus one, it's minus one and a half, but absolutely it gives me one and a half. So it's one and a half minus one and a half equal to zero, which is exactly what we were looking for. So if at the point x equals to minus one half, this is zero and everything to the left of it would be less than zero. And incidentally everything to the right will be greater than zero. Now, just as an illustration, let's do it graphically. Just to confirm that we are right. So first we have to build the graphs of this and this and then subtract one from another, right? All right, so this would be y equals absolute value of x. Y, because for positive x it's supposed to be the same as y equals x. For negative x it would be the same as y equals minus x. So I just combine these two pieces of two graphs and that's my graph y equals absolute value of x. Now, plus two, it means I have to shift the graph to the left by two units, right? This is one, this is two, this is minus one, this is minus two. So my graph of this particular element, absolute value of x plus two, should look like this. This is original which is just used to shift it. So this is y equals absolute value of x plus two. Now, x minus one means the graph should be shifted to the right by one. And the graph would be this, obviously. This is y equals absolute value of this. Now, we don't need our original absolute value of x. We used it completely to shift it left and right. Now, we have these two graphs and we have to subtract one from another. From this graph, we have to subtract this. Again, the best thing is to use the key points, exactly where the graph changes the behavior from decreasing to increasing. Because in between these points, graph behaves as a linear function, right? Because I can always say that above one, it's yx plus two minus x minus one. So I can drop the absolute value. And in between from minus two to two to one, I can drop this one, absolute value, just leave it as it is, and change the sign of this. It's still a linear function, so everywhere it's linear functions. So let's find what exactly the key points and then we'll think about it. What will be in case x is equal to one? This is zero and this is three, right? So if I subtract, this would be, I'll use the red. This would be my point of the new graph, which is a difference between these two. And that would be three. Now, minus two, minus two, this is equal to zero. Minus two, minus one, it's three, absolute value, three, minus, it will be minus three. Okay, so the graph would be here at minus three. So we have two points here, be here. Now, in between these points and outside of these points, the graph is a linear function. So I can just do it exactly. By the way, at zero, it's equal to two minus one one, right? So it would be here. So the graph would go like this. It's a linear function. And it's very easy to see right now that we have crossed the x-axis in between zero and minus one. So that's where minus one half actually is. What would be outside of this? Well, think about it this way. As I'm moving to the right, I'm adding something to this graph and I'm adding exactly the same thing to this graph, which means the difference between them, which is minus sign, should not change. So whatever the value was here, which is three, should be retained and the graph should go constant. And if you remember, the area C was like three, supposed to be less than zero and no solutions, etc., but three it was on the left side because it's a constant. Now, same thing here. If I move to the left, one graph is increasing and another graph is also increasing, but exactly by the same value, which means the difference between them should remain the same, whatever it was before, which is minus three, which means it should be constant minus three. And again, in area A, if you remember, we had this particular expression as equal to minus three, minus three less than zero, all this, which means all the x's on the left correspond to our equation. So this is a graphical confirmation that starting from this point and to the left, which is minus one half, exactly what we have basically derived, the graph is negative and that's exactly what we wanted to get. So this lecture was about how to approach inequalities which contain absolute value. Remember, if you have an absolute value of some expression, in this case it's linear expression, you always have to find the key point where it's equal to zero. And then, if you have many expressions, like in this case I have two, then mark all these critical points where each particular absolute value is equal to zero and consider your inequality on each interval, these points are dividing all the real numbers. Well, that's it for this lecture. Thank you very much. And probably I will include a certain number of interesting problems in the next lectures about inequalities. Thank you.