 In this example, I want to solve a linear inequality. So we have a linear expression for x plus 7, and we have another linear expression, 2x minus 3. And we want to solve this relationship when the left-hand side is greater than or equal to the right-hand side. When one has an inequality, you have to treat it a little bit different than equations. There are some similarities to it, but in some respect, there's more to it. Now, linear inequalities are simple enough that you probably are comfortable solving them very similar to how one solves a linear equation. The only difference is that when you multiply or divide by negative, you remember to switch the inequalities symbol around. And that's all there is to it. Well, that's great. If that made sense to you, I'm perfectly happy with that. But what I'm going to do is going to throw, well, I'm going to put a pause button on that. What I want to do in this example is present an alternative way of solving linear inequalities. And admittedly, it's going to look more complicated than it ought to be. And this is true. I'm not going to deny that fact. But we're going to be doing it in a slightly more complicated way for the following reasons. The strategies we do use to solve linear inequalities sort of the easy way, they're limited in their scope. They're only going to be basically useful for solving linear inequalities. What happens when we want to start solving inequalities involving absolute value, involving quadratic expressions, polynomial expressions, rational expressions, square roots and radicals, exponentials and logarithms, or any other type of function family? Turns out the simple approach to solving linear inequalities isn't going to help us out here. But there is a universal strategy we can use, which we're going to talk about here. And we're going to practice it on the linear inequalities because linear inequalities are simple enough that our other technique could be used to double check our answer. But if we're patient and try this more evolved approach, we're going to see and we get good at it now. When we start doing harder inequalities, we don't have to start from scratch there. We're actually in a pretty good position. So when it comes to solving inequalities in general, I have the following idea. If you want to solve some inequality, what you're going to do is you're going to first, you're going to first switch the inequality basically into an equality, switch into equation, and solve the equation. So that's like the first thing you want to do. Solve the associated equation. So if you ignore the inequality symbol, you have an equal sign to solve it. And so as this is a linear equation, we've learned how to do these things before. I would begin by trying to combine like terms. So looking at 4x plus 7 equals 2x minus 3, I would subtract 2x from both sides. The 2x cancels here. And this would then give us 2x plus 7 equals negative 3. And then minus 7 from both sides. The 7's cancel here. We get 2x equals negative 10. I would divide then both sides by the coefficient of 2. And this would then give us x equals negative 10 over 2, a.k.a. negative 5. So when you solve the linear equation, you get this number x equals negative 5. This negative 5 is going to be significant to us. The solution to the inequality is not negative 5. We actually expect when we solve inequalities that the solutions will be intervals because we have multiple x values. But this negative 5 is going to be significant. This negative 5 is what we're going to refer to as a marker. A marker, what do we mean by that? What we're going to be doing is we're going to be drawing the real axis for all the x-axis here, just as all the words online. And you're going to put a mark on it, a mark associated to negative 5. And negative 5 is then going to separate one side of the x-axis from the other. And to two intervals, you have negative infinity to negative 5. And you have negative 5 to infinity. We've now, basically, if we think of the real line as this infinite rope, I'm going to take my scissors and cut it at negative 5. And that's how we mark the line in these two sections. And so now what we need to do is investigate which of these intervals, because we have two intervals now, negative infinity to negative 5 and negative 5 to infinity, we have to then investigate which of these intervals belong to the solution set of the inequality. And one way to do this is simply to test it. The markers are significant, because the markers are the only places we can switch from being a solution to not being a solution. So as you transition from this interval to this interval, you're potentially switching from a realm of solutions to a realm of non-solutions or vice versa. What I mean is, if you are in this interval right here, negative 5 to infinity, it doesn't matter which number you pick. Every number bigger than negative 5 is either a solution to the inequality or it's not a solution. So we can pick a test point to see who belongs and who doesn't. So that's going to be step two. Once we've identified the markers, we're going to start testing values. And so we have to find a number which lives less than negative 5. I like to pick whole numbers to make the arithmetic a little bit easier. So if you want to be less than negative 5, I'm going to pick negative 6. That's sort of like the next one. Maybe a little bit better choice in hindsight could be like negative 10. Powers of 10 are usually easy to multiply by. It's like negative 10, negative 100. But I'll stick with the example you see on the screen here, negative 6 here. If I want to pick a number bigger than negative 5, I can pick the next integer, which is negative 4. But even easier than that, I'm going to pick 0. Whenever 0 is an option, I want to pick it, because that's the easiest arithmetic. So you'll notice, if we plug in negative 6 into the inequality on both sides, let's see if these things satisfy the inequality. I'm not sure yet. There's a question mark here, because I don't actually know. Is the left-hand side greater than or equal to the right-hand side? So when you plug in negative 6, what's going to happen? 4 times negative 6 is negative 24. Plus 7 is negative 17. That's the left-hand side. When we look at the right-hand side, negative 6 times 2 is a negative 12, minus 3 is a negative 15. And then double check, is negative 17 greater than or equal to negative 15? They're certainly not equal. And in fact, you're not going to expect equality except at the marker, because again, the marker came when we solved the equation. So you're actually going to expect greater than or less than. And here, you should remember, although 17 is bigger than 15, negative 17 is actually less than negative 15. So this is not a true statement. Negative 17 is not greater than negative 15. And therefore, that tells you that the interval, negative infinity to negative 5, is not part of the solution set. Now, I want to caution you, just because one side works, doesn't automatically mean the other side does work. We should verify this with another test form. We're going to try x equals 0 here. When you plug in 0 on the left-hand side, you get 4 times 0, which is 0, plus 7, which is 7. That's the left-hand side. The right-hand side, you take 2 times 0, minus 3, which is 0, minus 3, which is negative 3. And sure enough, 7 is greater than negative 3. This inequality is satisfied, which then tells us that this is part of the solution set. So numbers less than negative 5 do not satisfy the inequality. Numbers greater than negative 5 do satisfy the inequality. So negative 5 to infinity will be part of the solution set. But should negative 5 be part of the solution set? Is the marker itself included? Come back to the original inequality. Notice how we're greater than or equal to. If you have a greater, we found out that things greater than negative 5 are going to be okay in this situation. But we also found out that negative 5 itself will work. If you plug in negative 5, equality will happen and equality is acceptable. So when you have a greater than or equal to or a less than or equal to, that means your markers will be part of the final answer and you're going to put these square brackets as part of your interval because equality is allowed. And we've discovered by our test points that things greater than negative 5 will be there. And so our answer will be the interval negative 5 to infinity inclusive. That is negative 5 is part of the solution set. And this gives us a way of solving for an inequality. This works for any type of inequality, not just linear inequalities. We first solve the equation to find the markers and then we consider test points to determine which intervals do we include and which intervals do we not. Now you'll notice above here, I did draw a graph of a line. What line is being graphed right here? This line that you see on the screen is the line y equals 4x plus seven minus 2x minus three. That is if I move the right hand side to the left hand side, so we get the left hand side minus the right hand side is greater than or equal to zero. This is a common strategy we do. I didn't do it here, but this will be very useful in the future that you actually want to compare one of the sides to zero. And then if you graph that linear equation, you get something that looks like this. One of the advantages of comparing one side to zero is what is greater than or equal to zero mean? This would mean that you're above the x-axis. And so graphically, if we were to graph this line, the left hand side minus the right hand side, we'd see that we want all of those points that are above the x-axis, that's the right hand side of the graph. So if we take a graphical approach, we can avoid all of these test points that we did here because we used the graph instead. One of the advantages is if you know the graph, we can then very quickly find the solution set because we know this is an increasing function. So the positive side is going to be the side to the right of the x-intercept. But the downside is then we have to understand the graphs of these things. For a linear function, the graph is not too complicated, not a big deal, but as graphs get more complicated, you could see why test points could be advantageous for us. But I want to present both methods. You use test points or graphs to determine which solutions do we include. Now, if you do the graphical approach, I do emphasize you're going to want to compare one side to be zero so that we can then grab things that are above or below the x-axis, depending on whether it's greater than zero or less than zero. Let's do another example of this before we finish our lecture number, whatever we are on, number nine right now. Let's do another example. This is an example of actually a compounded inequality, sometimes called a sandwich inequality. This sandwich inequality just means it means that you can break this up into two pieces, right? So there's negative five is less than 3x minus two and then you also have to do 3x minus two is less than one. And when you have these two inequalities together, this is an and statement. You want this and this to be true. Both statements have to be true in order for this to work. Now, what we're going to do is we're going to solve the associated equations. Now, there's actually two equations here. There's three x minus two equals negative five, but there's also three x minus two equals one here. We actually get to solve these two equations simultaneously because it's the same arithmetic operations. To solve for x, we're going to add two to both sides. Now, the two will cancel on the left. Make sure you add two to negative five, which gives you negative three and add two to one, which gives you a three. Now we had to solve for x, we divide both sides by three, right? Negative three divided by one is a negative, sorry, negative three divided by three is a negative one. And then three divided by three is one here. So because of the compounded inequality, we're going to get two markers. We're going to get x equals negative one and x equals one. And so we place these two markers on the x-axis and this then gives us three intervals. Since there's multiple equations in play here, I'm not necessarily going to worry about inequality or graphing this thing. I'm just going to do this purely from a test point point of view because the test point, you don't have to understand the graph. You just plug in the number and see what happens. So for these test points, let's see, between negative one and one, there's zero, I like that. Zero's always a great choice. Bigger than one, I'll just take two. And less than negative one, I'll take a negative two, those seem simple enough. Let's try these things out here. If we plug negative two in the middle, we're going to get three times negative two which is negative six, minus two is a negative eight. Now negative eight is less than one, but negative eight is not bigger than five. So because the left one didn't work, it turns out this doesn't work at all. And so therefore we cannot include this interval. When you check the middle one when x equals zero, three times zero is zero, minus two is negative two. And sure enough, negative two is less than one, but it's also greater than negative five. Because both inequalities are satisfied, that means this interval passes the test. Negative one to one will be part of the solution set. And then if you check two here, three times two is six, minus two is four. Four is greater than negative five, so that one works, but four is not less than one. It's actually bigger than one. So this one fails as well. For an and statement, both have to be true in order for and to be true. So if I'm like, it's gonna say, okay, kids, let's go to the movies and get ice cream, right? If we go to the movies and I don't get them ice cream, that makes me a liar. If I get them ice cream and don't go to the movies, that makes me a liar. To keep my promise, I have to take my kids to the movie and buy them ice cream. So and is important. Both things have to work here. And as such, the solution here is going to be the interval negative one to one. So that was the only interval that passed the test. We are not gonna put brackets here because we have strict inequalities. We want less than and less than, right? Equality wouldn't work. If you plugged in say X equals one, you're gonna find out that you're gonna get one, like if you plugged in one right here, you're gonna get that negative five is less than one, which is less than one. That's not true, right? One is not less than one. You are not shorter than yourself, just like I'm not older than myself. The comparison doesn't exactly work there. So the markers are not gonna be included inside the solution. We're gonna get an open interval with parentheses here. Before we end this video, I do wanna mention one slight variation to this problem. Let's clear the board and start this thing all over again. So we had said that this inequality is the same thing as saying negative five is less than three X minus two. And we have three X minus two is less than one. So there's two statements here and it's connected by an and. Another word that you should be familiar with in mathematics is the idea of or, of or. Sometimes we have to deal with situation where there's an or statement, which or means that at least one of the things is true. This one has to be true or this one has to be true. But y'all have to be careful here when you have an or statement. In English, when we say things like, oh, should kids, should we go to the movies or should we go get some ice cream, right? It typically means exclusion, right? That you'll do one, not the other. In mathematics, that's not how we use the word or. When you use or, you mean one of the things, at least one of the things is true, but they both could be true. Like if I tell the kids, oh, tonight we're gonna go to the movies or we're gonna get ice cream. If we then go to the movies and then afterwards we get ice cream, who's gonna complain about that? You got both of your things. But in order for me to be an honest parent, I have to at least take them to the movies or I have to at least take them to the ice cream. If I don't do both, I'm not a liar. But if I do both, well then my kids are gonna think I'm a great dad, right? So with an or statement, at least one of the things has to be true. So when you come and check these test points, you're gonna get the same markers of negative one and one. When you look at the checkpoints here, with an or statement, you'll see that although this one doesn't work, this one does, at least one of them is true. So this will be included as part of the solution set. And the middle case, both of them worked because that's was the and solution. So that's great. But then the last one here, you see that, oh, this one works, but this one didn't, that's okay for an or statement, at least one of them is true. And so for this situation, if you had an or instead of an and, then your interval would actually be all real numbers, negative infinity to infinity. And so or does kind of change things. Now when you do an or statement, you're typically gonna be looking at these the other way around. Something that's a little bit more interesting would be this. If you switch your inequalities around like this, right, is negative five greater than three x minus two? Where does that happen? Like coming back to our intervals right here. So you'll know in this situation that negative five is greater than negative eight. So that's an or statement. Negative, negative eight's not. So I mean, if we're switching these inequalities around, right, oops, get rid of that. So if we want this direction that passes, if you switch this around, that fails. But an or statement would still get that one. So you would get this interval this time. If you flip the inequalities this time around, you'll see that neither one is true. Negative five is not greater than negative two, and it's not greater than one. So neither work, so you don't get that one. And then finally, if you switch the inequalities around, negative five is not bigger than four, but four is bigger than one. So you'd grab this one still. And so by having this or statement with some different inequalities, your solution would look something like negative infinity to negative one, union one to infinity. You get this gap, because you want this portion and this portion. So these are some examples of compounded inequalities. When it comes to linear inequalities, we don't usually do a lot of ors because they usually come in this form and actually there's an and. I wanted to make a mention of it right now because as we go in the future, you start looking at things like absolute values, inequalities, quadratic inequalities, or actually does come into play in those situations. So I wanted to give you some exposure to that right now when things are still fairly simple in comparison. And that's gonna bring us to the end of chapter, the end of lecture nine, which is our first section in chapter two. We'll talk some more about linear functions and of course in the next section. So take a look at that with the link. You should hopefully see you right now. Bye everyone.