Common Core Algebra I.Unit 2.Lesson 11.More Work with Compound Inequalities

Hello and welcome to another common core algebra one lesson. By eMac instruction. My name is Kirk weiler, and today we're going to be doing unit two lesson number 11 more work with compound inequalities. Before we begin, let me remind you that a worksheet and a homework assignment that go with this video can be accessed by clicking on the video's description. Or by going to our website at WWW dot E math instruction dot com. As well, don't forget about the QR codes on the upper right hand corner of the worksheets. Those will allow you with a tablet or a smartphone to scan the code and come right to this video. All right, let's begin. In the last lesson, we talked about compound inequalities. And we actually ended it by doing some problems that we're going to begin with today. All right? We're going to graph some inequalities that have compound inequalities that is. So two inequalities at once that are connected by either the word or the word and. So let's take a look at it. Exercise one asks us to graph each of the following compound inequalities on the number lines provided. For C and D, write the inequalities as a single statement. Well, we'll take a look at that in a second. Now remember, with or right, we want to graph all the numbers that make either one of these true. Doesn't have to make them both true. But it doesn't have to. So all the numbers that are less than one. Well, those are the numbers from one and moving to the left, right? Any number that I just shaded going forever to the left would solve that. Likewise, any number that is at four or to the right of it will make this true. So this is X is greater than greater than or equal to four. And this is X is less than one. And that's actually the graph of our inequality. Okay? Letter B, X is greater than 7. I don't remember open circle on the 7, greater than we shade to the right. X is less than negative two. So negative two open circle because we don't include the negative two and to the left. All right. Now, the scenario where we have and with and both must be true. Must about must be true. Apparently I'm not a very good speller. Anyway, that's okay. So let's think about this for a second, right? X is greater than negative three. That looks like this. X is less than 5. That looks like this. The numbers that make them both true are where they overlap. So all the stuff in here, right? Any number in here will be colored for both inequalities. So the correct graph looks like this. All right. No coloring it in. Now, a single inequality. There are often inequalities compound inequalities that involve and are written as a single inequality. Because notice, there's really only a single chunk of the number line graft. So we can actually write this in the following way. All right, now this is a little bit confusing. So let's make sure that you have it. This is something that can confuse students for years and years. So the sooner that you get comfortable with it, the better. Really, this is just saying that X is greater than negative three, while at the same time, X is less than 5. Notice they're both less than symbols, right? They're both less than symbols. Very often teachers will read this as X is greater than negative three and less than 5. Okay? But again, notice both less than symbols. You'll get crazy stuff from students like this. And this really doesn't make sense. This means X is greater than negative three while at the same time X is greater than 5. That's fine, but that would really just be X is greater than 5. So not that. Let's take a look at this one. X is less than or equal to 9, that would look like this. All right. X is less than or equal to 9. And X is greater than or equal to zero. All right, the overlap is right in here. Solid dots there, right? Those two are included. And what we would say then is zero is less than or equal to X less than or equal to 9. That's our single inequality. Typically, they're always written with the less than symbols. You could conceivably write it like this. But it is almost never written that way. That's right. To write it that way, but it's almost never written that way. So we'll cross it out. All right, I'm going to clear this out. Think about what I said. Think about what I wrote down. And then we're moving on. All right, here it goes. All right. Now we're going to kind of go in reverse. It says graph each of the following. First, rewrite as two inequalities involving the and connector. So when we see something like this, what it really means is that X is greater than or equal to negative four. That's this one. Now I realize all of a sudden I didn't use the less than I used the greater than. And X is less than 6. Right? So what does that look like? Well, X is less than 6, looks like this. X is greater than or equal to negative four looks like this. So here's where the overlap open circle here. Okay. The 6th does not make the compound inequality true. That's it. Likewise, this one, X is greater than or equal to negative 5. And X is less than or equal to 9. So again, kind of like this. Beautifully drawn like this, so the overlap is here. And fill it. All right. That's it. So really, this is more convention than anything else. Very often when we have chunks of the number line. Chunks of the number line, like this. Then very often, people will write those as a single inequality, a single inequality. Because even though there are two inequalities joined by and it is easier and takes less space to write them as a single one. A lot of students get confused about that though. In the next lesson, we'll be looking at something called interval notation and interval notation makes it much easier to write these things out. So you'll like that a lot. But we'll get to that in the next lesson. I'm going to clear out the text, and then we'll get into solving some solving some compound inequalities. Alma said systems of inequalities. And that will come a little bit later. All right. Let's take a look. Ah, one more before we do some solving. For each of the following graphs, write a compound inequality that describes all of the numbers shown graft. Now, notice here there are two. There are two distinct shaded portions. So really, what we have is we've got an or statement. Anytime you have two distinctly shaded portions, you have an OR. Right here, we have X is less than negative one, right? We're starting at negative one and every number less than it gets shaded. Here we have X is greater than or equal to 5, because we're starting at 5, and we're shaving every number, including the 5 greater than it. So our compound inequality, X is less than negative one, or X is greater than or equal to 5. On the other hand, here we have just a single portion shaded. That means that we have an and scenario going on. And scenario. Now we can write it as an and I'm going to write it two ways, right? Here, what we've got are all values of X that are greater than what, 6, 7, 8, negative 8, and then here we've got all numbers X less than two. So one way of writing this is X is greater than negative 8. And X is less than two. But what would be more common is writing it as that single inequality, negative 8 is less than X is less than two. So that would be more common. Okay? Now we can move on. I'm going to clear out the text so pause the video if you need to. Okay, here we go. Sorry about that. I moved on to the backside of the sheet too quickly. Now let's do it. Exercise four. Determine whether each of the following values of X falls in the solution set to this compound inequality show the work that leads to your answer. To each answer. All right? So I'd like you to pause the video right now. You should be able to do this. We did it last lesson. You should be able to do this to figure out if these values of X fall in the solution set or don't. So why don't you go ahead and do that? Pause the video. It might take a little while. I'm going to take you up to 5 minutes to do those. Okay, not each, just all together. All right. Let's go through it. And remember, we have an and here. So for an and write both have to be true. Both true. All right, so let me do this. For 5, let's take a look. 6 times 5 plus one greater than or equal to four. 30 plus one greater than or equal to four, 31, greater than or equal to four. That's true. Got to check the other one though. Negative two times 5 plus 8 is greater than negative 12. Negative ten plus 8 is greater than negative 12, negative two is greater than negative 12. You might have to think about that one a little bit, but that's also true. So the answer is yes, it is part. So ultimately speaking, when we eventually shade the thing down here, 5 better be shaded. Let's try negative three. All right, so we have 6 times negative three plus one greater than or equal to four. Negative 18 plus one is greater than or equal to four. Negative 17 is greater than or equal to four, that's false. Now I'm not going to even try the other one. Think about why that is just for a moment. All right, I'm not going to try the other one because the only way an and can be true is if they're both true. So as soon as I know that the first one is false, then I know because it's an and if it were an or totally different scenario. But because it's an and I know that it's not in the solution set. So remember, when we eventually solve in shade, the 5 better be shaded, but the negative three can't be. All right, let's do the ten. 6 times ten plus one is greater than or equal to four, 60 plus one, greater than or equal to four. 61 greater than or equal to four. That's true. Let's see, negative two times ten plus 8 is greater than negative 12. Sorry. Negative 20 plus 8 is greater than negative 12, negative 12 is greater than negative 12. Close, but that's false. So again, no. Negative 12 is not greater than negative 12. So tension will be shaded either. All right, let's solve this beast. All right, 6 X plus one is greater than or equal to four. I'm not going to even worry about the other. I'm going to use the additive property of inequalities to add negative one to both sides. And then I'm going to use the division property of inequality divide both sides by 6, watch out three divided by 6 is one half, right? So X is greater than or equal to one half. Okay, negative two X plus 8 is greater than negative 12, additive property of inequality, add negative 8 Bo sides. Negative two X is greater than watch out. This would be negative 20, divide both sides by negative two, watch out, watch out, watch out. We just divided by negative. We need to switch the inequality X is less than ten. So these two things are connected by and let's remember this. Okay, what do we have? One half would be right about here. So that's this. And then less than ten. Is this so our answer shaded on the number line would be this? Now, technically, we also should write down this. X, if we write it down as a single one, here's our answer. Or we could write down the following X is greater than or equal to one half, let's use the word and must use the word and must use the word and X is less than ten. All right, isn't that kind of cool? So we solved the two inequalities the way we normally would. We shade the way we normally would, and then we take the overlap. Okay? So pause the video. All right, here we go. And it's gone. All right, moving on. Let's do one with or. Or now remember with or the only way it's going to be false is if all of them are false, okay? All of them have to be false. So let's try negative 6. Let's do one half times negative 6 plus four. It's less than 5. Little fraction work here. One half times negative two is less than 5. And negative one is less than 5. Now that's true, and right now I'm not going to even test the other one. Again, think about why? Good. Hopefully you figured it out. If any part of an or statement is true, the overall thing is true, so yes. So negative 6 better be shaded. When all is said and done, negative 6 better be shade. Let's try zero. Let's see, one half zero plus four is less than 5. One half, four is less than 5. Two is less than 5. That's true. So again, I can say yes, I don't even have to check the other one. Let's do 8. One half, maybe I'll get lucky I'll never have to test the second one. That would seem weird. Let's see, one half of 12 is less than 5, 6 is less than 5. No, that's false. Now that doesn't mean it's overall false, it just means that part of its false, let's see the other part. It's the first time I have to test that. Negative two times 8 minus four is less than or equal to 14. Negative two times four is less than or equal to 14. Negative 8 is less than or equal to 14. That's true. So yes, so all of those things are going to have to be shaded. When I'm done with this problem, they all have to be shaped. So let's do it. Let's do exactly what we did before. Let's solve the two inequalities. Using R properties of inequality, first thing I'm going to do is multiply this one both sides by two. Because it's a positive. That's fine. X plus 5 is less than ten, then I'm going to use the addition property of inequality. And I'll get X is less than 5. All right, so that's one part of it. Let's see, let me go with this one now. Negative two times X minus four is less than or equal to 14. I'm going to divide both sides by negative two and watch out, right? That flips the inequality. So now it's greater than or equal to negative 7. I'm going to add four to both sides and that's going to give me X is greater than or equal to negative three. So let's look at this for a second. We got X is less than 5. That's this. And it not mean to draw the line through the circle there, but there's probably no way I'm going to be able to. There we go. We have X is greater than or equal to negative three. Now I want you to think about this. This is going to be true for any part that gets shaded at all. But if you think about that, we're not talking about the overlap, we're really talking about absolutely everything, right? Because numbers down here will get shaded by this one. Numbers up here will get shaded by this one and numbers in the overlap will be shaded by both. Now remember, that's not a problem. It's okay if they're both true. There's nothing wrong with that. So the correct answer is this. What's really neat is that it means that all numbers. Solve this compound. Inequality. It's my hybrid cursive print technique. It's not pretty, but more or less gets the job done. It's a little weird. All numbers make it work. That's the beauty of or it allows for a lot of answers. All right, I'm clearing this text out. Pause the video if you need to. Okay. Let's wrap it up. So today we worked a little bit more with compound inequalities. And we saw how oran and play app and how we can graph them. And sometimes represent inequalities involving ands with a single inequality, a compound with a single inequality. That's actually maybe the most important thing to come out of today with because you'll see it the most often as you move forward in math. But we'll work with it more. So it will reinforce it quite a bit. Especially in the next lesson. All right, for now though, I want to thank you. Thank you for coming for another common core algebra one lesson. Instruction. My name is Kirk weiland, and until next time. Keep thinking. And keep solving problems.