Common Core Algebra I.Unit 2.Lesson 6.More Linear Word Problems.by eMathInstruction

Hello and welcome to another common core algebra one lesson by E math instruction. My name is Kirk weiler, and today we're going to be doing unit two lesson number 6 more linear word problems with consecutive integers. Before we begin, let me remind you that you can find the worksheet and a homework that go along with this video by clicking on the video's description. As well, don't forget that on the right hand or upper right hand corner of every worksheet we've got our QR codes, which can be scanned with your smartphone or a tablet to bring you right to this video. All right, let's begin. In the last lesson, we looked at a number of different types of word problems. All right? Many of these had to do with real world scenarios. And we had to translate English into mathematical phrases or into mathematical expressions and equations. And then we went out and solved them. Today, though, we're going to concentrate on problems known as consecutive integer problems. So let's just review a few terms. All right, first, the integers. The integers of the set of whole numbers, both positive and negative, right? So those little dots at the end, the dot, dot, dot, dot, dot, dot, just means that it goes on forever. Maybe it's important to note what aren't integers. Things like one half, or two thirds, or irrational numbers, like pi. These numbers are integers. But the numbers that you see listed are. Now, consecutive integers, that's going to be the theme for today. Consecutive integers are a list of integers that are in order. All right? It's got to be at least two. Obviously, otherwise it wouldn't be plural. But it could be two, three, four, 5. We're going to be mainly dealing with two consecutive integer problems and three consecutive integer problems. But a list like 12 comma 13. Those are two consecutive integers. Three, four, 5. Those are three consecutive integers. Don't forget, we could have negatives, right? Negative 5, negative four, negative three, negative two. Those are consecutive integers. We also talk about things like consecutive events and consecutive odds. Now, I'll assume that you know what are evens and what odds. But two four 6 8 would be a great example of four consecutive even integers. Right? Negative 6 comma negative four would be two consecutive. Even integers. We could also have consecutive odds, like 7 9. That'd be a good example of two consecutive odd integers. Now watch out, we can even have kind of confusing consecutive auditors like negative one, one, three. If you don't believe me, right? Those are definitely consecutive. Take a look right up here. Here's my negative one. Then my one, then my three. Consecutive. Boom. Boom. Right? So it's important to just understand the terminology. Integers are whole numbers with the negatives. Consecutive integers are integers that are in a row that are in order from least to most. So I'm going to clear this out real quick, not like I did much writing here, but I've got to get rid of it anyway. And we are going to launch into some problems. All right, let's take a look. Exercise one. Let's work with just two consecutive integers first. Say we have two consecutive integers whose sum, that's an important term underline it. Whose sum is 11 less than three times the smaller integer. All right, letter a we're going to do exactly what we did in the last lesson. We're going to play around with these a little bit. To see, just to make sure that we have kind of a sense for it. Okay? So try a variety of combinations to see if you can find the correct pair of consecutive integers. Be sure to show your calculations. So I'm going to do one and then maybe have you do one. So let's try to and three. Let's see if those are the two correct consecutive integers. What does it say? It says there's some, so two plus three is 11 less than three times the smaller integer. Now this is important. Three times the smaller integer, that's easy enough. 11 less will be subtraction and it's important that it's this way and not 11 minus three times two. I know the 11 came first in the sentence, but when we say that a number is 11 less than another number, we're taking that number and subtracting 11. So let's see, is this a true statement? Well, here we have 5. Here we have 6 -11. Watch out. 6 -11 is negative 5. So that's in fact false, which means two and three aren't the correct choices. Pause the video now. Pick another two consecutive integers. Try any two you want 9 comma ten, 20, comma 21, get exotic, go negative, negative 11, negative ten. But try to and see if you come up with the right to. And then we'll go on and do it algebraically. All right, let's go forth. Now, again, I want to try to avoid doing the right pair. So let's try ten and 11. All right, so first thing we're doing is we're summing them ten plus 11. And we're seeing if we get 11 less than three times the smaller. All right, so here's their sum, and here's 11 less than three times the smaller. All right, so their sum is 21, and then we get 30 -11. 30 -11 is 19. That's close. You know, it's actually relatively close, but it's still false. Okay? So ten comma 11 isn't right. All right, let's do letter B now let's carefully set up let statements. So let's let N be the smaller integer. All right. Now, what's the larger one? Well, consecutive integers are always separated by one unit. So to get, let's say from ten to 11, we'd add one. To get from two to three, we add one. So N plus one is the larger integer. You could also say the first and the second, that's fine. What we have the smaller and the larger. Now let's translate what they gave me into a sense. Now, what I really want to do, and let's do this in a different color. Come up here for a second. Here's my end. Here's my N plus one. Here's my end, likewise N N plus one. N so let's just do exactly that. Let's take N let's add N plus one. Again, I'm putting it in parentheses because it is its own number. It's its own expression. And then we'll do three times the smaller -11. Right? Now, to solve the equation, what we're going to do is we're going to use the associative property of addition. And we're going to actually group these two together first. All right, giving me two N plus one equals three times N -11. I think I'll use the additive property of equality to subtract two N from both sides. That will give me one is equal to N -11. And then I'll use the additive property of equality again to add 11 to both sides. And get an equals 12. So we have it. What are the two numbers? Well, the smaller is 12. And the larger is 12 plus one. So 13. Now, one thing I like about consecutive integer problems. And I'm going to do the check in green. Is that the check sometimes can be pretty easily done. This said that if we find the sum of the two numbers, 12 plus 13. We will get 11 less than three times the greatest, or sorry, three times the smaller. So we get 25, three times 12 is 36, and in fact, 25 is 25. So we know we have the right answer. Okay? But look at how we've extended what we did in the last lesson. We play around with the problem set up by just guessing and checking. We keep track of what we're doing with the numbers, and eventually that's what we do with the variables and the expressions as well. Okay? So pause the video now if you need to, write down what you have to, and I'm going to clear out the screen. All right, here we go. Let's move on. Let's do some more. Okay. Exercise two. I'm thinking of three consecutive odd integers. When I add the larger two, the result is 9 less than three times the smallest of them. Hold one. What are the three consecutive odd integers? I'm going to switch black to blue, okay? And I'm going to just try this with three consecutive odd integers. Now that's important. Consecutive, odd integers. Okay? So I don't want to go with the ones that I know are right. I'm going to try to avoid that as always. So let me just glance at that. All right. So let's try some easy ones. 5, 7, 9. Let's just try these, okay? Now remember what we're going to do is we're going to keep track of what we're doing to these three things and then we're going to do exactly the same thing with the algebraic expressions. So what does it say? When I add the larger two, 7 plus 9, the result, right? This is the result is 9 less that's going to be subtraction. We're going to subtract 9, 9 less than three times the smallest. All right? Sum the larger two, 9 less than three times the smallest. Now, let's see if we actually did it right. Got the right three. And I kind of guaranteed we weren't going to. But there we go. That's false. All right, now the algebra. Let's lay the algebra out. Okay. So we're going to let N be equal to the smallest or the first. Integer. Now, if this is N, then what 7? Well, to get to the next odd we add two. We don't add one. We don't add three. We add two. So N plus two is the, I guess. Second integer. And now to get from 5 to 9, because we want to base it on the N, we have four. M plus four. Now this is very, very important. A lot of people get confused because they see odd integers, and they think, well, the first one's end, the second one should be N plus one, and the third one should be N plus three, because one and three are odd. But that's not the way it works if you think about odd numbers to go from one odd number to the next odd number. You add two, and then you add two again, add two again. So let's keep track of it, right? And again, let's do exactly what we did before. What was that 7? That 7 was N plus two. Again, I'm going to put it in parentheses. Okay, that 9, uh oh, hello. Not have that happen. Let's go back into here. I love it. All right, that 9 was my N plus four. Right? And that was equal to three times that was just N no parentheses really needed there, but it doesn't hurt to have them -9. Let's solve this beast, okay? We're going to use the associative property again of equality, not a associative property of equality, just the associative property of real numbers, to remove those parentheses. Now I'll do a little bit of commutative property, and I'll rearrange the two plus N and N plus two. Probably many of you might not write out all these steps, but I really want to justify my steps given our lesson that we had a few lessons ago. Now I'm going to do a little bit of associative property. There we go. So that's going to be two N plus 6 is three N -9. Now let's apply some properties of equality. Additive property of equality to subtract two N from both sides or add a negative two N additive property of equality to add 9 to both sides and get N equals 15. So what are the three consecutive odd integers? 15, 17, and 19. It doesn't look like a 9. 19. Now, I really like this problem. This is a problem you can definitely definitely think about reasonableness on. Number one, the answer's got to be an integer. Two, the answer's got to be an odd integer, right? So had I gotten N equals 12.3. Not an integer. If I had gotten N equals 18 eh wrong, not a non hatcher, right? So we should be able to look at these answers and really get a feeling for whether or not they even make sense. All right? And this one definitely does. So I'm going to clear this out. I want you to think long and hard about consecutive odd integers through this problem. Pause the video now if you need to. All right, here we go. Let's move on to the next problem. Okay, look at how look at this. Vast open space that we keep getting on these problems. Let's go back to blue. Blue it is. All right, three consecutive even integers. Three consecutive even integers. Have the property that when the difference between the first and twice the second is found. The result is 8 more than the third. Find the three consecutive even integers. All right, I'm going to do exactly what I did before. I'm going to try it with some, and then I'm going to translate. So three consecutive even integers. Again, I want to stay away from the right answers. So I'm going to try four, 6, and 8. All right, let's try four, 6, and 8. Three consecutive even integers. What does it say? The difference between the first, right? And twice the second. And that's important, right? Between the first and twice the second, the subtraction has to be in that order. The result is 8 more than the third. 8 more is plus 8. So that's a little bit confusing. This is the third and this is 8 more. Sorry if that's confusing. Well, that's the way that we set it up. Let's check to see if it's right. Remember, we have to do this multiplication first two times 6 is 12. 8 plus 8 is 16, four -12 is negative 8. And clearly that is false. So no worries there. Okay? But now, let's do some algebra. All right, I love testing these things out first or trying them out that gives me a basis for my algebra. Again, kind of going over to the red, right? Here, here and here, let's let N be equal to the first integer. Now what do we have? Well, to go from four to 6, we again add two. It's one of the things that confuses people the most about consecutive integer problems is that consecutive even in consecutive odd problems are set up in exactly the same way. So to go from four to 6, we add two and to go from four to 8, we add four. So then we're going to let N plus four be equal to the third integer. Okay? So now what did we do? We said the difference between the first, right? There's my N here's my N plus two. Minus two times N plus two really critical to have those parentheses there. Is equal to 8 more, there's my 8 more. Than the third. All right, so let's take a look at this. All right, we can use the distributive property now to distribute the negative two, and it's important that we distribute the negative two, so it'll be negative two N minus four. Here, we're just going to use the associative property to rewrite this side of the equation as N plus four plus 8. So we're going to group this addition together as opposed to this one. All right, we can combine these light terms by kind of thinking about it this way, and that will give me negative N minus four, right? Think about that for a second. Sorry, I glossed that over. N minus two N is negative one N here, of course, we'll just have N plus 12. Now how do we want to finish this? Well, if you like to avoid negatives, then what you might do is you might use the additive property of equality to add an end to both sides. That will give me negative four is equal to two N plus 12. Then maybe use the additive property of equality to add negative 12 to both sides. That'll give me negative 16 is equal to two N and then use the multiplicative property of equality to multiply both sides by one half or divide both sides by two and we get N equals negative 8. Now there's nothing wrong with that. You might say, oh, I got a negative. It's not reasonable. It's completely reasonable. It's an integer. It's an even, and that's all we were looking for. Now, watch out. This is a little tricky. I'm going to go back to blue. The first integer is negative 8. Second integer is negative 8 plus two, which is negative 6, and the third integer is negative 8 plus four. Which is negative four. Here are my three integers. They are not negative 8, negative ten, negative 12. They're negative 8, negative 6 negative four. And that, even that last little part of the problem can be kind of tricky. Okay? So pause the video now if you need to, and then I'm going to clear out the text. Okay, here we go. All right, let's do one more problem. The sum of four consecutive integers is negative 18. What are the four integers? Well, you know, there's some is negative 18. So I should probably try negative numbers. If I'm going to try this, let's try some negatives. So I don't know. Maybe I'll try negative ten, negative 9, negative 8, negative 7. This is pretty easy though. I'm just summing them. So, and the question is, does that equal negative 18? Well, negative ten plus negative 9 is negative 19 plus negative 8 is negative 27 plus negative 7 is negative 34. And that is false. So yet again, I guessed incorrectly, because I know the final answer, and I didn't want to guess correctly. All right, but here comes the algebra. All right, what do we do? Let's let N be equal to the first, let's let N plus one be equal to the second and plus two be equal to the third and M plus three be equal to the fourth. It's a little weird, I know, because N plus one is two N plus two is third N plus three is fourth. But now watch this. I'm going to add them up N plus N plus one. Plus N plus two. Plus N plus three and of course that's got to be negative 18. Now, watch me use some properties of real numbers. First, the associative property. So the associative property says I can get rid of all these parentheses because I can add in any order I want. All right, I would really have to do a lot of commutative property, some flip flop flip flop flip flop. Ultimately, though, what I'd have is N plus N plus N plus N plus one plus two plus three. So after doing the commutative property, again and again, I'd be there. Then I could use the associative property to say, well, I'm going to add those ends together and get four N, and I'm going to add those together and get 6. And now I can start using properties of equality. The additive property of equality, adding negative 6 to both sides. And then the multiplicative property dividing both sides by. Four. So my first integer is negative 6. Let me write them down in red. My next integer will be negative 6 plus one, watch out. That's negative 5, not negative 7. Then negative 6 plus two, negative four, negative 6 plus three, negative three. So there they are. All right. Nice problem to end on. Not too hard. Pause the video now, write down anything you need to and then we'll wrap it up. All right, here we go. So today's lesson really reinforced a lot of what we did in the previous lesson. In terms of setting up careful LUT statements, write trying things out, trying to establish patterns. Then establishing algebraic expressions and equations based on what we saw with just normal known numbers. All right. The big difference is today. We introduced some terminology, like consecutive integers, consecutive, even and jurors consecutive auditors. If you look at the problems we did though, they will be key to doing any type of consecutive integer problem, including ones later on that become a little bit more difficult. All right. Well, thank you for joining me for another common core algebra one lesson by email instruction. My name is Kirk weiler. And until next time, keep picking. I keep solving problems.