 Hi again, this is Chichu. Welcome to my channel. Now what we're going to do in this video is touch on a sort of a type of problem that comes your way starting off and basically in grade nine or so and then they make it more complicated every year. So come see you in grade nine, grade 10 and then more complicated stuff comes in grade 11 and 12, okay. And that, you know, the category of that problem that people see and people usually have a hard time with it initially until you see the pattern, right. And those problems are related to consecutive numbers, okay. And there's basically three different types of consecutive number questions that come your way. One of them at the beginning anyway in grade nine and grade 10 is generally it's just consecutive numbers, straight out consecutive numbers. The second type of problem that comes your way are even consecutive numbers and the other one are odd consecutive numbers, right. So what we're going to do is we're going to take a look at all three formats and we're going to lay them out here so you see, you know, what the difference between all three of them are and we're going to do one question from each one, okay, starting off with just straight out consecutive which is the easiest ones to do. And then we're going to, you know, in the third one, probably going to hit up stuff that, you know, grade 11 can do and sometimes you see this coming up in grade 12 with the more complicated stuff. If you have like three or four consecutive numbers in a row where you have to start doing some synthetic division, some of the stuff that we talked about in the language of mathematics towards the end of series 3B. So a lot of things, at least two of these questions that we're going to do, you're going to need some of the material we covered for factoring polynomials, trinomials and complex trinomials to a certain degree, okay. So basically let's lay out what we're talking about here. And the type of question that this information is related to is the type of question that you would have seen if you're in grade 9, latest by grade 10, you see this where they say, you know, the sum of two consecutive numbers is equal to a certain number, right, or the product of two consecutive numbers is equal to this, right. So if you get questions where you're talking about consecutive numbers, this is the way you should think about it, right. In mathematics, when we don't know something, we have to sort of put a place marker for that unknown, right. And for us in general, we use the letter x, you know, x marks the spot if you want to think about it that way, right. But x is basically your unknown, right. So if you're talking about two consecutive numbers, adding up, multiplying, you know, their reciprocal or whatever you want to do to them, right. If there's a problem coming to you with consecutive numbers and you don't know what those numbers are, the first thing you do is you call the first number x, right, because that's your unknown, right. So we're going to use x and the x represents a place marker, right. So the first type of question you end up getting are just consecutive numbers. And what we're going to do is we're going to actually just lay out all three of them first. And we're going to talk about how you set up the problems, and then we're going to do the problems, okay. So we're not going to do the problems as we move from one, you know, consecutive to even to odd. We're going to lay out everything. So you have a nice visual, nice understanding of what it is that we are talking about. Okay. So the first type of question we get are consecutive numbers. And this is the way you represent consecutive numbers. And if for those of you who are not sure what consecutive numbers are, consecutive numbers are basically numbers that follow each other, right. So one, two, three, four, five, those are consecutive numbers, 1000, 1001, 1002, 1003, those are consecutive numbers, right. So any numbers that are in sequence in integers right now, that's what we're talking about, right. And in general, what we are specifically only going to talk about is integers. Okay. So let's say you wanted to represent consecutive integer numbers, right. So how high up can I go all the way up here. That's good. So let's call this consecutive integers. Now the way you represent consecutive integers, as I mentioned, if you don't know what the first one is, just call it x, right. So if you don't know what the first number is, you're going to call it x. If they're supposed to be consecutive, following each other, right, the next number in the sequence, what should it be, right. There's two directions you can go. You can call, you can go from, you know, if they have to be consecutive, nothing says that they have to be consecutive going up. They could be consecutive going down, right. So what you could do for the next number to represent the next number, you can call it x plus one, or you could call it x minus one, right. So in general, like working, you know, representing the smallest number first and then working my way up. Okay. So what we're going to do for the next number, we're going to call it x plus one. So for example, if our first number was two, then the next number in sequence is going to be two plus one, which is three. The next number in sequence is going to be two plus two is four. The next number in sequence is going to be two plus three is five and so on and so forth, right. So these are the first type of questions you end up getting. And the type of question will, you know, we'll lay it out, we'll solve it later. The, you know, the type of question you get, they say, you know, the sum of two consecutive numbers is equal to 15, right. So if you don't know what the first number is, you say let x equal my first number, because that's what it is. It's a place marker. We don't know what it is. And then we say let our second number is going to be x plus one x plus one equal the second number. And this let thing is called a let statement. And we talked a little bit about this in in the language of mathematics when we're doing, when we're solving for quadratic equations, little bit more complex quadratic equations, because what let statements do, they allow you to simplify things and to set problems, set questions up where you can actually visualize and see how the thing flows, right. It's just, it's just a way for us to make things easier, right. So if you had, for example, two consecutive numbers, adding up to be 15, all you would do, you would say x plus x plus one is equal to 15, right. Let's do this one. Actually, it's so simple. So what you do, you say x, that's the first number. Or if you want here, we're laid out, you know, set it up so visually you actually see what's going on. You would say the first number plus the second number has to equal 15. That's what we said. The first number plus the second number has to equal 15. Two consecutive integers have to add up to be 15. So our first number is x, right. Our second number is x plus one. That has to equal 15. So 2x, x plus x is 2x. So 2x plus one is equal to 15. And then you bring the one over. So track one. And we talked about this. This video sort of kicks into the previous ASMR video we did and we talked about how to deal with an equal sign, right. How to move around an equal sign. That's essential. That's crucial. You have to know how to move around an equal sign before you can really do mathematics, right. You have to know your basic operations, addition, subtraction, multiplication, division, right, exponents and radicals. But to really be able to do mathematics, you need to know how to move around an equal sign. That's a given, right. That's the essential. That's one of the first steps you need to learn. So this becomes 2x is equal to 14. And then you divide by two, divide by two. Hopefully we're not off the board. We're going off the board. So 2x divided by two is x14 divided by two is seven. So our first number ends up being seven. Or not our first number. Let's say x is equal to seven because our first number might be something else, right. That's dark. Let's use rags to clean the board. My eraser is rough. So our first number, or our x, is going to be 14 divided by two is seven. Well x happens to be our first number. Our second number is x plus one. So seven plus one is equal to eight. So our second number is eight, right. That's two consecutive numbers that add up to give you, where was it? Well, I have to give you 15, right. So that's sort of a question that comes your way in grade, you know, eight, nine, right. Fairly simple consecutive word problem, consecutive integer word problem, right. Now the second type of problem we end up getting is consecutive even integers, okay. So very dirty. I just ended up cleaning the white board. So, you know, this is the first writing we're doing since cleaning it. So what's going to happen? It's going to look a little bit dirty after I give it a little white because the contrast between having black on there and the white in the background is a lot, right. Now the second type of problem we end up getting is consecutive even integers, right. And the type of questions you would get, the same type of question, right, that would say, oh, the sum of two consecutive even integers is equal to 16. You know, what are those numbers? Actually, I'll say 14. Okay, what are those numbers? And, you know, we'll end up doing one after we lay out what it looks like for consecutive even and consecutive odd, right. And we'll reduce, you know, we'll wipe this off and set our table up there and do a couple more problems, okay. So think about it for, before I write down what the answer is, just ask yourself how you would go about representing consecutive even integers, right. Because consecutive numbers was easy. If we don't know what the first number is, we're just going to call it x, right. That's our place marker. Our second number after that would have been x plus one, or it could have been x minus one, right. But I like adding, right. Above this, I could have just put down x minus one and work my way the other direction, right. Or just put x minus one here and work my way this way. The x was our place marker to represent the first number we didn't know. So how would you go about putting a place marker or representing an even number? Because if I just put down x, x could be anything, right. It could be odd as well. So I can't just put down x for consecutive even, right. So we're going to look for consecutive even. I can go up here. Let's call this consecutive even. Consecutive even integer specifically. Okay. Let's turn this into a table. So we need to put in a place marker here to represent an even number. And we can't just put x. If we put x, x can be positive or negative. Sorry, it could be odd or even, right. So x is not going to do it. And if we did put x, what would the next number be if they're consecutive even and consecutive even is, well, they have to be even and they have to be in sequence, right. So two, four, six, eight, those are consecutive even integers. Negative two, negative four, negative six, negative eight, those are consecutive even integers, right. Because integers could be positive or negative whole numbers, right. But if we just put an x and we say it's an integer, then it could be odd as well. So how do we make a number even? And this is a question that before I give the answer to anyone that I'm teaching this initially, I always pause, I ask them. So really answer this question to yourself. How do you represent an even number in a place marker, right. If you have to, in this box, right, in this unknown, how are you going to make sure that number is even, right. You got an answer. Now the best way to make a number even is pair it up, right. x plus x, no matter what x is, is going to be even, right. How does that work? Well, if I write down x plus x, I'm guaranteed to have that as an even number. Agreed? Try it out. One is odd. One plus one is two. That's even. Three is odd. Three plus three is six. That's even. Four is even. Four plus four is eight. That's still even, right. So if you add a number to itself, it becomes even. The simpler way to do this, as we talked about in a lot of videos, instead of just writing add it together, you multiply it by two, right. So the way you can write an even number is not just write down x, but two x. So our first number is going to be two x. Now keep this in mind that x is not our number, our first number. Two x is our first number. That's really important. People, you know, you've seen the problems. We end up solving these things. We end up solving for x and a lot of people make the mistake of calling x the number. x is not number. Two x is the number, right. So if two x's are first consecutive even integer that we're looking for in some kind of problem that we're given, you know, two consecutive numbers, multiply to equal something or add to equal something, then how would you represent the second number? Now if you start off, let's say your first number was four, to get to the next number you either add to or subtract, right. So the next number in our sequence would be two x instead of plus one, it'd be plus two, right. Two x plus two or minus two, but again I like working up instead of working down, right. So if our first number is two x, our next number is going to be two x plus two. The next number after that is going to be two x plus another two, which is four. So it's going to be two x plus four and then two x plus six and so on, right. That's how you represent consecutive even integers. Your first number is two x, your second number is two x plus two, two x plus four, two x plus six and so on. So on that mind frame, right, in that mindset, how would you represent consecutive odd integers, right. Consecutive, how do we represent consecutive odd integers? Do you have an answer? You can't just multiply it by three, right. Because if you multiply it by three, sometimes it's going to be odd, sometimes it's going to be even. One times three is three, that's odd, okay. Two times three is six, that's even. So it turns out, I hope you've thought about it, I hope you've tried some answers, right. Because it's really important to think about problems and try to see how you can represent them mathematically instead of being given the solution, right. But the way to represent consecutive odd numbers or the way to guarantee that number is going to be odd because in this box, in this location, we need to have an odd number. We have to guarantee for it to be odd and we can't just put down x, because x, if it's an integer, then it could be odd or even, again just the same way that the even was, right. So how do we make a number odd? The way we make a number odd, it turns out the easiest way to make a number odd is to make it even first and then add one, okay. So all we're going to do is basically go here, let me write it out the same way we did before, we can go x plus x, which is just two x, and then add one, right. Pair them up, add one, just think about it, right. If you're putting a group of people together to guarantee that you have an odd number of people, pair everybody up and then bring in one person, or take one person on your fire. So the best way to represent an odd number is to make it even first, make it two x, or subtract one if you want, right. For this one actually I do sometimes go in the other direction, if there's only one or two consecutive numbers I'm looking for, it's consecutive odd numbers I'm looking for, because the numbers are sort of smaller when you start dealing with them, okay. But for now just to be consistent, let's just work up, we're just going to keep on adding. So if our first number, remember our first number is not going to be x, our first number is two x plus one, so if our first number is two x plus one, what's our second number? Well it's going to be two x plus three, right. We're going to add two more to this, right. Because if you're sitting in an odd number, let's say seven, to go to the next odd number in the sequence, right, is going to be, you have to add two from seven, next one is nine, so seven plus two is nine, right. So every time in the steps you're going to add two, it's the same things here, every time you go down you're going to add two, over here every time you go down you're going to add two. So our next number ends up being two x plus three, two x plus five, two x plus seven, dot, dot, dot, so on and so forth, right. Easy, know this chart. If you're in grade eight, nine, 10, 11 and 12, these questions come to you throughout all of high school, right. If you're lucky because it they sort of give you an appreciation of how to deal with numbers, how to deal with problems, how to represent things mathematically, okay. Now I say we're going to do all the problems in one shot but we already did the problem for this, right. If you want should we do another one for this? We could do another one for this, let's do another one for this, okay. So let's represent these, I'm going to erase this and put them up top and then we're going to do that statement and solve them really quickly, condensed version, okay. So let's take this down. I need to get, I need to upgrade my chalkboard, I think this thing has or my whiteboard, I think this thing has his lifespan. So what we're going to do is we're going to transfer what we have on the table up here, condensed version, right. So consecutive even or just consecutive numbers, right. Con, it was x our first number, x plus one or second number and so on, right. Consecutive even, con even was 2x is our first number, right. Our second number is going to be 2x plus 2 and so on. Consecutive odd, consecutive odd is 2x plus 1, 2x plus 3 and so on. So let's do one of these guys first and we're going to lay it out properly, okay. And then we're going to do it and we're going to follow the same format for the next two. So let's say you had the following question. Let's say they gave you a question and said the sum of two consecutive integers is equal to 35. What are those numbers? So first thing you do when you're solving a word problem like this, you end up using a let statement. I mentioned the let statements come into play in mathematics to simplify problem, to simplify equations for us, right. So what we're going to do, we're going to say let, right. We're going to say let x equal the first number, first. Conversely, we could also write the sentence like this. Now just keep this in mind. I might switch up between the two formats, but you could also say let the first number equal x, right. I could have just written it this way. Let first number equal, it's just like as we talked about in the previous equation when we're talking about the equal sign. One side equals the other side. It doesn't make a difference if I write it on this side or write it on this side, right. I could just flip the whole thing around, right. If this is equal to this, then this is equal to this. And reading from left to right or right to left really doesn't make a difference, okay. So you have to know how to move around an equal sign. Crucial, crucial, crucial, right. So we let the first number equal x. Our second number is going to be x plus 1. So let second number equal x plus 1. The problem again was this. Let the sum of two integers equal 35. What are those numbers? So if we let the first number equal x, and we let the second number equal x plus 1, then the sum of two integers is equal to 35 means two consecutive integers is equal to 35 means that the first number plus the second number has to equal 35. Well our first number is x plus x and our second number is x plus 1, right. So their sum, if you add them up, has to equal 35. So the way you lay this out is you go x or again let's lay this out how your mind process should work. You would say the first number plus the second number has to equal 35, right. You just lay out the problem. Well our first number is x plus x plus 1. That's our second number has to equal 35. x plus x is 2x plus 1 equals 35. Bring the 1 over minus 1. 2x is equal to 34. Divide by 2. Divide by 2. So x is equal to 17. Our first number is 17, right. What's our second number? Our second number is x plus 1. So our second number is 17 plus 1. So our first number ends up being 17. Our second number ends up being x we're going to put there. 17 we're going to put here. 17 plus 1 is 18. And those are our two consecutive integers. 17 plus 18 is equal to 35. So we solve the first type of problem we get. And this problem comes to you in grade 8, 9, 10, depending on that. That's two consecutive even. So consecutive even problem let's make it a little bit more difficult. We're not just going to go to the sum of two consecutive even integers is equal to something, right. We could. It'd be easy. Let's say a problem easy version of this question would be the sum of two consecutive even integers is equal to 14. What are those numbers? Wow. Let's do it. It'll be really quick. Okay. So again, the problem is the sum of two consecutive even integers is equal to 14. What are those numbers? So what we're going to do, we're going to say let the first number equal. Well, we know that the first number has to be equal to 2x. It's not x anymore, right. We set up the table went through it. So the first number is 2x. Okay. So that first number, I guess I should put number here equal to x, then let the second number equal is going to be either 2x plus 2 or 2x minus 2, right. So for first unknown, even integer is 2x. The next number, because they have to be consecutive, you can either subtract or add to. We're going to add to. So it's going to be 2x plus 2. So the sum of two consecutive even integers is equal to 14. The first number plus the second number has to equal 14. Then 2x plus 2x plus 2 has to equal 14. That's our first number. That's our second number has to equal 14. So this is going to be 4x. And I'm going to bring the two over in one shot minus 2 has to equal 12, right. Then you divide both sides by 4. Divide by 4. Divide by 4. So x is equal to 3. Okay. Now remember, x is not our number. 3 is not our number. Should be obvious, because it's not even. We're looking for two consecutive even integers that are up to give us 14. So x is not our number. Keep this in mind when you're solving these types of problems. Our first number is, in the last statement, our first number is 2x. So first number number is going to be 2 times 3. We just substitute 3 in for x. Well, 2 times 3 is 6. Our second number is 2x plus 2. So it's going to be 2 times 3 plus 2. 2 times 3 is 6 plus 2 is 8. Our two consecutive even integers that are up to give us 14 is 6 and 8. Those are two consecutive even integers. Let's do one consecutive even integer problem that's a little bit harder, okay. And you see this in grade 11, usually, and sometimes in grade 10. Let's say they ask you a question. That's the following. The sum of the reciprocal of two consecutive even integers is 5 over 12. More difficult. It's still using the concept of consecutive numbers, right. Something that's introduced in grade 8 and 9, but now they add a reciprocal to it. We're dealing with fractions and the way to solve for this, the techniques usually learn in grade 10, right. So by just adding one word, which is the reciprocal of two even consecutive numbers, right, it kicks up the difficulty level, right. So how would you do this? The sum of the reciprocal of two consecutive even numbers is 5 over 12, okay. So first thing we do is we do our let statement again. We say let the first number, let first number equal 2x because it is the second number equal 2x plus 2. Now the sum of the reciprocal of two consecutive even integers is 5, okay. Let's lay it out. Reciprocal means flip it, right. So if our first number was 2x, the reciprocal of 2x is 1 over 2x. I might try to do this whole thing in one shot without skipping too many steps, but we'll see, right. So we don't adjust the thing over a little bit. That's our let spin, but equations I'm going to set up here and maybe we can do a little bit of work here as well, okay. So again, the sum of the reciprocal of two consecutive even integers is 5 over 12, okay. So reciprocal means flip it. If our first even integer is 2x, then the flip of that, the reciprocal of that is 1 over 2x. Hopefully that's good enough for you to see, right. It's the sum of the reciprocal of both numbers. So the reciprocal of the second number is going to be 1 over 2x plus 2. This has to equal 5 over 12. That's our equation. We just laid it out. The rest of this is just algebra until you get to the end and then you have to go back to your let statement and plug the x in and figure out what your numbers are, right. But we've done the difficult part to a certain degree, right. Now we just got the grunt work to do, right. The sum of the reciprocal 1 over 1 over has to equal 5 over 12. So now we just have to solve this and we talked about this in the previous equation, in the previous video where I'm talking about the equal sign, right. How to solve equations. You multiply this by the common denominator, right. And the common denominator, now I'm going to make this a little bit easier because I'm going to factor out 2 out of this. So this guy is really 2x plus 1, right. Whenever you try to add rational functions and multiply rational functions and deal with rational functions, the best thing to do is factor everything first, right. Because we can see the commonality between each one. So I'm going to multiply this whole equation by the common denominator is going to be 12 times I need an x and I need x plus 1. And as you can tell x and x plus 1 just comes in from that, right. But I would just factor out 2. So if we multiply this whole equation by that, we end up getting 2 reduces 12 down to 6. x kills x. So this side becomes or this term becomes 6 times x plus 1 plus 2 reduces 12 down to 6. x plus 1 kills x plus 1. So it's just 6x that multiplies that. So this becomes 6x is equal to 12 kills 12. So x plus 1, x times x plus 1 multiplies the 5. So this side becomes 5x times x plus 1. Now what we've got to do is we've got to crunch crunch, right. Simplify the expression on the side, simplify the expression on the side, right. Combine like terms if you can. Or if you know how to do this, you know your algebra well. You can do multiple steps in one shot, right. So first thing I do I usually almost always get rid of my brackets first, right. So this is going to multiply in here, multiply in here, multiply in here, multiply in here. This becomes 6x plus 6 plus 6x is equal to 5x squared plus 5x. Now what I'm going to do, I'm going to, because we're short on space, I'm going to do multiple steps and, you know, multiple movements in one shot in one step. That will get our equation here. I'm going to factor it here and solve for it, okay. Hopefully we can fit it all in. But what I'm going to do is this and this I'm going to bring over because I want my x squared to be positive, right. So 6x plus 6x is 12x. So this is going to be minus 12x. And I'm going to bring the 6 over and it becomes minus 6, right. So this side is 5x squared. 5 minus 12 is negative 7x minus 6 is equal to 0. And if you don't know, you know, you're having a hard time with this, you don't know this stuff. This is just combining like terms, solving equations. And we did a lot of this stuff in, you know, series 3a and 3b for the language of mathematics. I spent a couple of years going through these things. So I'm not going to go through all the little details, because explain all the little steps because we've covered a lot of that. And it took two years to cover, right. So there is a lot of stuff that I can't cover in this single video, okay. But if you want to know how to do these, if you don't know already, that's where you want to be. Now I'm going to make a little bit of room here. I'm going to erase this. But we already know how we got those, right. And we're going to factor this up here, okay. Now this is a complex trinomial. So what I'm going to do is get rid of the number in front of the x squared. Now I'm going to use the four step method to solve this. If you've learned this, the odds are they, you know, in school they taught you decomposition and I don't like decomposition. I like the method that I've shown before, which is the four step method. So if you want to learn that method, know exactly what I'm doing. Take a look at that video. But what I'm going to do is I'm going to grab the five, multiply it here. So this becomes x squared minus 7x minus 30. So this becomes x squared minus 7x minus 30 is equal to zero. And now I factor this just like a simple trinomial. Two numbers are multiplied to give me negative 30 and add to give me negative seven. That's negative 10 and three, right. So this becomes xx minus 10 plus three. And then what do I do in this method of solving complex trinomials? I take the five here and I drop it in front of the x's. So this step becomes five comes back in front five x minus 10 five x plus three is equal to zero. Now in this factoring technique of this four step method, where I factor complex trinomial, there's one other step that we need to do, which takes us to the end with the thing, with the expression being fully factored is once we reach this level and we drop the A term, the five term from the x's, from each term we take out a GCF and dump it. It's the only time really you dump the GCF, right. So over here between five x and 10 the GCF is five. So I take a five out of this expression and dump it, right. So this becomes x minus two and there is no GCF there, so stays the same. So this becomes five x plus three is equal to zero. Okay. Now we've reached a level again, we talked a lot about the sin series 3 and 3B and we put together the video in ASMR math, which is the power of zero, right. This is the power of zero. This is what zero allows us to do now, which is basically if we have two things or more things, multiply together to give us zero, then at least one of them has to be zero and since we don't know which one is zero, what we do, we split them up and set each one equal to zero, right. We talked a lot about that in that video, very important, very important, right. So what we can do now is just say, oh x minus two has to equal zero, equal zero and five x plus three has to equal zero and then we just isolate x, just do algebra. So we have x is equal to two and x is equal to negative three over five, right. Those are our x's, right. Now doing this in the original problem, the question was the sum of the reciprocal of two consecutive even integers is equal to five over 12, right. So they have to be integers. Now negative three over five is not an integer, it's a rational number, right. So once we reach this point, what we can do is just kill this one, right, because it can't be that. It's not an integer, negative five or negative three over five is not an integer. So the only possible answer is x is equal to two, right. So we ended up getting x is equal to two. Is that our number? Is that what we're looking for? That's x. So as soon as we end up solving a problem like this, we get to the end, we've solved for x, right. What we need to do is look back to our last statement, ask ourselves what is it that we're trying to solve? What is it that we're trying to get the answer for? Are we just trying to find out what x is? No, we're trying to find out what these two well these two they are positive with these two positive even consecutive integers are. So if x is equal to two, then our first number was two times x. So it's going to be our first number is going to be two times two, which is four. Our next number is two times two. I don't know the room here. Two times two is four plus two is six, right. Six. So the two numbers, the two even consecutive integers that if we take the reciprocal of and add together to give us five over 12 are four and six. That's a consecutive even word problem, a problem you would get in grade. You can get it in grade 10. I would give it in grade 10. Let's do a consecutive odd integer problem and we're going to make it a little bit more difficult. Well, actually, no, it's not going to be more difficult. The adding and subtracting fractions is usually more difficult. And keep in mind, these problems just don't have to be two consecutive even integers, right? It could be three, four, five. It could be a whole bunch of consecutive integers that add, multiply, divide, do whatever it is you want them to do to give you a certain result, right? And every time you add one more consecutive integer, well, on the next level, it would be much harder, right? If I had three consecutive even integers, the reciprocal love that add to give me a certain something, then it would have been one over two x plus one over two x plus two plus one over two x plus four equals something. And that becomes more difficult. That you're going to need synthetic division, something that you cover in grade 12, right? So as soon as you start getting into the reciprocal of three consecutive integers or three even or odd consecutive integers, you're sort of talking about grade 12 type of word problems because you need synthetic division to be able to break down that polynomial, right? Again, stuff we've talked about in series 3b towards the end of it, we spend a fair bit of time because synthetic division takes a fair bit of work to do. Let's do consecutive odd word problem and you'll see how things vary. So for the consecutive odd problem, let's assume we had the following question. Let's say they asked you and instead of doing adding and subtracting, you could have done subtracting the previous questions too, right? But instead of having adding and subtracting, let's talk about multiplication, right? What if they asked us to find the numbers if the product of two consecutive odd integers was 63? So the problem would be the sum of the product of two consecutive odd integers is 63. So how would we lay this out? Well, we start off with our let statement again, right? So let the first number equal to x plus 1. That's our first number, right? We've already laid it out. If we know the table, then these types of word problems become easier, right? So our first number is 2x plus 1. Let the second number equal 2x plus 3. Simple as that. The same way we did the previous two, or the previous three, I guess, because we did one, we cheated, we did one, actually four, but we did, I guess, two for this and two for this, right? We're only going to do one for the odd. So the product of two consecutive odd integers is 63, okay? Well, we got our first number is 2x plus 1, and our second number is 2x plus 3, right? Their product has to be equal to 63. So product means multiplication. So you just multiply them together and set them equal to 63, right? So 2x plus 1 times 2x plus 3 has to equal 63. And again, sirs 3 and 3b. You get these problems. This question could come to you in grade 10. Grade 10 or grade 11. Easy grade 11, hard grade 10, or normal grade 10. This question would be too easy for grade 12, or it would just be a warmer, warm-up, right? So all we've got to do now is just foil this thing, right? And we've talked about foiling a lot. You should know how to do this. And just a heads up, the difficulty level of this goes up to grade 12 if we say the product of three consecutive odd integers is equal to a certain number. Because with the third one, all we would do if there was three numbers, we'd add 2x plus 5. And all we do is just have these two, we'd have 2x plus 5 in there. All three of them multiply together to equal a number. But for now, all we've got to do is just foil this thing out, right? Just multiply this and this. Just multiply this and this. So 2x times 2x is going to be 4x squared. 2x times 3 is going to be 6x plus 2x plus 3 is equal to 63. Right? And then I'm just going to combine like terms, move everything to one side because this is a quadratic, I have an x squared term and an x. I can't just separate them. I can just come, you know, isolate x on one side and set it equal to something. I need to be able to put everything to one side and factor, right? So we're going to combine everything. 2x plus 6x plus 2x is 8x. So this is going to be 4x squared plus 8x. And I'm going to bring the 63 over minus 63 minus, oops, minus 63, not three, minus 63 is equal to zero. Right? Oh, minus 60 is equal to zero. Three minus 63 is 60. And then what we're going to do is, you know, when we're trying to simplify, we're always looking to, looking for GCF, right? The GCF is really important, really important factoring technique to know, right? And we're going to notice, and we're always going to try to simplify, right? So we're all, we're going to notice that 4 divides into everything. So we're going to divide out the whole equation by four, right? Just to kick it down a notch, right? So divide everything by four, right? So this is going to be x squared plus 2x minus 15 is equal to zero. Okay. Four kills four. Four goes into eight twice. Four goes into 60 15 times, right? Now what we do, we factor this like a simple trinomial. Two numbers are multiplied to give you negative 15 and have to give you positive two, positive five and negative three, right? So we break this down, x, x plus five minus three. And now again, we use the property of zero, right? The only way to multiply things together to give you zero of at least one of them is equal to zero, so we can set each one equal to zero because we don't know which one is zero or they could all be zero, right? So we split this up. I'm going to do this here. Or do I have to do it up here? So this is going to be x plus five is equal to zero and x minus three is equal to zero. So this is going to be x is equal to negative five and this is going to be x is equal to three, right? So I got two numbers, two x's as my solution so far for solving this quadratic, right? And they're both integers, so they both could be legit, right? And they are both legit. So what we need to do now is go back to our let statement and figure out what our two numbers are because these aren't necessarily our numbers, right? Or sorry, these aren't necessarily our numbers, right? These are our numbers. These are just the x's. So what's our first number? So what we do is because we got two different answers, there's two different groupings of numbers that's going to happen. What we're going to do is we're going to substitute negative five in for x first and find the grouping of those two numbers and then we're going to substitute in three for the x and find those two numbers, right? Something neat's going to happen. So let's try negative five first. If it's negative five, then all we do is we put an x is equal to negative five to find the first number. The first number is two x plus one. So our first number, should we erase this? Let's say we had it for this one. Let's do it all in here so we don't erase anything, right? If we're going to do it for this one, our first number is going to be two x plus one, which is going to be two times negative five plus one, which is going to be negative 10 plus one, which is negative nine, right? That was our first number for x is equal to five, negative five. And what we're going to do, we've got to find the second number associated with x is equal to negative five. So our second number is going to be two times negative five plus three, right? Because that's our second number, which is going to be negative 10 plus three, which is going to be negative seven. So our first two odd integers that multiply, consecutive integers that multiply together to give us 63 are negative seven and negative nine, right? Negative seven times negative nine is 63, okay? So that's our first group of numbers. We're going to have to erase this. So let's erase this. Our first number is associated with x is equal to negative five. So x is equal to negative five gives us negative nine and negative seven. That's our first number and our second number. Now we have to find what those two numbers are when x is equal to three. Well, if x is equal to three, our first number is going to be two x plus one, which is going to be two times three plus one, two times three, seven plus one, seven. Our second number is going to be two x plus three, right? Two times three plus three, two times three is six plus three is nine. So our second set of numbers is seven and nine, right? Which is neat because we didn't say they had to be positive odd integers. We just said they had to be integers, consecutive anyway, odd integers. So the numbers could be negative nine and negative seven and seven and nine and nine and seven. Those are the type of problems in general we get in high school regarding consecutive integers, just straight off consecutive, even consecutive, odd consecutive. And you could get variations on this that could make it a lot harder, right? But this should give you a really good idea of how you set up these problems, which is really the main part, the hardest part of any work problem is setting it up. I hope that out. That should allow you to deal with almost any type of consecutive integer work problem you end up getting. And keep in mind, always, not necessarily always write out your let statement for the easy ones. It should be obvious. But if you start getting hard problems, use let statements. And whenever you solve for x, when you get to the end and you solve for x, right? Always look back to your let statements and ask yourself what it is that you're trying to solve for. Were you just trying to solve for x? Or are you going to use x to get your answer? And in this case, we're using x to get our answer again. I hope you enjoyed. I'll see you guys in the next video. Bye for now.