It is Algebra. I think let's skip maybe one minute to do a warm-up question. The integers a, b, c, and d satisfy a, b equal to 2cd. Which of the following numbers cannot be the value of the product a, b, c, d? If you have the answer, please feel free to send it through the chat, or I put a poll here. It's a warm-up question, so we'll do it quickly. Cannot be the value, so question number 12. Usually, just a reminder for each class that we try to only pick sort of difficult questions, four and five points, because you want to illustrate important concepts and talk about problem-solving strategies. So don't be discouraged if you see the questions are sort of more difficult than a typical exam. It's purposely designed that way. a, b, c, d, so if we replace a, b equal to 2cd, we have a 2cd squared, and a, b, c, d are integers. That means this product is twice a squared number. And we check everything over here. 50 is 2 times 25, which is a squared number. So 50 is actually good. 100 is 2 times 50, and 50 is not a squared number. So that means b cannot be the value of this product. This is 2 times 100. You can really go through or just stop here, because b is obviously not the right answer. OK, so let's go to our main lessons. So I was just talking to Andrew before, and by this point, most of you probably have taken Algebra 1. And then when we talk about solving system of equations, there's a huge difference between whether the equations are linear or non-linear. So the common techniques for solving both linear and non-linear equations are isolation, substitution, eliminations. So if we have a system of linear equations, then we pretty much know what to do. Usually, the number of equations is the same as the number of variables. And then we just use linear combinations of the equation to isolate for the variables we are looking for. For example, here, if we have a system of three variables, and then I look at it here, my third equation has only y and z. So I want to have another equation with y and z for 1 and 2. That means I need to eliminate x. So look at the coefficient over here. I would multiply the first equation by 3 and multiply the second equation by 2. And that way, I have the same coefficient of x. I can subtract this from this. And I have phi y minus z equal to 8. I combine these two, adding them, gives me y equal to 11. I can find y. I can find z. And I can find x, right? So hopefully, you've seen a lot of this in lower grades, right? We have a lot more linear equation in set A we ran in the fall. And in this set B, the focus is more on nonlinear. But I think it's good to review this. And then when we're solving word problems, some useful tips are effective choice of variables. Because oftentimes, we have many unknowns. But we maybe have to find that there's usually a smart way of assigning variables to unknowns that would significantly reduce the amount of computation you have to do. And in many situations, we do not have to assign variables to every unknowns because they are related, right? We do a problem like that. And then another thing to keep in mind is that we can always make use of symmetry. If we can, you know, the problem has that symmetry. For example, in this system of equations, the right-hand sides are different. But obviously, we have a symmetry on the left-hand side. So very common techniques that we add up all the equations so we can find the sum of x, y, and z. And we combine it with any of these equations to find for the variables, right? So I think that's pretty much we know. Linears is something that we don't have to do. That's the lesson. So let's quickly do one question with this. Anna wants to write a number in each of the square of the grid so that the sum of the four numbers in each row and the sum of the four numbers in each column are the same. She has written some numbers as shown. What numbers does she write in the shaded square? So the question says that the sum of the numbers in each row and each column, they are all the same. And then you can quickly draw it on a piece of paper, 4 by 4 square. And until I give you the next question, 27, 2020, pretty recent. Very good. I started hearing answers. So I have an answer. You can enter to the polls question sent to the chat and answer and Andrew and I will answer you. So the first thing is that you realize it's actually, it could be a linear system of equation, right? We have a one, two, three, four, five, six, seven, and then we have four rows and four columns, that's eight of them, but we just equate them. So exactly, we would have enough equation to solve for the variables, right? That would be the observation, but remember the tips that sometimes we have too many unknowns and we do not need every variable for each unknown. So this is an example, I will do couple of correct answers, so that's very good. For example, this row has only one number missing, so let's call it A. So I put A over here, then I can right away calculate that the sum of this row is 10 plus A, right? And then the second row also one missing number, but because it also 10 plus A, that means this is 10 already, and that means this one has to be A minus two, such that add together, it also become 10 plus A, right? So you sort of continue this, we don't know what to do here, so we don't do anything. Look at this one again, it also has to be A plus 10. So this is eight, seven, so A plus 15, so that mean I have to subtract away five in this cell, and this one is three, eight, actually I don't even need to, I just realized I don't need this cell, three, eight, 11, 15, so this one also A minus five, and this one A and nine, and that means I need exactly one here. So I'm all set because over here, I have eight minus five, that's three, so A plus three, that means I need seven over here, so that is gonna be a 10 plus A. Of course, you can fill up, this one's gonna be, this is eight, seven, so that's gonna be, what is it, six, right? And then seven versus eight, yeah. It should be consistent because we know that we have enough equations and enough variables to solve for that. So that's the idea, this one example is when you do not have to do variables for every unknowns. But in action, this is a set of linear equations. Okay, it's good, so let's go to the next topic. When we enter the realm of nonlinear, things are a bit more complicated. The first thing is the simplest sort of nonlinear you've seen is quadratic equations. We don't have in lesson, but you see quite a number of quadratic equations in level seven and eight, so usually the numbers are pretty good. You don't ask you to do like for some very odd numbers, so for example, you can, for this one, you can x squared minus eight x plus 15, you can use factorizations, factor into product of x minus three, x minus five, and you can find the roots is three and five. Or you use completing the square method, you realize that x squared minus eight x plus 16 is perfect square x minus four, so we have to separate one, and then that means x minus four squared of one, and then you also can find the root three and five. So it depends on the question, one could be easier than other. I don't think you need to use a general formulas for the roots, but again, we know everything about quadratic equations. Sometimes you use some algebraic manipulations, and one of the most useful identities is a squared minus b squared is a plus square a minus b. And then, of course, we can also use the techniques, the same techniques, substitution, eliminations, and some clever manipulation to extract the desired variables. And a lot of the time for question with constraints, and remember to use inequalities, yeah, when you have constraints, and when you deal with integers or things like that. Okay, so we're gonna have a bunch of numbers with nonlinear equations, so let's go for it. First one here, number y is defined as the sum of the digits of number x, and z as a sum of the digits of number y. How many natural number x satisfy the equation x plus y plus z equal to 60? So y is sum of x and z. So when I read this equation like this, and when we, you know, we're trying to make sense of the equation, it's always useful to start with some example, right? Suppose x is 123, for example, or maybe, and y is the sum of digits of x is gonna be 18, and z is gonna be nine, yeah? So I give one example to clarify what the question means. Question 29, it's a bit long, so I give you a couple of minutes to solve these questions. Okay, a little bit of hint, I have one answer, C. Actually, it's D. You can actually find all these numbers. The question asked how many, but actually, to be sure, we have to find all these numbers. So maybe I'm missing one. More interesting questions to go, so let's sort of quickly get started. We talk about inequality and constraints, right? Apparently, x has to be bigger than 9, right? Because if it's smaller than 9, it's not going to be 999, it's 27, it's too small. And also, we realize that x has to be bigger than 9, and then it obviously has to be smaller than 60, right? So that means it's a a two-digit numbers, right? So sort of like a b, you can just sort of very common techniques, you can do two-digit numbers by a b, and then y is going to be a plus b, right? And then z would be sum of these numbers. But then over here, we're going to start using a little bit of inequalities. a plus b would be maximum is going to be 18, and z as a sum of all the numbers, the digit of the numbers that are, you know, at most 18, then z is going to be, you know, less than and smaller than 9. And a and b, because these two are smaller than 18 plus 9, which is 27, so you take 60 minus 27, that would give us 33. So we know that a b is sort of much, has to be at least 33, right? I mean, this is probably not the hardest problem with inequality constraint, but if you use the ideas of why these tools are helpful. So we know that the number has to be at least 33, and then I think for now, probably you can just do trial and error. For 30, you can go up to 39, and then plus 12 and plus 3, that's x, y, z, right? x, y, z. And that stays smaller than 60, so maybe not so good, you can do a few more. And then for x, you go to 4, then we can do 40 plus 4 plus 4, this is no good, 41 plus 5. You can also use, sort of, do more inequalities, but in a sense, in this way it might be a bit easier, because the numbers are small. We can just do, you know, trial and errors here with quick calculation, we probably can get the answer. This is 44 plus 8 plus 8. Okay, this one is good, it's good, it's actually 60, so we have the first number, and you can continue 45. Actually, the odd, you can realize that the odd number has to be out, because, you know, this y and z are the same, so all the odd numbers are out. No, not quite up until the last one. Let's just do 45 plus 9 plus 45 plus 9 plus 9, this one also I would not have to calculate it, 46 plus 10 plus 1, also odd, so I don't have to do anything, probably 7 plus 11 plus 2, this one actually go to 60, right, so you do a few more, maybe 8 plus 12, 3, also not, 49 plus 13 plus 4 and 50 plus 5 plus 5, this one also go to 60, and then 51 plus 6, this one get a bit bigger, so 50, again, and then basically when you go with 61, you see that you're going to be with 60. You can use more constraint, more inequalities, but as I said, it's probably easier just to do trying errors, so we'd have three numbers in this case. That would be question number two. Let's go to the next one. Okay, so again, we talk about linear and non-linear transformation in these questions. A machine transforms a given number x different from 0 into one of the numbers x plus 3, x minus 2, 1 over x or x square. The number 1.99 was entered into this machine and three such transformations were made, the first on this number and the next two on the consecutive results. Let y be the greatest number that could be obtained in that way, then which of the choices is correct? So we have three transformations, that means we start with 1.99, right, and then remember that we have to do three, so go to a, go to b, and go to c, and these are the transformations that we can choose. I didn't say that we cannot choose the same thing, right, as long as we choose from this set of four transformations, and then we want to maximize c, so pretty much we have to see what transformation we need to use, right. Yeah, it's not super important, but just the idea that if it's linear, you won't be able to make something so big, right, because the chain is sort of gradually, and to make something really big, we need like nonlinear transformations, like a parabola, x square, or hyperbola, like something like that, to make something from small, and like somehow we sort of exponential increasing. So, I'll be able to give people a vein now. I relaunched the pose, I have some answer, you know, take some time. Look at the transformation and they are very different in nature, right? They are very different in nature, x plus 3 and x minus 2 are linear. This one, if you want to get x, y big, then x also has to be big, I mean, the increase is better than linear because it's square, but it still has to be big. But this one is interesting one, 1 over x, when x is small, y can be very big, right? So how to combine these to get the largest possible answer? OK, good. I think we can slowly start it. So if we start with 1.99, and then at first we want to, for example, I want to apply one of the nonlinear one, I take 1 divided by 1.99, right? So that one is like approximately 1 over 2, right? And this one, 1 over 2, if I add 2 and 3, it's not that much. Even if I add 3, it's going to be 3.5. And when I square it, it doesn't look like it's going to be very good, right? And then, so the point is that I want something that's very small, and then I can take the reciprocal of it. It's going to get very big. And then I can even magnify that effect by squaring it, right? So interesting, I got A, B, C, and E, but none of these correct answers, so that means it's an interesting question. So we're going to try this 1 plus 99. How to make it small, for example? How to make it small? I'm going to subtract it by 2, right? And then I get 0.01, minus 0.01, right? But it really doesn't matter, because if I square this thing, I square it, I'm going to get, this is 0.01. Oh, OK, I'm sorry. Then I probably take the reciprocal of it, which is going to be minus 100, right? And then if I square this one, I square, I will get 10,000, 10 to the fourth. Or like, I mean, if I do, it could subtract minus 0.2, and 0.01. And if I square this one, I square, I also have 0.0001. I make it even smaller, right? Because when you square a number that's smaller than 1, it becomes smaller than itself. And then that's what I'm looking for, because I take the reciprocal that it also 10 to the fourth. Andrew, is it? I didn't make a mistake here, right? Because when I square it, you get rid of the negative side. So any of these two transformations would give you 10 to the fourth. And there's only a few choices, so we can pretty much show that 10 to the fourth is the maximum possible answer. So we cannot get as high as the 20,000, and we can get higher than this, so it's D. So sometimes graph is helpful. I don't know in this case, but in a lot of cases, the phi graphs are very helpful. And then we need to move to word problems. So in earlier grades, you probably have done a lot of word problems, like percent ratio, proportion, multi ratios, percent increases and decreases, tons of word problems. And most of the time, yeah, I mean, there's some nonlinear, but most of the time, this could lead to a nonlinear equation. So I think if you move to this level, probably you are very comfortable with this. We had some problem in set A, but yeah, our focus can be a little bit different. And then we have a lot of questions. It's more like a flavor in math kangaroo problems that involves velocities. And these have to pay special attention to two things. It's called frames of reference. Some of you probably started taking high school physics. But if not, then something kind of intuitive you need to use that when two objects move in the same direction, then they have to, if you want the relative speed is a difference between the two speeds, right? And when they move in opposite direction, their relative speed is the sum of the two speeds. And also the question of velocity also become interesting when velocity means speed and direction is a vector quantity. So sometimes you have objects and people move in different directions. They could be perpendicular to each other. Remember that the first question we did in lesson number one. Or sometimes they go on a circular path and chasing each other. So in those question, it's very helpful to draw a diagram. So let's see what we can do with question involves velocity. The first one is actually a very practical situation. At 9 PM, the driver noticed that he is driving at the speed of 100 kilometers per hour. At that speed, he would have enough fuel to drive 80 kilometers more. The closest gas station where he can fill his tank is 100 kilometers away. After noting that fuel efficiency is, I think, inversely proportional to the speed of the car, the driver decided to reach the station in the shortest time. At what time will he reach the gas station? So let's spend some time to analyze the questions together. So this is 9 PM. And the driver is here. He's driving at speed of, since this is like 80 kilometers away from the starting point, 80 kilometers away. And his speed is 100 kilometers per hour. But then he realized that what he has to do, he has to go a distance of 100 kilometers. He had to reach over here. So if he drive at this high speed, he would not have fuel to go over here. And what does it mean? That means he has to fuel efficiency. That means he had to reduce the speed of the car, right? We probably can go with V2. Don't know what it is, but we know that it's V2. And then if he go with this speed, he would be able to cover the distance of 100 kilometers. Oh, the last thing is that he want to risk it in the shortest time. And what does it mean? It means he wanted like, you know, runs out, you ride the smart and like nothing left pretty much intuitive, but I think that's what the question means so, you know, take some time. Oh, I just wanted to make sure some of you probably got the correct answer, but be careful because it's p.m. so that means, you know, if we already say p.m. and a.m. that means we use the 12-hour clock, right? So I know it's sufficient to disqualify these two answers, but you should be suspicious with these two answers because it's a 12-hour clock. So this is a 24-hour clock, right? So be careful with d and d, d and e if that's one of your choices. So this is one of the, you know, situation in which you have a word problems and the first thing that we have to convert all of this into an equation, right? Okay, good. It looks like you are getting on track. At least wait 30 more seconds to see if it answers. this way. One more hint because I want to give everyone the chance to do it. When we say that the speed is proportional, inversely proportional to the fuel efficiencies, that means if you take the ratio of v1 over v2, right, that should be like inverse of d2 over d1, right. And like in physics, sometimes you sort of argue like this is a really old information that I have, what can I, what equation I can write from here, like not many equations that I can write, right. So if you sort of get stuck with the math, you can, you know, invoke your physical intuition and pretty much that's what you can write, given these quantities and information. And that means v1 is bigger than v2, so d2 is, yes, I think that's what it means. Okay, so d2 bigger than v1, so you kind of substitute the numbers into here, v1 should be 100 divided by v2, right, and that would be 100. d2 is 100 divided by 80. And from here, we actually figure out that v2 is 80 kilometer per second. So usually, like, they make the number easy, so you don't have to do a lot of computation. Yes, so v2 will 80. Any more answer? What would be, would be the arriving, arrival time? Okay, good, more answer. So that means d2, right, is not like this time, but the actual arrival time is going to be, remember, distance, I'm sorry, time is distance divided by speed, so it's going to be 100, you have to take d2 over v2, you have to take 100 divided by 80, which is 1.25, which is one and a quarter, and a quarter of an hour is 50 minutes, right, so the answer is b. Okay, good. Oh, I'm sorry, I missed some of the messages. I will try to answer while you are doing the other questions. Good, let's go to the next one. This is a longer question. Oh, okay. One meter long. How long was the pipe? Okay, so it's a long question, question number 30, and just like the other questions, I usually like to, you know, draw some pictures if that helps. So, you know, I give you one minute to see what pictures you can draw, and I will start drawing. Okay, so let's see what the question means. So you have a tractor, like a tractor's pulling a pipe, right? It's this long pipe, let's call it like L. And then if V, if Yuko goes along in the same directions, then he can go like this way. Oh, the tractor is going this way, so let's say that it has velocity Vy, Vt of the tractor, and Vy, velocity of Yuko, if Yuko's direction, then after that he'll measure like how long he's walked along the road and that actual distance would be, because this guy is moving, so by the time he moves over here, the distance that he walks is 140 meters, right? So when you read the question, you don't actually see what 140 paces is not his velocity, his speed, but after he walked from one end to the other starting point to the end, because that time the tractor was over here, then he sort of jumped, and then each jump is at one meters, and he measures that the distance is 140 meters, because that's what it means. On the other hand, if the tractor is here, also pulling the pipe, right? Also the pipe has length V and it goes direction Vt, but this time Yuko, he goes in the opposite direction, right? He goes in the opposite direction, so he probably starts from here and then by the time he, this one he sees the end, then the tractor already moved here, but now this distance is shorter, it's only 20 meters, and that makes sense, right? Because they go in the opposite direction, so he only walks a shorter distance to get to the other end of the pipe. Hope it makes sense. Okay, so let's continue with this. I have some answer, but this is a difficult question, so let's try to do it together. We go this way. So now, if we talked about, remember we talked, if we go in the same direction, so the relative speed, so suppose that the pipes stay, you know, the trucks, the tractor stays the same. So relative to the truck, y actually only move with velocity of Vy minus Vt, right? Definitely Vy has to be bigger than Vt. And now the time it takes for him to go from one end to the other would be T1, right, is equal to L, the length of the pipe, divided by Vy minus Vt. Okay, I hope this makes sense, right? If it doesn't make sense, let us know. And then now, but with respect to the road, right, now what's the actual distance he travels along the road? This is the time that he walks, and his velocity is y, so that means the distance that he walks on the road, which is 140 meters, that's the distance on the road, that one's going to be just equal to Vy times T1, right? It's complicated, but that's what it means. And similarly over here, the time it takes for him to walk from one end of the pipe to the other end is going to be T2, equal to the length of the pipe, but divided by the new speed. Now the relative speed between Yuko and the tractor is going to be the sum of the two speeds, Vy plus Vt, right? And the distance he actually travels on the ground is 20 meters, it's going to be equal to his velocity, because this velocity is with respect to the road, so that's, it's called relative speed relative speed times T2, okay? So now you have a set of one, two, three, four equations, right? And we need to find what L is. So now we did the initial step of turning a word equation into a set of, you know, I'm sorry, word problem into set of equations, and now it's when your skills in algebra and manipulation comes into handy. So I give you a couple of more minutes, it's kind of interesting set of equations to see if you can find out the answer, what L should be. Okay, wouldn't be the velocity multiply t1. Okay, so I have a question about this equation. So this one, the time it takes t1, I take is because we know that he travels from one end of the pipe to the other end of the pipe. So now, in order to get the first set of this equation here, we can imagine that this tractor is not moving. If the tractor is not moving, then he would move with speed of vy minus vt. I hope this one makes sense. It's like when you go with a boat along the river, right? Then you would, when you go down the river, the speed would be your speed of the engine plus the speed of the water. And then we go upstream, your speed would be speed of the engine minus the speed of water. So in this same case, you assume that the tractor is not moving, then he only moves with the speed vy minus vt. And that would tell us how much time he actually walks. But then when he measures the actual distance on the ground is 140 meters, that actual distance we have to calculate using his speed relative to the ground, which is vy. I hope it answers the question because I have it in the chat. If anything not clear, feel free to stay after class. So now we have these two set of equations. And if you look at over here, we don't have to solve for all the variables, but we look at this proportion of two and four. From two and four, right? We know that T1 is equal to 72. I hope that makes sense. From two and four, we can realize that T1 equal to 72. And then if you go from one and three, because T1 is 72, that means vy plus vt, vy plus vt. This one gonna be seven times of vy minus vt. So that is a lot of manipulations here, equal to seven times of vy minus vt. So you group all of them. So another like just common technique, we group all y together and t together, we have seven, I'm sorry, we have eight vt is equal to six vy. And that makes sense because the truck should be, vt is gonna be three fourth of vy, right? Because the truck, the speed should be smaller than vy. And now you kind of replace into one, you would have T1 from one here is gonna be y divided by vy minus vt, which is a quarter of vy, right? A quarter of vy. And that means L is gonna be a fourth of vy times T1. But vy times T1 is 140 meters. So y is gonna be 140 meters divided by four. And the answer is B, 35 meters, right? So I admit that this is a very difficult questions for the exam because you need all of that. But the point is that even after you have the decisive equations, you know, you have to do the right manipulation to get to the answer. So a difficult question, but if you had any question, feel free to stay after class and I will, I will go through that with you. Yeah, we'll have another good questions coming up. So let's, and Joe, could you please help with these questions? Second, number of integers are written on a blackboard, including the number 2018. The sum of all these integers is 2018. The product is also just 18. Which of the following could be the number of integers written on the blackboard? Do you see the polls, Andrew? Okay, good. I just want to make a comment that after you find the number of integers, don't forget 2018. You're probably choosing the other numbers and have to add one to get to the total numbers. Okay, so something important for this question is to remember that because we're using integers, the product, the absolute value of the product can only stay the same, become 0, or get larger. So we have to think of a combination of integers that leaves the product the same, but also has a sum of 0. I'm just trying what Andrew said, sum of all other numbers is zero, right, because, and then. Okay, so the first thing to do is we need to look for a combination of integers with a product of 1. And we can pretty intuitively know that in order for that to happen, we need to only use 1s, right? So we can have 1 times 1 times 1. But then we can see that the sum here would be positive, so it would be 3. So we actually need to mix in some negative 1s. So in order to have a sum of 0s, we have to have negative 1s and 1s. For the product to be 1, we need to have an even number of negative 1s. That means that we can only have these numbers in groups of 4. Because the product of these is 1 and the sum is 0. And from this, we know that we can only have a multiple of 4 for the rest of the numbers. So we know that we can express the number of integers as 4 times n plus 1, where n is an integer. And if we look at the answer choices, we can see that the only answer choice that works for this is 2017. Because the other ones all are multiples of 4 plus 1. So we get the answer as 2017. Great, thank you, Andrew. So we have to make sure that they have to be the same number of minus 1 and 1, right? And then they have to be even, because minus 1, 2, an even power is going to give us 1. And then that means 1 for n. So maybe some of you got 2016, probably you forgot to add 1. And then just a number divided by 4 if the last two digits is a multiple of 4. So the answer is 2017. Okay, good. One more question. Great answer. Very good. So I give you, we just have enough time for this. So let's see if we can find it out. X, Y, and Z are positive real numbers. Now they are not integers as real. And then with this constraint and the sum is equal to 20. So which of the statements below is always true? Which of the statements below is always true? Just sort of a note on the logic, to prove that a statement is all the way through, then we have to prove for all cases generally, but to show that it's not all the way through, you only need to find a counterexample. So you go through all of it, and to see if you can find a counterexample, then you can. So for example, someone says that A is all the way through, then if someone else can find a counterexample, um, okay, good, we have a lot of answers, so we wait a little bit more, maybe one minute, and I can go to the solution. Okay, so this is a long question, so I already typed up the solution here. So for example, if the first one, if A is true, we just have to find a counterexample. So 99 is like around 100, right? And that means like you have a 10 and 10 would give you 20 and the other one could be small. But the point is that we can make, we can make Z very, very small, right? Oh, no, no, I'm sorry. We can make like this one sort of very closer to, Z is in, I mean, we can just get rid of it because we only care about X and Y. And X and Y can get arbitrarily close to 100 because we can choose them arbitrarily close to 10, leaving Z arbitrarily small as long as it's bigger than zero. So that means this one, you can make it very, very close to 100 and that would be bigger than 99. I mean, even if you don't buy the actual numbers, that would be the reasoning. And for the second one, X times Y bigger than one, this one, we know that we can really refute this because if you multiply by a big number, X is bounded by 20, so X is at most 20. So if we multiply something very small, close to zero, that would bring it down to a very small number. So for example, you choose this, still satisfy Y bigger than Z, and that makes the products much smaller than one, right? This one actually is the hardest. Let's go to D first. X times Y bigger than 75, but for 75, it's kind of easy to find a counterexample, 10, 7.5, and 2.5, because we know that Y has to be bigger than Z, so Z would be easy. Then in the exam, it's just, you know, choose a low-hanging fruit, whatever you can do, do it. C, actually a harder one because Y, when I know that X times Y, I want to find X times Y equal to 25. So I could see that, oh, maybe 10 times 2.5 is 25, but then Z is 7.5, which is smaller than Y, so that would not, you know, satisfy this. So I have to sort of find a number, so it's gonna be X, and Y is gonna be 25 over X, and Z is gonna be equal to 20 minus X, minus 25 over X, and I have to sort of play a little bit with this number to make sure that Y bigger than Z, so this will be the hardest one to prove. But if you haven't any doubt, it's only between C and E, and you can, you know, regularly eliminate A, B, and D, but, you know, that one, this one is a little bit harder, but you can also get an example. So any of these, it is possible, and the answer is E, okay? Actually, in the end, you got the answer, so that's good. So just wrap up for algebra. When we have a system linear equation, we know exactly what to do, just use linear combination of them, and then we have a nonlinear, then, you know, algebra factorizations, and, you know, this is for quadratic equations. When you have a constraint, especially for a question where you have a lot of variables, but very few informations, and always think of, you know, using inequalities to narrow down the range of the solutions. And then for a lot of word problems, be careful, like find a smart way to choose a variable so that you can reduce the amount of computations that you have to, you know, conduct. And then common word problems, you may need to review ratio, percent, proportion, velocity. You can find a ton of these questions in lower grades. And finally, you know, it is not like something that is a special, you know, emphasis on math can go, you don't have a lot of complicated equations, but the ability to manipulate algebra expressions, and, you know, like the question with velocity, when you have a ton of equations, how to sort of quickly manipulate them, combine them to attract the right variable is something that is a very good skill that you need to brush up, then that's a good time to do so. So next week, I think the topic's gonna be, I think number theory, and the class ends here, but we're gonna stay for a while. So if you have any questions, feel free to ask, and we are happy to answer.

algebraic problems

linear equations

nonlinear equations

word problems

quadratic equations

problem-solving strategies

substitution

elimination

systems of equations

number theory