Let's start, welcome to another webinar session. Today we are going to talk about algebraic thinking, algebraic equations, how to create algebraic equations for mathematical problems and so forth. Let's see, for the warm up, I will just start since many of our friends get the same answer anyway. It says four chocolate bars cost $6 more than one chocolate bar. Well, let's say what's the cost of the one chocolate bar? Let's say it's X. You know, in many cases we are using X as a variable anyway. So the cost for four chocolate bars should be 4X then. Then it says this one cost, the cost of four chocolate bar is equal to, it says $6 more than one chocolate bar, basically more than one chocolate bar. Six plus more, six plus X or X plus six, it's going to be the same thing anyway. So we're gonna find X here. Well, as you see, we have X terms on the both sides. We should focus to cancel the smaller X. We have 4X on the left, 1X on the right. That means we should cancel the smaller one, which is 1X. So 4X minus 1X give us 3X is equal to six. At the end, our goal is to find only 1X, not 3X. So at the end, we can just divide by three. That give us X equals two, guys. I hope it does make sense. If it is not, please ask us via chat. You can send Soham to message or just chat to everyone. Then you can check and share with us. All right, with that silence, I believe we are good, actually. You don't need an explanation, I guess. Okay, guys. Just for some of the new friends, I'm gonna explain four-step problem solving strategy really quick. Make sure, guys, you understand the problem. Make sure they know what they are asking. From the previous question, I just underlined some of the words. To make sure, I kind of focus what I need to find. What's my goal in the problem? So everyone's asking, plan how to solve the problem. Definitely, are you gonna use algebraic baselines, like from this topic, we're gonna talk about algebraic thinking, or are you gonna create some sort of pictures or graph or organizer, whatever works. So what do you need to solve this problem? This is really important also for some of the questions. Let's say we have like given some of the hard algebraic expression. It's really hard to figure out what to do, but instead, when you just, instead of just jumping in, when you just try to reread and try to understand what they are asking, maybe the question is not that difficult at all. You just need to figure out one little piece, they will go from there. Another step is you need to carry out your plan, carefully complete your calculation and organize your thoughts and steps. Make sure guys, you don't do any calculation mistakes, because those answer choice also kind of design in a certain manner. That means when you do little calculation mistakes, you may find one of the answer choices, which is wrong, but you thought it's right, then you just choose this one, then it's wrong. Even you know the topic, because you didn't pay attention well, the question is gone, that's really sad. Look back and check and reflect, does this answer make sense? You know, when they ask you to find, let's say number of people in the carnival, you shouldn't get 105.7. You know, you gotta find some whole number. Even you cannot say, there are negative 300 people attend that event. No, it says positive as well. Just make sure whatever you got, it makes sense guys. I really wanna believe that there are no question about this piece, otherwise you would ask that to Soham, so we can just check that. Okay, as you see, today we're gonna talk about algebraic thinking here. Algebra is a variable, like typically a letter, as you remember from the first slide, just use X. That letter to stand in the place of an unknown or changing value. If we know X guys, then we can just create some sort of equation in terms of the X. If there is no X, well, in the elementary level, we just, we used to put the box or stuff. Even the box, you know, we have to use some sort of object. Without X, without unknown, you cannot really create the equation, it's gonna be meaningless. Determining numerical values for the variables helps to solve mathematical problems, of course, if it's possible, by the way. Variables can be used in equations or inequalities. You know, when you see the variables in inequalities, you shouldn't just kinda see what's going on. You should just able, you will be able to solve that. Anyway, frequently multiple variable and equations can be created to describe a problem statement and must be solved together to meet all the required conditions. It's kind of for solving some equation. There are tons of different ways to solve that, anyway. In most cases, adding to both sides of an equation, multiplying both sides or applying powers to both sides will not make an equation untrue. You know, when you have the scale balance, so, oops. So, whatever you add in one side, like let's see, just put plus five pounds, whatever. You need to add same five pounds to other side to keep the balance, otherwise you will not be able to do that. And those balance represents the equation from real life world, you know. That's the reason they say, if you multiply both sides by the same number, that's fine, you can do this. If you divide, yeah, you can get the powers, you can even get the square root of both sides if it is possible. Just make sure you consider plus or minus sign, but you can do all of those steps to cancel. These operations allow you to solve for the variable and determine the answer to math problems. So, in that case, guys, we need to also think of what is solving means. Guys, solving means, basically, when they say solve that equation, you are going to find that variable. It's X, Y, Z, whatever. But finding variable means, you know, you need only one variable, you need only one X. They are going to give you a bunch of X, plus, minus, some numbers in the square, cube, whatever. But at the end, your goal is to make it simple and you need to have only one X. This is your goal. To be able to reach out to your goal, you are literally, you know, adding or subtracting the same thing from the equation, you know, or just doing the opposite operation, whatever is necessary. Like, let's say, if they have, like, two X, plus six, is equal to, for example, like, 20. In that case, if you think of PEMDAS, you would do, you know, first multiplication, then addition or subtraction. But since when solving means you've got to do inverse operations, you've got to focus add or subtract to be able to cancel that plus sign first. Remember, we do the opposite of PEMDAS, basically, when you solve equations. You need to focus add or subtracting first. Cancel that six first. Then you should have two X is equal to 14. Then what? Well, we have two times X, but remember, we had only one X we need to find. Then, since we both know opposite of multiplication is division, you know, we have two times X, and the opposite of divide, opposite of multiplication is divide, then you've got to divide that, then find the answer, whatever. So as you see, you need to isolate that value. You have to have only one X value, even if it's given like a bunch of extra number term or calculate it next to X. I really want to believe that there is no question about this piece. If there were, you would ask, I believe. Okay. An algebraic term is composed of constants and or variables that are multiplied together. For example, 71 is one of algebraic term. Six X is another term, because six and X together here, you've got to think about they are the complete whole. 14 times B, they are also together, that means 14B is also another algebraic term. Seven X squared is a term, and three X, Y, these are some of the examples of algebraic terms, guys. Algebraic terms are separated by additional subtraction in expressions, equations, and inequalities. Similar algebraic terms, also called like terms, contain the same variable to the same powers. You can only combine terms if they are similar. For example, we can say three X plus two X is five X. Yes, you can say, but sir, three and two are different. These are just coefficients, guys. But at the end, we know that these are the X terms. Since the first term is X term, depends on X, another term is also X term, we are going to be able to do the calculations, add, subtract, whatever, but you cannot combine four X plus five Y, okay? Combine means you just got to add them, you know, because you will be able to multiply them, but you cannot just add or subtract them. You cannot combine X plus X squared because the powers are different. As you see, even you see X, still we can do nothing. You can just leave it like that at the end. You solve the equation and you get that. You don't need to. I mean, you can factor, but this is not discussion of right now. One way to remember is, for example, apples, apples, you know, you can combine orange to oranges, that's it. So we can think of three apples plus two apples is five apples, yeah. This is an application of factoring or distribution property. So when we have three X plus two X, since we have X terms, we can keep X out and say three plus two is five. Then we can say that total, that sum is going to be five X. But when you cannot add four X and minus six X Y, but you can basically factor that, you know, you can look for the both of the terms. When you factor, first you got to look at numbers. Both of them are even numbers. That means both of them are divisible by two. I am talking about dot, that's six and four. Then we are going to look for X and Y values. If any variable has in the both of the expressions, which is X here, you got to keep X out as well. Then you got to think of what is four X divided by two X. It's basically two, it's inside. Then you will think of what is negative six X Y divided by two X. First of all, you can just, you know, cancel negative six and two, which is negative three. Then X Y divided by X, you have left over only Y, so you get three Y basically. You can factor as you see. You could, we couldn't add them, but at least we factor them. I hope it is clear guys. Meanwhile, if you have any questions about any of those, you can just ask me, okay? Okay, we are almost there guys, no worries. Little bit more information. Adding a term like basic constant or variable to both sides of an equation, how it works. Let's just check that in that example. You see that we have two X plus five, okay? That one is equal to 21, but we won't know that at the end, we should find only X. You can guess, I mean, you can just see what X is 10, what X is 20 or whatever, but it's going to take some time. So I don't recommend that. Just try to use algebraic ways to solve that. Well, if you think of inverse of PEMDAS again, you got to work with additional subtraction as a first term, first thing to focus. You got to cancel that plus five by doing minus five as it's here as well. Remember, you can add the same term to both sides here like negative five, or you can take out five from both sides. It's going to the same thing because if you do the same thing on the both sides, that's totally fine. You are mixing up with, you know, you're not messing with the equation guys. That's totally fine. That's totally reasonable. Anyway, that yields two X is going to 16 because 21 minus five is 16. Well, still we have two times X, remember, but we got to find only one X. Then we got to multiply both sides of an equation by one half, or you can divide both sides by two. At the end, we should get X equals eight guys. I hope it does make sense. For those type of case, what I recommend guys, I recommend you just create something similar, but instead of two X plus five, you just do like seven X plus, or like seven X minus nine, whatever. Then that value is equal to nine. Then try to solve this one. You will still use the same method, you know, either add or subtract first and divide or multiply second. But if you just do your own practice, it's going to be more helpful instead of just I am telling you. Okay, combined equations can help eliminate variables and allow solving them one at a time. Solving them one at a time. For example, we have four X minus three Y is equal to 14, and we have three X plus three Y is equal to 21. Well, we can combine them. We can just keep them together. So in that case, what would you get here guys? Look at here. Look on the left piece, left part, we have only X and Y terms. We just combine both of the equations. We just add those equations. Four X plus three X, you know, here X terms, and minus three Y plus three Y. Well, as even without doing anything, you realize that those Y terms cancel each other. Well, since we said guys, left part, we just combine left part. We got to combine right part of the equation as well. That gives us 14 and 21. That 14 and 21 are here. We just add those numbers. When we simplify like terms, we have left over here four X plus three X, total seven X. Since these are the same, like these are the X terms, we can simplify, otherwise we couldn't simplify or combine, okay? So 14 and 21, these are the same type of terms. We call them like terms. These are just a number, just an integer. Then we combine, basically add them. If we had like 14 plus 21 Y, we wouldn't do anything because one of them is just number. One of them is just, you know, Y term. So we were able to do that. Anyway, at the end, when you solve this system equation, you should get X equals five. Then we need to use that. We are in one of the original equation to get Y because when you solve some equations, that means you are not only finding one of the value. You need to find both of the X and Y values. When you find both of the variables, that means you are done solving that. Please be careful. Please be careful of what they are asking when you read the question. There are numerous ways to use variables, analyze operations to solve problems. And let's try some of the contest problems. All right, here is my first gift for you guys. And I'm gonna start timer. Bye-bye. It has been about half a minute. No, one person said 14. Oh, we got 14 as an answer. Okay. Which has been more than one minute guys. Another person says 14 and another one. Okay. So far 14 gets 3 votes. If we can get one more answer, maybe. Okay, class, we have... One more says 14. Okay, I was gonna say, we have definitely more than three people in that classroom, so... We get four friends says 14. Let's start. Alright, guys. First thing, you just, you know, read the problem. The sum of the volume of three pictures, let's just say, that's P, represents by pictures, and two bottles is equal to 16 liters. Okay. The volume of each of the pictures is two times greater than the volume of each of the bottles. Okay. As you see, we have been given two different equations so far. The sum of the volumes of two pictures and three bottles is... As you see, guys, at the end, we do not just find... They don't ask us to find only one picture and one bottle. It's going to be two pictures and three bottles. That was the reason... That's the reason I was saying that. Make sure, guys, you understood what they are asking. The total of two pictures plus three bottles is what? You got to focus on that. Anyway, so let's just... From that given information, let's just try to create some sort of algebra equation. Well, we have the volume of three pictures. That means, like, three times P, you know, plus... We have two bottles. Let's say, B represents four bottles. Let's just use capital letters. Two bottles is equal to 16. Gotcha. Gotcha. We have one more. It says the volume of each of the picture is... That means P is equal to... Is two times greater than the volume of each of the bottles. Okay. That means two times the bottle. So... Then what would you do? At the end, we are going to find what is 2P plus... 3B. This is what we are going to find, guys. Let's just find first P and B. We have already given, guys, P is equal to 2B. If that is the case, can I just plug that P... In the equation, first equation here. So what I am trying to say here is we have three times P. Instead of putting P, I can just say three times 2B. Then we have another 2B. And it gives us 16. Well... After this, guys, we just got to keep going. Three times two is six. We have 6B. Plus 2B is equal to 16. We have like terms as we have B terms. That means they are the same type of terms. We can just add them. 8B is equal to 16. Then... We need to find only one B, guys. Not 8 times B. We got to divide by 8. That gives us B is equal to 2, guys. I hope this part is clear. Anyway... I mean, I know you would ask a question if you kind of struggle in one piece. After we found B, guys, as you see, we are not even done yet. We are going to find P as well. There are two different equations. The first one says 3P plus 2B is equal to 16. Another one says P is equal to 2B. Well... When you look at those, both of the equations... You got to make the decision, guys, which one you want to use. I would use the second one because that one seems easier, you know. Then, when I try to find P, I can just plug that B here in the equation. 2 times 2. That means P is equal to 4, okay? The volume of the picture is 4, basically. But at the end, we got to find volume of two pictures. We have 2 times P plus 3B. I will just plug the values. 2 times 4 plus... We have 3 times 2. This one is 8 plus 6. Oops. Should be 14, guys. Are there any questions, guys? I believe you asked me already, so I'm going to move on. Alright, check number 20, I mean number 2 plus year 2020 and it was question number 10 for that year. Okay, I stopped the timer. One person says 43. Okay. Okay, 43, let's see. All right guys, it has been 1 minute and 15 seconds, let's see. Another person says 43. Oh, okay. We have 2 votes for 43 so far. Another person says 43, maybe one more or I'm gonna think the answer should be 42, let's see. No more answer, I guess. I'm going to start then. All right. Okay, so... A number is written in each of the cells of a 3x3 square. As you see, the numbers are not visible because they are covered by ink. Gotcha. However... The sum of the numbers in each row and the sum of the numbers in two columns are all known as shown by the errors in the diagram. What is the sum of the numbers in the third column? Basically, we are going to find the question mark there. So look, again, there are all different types of ways for this one. If you... Look, even the name of the topic is algebraic thinking. If you want to say, you know, sir, I can just say, let's say this is A, this is B, and this is C. Basically, I should be able to say A plus B plus C is 24. Then D, E, F. When you add them, D plus E plus F is 26. And G, H, K. G plus H plus K is 40. Then how can we find C plus F plus K? Look, you do not necessarily do that, guys. Just think about that. In that three by three square, in that box, we have total nine numbers. Yes. So if you look at the sum of those rows here, they have to use all of those nine numbers to be able to find that sum. There is no other case, you know. When you say 24 plus 26 plus 40, guys, that means you are finding the sum of every single number here. A plus B plus C plus plus plus until K. But, as you know, we are going to find C plus F plus K, this piece. This piece we're going to find. We don't know that yet. But at least, guys, we know the sum of other numbers. At least we know A plus D plus G, which is 27. And also we know B plus E plus H, which is 20. And let's say that question mark is X, the sum of that column is X. OK. We both know that, guys. Either you add those rows or either you add those columns at the end, you have total of nine same numbers. I am saying same numbers, guys, because you never ever use other numbers anyway. Well, in that case, if we have same total of every single case, can I say that they are equal to each other? Yes. You see, we do not necessarily add A plus B plus C or whatever. It's just waste of time. Don't worry about that. That's the reason that I was just saying that for some questions, guys. Remember the lesson instruction. You do not necessarily read the question and just jump into solution. No. You read the question and try to figure out what would be best and easiest way to tackle the problem. You are kind of planning in your mind. That's really important. You see, in that case, we are not necessarily worried about what's A plus B plus C or plus D plus E plus F. No one cares about that. And at the end, you can do some sort of canceling. Remember from the scale, you can just cancel. You can add or subtract the same number from both sides. Well, even from here, guys, let me give you another trick. You do not necessarily add the numbers first because look at every single number. We have 26 here and 27 here. They are really close numbers. Basically, if we take out 27 from both sides, look why I'm doing that? Because we got to cancel every single number next to X. So they are gone. So here we have negative 1 left. It's fine. We can deal with that later. We are good. Instead of adding them first, you know. So what about here? We have 20 here and we have 24 here. Well, you know, they are close to each other. We can cancel that 20 by subtracting 20 from both sides. Then we have 4 left. At the end, as you see, we have 4 minus 1 plus 40 is equal to X. I really want to believe that you all know that 4 minus 1 is 3 plus 40 is 14. I hope it does make sense, guys. If it's not, please let me know. Look like we are good. I'm going to move on now. I haven't seen any message. Yeah, OK, we are good. Cool. See you on the next one. All right, guys. Number 3 is your next challenge. Then we've got a start time. okay guys it has been two minutes then maybe we can just go one person who says A, one person who says D, and one person who says B. okay we have so far A, B, and D okay that's interesting one of the two friends found the same answer it's fine less statement or maybe it just seems a little tough to you remember guys you should use the problem-solving strategies we have been discussing if it's possible draw some sort of you know graph table or whatever or picture maybe and one person says B. all right again you do not have only four friends here but at least we get some answers let's see what was the answer guys so look we have given A over B is 9 over 4 and B over C is 5 over 2, 5, 5, 5 anyway then we gotta find A minus B divided by B minus C. look here if we had like let's say A over B over C like I'm just making up numbers 4 over 7 over let's say 5 then you would say if A is 4x, B is 7x, C is 5x then we can just you know plug them in that's A minus B and B minus C we could find some sort of answer but as you see A over B I'm gonna make a chart you and I can create another chart okay so A, B and B over C. so so far guys A over B is 9 over 4 and B over C is 5 over 3 and you see if you start with you know first proportions we have B is 4 versus B is 5 if we had same B values we would have like kind of more better ideas but not yet well if it is not the case guys let's just use equivalent fractions what I meant instead of having 9 over 4 guys we can just expand like say just multiply both sides by 2 you can say 18 over 8 you know we can keep going like for couple of them we can multiply each of them 12 we can keep multiplying 16 or the 5, 20, 54 over 24 that should be enough so what about this part guys we can have 5 over 3 or 10 over 6 you know 15 over 9, 20 over 12, 25 over 15 and so far as you see guys after you create couple of A and B ratios I recommend you to check both of the B columns so we have 4 through 24 of multiple 4s and here 5 through 25 of multiple 5s I believe you are going to figure out guys this 20 and this 20 are the same numbers well when we get that then we are good what I mean guys so look when we change the column we can just see easily A over if A over B is 45 over 20 then B over C can be 20 over 12 even we can combine both of those you know proportions we can say well that means A over B over C gives us guys 45 over 20 over 12 basically so as you see we combine them from that information I can easily say that A is 45 X B is 20 X and C is 12 basically A is multiple of 45 at the same time B is multiple of 20 and C is multiple of 12 then what well we can just plug in those X values here in the expression they say A minus B well we have 45 minus 20 you divide by we have B minus C so B was 20 and C is 12 so we get 45 minus 20 is 25 guys and 20 minus 12 is 8 so this is going to be your final answer. Any question for me about this one? yeah I don't know what that means no we can move on. I kind of guys try to explain extra because well I was expecting more answer from this one maybe this is the one of the new type of question here is I don't know anyway okay number four here is your new gift enjoy and one person says 30 okay and another person he says 30 as well okay we have classified So, we have only, only two answers so far. Okay, two more, say 30. Oh, gotcha, let's finish, then we can start. Right now, 30 get, oops, total four votes. That's what's gonna happen. All right, so, as it's like a whole paragraph question and we just compare all of those different aliens with like green, orange, or blue color, it looks like we are going to work with a lot of algebraic equations. So, group of aliens was traveling in a spaceship, each of them was dressed in a jumpsuit that was green, orange, or blue. Let's say capital G represents number of green. Orange, I wouldn't choose all because it looks like a zero, I can just make R for orange and B for blue. Then what? Each alien dressed in a green jumpsuit had two tentacles, okay. Each of them dressed in an orange has three and each alien dressed in a blue has five tentacles, okay. This is the first thing and it says all aliens together had 250. Look, let's see, green one says two, so we have two green, basically, plus, this is not six guys, this is G by the way. And we have how many orange? Orange has three, okay, three orange. And how many blue? And blue has five, okay, we have five blue. And the total tentacles is 250 as you see. Well, we have one, two, three different variables and we have only one equation. There is no way guys we can solve this. If you have only one variable, only one unknown value, you have to have at least just one equation that would work. If you have two variables, you should have at least two different equations to solve than you're able to. If you have one, two, three different variables, you have to have three different equations to solve this. I mean, you can try to guess and try to figure out something, but I am talking about algebraic ways to solve. That's the reason I am saying you need at least this amount of equations so we can just solve it. Anyway, that means if you get only this equation, we are done, we just start. Okay, there were as many aliens in a green jumpsuit as those with orange. Well, that's good. They are saying G is equal to R from that sentence. Okay, and there were 10 more aliens dressed in blue than green. Well, that means guys, there were 10 more dressed in blue than green. That means there were 10 more in blue than green. When you compare blue and green, blue should be 10 more. That means green plus 10 should be blue, okay? That's what we can do. Right now, we get one, two, three different equations. We got to do what? We got to solve it. Well, we have that one is already started. Two green, we have green is equal to orange. I mean, yeah, let's just plug it here and we can say since we have five blue and we can convert that blue also in terms of green, let's just plug them in all in terms of green. That seems the most logical way. Anyway, so we have already the two green. I'm going to keep it two green plus instead of two R, I'm going to put, you know, just plug that in and plus we have this one. I'm going to plug that one as well, 250. Let's use different colors. What I meant three times R, I just switch R and G. I make it three green, you know, here, not 36 by the way, three green. Gosh, that's green. Okay, whatever. And since blue is equal to green plus 10, I'm going to make it five then green plus 10. All right. Then it's look like we got to do a lot of, you know, combining like terms and distribution or whatever. So first thing, since we see that parenthesis, we should distribute here. We get two green plus three green, five green is already plus five times G, five G, another five G coming plus five times 10 is 50 is equal to 250 where five G plus five G is 10 G plus 50 is equal to 250 guys. Then you got to take out 50 from both sides, you know, that means 10 G is equal to 200. And of course, I will, you know, set, you got to divide both sides by 10. That means guys, G is equal to 10. We found number of aliens who dress in green since G represents green. But as you see, we couldn't find the target number. They asked us to find, you know, number in blue here at the last sentence. Well, we have another equation here. It says G plus 10 is equal to blue. And we should just say that 20 plus 10 is equal to 30. I just applied this, you know, same equation. Then we get 30 represents the number of blue events. It was long and it requires a lot of explanation. We got this. I'm just waiting. If you have any questions, just ask us via chat. Okay, we are good. We can move on. I will. Let me go to the next one, guys. Let's see what you will get from this one. all right guys it has been two minutes well so long do we get many answers or not yet okay we can be a little bit more than okay now someone answers a okay anyone else And one more person answers A. Look at that, all right. Anyone else for A or what about others? Anyone says B, C, D? Nothing? This sum of indicate the total weight of the figures, okay. Two figures of the same shape have the same weight, okay. If P is less than Q is less than R, then which one should be right? Let's start guys, you don't have too much time left anyway. Let's see if those friends who find A is correct. So look, it says P is less than Q is less than R. If you have inequalities like that guys, look, there are some different cases like that inequality represents P is less than Q, okay. Also Q is, oops, what type of Q is that? Also Q is less than R. Also guys, we can say P is less than R. We will try to see if any of them works. So let's just compare, we have P and Q. Compare P and Q. Well, we have same of those squares, they are gone. We have two triangle is less than two circle. If guys, two triangle less than two circle is not 20 by the way, you just cancel those two, basically divide by two. We know that the weight is that circle, I mean, that's triangle here at the top I am putting, less than weight of circle. This is the first thing you can find. Let's just keep working. We have been also, they also say Q is less than R. Where is Q and R? We have Q and R here. Let me erase those piece, if we can find something, Q and R. So these are gone as you see. Well, we have circles and square and another triangle. I wouldn't bother too much, because we have three different to compare. We have also P and R. P is here, R is here. Well, we can cancel one of square here. We can cancel one of the triangle here. At the end, we get triangle and that square. We found that triangle is less than square. All right, in terms of that given information, you gotta check them one by one until you find that. The first one says P is less than S is less than Q. It says P is less than S. When you compare P and S guys, so one of the, oops, let's just erase those. One of the square is already gone. One of the little triangle gone. We gotta compare triangle and circle. Yes, we both know triangle is less than circle. That's correct. P is less than S. Also, you can try to compare S and Q if it's possible. Yes, we can try that too, really quick. S and Q. We just gotta continue the same inequality. So one of the square is gone. One of the circle is gone. Then it says S is less than Q. Yes, we found, as you see, that little triangle is less than that little tiny circle. Then S is also less than Q. That means A is already right. If you find the right answer, you don't necessarily check others, guys, okay? Because we have been working for every single of those. I mean, we have been comparing a lot of those shapes anyway, so it should be fine. Any question for me? Okay, no one says anything. All right, guys, that means I'm gonna see you next time. All right, and until next time, take care, bye.

algebraic thinking

solving equations

problem-solving strategy

variables

simplification

algebraic terms

interactive learning

equation setup

practice problems

collaboration