 So let's think about those tape diagrams and see how it can use them to solve algebraic problems. So the basic problem with algebra is to find an unknown number given some fact about it, given some factor set of facts about it. So for example I might want to find a number given that I know that x plus 11 equals 20. Or maybe I want to find a number given that I know that 15 minus y is equal to 8. Or again maybe I want to find the unknown number given that z minus 8 is equal to 12. And you might think for a minute, when you were first introduced to algebraic problems like this, once upon a time, really only about 150 years ago, these were college level problems. You didn't see algebra until you were in college and by college we are talking about places like Yale, Harvard, all of the major universities. College was the time that you first got introduced to algebraic problems like this. Well, more recently the algebra has been moving earlier and earlier in the curriculum. At one point it was a high school topic. Most of us were probably introduced to algebra in the middle school, but recent research shows that provided that children have an understanding of the basic concepts of arithmetic, and that means not that they know how to do arithmetic, but they actually understand what it is they're doing. If they have this basic understanding of arithmetic, you can start to introduce algebraic problems like this as early as the first or second grade. Now there's a proviso here. You can introduce problems like this. You can have them solve problems like this, but you can't teach them the same way that we were taught algebra by following a sequence of steps. That doesn't do any good. Instead the solutions to these rely on an understanding of the basic concepts of arithmetic. And a little bit of help here is going to be gained if we use a tape diagram. So let's consider that problem, x plus 11 equals 20. And we want to solve this problem using a tape diagram. Tape diagrams, the key to using them is to remember three things. Every time I write down an equation, so there's an equal sign, an equation is something equals something else. Every time I write down an equation, that corresponds to the idea that something and something else are the same thing. And with a tape diagram, what that means is that the equation is a claim that two tapes have the same length. Now those tapes are going to be the expressions on either side of the equal sign. And so sometimes I might only have one thing, so here's a 20. And on the other side I have this x plus 11. And the expression I have here, well, every term is then going to be some component of the tape. So here I have two terms x and 11. And the tape itself is going to be the result of the arithmetic operations in the expression that I have. So here I have some arithmetic operation that gives me a tape corresponding to x plus 11. I have a tape that has length 20. Let's go ahead and set that up. So again, my algebraic equation corresponds to two tapes of equal length. There they are. And one of my tapes has length 20. Again, one side of the equation is 20, so one of the tapes is length 20. And the other side is going to correspond to this expression x plus 11. So let's invoke our understanding of the basic concepts of arithmetic. When I add my set theoretic definition, my understanding of arithmetic is that I'm combining two things. And in this particular case, what I seem to be combining is a thing of length x and a thing of length 11. Now, since this x plus 11 has to be the same as 20, that means that this tape here has to be formed by putting together x and 11. So this remainder here has to be my 11. And so there's my tape diagram representation of the equation 20. 20 is equal to x plus 11. And at this point, it's logic. It's not even algebra. Well, we have to do a little bit of arithmetic. But the idea is I have this thing of length 20. And it's the same length as this. Now, I know this part's 11. So what's this part? Here's what I'm looking for. And so what I might do is I might remove 11 from both of these. It's gone. And this used to be 20. Well, this thing left over here must be 9. It's the same size as this piece over here. So that tells me that x and 9 have to be the same thing. And there's my solution to the algebra problem. And you'll note I didn't do anything with this equation that corresponds to what you might think about as the standard rules for algebra. I drew the tape diagram and I used a little bit of logic. Now, this is actually a good one to illustrate some of the advantages of this. So here's a problem 15 minus y equals 8. And this is a problem that most beginning algebra students at the middle school, high school, and college level actually struggle with because this is a problem. This is not an easy problem. And many people get this one wrong. But a second grader can do it if they understand what subtraction means. So let's go ahead and think about this is our tape diagram. So again, an equation can be represented by two tapes of equal length. And one of those tapes has length 8. And the other tape is going to correspond to 15 minus y. Well, let's think about that. 15 minus y. Well, when I'm subtracting, what I'm doing is I'm removing an amount from something else. So this 15 minus y sounds like I started with a tape of length 15. Well, if this is 8, 15's got to be bigger. I must have started with a bigger tape. And then I've removed a portion of length y. So I'm going to take my bigger tape. I'm going to hack off a section of length y. That's gone. And what's left is my 15 minus y. And again, there's an equality there. So that says the tape of length 8 has the same length as the tape of length 15 minus y right there. And again, same question as before. What did I take away? What did I remove here? Well, if I know this whole thing is 15 and this whole thing is 8, I can count up from 8 to 15 to figure out what's left. And so I can count up by 7. And that's going to take me to 8 plus 7, 15. And again, this 7 and this y are the same size. So that tells me that y is equal to 7. How about that third problem, z minus 8 equals 12. And again, we can represent our equation using two tapes of equal length. Let's go ahead and build that tape. This is z minus 8. So again, that's a subtraction. That is, we're removing some amount. And so maybe I'll start with a tape of length z. Don't know what it is. And then minus subtract a bit of length 8. So I'll hack that portion off. And here's my 8. Got rid of it. What's left over is z take away 8. And again, the tape diagram corresponds to the equality in length of two tapes. z minus 8, here it is. And 12 must be a tape of the same length. So I'll put that in there. And well, I want to find what z is. Well, I should be able to see that here, this whole thing, z, is tape of length 12 plus that last bit, tape of length 8. And so my entire tape, 12 plus 8, is going to be 20. And there's my solution.