 So now that we have a definition of division as the inverse of multiplication, we're ready to do a little bit more algebra. So let's consider a new problem, solve for x, 30 divided by x is equal to 5. And again, to highlight the similarity between arithmetic and algebra, we're going to try and use three different methods that do not use the so-called rules of algebra. So let's take a look at that. We might start by considering our definition of division. So what is that? Well, let's go ahead and set that down for reference. Remember, our definition of division is based on a definition of multiplication. If I have a product a times b equal to c, then I can reverse that and rewrite it as a division a is equal to c divided by b. And let's go ahead and compare our two statements. And so the first thing we might note here is that this 30 is taking the role of c. So I'll replace it. The b here is taking on the role of x. So again, I'll replace it. And then this a here, this a here is taking on the role of 5. So I'll replace it there. So now I have a multiplication statement that corresponds to a division statement. So if 5 times x equals 30, and here you can see why we don't really like using the syntax cross to indicate multiplication. But in this case, we're kind of stuck with it. So 5 times x is equal to 30, then 5 is 30 divided by x. And here's why the conversely part is important, because conversely means that if I have this, then I also have 5 times x equal to 30. So that gives me a new statement by the definition of division. Of course, I haven't actually solved for x yet, but I can apply the definition of division a second time and use commutativity and reverse the order of the product. So by commutativity, this 5 times x becomes x times 5. And by a second application of the definition of division, if x times 5 equals 30, then x is equal to 30 divided by 5. And that's just an arithmetic operation I can do without comment. So x is equal to 6. And again, just as a note, if we want to show our solution without using the rules of algebra, the actual solution, based on our definition of division, properties for arithmetic, definition of division again, is the portion shown in green. All right, so is there a different way we could look at this? Sure, we might take a look at this as a partative division problem. So by our partative to interpretation of division, 30 divided by x equals 5 means that what I've done is I divided 30 into some number of parts, each of which has a size of 5 units. And since I don't know how many parts there are, but I do know how big each part is, I can begin marking off parts of size 5 on a tape diagram. So I'll go ahead and set down a tape that represents my 30. And I'll start setting down parts of size 5. So I'll break off sides of size 5. So here's going to be break off 1 and put down a second and a third. Let's check that's 5, 10, 15. Let's keep going 20, 25, and 30. And there's all of our parts of size 5. And altogether, all of these make up our original 30. And I can count how many parts do I have? One, two, three, four, five, and six. So 30 has to be divided into six parts of size 5. And so that tells me that x is equal to 6. And again, our solution is the portion in green together with the tape diagram that shows how we got our numerical value. Well, remember, there is a second way of interpreting division, which is as a quotitive division. And so in this case, what I've done is 30 divided by x. So I'm going to do this as a quotitive division. I've divided 30 into pieces of size x. And at the end of it, I've gotten 5 such pieces. So let's go ahead and do a similar solution using a tape diagram. We'll start with our tape diagram representing 30. I know that I have five pieces, each of size x. So there's my one, two, three, four, five pieces. And the question is how big is each part? And so I have five pieces here, all together there, equal to 30. So that means each one of these must be of size 6. And since I've divided 30 into five pieces, each of size 6, then my unknown is going to be equal to 6. And again, the solution, this third solution, not using the rules of algebra, but rather using an arithmetic process, the tape diagram to show the division and the wording around it. The answer that I want to give is the portion that's highlighted in green.