 Alright, so let's consider multiplication with mixed numbers. And one important concept here is that of an improper fraction, this is going to occur whenever we have a numerator exceeding the denominator, and one of the common ways of looking at multiplication and division with mixed numbers is to convert them into improper fractions. So in general, the mixed number, pole number A plus fraction BC becomes the improper fraction A times C plus B over C. And this is a formula that emerges from considering what A is as a fraction with denominator C. And questions here aren't too exciting to convert the improper fractions from mixed numbers to pole numbers. So this is 3 and 2 7, so that's 3 times 7 plus 2, that gives us the improper fraction 23 over 7. And then we have something like 15 and 5 12, so we do 15 times 5 is 15 times 12 plus 5. And after struggling through the arithmetic, there we get the improper fraction 185 over 12. Now the reason we would care is that one way we have of multiplying mixed numbers is since we know how to multiply two fractions, if I convert a mixed number into a fraction, I can do the multiplication that way. So for example, take the product 3 and 2 7 times 2 and 3 4, and here's the observation to make. Now the numbers are relatively small, 3 times 7, not too bad, plus 2, of course, easy. 2 times 4, not too bad, plus 3. So it's not too difficult to convert these to improper fractions. So 3 and 2 7, that's 23 over 7, 2 and 3 quarters, that's 11 over 4, there's my two improper fractions, and I now can multiply the two fractions. A little bit more work, 23 times 11, I have to figure out what that is, but that's 23 over 28, and here's the catch. Again, the dialect should be the same. So we've asked the question in the dialect of mixed numbers, this has to be reduced to a mixed number. So what do we do? Well, we divide 28 into 253, and after a bit of work, we find that's equal to 9 and 128. Now, again, quick reminder of this concept of adaptive expertise. You have adaptive expertise if you switch from one strategy to another, whichever works best for a particular problem. The difficulty is if you only have one tool, you have to use it, no matter how painful it will be. If you only have a hammer, you've got to treat every problem as a nail. So if I have this nice tame multiplication problem, 1 and 1, 6 times 1 and 3, 8, well, by improper fractions, that's not too bad. 1 times 6 plus 1, that's 7, 6, 1 times 8 plus 3, that gives us 11, 8, and again, multiplication of two fractions, not too bad, 77 over 48. We do have to do that reduction back to a mixed number, but again, in this case, not terrible, 1 and 29, 48. But let's take a slightly more involved problem, 7 and 1, 6 times 4 and 3, 8. Well, if I know how to do improper fractions, that's not too difficult. 7 times 6 plus 1, 43, 6, 4 times 8 plus 3, 35, 8. I multiply the two together, and I had this point have to do a lot of work to recover that as a whole number, plus a fractional amount. In other words, I want to give my answers a mixed number. Fortunately, we don't just have a hammer. We know how to multiply in general. We don't have to multiply this as two fractions. And this relies on the fact that our mixed number, any mixed number, can be viewed as a whole number plus a fractional part. So, well, let's use the area model. That's a nice model for multiplication. 7 into 6 times 4 and 3, 8. Well, I'll split that. That's 7 and 1, 6, 4 and 3, 8. And, well, that's an easy multiplication here because that's just 4 times 7. And I know how to do that. I might go on to this next box. This is 1, 6 by 4. Again, relatively easy multiplication. That's 4 over 6. And while we're at it, let's go ahead and reduce that to 2 thirds. We have one more multiplication here. This is 7 by 3, 8. That gives us 21 over 8. Looking ahead, we know that we want our final answer as a mixed number. So, we might as well convert that into the mixed number 2 and 5, 8. And then, finally, we'll take care of this last area here. That's 1, 6 by 3, 8. So, I can compute that area directly, 3, 48. And again, might as well reduce that while we can. That reduces to 1, 16. Remember, for the area model of multiplication, the area is the product. So, I'm going to add all these pieces together to find what the product, 7, 1, 6 times 4 and 3, 8 is. So, I'm going to add those together. That's a 28 plus 2 thirds plus 2 and 5, 8 plus 1, 16. I can add the whole numbers, 28 and 2. I can add this bunch of fractions here. And I might start out by noting that these two almost have an easy common denominator. That's 10, 16, 11, 16. I'd have to do a little bit of work here, 3 and 16, 48 and so on. So, again, get the improper fraction, 65, 48. Split it, combine, and write our final answer, 31 and 17, 48.