Week Three: Chip Models

During the last and final week of this unit, we got very hands on! The way we got involved was by using chip models, or small colored chips, to determine integers.

The first thing to know about these chips is that they are yellow on one side and red on the other. Red means it is negative numbers and yellow means positive numbers. A fun, and easy way our group used to remember the difference was that red represented blood which is always bad, or negative. Whereas yellow is like sunshine, meaning positive or happy!

Just for the sake of simply understanding, in the visuals the blank (white/open) circles are going to be negative, supposed to be the red ones. The dark, filled in circles are going to be positive, supposed to be yellow ones.

For our first example we have the addition problem of 4+(-3)=?

We started it with two groups, one is positive 4 chips (black) and the other is the negative 3 chips (white).

The next step is to combine the two groups into one...

We now have one groups of 4 positive chips and 3 negative chips...

This next step is to match up the positive chips with however many negative chips there are. In this example we have 3 negative chips, which means we will match up 3 positive chips with it. We have done so by placing a box around the 3 negative and 3 positive chips. By matching them up we are cancelling the negative ones out to reach zero to help us reach our answer.

We are now only left with one positive chip...which means... 4+(-3)=1. Our final answer is Positive One! 

Week Two: Area Models, Partial Products, and Ratio Tables

During the second week of the new unit, we continued on with finding easier ways to solve mulitplication and division problems. Three main student-friendly ideas we learned about were area models, partial products, and ratio tables. I will break down each one here.

Area Model: 

The picture above describes how to solve the problem 19x48 with an area model. What to do to solve it in this manner is to draw out the multiplication problem in an area model box. Then, you simplify each entity (19 and 48) in the box to make it easier. In the picture, I broke down 48 into 20, 20, and 8. As well as breaking down 19 into 10 and 9. You then draw lines to divide the box into those numbers. After it's divided, you multiply the "length" and "width" to solve the answer in the smaller box. The answers for the smaller boxes within the large area model were 200, 200, 80, 180, 180, and 72. Once you have all the smaller solutions, you add those to find the final total. The final total of 19x48 was 912. This is a great way to create a simple visual for a difficult multiplication problem.

Partial Products: 

In this photo above, it displays the way of solving the same problem (19x48). It looks like a normal stacking method of solving a multiplication problem...but instead of borrowing you just put the total below. For example, in the picture, you put below the line the total answer of 9x8 which is 72. You do this for each part of the problem, and then add up the total. It is a much easier way than the traditional borrowing method.

Ratio Tables: 

The way to solve the same problem (again...) 19x48 with ratio tables is so so easy!! You make a table of the multiplication tables of one of the parts of the problem with just a few multiples. Such as using 1, 2, 5, and 10 and mulitplying it by 48. So, you have options of 1x48=48, 2x48=96, 5x48-240, and 10x48=480. The problem is 19x48, so you just have to add up the multiples to get to 19...what I did was I chose 10+5+2+2=19. So I added the answers from the ratio table, 480+240+96+96= 912. I got thw same answers using this way as the area model and partial products ways. Again, this is such a simple way to solve a difficult multiplication problem for a young student. It is always a good idea to put visuals (tables, area models, etc.) when solving a difficult problem.

Week One: Multiplication

During the first week of the second unit we dove right in and got started working with multiplication. We began to break down how difficult it is for students to solve multiplication or division problems when they don't even know the basics. 

We started off by learning that if a student can understand the multiplication facts of 2's, 5's and 10's, they should then be able to solve any other problem with no problem. These are called Derived Facts. An example of this is... 

This problem explains that if a student has the basics (2's, 5's, and 10's) memorized, then a problem of 9x3 (example) should not be hard. This picture shows the problem 9x3=?. To solve this, the student would understand that they are trying to figure out what 9 groups of 3's are or 3 groups of 9's. But, in this case we are using 9 groups of 3's. 

The student then knows that since 10x3=30...which is 10 groups of 3's...so the next step is to take away one group of 3 to get to only 9 groups. 

They successfully solved 10x3 which equals 30. So, they take away one group of 3, 30-3=? which equals 27. Therefore... 9x3=27! 

This problem shows that the student understood their multiplication facts of 10's and then went from there to solve the problem correctly.

This week was very insightful! I learned that you have to start with the basics of multiplication and then work your way up. Sometimes I get flustered with solving difficult multiplication problems and I now know that it is ALWAYS okay to go back to where you started and go from there. I can't wait to know what we will learn next week!