Before we started the activity we discussed the value of each of the things below. In hindsight, I should not have used pennies because we know the value of a penny. I asked students to pretend that they didn't. We realized that we would talk until we were blue in the face, we still wouldn't know the exact value of each object. That's when we decided to use variables for each item.

The value of a penny = P

The value of a button = B

The value of a paper clip = C

One by one the students came to the front of the room to either place items in the box (I used orange paper in the photos so it would show up better) or take items out. I limited them to only one type of item.

The first student placed three pennies in the box. I asked the students what the total value of all the items in the box were, they told me 3P, so we wrote 3P on our papers.

The next student came forward and put four paper clips in the box, so we wrote + 4C.

Now we have 3P + 4C. I asked if we should make that 7PC. We agreed that wouldn't make sense (get it? sense/cents?? Never mind).

The third student took 2 pennies out of the box. Now we have 3P + 4C - 2P.

The last student for this round of the activity added 5 buttons to the box. 3P + 4C - 2P + 5B

Keep in mind that the students can't see what's in the box. I asked students to simplify the expression and they easily came up with 1P + 4C + 5B. I asked if it would be a different set of items in the box if we wrote 4C + 1P + 5B. They responded that the order didn't matter. Yay! The Commutative Property that we just learned last week!

Now, on to the distributive property.

After we did a few round with the single items, I brought out some Ziploc bags with items prepackaged.

I had 5 bags with 3 buttons and 2 paper clips,...

...5 bags with 4 pennies and 1 button, ...

... and 5 bags with 2 pennies and 1 paper clip.

The students can either put in or take out single items or Ziploc bags. The first student put in 3 buttons. 3B.

The next student placed in two bags that have 3 buttons and 2 paper clips.

3B + 2(3B + 2C)

The third student took out 1 bag with the 3 buttons and 2 paper clips.

3B + 2(3B + 2C) - 1(3B + 2C)

The students simplified the expression while I emptied the bags into the box to get a total count.

When students came up with answers that did not match the content in the box, we discussed how the mistake was made and why it didn't work. Like the student who distributed a 1 rather than a -1.

We also discussed that since the 2nd students added 2 bags and the 3rd student took one back out, it was as if only 1 bag was added. Looking at that algebraically: 3B +

**(3B + 2C)**__2__**(3B + 2C). Combine like terms with parenthesis to get 3B +**__- 1____1__(3B + 2C), then distribute. Same correct answer, interesting. Does this work all the time or just this problem?
Extension questions:

Write an expression, ask the students to explain what someone would have had to put in or take out of the box to make that expression, then finally simplify.

I like the questions that I Speak Math asked at the end of her video (I always forget to emphasize terminology). What are the coefficients? What are the terms? What is the whole thing called? What are the constants? Etc.