A Nice Algebra Puzzle

Last week I went into a Grade 9 class that had just begun to learn about simplifying polynomials by collecting like terms. I had an idea for a task that I thought would help them with this so began by showing them this pyramid.

I explained that the numbers in two adjacent squares add to give the number in the square directly above them. With this information, I split them into visibly random groups of 3 and had them work at whiteboards to find four numbers that go in the bottom row that would give 54 in the top square.

After completing this, I then gave them this pyramid:

I wanted to see how comfortable they were with collecting like terms before giving them something more thought-provoking. Some were able to complete this symbolically and others were happy to use algebra tiles to help their thinking:

I then gave them this task:

By now, I could hear how adept the students were at collecting like terms and was impressed at the different ways they went about solving the task:

Next, I gave them this task with the restriction that all the terms on the bottom row had to be different:

Again, I was really pleased how they went about solving this (some symbolically, some using algebra tiles again) and by listening to the students talk, I could tell that they were really understanding this. I even overheard a few groups say how much fun the task was!

As this had taken less time than I expected, I had to then think quick and come up with an extension. I asked them that if the value of the top brick was 57, what would the value of the four bottom bricks be?

This required a bit of clarification for two or three groups, but once they understood it, it allowed them to demonstrate their algebraic skills in solving equations and substitution:

It was great to see all the different solutions as well as see that in some cases, the different terms actually resulted in the same value once the substitution was made.

I finished by asking them to figure out:

a) the side lengths of an isosceles triangle, given that the perimeter is 120 cm and that the two equal sides are double the size of the other other side

b) the dimensions of a rectangle given that the perimeter is 1000 cm and that the length is triple the width.

For the first problem, they did this by trial and improvement, so I walked them through how to set this up algebraically. I was pleased to see them al use this approach for the rectangle problem.

I was really happy with the way they remained engaged throughout these tasks, especially as it was a Friday afternoon and would definitely use these again.