Multiple Solutions Part 2: Celebrating Different Methods (Lesson 1)

Last year I taught two lessons on constructing equations to a top set year 8 class. In the first lesson I set 3 questions for the students to attempt in any way they saw fit.  It’s important to note that I did not tell them that they could construct equations to solve the problems. I wanted them to do whatever felt natural rather than leading them into a certain strategy.  The questions I set are shown below:

1)     I think of a number, add 7, multiply by 3, subtract 3, divide by 6, then multiply by 12. The answer is 72. Find the number I was thinking of.

2)     The number in each square is the sum of the numbers on either side. Find the missing numbers in the circles.


3)     Adrienne has three times as many dollars as Karim. Adrienne gives Karim $20. She then has $5 more dollars than Karim. How much does Adrienne have now?

While they were doing the problems I circulated around the room to get a sense of the strategies being used. As I suspected all but one of the pairs started with an algebraic approach. I didn’t anticipate however that some of them would struggle to write down the correct equation. These students reverted to what they had learnt in year 7 and worked back from 72 using inverse operations.

I helped the one pair who persisted with the algebraic approach by leading them into the structure shown below. In this way, the pair would later be able to explain why the inverse operation technique worked to the rest of the class.

([3(x + 7) – 3]/6) × 12 = 72

Once every pair had moved onto question 2 we stopped to discuss the solution to question 1. I praised the entire group for finding some interesting strategies and asked a pair who had used the inverse operation technique to explain their reasoning. Once they had given an adequate explanation I asked the pair who’d struggled on the algebraic technique to explain their strategy on the other side of the board through slowly constructing the equation.

It became immediately obvious to the class that the algebraic technique was in fact identical to the inverse operation technique. The main advantage of the algebraic technique being that it is elegantly communicated and provides an explanation as to why the inverse technique works.

On question 2 most of the pairs used a numerical trial and error technique and every single pair arrived at the correct solution quite quickly. Again, once I saw that everyone had finished I asked one pair who had used a structured table to order their trials to come to the board and then another pair who had successfully used an algebraic strategy shown below.

(30 – x) + (23 – x) = 25

53 – 2x = 25

x = 18            etc.

(Note: I neglected to show the approach of using three different variables and then constructing 3 seperate equations at this point since I wanted the idea of using different algebraic approaches to hit them in the next lesson.)

The pair made a decent attempt to explain why the bottom left and right circles were 30-x and 23-x but struggled to get it across to the rest of the class. I therefore intervened and used a diagram to help explain it.

Once this was verified, which is indeed the most important part of this strategy, the students were easily able to solve the equation and find the three missing values. It’s important to note that we did discuss that the positioning of x could have been in any one of the three circles.

At this point in the lesson it was blindingly obvious that the students were impressed and intrigued by the algebraic strategies and were enthused to try to use an algebraic strategy for the last question.

All pairs with the exception of one started by trying to construct an equation to solve the problem. Many of them became quickly confused with the algebraix approach and reverted back to trial and error until they eventually found a solution. One pair however persisted with an algebraic approach until they derived two simultaneous equations to solve the problem. Incidentally this pair was made up of two girls who’ve recently moved to Brazil from Japan and South Korea (This is something I will come back in an upcoming post). You may think at this point that it’s possible that one of them could have seen a similar question before. I do feel that they struggled on the problem and am confident that this was not the case.

I asked them to explain their strategy on the board which was written as follows:

A = 3K

A – 20 = K + 5

3K – 20 = K + 5

K = 12.5 and therefore A = 37.5

So Adrienne now has $17.50.

I congratulated the pair on their solution and so did the rest of the class. I proceeded by congratulating the entire class on their strategies, making it clear that I did not favour the algebraic approach over the numerical. I then explained how a mathematician would lay out the solution above by labelling the equations by using numbers 1 and 2 for example, and stating that they were then going to substitute equation 1 into equation 2.

After the lesson I was extremely happy that one pair had used this method. Not only because it’s a hard problem to solve for any year 8 but also because I’d spent time before the lesson thinking of an algebraic approach that I felt would be easier for the students to understand. We therefore had two different algebraic techniques for the same problem that we could discuss at the start of next lesson for both question 2 and question 3 – that’s where the real fun begins! (Still to Come – Multiple Solution Part 3: Celebrating different methods (lesson 2))

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