I was going to write this blog post about moving from guess and check to an algebraic solution method, and why you might want to use algebra.

Then I was reading through some of the Guess and Check submissions we got, and thinking, “wow, that looked pretty effortless.” One student, Jack M. from Rosemont School of the Holy Child, pointed out that the solution has to be a multiple of 3 between 20 and 40. If you start your guess and check with some of that reasoning, you know you won’t have to do very many guesses. And if you’re good at adjusting a guess up or down based on your results, you can get the answer pretty quickly. Jack only needed two guesses: he started with the first multiple of 3 above 20, which is 21, and got to 24 on his second guess!

Rashmi R. from West Woods Upper Elementary School got the answer in 5 systematic guesses (he didn’t focus on the multiple of 3 idea, but was still very efficient). He guessed 10 CDs were sold, then 20, then 30, then 25, then 24. Each time he used the data from the previous guess to think about if he needed to increase his guess or decrease to something between the previous two guesses.

Adam S. from Highlands Elementary School may have had a lucky first guess or he may have thought hard about a reasonable starting number… he doesn’t say. Either way, he started with 25 as first guess, realized it was too high, adjusted to 23 for his second guess, realized it was too low, and got to 24 on his third guess. Take a look at his work, below:

First, I decided to use the stratergy guessing and checking because I needed to find out how many CD’s were sold each day. Next, I completed the equation shown here:

40-25=15×2=30-25=5×3=15-25= -10

Because I needed to find out how many CD’s were sold each day and the amount of CD’s left on the first day were doubled, sold the same amount as last time then were tripled then left with no more.

Next I did the equation:

40-23=17×2=34-23=11×3=33-23= 10

I did this because I needed to find out how many CD’s were sold each day and the amount of CD’s left on the first day were doubled, sold the same amount as last time then were tripled then left with no more.

I knew I was getting closer but I needed to go higher so I tried selling 24 copies each day:

40-24=16×2=32-24=8×3=24-24=0

I knew that selling 24 copies each day was the correct answer because after 3 days, there were no more copies on the shelf.

People reading this might be noticing that what Adam calls equations aren’t really equations… 40 – 24 = 16, but 40 – 24 ≠ 16 * 2, because 16 ≠ 32. But if you read the equal signs as more like “then that answer” signs than equal signs, you can see the steps Adam did. I’m going to use a diamond symbol to replace the 23, 24, and 25 because I want to show how it’s the same steps each time. You can look back and see how each time I put the symbol Adam put a 25 in his first “equation,” a 23 in his second, and a 24 in his third. I’m also using * for multiply, so no one gets confused with the letter x.

Adam’s steps: 40 – ◊, then that answer *2, then that answer – ◊, then that answer *3, then that answer – ◊.

At the end, Adam checks if the result is equal to 0.

I used algebraic notation (kinda, I used a symbol to replace a changing numerical value for the same quantity) to show what Adam did because I wanted to communicate his work in a particular way. That’s a good reason to use algebra. Adam didn’t use algebra, and so I wondered why. If his work can be communicated algebraically, why wouldn’t he go ahead and do that? You can usually get to the answer pretty efficiently once you can set up an algebraic expression of it.

One reason is, Adam could get to the answer in only three guesses. He didn’t need to keep guessing and guessing. The answer was an integer, and a pretty reasonable one. And guessing meant that he could keep plugging numbers in at each step, rather than dealing with substituting variables. If you read my expression, 40 – ◊, then that answer *2, then that answer – ◊, then that answer *3, then that answer – ◊, there’s a lot of sort of awkward, “then that answer.” In Adam’s work, the answers are numbers that can be easily plugged in to the next expression.

For algebra to be a better solution method than guess and check, the pain of guessing over and over has to outweigh the pain of dealing with, “then that answer” over and over.

To formally algebra-ize my expression, here are the steps I would take:

40 – ◊, then that answer *2, then that answer – ◊, then that answer *3, then that answer – ◊.

first “answer”: 40 – ◊

that answer times 2: 2(40 – ◊)

that answer minus ◊: 2(40 – ◊) – ◊

that answer times 3: 3(2(40 – ◊) – ◊)

that answer minus ◊: 3(2(40 – ◊) – ◊) – ◊

Then set that mess equal to 0: 3(2(40 – ◊) – ◊) – ◊ = 0

Solving that, I made three arithmetic mistakes (no, actually 4, I realized when I looked back at the cross-outs on my scratch paper!) Maybe guess and check was more efficient.

Now, you might interpret this whole blog post as me being against teaching algebra, or suggesting that algebra is not a terribly powerful tool. I don’t mean that at all… I know algebra is powerful, important to be good at, and worth learning. I also think it’s important, as they say in the Common Core Mathematical Practices, to “use appropriate tools strategically.” I like to know that students are using strategies that are efficient and effective for them, and that they get good at all kinds of strategies.

In Adam’s case, I’d like to know, for example, whether he is comfortable enough with the concept of substitution that he could algebra-ize his solution if he wanted to. And I’d like to know what the threshold is for him to switch to an algebraic method when lots of substitution is involved. What if he had to guess 4 times? 14 times? 140 times?

As a teacher, I want to make sure that I ask students to do problems in which guess and check is truly onerous, so they experience the power of algebra and some prodding to develop more general techniques. I also want to provide problems that can be solved authentically many different ways, to have good conversations about “using effective tools strategically,” to compare the similarities and differences in different methods, and to practice tricky algebra skills in easy, checkable contexts.

Speaking of algebra, I thought I’d leave you with two examples of how students with algebraic approaches organized the many, many quantities in this problem. Some of those approaches are pretty complex, but I like how you can watch the students make sense of the problem using algebraic tools. They didn’t just pull the algebra in after they figured out what the heck was going on in the problem.

Cary D., from Birch Wathen Lenox School24 copies =nI figured this problem out using a series of equations.

Let n= number of copies sold

Let l= leftover copies

Let p= number of copies left in caseFor Monday night I figured out, 40-n=l

For Tuesday morning I figrued out, 2l=

For Tuesday Night I figured out, 2l-n=p

For Wednesday Morning I fligured out, 3p

For Wednesday night I figured out, 3p-n=0Then I thought and eliminated unnesseary equations like wednesday morning and tuesday morning.

The I wrote 40-n=l and 2l-n=p out and figured out an equation and substitued 40 in

2(40-n)- n=p

80-3n=pThen I wrote 80-3n=p and 3p-n=0 out (wednesaday night equation)

From this I substitued the p in wendesday night’s equation out and put 80-3n in.

3(80-3n)-n=0 Then I solved

240-9n-n=0

240-10n=0

+10n +10n

240 =10n

10 10

24 copies =n

And

Renuka D., individual submitterI made a table and entered quatities as I read the question

stock = N Display Sold Mon N-40 40 M Tues 2(40-M) M Wed 3*(2(40-M)-M) M At the end of Wednesday the display had

3*(2(40-M)-M)-M = 0

3*(80-2M-M)-M = 0

240-6M- 3M-M = 0

240 = 10M

M= 24

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