Marveling At The Historical

Math Oldies But Goodies

  • About This Blog

    This blog is mostly about math procedures in textbooks dated from about 1825-1900. I’m writing about them because some of the procedures are exquisite and much more powerful, and simpler, than some of the procedures in current text books. Really!

    I update this blog as frequently as possible ... every 2-3 days. And, if you are a lover of old texts and unique procedures, you might want to talk to me about them, at I’m not an antiquarian; the books I have are dusty, musty, brown-paged scribbled-in texts written by authors with insights into how math works. Unfortunately, most of their procedures have vanished. They’ve been overcome by more traditional perspectives, but you have to realize that at that time, they were teaching the traditional methods.

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Posts Tagged ‘equations’

Heron’s Area of a Triangle

Posted by mark schwartz on October 24, 2016


This is a brief story about a fun event that almost always happens when discussing the area of a triangle. The formula for the area is A = 1/2 bh, where ‘b’ is the base and ‘h’ is the height.

The Story

Simple enough if it’s a right triangle and the base and height can readily be seen or calculated from a2 + b2 = c2. But what happens when it’s not a right triangle? Well, one has to wiggle around a bit and do a few more calculations to determine the height, but it can be found.

But being an instructor that likes to stretch students thinking and imaginations, I draw a very scalene triangle with sides of 4, 8, and 10 and present them the task of finding the area. As I roam the room watching them work and listening to their grumblings, I ultimately have them stop and present to them Heron’s formula. I don’t bother with the derivation and for those that are interested in knowing it, I recommend Googling it.

The formula is A = 1/4 the square root of (a + b + c)( ̶ a + b + c)( a ̶ b + c)( a + b ̶ c).

This is a really nifty formula because of the pattern in it – add all the sides together, then in the next three parentheses, just negate each side, one at a time, in order. And, given that there are 4 parentheses, just divide the square root of this product by 4.

In class, I first apply the formula to a 3, 4, 5 right triangle to demonstrate how it works. Then we play with a few other right triangles to get comfortable with the formula. Then we return to that weird scalene triangle. Using Heron’s formula, we get an area of 15.1987 … and now the fun begins!

Someone typically asks “how do you know this is correct?” So, this sets us up to explore how to find the area if we didn’t have Heron’s formula; so we do all the Algebra necessary to demonstrate that indeed 15.1987 is correct. In one class, this whole presentation got applause and I take that as their having had fun with it.


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Must We Filter Students Through the Math Sieve?

Posted by mark schwartz on October 19, 2016

Deborah Blum in The Best American Science Writing, 2011 (page 184) cites a California Institute of Technology science historian as saying “K-12 science classes in the United States are essentially designed as a filtration system, separating those fit for what he called ‘the priesthood of science’ from the unfit rest of us.”

I believe the same can be said for math classes. Of course, I can assume that math was included in science, but to be very specific about it, math actually seems to be a more severe filter than general science. Today, many science classes involve students in exploration and experimentation and some of the valuable lessons of accurate measurement, recording and analysis. And, some of these activities include the necessity of math. But, when doing math in a vacuum, unrelated to an activity – in essence, the math part of the activity is secondary – the filtering action seems more apparent.

For example, in today’s texts there are typically sections on “applications”. There are even entire texts dedicated to applications and these applications show the students how math is in our everyday activity – sports, statistics, banking, calculating interest, taxes, consumption, measurements of all kinds. And this is fine. But, it’s still done in the context of filtering those who have an aptitude for it from those who don’t because …

Texts still tend to present formulae and algorithms and teachers say “this is how to do it”. In essence, teachers are saying “here’s how to do it” rather than asking “how do you imagine how this can be done?”   We don’t ask students to generate their own conception of how to solve the problem, most likely because we believe they can’t or don’t. However, many math researchers of early childhood “math” capability have found that even before entering elementary school, most children are already identifying quantitative relationships, imagining algorithms that help them understand the relationships, verifying that their conception will always work, and subsequently and repeatedly, altering their algorithm if their conceptions don’t work. It’s sort of a fundamental, built-in scientific approach to what’s going on around them. So, having created their own quantitative environment, what happens not only to the environment but also – more critically – their formulating such systems when the teacher, the text, and “schooling” provides the algorithms for them? Who needs to continue exploring the pieces of the puzzle when a solution methodology is already provided? Further, if a student in elementary school proposes a solution differing from the text, is the teacher prepared to explore that proposal to its end and see if indeed it may be worthy of consideration?

When math is taught, it in essence teaches students not to think about the relationships. The tendency – and the pedagogy – is to teach students how not to think about it because we proffer the historically valid rule, procedure, formula or algorithm which allows them to get to the answer in the most efficient way (“rule” will be used from now on to summarize procedures, formula, algorithms, etc.). Why mess around with inefficient or erroneous methods? Just give them the rule and have them practice it. Well, this does two things: first of all, practice doesn’t make perfect, rather perfect practice makes perfect and second, it suppresses what seems to be a natural urge to play with the information presented and explore the quantitative relationships that might be there.

Let’s address the practice concept for a moment. A common phrase touted by math instructors is “math is not a spectator sport” or “you don’t learn math by watching others do it.” There is some validity to this, but there is also the reality that as Yogi Berra commented “you can observe a lot just by watching.” But the question is, what is it that students should be observing? Watching a math instructor use a predetermined rule to solve a pre-established problem and then ask students to mimic this activity may actually work for some students. But, in a broader sense, what is it that we want students to learn when we teach math?

This is not a simple question and doesn’t have a simple answer. Most likely, the answer is to get students to be able to do the indicated calculation or solve the problem. But is that what is intended for them to learn? Should the lesson be about applying a rule or about exploring the quantitative relationship? Rather, it’s establishing a context in which the student can imagine alternative rules and test those rules for reliability and validity. And what are we, as instructors to do, if a student discovers a less efficient but comparably valid rule? Here’s where we run into the range of expectation of the instructor as well as the training and experience of the instructor.

Going back to the premise of math learning as a filter system, it seems reasonable to assume that all students, those who can attain the priesthood and those who can’t, could manage in a system that allows and prompts for exploration, rather than being given the rules. It would still act as a filter system, but the real key is that those not destined for the priesthood would gain a better grasp of quantitative and mathematical relationships. Basically, it is math learning by doing but the “doing” is now differently defined.

Here’s something that happened in class one day. We were just beginning to work with simple equations in an introductory Algebra class. The text approached setting up the equation by making a statement which could be directly translated to an equation. This has become a typical introductory approach. For example, the student is asked “if you take a number, double it and add 1, the result will be 5. What is the number?” The expectation is that the student will write “x”, then double it by writing “2x”, then add 1 by writing “2x + 1” and then showing that 2x + 1 will have a result of 5 by writing the equation 2x + 1 = 5.

As I moved around the room watching and helping students work through this translation, this is what I saw on one student’s paper:

P   P   P   P   X                   The answer is 2.

I asked her how she got 2 as an answer and it went something like this: I knew there were 5 pieces when I got done, so I wrote “P” five times. But since one was added, I had to take one away. So, one of the “Ps” became an “X”. Then, since the number was doubled, I had to take half of it, so half of the 4 “Ps” that are left gave me 2.”

This is perfect logic and a valid way to reason through to the answer. In essence, she saw that the process could be reversed and mapped it. It doesn’t, however, meet the intended goal of

having a student construct and then solve an equation. What is an instructor to do? Consider that in the future, this student might be asked to solve the equation 4 ─ 2(2x + 1) = 3x + 5. Can this equation be solved using this student’s strategy? Yes, but not as efficiently as the traditional equation solving strategy. What happens to this student’s sense of self, sense of algebra and equations, and sense of quantitative relationships if, as an instructor, I have to say “no, that’s not the way to do it.”?

And the issue isn’t only the student; it’s the pedagogy. It seems that the pedagogy is probably more the issue because it doesn’t allow students to try out various strategies and come to the realization that their strategy works for some equations but not all equations. They now have a choice. They can learn several strategies and tailor the strategy to the circumstance, or accept the traditional pedagogy which offers an efficient method for solving equations of all types. It may be contended that if the student builds a library of different strategies for different equations, that it may be a big library and there may be an equation not amenable to one of the strategies. I would reply that a strategy developed and employed by a student is likely to be better remembered, and modified as necessary, than one that is presented and never “owned”.

There are ways of approaching the solution of equations which allow for the type of visual representation that this student used. Further, equations can be solved using objects and images or both; no paper and pencil need be used – at least not at first. All students could be started with this student’s approach and as the equations become more sophisticated, it could be noted that an alternative strategy needs to be used for these more sophisticated types of equations. Starting with their conceptions may well result in their all coming to the conclusion that the most efficient strategy – the classic traditional strategy – is most favorable. However, consider that getting to this point would take more time, yet that time is valuable in establishing students’ capability to imagine alternative methods, compare and contrast them, and conclude which is best. Further consider that when students are taught, for example how to solve systems of equations, texts and instructors teach the substitution and the addition method, and sometimes even matrix and determinants. We bother to do this because, with some examination before plunging into the solution, it may be determined that one method is better than the other, under the circumstance. So, why not allow students to use their methods as well as they work their way through solving the problem?

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Homework: Solve This Equation 4 Ways

Posted by mark schwartz on October 2, 2016


I once had a class of pre-service and in-service teachers and one of our discussions led us to explore using one’s imagination to work through a math problem, rather than relying on the standard, traditional algorithm. I asked if all of them had taken Algebra and were comfortable solving equations. As it turns out they, collectively, were quite adept at it. So, the following story shows not only how adept they were but also how imaginative they were.

The Story

First let me note that they worked in groups, so the following homework assignment was started in class and I urged them to swap email addresses so they could work on it at home, which they did. Here’s the homework: you are to solve 4 ─ 2(x ─ 1) = 8 + 6(x ─ 1) 4 different ways and annotate the steps in your solution. We reviewed the traditional algorithm in class, so it didn’t count, but I did include it in the following list of solutions.

At our next class, I asked them to write their alternative solutions on the board and although there were many similar solutions, listed below are some of the unique ones. As we discussed these, the question was asked “when teaching math, should we ever have students do an exercise like this?”

Some felt we should; some felt otherwise. The critical difference was whether or not their school’s curriculum would allow for this. They liked the idea of using imagination and had fun doing the exercise but they collectively concluded that students in elementary and secondary school should stick to the traditional. I asked them that if a student, unprompted and independently, worked an equation – or solved any other mathematical expression – in an unorthodox way, what would they do?

It was a good discussion with no clear closure about what to do with the unorthodox student. Given the current press in Common Core Math, where students are to express their reasoning, it seems that accepting unorthodox solutions might be reasonable, but on the other hand a student may have an “aha” moment and can’t clearly articulated how the solution was found. It’s an interesting challenge for today’s teachers.

Here are some of the results of their work, annotated. Follow what they did because clearly imagination was in play.


4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 ─ 2x + 2 = 8 + 6x ─ 6

6 ─ 2x = 2 + 6x

4 = 8x

1/2 = x


1st Alternative (only a slight difference, but a difference):

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

2 ─ (x ─ 1) = 4 + 3(x ─ 1)       divide all terms by 2

2 ─ x + 1 = 4 + 3x ─ 3             do the indicated multiplications on both sides

3 ─ x = 1 + 3x                           combine like terms on both sides

2 ─ x = 3x                               subtract 1 from both sides

2   = 4x                                    add x to both sides

1/2 = x                                       divide both sides by 4 and simplify answer


2nd Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

0 = 4+8(x─1)                           subtract 4 ─ 2(x ─ 1) from both sides

0 = 4 + 8x ─ 8                         do indicated multiplication

0 = ─4 + 8x                             combine like terms

4 = 8x                                       add 4 to both sides

1/2 = x                                       divide both sides by 8


3rd Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 = 8 + 8(x ─ 1)                       add 2(x ─ 1) to both sides

─4 = 8(x ─ 1)                          subtract 8 from both sides

─ 1/2 = x ─ 1                            divide both sides by 8

1/2 = x                                      add 1 to both sides


4th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4/(x ̶ 1) ̶ 2 = 8/(x ̶ 1) + 6    divide every term by (x ─ 1)

─ 2 = 4/(x ̶ 1) + 6                  subtract 4/(x ̶ 1) from both sides

─ 8 = 4/(x ̶ 1)                       subtract 6 from both sides)

─ 8(x ─ 1) = 4                          multiply both sides by (x ─ 1)

─ 8x + 8 = 4                            do the indicated multiplication on the left side

─ 8x = ─ 4                                 subtract 8 from both sides

x = 1/2                                     divide both sides by ─ 8


5th Alternative

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

1/2 ─ (1/4)(x ─ 1) = 1 + (3/4)(x ─ 1)     divide every term by 8

1/2 = 1+ x ─ 1                                          add   (1/4)(x ─ 1) to both sides

1/2 = x                                                    combine terms on the right side


6th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4 ─ 2x = 8 + 6x                   let x = x ─ 1

─ 4 = 8x                               subtract 8 from both sides; add 2x to both sides

─ 1/2 = x                               divide both sides by 8 and simplify

─ 1/2 = x ─ 1                         let x ─ 1   = x (“reverse” first operation)

1/2 = x                                   add 1 to both sides


7th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

4(x + 1) ─ 2(x ─ 1) (x + 1) = 8(x + 1) + 6(x ─ 1)(x + 1)     multiply every term by (x + 1)

4x + 4 ─ 2x2 + 2 = 8x + 8 + 6x2 ─ 6                                do all indicated multiplications

─ 2x2 + 4x + 6 = 6x2 +8x +2                                            combine like terms on both sides

─ x2 + 2x + 3= 3x2 +4x + 1                                              divide all terms by 2

(─x + 3)(x + 1) = (3x + 1)(x + 1)                                  factor both sides

─x + 3 = 3x + 1                                                              divide both sides by (x + 1)

3 = 4x + 1                                                                       add x to both sides

2 = 4x                                                                            subtract 1 from both sides

1/2 = x                                                                          divide both sides by 4 and simplify


8th Alternative:

4 ─ 2(x ─ 1) = 8 + 6(x ─ 1)

16 ─ 16(x ─ 1) + 4(x ─ 1)2 = 64 + 96(x ─ 1) + 36(x ─ 1)2                  square both sides

16 ─ 16x + 16 + 4x2 ─ 8x + 4 = 64 + 96x ─ 96 + 36x2 ─ 72x +36     do all indicated operations

4x2 ─ 24x + 36 = 36x2 + 24x + 4                                                       combine like terms

x2 ─ 6x + 9 = 9x2 + 6x + 1                                                                 divide every term by 4

0 = 8x2 + 12x ─ 8                                                             subtract x2 ─ 6x + 9 from both sides

0 = 2x2 + 3x ─ 2                                                                divide every term by 4

0 = (2x ─ 1)(x + 2)                                                           factor 2x2 + 3x ─ 2

1/2 = x                                                                               setting both factors equal to 0 gives…

(x = ─ 2 is an extraneous root

Introduced when both sides were


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Unequations Buzz

Posted by mark schwartz on August 11, 2016


Had a thought. Simple one-variable 1st degree equations, by definition, state that there is a bunch of stuff “here” that equals a bunch of stuff “there”. For example, 2(3x ̶ 1) = 5(x + 1). What is meant by “equal”? Looking at this equation, obviously the two bunches of stuff are not equal! What this statement means is that if you can find the value of the variable “x”, replace the “x’ with that value in both sides of the equation and evaluate both sides, the value on both sides of the equation sign will be equal. Thus, that’s why one solves for the value of “x”.

The fundamental rule for solving equations is “whatever you do to one side of the equation, you do to the other side.” This, in essence, maintains the equality. My thought was that rather than start with an equality and burp out the rules, start with an unequation and have students play with it to find out how to make it an equation. However, we won’t use paper and pencil; we’ll use poker chips.

The Story

In order to solve an equation of this order, students need to know a lot of stuff – identification of terms, order of operations, distributive law, the four basic operations with signed numbers and to verify their answer, substitution of a value for the unknown and of course the basic rule of “whatever you do to one side of the equation, you do to the other side.”

Solving unequations is simpler and is a kinesthetic, visual way to have students play with all those things which, in my view, expands their conception of equations. In many instances, I’ve seen students who know all the elements but somehow can’t blend them together to solve equations. Here’s how unequations work.

Each group of students (2 or 3 to a group) gets a handful of white poker chips and each chip has a positive on one side and a negative on the other. You can use other markers if you choose.

I ask them to put 1 to 5 chips in each pile but the total value in each pile can’t be the same. Two questions that always comes up are (1) can we put positives in one pile and negatives in the other and (2) can we put positives and negatives in the same pile? So, right away, they’re thinking about this exercise; they’re engaged. We have a discussion about this and although they don’t yet know what to do with these 2 piles (although some guess they’re equations), I let them determine what is allowable. So again, right away they “own” this exercise because they have determined what’s allowable. By the way, the discussion about what is allowable has many branches and typically includes a lot of “what if” banter. I just listen.

Once this is resolved, I then ask them to label the pile on the left “A” and the pile on the right “B”. This also is fun because there typically is someone who stacks the piles vertically rather than horizontally, so I simply say the pile furthest from you is A and the pile closest is B.

When everyone is ready I then ask them to do something to their pile A such that the total value in both piles is equal. This is also a fun point in the exercise for classes that allow positives in one pile and negatives in the other, but overall the buzz within each group again is one of the goals of this exercise. When this is done, I ask them to return to their original piles and then I ask them to do something to their pile B such that the total value in both piles is equal.

In both cases, I ask them if there was only 1 way to make the piles equal. Buzz, buzz again and the consensus was yes.

The next question to them was do something to both piles at the same time such that the total value in both piles is equal. This really generates buzz and questions to me, which I say I’ll answer later. The reason I won’t answer is that I want them to explore how this works. What they discover is that there is an unlimited number of ways to do this. For example, if A = 2 and B = 4, add 5 to A and 3 to B and both piles equal 7. There usually is an “aha” moment when they realize that as long as the difference between the two numbers added to A and B is 2, the total value will always be equal. Some also discover that unequal amounts can be subtracted from both piles and further that two numbers differing by 2 can result in an equal value in both piles. And there’s another “aha” moment – the total value in both piles can be negative if both were positive at first! And what’s more, zero is a valid value!

So, we played with these 3 options for a while and there was discussion all along about not only what was allowable but also the range of answers under the different conditions. Then we moved to equal piles to begin the exercise.

I ask them to adjust their piles so that there is an equal number in both piles. This then brings up the issue of their rule allowing positives in one pile and negatives in the other, if they allowed this. They realize they have to rule it out. But I then ask if they can have an equal value in both piles while having positives and negatives in the same pile. Can the total in both piles be positive or negative? Buzz, buzz and the conclusion is that it’s ok but this comes after a lot of discussion and this really gets them going about signed numbers. For example, if they are to have 3 positives in both piles to begin, they could put 4 positives and 1 negative, or 6 positives and 3 negatives or … here it goes again with an unlimited number of both as long as the total is 3.

So, I ask them to consider there beginning equal value in both piles and typically they make it simple – either all positive or all negative and they do this partly – they tell me – because they don’t know what I’m going to ask them to do. At this point, the equation question arises and I have to admit that we’re headed in that direction. After playing with this for a while, the class concludes (again) that there is an unlimited number of values that can be added or subtracted to maintain the inequality.

The next step is to give each group a few blue chips. What the group is asked to do is have one person look away of shut their eyes while the others in the group do two things: (1) set up two piles with an equal number of chips in both and (2) remove a certain number of chips from one of the piles and place a blue chip in that pile. In essence, create a simple equation. When they are done setting it up, the closed-eye person is to look at what they’ve done and answer the question: what must you replace the blue chip with in order to make the piles have equal value?

Do each of these exercises until the class seems comfortable with all the ideas that got buzzed about.

At this point, if you’d like to extend this 2-pile concept to work with introducing work with equations, see Chipping Away at Equations in this blog. It links up with this posting and together it gives students a different view of equations.

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Algebraic Fishiness

Posted by mark schwartz on July 22, 2016


In an Algebra class, I gave the following two problems as an in-class assignment. We had just finished the material on solving equations and we had spent considerable time on what are called application problems. I decided to stretch their imaginations a bit by providing these two problems which are quite different from anything they had seen in the text. This, again, was a group activity and I roamed around the classroom watching the in-group strategies develop. If I hadn’t done this, I would have missed a group’s interesting insight.

The Story:

Here are the two problems.

  1. This problem is from Algebra by Davies, published by Barnes and Company, NY, NY, 1858.

A fish was caught whose tail weighed 9 pounds. His head weighed as much as his tail and half his body; his body weighed as much as his head and tail together: what was the weight of the fish?

  1. This problem is from An Introduction to Algebra by Colburn published by Hilliard, Gray and Co., 1839.

There is a fish whose head is 4 inches long, the tail is twice the length of the head, added to 2/5 of the length of the body, and the body is as long as the head and the tail both. What is the whole length of the fish?

Please note that these two fish problems represent a problem that was quite common in texts of that era. Conceptually, it’s the same problem, one phrased as a weight problem and the other as a length problem.

There was some grumbling when the class first read these problems. They hadn’t seen this type of self-referential relationship in a problem. They questioned if these were trick questions; some said it was the same problem so why give it twice; someone asked if it was the same fish, and I heard a few other choice comments. I suggested they take a deep breath and think about it and further, once they have solved one, let me see what you’ve done before moving on to the second problem.

On occasion, I gave Socratic help.

One group called me over and showed me the solution for the first problem, which was correct. But their setup for the problems was unique.

Their setup was:

T = 9                                               H = 4

H = T + 1/2(B)                              T = H + 2/5(B)

B = H + T                                       B = H + T

What’s unique about this is that they identified the relationship of the unknowns for both problems before setting up the equation for either of them. I asked them why they took this approach. The response was that the problems looked so much alike that they thought maybe there was one solution strategy, which as it turns out is true. This setup, in their minds, verified that point. In both cases, they started the solution with B = H + T. These are their solutions, but I tidied them up a bit (without changing any steps in the solutions) and showed them this way so it’s easier to see their parallel solutions. The complete solutions for both were:

For the weight problem:                   For the length problem:

B = H + 9                                            B = H + 4

H = 9 + 1/2(H + 9)                            T = 8 + 2/5(4 + T)

2H = 18 + H + 9                                 5T = 40 + 8 + 2T

H = 27                                                 3T = 48

T = 16

B = 27 + 9 = 36                                   B = 4 +16 = 20

Fish = 72 pounds                                Fish = 40 inches

They went a little further and decided that it was the same fish; that a 40 inch fish could weigh 72 pounds based on one of the group members being an avid fisherperson. Neat group!

In summary, this group of students used their imaginations, which resulted in their seeing not only the similarity in the problems but also the similarity in the solution strategy. It’s a testament to group activity and “aha” moments.

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Chipping Away at Equations

Posted by mark schwartz on April 15, 2016


This is written to demonstrate how an instructor can introduce equations using a manipulative approach. If you’re not an instructor, you can learn yourself. It takes a lot of words to describe a little activity, so be patient and realize that the activity helps students understand a lot about equations that otherwise might not be seen. I always have students work in groups of 3 or 4, particularly with this exercise. They got a chance to visualize that there are a variety of ways to find the answer; in essence, although there are traditional “rules” for solving equations, there are always alternative paths.

The Story:

When this activity is first introduced, do not identify it as solving equations. It’s simply an activity about keeping two piles of objects at equal value. As you walk around the room the first time, you can give each group of students a set of poker chips; some blue, some red and some white ones, having already used a magic marker to put a “+” on one side and a “ ̶ “ on the other side of the white ones. You have to develop a “patter” about what they are about to do, but don’t use the word “equation”, even if someone in the class does; and typically someone does.

The other nice thing about this is that you are able to roam around the classroom and observe what is being done, correcting where necessary, answering questions, giving support to student efforts and seeing some novel ways to get to the answer.

One thing which must be presented, or reviewed if presented earlier, is the idea that one positive added to one negative equals zero. A good example is to identify a student and state that they have no money. Then, give them a dollar which is a plus one for them. Then have the student return the dollar, which is a minus for them. Point out that one positive and a one negative together result in a value of zero given to the student. Of course, you can use any method you chose to demonstrate this point.

The Activity

Once you have distributed the chips, the students will play and tinker with them. When things calm down, ask each group to lay out two equal piles of chips using the white positive chips (it’s OK if each group has a different amount but more than 1, and use less than half of them). The important thing to bring out at this point is that the value of the two piles are equal. Remind them that if they have 2 positive chips in both piles, that the value of the piles is equal. If there is 1 positive chip in one pile and 2 negative chips in the other, the value of the piles is not equal.

Review the idea of one positive and one negative together having a value of zero. Also, note that this is true for the red and blue chips, treating the blue ones as having an unknown positive value and the red ones as having an unknown negative value. Write this somewhere on the board and leave it there for the entire class.

Tell the class that the point of this activity is: whatever they do to one pile of chips, they are to do the same thing to the other. What can be done is to either put chips into the pile or take chips out of the pile. The reason: the goal is to keep the value in both piles equal. Demonstrate this on the board by drawing two piles of chips, then add a number of positives in both piles then add the same number of negatives to both piles (this may bring up an issue; the number of negatives equal to the number of positives? yes). And now since there has been an equal number of negatives and positives added to both piles, the final value of both piles is …?? (this should bring them back to the value with which they started).

Try some more. Don’t have the students clear the piles. (if you’re not asked about this, state it. In some cases you clear both piles from the table, in some cases you keep both piles). Ask them to add two blue chips to one of the piles. What do they have to do now to make a pile of equal value? Keep stressing “equal value.” Verify that each group now has two blue chips in both piles. Keep the piles.

Next, ask them to put a number of negative chips in each pile but the number of negative chips must be equal to or less than the number of positive chips.

Hesitate for a moment and see if any students ask about having positives and negatives in the same pile and can an equal number of both becomes “zero”? If there is no prompt from the class, prompt them about one positive and one negative added together being a zero. Point out that if it’s a zero, it can be removed from the pile because removing zero from a pile doesn’t change its’ value. But, the zero can be removed from both piles. Given this, when they look at their two piles, both should contain an equal number of white positive chips (unless they started with one white positive, in which case they will only have 2 blue chips). Roam the room to verify that everyone gets this point; verify and correct where needed.

Have each group remove their piles and have them build a new pair of piles each containing 5 white positives. Now ask them what would need to be done to both piles to have the value of each pile be 2 positives?

This should lead to questions and a discussion. If it doesn’t come from the students, you have to make a point and discuss how there are two different ways to do this. One way is to simply take 3 positives from both piles. The second way is to add 3 negatives to both piles, and then remove 3 zeroes from both piles. This may seem odd to some students, but it’s an important operation to understand. Why bother adding stuff and then removing stuff, when simply taking 3 positives away will do it? In essence, it’s the same and it can be shown that 5 positives minus 3 positives is the same as 5 positives plus 3 negatives.

This can be show on the board as algebraic expressions: +5 ̶ ( + 3 ) = +5 + ( ̶ 3 ), since we have the incantation “to subtract, add the opposite”. This statement resolves to 5 ̶ 3 = 2.

Offer other examples to make sure the class sees this point. As an example, put 3 negatives in both piles. Have students use both procedures to bring the value of both piles to one negative. Again, cruise the room and observe and help. Have several more examples ready and have students again use both procedures.

One more example at this point. Have the class put an equal number of one to five positives in both piles and do what needs to be done to finish with two negatives in both piles. Have a second example ready (and maybe a third), starting with an equal number of red chips in both piles and do what needs to be done to finish with one blue chip in both piles.

Here’s a review of the KEY POINTS:

  • both piles have equal value;
  • whatever is added or subtracted from one pile, you must do the same to the other to maintain the equal value;
  • in a pile, one positive and one negative together equal zero and can be removed from a pile;
  • one blue chip (a “positive”) and one red chip (a “negative”) in a pile together equal zero and can be removed from the pile.
  • if you start with a bunch of red, blue, positive and negative chips in both piles, you can put in or take out of each pile the same thing until you reach the goal, which is: there are only blue chips (one or more) in one pile and either all positives, all negatives or no chips in the other pile.
  • No red chips should remain in either pile. (although later, this can be modified)

Before going on, ask if there are any questions; ask if the ideas are clear; ask if the idea of zero is clear; ask if they know what to do with the chips. Review the KEY POINTS above. Try to satisfy yourself that everyone understands how this activity works, because now we’re going to introduce another idea.

Ask them to clear any piles from the table and to put 2 blue and 3 positives in one pile and 1 blue and 4 positives in the other. The goal: add or subtract from both piles whatever needs to be done to wind up with blue chips in one pile and everything else in the other pile. Show this on the board by drawing two piles on the board with “+”s and “b”s and find the answer.

b b + + +                                                                   b + + + +

The reason for putting this one the board is that the class may not remember the original setup and the original setup will be used after they reach the goal.

They should come up with 1 blue in one pile and 1 positive in the other. Again, cruise the room observing, asking and answering questions and helping where needed. Once all groups have done this, ask them how much is the blue chip worth? Then ask: how can you demonstrate this is true – that the value of both piles is equal?

Wait for the class to discuss and debate this. It is most likely that someone will offer the answer and if not, discuss it and provide the option of replacing every blue chip with its value in the original setup and see if the value of both piles is equal.

Note to the class that although each group will start with the same problem and wind up with the same answer, there may be more than one way to get from start to finish.

At this point, you are urged to have ready a set of problems to present. I’ll give you some but you might want to generate some more. Present the problems such that the students have to play with the adding and subtracting of positives and negatives and blues and reds. The first problem should cause some questions from the class, so be prepared (if needed) to hold an all-class discussion … but ease the class into it. Then try a problem starting with positives and negatives and only red chips. Write these sample problems on the board, one at a time, without the answers and leave them on the board, leaving some space between them for writing between them. Include any additional problems you created. Step through the first one with the class.

1 pile has 3 blue, 1 red, and 3 negatives; the other pile has a blue, 2 positives and 1 negative.    Answer: 1 blue = 4 positives


1 pile has 2 blue and 3 negatives; the other pile has 1 red and 3 positives.     Answer: 1 blue = 2 positives


1 pile has 2 blue, 2 positives and 3 negatives; the other pile has 1 blue and 3 positives. Answer: 1 blue = 4 negatives.


1 pile has 2 blue, 2 negatives and 1 red; the other pile has 2 reds and 1 positive.   Answer: 1 blue = 1 positive


1 pile has 3 red; the other pile has 2 blue, 7 positives and 2 negatives.   Answer: 1 blue = 1 negative


1 pile has 4 negatives; the other has 2 red and 2 negatives:     Answer: 1 blue = 1 positive


The next problem may (should!) prompt questions from the class. If no one asks, then wait and see what happens, but typically at least one student will ask and then you can make a mini-presentation to the class. It all revolves around what they will do with the phrase “group of”. Basically, rather than reading 2( x + 1 ) as two times the quantity of x + 1, it will be read as two groups of one positive unknown and one positive. More on this later. Demonstrate this with problem 6.


1 pile has 2 positives and 3 groups of one blue and 1 negative; the other pile has 2 groups of one blue and 1 positive.  Answer: 1 blue = 3 positives.


To repeat, I would recommend that you generate a few more problems and exclude any problems where you have to split a blue chip. For example, if the final outcome were 2 blues in one pile and 1 positive in the other, 1 blue would be equal to half of the 1 positive. This idea was explored in a later class.

And the next step is …

Note to the class that if they hadn’t already realized it, they’ve been doing equations. The problems they solved using chips were really equations, but the problems weren’t presented as equations; they were presented as “build two piles”, then add or subtract chips from both piles to reach the goal of just blue chips in one pile.


Look at one of the problems you did … “1 pile has 2 blue and 3 negatives; the other pile has 1 red and 3 positives.” If this is written in algebraic language:

2 blue is the same as 2x (recalling that a blue chip represents one positive unknown) and

3 negatives can be written as ̶ 3, so one pile can be written algebraically as 2x ̶ 3.

The other pile has 1 red, which can be written as “̶ x” (recalling that a red chip represents one negative unknown) and

3 positives can be written as +3, so this pile can be written as   ̶ x + 3.

Since the piles are of equal value, algebraically it can be written as 2x ̶ 3 =   ̶ x + 3.

You may demonstrate with other problems if you like, but the key is to state that when the class did the problems before, the problems weren’t given as equations which had to be set up as 2 piles of chips. But, now the problems will be given as equations and will be “translated” to chips and solved accordingly. Briefly review that one positive means a white positive, one negative means a white negative, one positive unknown means one blue chip and one negative unknown means a red chip.


Here’s one example to demonstrate to the class and have them do it with you. Write the equation on the board and talk through how it “translates” to 2 piles of chips, writing the chip version on the board too. As you do it, emphasize the word “group”. Expect questions.

Given the equation, 3 + 2 ( x + 1 ) = 4 + 3 ( x ̶ 1) read it as “put 3 positives in a pile and in that same pile, put in 2 groups of a positive unknown and one positive. In the other pile put in 4 positives and 3 groups of a positive unknown and one negative.” (the words can vary a bit ). Now that the equation is in chip form, students are to add or subtract whatever needs to be done to both piles, the goal of which is to end with all the blue chips in one pile and all else in the other pile, recalling that no red chips are to be in either. Walk around and observe and help and applaud where needed. Have the class substitute the value of the blue back into the original chip statement and verify that their solution is correct. After the class has done it, do it on the board as well. The answer is one blue chip = 4 positives.

Some students may want to consider doing, for example, the multiplication of (x + 1) by 2 and put in 2 unknowns and 2 positives. Algebraically, this is correct but the idea here is to get students to slow down and read the parentheses as a grouping symbol and consider how many of that group is to be put in or taken out of the pile. I believe that conceptually it gives students a deeper understanding of the relationship between knowns and unknowns in an equation.

After the demonstration you may be asked for a second demonstration, so prepare a second example to work through with the class. If not, continue with the examples below; and again, you might want to prepare a few more (or take some out of a textbook but be careful to avoid any fractional answers). Announce that they are to find the value of one blue chip and then verify their answer by substituting the value of one blue chip back into the original setup. (note: in one of my classes, several groups set up duplicate chip setups rather than set the problem up a second time after solving the problem. You might suggest this).

2 + 3 ( x ̶ 1 ) = 2 ( x + 1 )                 Answer: one blue = 3 positives

2x + 2 ( x ̶ 1 ) = 2 ( x + 1 )              Answer: one blue = 2 positives

( 2x ̶ 1 ) = 1 + 3 ( x ̶   2 )                 Answer: one blue = 3 negatives

( ̶ x ̶ 1) = 2 ( 2x + 2 )                    Answer: one blue = 1 negative

This next one should cause some students to ask what to do, and if a lot of the class seems puzzled, then hold a mini-session and have them work through what to do. Some will be tempted to do algebra first, but don’t allow this. Have the class work through the meaning of the statement and determine what to do. For example, this is read as “take out one group of one positive unknown and one negative” and what the class will realize (hopefully) is the need to add a zero group to enable the action to happen.

̶  ( x ̶ 1 ) = 2 ( 2x ̶ 2 )                     Answer: one blue = 1 positive

IN SUMMARY: There are several things to consider. First, this activity may take more than one class period so consider where you are in the activity and where a reasonable “break” in content might be. Second, after this exercise, continue with the traditional solution methodology and include problems with fractional answers, and zero as an answer and equations where the unknown “disappears”.

Again, the entire activity of solving equations with chips is to get students to slow down, think about the relationships in the equation, and have a visual and kinesthetic reference point for the seemingly abstract activity of solving equations. Some groups tried solving the exercises using different sequences of operations! As students work through the traditional methodology you might find, as I did, that they refer to the chip activity and even some may say, as I heard, that solving equations make sense and is easy.

You can leave a comment below if you’d like to … it would be appreciated.


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Hannah Solves a Problem

Posted by mark schwartz on April 1, 2016


This posting shows just how imaginative students can be. We had been studying algebra in a Contemporary Math class and the text we used had the simplest of algebra problems. This course is a survey of a range of topics; not an in-depth study of any topic. I decided that it would be interesting to see if the algebra learned in this class (as well as some students in class having had algebra previously) would work as well with older problems, meaning a problem taken from one of my 1800s collection of math texts. Could the students visualize how to address this problem because it’s a series of actions and yikes!! … it has fractions in it. I said to use the strategy of their choice and if they wanted to work alone rather than in their usual group that was okay.

The Story:

A problem presented in class from Introduction to Algebra (W.C. Colburn, 1839, p. 40) was:

A person in play lost 1/4 of his money, and then won 3 shillings; after which he lost 1/3 of what he then had; and this done, found that he had but 12 shillings remaining; what had he at first?

The expectation was that students would apply a problem-solving strategy we had studied, namely, identify the unknown, identify the relationships and write and solve an equation. Algebraically, the problem sets up as a fussy equation:

x – x/4 + 3 – (x – x/4 + 3)/3 = 12


Hannah, however, presented the following:

6   6   6                 18 − 3             5     5   5   5                 20

When asked how she came to this answer and the meaning of the string of numbers, her explanation was (somewhat paraphrased):

I worked backwards. I started with the “remaining” 12, and 12 was 2/3 of something since 1/3 was taken away. If it was 2/3 then 2 equal parts of the 3 equal parts of 12 is 6 each, so the three 6’s gives 18. He gained 3 shillings, so I had to take three shillings away, so I got 18 ̶ 3 or 15. And then I did the same thing with 15 that I did with 12. Since 15 was 3/4 of something since 1/4 was taken away, 15 is 3 equal parts of the 4 things, or 3 equal parts of 5 each. Since there are 4 equal parts to start, I added one more 5, and the answer is 20.

Hannah, by the way, was very proficient in solving equations and very good at applying traditional problem solving strategies. She just chose to play with this problem this way. If I were a rigid traditionalist and unaware of her math skills, I would have to have concluded that Hannah can’t do Algebra. Rather, she reinforced the idea that there are alternative paths to solving problems and allowing students some freedom provides some novel and interesting outcomes.

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