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 markdotmath@gmail.com. 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.

Chipping Away at Equations

Posted by mark schwartz on April 15, 2016

Introduction:

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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