5

The Rule of Dark Position

… It maie bee thoughte, to bee a rule of wonderful inuention, that teacheth a manne at the firste worde, to name a true number, before he knoweth resolutely, what he hath named… and therefore the name is obscure, till the worke doe detect it, I thinke this rule might well bee called, the rule of darke position, or of straunge position…

—Robert Recorde (ca. 1510–1558), The Whetstone of Witte

Throughout its long history, algebra has used such terms as roots (jidhr in Arabic, radix in Latin), thing (shay in Arabic, cosa in Italian, coss in German), as much as so much (yāvattāvat in Sanskrit), unknowns, and variables in the service of finding the answers to what we presently call word problems (or story problems).¹ Such problems have consistently been the definitive arena in which to demonstrate to students the applicability of elementary algebra. Their continued use is not without controversy (see Chapter 6).

There are the standard types of word problems that include finding numbers with certain properties, piece length problems, motion in water, in air, or on land, finding unknown angle measures, finding unknown ages, finding amounts invested in various interest-bearing accounts, work rate problems, and mixture problems. Then there are others that involve a twist or two on the standard fare.

As a teacher, I often tell my students not to get too down on themselves when they encounter “that next word problem” down the road that trips them up—and that being stumped, and learning how to handle it, is also important to their mathematical education. Furthermore, I tell them that it takes a good while to achieve the fluency necessary to apply algebra in such situations and that even mathematicians, themselves, are never totally safe from being initially puzzled by some word problem out there on the horizon.

It’s not unlike walking incident free for years and still having that one moment on that one day when you stumble suddenly, falling flat and hard (papers flying everywhere) in front of your coworkers. It’s just the way it is.

So, what then can we hope to accomplish in a brief chapter like this on word problems? Certainly not the operational mastery of such problems; that is way too large a project to attempt here. We won’t even cover every type of word problem mentioned above, much less all of the other types out there. Such a project on its own could easily fill a book or two.

Our goals must be more modest than this, yet at the same time not small. As such, we will keep our focus here primarily conceptual as before—visiting only a few interpretive centers along the way—in our continued attempt to give readers more views of the sweeping possibilities out there in the algebraic countryside.

Let’s get started.

MERGING TERMINOLOGIES

WORD PROBLEM 1: If triple a number added to 40 gives 253, what is the number?

This is a classic type of exercise that many newcomers to algebra are likely to encounter. Let’s first look at it from the viewpoint of the two dramas introduced in Chapter 3.

TWO DRAMAS VIEWPOINT

This problem has variation in it: triple a number added to 40. And like the student trying to guess at the correct number, we can quickly see this as follows:

• Guess that 5 is the number; then triple 5 added to 40 gives 15 + 40 or 55.

• Guess that 10 is the number; then triple 10 added to 40 gives 30 + 40 or 70.

• Guess that 22 is the number; then triple 22 added to 40 gives 66 + 40 or 106.

We can keep trying to pick numbers and hope that we happen upon the one that works. This is reminiscent of the condition we faced in the last chapter with the situation tables for hamburger meals, and we will deal with it the same way by now pivoting to an algebraic mindset.

Let’s represent this variation as an algebraic expression. We need to tag the thing that can vary in the problem by a letter, let’s say x, which means that triple a number added to 40 translates to 3x + 40. This expression captures the ensemble of all possible values. This is the first drama. Using the expression to place the guesses in a diagram yields this:

Out of all of the infinitely many values possible, we want the specific number that makes the expression become 253. Sound familiar? This is simply the second drama revisited, which leads to the equation 3x + 40 = 253. Let’s solve this via reduction diagrams:

Thus, the number that works is 71. Checking our math, 3(71) + 40 = 213 + 40 = 253.

TRADITIONAL METHOD

If triple a number added to 40 gives 253, what is the number?

Our experience with enough of these problems tells us that this is a problem in which we are trying to find an unknown number, and that the techniques of algebra—namely, setting up and solving an equation—can be employed directly to find this number. So instead of experimenting with a few numbers to get a feel for the situation, we can dive right in and try to set up an equation directly.

The idea is to represent the unknown or the thing you are trying to find by x (other letters will work as well) and then translate the relationship in the word problem into an equation in x. This allows the rephrasing of the question as, “If triple an unknown number added to 40 gives 253, what is the unknown number?” This yields as before the equation 3x + 40 = 253. The solution follows to yield 71 as the unknown number.

POINTS OF VIEW

Before moving on to the next word problem, it will be prudent to first pay nuanced attention to the two different methods discussed in the previous section. Though they are clearly related, there is a world of difference in how their interpretations play out. The method involving the two dramas may be called the “viewpoint of variable expressions,” whereas the traditional method may be called the “viewpoint of the unknown position or value.”

To get an appreciation of each method as a distinct viewpoint, let’s consider the differences between (1) traveling to an address using local road maps with an eye to better understanding the overall neighborhood and (2) traveling to an address by following instructions from a GPS route finder with the goal of only getting there. The former can give an expansive view of the surrounding region and its place in the greater metropolitan area—allowing the traveler to visualize which roads are connected, see the overall geometry of the road networks, and understand where some of the parks and bodies of water are located—while the latter is strictly concerned with reaching the correct destination.

Both activities may involve some of the same actions—such as traveling on the same street, making the same turns, and waiting at the same stop signs—but the viewpoints and takeaways are totally different.

This approximates some of the differences between the more general “variable expressions” viewpoint and the more specific “unknown values” viewpoint. Both approaches ultimately involve solving the same equation, but the “variable expressions” viewpoint considers the expression (3x + 40) as a generator of a whole terrain of possibilities of which 253 is the one that we are interested in—whereas the “unknown values” viewpoint looks at the word problem as specifying a relationship that we can encode by the equation 3x + 40 = 253 and then seeks to answer that specific question with less concern about the more general situation. It requires no awareness of any other values that 3x + 40 can take on.

As we know from Chapters 3 and 4, the “unknown values” viewpoint is closer to the one that dominated much of the early history of algebra—certainly from the time of Al-Khwarizmi in the ninth century up to Viète in the late sixteenth century. This viewpoint has been employed under different guises such as finding “the root,” “the thing,” or “the cossicke number.”² In fact, in the earlier to middle parts of the sixteenth century, many for a time called the entire subject of algebra the Cossick Art or the Rule of the Coss, with the practitioners themselves sometimes being called cossists.³

Robert Recorde likened the practice of introducing an unknown into the process to the creation of a dark spot in the problem—boldly calling it “the rule of dark position.”

Even today, the “unknown values” viewpoint tends to be the predominate method taught in elementary algebra courses. And this is understandable to a degree, since solving word problems by thinking of a letter as storing a piece of unknown information often feels more representative and to the point than does thinking of the letter as storing the possibility of many values—with only one of them being the value we want.

Here, we will subscribe primarily to the “unknown values” point of view, as well. However, it is also useful to keep the “variable expressions” viewpoint in mind, as has been encapsulated in our description of the two dramas. We will freely dial into that more expansive viewpoint when it is convenient and/or illustrative to do so.

DEVELOPING IN THE DARKROOM

WORD PROBLEM 2: A statue is 139 years old. It is four years older than five times the age of an office building downtown. Find the age of the office building.

We’ll start by asking, what is it that we are trying to solve for? The problem tells us directly that it is the age of the office building, which is presently unknown to us. We don’t, however, let this ignorance stop us and forge ahead anyway by representing the age of the building as x. Injecting this equivalence (age of the building = x) into the word problem makes it become, “A statue is 139 years old. It is four years older than five times x.”

Algebra allows us to immediately translate this into “139 years is 4 years older than 5x” or in symbols alone as 139 = 4 + 5x (or 5x + 4 = 139). Setting up the algebraic equation here is the decisive link, which I sometimes call the “jump for joy stage.” Once this stage is reached, the systematic algorithms for solving equations allow us to then maneuver in for the attack. In this case, we subtract 4 from both sides first, which gives 5x = 135, and then divide both sides by 5, which yields x = 27. This tells us that the office building is 27 years old, and we are done.

In this viewpoint of the unknown value, we never looked at 5x + 4 as a general variable expression producing a lot of different values. We don’t care about those other values and didn’t consider them. We only cared about the expression as being part of the specific equation 5x + 4 = 139 and then solving it accordingly.

WORD PROBLEM 3: A 570-foot rope is cut into three pieces. The second piece is four times as long as the first, and the third piece is ten feet longer than twice the first piece. How long are the three pieces?

TRADITIONAL METHOD

Here, we have three unknowns: the lengths of the first piece of rope, the second piece, and the third piece. This may seem like a much harder problem than the ones above it, but we are in luck, owing to the fact that the lengths of the three pieces are related to each other in the following way:

• The second piece is four times as long as the first.

• The third piece is ten feet longer than twice the first.

These relationships mean that we really have to find only one unknown value because once we learn it (the length of the first piece), the other two unknown lengths can be calculated directly from that number. Let’s choose to represent the length of the first piece by x. Once we do this, we obtain the following algebraic relationships:

• The length of the first piece = x.

• The length of the second piece = 4x (four times as long as the first).

• The length of the third piece = 10 + 2x (ten feet longer than twice the first).

Now we just need to find the decisive link—an equation that encapsulates the relationship between the three pieces and the full length of the 570-foot rope. This follows immediately by realizing that adding up the lengths of the three pieces should give us back the 570 feet of rope (assuming no loss of rope in the cutting). Thus, we have

length of first piece + length of second piece + length of third piece = 570

or, using their algebraic names,

x + 4x + 10 + 2x = 570.

(Jump for joy stage!)

Simplifying the left-hand side of the equation by combining like terms yields 7x + 10 = 570. Solving this via reduction diagrams gives the following:

Thus, the first piece of rope is 80 feet long. Using this value sets off a cascade that allows us to find the other two pieces:

• First piece = x, which becomes 80 feet.

• Second piece = 4x, which becomes 4 times 80 feet or 320 feet.

• Third piece = 10 + 2x, which becomes 10 feet + (2 times 80 feet) = 170 feet.

• Adding up the lengths of the three pieces gives 80 + 320 + 170 = 570 feet.

We have our answer. Once we set the process in motion, it’s as though the three unknowns—the three dark positions—develop right before our eyes. Harvard mathematician Barry Mazur describes his early fascination with this process as follows:

To work out those simple queries… is rather like seeing a concrete visual image develop out of a blank nothing on photographic paper in a darkroom tray. You start with something you deemed X, and at the end of the process you discover X to be concrete, some particular number. There is a sense of power in this (… the early algebraists were very aware of this unexpected power).⁴

TWO DRAMAS VIEWPOINT

Looking at this problem through the lens of the two dramas, we see that the three pieces of rope are related in the following way: the second piece is four times as long as the first and the third piece is ten feet plus twice the length of the first. There is an “ensemble of situations” that satisfy these relationships between a set of three pieces.

For instance, if the first piece of rope is 5 feet long, the second piece would be four times longer (or 20 feet long) and the third piece would be ten feet more than twice 5 feet (or 20 feet long). If we add these three lengths, they combine to form a 45-foot length of rope, which is much shorter than the required length of 570 feet of rope. Thus, it is not the situation that answers the question in Word Problem 3.

Let’s represent the stated relationships between the three pieces of rope in the word problem as three bits of information in the following way: (first piece, second piece = four times the first piece, third piece = ten feet plus twice the first piece), or for the specific case in the last paragraph, (5 feet of rope, 20 feet of rope, 20 feet of rope). We will call each set of specific information about the relationships between the three pieces of rope a triplet. Because calculating each triplet depends on the length of the first piece, we will denote the first piece as x. Doing so yields the following:

The last one is the variable triplet.

Though this looks different from the single-variable expressions we have seen so far, the variable triplet does the same task, which is to represent (in an algebraic expression) the cloud of all possible triplet combinations. Let’s consider a few:

• First piece is 5 feet (x = 5): (x, 4x, 10 + 2x) → (5, 4(5), 10 + 2(5)) → (5 feet, 20 feet, 20 feet).

Sum of the three pieces (length of rope) = 5 + 20 + 20 = 45 feet.

• First piece is 12 feet (x = 12): (x, 4x, 10 + 2x) → (12, 4(12), 10 + 2(12)) → (12 feet, 48 feet, 34 feet).

Sum of the three pieces (length of rope) = 12 + 48 + 34 = 94 feet.

• First piece is 35 feet (x = 35): (x, 4x, 10 + 2x) → (35, 4(35), 10 + 2(35)) → (35 feet, 140 feet, 80 feet).

Sum of the three pieces (length of rope) = 35 + 140 + 80 = 255 feet.

• First piece is x feet: (x, 4x, 10 + 2x).

Sum of the three variable pieces (variable length of rope) = x + 4x + 10 + 2x = 7x + 10 feet.

This ensemble of values summarizes and displays pictorially as follows:

The scenario with x = 80 and triplet (80, 320, 170) corresponds to the solution for the conditions described in Word Problem 3.

Now you may be asking: What is the value of looking at the problem this way? It certainly seems a lot longer and more drawn out than using the “unknown values” viewpoint. But what this viewpoint highlights is the capacity of a group of parameters to once again capture an entire class of problems. That is, it shows us that there are infinitely many word problems corresponding to the relationships between the three pieces as stated Word Problem 3. Let’s extract a few of them from the ensemble diagram:

• A rope 45 feet long is cut into three pieces. The second piece is four times as long as the first, and the third piece is ten feet longer than twice the first piece. How long are the three pieces? Answer: 5 feet, 20 feet, and 20 feet.

• A rope 255 feet long is cut into three pieces. The second piece is four times as long as the first, and the third piece is ten feet longer than twice the first piece. How long are the three pieces? Answer: 35 feet, 140 feet, and 80 feet.

• A rope 56,010 feet long is cut into three pieces. The second piece is four times as long as the first, and the third piece is ten feet longer than twice the first piece. How long are the three pieces? Answer: 8000 feet, 32000 feet, and 16010 feet.

We’ve seen similar conditions before. The act of an individual cutting 570 feet of rope into three pieces according to the specifications of the word problem, has infinitely many other situations that “rhyme” with it. It represents a single instance of a symphony of situations that could involve many individuals cutting different-sized ropes into three pieces. Pieces that are related in the same way as described in the word problem. Whether the length of rope is 255 feet or 56,010 feet, algebra allows us to solve them all by setting up the same type of equation. The equations for these two specific situations would respectively wind up as 7x + 10 = 255 and 7x + 10 = 56010.

It is worth pointing out that because the length of rope is constant for a particular word problem, but can change value from word problem (scenario) to word problem (scenario), it acts like both a constant and a variable—namely, like a parameter. This means that we can apply big algebra here, if we’d like, and represent the entire class of situations in one grand word problem. Let’s represent the length of rope by a letter earlier in the alphabet, say d:

A rope d feet long is cut into three pieces. The second piece is four times as long as the first, and the third piece is ten feet longer than twice the first piece. How long are the three pieces? The general equation simplifies to 7x + 10 = d.

We can obtain each of the four word problems in this section from this general one by replacing d by 570, 45, 255, and 56010, respectively.

The “variable expressions” viewpoint has given us a “road map of the neighborhood” for this category of word problems. Moreover, this point of view also hints at the prospect that other word problems we may encounter will have this same circumstance play out—of being a single instance of an infinite assemblage of “rhyming” word problems. It naturally exposes these possibilities. The “unknown values” viewpoint, though solving the problem more clinically, does not as easily reveal the more general outline of these problems.

We will keep the prospect of these more sophisticated numerical symphonies in mind as we move forward.

A TASTE OF GEOMETRY

WORD PROBLEM 4: In a triangle, the measure of the first angle is 5 degrees more than the measure of the third angle. The second angle is 15 degrees more than three times the third angle. Find the measure of each angle.

We are clearly trying to find the measure of the three angles, all of which are currently unknown to us. As the measures of the first two angles are related in a clearly specified way to the measure of the third angle, we will represent this third angle by the unknown x. If we do this, we obtain the following algebraic relationships (reading from the third angle up):

• First angle (in degrees) = x + 5 (five degrees more than the measure of the third).

• Second angle (in degrees) = 15 + 3x (fifteen degrees more than three times the third).

• Third angle (in degrees) = x.

The steps to this point have been similar to those in Word Problem 3, but now we’re missing something—how to connect this information to an equation. It is implicitly embedded in the problem, but a crucial fact about triangles is needed to see it. What is it?

It is the property that if we add up the degree measures of each of the three angles in any triangle, regardless of shape or size, we will always obtain the same value of 180 degrees.⁵ This is a well-known and basic geometric fact about triangles, but it is actually quite remarkable when you consider that there are infinitely many possible triangles out there.

Similar universal truths are known about other familiar shapes—you may know that the circumference of a circle of any size can be approximated by the simple calculation of 3.14 times the diameter, or more exactly by the expression pi times the diameter or πd.

We can use this general property of triangles to set up the equation: first angle + second angle + third angle = 180. Translating into algebraic language gives , which when simplified gives 5x + 20 = 180. The following reduction diagram takes us to the solution:

Once we know that the third angle has a measure of 32 degrees, we can calculate the other two (again from the third angle up):

• First angle = x + 5, which becomes 32 + 5 = 37 degrees.

• Second angle = 15 + 3x, which becomes 15 + 3(32) = 111 degrees.

• Third angle = x, which becomes 32 degrees.

• Adding up the three angle measures gives 37 + 111 + 32 = 180 degrees.

WORD PROBLEM 5: A rectangle has a perimeter of 2434 feet. What are the dimensions of the rectangle if its length is 151 feet less than three times its width?

The dimensions of the rectangle here mean its length and its width. The problem additionally tells us how these two are related with the length being defined in terms of the width. Thus, we will let the width be represented by x. This yields the following:

• Width = x.

• Length = 3x – 151 (151 feet less than three times its width).

Note that composing our expression for the length requires a little more attention to detail here, because the order in which we subtract values makes a difference in the result. After all, 20 – 8 = 12 is not equal to 8 – 20 = –12.

This difference in the order of subtraction is usually simpler to catch in arithmetic due to the fact that the two numbers can be combined right away, but in algebra where different terms in expressions can’t always be combined right away, it becomes easier to miss an initial error. That is, if you use 151 – 3x for the length instead of the correct form 3x – 151, there is no immediate alarm bell going off that will allow you to catch the error quickly as there might be with simple arithmetic. The error only becomes apparent at the end of the procedure when you obtain a negative value for x (or the width), which for a physical length should be a positive number.⁶

As before, we need some additional knowledge of geometry in order to create our equation. This knowledge once more is implicitly given in the problem and requires that we know how to calculate the perimeter of a rectangle.

Just as a reminder, the perimeter (circumference) of an object, such as a rectangle (circle), is the distance around the boundary of that object. For the rectangle, we have the following:

If we start at the point in the lower left corner and walk around the boundary of the rectangle in a counterclockwise direction back to the starting point, we will cover a distance given by L + W + L + W, which simplifies according to the algebraic rules as 2L + 2W. Thus, we have that the perimeter = 2L + 2W.

In the word problem, we know that the perimeter is 2434 feet, and thus we obtain the equation 2L + 2W = 2434. In “x” language this translates through W = x and L = 3x – 151 to

2(3x – 151) + 2x = 2434,

which after multiplying by 2 via the distributive property simplifies to

6x – 302 + 2x = 2434

or

8x – 302 = 2434.

Solving this using the standard techniques we’ve learned gives x = 342. This sets off the cascade:

• Width = x, which becomes 342 feet.

• Length = 3x – 151, which becomes 3(342) – 151 = 1026 – 151 = 875 feet.

• Checking the perimeter gives 2(875) + 2(342) = 1750 + 684 = 2434 feet.

These geometrical problems show that sometimes finding a solution requires knowledge outside of what’s explicitly stated in the problem. This may seem obvious to say, but it is still useful to keep in mind when tackling these types of situations—as keeping such an awareness on the table can often give one a more flexible approach to solving a problem.

We’ve also witnessed here a fruitful partnership between algebra and geometry. In the seventeenth century, this partnership would turn into an exciting marriage of the subjects to form the gem known as analytic geometry. And though beyond the scope of this book, this creation of Pierre Fermat, René Descartes, and others would turn out to be a conceptual advancement of the absolute first rank. Algebra and geometry have worked together happily ever since.

PACKAGED CURRENCIES

WORD PROBLEM 6: Given a collection of dimes and quarters worth $34.75, how many of each type of coin must there be if the number of dimes is twelve more than triple the number of quarters?

Here, we are clearly trying to find out how many dimes and quarters we must have to total exactly $34.75, while at the same time having the number of dimes be 12 more than triple the number of quarters.

Though it is certainly possible to have various combinations of quarters and dimes that total $34.75 (137 quarters and 5 dimes, or 111 quarters and 70 dimes, for instance), the task is also to make the combination satisfy the specified relation. The relation in the problem sets as a condition that the number of dimes exceed the number of quarters by a specified amount, and neither of these scenarios satisfy this.

Given that we have many different combinations—of dimes and quarters—to choose from, we will let the “electrical soil” of algebra do its job by helping us ferret out the correct one. However, this problem is different in some ways from the previous problems in this chapter. Dimes and quarters each have a distinct monetary value, so simply knowing the number of dimes and quarters is not enough to know whether they will collectively be worth $34.75. We have to multiply each coin by its worth (10 cents for each dime and 25 cents for each quarter) to obtain the total amount of money they are worth, so the dimes and quarters here each represent packaged currency.

To get more of a flavor for the problem before we move into how algebra will help us solve it, let’s guess that there are 5 quarters. To satisfy the condition of the problem, the number of dimes must be twelve more than triple the number of quarters—meaning we must triple 5 to obtain 15, and then add 12, which gives us a total of 27 dimes.

If this problem were like the rope length or angle problems, then we could directly add 27 + 5 to obtain 32 of something. But 32 of what? Upon closer inspection, this clearly doesn’t make sense because we are talking about different types of coins, each with its own currency value. So, before we can test this answer to see if the conditions of the problem are met, we must convert the coins into their monetary values.

In this case, 5 quarters becomes 5 times 25 cents, and 27 dimes becomes 27 times 10 cents, which total 125 + 270 = 395 cents, or $3.95. Clearly, our guess is off the mark, as this is much lower than $34.75.

Let’s take another crack at it by guessing 30 quarters this time. Here, the number of dimes (twelve more than triple the number of quarters) must be “triple 30 plus 12,” or 3(30) + 12 = 102 dimes. The monetary value of these coins is 30 quarters + 102 dimes = 30(25 cents) + 102(10 cents) = 750 + 1020 = 1770 cents or $17.70, which is a significant improvement on our first guess but still low of the mark.

This time let’s use algebraic thinking and guess that there are x quarters. This means that the number of dimes must be “triple x plus twelve” or 3x + 12. Proceeding as before, the monetary value is now

x quarters + (3x+12) dimes = x(25 cents) + (3x+12)(10 cents),

which after putting the currency values in front becomes

25x + 10(3x + 12).

This variable expression is equipped to give the monetary values (in cents) of all combinations that satisfy the specified relationship between quarters and dimes in the word problem. Here is the symphony diagram corresponding to our two previous guesses:

The question of the problem can now be rephrased as follows: Out of all the possible monetary values generated by the expression 25x + 10(3x + 12), what is the one that yields a value of 3475 cents ($34.75)? This leads directly to the equation 25x + 10(3x + 12) = 3475. (Jump for joy stage!) And once we have our equation, the problem is then on the algebraic grid.

The following reduction diagram shows how the rules allow us to maneuver to the solution:

There are 61 quarters, which cascades (via 3x + 12) to 3(61) + 12 or 183 + 12 = 195 dimes.

If we check our work, 61 quarters and 195 dimes give the correct amount: 61(25) + 195(10) = 3475 cents, or $34.75.

Let’s do one more example, this time with paper currency.

WORD PROBLEM 7: A total of $17,252 is divided up into one-, five-, and ten-dollar bills. If the number of fives is seven times the number of ones and the number of tens is quadruple the number of ones, how many of each type of bill are there?

We are trying to find the number of one-dollar bills, five-dollar bills, and ten-dollar bills satisfying the specific relationships given to us by the problem and totaling $17,252. Because the number of fives and of tens are defined in terms of the number of ones, we will let the latter be x. This gives the following:

• Number of one-dollar bills = x.

• Number of five-dollar bills = 7x (seven times the number of one-dollar bills).

• Number of ten-dollar bills = 4x (quadruple the number of one-dollar bills).

Based on the previous problem, we know that each of these bills is packaged currency and must therefore be converted into their respective dollar values of $1, $5, and $10 when we make the decisive step of placing them into an equation that reflects the situation. Doing so yields

$1(number of one-dollar bills) + $5(number of five-dollar bills) + $10(number of ten-dollar bills) = $17252,

which translates algebraically to

1(x) + 5(7x) + 10(4x) = 17252.

After multiplying, this simplifies to

x + 35x + 40x = 17252

or

76x = 17252.

Dividing both sides of the equation by 76 gives us a solution of x = 227. We cascade this result to discover the number of five- and ten-dollar bills:

• Number of one-dollar bills = x, which becomes 227.

• Number of five-dollar bills = 7x, which becomes 7(227) = 1589.

• Number of ten-dollar bills = 4x, which becomes 4(227) = 908.

The total dollar value is 1(227) + 5(1589) + 10(908) = $227 + $7945 + $9080 = $17252.

CONCLUSION

In the first line of his landmark ninth-century book—Al-Kitāb al-Mukhtaṣar fī Hisāb al-Jabr wa’l-Muqābala (The Compendious Book on Calculation by Completion and Balancing)—Al-Khwarizmi stated, “When I considered what people generally want in calculating, I found it always is a number.”⁷ This seemingly simple statement, from the standpoint of today, implied far more diversity in Al-Khwarizmi’s era.

Modern views of algebra fixate on the relationships between numbers (constants) and different types of fundamental variations (x, x³, x⁵, and x⁸, etc.), but at the time Al-Khwarizmi wrote these words, and all the way up through the sixteenth century, mathematicians thought of fundamental variations as numbers in their own right—just of a different species or type.⁸

Al-Khwarizmi states a few paragraphs later that, for the problems he treats, there are three types (species) of numbers: squares, roots, and simple numbers. He also mentions that these may be joined together to create certain compound species.⁹ In the language we have been using, this translates to the three fundamental types: square variations, linear variations, and constants, respectively, represented symbolically by x² forms (such as 5x², –8x², etc.), x forms (such as 76x, 7x, etc.), and constants (such as 3, 10, etc.)—complete with their possible combinations (5x² + 7x + 5, 7x + 10, 11x² + 2x, etc.).

Interestingly, all of the word problems we have considered in this chapter reduce to expressions involving only x variations and constants. No square variations occur at all. More involved word problems could, of course, require square variations and/or even more complicated ones.

Now in closing, how do we summarize the many strands running through this chapter?

Clearly, the most central of the themes and the one that ties all of the different strands together is the notion of “method.” Of method, Alfred North Whitehead states, “The science has not been perfected, until it consists in essence of the exhibition of great allied methods by which information, on any desired topic which falls within its scope, can easily be obtained.”¹⁰

Here, we have problems considering a variety of topics such as rope lengths, angle measures, ages of buildings, number of coin or bill denominations, and the dimensions of rectangles. Yet their resolutions in algebra all reduce to equations of the same basic types involving terms that are multiples of x plus additional constants. Furthermore, standard methods are involved in handling them no matter the original context of the problem. This is information that we might have completely missed out on but for the systematic techniques of algebra.

In other words, even if we had been able to find the answers to each of these word problems through arithmetical ingenuity alone, the techniques so developed for each would have most likely seemed disconnected and haphazard—thus making it hard to generalize them and educate others in their use. However, the methods of algebra unite the word problems in such a way and to such a degree that their commonalities become more transparent, which can help us to gain new insights. Such unification also opens up new horizons for making the subject accessible to larger audiences of people—thus extending the reach of the subject.

To get a feel for what we mean by “method,” let’s go back to arithmetic for a moment and consider these three problems:

1. Given a collection of 912 objects, how many times can we take away 16 objects from the collection until it is exhausted?

2. Given 912 people, if we organize them into teams of 16 players, how many teams will there be?

3. Given $912, if we partition or distribute this amount equally between 16 people, how much money will each person receive?

These each describe a different action and their answers are interpreted in different ways, but they are all solvable by one and the same method, namely division. To answer each, all we have to do is calculate . This can be computed through the process of long division as follows:

We obtain 57, which is interpreted, respectively, as 57 times, 57 teams, and $57.

Here, the meaning in each of the three individual problems, briefly, gets turned over to the syntax and maneuvers of long division. And in the same way, the meanings in our word problems get briefly turned over to the syntax and rules of algebra—such as setting up and simplifying expressions and solving equations—to great advantage.

In the previous chapter, we talked about how Viète saw that a wide variety of problems deep down were really problems about algebra, just cloaked in a different form. On a much smaller scale, this insight has been on display in this chapter: Our distilling of a word problem down to its quantitative essence—into the form of a variable expression and an equation—can be looked upon as an uncloaking of the problem to its bare algebraic form. Once placed in this form, it is as if the problem loses the weight and ambiguity of language and becomes decisively more transparent to handle. Algebra’s development of this process to almost an act of routine exhibits symbolic maneuver at its first class finest.

Finally, let’s revisit a statement made in Chapter 1:

Languages, in general, give us this wide-ranging ability to describe lots of objects and ideas with a relatively small glossary of words. Taking these words, then, in combination to form sentences—language expressions—gives us the breathtaking ability to describe nearly everything that we experience in life or are able to think about in the world around us. We seek the same in the world of numerical variations.

Hopefully, this chapter has given a small glimpse of this lofty ambition in practice. Basically, through two fundamental types of expressions (our “small glossary”) in the form of multiples of x (such as 3x, 7x, 5x, 55x, 76x, etc.), basic real numbers (such as 10, 40, 120, 3475, 17252, etc.), and their combinations, we can handle all of the variety of word problems presented in this chapter and infinitely many more problems of a similar constitution. Though the picture isn’t close to being complete—due to the existence of other, more complicated variations that can’t be described by the simple ones we have discussed in this book—hopefully the idea of what mathematicians want to achieve, in respect to this ambition, is a little bit clearer based on our discussions here.

Now the time has come to briefly switch gears and take a look into the external environment surrounding algebra. Throughout the first five chapters of the book, we have conceptually explored in some detail fundamental topics internal to the subject, and sought to establish connections, where possible, between them and other powerful areas of human activity and expression. Next, we turn our attention to algebra in education, an area that historically has been the source of much emotional fire and passionate commentary. In Movement 3, we take a brief journey into the fascinating world of this highly important ecosystem.