Equations and Motions

By this happy invention, modes of investigation at once difficult and disconnected, and dependent for success, in each particular case, on the skill and ingenuity of the inquirer, and often on accident, are reduced to a simple and uniform process.

—Charles Davies (1798–1876), Elements of Analytical Geometry and of the Differential and Integral Calculus

3

Numerical Forensicsⁱ

The truth must be quite plain, if one could just clear away the litter.

—Miss Jane Marple in Agatha Christie’s A Caribbean Mystery

The letter is susceptible of operations which enables one to transform literal expressions and thus to paraphrase any statement into a number of equivalent forms. It is this power of transformation that lifts algebra above the level of a convenient shorthand.

—Tobias Dantzig (1884–1956), Number: The Language of Science

A planet orbiting around the sun is in a state of constant motion. And while the elliptical loop it travels on remains relatively unchanged over short periods of time, its position on that loop is shifting all the time—creating in the process its own numerical symphony. Like the airplane from Chapter 1 that constantly changes its location over a four-hour period, so too does a planet constantly change its position during its orbit.

If it helps, think of a car traveling at the constant speed of 75 miles per hour on Interstate 15 South from Salt Lake City, Utah, to Las Vegas, Nevada. The route along the freeway is determined but the car’s location on that route is constantly changing as it travels from north central Utah to southern Nevada.

All of this means that we now in effect have two dramas playing out before us:

1. The entire path or course: The ensemble of all possible locations of the planet or the car (namely, the entire orbital path or the entire stretch of I-15 between the two western cities).

2. The time when a specific location on the path or course is reached: The time of year when the planet reaches a specific location on its orbit, or the time after leaving Salt Lake City when the car reaches a specific spot on the freeway.

This “rhymes” with the situation in which we now find ourselves. Consider the variable expression 3x + 7. Like the expressions we’ve seen before, this one can store and generate an entire ensemble of values. This is the first drama.

The expression acts like an organizing principle for those values—much like an orbit or a highway does for the respective objects that travel along their paths. A few of the possible values in this ensemble are listed here:

Six of the possible values that 3x + 7 can generate

Sometimes we are more interested in locating the specific value of the variable x that causes the expression 3x + 7 to yield a given number. This is the second drama.

Here are the values of x that produce each of the numbers just shown:

Six specific values of the variable x for the expression 3x + 7

Metaphorically, the expression 3x + 7, with its entire ensemble of all possible values, corresponds to the entire orbit of the planet or to the entire stretch of freeway between Salt Lake City and Las Vegas. We can think of all the possible values that 3x + 7 can take on as the “path” of 3x + 7, or better yet as the “cloud of possible values” of 3x + 7.

We have discussed expressions generating ensembles of values in some detail in the first two chapters. This is one side of the coin. Now we want to focus on the other side of the coin, the second drama that involves locating the value of x which causes 3x + 7 to generate the specific number that we want from the cloud of possible values.

We see this at work in a lot of the problems that we’ve already encountered. For example, we know that the expression 16h describes the total pay of a person who earns $16 an hour after having worked h hours. In Chapter 1, we chose different values of h to calculate the employee’s pay according to how much time they worked, and this allowed us to answer straightforward questions like how much the employee earned if they worked 68 or 237 hours. But what if the employee has a different sort of question? Suppose they have an upcoming expense in the amount of $3792. How many hours would the employee have to work in order to earn $3792?

This type of question is often a harder one to answer and will, in general, require us to perform additional maneuvers: different, in some cases, from those described earlier.

The development and standardization of these maneuvers comprises a huge portion of the “elementary” algebra learned by students up and through calculus. A central goal in this chapter is to understand why the techniques used to answer some of these questions took the forms that are taught in schools today.

THE TWO DRAMAS: VARIABLE EXPRESSIONS VS. BARE VARIABLES

Before we go into the intricacies surrounding the second drama, we need to directly address what can often be a source of confusion. In the case of a planet’s orbit or a car on a highway, there is a clear distinction between what the first drama and second drama look like. We list these in the next table.

	First Drama	Second Drama
A	Entire elliptical orbit of planet	Time of year when a specific location on the orbit is visited
B	Entire path of I-15 from Salt Lake City to Las Vegas	Time after leaving Salt Lake City when the car visits a certain spot on the freeway somewhere in Utah, Arizona, or Nevada

This distinction becomes less clear when we deal with variable expressions whose ensembles are purely numerical (and not as easy to visualize as an orbit or stretch of road), such as is the case of 3x + 7 or 16h.¹

	First Drama	Second Drama
C	All values stored or generated by 3x + 7	Specific value of x that makes 3x + 7 generate a certain value
D1	All values stored or generated by 16h	Specific value of h that makes 16h generate a certain value
D2	All possible hourly earnings from $16 an hour	Specific number of hours needed to work to yield a particular sum of money

As we move forward, we need to make our analysis of the second drama a bit more operational, so we take note of the fact that the central objects in the first drama are the variable expressions themselves (e.g., “3x + 7” or “16h”) along with the values that they can produce (metaphorically, the ensembles or clouds of possible values), whereas the focus of the second drama is the search for a specific value of x or h that yields the desired value or location. We can think of the x or the h in these cases as a bare or undressed variable.

Applying this to some of the variable expressions we used earlier yields the following:

Variable Expression	Bare or Undressed Variables
450x	x
16h	h
16h2	h
6x3 + 7y3	x, y
100x + 2013 + z – y	x, y, z

Here, the bare variables operationally correspond to the letters (as alphabetic characters) used in each variable expression.

A SIMPLE BUT REVOLUTIONARY IDEA

The central question of this chapter now becomes: Given an algebraic expression (such as 16h, 3x + 7, or x² – 12x + 100), how do we find specific values for h or x that cause the expression to yield a given number? The following tables show how we might visualize this question for different desired results:

Algebraic Expression	Question	Value That Works	Verification
16h	When does 16h yield 64?	h → 4	16(4) → 64
3x + 7	When does 3x + 7 yield 64?	x →19	3(19) + 7 → 57 + 7 → 64
x2 – 12x + 100	When does x2 – 12x + 100 yield 64?	x → 6	62 – 12(6) + 100 → 36 – 72 + 100 → 64

A: Expressions, questions, and answers

Algebraic Expression	Question	Value That Works	Verification
16h	When does 16h yield 1520?	h → 95	16(95) → 1520
3x + 7	When does 3x + 7 yield 2473?	x → 822	3(822) + 7 → 2466 + 7 → 2473
x2 – 60x + 1200	When does x2 – 60x + 1200 yield 300?	x → 30	302 – 60(30) + 1200 → 900 – 1800 + 1200 → 300

B: Expressions, questions, and answers

From looking at these expressions, we see that it is a relatively straightforward process to verify that the stated value works once we know what it is—it reduces to a simple calculation. But how do we find the value of the bare variable we’re looking for when it hasn’t been given to us?

In order to address this question, our first task will be to free it from the shackles of ordinary language and frame it in more maneuverable forms that are hopefully also straightforward to understand. The hope then is that we can learn how to maneuver these forms in systematic ways that allow us to calculate the answer. We saw the power of maneuverability in the previous chapter. It will prove similarly decisive here.

To those who are very comfortable with handling these types of problems, what we are doing may seem like a roundabout way to approach this subject. But history shows us that finding specific numerical values—for bare variables—that satisfied certain conditions was by no means a trivial question. Despite the fact that certain parts of algebra date back to antiquity, it wasn’t really until the mid-1500s and into the 1600s, during the rise of the symbolic algebra era, that mathematicians acquired new insights that enabled them to develop more user-friendly methods to answer these questions: methods that allowed both for a comprehensive theory to emerge and for educators to develop systematic ways to teach the subject to the masses.

Let’s start by considering the variable expression x + 4 with its attendant ensemble of values and asking what value of x will cause the expression to yield the number 14. Arithmetic tells us that this works when x is 10 as 10 + 4 sums to 14.

Unfortunately, we won’t always be able to answer so intuitively questions like these. For instance, can you determine on sight the value of x in 3x + 7 that will yield 8545? Most likely not. Consequently, we will need to devise techniques that will allow us to project our knowledge beyond the horizon of what we can initially see.

You are already familiar with this type of situation in arithmetic. For instance, though the multiplications 4 · 2 or 3 · 20 might be easy mental math for most, this is not the case for products such as 56 · 45 or 345 · 253. Techniques learned from our school days, however, allow us to systematically compute, in writing, these latter two products. These techniques require use of a times table and one of the recipes for long multiplication. Once learned, these additional strategies give us the ability to project our knowledge based on the evidence in front of us to produce the information that we want to know and that is presently unknown to us.

This is precisely what we want in the algebraic case as well.

It is easy nowadays to take the equals sign (=) for granted, but throughout most of the history of algebra, it was not used at all. It was introduced to the world in the book The Whetstone of Witte (the sharpening of wit), written in 1557 by the Welsh mathematician Robert Recorde. Recorde said he chose the symbol “bicause noe .2. thynges, can be moare equalle” than two parallel lines.² The notation didn’t immediately catch on—taking nearly a full century more before becoming part of the common mathematical vocabulary.

More important than the sign itself, however (as other signs could have worked just as well), was the revolutionary idea for which it was the “icing on the cake.” The idea was that we could place a wide array of problems involving finding or detecting unknown values on a general platform that was much easier to follow than anything else before it—with the crucial benefit that this entire platform could then be operated on in well-scripted and highly visible ways as a mathematical object itself, almost like the very numbers and variables, themselves, that made it up.³

It was to prove to be one of the most important operational advancements in the history of all of mathematics, rivaling the introduction of place-value systems into arithmetic—which is saying something.

Using this platform in regard to the question involving finding what value of x makes x + 4 become 14, we obtain

x + 4 = 14.

This object is different from the algebraic expressions in the previous chapter. How so? Well, first is the way in which we choose to interpret it, which is as an algebraic (and more operational) representation of the question involved in the second drama (i.e., for what value of x does x + 4 yield 14?). The first drama would be more interested in the general ensemble of values produced by the expression x + 4 (not just the value 14) and what the expression can tell us overall about a particular type of variable behavior.

Second, it is an expression that comes as a packaged deal, consisting of two distinct commodities from the start: one commodity “x + 4” on the left-hand side of the equals sign and another commodity “14” on the right-hand side. And though these two commodities do not initially seem to interact with one another, we shall soon see that they can be co-opted to work together in concert in ways that will ultimately reveal to us the information we seek.

REPHRASING THE QUESTION

From here on out, we will call expressions such as x + 4 = 14 equations. The type of equations that we will analyze in this chapter are further distinguished from earlier expressions with equals signs by the fact that they are usually conditional statements—sometimes true and most times false.

To illustrate this, let’s look closer at the equation x + 4 = 14. It has variation built in it, so in principle we could plug in different values for x. Let’s see what it looks like for x = 0, 3, 10, 25, and –20:

Value of x	x + 4 = 14 Becomes:	True?
x = 0	0 + 4 = 14	No
x = 3	3 + 4 = 14	No
x = 10	10 + 4 = 14	Yes
x = 25	25 + 4 = 14	No
x = –20	–20 + 4 = 14	No

Of the five values tried, 10 is of course the only one that works, and furthermore, of all the infinitely many values that we can plug into this equation, 10 is the only one that makes the equation true—every other one will fail just as 0, 3, 25, and –20 did. There are no other real numbers out there hiding behind a tree that will make both sides of this equation equal.

The fact that there are usually only specific values out of a sea of possibilities that make an equation true means that we’ll need to do some investigative work to identify those values.

Values that make an equation into a true statement, as 10 does for x + 4 = 14, are called solutions. We would say that 10 is the solution to the equation x + 4 = 14, whereas the numbers 0, 3, 25, and –20 are not solutions. Note that 10 would not be a solution to the equation 2x + 8 = 14 because it produces the statement 2(10) + 8 = 14, which is false—the solution to this latter equation is the value 3.

Here are a few examples of what rephrasing questions surrounding the second drama into the new platform of equations and solutions looks like:

Variable Expression	Question	Question in Equation Form: Find the Solution to:	Answer in Solution Form
16h	When does 16h yield 64?	16h = 64	h = 4
3x + 7	When does 3x + 7 yield 64?	3x + 7 = 64	x = 19
x2 – 12x + 100	When does x2 – 12x + 100 yield 64?	x2 – 12x + 100 = 64	x = 6

Questions using the platform of equations and solutions

To effectively deal with the second drama, then, requires us to learn how to find solutions to equations. Elementary algebra supplies a most effective recipe for the simplest types.

MOST REMARKABLE PROPERTIES

Let’s now take the equation x + 4 = 14 and subject it to some arithmetic. We will not play favorites and will abide by the rule that whatever changes we introduce to one side of the equation must be done to the other side as well. The following table shows the results of four such calculations with this equation:

Operation on Original Equation (x + 4 = 14) in Words	In Equation Notation	Equation After Simplifying Left- and Right-Hand Sides
Add 20 to both sides	20 + x + 4 = 14 + 20	x + 24 = 34
Subtract 2 from both sides	x + 4 – 2 = 14 – 2	x + 2 = 12
Multiply both sides by 5	5(x + 4) = 14(5) or 5x + 5 · 4 = 70	5x + 20 = 70
Divide both sides by 2	or

Notice that each operation changes the way the original equation looks—actually converting it into a new equation. Yet, if we replace x by 10 in each of the new equations, we still get a true statement—meaning 10 remains the solution. Remarkably, the solution to the original equation seems completely unaffected by the changes we made!

What would happen if we altered one side of the equation and not the other? If we were to add 20 to the left-hand side of the original equation (obtaining x + 24) and leave the right-hand side of the equation as 14, we would create the new equation x + 24 = 14. And 10 is not a solution now because 10 + 24 = 14 is not true.

Try subtracting 2, multiplying by 5, or dividing by 2 only on the left-hand side of the original equation, and you’ll see that 10 will fail as a solution in those cases as well. By changing one side of the equation and not the other, we have made an unbalanced adjustment to the equation itself, and this is why it has a different solution.

The properties exhibited in the previous table turn out to be true of equations in general. We state them as a general rule:

Adding the same number to, subtracting the same number from, multiplying the same number (excluding zero) to, or dividing the same number (excluding zero) out of both sides of an algebraic equation may change the way the original equation looks, but it doesn’t change the solution. That is, the solution to the original equation will also be a solution to the new equation.

Think of the solution as the key that opened the original equation and after the changes still unlocks the new equation. This means that there is an entire ensemble of equations that have cosmetic differences in form but turn out not to be different in content—in that they have the same solution. The following diagram summarizes this using our recent results:

Five cosmetically different equations that have the same solution of 10

There is something truly significant going on here. Remember that each of the equations on the perimeter was obtained from the original equation in the center through a simple arithmetic operation applied to both sides of the equals sign. There are an infinite number of equations that could be generated through arithmetic intervention—for instance, instead of multiplying both sides of this equation by 5, you could multiple both sides by 6, 75, 2000, and so on.

If you perform several of these operations in succession, you can transform a simple equation into one that looks quite different from the original, such as

Yet the solution to this more complicated-looking equation is still 10 because it has been derived from x + 4 = 14 using permissible rules (multiply both sides of x + 4 = 14 by 200, then divide both sides of that result by 7, and finally add 85 to both sides).

Moreover, if we return to our goal of answering questions about finding specific values for a variable that cause an expression to yield a given result, we see that the following questions relating to the second drama all have the same answer:

• When does the ensemble generated by x + 4 yield 14?

• When does the ensemble generated by x + 24 yield 34?

• When does the ensemble generated by x + 2 yield 12?

• When does the ensemble generated by 5x + 20 yield 70?

• When does the ensemble generated by yield 7?

• When does the ensemble generated by yield 485?

This naturally brings up a question: Because we can convert a given equation into many cosmetically distinct versions of itself, all of which have the same solution, it raises the prospect of whether it is possible to do this in a way such that each conversion gives us a simpler-looking equation than the one before it—culminating, ultimately, to a form in which the solution reveals itself to us for free.

UNDRESSING EQUATIONS

Let’s start again with x + 4 = 14. But now instead of changing both sides of the equation with random arithmetic, we are going to be more strategic in how we choose to operate, with an eye toward obtaining as the final step the bare or undressed variable. We will for the time being call this method undressing the equation.

In order to get x by itself in this equation, all we need do is get rid of the 4, but we have to do it legally following the rule we’ve established: What is done to one side must also be done to the other side. Here, simply subtract a 4 from the components on both sides of the equals sign. This yields the following:

Operation on x + 4 = 14 in Words	In Equation Notation	Equation After Simplifying
Subtract 4 from both sides	x + 4 – 4 = 14 – 4	x + 0 = 10, which simplifies to x = 10

Indeed, in this case the final step after simplifying results in the value that will make the equation true. The solution won’t always be so self-evident, however, so let’s now take a look at how this process enables us to find the solution in cases where the value is not immediately clear to us.

Remember our earlier question about finding the value of x in 3x + 7 that will yield 8545? The equation 3x + 7 = 8545 is more dressed up, and its solution is not immediately obvious. Undressing it will require two steps this time. We need to free x from both the 7 added to it and the 3 multiplied by it:

	Operations on 3x + 7 = 8545 in Words	In Equation Notation	Equation After Simplifying
1	Subtract 7 from both sides	3x + 7 – 7 = 8545 – 7	3x + 0 = 8538, which simplifies to 3x = 8538
2	Divide both sides of simplified equation (3x = 8538) by 3		1x = 2846 or x = 2846

To further enhance readability, we will now employ the following equivalent vertical format:

The value of the bare variable gives us the solution to the original equation, and when we substitute 2846 for x in 3x + 7, we see that it does generate 8545. But something vastly different has occurred now in that we didn’t arrive at the solution through trial and error—rather we found a systematic way to carve it out of the original equation! Moreover, we were able to do so in only two maneuvers (getting rid of the 7 from the left-hand side and then getting rid of the 3).

We could have also successfully undressed (solved) the equation by dividing by 3 first and then subtracting 7. However, this would have led to a messier situation involving fractions, so our chosen order of operating is more straightforward.

Another way to look at this procedure of undressing (resolving or solving) an equation is as a process that reduces the original equation to a form in which the solution presents itself as clear as day.

This smells tantalizingly similar in spirit to reducing fractions to simplest form. Consider the following reduction of

The solving of this equation is a type of symbolic maneuver that has allowed us to project our knowledge beyond what was immediately known to us in order to discover something new. It is different in the details but similar in spirit to our efforts in the first two chapters of simplifying algebraic expressions for greater clarity and understanding.

The equation conveniently contains all of the evidence needed to solve it—that is, all of the information we need is encoded in its structure. All we need do, as Agatha Christie’s Miss Marple says, is “clear away the litter” and “the truth will be quite plain.” This will generally be the case for the equations that we will solve in this book.

Let’s try another equation. Here’s an example of how to solve 6x – 2450 = 3694 using the vertical format:

In practice, much of the captioning is left out without losing computational content, although some conceptual content may be lost. Even the arrows and words are ultimately not needed either, as the right-hand side in the following shows:

Solving these equations then involves choosing carefully crafted arithmetic operations that allow us to isolate the x (or undress it) on one side of the equation. This process of solving the equation by reducing it to its simplest terms was called resolving the equation (or the resolution of equations) in many of the textbooks of the eighteenth and nineteenth centuries.⁴

What happens then, we might ask, if there are variables on both sides of an equation?

VARIABLES ON EACH SIDE

Let’s see if we can extend the techniques we’ve developed so far to solve this equation:

15x – 2247 = 12x + 2145.

In order to simplify, we will need to maneuver the variable components to the same side of the equation so that we can combine them into a single term. One possible way to go about this is to first subtract 12x from both sides, then add 2247 to both sides, and lastly divide both sides by 3. This plays out as follows:

Let’s verify that 1464 is the solution to this equation by substituting this value for x into both sides of the original equation:

15x – 2247 = 12x + 2145

becomes

15(1464) – 2247 = 12(1464) + 2145

or

21960 – 2247 = 17568 + 2145,

which gives

19,713 = 19,713.

Once more, strategic maneuvers of the original equation, this time with an additional step, allow us to systematically solve a problem whose answer is by no means obvious up front. The reduction diagram illustrates this process in an abbreviated fashion:

With practice, the mechanical steps to solve equations like these become a matter of routine—so routine, in fact, that it can be easy to forget that the questions we are answering are actually quite sophisticated.

In the case of 15x – 2247 = 12x + 2145, we have two variable expressions, each generating their own ensemble or cloud of values. For infinitely many choices of x, these two expressions give completely different results:

Value of x	15x – 2247 Becomes:	12x + 2145 Becomes:	Expressions Equal?
x = 0	0 – 2247 = –2247	2145	No
x = 3	45 – 2247 = –2202	36 + 2145 = 2181	No
x = 10	150 – 2247 = –2097	120 + 2145 = 2265	No
x = 500	7500 – 2247 = 5253	6000 + 2145 = 8145	No
x = –20	–300 – 2247 = –2547	–240 + 2145 = 1905	No

The question is, for what value of x in the cloud of infinitely many possibilities will these two expressions yield the same result? If we tried this by trial and error, it could take many of us a very, very long time. However, by solving this equation, we were able to obtain the answer to this question in only three steps.

Medieval algebraists could handle questions like this, too. However, they viewed what they were doing in a different way and used words to demonstrate their solutions rather than mathematical symbols. Using modern English words, such a problem might have been posed and solved in writing as follows:

Problem: Given equal amounts of cash in fifteen accounts from which if you remove two thousand two hundred forty-seven dollars you have the same amount as if you had the cash equally distributed in twelve accounts to which you increased that total by two thousand one hundred forty-five dollars, how much cash must there be in each account?

Solution: The computation is thus: Take a given account and call it a thing. From the fifteen take away twelve, which leaves three things. Increase two thousand two hundred forty-seven dollars by two thousand one hundred forty-five, and you obtain four thousand three hundred ninety-two dollars. Now take the four thousand three hundred ninety-two dollars and split it into three portions, which gives one thousand four hundred sixty-four dollars for each portion. This then is the thing or the amount of cash in each account.

This type of presentation of problem solving, where words are fundamental to the demonstration of the solution process, is called rhetorical algebra. The substitution of the word “things” for “accounts” here is similar to the process in which we use x to represent the unknown, and it more naturally separates the different types of entities. Dollars can be added to or subtracted from dollars, and things can be added to or subtracted from things.

Our modern platform of equations allows for the separation, in writing, of the solution process from the context in which the problem is delivered. That is, a word problem involving cash in 15 accounts could be written similarly to the rhetorical example above, but our solution method can still employ the x’s and numerals we used earlier—making no reference at all to accounts, things, or dollars. It is only at the end, after obtaining the solution, that we can decide to interpret that number in the context of the original problem.

This separation turns out to be both a great strength and a potential weakness of our modern mathematical methods. It is a strength because it allows us to isolate, simplify, and study in great depth the craft of solving equations without overburdening our working memory with the additional weight of interpretations. We can literally turn equation solving into a purely mechanical process with little regard for what each solution means and why we’re looking for it.

Yet this very ability to isolate and automate equation solving can become a grave weakness if we do so at the expense of conceptual understanding. In order to foster a greater appreciation of algebra, we need to understand why the equations we’re solving matter. And the frustration with the subject many of us feel as students stems from this historical disconnect.

FACES OF THE EQUALS SIGN

In Chapter 1 and Chapter 2, we were more concerned with using the equals sign (=) than understanding its history, but in this chapter we’ve delved deeper into its introduction in 1557 and the conceptual ideas for which it was, in a sense, the herald. This shift in concern can be looked upon, in a way, as representative of the distinction between the different types of equations and different uses of the equals signs. These different uses can be the source of much confusion for many students of algebra:

	Equation Name	Example	Chapter
1	Conditional Equation	x + 4 = 14	Chapter 3
2	a. Identity Equation (with variables)	3(x2 + x) + 7x2 = 3x2 + 3x + 7x2 = 10x2 + 3x	Chapter 2
	b. Identity Equation (with variables)	x + y = y + x	Chapter 2
3	Identity Equation (with numbers only)	8 + 12 = 12 + 8	Chapter 1

Various types of elementary algebraic equations

CONDITIONAL EQUATION

A conditional equation is an equation that is not true for every possible value of x. This type of equation takes its name because its validity is conditional upon a specific value of x that we call the solution to the equation. We’ve worked with many conditional equations already—as we discovered earlier in this chapter, the only value of x that makes the equation x + 4 = 14 true is 10.

As we proceed to solve equations of this type, the goal is to find the numerical value of the bare variable (x = a number). This requires that choreographed arithmetic changes be made simultaneously to both sides of the equals sign until the solution is carved out. How to successfully do this has been the procedural focus of this chapter.

IDENTITY EQUATION (WITH VARIABLES)

Identity equations are equations that are true for every value of x that can be chosen—not conditional based on a specific value of x. You can think of them as representing equivalent expressions or values. For example, the three expressions in 3(x² + x) + 7x² = 3x² + 3x + 7x² = 10x² + 3x are equivalent, in the sense that they are equal for every real number that can be chosen for x. We show this for the examples x = 0 and x = 10:

We will get the same three equalities of A, B, and C for all other values chosen for x. Try it yourself for x = 7 and x = 25. You should obtain 511 and 6325, respectively, for A, B, and C.

In the case of identity equations, we are more interested in the relationships between equal expressions than we are in finding specific values of x that make the equation true (as they all already do that). The goal is either to simply state the form of the equivalence (x + y = y + x) or to maneuver the entire expression to a simpler form that will aid in clarity and/or usability, as is the case with the simplifications done in the table and with others discussed in Chapter 2.

For the latter goal, we usually start out with a single expression that stands alone without need of an equals sign, then add in equals signs as we begin to simplify it. It is almost as if the initial expression flows into the other simpler expressions, and we use the equals sign to express this flow as well as the equivalence itself.

This is in contrast to conditional equations, where both sides of the equation change together in harmony as we attempt to solve for or isolate the variable.

Sequences of maneuvers that involve simplifying identity equations are sometimes called derivations, especially when they also employ scientific and engineering principles. This is where we can get a chain of many expressions, like the three in the previous table linked together by equals signs. In the sciences and engineering, sometimes well-justified approximations that further simplify expressions may also be involved in the chain of reasoning.

IDENTITY EQUATIONS (WITH NUMBERS ONLY)

For these, a simple arithmetical equivalence or derivation is stated with no variation. These are often single arithmetic statements, such as 1 + 1 = 2 or 11 · 7 = 77. Or they can be compound derivations, such as 8 + 7 · 5, which we often simplify in writing as follows: 8 + 7 · 5 = 8 + 35 = 43. Another example is 6(9 – 5) + 2(7 + 8), which we may write as 6(9 – 5) + 2(7 + 8) = 6(4) + 2(15) = 24 + 30 = 54. As before, the sense is of one numerical expression flowing into another. We are not making choreographed changes to both sides of the equals signs—our main goal, as with the examples here, is simply to reduce the numerical expression down to a single number.

It is worth pointing out that, with the exception of the fact that an equality of some sort is expressed, there is no universal agreement for using the term equation in elementary algebra textbooks. Some definitions include all of these varieties, whereas some only admit the conditional ones.

Since the conditional equations are included in all interpretations, this is the most common interpretation meant when the term equation is used.

Often students will see or create an equals sign and presume that they need to solve an identity equation (often called an identity) where no solution is called for. In cases where you aren’t sure what to do, try to remind yourself of the context and ask whether this is an equation that is true for all numbers or whether it is true for some numbers and false for others. It may help to test it for a few random cases—just make sure to include choices besides 0 and 1. If it is conditional, then attempting to solve the equation may be appropriate.

Otherwise, all that may be required is to allow one step to flow into the next until a simpler form of the expression is obtained.

The various interpretations of equations and the equals symbol brings to mind polysemous and homonymous words. A polysemous or homonymous word is one that has multiple meanings. The following words have multiple meanings in English:

Bank:⁵

• An establishment for the custody, loan, exchange, or issue of money, etc. (noun).

• The rising ground bordering a lake, river, or sea (noun).

• To bounce (a ball or shot) off a surface (such as a backboard) into or toward a goal (verb).

• To incline laterally (verb).

Table:⁶

• A piece of furniture consisting of a smooth, flat slab fixed on legs (noun).

• A systematic arrangement of data usually in rows and columns for ready reference (noun).

• To decide not to discuss (something) until a later time (verb).

The way we handle such words is by understanding the context in which the word is used, and with practice most users of English have no problems with properly interpreting their meaning. And the same can be said of the various uses of the equals sign. Much as with language, we can learn to interpret that information to understand how the equals sign is being used in a given context.

CONCLUSION

Given the nature of our discussion in this chapter, perhaps it will be useful if we conclude with an example.

Let’s say a salesperson contemplating renting a car observes that the costs to rent for 2 days, 3 days, 7 days, and 11 days, respectively, are $70, $90, $170, and $250. If this were you, you might want to know how these prices are generated and their relationship to the rental period. You might also have more specific questions, like what the cost would be to rent a car for a month or how much time you could afford to rent on a certain budget.

The rule can be found in the variable expression 20x + 30, where x represents the number of days rented.⁷ Verify that it gives the correct costs for x = 2, x = 3, x = 7, and x = 11.

So, in answering the first question, we now have a variable expression that can generate the cost of any other rental period the salesperson might want to know—that is, it contains the ensemble of all possible rental prices based on the number of days rented. This is the first drama.

All of this assumes, of course, that the rental car company doesn’t offer discounts for rentals over a certain number of days. If they do, then another, more complicated expression would be needed to model these changes.

Once we know the formula, we can easily calculate the cost for any rental period. If, for instance, the salesperson plans to rent the car for 28 days, all they would need do is set x = 28. This transforms our expression 20x + 30 into (20)(28) + 30 = 560 + 30 = 590, which interprets as $590 to rent the car for 28 days.

If, however, they want to know how much time they can afford within a specific budget, say $900, we’ll need to rephrase the question as an equation to identify the specific value of x where 20x + 30 = 900.⁸ This is the second drama. Algebra supplies us with the framework to find the answer by solving the equation 20x + 30 = 900. The following reduction diagram gives the solution:

A $900 budget will therefore allow the salesperson to afford a 43.5 day rental, but because half-day rentals are rarely allowed, the longest they can afford to rent the car is 43 days. Note that rounding 43.5 up to 44 days would put the costs $10 over what they can afford.

In many problems that require algebra for clarity, there can be great interplay between the two dramas. Historically, however, it was the second drama of finding specific unknown values that dominated the early history of algebra—with most of it being presented in the rhetorical style using words, as discussed earlier.

It appears that it was only after the advent of more purely symbolic ways to represent algebra that the first drama of systematically representing varying behavior with algebraic expressions was discovered to be an area worthy of study in its own right. This sixteenth- or seventeenth-century breakthrough was critically aided by the spectacular uses of such expressions by scientists, philosophers, and mathematicians, such as Pierre de Fermat, René Descartes, Isaac Newton, and Gottfried Leibniz, who sought to understand and more deeply analyze the problems of motion and the geometrical properties of curves—where they thought of an object’s movement as a symphony of changing numerical positions that could be described algebraically or geometrically.

Today, issues relating to the first drama fall under the study of functions—where calculus ultimately came to play a transformational and landmark role. Issues relating to the second drama come under the study of equations—which ultimately led to intellectual advancements every bit as significant and revolutionary as those from the first.

In a very real sense, a great deal of modern mathematics can trace its origins to problems emanating from these two important veins.

With any luck, this chapter has helped you catch a few more glimpses of the vast conceptual continent called algebra. And just as intensely as the subject can mystify and strike dread in the uninitiated, so too can it intensely tempt and mesmerize many a seasoned explorer of its terrain. These polar viewpoints can become highly problematic when the capabilities of algebra and mathematics in general are overestimated, taken on blind faith, or used to intimidate.

Footnote

ⁱ The idea for the chapter title came from the article “The Math Myth That Permeates ‘The Math Myth’” (Devlin 2016).