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Section 33.1 Key Words are (Mostly) Evil

Students tend to dislike word problems. This is probably because word problems ask students to do something more difficult than pure calculations. Word problems ask students to understand information and then translate it into a form that allows them to use their mathematical toolbox.

In order to help students with this transition, some teachers started to introduce rule-based patterns into reading word problems. The problem is that this has led to students thinking incorrectly about the process of thinking through word problems. This approach is sometimes called "key words" and you can find lists of "key words" that are categorized by operation.

The problem with this is that these lists are (mostly) evil. There is a grain of truth in them, but it sells students short of an actual understanding of how to think through word problems. It causes students to read word problems in ways that are sometimes incorrect and does not advance the cause of mathematical thinking.

Activity 33.1.1. The Problem with Key Words.

Here is a classic example: "Bob has 5 rocks. Alice has 3 more rocks than Bob. How many rocks does Alice have?" The common thought process for people is that "more" tells us we need to add, so Alice has 8 rocks. The answer is correct, but the reasoning is wrong. But to understand why the reasoning is wrong, we need to look at a slightly different problem.

Try it!

Alice has 5 rocks. Alice has 3 more rocks than Bob. How many rocks does Bob have? Explain your reasoning.

Solution.

Since Alice has 3 more rocks than Bob, Bob has a smaller number of rocks compared to Alice. So if Alice has 5 rocks, we need to subtract to get the number of rocks that Bob has, which means Bob has 2 rocks.

In the example, the mathematical operation that was required was subtraction, but "more" means to add (if you believe in the rule-based key word approach). And so this leads us to the question of what's really happening in the word problem.

Activity 33.1.2. Mathematical Relationships.

In most word problems, the words do not directly describe mathematical operations that you need to perform. They are not instructions. Rather, they are representations of mathematical equations using words. In the example above, the relationship is "Alice has more rocks than Bob." We can translate this into the following:

\begin{equation*} \begin{array}{c} a = \text{The number of rocks Alice has} \\ b = \text{The number of rocks Bob has} \\ \\ a = b + 3 \end{array} \end{equation*}

There are some key points to remember. The first is that the variables used are all defined. This is an important element of communication. If you introduce a symbol, you need to be clear what the symbol means. Furthermore, it's not acceptable to say that \(a\) represents "Alice." Variables are symbols that represent a quantity or a mathematical expression. Alice is a person, not a number. So it is good to get into the habit of being accurate in describing your variables.

Once we have the relationship established, we can then move to plug in the variables. If we are told that "Bob has 5 rocks" then we can set \(b = 5\) and solve:

\begin{equation*} \begin{aligned} a \amp = b + 3 \\ 5 \amp = b + 3 \amp \eqnspacer \amp \text{Substitute $a = 5$} \\ b \amp = 2 \amp \amp \text{Subtract $3$ from both sides} \end{aligned} \end{equation*}

Or we are told that "Alice has rocks" then we can set \(a = 5\) and solve:

\begin{equation*} \begin{aligned} a \amp = b + 3 \\ 5 \amp = b + 3 \amp \eqnspacer \amp \text{Substitute $a = 5$} \\ b \amp = 2 \amp \amp \text{Subtract $3$ from both sides} \end{aligned} \end{equation*}

The actual decision to add or subtract is based on the relationship that is established by the words, not through identifying a specific word as if it were an instruction. This distinction once again highlights mathematical thinking. You're not looking for a specific rule or algorithm to follow, you're looking to understand a relationship so that you can apply the appropriate tool to solve your problem.

If you are unsure about the mathematical relationship you've created, it can always be checked by plugging in a couple numbers. There is clarity in examining specific values that can sometimes be lost when looking at symbols. For example, when reading \(a = b + 3\) some students will think that \(b\) is the bigger value "because 3 is being added to it." But this is not true, and the clearest way to see it is to just pick a value of \(b\) and see what the formula actually gives as the result.

Try it!

Annika is 4 inches taller than Carlos. Write an equation that describes this relationship.

Solution.
\begin{equation*} \begin{array}{c} a = \text{Annika's height in inches} \\ c = \text{Carlos' height in inches} \\ \\ a = c + 4 \end{array} \end{equation*}

Activity 33.1.3. Solving a Full Word Problem.

Once a mathematical relationship is established, it then becomes the foundation for solving a word problem based on that relationship.

Try it!

Annika is 70 inches tall. She is also 4 inches taller than Carlos. How tall is Carlos?

Solution.
\begin{equation*} \begin{aligned} a \amp = c + 4 \\ 70 \amp = c + 4 \amp \amp \text{Substitute $a = 70$} \\ c \amp = 66 \amp \amp \text{Subtract $4$ from both sides} \end{aligned} \end{equation*}

Students shy away from word problems because of the extra step of effort that's required to translate the words into mathematical symbols. However, that extra step is the step that covers the gap between "real life" and mathematics. Making that connection is an important piece of the puzzle for mathematical thinking.

In addition to having mathematical relationships described by words, there are situations where the relationship is described by a chart of values.

\begin{equation*} \begin{array}{c|c} x \amp y \\ \hline 1 \amp 2 \\ 2 \amp 4 \\ 3 \amp 6 \end{array} \end{equation*}

After some thought, the pattern that you will likely see is that \(y = 2x\text{.}\) And you can check this by plugging in values.

It turns out that there are all sorts of formulas that would match those values. If you were to use the equation \(y = -\frac{1}{6} x^3 + x^2 + \frac{1}{6} x + 1\text{,}\) you would find that it also works! However, for these problems we are going to stick with the "obvious" answer and not worry about more complicated relationships.

Activity 33.1.4. Finding a Mathematical Relationship.

When trying to find a formula that matches a given table, there is no magic formula. And while there is a computational method if you can guess the "form" of the equation, it's better to develop intuition. The relationships that you will be given to work with are going to be relatively simple, and you should be able to come up with them by trying to think logically and in an organized manner.

Try it!

Find a formula that relates and based on the following chart.

\begin{equation*} \begin{array}{c|c} x \amp y \\ \hline 1 \amp 9 \\ 4 \amp 6 \\ 8 \amp 2 \end{array} \end{equation*}
Solution.
\begin{equation*} x + y = 10 \end{equation*}