Chapter 1 — Fundamental Concepts¶
1.6 — Solving Problems with Equations¶
The difficulty of a word problem rarely lies in the algebra; it lies in the translation that precedes it.
In §1.5 the equation was already written. In practice it almost never arrives that way: it has to be built from a problem told in ordinary language, and that step of translation, from words to symbols, is usually more delicate than the solution that follows. This section adds no algebraic techniques. It proposes a method for setting up a problem correctly and applies it to the most frequent types of problem.
It is worth keeping in mind from the outset what an equation set up from a problem is. It is a model: a symbolic description that retains some relations between the quantities and leaves others aside. The relations it retains always rest on assumptions, such as that a speed is constant or that two volumes add, and the answer is worth what those assumptions are worth. Mathematics guarantees that the equation is correctly solved; it does not guarantee that the equation describes the problem well. That second guarantee is given by whoever sets it up.
§1. A four-step method¶
No procedure replaces understanding the problem, but an order of work greatly reduces the margin of error. In each problem it is useful to distinguish four moments.
Identify the unknown. Before writing a letter, we decide precisely what it represents: not "the variable" in the abstract, but a concrete quantity in the problem, with its units. For example, "\(x\) is the number of liters added", and not simply "\(x\) is the liquid". A badly chosen unknown is often the hidden cause of a wrong setup.
Translate the relations into equations. Each phrase of the statement that links quantities becomes an equality. Expressions such as "twice", "the sum of" or "\(20\%\) more than" have a direct translation: \(2x\), \(x + y\) and \(x + 0.2x\). It pays to make this translation phrase by phrase, without jumping to an equation written from memory, because this is where most errors slip in. In this step the assumptions of the model are also made explicit.
Solve. Here, and only here, the techniques of §1.5 come in.
Check against the statement. Each solution is tested against the original problem, not only against the equation. This check does two jobs. On the one hand, it discards values that solve the equation but make no sense in the problem: a negative time, a non-integer number of people, a length greater than the total available. It is a different check from the one for extraneous solutions in §1.5: there the question was whether the number satisfies the original equation; here it is whether the equation, once solved, still describes the problem. On the other hand, it detects translation errors. An error in the second step produces a wrong equation that may be perfectly solved, and checking the answer in that same equation does not reveal it; checking it against the statement does.
Reminder. Identify the unknown, translate, solve and check against the statement. Most errors occur in the translation, not in the solution.
§2. Mixture and percentage problems¶
In these problems, quantities of different composition are combined to obtain a final quantity of intermediate composition. Their characteristic equation equates the amount of one component before and after mixing. A concentration of \(10\%\) means that one tenth of the volume is the component we are interested in, so that in \(V\) liters there are \(0.10V\) liters of that component.
Example. We have one acid solution at \(10\%\) and another at \(30\%\). How many liters of the second must be added to \(5\) liters of the first to obtain a mixture at \(18\%\)? Let \(x\) be the number of liters of the \(30\%\) solution. We assume that volumes add on mixing, so that the mixture has \(5 + x\) liters; for real liquids this is an approximation, often a very good one. Pure acid is neither created nor destroyed, only redistributed. We therefore equate the acid contributed by each solution with the acid in the final mixture:
Solving, \(0.5 + 0.3x = 0.9 + 0.18x\), whence \(0.12x = 0.4\) and \(x = \frac{10}{3}\), that is, about \(3.33\) liters. Let us check against the statement. The mixture has \(5 + \frac{10}{3} = \frac{25}{3}\) liters and contains \(0.5 + 0.3 \cdot \frac{10}{3} = 1.5\) liters of acid; the proportion is \(1.5 \div \frac{25}{3} = 0.18\), the \(18\%\) required. Moreover, the amount added is positive, as it must be.
Reminder. In a mixture problem, the equation equates the amount of one and the same component before and after mixing.
§3. Joint work problems¶
When two agents, whether people, machines or faucets, perform the same task at different rates, it is better to work with the rate of work rather than with time. If an agent takes \(a\) hours to complete the task working at a constant rate, in one hour it does \(\frac{1}{a}\) of the task, and in \(t\) hours it does \(\frac{t}{a}\). If two agents work at the same time without hindering each other, what they do together is the sum of what each one does. That is the assumption of the model: rates add, times do not.
Example. One faucet fills a tank in \(4\) hours; another, in \(6\). How long do the two take together? Let \(t\) be the time sought, in hours. In that time the first fills \(\frac{t}{4}\) of the tank and the second \(\frac{t}{6}\), and between them they fill it exactly once:
Despite the fractions, this is a linear equation: the unknown is in the numerators, and the denominators are numbers. Multiplying both sides by \(12\), which is not zero, we obtain the equivalent equation \(3t + 2t = 12\), whence \(5t = 12\) and \(t = \frac{12}{5} = 2.4\) hours, that is, \(2\) hours and \(24\) minutes. Had we set up the equation with the rates, \(\frac{1}{4} + \frac{1}{6} = \frac{1}{t}\), the unknown would be in a denominator and it would be a rational equation like those of §1.5; the result is the same, with the additional condition \(t \neq 0\).
The result is less than the time of the faster faucet, and this is no accident. In general, if the individual times are \(a\) and \(b\), the joint time \(t\) satisfies \(\frac{1}{t} = \frac{1}{a} + \frac{1}{b}\). Since \(\frac{1}{b} > 0\), we get \(\frac{1}{t} > \frac{1}{a}\). Multiplying this inequality by the positive number \(at\), which preserves its direction by the order axioms of §1.1, we obtain \(a > t\); likewise, \(t < b\). This bound is a quick check of the setup: if the joint time came out greater than either individual time, the translation would be wrong.
Reminder. In joint work problems, rates add, not times. The joint time is always less than that of each agent separately.
§4. Motion problems¶
These problems rest on the relation \(\text{distance} = \text{speed} \times \text{time}\), valid for motion at constant speed, which is the usual assumption. Their difficulty is usually not algebraic but a matter of organizing the data. It almost always helps to build a table with one row for each moving object or each stretch, and columns for speed, time and distance, and to let the table suggest the equation.
Example. Two cyclists leave the same point in opposite directions, one at \(15\) km/h and the other at \(20\) km/h. How long until they are \(70\) km apart? Let \(t\) be the time in hours. In that time the first covers \(15t\) kilometers and the second \(20t\), and since they move apart in opposite directions, the distance between them is the sum:
The symbol \(\Longrightarrow\) indicates that each equation follows from the previous one; here each step is also reversible, in the sense of §1.5. In two hours, the first covers \(30\) km and the second \(40\) km, and \(30 + 40 = 70\), as the statement requires.
Reminder. In motion problems, a table with speed, time and distance for each moving object organizes the data and suggests the equation.
§5. Geometry problems¶
Here the relation between the quantities is translated not from a phrase but from a known geometric formula: an area, a perimeter, a volume. Often the resulting equation is one of those already solved in §1.5. In particular, an area problem frequently leads to a quadratic equation with two solutions, of which not both necessarily make geometric sense.
Example. A rectangle is \(4\) cm longer than it is wide, and its area is \(96\) cm². Let \(x\) be the width in centimeters; the length is \(x + 4\), and
where we factored as in §1.3. By the zero product property of §1.5, the solutions are \(x = 8\) and \(x = -12\). Here checking against the statement is decisive: a width cannot be negative, so \(x = -12\) is discarded. It is not discarded for being incorrect, since it satisfies the equation, but because it describes no rectangle. The answer is a width of \(8\) cm and a length of \(12\) cm, and indeed \(8 \cdot 12 = 96\).
Reminder. A solution of the equation may fail to be a solution of the problem. Discarding it is part of solving the problem, just as much as finding it.
With the method of setting up problems now established and practiced on the most frequent cases, the chapter moves on to a natural generalization of the equation: the inequality, where instead of exact values one looks for a range of possible values.