Applications of Equations with Algebraic Fractions

Many real-world situations involve quantities that are naturally described as fractions: a rate of work, a speed over a changing distance, or a share of a mixture. When a word problem asks you to combine two or more of these fraction-based quantities, the result is usually an equation containing algebraic fractions, also called a rational equation. Solving the problem means translating the story into that equation and then solving it using the same techniques covered in solving rational equations.

These applications show up under many names — rational equation word problems, work-rate problems, distance-rate-time problems — but they all follow the same pattern: identify the unknown, write each piece of information as a fraction, and combine the fractions into a single equation that models the situation.

Before jumping into specific problem types, it helps to work through the same four steps every time:

1. Define the variable. Decide exactly what unknown quantity \(x\) represents, including its units (hours, mph, liters, and so on).
2. Write each fractional quantity in terms of \(x\). A rate is often \(\frac{1}{\)time\(}\), and a time is often \(\frac{\)distance\(}{\)speed\(}\).
3. Combine the fractions into one equation. This step usually requires you to add or subtract rational expressions that share or need a common denominator.
4. Solve and check. Clear denominators, solve the resulting equation, and reject any solution that is negative, undefined, or does not make sense for the situation.

Because you are multiplying through by variable denominators, it is important to remember which values of \(x\) are not allowed in the first place — the same restrictions you would find when you simplify a rational expression and identify its restrictions.

In a work problem, "rate" means the fraction of a job completed per unit of time. If a task takes \(a\) hours to finish alone, the rate of work is \(\frac{1}{a}\) of the job per hour. When two workers or machines act together, their rates add.

Example 1: Pipe A can fill a tank in 6 hours. Pipe B can fill the same tank in 9 hours. How long would it take to fill the tank if both pipes run together?

Let \(t\) be the number of hours it takes both pipes working together. Pipe A's rate is \(\frac{1}{6}\) of the tank per hour, and Pipe B's rate is \(\frac{1}{9}\) of the tank per hour. Working together for \(t\) hours, they complete one whole tank, so:

\(\frac{1}{6} + \frac{1}{9} = \frac{1}{t}\)

The lowest common denominator of 6, 9, and \(t\) is \(18t\). Multiplying every term by \(18t\) clears the fractions:

\(3t + 2t = 18\)

\(5t = 18\)

\(t = \frac{18}{5} = 3.6\) hours

Since \(t = 3.6\) is positive and does not make any denominator zero, it is a valid answer: together the pipes fill the tank in 3.6 hours.

Distance-rate-time problems use the relationship \(\)time\( = \frac{\)distance\(}{\)rate\(}\). When a trip has two legs — for instance, traveling with and against a current or wind — each leg contributes its own fraction to the total time, and the two fractions are added or set equal to a known total.

Example 2: A boat travels 60 miles downstream and then 60 miles back upstream on a river with a current of 5 mph. The round trip takes 5 hours. Find the boat's speed in still water.

Let \(x\) be the boat's speed in still water, in mph. Downstream, the current adds to the boat's speed, so the boat travels at \(x + 5\). Upstream, the current works against the boat, so it travels at \(x - 5\). Using time = distance divided by rate for each leg and adding the two times:

\(\frac{60}{x+5} + \frac{60}{x-5} = 5\)

Multiply every term by the common denominator \((x+5)(x-5)\):

\(60(x-5) + 60(x+5) = 5(x+5)(x-5)\)

\(60x - 300 + 60x + 300 = 5(x^2 - 25)\)

\(120x = 5x^2 - 125\)

\(5x^2 - 120x - 125 = 0\)

Dividing every term by 5 gives \(x^2 - 24x - 25 = 0\), which factors as \((x - 25)(x + 1) = 0\), so \(x = 25\) or \(x = -1\). A boat speed cannot be negative, so \(x = -1\) is rejected, leaving the still-water speed as 25 mph.

The figure below shows how the time for a fixed 240-mile trip changes as speed increases — a useful picture of why the equation \(t = \frac{240}{x}\) behaves the way it does in these problems.

Because clearing denominators can introduce solutions that do not satisfy the original equation, every application problem needs a final sense check, not just an algebra check. Reject any solution that:

• makes a denominator in the original equation equal to zero (an extraneous solution),
• is negative when the quantity represents a time, distance, speed, or amount of work,
• is not a whole or sensible number when the context requires one (for example, a number of workers).

If a problem involves several fractions being combined before you solve, it can also help to first tidy the expression using the ideas behind simplifying complex fractions with variables, so the equation you actually solve is as simple as possible.

When you see a new word problem, look for the phrase that tells you what is being combined: "working together," "against the current," "per hour," or "of the job." That phrase usually tells you which two fractions need to be added, subtracted, or set equal to a total. Write the variable definition first, in words, before writing any equation — most errors in these problems come from an unclear variable, not from the algebra itself.

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