Math Core

A rational equation is an equation that contains rational expressions, like $\dfrac{5}{x} - \dfrac{1}{2} = \dfrac{3}{2x}$ . Equations like this come up whenever rates, averages or shared work are involved. The strategy is simple: clear the fractions, solve the polynomial equation that remains, and then check for answers that don't really work.

Clearing the fractions

Fractions are what make these equations hard, so get rid of them. Multiply every term on both sides by the least common denominator (LCD). Each denominator divides the LCD, so every fraction cancels down to a polynomial.

Worked example: Multiplying by the LCD

Solve $\dfrac{5}{x} - \dfrac{1}{2} = \dfrac{3}{2x}$ .

The denominators are $x$ , $2$ and $2x$ , so the LCD is $2x$ . Note that $x \ne 0$ . Multiply every term by $2x$ :

\begin{aligned} 2x \cdot \frac{5}{x} - 2x \cdot \frac{1}{2} &= 2x \cdot \frac{3}{2x} \\ 10 - x &= 3 \\ x &= 7. \end{aligned}

Check: $\dfrac{5}{7} - \dfrac{1}{2} = \dfrac{10 - 7}{14} = \dfrac{3}{14}$ , and $\dfrac{3}{2(7)} = \dfrac{3}{14}$ . The solution is $x = 7$ .

Proportions

When the equation is one fraction equal to another, $\dfrac{a}{b} = \dfrac{c}{d}$ , multiplying both sides by $bd$ gives $ad = bc$ . This shortcut is called cross-multiplying. It is just the LCD method for this special shape.

Worked example: Cross-multiplying

Solve $\dfrac{3}{x + 2} = \dfrac{5}{x - 2}$ .

Cross-multiply:

\begin{aligned} 3(x - 2) &= 5(x + 2) \\ 3x - 6 &= 5x + 10 \\ -16 &= 2x \\ x &= -8. \end{aligned}

Check: $\dfrac{3}{-6} = -\dfrac{1}{2}$ and $\dfrac{5}{-10} = -\dfrac{1}{2}$ . The solution is $x = -8$ .

Extraneous solutions

Multiplying by an expression that contains $x$ can introduce false solutions. The new polynomial equation is defined for every $x$ , but the original equation is not defined where a denominator is zero. If your algebra produces one of those forbidden values, it is an extraneous solution and must be thrown out.

Solving a rational equation

Factor each denominator and list the excluded values (where any denominator is zero).
Multiply every term on both sides by the LCD.
Solve the resulting polynomial equation.
Reject any answer that is an excluded value. Check the rest in the original equation.

If every candidate is rejected, the equation has no solution.

Worked example: One real solution, one extraneous

Solve $\dfrac{1}{x - 2} + \dfrac{x}{x + 2} = \dfrac{4}{x^2 - 4}$ .

Since $x^2 - 4 = (x - 2)(x + 2)$ , the excluded values are $x = 2$ and $x = -2$ , and the LCD is $(x - 2)(x + 2)$ . Multiply every term by the LCD:

\begin{aligned} (x + 2) + x(x - 2) &= 4 \\ x^2 - x + 2 &= 4 \\ x^2 - x - 2 &= 0 \\ (x - 2)(x + 1) &= 0. \end{aligned}

The candidates are $x = 2$ and $x = -1$ . But $x = 2$ is excluded (it makes $\dfrac{1}{x - 2}$ undefined), so it is extraneous.

Check $x = -1$ : the left side is $\dfrac{1}{-3} + \dfrac{-1}{1} = -\dfrac{4}{3}$ and the right side is $\dfrac{4}{1 - 4} = -\dfrac{4}{3}$ . The only solution is $x = -1$ .

Common mistake

Never skip the check. An answer like $x = 2$ in the example above comes out of perfectly correct algebra, yet it is not a solution. Write the excluded values down at the very start so you can't forget them at the end.

You can see why this happens on a graph. Solving $\dfrac{1}{x - 2} + \dfrac{x}{x + 2} = \dfrac{4}{x^2 - 4}$ means finding where the two sides' graphs meet. At $x = 2$ both sides have a vertical asymptote, so there is no intersection there at all.

Work and rate problems

If a job takes someone $t$ hours, they complete $\dfrac{1}{t}$ of the job per hour. When people work together, their rates add:

\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{t_{\text{together}}}.

Worked example: Painting a fence

Ana can paint a fence alone in $5$ hours. With Ben helping, the job takes $3$ hours. How long would Ben take alone?

Let $b$ be Ben's time in hours. Rates add:

\frac{1}{5} + \frac{1}{b} = \frac{1}{3}.

Multiply by the LCD, $15b$ :

\begin{aligned} 3b + 15 &= 5b \\ 15 &= 2b \\ b &= 7.5. \end{aligned}

Ben would take $7.5$ hours alone. That's reasonable: he's slower than Ana, and together they are faster than either one.

Distance problems work similarly, using $\text{time} = \dfrac{\text{distance}}{\text{rate}}$ . When two trips take the same time, set their time expressions equal and solve the proportion.

Tip

Sanity-check every word problem answer. A combined time must be less than each person's solo time. A current's speed must be less than the boat's speed. A negative time or distance is always extraneous in context.

Practice

Practice 1

Solve $\dfrac{6}{x} = \dfrac{3}{4}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

Solve $\dfrac{4}{x - 1} = \dfrac{2}{x + 3}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 3

Solve $\dfrac{1}{x} + \dfrac{1}{2x} = \dfrac{1}{4}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Find all solutions of $x + \dfrac{6}{x} = 5$ .

Separate answers with commas, e.g. 2, -5

Practice 5

Solve $\dfrac{x}{x - 2} = \dfrac{2}{x - 2} + 3$ .

Practice 6

Solve $1 + \dfrac{3}{x - 1} = \dfrac{6}{x^2 - 1}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Pipe A can fill a pool in $6$ hours. With pipe B also running, the pool fills in $4$ hours. How many hours would pipe B take to fill the pool alone?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

A boat moves at $12$ miles per hour in still water. It travels $45$ miles downstream in the same time it takes to travel $27$ miles upstream. What is the speed of the current, in miles per hour?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.