## 8.4 – Rational Equations

### Learning Objectives

• (8.4.1) – Solving rational equations by clearing denominators
• Identify extraneous solutions in a rational equation

# (8.4.1) – Solving rational equations by clearing denominators

Equations that contain rational expressions are called rational equations. For example, $\displaystyle \frac{2x+1}{4}=\frac{7}{x}$ is a rational equation. Rational equations can be useful for representing real-life situations and for finding answers to real problems. In particular, they are quite good for describing a variety of proportional relationships.

One of the most straightforward ways to solve a rational equation is to eliminate denominators with the common denominator, then use properties of equality to isolate the variable. This method is often used to solve linear equations that involve fractions as in the following example:

Solve  $\displaystyle \frac{2}{3}x - \frac{5}{6} = \frac{3}{4}$ by clearing the fractions in the equation first.

$\begin{eqnarray*} \frac{2}{3}\,x - \frac{5}{6}\,= \frac{3}{4}\,& & \text{ Multiply}\,\text{ each} \text{ term}\,\text{ by}\,\text{ LCD}, 12\\ & & \\ \frac{2 (12)}{3}\,x - \frac{5 (12)}{6}\,= \frac{3 (12)}{4}\,& & \text{ Reduce}\,\text{ fractions}\\ & & \\ 2 (4) x - 5 (2) = 3 (3) & & \text{ Multiply}\\ 8 x - 10 = 9 & & \text{ Solve}\\ \underline{+ 10 + 10}\,& & \text{ Add}\,10 \text{ to}\,\text{ both} \text{ sides}\\ 8 x = 19 & & \text{ Divide}\,\text{ both}\,\text{ sides}\,\text{ by}\,8\\ \overline{8}\, \overline{8}\,& & \\ x = \frac{19}{8}\,& & \text{ Our}\,\text{ Solution} \end{eqnarray*}$

We could have found a common denominator and worked with fractions, but that often leads to more mistakes. We can apply the same idea to solving rational equations.  The difference between a linear equation and a rational equation is that rational equations can have polynomials in the numerator and denominator of the fractions. This means that clearing the denominator may sometimes mean multiplying the whole rational equation by a polynomial. In the next example, we will clear the denominators of a rational equation with a terms that has a polynomial in the numerator.

### Example

Solve the equation: $\displaystyle 3 x - \frac{1}{2}\,= \frac{1}{x}$

In the next two examples, we show how to solve a rational equation with a binomial in the denominator of one term. We will use the common denominator to eliminate the denominators from both fractions. Note that the LCD is the product of both denominators because they don’t share any common factors.

### Example

Solve the equation $\displaystyle \frac{8}{x+1}=\frac{4}{3}$.

### Example

Solve the equation: $\displaystyle \frac{5 x + 5}{x + 2}\,+ 3 x = \frac{x^2}{x + 2}$

In the video that follows we present two ways to solve rational equations with both integer and variable denominators.

### Identify Excluded Values and Extraneous Solutions

Some rational expressions have a variable in the denominator. When this is the case, there is an extra step in solving them. Since division by 0 is undefined, you must exclude values of the variable that would result in a denominator of 0. These values are called excluded values. Let’s look at an example.

### Example

Solve the equation $\displaystyle \frac{2x-5}{x-5}=\frac{15}{x-5}$.

In the following video we present an example of solving a rational equation with variables in the denominator.

You’ve seen that there is more than one way to solve rational equations. Because both of these techniques manipulate and rewrite terms, sometimes they can produce solutions that don’t work in the original form of the equation. These types of answers are called extraneous solutions. That’s why it is always important to check all solutions in the original equations—you may find that they yield untrue statements or produce undefined expressions.

### Example

Solve the equation $\displaystyle \frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}$.

Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution.

Solve the following rational equation: $\displaystyle \frac{-4x}{x - 1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}$.

### Example

Solve $\displaystyle \frac{3x+2}{x - 2}+\frac{1}{x}=\frac{-2}{{x}^{2}-2x}$.