## Solve Rational Equations

### Learning Outcomes

• Solve rational equations by clearing denominators
• Identify extraneous solutions in a rational expression

Equations that contain rational expressions are called rational equations. For example, $\frac{2x+1}{4}=\frac{x}{3}$ 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 and 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  $\frac{1}{2}x-3=2-\frac{3}{4}x$ by clearing the fractions in the equation first.

Multiply both sides of the equation by $4$, the common denominator of the fractional coefficients.

$\begin{array}{c}\frac{1}{2}x-3=2-\frac{3}{4}x\\ 4\left(\frac{1}{2}x-3\right)=4\left(2-\frac{3}{4}x\right)\\\text{}\\\,\,\,\,4\left(\frac{1}{2}x\right)-4\left(3\right)=4\left(2\right)+4\left(-\frac{3}{4}x\right)\\2x-12=8-3x\\\underline{+3x}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{+3x}\\5x-12=8\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\underline{+12}\,\,\,\,\underline{+12} \\5x=20\\x=4\end{array}$

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 term that has a polynomial in the numerator.

### Example

Solve the equation $\frac{x+5}{8}=\frac{7}{4}$.

In the next example, 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 do not share any common factors.

### Example

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

### Example

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

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

## Excluded Values and Extraneous Solutions

As you have seen, 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 us look at an example.

### Example

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

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

You have 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 do not work in the original form of the equation. These types of answers are called extraneous solutions. That is 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 $\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}$.

## Summary

You can solve rational equations by finding a common denominator. By rewriting the equation so that all terms have the common denominator, you can solve for the variable using just the numerators. Or, you can multiply both sides of the equation by the least common denominator of all fractions so that all terms become polynomials instead of rational expressions.

An important step in solving rational equations is to reject any extraneous solutions from the final answer. Extraneous solutions are solutions that do not satisfy the original form of the equation because they produce untrue statements or are excluded values that cause a denominator to equal $0$.