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# Rational Equations

### Learning Objectives

• Methods for solving rational equations
• Solve rational equations by clearing denominators
• Identify extraneous solutions in a rational equation
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  $\frac{2}{3}x - \frac{5}{6} = \frac{3}{4}$ 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: $\displaystyle 3 x - \frac{1}{2}\,= \frac{1}{x}$

Answer: $\begin{eqnarray*} 3 x - \frac{1}{2}\,= \frac{1}{x}\,& & \text{ Multiply}\,\text{ each} \text{ term}\,\text{ by}\,\text{ LCD}, (2 x)\\ 3 x (2 x) - \frac{(2 x)}{2}\,= \frac{(2 x)}{x}\,& & \text{ Reduce} \text{ fractions}\\ 6 x^2 - x = 2 & & \text{ Distribute}\\ \underline{- 2 - 2}\,& & \text{ Subtract}\,2 \text{ from}\,\text{ both} \text{ sides}\\ 6 x^2 - x - 2 = 0 & & \text{ Factor}\\ (3 x - 2) (2 x + 1) = 0 & & \text{ Set}\,\text{ each}\,\text{ factor} \text{ equal}\,\text{ to}\,\text{ zero}\\ 3 x - 2 = 0 \text{ or}\,2 x + 1 = 0 & & \text{ Solve}\,\text{ each} \text{ equation}\\ \underline{+ 2 \quad + 2}\, \underline{- 1 - 1}\,& & \\ 3 x = 2 \hspace{3em}\,2 x = - 1 & & \\ x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,& & \text{ Check} \text{ solutions}, \text{ LCD}\,\text{ can}' t \text{ be}\,\text{ zero}\\ 2 \left( \frac{3}{2}\,\right) =2 \left( - \frac{1}{2}\,\right) = - 1 & & \text{ Neither}\,\text{ make}\,\text{ LCD}\,\text{ zero}, \text{ both}\,\text{ are} \text{ solutions}\\ x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,& & \text{ Our} \text{ Solution} \end{eqnarray*}$

[latex-display]\displaystyle x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,[/latex-display]

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

Answer: Clear the denominators by multiplying each side by the common denominator. The common denominator is $3\left(x+1\right)$ since $3\text{ and }x+1$ don't have any common factors.

$\begin{array}{c}3\left(x+1\right)\left(\frac{8}{x+1}\right)=3\left(x+1\right)\left(\frac{4}{3}\right)\end{array}$

Simplify common factors.

$\begin{array}{c}3\cancel{\left(x+1\right)}\left(\frac{8}{\cancel{x+1}}\right)=\cancel{3}\left(x+1\right)\left(\frac{4}{\cancel{3}}\right)\\24=4\left(x+1\right)\\24=4x+4\end{array}$

Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.

$\begin{array}{c}24=4x+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\20=4x\,\,\,\,\,\,\,\,\\\\x=5\,\,\,\,\,\,\,\,\,\end{array}$

Check the solution in the original equation.

$\begin{array}{r}\,\,\,\,\,\frac{8}{\left(x+1\right)}=\frac{4}{3}\\\\\frac{8}{\left(5+1\right)}=\frac{4}{3}\\\\\frac{8}{6}=\frac{4}{3}\end{array}$

Reduce the fraction $\frac{8}{6}$ by simplifying the common factor of 2:

$\large\frac{\cancel{2}\cdot4}{\cancel{2}\cdot3}\normalsize=\large\frac{4}{3}$

[latex-display]x=5[/latex-display]

### Example

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

Answer: $\begin{eqnarray*}\frac{5 x + 5}{x + 2}\,+ 3 x = \frac{x^2}{x + 2}\,& & \text{ Multiply} \text{ each}\,\text{ term}\,\text{ by}\,\text{ LCD}, (x + 2)\\ & & \\ \frac{(5 x + 5) (x + 2)}{x + 2}\,+ 3 x (x + 2) = \frac{x^2 (x + 2)}{x + 2} & & \text{ Reduce}\,\text{ fractions}\\ & & \\ 5 x + 5 + 3 x (x + 2) = x^2 & & \text{ Distribute}\\ 5 x + 5 + 3 x^2 + 6 x = x^2 & & \text{ Combine}\,\text{ like}\,\text{ terms}\\ 3 x^2 + 11 x + 5 = x^2 & & \text{ Make}\,\text{ equation}\,\text{ equal} \text{ zero}\\ \underline{- x^2 - x^2}\,& & \text{ Subtract}\,x^2 \text{ from}\,\text{ both} \text{ sides}\\ 2 x^2 + 11 x + 5 = 0 & & \text{ Factor}\\ (2 x + 1) (x + 5) = 0 & & \text{ Set}\,\text{ each}\,\text{ factor} \text{ equal}\,\text{ to}\,\text{ zero}\\ 2 x + 1 = 0 \text{ or}\,x + 5 = 0 & & \text{ Solve}\,\text{ each} \text{ equation}\\ 2 x = - 1 \text{ or}\,x = - 5 & & \\ x = - \frac{1}{2}\,\text{ or}\,- 5 & & \text{ Check}\,\text{ solutions}, \text{ LCD}\,\text{ can}' t \text{ be}\,\text{ zero}\\ - \frac{1}{2}\,+ 2 = \frac{3}{2}\,- 5 + 2 = - 3 & & \text{ Neither} \text{ make}\,\text{ LCD}\,\text{ zero}, \text{ both}\,\text{ are} \text{ solutions}\\ x = - \frac{1}{2}\,\text{ or}\,- 5 & & \text{ Our}\,\text{ Solution} \end{eqnarray*}$

Answer [latex-display]x = - \frac{1}{2}\,\text{ or}\,x = - 5[/latex-display]

In the video that follows we present two ways to solve rational equations with both integer and variable denominators. https://youtu.be/R9y2D9VFw0I

## 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 $\displaystyle \frac{2x-5}{x-5}=\frac{15}{x-5}$.

Answer: Determine any values for x that would make the denominator 0.

$\frac{2x-5}{x-5}=\frac{15}{x-5}$

5 is an excluded value because it makes the denominator $x-5$ equal to 0. Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for x.

$\begin{array}{r}2x-5=15\\2x=20\\x=10\end{array}$

Check the solution in the original equation.

$\begin{array}{r}\frac{2x-5}{x-5}=\frac{15}{x-5}\,\,\\\\\frac{2(10)-5}{10-5}=\frac{15}{10-5}\\\\\frac{20-5}{10-5}=\frac{15}{10-5}\\\\\frac{15}{5}=\frac{15}{5}\,\,\,\,\,\,\,\,\,\end{array}$

[latex-display]x=10[/latex-display]

In the following video we present an example of solving a rational equation with variables in the denominator. https://www.youtube.com/watch?v=gGA-dF_aQQQ&feature=youtu.be 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 $\displaystyle \frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}$.

Answer: Determine any values for m that would make the denominator 0. $−4$ is an excluded value because it makes $m+4$ equal to 0. Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for m.

$\begin{array}{l}16=m^{2}\\\,\,\,0={{m}^{2}}-16\\\,\,\,0=\left( m+4 \right)\left( m-4 \right)\end{array}$

$\begin{array}{c}0=m+4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,0=m-4\\m=-4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,m=4\\m=4,-4\end{array}$

Check the solutions in the original equation. Since $m=−4$ leads to division by 0, it is an extraneous solution.

$\begin{array}{c}\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}\\\\\frac{16}{-4+4}=\frac{{{(-4)}^{2}}}{-4+4}\\\\\frac{16}{0}=\frac{16}{0}\end{array}$

$-4$ is excluded because it leads to division by 0.

$\begin{array}{c}\frac{16}{4+4}=\frac{{{(4)}^{2}}}{4+4}\\\\\frac{16}{8}=\frac{16}{8}\end{array}$

[latex-display]m=4[/latex-display]

### More Rational Equations

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: $\displaystyle \frac{-4x}{x - 1}+\frac{4}{x+1}=\frac{-8}{{x}^{2}-1}$.

Answer: We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, ${x}^{2}-1=\left(x+1\right)\left(x - 1\right)$. Then, the LCD is $\left(x+1\right)\left(x - 1\right)$. Next, we multiply the whole equation by the LCD.

$\begin{array}{l}\left(x+1\right)\left(x - 1\right)\left[\frac{-4x}{x - 1}+\frac{4}{x+1}\right]\hfill&=\left[\frac{-8}{\left(x+1\right)\left(x - 1\right)}\right]\left(x+1\right)\left(x - 1\right)\hfill \\ -4x\left(x+1\right)+4\left(x - 1\right)\hfill&=-8\hfill \\ -4{x}^{2}-4x+4x - 4\hfill&=-8\hfill \\ -4{x}^{2}+4\hfill&=0\hfill \\ -4\left({x}^{2}-1\right)\hfill&=0\hfill \\ -4\left(x+1\right)\left(x - 1\right)\hfill&=0\hfill \\ x\hfill&=-1\hfill \\ x\hfill&=1\hfill \end{array}$
In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution.

### Example

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

Answer: $x=-1$, ($x=0$ is not a solution).

### Try It

[ohm_question]44861[/ohm_question]