### Learning Objectives

- Solve rational equations
- Use rational functions
- Solve a rational equation for a specific variable

Before you get started, take this readiness quiz.

- Solve: $\frac{1}{6}x+\frac{1}{2}=\frac{1}{3}.$

If you missed this problem, review Example 2.9. - Solve: ${n}^{2}-5n-36=0.$

If you missed this problem, review Example 6.45. - Solve the formula $5x+2y=10$ for $y.$

If you missed this problem, review Example 2.31.

After defining the terms ‘expression’ and ‘equation’ earlier, we have used them throughout this book. We have *simplified* many kinds of *expressions* and *solved* many kinds of *equations*. We have simplified many rational expressions so far in this chapter. Now we will *solve* a rational equation.

### Rational Equation

A **rational equation** is an equation that contains a rational expression.

You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.

### Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution to a rational equation.

### Extraneous Solution to a Rational Equation

An **extraneous solution to a rational equation** is an algebraic solution that would cause any of the expressions in the original equation to be undefined.

We note any possible extraneous solutions, *c*, by writing $x\ne c$ next to the equation.

### Example 7.33

#### How to Solve a Rational Equation

Solve: $\frac{1}{x}+\frac{1}{3}=\frac{5}{6}.$

Solve: $\frac{1}{y}+\frac{2}{3}=\frac{1}{5}.$

Solve: $\frac{2}{3}+\frac{1}{5}=\frac{1}{x}.$

The steps of this method are shown.

### How To

#### Solve equations with rational expressions.

- Step 1. Note any value of the variable that would make any denominator zero.
- Step 2. Find the least common denominator of
*all*denominators in the equation. - Step 3. Clear the fractions by multiplying both sides of the equation by the LCD.
- Step 4. Solve the resulting equation.
- Step 5. Check:
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.

We always start by noting the values that would cause any denominators to be zero.

### Example 7.34

#### How to Solve a Rational Equation using the Zero Product Property

Solve: $1-\frac{5}{y}=-\frac{6}{{y}^{2}}.$

Solve: $1-\frac{2}{x}=\frac{15}{{x}^{2}}.$

Solve: $1-\frac{4}{y}=\frac{12}{{y}^{2}}.$

In the next example, the last denominators is a difference of squares. Remember to factor it first to find the LCD.

### Example 7.35

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

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

Solve: $\frac{5}{y+3}+\frac{2}{y-3}=\frac{5}{{y}^{2}-9}.$

In the next example, the first denominator is a trinomial. Remember to factor it first to find the LCD.

### Example 7.36

Solve: $\frac{m+11}{{m}^{2}-5m+4}=\frac{5}{m-4}-\frac{3}{m-1}.$

Solve: $\frac{x+13}{{x}^{2}-7x+10}=\frac{6}{x-5}-\frac{4}{x-2}.$

Solve: $\frac{y-6}{{y}^{2}+3y-4}=\frac{2}{y+4}+\frac{7}{y-1}.$

The equation we solved in the previous example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. In the next example we get two algebraic solutions. Here one or both could be extraneous solutions.

### Example 7.37

Solve: $\frac{y}{y+6}=\frac{72}{{y}^{2}-36}+4.$

Solve: $\frac{x}{x+4}=\frac{32}{{x}^{2}-16}+5.$

Solve: $\frac{y}{y+8}=\frac{128}{{y}^{2}-64}+9.$

In some cases, all the algebraic solutions are extraneous.

### Example 7.38

Solve: $\frac{x}{2x-2}-\frac{2}{3x+3}=\frac{5{x}^{2}-2x+9}{12{x}^{2}-12}.$

Solve: $\frac{y}{5y-10}-\frac{5}{3y+6}=\frac{2{y}^{2}-19y+54}{15{y}^{2}-60}.$

Solve: $\frac{z}{2z+8}-\frac{3}{4z-8}=\frac{3{z}^{2}-16z-16}{8{z}^{2}+2z-64}.$

### Example 7.39

Solve: $\frac{4}{3{x}^{2}-10x+3}+\frac{3}{3{x}^{2}+2x-1}=\frac{2}{{x}^{2}-2x-3}.$

Solve: $\frac{15}{{x}^{2}+x-6}-\frac{3}{x-2}=\frac{2}{x+3}.$

Solve: $\frac{5}{{x}^{2}+2x-3}-\frac{3}{{x}^{2}+x-2}=\frac{1}{{x}^{2}+5x+6}.$

### Use Rational Functions

Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.

### Example 7.40

For rational function, $f\left(x\right)=\frac{2x-6}{{x}^{2}-8x+15},$ ⓐ find the domain of the function, ⓑ solve $f\left(x\right)=1,$ and ⓒ find the points on the graph at this function value.

For rational function, $f\left(x\right)=\frac{8-x}{{x}^{2}-7x+12},$ ⓐ find the domain of the function ⓑ solve $f\left(x\right)=3$ ⓒ find the points on the graph at this function value.

For rational function, $f\left(x\right)=\frac{x-1}{{x}^{2}-6x+5},$ ⓐ find the domain of the function ⓑ solve $f\left(x\right)=4$ ⓒ find the points on the graph at this function value.

### Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.

$\begin{array}{cccccc}& & & \hfill m& =\hfill & \frac{y-{y}_{1}}{x-{x}_{1}}\hfill \\ \text{Multiply both sides of the equation by}\phantom{\rule{0.2em}{0ex}}x-{x}_{1}.\hfill & & & \hfill m\left(x-{x}_{1}\right)& =\hfill & \left(\frac{y-{y}_{1}}{x-{x}_{1}}\right)\left(x-{x}_{1}\right)\hfill \\ \text{Simplify.}\hfill & & & \hfill m\left(x-{x}_{1}\right)& =\hfill & y-{y}_{1}\hfill \\ \text{Rewrite the equation with the}\phantom{\rule{0.2em}{0ex}}y\phantom{\rule{0.2em}{0ex}}\text{terms on the left.}\hfill & & & \hfill y-{y}_{1}& =\hfill & m\left(x-{x}_{1}\right)\hfill \end{array}$

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through the point $\left(2,3\right).$ We will add one more step to solve for *y*.

### Example 7.41

Solve:$m=\frac{y-2}{x-3}$ for $y.$

Solve: $m=\frac{y-5}{x-4}$for $y.$

Solve: $m=\frac{y-1}{x+5}$ for $y.$

Remember to multiply both sides by the LCD in the next example.

### Example 7.42

Solve: $\frac{1}{c}+\frac{1}{m}=1$ for *c*.

Solve: $\frac{1}{a}+\frac{1}{b}=c$ for *a*.

Solve: $\frac{2}{x}+\frac{1}{3}=\frac{1}{y}$ for *y*.

### Media

Access this online resource for additional instruction and practice with equations with rational expressions.

### Section 7.4 Exercises

#### Practice Makes Perfect

**Solve Rational Equations**

In the following exercises, solve each rational equation.

$\frac{6}{3}-\frac{2}{d}=\frac{4}{9}$

$\frac{3}{8}+\frac{2}{y}=\frac{1}{4}$

$1+\frac{4}{n}=\frac{21}{{n}^{2}}$

$1-\frac{7}{q}=\frac{\mathrm{-6}}{{q}^{2}}$

$\frac{8}{2w+1}=\frac{3}{w}$

$\frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{{y}^{2}-81}$

$\frac{9}{a+11}-\frac{6}{a-11}=\frac{6}{{a}^{2}-121}$

$\frac{2}{s+7}-\frac{3}{s-3}=1$

$\frac{w+8}{{w}^{2}-11w+28}=\frac{5}{w-7}+\frac{2}{w-4}$

$\frac{y-5}{{y}^{2}-4y-5}=\frac{1}{y+1}+\frac{1}{y-5}$

$\frac{c+3}{12c}+\frac{c}{36}=\frac{1}{4c}$

$\frac{m}{m+5}=\frac{50}{{m}^{2}-25}+6$

$\frac{p}{p+7}-8=\frac{98}{{p}^{2}-49}$

$\frac{r}{3r-15}-\frac{1}{4r+20}=\frac{3{r}^{2}+17r+40}{12{r}^{2}-300}$

$\frac{t}{6t-12}-\frac{5}{2t+10}=\frac{{t}^{2}-23t+70}{12{t}^{2}+36t-120}$

$\frac{5}{{x}^{2}+4x+3}+\frac{2}{{x}^{2}+x-6}=\frac{3}{{x}^{2}-x-2}$

$\frac{2}{{x}^{2}+2x-3}+\frac{3}{{x}^{2}+4x+3}=\frac{6}{{x}^{2}-1}$

**Solve Rational Equations that Involve Functions**

For rational function, $f\left(x\right)=\frac{x-2}{{x}^{2}+6x+8},$ ⓐ find the domain of the function ⓑ solve $f\left(x\right)=5$ ⓒ find the points on the graph at this function value.

For rational function, $f\left(x\right)=\frac{x+1}{{x}^{2}-2x-3},$ ⓐ find the domain of the function ⓑ solve $f\left(x\right)=1$ ⓒ find the points on the graph at this function value.

For rational function, $f\left(x\right)=\frac{2-x}{{x}^{2}-7x+10},$ ⓐ find the domain of the function ⓑ solve $f\left(x\right)=2$ ⓒ find the points on the graph at this function value.

For rational function, $f\left(x\right)=\frac{5-x}{{x}^{2}+5x+6},$

ⓐ find the domain of the function

ⓑ solve $f\left(x\right)=3$

ⓒ the points on the graph at this function value.

**Solve a Rational Equation for a Specific Variable**

In the following exercises, solve.

$\frac{I}{r}=P$ for $r.$

$\frac{x+5}{2-y}=\frac{4}{3}$ for $y.$

$m=\frac{n}{2-n}$ for $n.$

$\frac{3}{s}+\frac{1}{t}=2$ for $s.$

$\frac{6}{x}+\frac{2}{3}=\frac{1}{y}$ for $y.$

$r=\frac{s}{3-t}$ for $t.$

$\frac{R}{T}=W$ for $T.$

$c=\frac{2}{a}\phantom{\rule{0.2em}{0ex}}+\frac{b}{5}$ for $a.$

#### Writing Exercises

Your class mate is having trouble in this section. Write down the steps you would use to explain how to solve a rational equation.

Alek thinks the equation $\frac{y}{y+6}=\frac{72}{{y}^{2}-36}+4$ has two solutions, $y=\mathrm{-6}$ and $y=4.$ Explain why Alek is wrong.

#### Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ On a scale of $1-10,$ how would you rate your mastery of this section in light of your responses on the checklist? How can you improve this?