8.6: Solve Rational Equations
- Page ID
- 18982
By the end of this section, you will be able to:
- Solve rational equations
- Solve a rational equation for a specific variable
Before you get started, take this readiness quiz.
If you miss a problem, go back to the section listed and review the material.
- Solve: \(\frac{1}{6}x+\frac{1}{2}=\frac{1}{3}\).
If you missed this problem, review Exercise 2.5.1. - Solve: \(n^2−5n−36=0\).
If you missed this problem, review Exercise 7.6.13. - Solve for y in terms of x: 5x+2y=10 for y.
If you missed this problem, review Exercise 2.6.22.
After defining the terms expression and equation early in Foundations, 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 rational equations.
The definition of a rational equation is similar to the definition of equation we used in Foundations.
A rational equation is two rational expressions connected by an equal sign.
You must make sure to know the difference between rational expressions and rational equations. The equation contains an equal sign.
\[\begin{array}{cc} {\textbf{Rational Expression}}&{\textbf{Rational Equation}}\\ {\frac{1}{8}x+\frac{1}{2}}&{\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}}\\ {\frac{y+6}{y^2−36}}&{\frac{y+6}{y^2−36}=y+1}\\ {\frac{1}{n−3}+\frac{1}{n+4}}&{\frac{1}{n−3}+\frac{1}{n+4}=\frac{15}{n^2+n−12}}\\ \nonumber \end{array}\]
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.
Here is an example we did when we worked with linear equations:
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| We multiplied both sides by the LCD. |  |
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| Then we distributed. |  |
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| We simplified—and then we had an equation with no fractions. |  |
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| Finally, we solved that equation. |  |
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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.
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.
How to Solve Equations with Rational Expressions
Solve: \(\frac{1}{x}+\frac{1}{3}=\frac{5}{6}\).
- Answer
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Solve: \(\frac{1}{y}+\frac{2}{3}=\frac{1}{5}\).
- Answer
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\(−\frac{15}{7}\)
Solve: \(\frac{2}{3}+\frac{1}{5}=\frac{1}{x}\).
- Answer
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\(\frac{15}{13}\)
The steps of this method are shown below.
- Note any value of the variable that would make any denominator zero.
- Find the least common denominator of all denominators in the equation.
- Clear the fractions by multiplying both sides of the equation by the LCD.
- Solve the resulting equation.
- 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.
Solve: \(1−\frac{5}{y}=−\frac{6}{y^2}\).
- Answer
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Note any value of the variable that would make any denominator zero. 
Find the least common denominator of all denominators in the equation. The LCD is \(y^2\) Clear the fractions by multiplying both sides of the equation by the LCD. 
Distribute. 
Multiply. 
Solve the resulting equation. First write the quadratic equation in standard form. 
Factor. 
Use the Zero Product Property. 
Solve. 
Check. We did not get 0 as an algebraic solution. 
Solve: \(1−\frac{2}{a}=\frac{15}{a^2}\).
- Answer
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5, −3
Solve: \(1−\frac{4}{b}=\frac{12}{b^2}\).
- Answer
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6, −2
Solve: \(\frac{5}{3u−2}=\frac{3}{2u}\).
- Answer
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Note any value of the variable that would make any denominator zero. 
Find the least common denominator of all denominators in the equation. The LCD is 2u(3u−2). Clear the fractions by multiplying both sides of the equation by the LCD. 
Remove common factors. 
Simplify. 
Multiply. 
Solve the resulting equation. 
We did not get 0 or \(\frac{2}{3}\) as algebraic solutions. 
Solve: \(\frac{1}{x−1}=\frac{2}{3x}\).
- Answer
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−2
Solve: \(\frac{3}{5n+1}=\frac{2}{3n}\).
- Answer
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−2
When one of the denominators is a quadratic, remember to factor it first to find the LCD.
Solve: \(\frac{2}{p+2}+\frac{4}{p−2}=\frac{p−1}{p^2−4}\).
- Answer
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Note any value of the variable that would make any denominator zero. 
Find the least common denominator of all denominators in the equation. The LCD is (p+2)(p−2). Clear the fractions by multiplying both sides of the equation by the LCD. 
Distribute. 
Remove common factors. 
Simplify. 
Distribute. 
Solve. 
We did not get 2 or −2 as algebraic solutions. 
Solve: \(\frac{2}{x+1}+\frac{1}{x−1}=\frac{1}{x^2−1}\).
- Answer
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\(\frac{2}{3}\)
Solve: \(\frac{5}{y+3}+\frac{2}{y−3}=\frac{5}{y^2−9}\)
- Answer
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2
Solve: \(\frac{4}{q−4}−\frac{3}{q−3}=1\).
- Answer
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Note any value of the variable that would make any denominator zero. 
Find the least common denominator of all denominators in the equation. The LCD is (q−4)(q−3). Clear the fractions by multiplying both sides of the equation by the LCD. 
Distribute. 
Remove common factors. 
Simplify. 
Simplify. 
Combine like terms. 
Solve. First write in standard form. 
Factor. 
Use the Zero Product Property. 
We did not get 4 or 3 as algebraic solutions. 
Solve: \(\frac{2}{x+5}−\frac{1}{x−1}=1\).
- Answer
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−1, −2
Solve: \(\frac{3}{x+8}−\frac{2}{x−2}=1\).
- Answer
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−2, −3
Solve: \(\frac{m+11}{m^2−5m+4}=\frac{5}{m−4}−\frac{3}{m−1}\).
- Answer
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Factor all the denominators, so we can note any value of the variable the would make any denominator zero. 
Find the least common denominator of all denominators in the equation. The LCD is (m−4)(m−1) Clear the fractions. 
Distribute. 
Remove common factors. 
Simplify. 
Solve the resulting equation. 
Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution. There is no solution to this equation.
Solve: \(\frac{x+13}{x^2−7x+10}=\frac{6}{x−5}−\frac{4}{x−2}\).
- Answer
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no solution
Solve: \(\frac{y−14}{y^2+3y−4}=\frac{2}{y+4}+\frac{7}{y−1}\).
- Answer
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no solution
The equation we solved in Example had only one algebraic solution, but it was an extraneous solution. That left us with no solution to the equation. Some equations have no solution.
Solve: \(\frac{n}{12}+\frac{n+3}{3n}=\frac{1}{n}\).
- Answer
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Note any value of the variable that would make any denominator zero. 
Find the least common denominator of all denominators in the equation. The LCD is 12n. Clear the fractions by multiplying both sides of the equation by the LCD. 
Distribute. 
Remove common factors. 
Simplify. 
Solve the resulting equation. 
Check. n=0 is an extraneous solution. 
Solve: \(\frac{x}{18}+\frac{x+6}{9x}=\frac{2}{3x}\).
- Answer
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−2
Solve: \(\frac{y+5}{5y}+\frac{y}{15}=\frac{1}{y}\).
- Answer
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−3
Solve: \(\frac{y}{y+6}=\frac{72}{y^2−36}+4\).
- Answer
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Factor all the denominators, so we can note any value of the variable that would make any denominator zero. 
Find the least common denominator. The LCD is (y−6)(y+6). Clear the fractions. 
Simplify. 
Simplify. 
Solve the resulting equation. 
Check. y=−6 is an extraneous solution. 
Solve: \(\frac{x}{x+4}=\frac{32}{x^2−16}+5\).
- Answer
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−4, 3
Solve: \(\frac{y}{y+8}=\frac{128}{y^2−64}+9\).
- Answer
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7
Solve: \(\frac{x}{2x−2}−\frac{2}{3x+3}=\frac{5x^2−2x+9}{12x^2−12}\).
- Answer
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We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD. 
Note any value of the variable that would make any denominator zero. 
Find the least common denominator.The LCD is 12(x−1)(x+1) Clear the fractions. 
Simplify. 
Simplify. 
Solve the resulting equation. 
Check. x=1 and x=−1 are extraneous solutions.
The equation has no solution.
Solve: \(\frac{y}{5y−10}−\frac{5}{3y+6}=\frac{2y^2−19y+54}{15y^2−60}\).
- Answer
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no solution
Solve: \(\frac{z^2}{z+8}−\frac{3}{4z−8}=\frac{3z^2−16z−68}{z^2+8z−64}\).
- Answer
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no solution
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.
We’ll start with a formula relating distance, rate, and time. We have used it many times before, but not usually in this form.
Solve: \(\frac{D}{T}=R\) for T.
- Answer
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Note any value of the variable that would make any denominator zero. 
Clear the fractions by multiplying both sides of the equations by the LCD, T. 
Simplify. 
Divide both sides by R to isolate T. 
Simplify. 
Solve: \(\frac{A}{L}=W\) for L.
- Answer
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\(L=\frac{A}{W}\)
Solve: \(\frac{F}{A}=M\) for A.
- Answer
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\(A=\frac{F}{M}\)
Example uses the formula for slope that we used to get the point-slope form of an equation of a line.
Solve: \(m=\frac{x−2}{y−3}\) for y.
- Answer
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Note any value of the variable that would make any denominator zero. 
Clear the fractions by multiplying both sides of the equations by the LCD, y−3. 
Simplify. 
Isolate the term with y. 
Divide both sides by m to isolate y. 
Simplify. 
Solve: \(\frac{y−2}{x+1}=\frac{2}{3}\) for x.
- Answer
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\(x=\frac{3y−8}{2}\)
Solve: \(x=\frac{y}{1−y}\) for y.
- Answer
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\(y=\frac{x}{1+x}\)
Be sure to follow all the steps in Example. It may look like a very simple formula, but we cannot solve it instantly for either denominator.
Solve \(\frac{1}{c}+\frac{1}{m}=1\) for c.
- Answer
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Note any value of the variable that would make any denominator zero. 
Clear the fractions by multiplying both sides of the equations by the LCD, cm 
Distribute. 
Simplify. 
Collect the terms with c to the right. 
Factor the expression on the right. 
To isolate c, divide both sides by m−1. 
Simplify by removing common factors. 
Notice that even though we excluded c=0 and m=0 from the original equation, we must also now state that \(m \ne 1\).
Solve: \(\frac{1}{a}+\frac{1}{b}=c\) for a.
- Answer
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\(a=\frac{b}{cb−1}\)
Solve: \(\frac{2}{x}+\frac{1}{3}=\frac{1}{y}\) for y.
- Answer
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\(y=\frac{3x}{6+x}\)
Key Concepts
- Strategy to Solve Equations with Rational Expressions
- Note any value of the variable that would make any denominator zero.
- Find the least common denominator of all denominators in the equation.
- Clear the fractions by multiplying both sides of the equation by the LCD.
- Solve the resulting equation.
- Check.
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.
Glossary
- rational equation
- A rational equation is two rational expressions connected by an equal sign.
- 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.