Rational Expressions and Equations

67 Solve Rational Equations

Learning Objectives

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.

  1. Solve: \frac{1}{6}x+\frac{1}{2}=\frac{1}{3}.
    If you missed this problem, review (Figure).
  2. Solve: {n}^{2}-5n-36=0.
    If you missed this problem, review (Figure).
  3. Solve for y in terms of x: 5x+2y=10 for y.
    If you missed this problem, review (Figure).

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.

Rational Equation

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}{cccc}\hfill \mathbf{\text{Rational Expression}}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\mathbf{\text{Rational Equation}}\hfill \\ \hfill \phantom{\rule{1.95em}{0ex}}\frac{1}{8}x+\frac{1}{2}\hfill & & & \hfill \phantom{\rule{3.1em}{0ex}}\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}\hfill \\ \hfill \phantom{\rule{2.5em}{0ex}}\frac{y+6}{{y}^{2}-36}\hfill & & & \hfill \phantom{\rule{4.65em}{0ex}}\frac{y+6}{{y}^{2}-36}=y+1\hfill \\ \hfill \phantom{\rule{2.5em}{0ex}}\frac{1}{n-3}+\frac{1}{n+4}\hfill & & & \hfill \phantom{\rule{5em}{0ex}}\frac{1}{n-3}+\frac{1}{n+4}=\frac{15}{{n}^{2}+n-12}\hfill \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:

. .
We multiplied both sides by the LCD. .
Then we distributed. .
We simplified—and then we had an equation with no fractions. .
Finally, we solved that equation. .
.

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.

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.

How to Solve Equations with Rational Expressions

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

Solution

The above image has 3 columns. It shows the steps to find an extraneous solution to a rational equation for the example 1 divided by x plus one-third equals five-sixths. Step one is to note any value of the variable that would make any denominator zero. If x equals 0, then I divided by x is undefined. So we’ll write x divided zero next to the equation to get 1 divided by x plus one-third equals five-sixths times x divided by zero.Step two is to find the least common denominator of all denominators in the equation. Find the LCD of 1 divided by x one-third, and five-sixths. The x is 6 x.Step three is to clear the fractions by multiplying both sides of the equation by the LCD. Multiply both sides of the equation by the LCD, 6 x to get 6 times 1 divided by x plus one-third equals 6 x times five-sixths. Use the Distributive Property to get 6 x times 1 divided by x plus 6 x times one-third equals 6 x times five-sixths. Simplify – and notice, no more fractions and we have 6 plus 2 x equals 5 x.Step 4 is to solve the resulting equation. Simplify to get 6 equals 3 x and 2 equals x.Step 5 is to check. If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation. We did not get 0 as an algebraic solution. We substitute x equals 2 into the original equation to get one-half plus one-third equals five-sixths, then three-sixths plus two-sixths equals five-sixths and finally, five-sixths equal five-sixths.

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

-\frac{15}{7}

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

\frac{15}{13}

The steps of this method are shown below.

Solve equations with rational expressions.
  1. Note any value of the variable that would make any denominator zero.
  2. Find the least common denominator of all denominators in the equation.
  3. Clear the fractions by multiplying both sides of the equation by the LCD.
  4. Solve the resulting equation.
  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.

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

Solution
.
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}}.

5,-3

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

6,-2

Solve: \frac{5}{3u-2}=\frac{3}{2u}.

Solution
.
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 is2u\left(3u-2\right).
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}.

-2

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

-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}.

Solution
.
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\left(p+2\right)\left(p-2\right).
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\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}-2 as algebraic solutions.
.

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

\frac{2}{3}

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

2

Solve: \frac{4}{q-4}-\frac{3}{q-3}=1.

Solution
.
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\left(q-4\right)\left(q-3\right).
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.

-1,-2

Solve: \frac{3}{x+8}-\frac{2}{x-2}=1.

-2,-3

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

Solution
.
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\left(m-4\right)\left(m-1\right).
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}.

no solution

Solve: \frac{y-14}{{y}^{2}+3y-4}=\frac{2}{y+4}+\frac{7}{y-1}.

no solution

The equation we solved in (Figure) 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}.

Solution
.
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 is12n.
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}.

-2

Solve: \frac{y+5}{5y}+\frac{y}{15}=\frac{1}{y}.

-3

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

Solution
.
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\left(y-6\right)\left(y+6\right).
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.

-4,3

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

-7,8

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

Solution
.
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\left(x-1\right)\left(x+1\right)
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{2{y}^{2}-19y+54}{15{y}^{2}-60}.

no solution

Solve: \frac{z}{2z+8}-\frac{3}{4z-8}=\frac{3{z}^{2}-16z-6}{8{z}^{2}+8z-64}.

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\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}T.

Solution
.
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.

w=\frac{A}{l}

Solve: \frac{F}{A}=M for A.

A=\frac{F}{M}

(Figure) 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}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}y.

Solution
.
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.

x=\frac{3y-8}{2}

Solve: x=\frac{y}{1-y} for y.

y=\frac{x}{1+x}

Be sure to follow all the steps in (Figure). 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\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}c.

Solution
.
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\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}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.

a=\frac{b}{cb-1}

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

y=\frac{3x}{6+x}

Key Concepts

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

Practice Makes Perfect

Solve Rational Equations

In the following exercises, solve.

\frac{1}{a}+\frac{2}{5}=\frac{1}{2}

10

\frac{5}{6}+\frac{3}{b}=\frac{1}{3}

\frac{5}{2}-\frac{1}{c}=\frac{3}{4}

\frac{4}{7}

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

\frac{4}{5}+\frac{1}{4}=\frac{2}{v}

\frac{40}{21}

\frac{3}{7}+\frac{2}{3}=\frac{1}{w}

\frac{7}{9}+\frac{1}{x}=\frac{2}{3}

-9

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

1-\frac{2}{m}=\frac{8}{{m}^{2}}

-2,4

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

1+\frac{9}{p}=\frac{-20}{{p}^{2}}

-5,-4

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

\frac{1}{r+3}=\frac{4}{2r}

-6

\frac{3}{t-6}=\frac{1}{t}

\frac{5}{3v-2}=\frac{7}{4v}

14

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

\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{{x}^{2}-16}

\frac{-4}{5}

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

\frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{{z}^{2}-100}

-\frac{1}{3}

\frac{9}{a+11}+\frac{6}{a-11}=\frac{7}{{a}^{2}-121}

\frac{1}{q+4}-\frac{7}{q-2}=1

\text{no solution}

\frac{3}{r+10}-\frac{4}{r-4}=1

\frac{1}{t+7}-\frac{5}{t-5}=1

-5,-1

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

\frac{v-10}{{v}^{2}-5v+4}=\frac{3}{v-1}-\frac{6}{v-4}

\text{no solution}

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

\frac{x-10}{{x}^{2}+8x+12}=\frac{3}{x+2}+\frac{4}{x+6}

\text{no solution}

\frac{y-3}{{y}^{2}-4y-5}=\frac{1}{y+1}+\frac{8}{y-5}

\frac{z}{16}+\frac{z+2}{4z}=\frac{1}{2z}

-4

\frac{a}{9}+\frac{a+3}{3a}=\frac{1}{a}

\frac{b+3}{3b}+\frac{b}{24}=\frac{1}{b}

-8

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

\frac{d}{d+3}=\frac{18}{{d}^{2}-9}+4

2

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

\frac{n}{n+2}=\frac{8}{{m}^{2}-4}+3

1

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

\frac{q}{3q-9}-\frac{3}{4q+12}
\phantom{\rule{1.5em}{0ex}}=\frac{7{q}^{2}+6q+63}{24{q}^{2}-216}

\text{no solution}

\frac{r}{3r-15}-\frac{1}{4r+20}
\phantom{\rule{1.5em}{0ex}}=\frac{3{r}^{2}+17r+40}{12{r}^{2}-300}

\frac{s}{2s+6}-\frac{2}{5s+5}
\phantom{\rule{1.5em}{0ex}}=\frac{5{s}^{2}-3s-7}{10{s}^{2}+40s+30}

\text{no solution}

\frac{t}{6t-12}-\frac{5}{2t+10}
\phantom{\rule{1.5em}{0ex}}=\frac{{t}^{2}-23t+70}{12{t}^{2}+36t-120}

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

\frac{C}{r}=2\pi \phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}r

r=\frac{C}{2\pi }

\frac{I}{r}=P\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}r

\frac{V}{h}=lw\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}h

h=\frac{v}{lw}

\frac{2A}{b}=h\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}b

\frac{v+3}{w-1}=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}w

w=2v+7

\frac{x+5}{2-y}=\frac{4}{3}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}y

a=\frac{b+3}{c-2}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}c

c=\frac{b+3+2a}{a}

m=\frac{n}{2-n}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}n

\frac{1}{p}+\frac{2}{q}=4\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}p

p=\frac{q}{4q-2}

\frac{3}{s}+\frac{1}{t}=2\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}s

\frac{2}{v}+\frac{1}{5}=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}v

w=\frac{15v}{10+v}

\frac{6}{x}+\frac{2}{3}=\frac{1}{y}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}y

\frac{m+3}{n-2}=\frac{4}{5}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}n

n=\frac{5m+23}{n}

\frac{E}{c}={m}^{2}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}c

\frac{3}{x}-\frac{5}{y}=\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}y

y=\frac{20x}{12-x}

\frac{R}{T}=W\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}T

r=\frac{s}{3-t}\phantom{\rule{0.2em}{0ex}}\text{for}\phantom{\rule{0.2em}{0ex}}t

t=\frac{3r-s}{r}

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

Everyday Math

House Painting Alain can paint a house in 4 days. Spiro would take 7 days to paint the same house. Solve the equation \frac{1}{4}+\frac{1}{7}=\frac{1}{t} for t to find the number of days it would take them to paint the house if they worked together.

2\frac{6}{11} days

Boating Ari can drive his boat 18 miles with the current in the same amount of time it takes to drive 10 miles against the current. If the speed of the boat is 7 knots, solve the equation \frac{18}{7+c}=\frac{10}{7-c} for c to find the speed of the current.

Writing Exercises

Why is there no solution to the equation \frac{3}{x-2}=\frac{5}{x-2}?

Answers will vary.

Pete thinks the equation \frac{y}{y+6}=\frac{72}{{y}^{2}-36}+4 has two solutions, y=-6\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}y=4. Explain why Pete is wrong.

Self Check

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

This table has three rows and four columns. The first row is a header row and it labels each column. The first column is labeled "I can …", the second "Confidently", the third “With some help” and the last "No–I don’t get it". In the “I can…” column the next row reads “solve rational equations”. The next row reads, “solve rational equations for a specific variable”. The remaining columns are blank.

After reviewing this checklist, what will you do to become confident for all objectives?

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.

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Elementary Algebra by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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