Solve Rational Equations

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.

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

\begin{matrix} Rational Expression & 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} \end{matrix}

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.

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

How to Solve Equations with Rational Expressions

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

Solution

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 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.

The steps of this method are shown below.

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

Solution

	$A mathematical equation is displayed, showing the fraction 5 over the expression 3u minus 2, set equal to the fraction 3 over 2u.$
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 $2 u (3 u - 2)$ .
Clear the fractions by multiplying both sides of the equation by the LCD.	$An algebraic equation showing 2u(3u-2) multiplied by a fraction on both sides of the equality, with the fractions being 5/(3u-2) on the left and 3/(2u) on the right.$
Remove common factors.
Simplify.
Multiply.
Solve the resulting equation.
We did not get 0 or $\frac{2}{3}$ as algebraic solutions.

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

	$A mathematical equation is displayed, showing a sum of two fractions on the left side equal to a fraction on the right side. The equation is (2/(p+2)) + (4/(p-2)) = (p-1)/(p^2-4).$
Note any value of the variable that would make any denominator zero.	$An algebraic equation is shown, featuring the sum of two fractions, 2/(p+2) and 4/(p-2), equaling a third fraction, (p-1)/((p+2)(p+2)), with the conditions p not equal to -2, p not equal to 2.$
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.	$An algebraic equation showing (p+2)(p-2) multiplied by the sum of two fractions on the left side, equaling (p+2)(p-2) multiplied by a single fraction on the right side. The common multiplier (p+2)(p-2) is in red.$
Distribute.	$A mathematical equation where a common factor, (p+2)(p-2), highlighted in red, is multiplied across three terms involving fractions. This step is typically used to clear denominators in rational equations.$
Remove common factors.
Simplify.
Distribute.
Solve.


We did not get $2 or - 2$ as algebraic solutions.

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

Solution


Note any value of the variable that would make any denominator zero.	$A mathematical equation showing the sum of two fractions equaling one, with specified restrictions for the variable q. The equation is 4/(q-4) + 3/(q-3) = 1, where q is not equal to 4 or 3.$
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.	$An algebraic equation showing a step where a rational expression is multiplied by the common denominator (q-4)(q-3) on both sides to clear the fractions.$
Distribute.
Remove common factors.	$A mathematical equation illustrating the simplification of algebraic expressions by canceling common factors. The image shows (q-4) and (q-3) being canceled out in the numerators and denominators of the fractional terms.$
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: $\frac{m + 11}{m^{2} - 5 m + 4} = \frac{5}{m - 4} - \frac{3}{m - 1} .$

Solution


Factor all the denominators, so we can note any value of the variable the would make any denominator zero.	$The image displays a complex algebraic fraction (m+11)/((m-4)(m-1)) rewritten as the difference of two simpler fractions: 5/(m-4) - 3/(m-1), with restrictions m!=4 and m!=1.$
Find the least common denominator of all denominators in the equation. The LCD is $(m - 4) (m - 1)$ .
Clear the fractions.	$An algebraic equation showing both sides multiplied by (m-4)(m-1) to clear denominators, simplifying the expression involving fractions with variables m.$
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.

The equation we solved in Example 6 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: $\frac{n}{12} + \frac{n + 3}{3 n} = \frac{1}{n} .$

Solution

	$A mathematical equation is displayed on a white background. The equation is n/12 + (n+3)/(3n) = 1/n, featuring fractions with variables in both numerators and denominators.$
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 $12 n$ .
Clear the fractions by multiplying both sides of the equation by the LCD.	$A mathematical equation is shown with 12n multiplying a sum of fractions (n/12 + (n+3)/(3n)) on the left, and 12n multiplying a fraction (1/n) on the right, all in black text with '12n' in red.$
Distribute.
Remove common factors.
Simplify.
Solve the resulting equation.



Check.
$n = 0$ is an extraneous solution.

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 $(y - 6) (y + 6)$ .
Clear the fractions.	$A mathematical equation demonstrating the process of clearing fractions by multiplying both sides of the equation by the least common denominator, (y-6)(y+6).$
Simplify.
Simplify.
Solve the resulting equation.




Check.
$y = - 6$ is an extraneous solution.

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

Solution


We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD.	$A mathematical equation shows the subtraction of two algebraic fractions set equal to a third algebraic fraction. The equation is x over 2(x-1) minus 2 over 3(x+1) equals (5x^2 - 2x + 9) over 12(x-1)(x+1).$
Note any value of the variable that would make any denominator zero.	$A mathematical equation featuring fractions: x/(2(x-1)) - 2/(3(x+1)) = (5x^2 - 2x + 9)/(12(x-1)(x+1)), with the conditions x ≠ 1 and x ≠ -1.$
Find the least common denominator.The LCD is $12 (x - 1) (x + 1)$
Clear the fractions.	$An algebraic equation showing the multiplication of both sides by 12(x-1)(x+1) to eliminate fractional terms, a common step in solving rational equations.$
Simplify.
Simplify.
Solve the resulting equation.



Check.
$x = 1$ and $x = - 1$ are extraneous solutions. The equation has 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: $\frac{D}{T} = R for T .$

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.	$A mathematical equation shows D/R = RT/R, with the variable R in the denominator of both fractions highlighted in red, indicating a step in solving or manipulating the equation.$
Simplify.

Example 11 uses the formula for slope that we used to get the point-slope form of an equation of a line.

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

Solution


Note any value of the variable that would make any denominator zero.	$A mathematical equation displays the slope m as a fraction: m = (x-2) / (y-3), with the condition that y is not equal to 3.$
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.	$A mathematical equation shows 'ym over m equals x minus 2 plus 3m over m' with the variable 'm' in the denominator highlighted in red. The numerator of the first fraction is 'ym', and the denominator is 'm'. The numerator of the second fraction is 'x - 2 + 3m', and the denominator is 'm'.$
Simplify.

Be sure to follow all the steps in Example 12. It may look like a very simple formula, but we cannot solve it instantly for either denominator.

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

Solution

	$A mathematical equation showing the sum of two fractions, 1 over c plus 1 over m, equaling 1, with the instruction to solve for c.$
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, $c m$ .
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 $m \neq 1$ .

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.

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

Solution

$10$

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

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

Solution

$\frac{4}{7}$

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

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

Solution

$\frac{40}{21}$

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

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

Solution

$−9$

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

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

Solution

$−2, 4$

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

$1 + \frac{9}{p} = \frac{−20}{p^{2}}$

Solution

$−5, −4$

$1 - \frac{7}{q} = \frac{−6}{q^{2}}$

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

Solution

$−6$

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

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

Solution

$14$

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

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

Solution

$\frac{−4}{5}$

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

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

Solution

$- \frac{1}{3}$

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

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

Solution

$- 2, - 1$

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

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

Solution

$−5, −1$

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

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

Solution

$no solution$

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

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

Solution

$no solution$

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

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

Solution

$−4$

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

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

Solution

$−8$

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

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

Solution

$2$

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

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

Solution

$1$

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

$\frac{q}{3 q - 9} - \frac{3}{4 q + 12}$
$= \frac{7 q^{2} + 6 q + 63}{24 q^{2} - 216}$

Solution

$no solution$

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

$\frac{s}{2 s + 6} - \frac{2}{5 s + 5}$
$= \frac{5 s^{2} - s - 18}{10 s^{2} + 40 s + 30}$

Solution

$no solution$

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

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

$\frac{C}{r} = 2 π for r$

Solution

$r = \frac{C}{2 π}$

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

$\frac{V}{h} = l w for h$

Solution

$h = \frac{v}{l w}$

$\frac{2 A}{b} = h for b$

$\frac{v + 3}{w - 1} = \frac{1}{2} for w$

Solution

$w = 2 v + 7$

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

$a = \frac{b + 3}{c - 2} for c$

Solution

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

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

$\frac{1}{p} + \frac{2}{q} = 4 for p$

Solution

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

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

$\frac{2}{v} + \frac{1}{5} = \frac{3}{w} for w$

Solution

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

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

$\frac{m + 3}{n - 2} = \frac{4}{5} for n$

Solution

$n = \frac{5 m + 23}{4}$

$\frac{E}{c} = m^{2} for c$

$\frac{3}{x} - \frac{5}{y} = \frac{1}{4} for y$

Solution

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

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

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

Solution

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

$c = \frac{2}{a} + \frac{b}{5} for a$

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.

Solution

$2 \frac{6}{11}$ 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 $\frac{3}{x - 2} = \frac{5}{x - 2}$ ?

Solution

Answers will vary.

Pete thinks the equation $\frac{y}{y + 6} = \frac{72}{y^{2} - 36} + 4$ has two solutions, $y = −6 and y = 4$ . 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?

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.