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12.13: Applications Using Rational Equations

Difficulty Level: At Grade Created by: CK-12
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Practice Applications Using Rational Equations
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Consider a flying airplane. If it takes the airplane the same amount of time to travel 3000 miles in calm air as to travel 2000 miles against a 100 mph wind, could you determine the speed (in mph) of the plane in calm air? After completing this Concept, you'll be able to use rational equations to solve real-world applications like this.

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CK-12 Foundation: 1213S Solve Applications Using Rational Equations


A motion problem with no acceleration is described by the formula \begin{align*}distance = speed \times time\end{align*}distance=speed×time. Since speed is a measure of rate, the formula may also appear as \begin{align*}distance = rate \times time\end{align*}distance=rate×time. These problems can involve the addition and subtraction of rational expressions.

Example A

Last weekend Nadia went canoeing on the Snake River. The current of the river is three miles per hour. It took Nadia the same amount of time to travel 12 miles downstream as it did to travel 3 miles upstream. Determine how fast Nadia’s canoe would travel in still water.


Define variables:

Let \begin{align*}s =\end{align*}s= speed (rate of travel) of the canoe in still water

Then, \begin{align*}s + 3 =\end{align*}s+3= the speed of the canoe traveling downstream

\begin{align*}s - 3 =\end{align*}s3= the speed of the canoe traveling upstream

Construct a table:

Direction Distance (miles) Rate Time
Downstream 12 \begin{align*}s + 3\end{align*}s+3 \begin{align*}t\end{align*}t
Upstream 3 \begin{align*}s - 3\end{align*}s3 \begin{align*}t\end{align*}t

Write an equation:

Since \begin{align*}distance = rate \times time\end{align*}distance=rate×time, we can say that \begin{align*}time=\frac{distance}{rate}\end{align*}time=distancerate.

\begin{align*}\text{The time to go downstream is:} t=\frac{12}{s+3} \\ \\ \text{The time to go upstream is:} t=\frac{3}{s-3} \\ \\ \text{Since the time it takes to go upstream and downstream are the same, we have:} \frac{3}{s-3}=\frac{12}{s+3}\end{align*}The time to go downstream is:t=12s+3The time to go upstream is:t=3s3Since the time it takes to go upstream and downstream are the same, we have:3s3=12s+3

Solve the equation:

\begin{align*}\text{Cross-multiply:} 3(s+3)=12(s-3) \\ \\ \text{Simplify:} 3s+9=12s-36 \\ \\ \text{Solve:} s=5 \ mi/h\end{align*}Cross-multiply:3(s+3)=12(s3)Simplify:3s+9=12s36Solve:s=5 mi/h

Check: Upstream: \begin{align*}t=\frac{12}{8}=1 \frac{1}{2} \ hour\end{align*}t=128=112 hour; downstream: \begin{align*}t=\frac{3}{2}=1 \frac{1}{2} \ hour\end{align*}t=32=112 hour. The answer checks out.

Example B

Peter rides his bicycle. When he pedals uphill he averages a speed of eight miles per hour, when he pedals downhill he averages 14 miles per hour. If the total distance he travels is 40 miles and the total time he rides is four hours, how long did he ride at each speed?


Define variables:

Let \begin{align*}d =\end{align*}d= distance Peter bikes uphill at 8 miles per hour.

Construct a table:

Direction Distance (miles) Rate (mph) Time (hours)
Uphill \begin{align*}d\end{align*}d 8 \begin{align*}t_1\end{align*}t1
Downhill \begin{align*}40 - d\end{align*}40d 14 \begin{align*}t_2\end{align*}t2

Write an equation:

We know that \begin{align*}time=\frac{distance}{rate}\end{align*}time=distancerate.

\begin{align*}\text{The time to go uphill is:} && t_1=\frac{d}{8} \\ \\ \text{The time to go downhill is:} && t_2=\frac{40-d}{14} \\ \\ \text{We also know that the total time is} \ 4 \ \text{hours:} && \frac{d}{8}+\frac{40-d}{14}=4\end{align*}The time to go uphill is:The time to go downhill is:We also know that the total time is 4 hours:t1=d8t2=40d14d8+40d14=4

Solve the equation:

\begin{align*}\text{Find the lowest common denominator:} && \ \text{LCD}=56 \\ \\ \text{Multiply all terms by the common denominator:} && 7d+160-4d=224 \\ \\ \text{Solve:} && d=21.3 \ mi\end{align*}Find the lowest common denominator:Multiply all terms by the common denominator:Solve: LCD=567d+1604d=224d=21.3 mi

Check: Uphill: \begin{align*}t=\frac{21.3}{8}=2.67 \ hours\end{align*}t=21.38=2.67 hours; downhill: \begin{align*}t=\frac{40-21.3}{14}=1.33 \ hours\end{align*}t=4021.314=1.33 hours. The answer checks out.

Example C

A group of friends decided to combine their money and buy a birthday gift that cost $200. Later 12 of the friends decided not to participate any more. This meant that each person paid $15 more than their original share. How many people were in the group to begin with?


Define variables:

Let \begin{align*}x =\end{align*}x= the number of friends in the original group.

Make a table:

Number of people Gift price Share amount
Original group \begin{align*}x\end{align*}x 200 \begin{align*}\frac{200}{x}\end{align*}200x
Later group \begin{align*}x - 12\end{align*}x12 200 \begin{align*}\frac{200}{x-12}\end{align*}200x12

Write an equation:

Since each person’s share went up by $15 after 12 people refused to pay, we write the equation \begin{align*}\frac{200}{x-12}=\frac{200}{x}+15\end{align*}200x12=200x+15

Solve the equation:

\begin{align*}\text{Find the lowest common denominator:} && \text{LCD} =x(x-12) \\ \\ \text{Multiply all terms by the LCD:} && x(x-12) \cdot \frac{200}{x-12}=x(x-12) \cdot \frac{200}{x}+x(x-12) \cdot 15 \\ \\ \text{Cancel common factors and simplify:} && 200x=200(x-12)+15x(x-12) \\ \\ \text{Eliminate parentheses:} && 200x=200x-2400+15x^2-180x \\ \\ \text{Get all terms on one side of the equation:} && 0=15x^2=180x-2400 \\ \\ \text{Divide all terms by} \ 15: && 0=x^2-12x-160 \\ \\ \text{Factor:} && 0=(x-20)(x+8) \\ \\ \text{Solve:} && x=20, x=-8\end{align*}Find the lowest common denominator:Multiply all terms by the LCD:Cancel common factors and simplify:Eliminate parentheses:Get all terms on one side of the equation:Divide all terms by 15:Factor:Solve:LCD=x(x12)x(x12)200x12=x(x12)200x+x(x12)15200x=200(x12)+15x(x12)200x=200x2400+15x2180x0=15x2=180x24000=x212x1600=(x20)(x+8)x=20,x=8

The answer that makes sense is \begin{align*}x = \mathbf{20}\end{align*} people.

Check: Originally $200 shared among 20 people is $10 each. After 12 people leave, $200 shared among 8 people is $25 each. So each person pays $15 more. The answer checks out.

Watch this video for help with the Examples above.

CK-12 Foundation: Solving Applications Using Rational Equations

Guided Practice

Carrie is a runner. When she runs uphill she averages a speed of 2 miles per hour, when she runs downhill she averages 5 miles per hour. If she runs through the hilly streets of San Francisco for 4.5 hours and the total distance she travels is 13.5 miles, how long did she run uphill and how long did she run downhill?


Define variables:

Let \begin{align*}t_1 =\end{align*} time Carrie runs uphill.

Let \begin{align*}t_2 =\end{align*} time Carrie runs downhill.

Let \begin{align*}d =\end{align*} the distance Carrie runs uphill.

Construct a table:

Direction Distance (miles) Rate (mph) Time (hours)
Uphill \begin{align*}d\end{align*} 2 \begin{align*}t_1\end{align*}
Downhill \begin{align*}13.5 - d\end{align*} 5 \begin{align*}t_2\end{align*}

Write an equation:

We know that \begin{align*}time=\frac{distance}{rate}\end{align*}.

\begin{align*}\text{The time to go uphill is:} \qquad \qquad \qquad \qquad \qquad \quad t_1=\frac{d}{2}\!\\ \\ \text{The time to go downhill is:} \qquad \qquad \qquad \qquad \qquad t_2=\frac{13.5-d}{5}\!\\ \\ \text{We also know that the total time is} \ 4.5 \ \text{hours:} \qquad \ \frac{d}{2}+\frac{13.5-d}{5}=4.5\end{align*}

Solve the equation:

\begin{align*}\text{Find the lowest common denominator:} \qquad \qquad \qquad \qquad \ \text{LCD}=2\cdot 5=10\!\\ \\ \text{Multiply all terms by the common denominator:} \qquad \qquad 5d+27-2d=45\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \qquad \qquad \qquad d=6 \ mi\end{align*}

Since \begin{align*}d=6\end{align*} is the distance Carrie to ran uphill, then \begin{align*}13.5-d=13.5-6=7.5\end{align*} is the distance Carrie ran downhill.

Check: Uphill: \begin{align*}t=\frac{6}{2}=3 \ hours\end{align*}; downhill: \begin{align*}t=\frac{13.5-6}{5}=1.5 \ hours\end{align*}. The answer checks out.

Explore More

For 1-4, solve for \begin{align*}x\end{align*}.

  1. \begin{align*}\frac{3x^2+2x-1}{x^2-1}=-2\end{align*}
  2. \begin{align*}x+\frac{1}{x}=2\end{align*}
  3. \begin{align*}3+\frac{1}{x+1}=\frac{2}{x}\end{align*}
  4. \begin{align*}\frac{1}{x}-\frac{x}{x-2}=2\end{align*}

For 5-10, solve the following applications.

  1. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages 7 miles/hour and when he walks he averages 3.5 miles per hour. If he walks and jogs a total of 6 miles in a total of 1.2 hours, how far does he jog and how far does he walk?
  2. A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat’s speed in still water is 20 miles per hour. Find the speed of the water.
  3. Paul leaves San Diego driving at 50 miles per hour. Two hours later, his mother realizes that he forgot something and drives in the same direction at 70 miles per hour. How long does it take her to catch up to Paul?
  4. On a trip, an airplane flies at a steady speed against the wind and on the return trip the airplane flies with the wind. The airplane takes the same amount of time to fly 300 miles against the wind as it takes to fly 420 miles with the wind. The wind is blowing at 30 miles per hour. What is the speed of the airplane when there is no wind?
  5. A debt of $420 is shared equally by a group of friends. When five of the friends decide not to pay, the share of the other friends goes up by $25. How many friends were in the group originally?
  6. A non-profit organization collected $2250 in equal donations from their members to share the cost of improving a park. If there were thirty more members, then each member could contribute $20 less. How many members does this organization have?

Answers for Explore More Problems

To view the Explore More answers, open this PDF file and look for section 12.13. 


Rational Expression

Rational Expression

A rational expression is a fraction with polynomials in the numerator and the denominator.

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Difficulty Level:
At Grade
Date Created:
Oct 01, 2012
Last Modified:
Apr 11, 2016
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