12.13: Applications Using Rational Equations
What if you took an airplane trip? The current of the jet stream is 100 miles per hour. It took you the same amount of time to travel 3000 miles without the jet stream as it did to travel 2000 miles with the jet stream. How could you determine the speed of the plane in calm air? After completing this Concept, you'll be able to solve real-world applications like this one using rational equations.
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CK-12 Foundation: 1213S Solve Applications Using Rational Equations
Guidance
A motion problem with no acceleration is described by the formula \begin{align*}distance = speed \times time\end{align*}. 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.
Solution
Define variables:
Let \begin{align*}s =\end{align*} speed of the canoe in still water
Then, \begin{align*}s + 3 =\end{align*} the speed of the canoe traveling downstream
\begin{align*}s - 3 =\end{align*} the speed of the canoe traveling upstream
Construct a table:
Direction | Distance (miles) | Rate | Time |
---|---|---|---|
Downstream | 12 | \begin{align*}s + 3\end{align*} | \begin{align*}t\end{align*} |
Upstream | 3 | \begin{align*}s - 3\end{align*} | \begin{align*}t\end{align*} |
Write an equation:
Since \begin{align*}distance = rate \times time\end{align*}, we can say that \begin{align*}time=\frac{distance}{rate}\end{align*}.
\begin{align*}\text{The time to go downstream is:} \qquad \qquad \qquad \qquad \qquad \qquad \quad \qquad \qquad \qquad \qquad \qquad t=\frac{12}{s+3}\!\\ \\ \text{The time to go upstream is:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad \qquad \qquad \qquad \qquad t=\frac{3}{s-3}\!\\ \\ \text{Since the time it takes to go upstream and downstream are the same, we have:} \qquad \frac{3}{s-3}=\frac{12}{s+3}\end{align*}
Solve the equation:
\begin{align*}\text{Cross-multiply:} \qquad \qquad \qquad \qquad 3(s+3)=12(s-3)\!\\ \\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \ \ 3s+9=12s-36\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad s=5 \ mi/h\end{align*}
Check: Upstream: \begin{align*}t=\frac{12}{8}=1 \frac{1}{2} \ hour\end{align*}; downstream: \begin{align*}t=\frac{3}{2}=1 \frac{1}{2} \ hour\end{align*}. 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?
Solution
Define variables:
Let \begin{align*}d =\end{align*} 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*} | 8 | \begin{align*}t_1\end{align*} |
Downhill | \begin{align*}40 - d\end{align*} | 14 | \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}{8}\!\\ \\ \text{The time to go downhill is:} \qquad \qquad \qquad \qquad \qquad t_2=\frac{40-d}{14}\!\\ \\ \text{We also know that the total time is} \ 4 \ \text{hours:} \qquad \ \frac{d}{8}+\frac{40-d}{14}=4\end{align*}
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \qquad \qquad \qquad \qquad \ \text{LCD}=56\!\\ \\ \text{Multiply all terms by the common denominator:} \qquad \qquad 7d+160-4d=224\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ \qquad \qquad \qquad d=21.3 \ mi\end{align*}
Check: Uphill: \begin{align*}t=\frac{21.3}{8}=2.67 \ hours\end{align*}; downhill: \begin{align*}t=\frac{40-21.3}{14}=1.33 \ hours\end{align*}. The answer checks out.
Example C
A group of friends decided to pool together 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?
Solution
Define variables:
Let \begin{align*}x =\end{align*} 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*} | 200 | \begin{align*}\frac{200}{x}\end{align*} |
Later group | \begin{align*}x - 12\end{align*} | 200 | \begin{align*}\frac{200}{x-12}\end{align*} |
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*}
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \qquad \quad \text{LCD} =x(x-12)\!\\ \\ \text{Multiply all terms by the LCD:} \qquad \qquad \qquad 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:} \qquad \qquad \ 200x=200(x-12)+15x(x-12)\!\\ \\ \text{Eliminate parentheses:} \qquad \qquad \qquad \qquad \qquad \ 200x=200x-2400+15x^2-180x\!\\ \\ \text{Get all terms on one side of the equation:} \qquad 0=15x^2=180x-2400\!\\ \\ \text{Divide all terms by} \ 15: \qquad \qquad \qquad \qquad \quad \ \ 0=x^2-12x-160\!\\ \\ \text{Factor:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 0=(x-20)(x+8)\!\\ \\ \text{Solve:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \quad \ x=20, x=-8\end{align*}
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
Vocabulary
- Since \begin{align*}distance = rate \times time\end{align*}, we can say that \begin{align*}time=\frac{distance}{rate}\end{align*}.
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 is running through the hilly streets of San Francisco and the total distance she travels is 13.5 miles and the total time she run 4.5 hours, how long did she run uphill and how long did she run downhill?
Solution
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.
Practice
For 1-4, solve for \begin{align*}x\end{align*}.
- \begin{align*}\frac{3x^2+2x-1}{x^2-1}=-2\end{align*}
- \begin{align*}x+\frac{1}{x}=2\end{align*}
- \begin{align*}-3+\frac{1}{x+1}=\frac{2}{x}\end{align*}
- \begin{align*}\frac{1}{x}-\frac{x}{x-2}=2\end{align*}
For 5-10, solve the following applications.
- 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?
- 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 current.
- 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?
- 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?
- 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?
- 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?
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Here you'll learn how to solve distance problems and other real-world applications that involve rational equations.