12.11: Applications of Adding and Subtracting Rational Expressions
What if it took you 3 hours to clean your house by yourself and it took your brother 2 hours to clean it by himself? How long would it take both of you to clean the house if you were working together? After completing this Concept, you'll be able to solve applications like this one involving addition and subtraction of rational expressions.
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CK-12 Foundation: 1211S Solve Applications by Adding and Subtracting Rational Expressions
Guidance
In the previous concept, you learned how to add and subtract rational expressions. In this concept, you will use those tools to solve real-world problems.
Example A
In an electrical circuit with two resistors placed in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistance: \begin{align*}\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}\end{align*}
Solution
\begin{align*}\text{Let's simplify the expression} \ \frac{1}{R_1}+\frac{1}{R_2}.\!\\ \!\\ \text{The lowest common denominator is} \ R_1R_2, \ \text{so we multiply the first fraction by} \ \frac{R_2}{R_2} \ \text{and the}\!\\ \!\\ \text{second fraction by} \ \frac{R_1}{R_1}: \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \frac{R_2}{R_2} \cdot \frac{1}{R_1} + \frac{R_1}{R_1} \cdot \frac{1}{R_2}\!\\ \!\\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \frac{R_2 + R_1}{R_1R_2}\!\\ \!\\ \text{The total resistance is the reciprocal of this expression:} \qquad R_{tot}=\frac{R_1R_2}{R_1+R_2} \quad \mathbf{Answer}\end{align*}
Example B
The sum of a number and its reciprocal is \begin{align*}\frac{53}{14}\end{align*}. Find the numbers.
Solution
Define variables:
Let \begin{align*}x\end{align*} be the number; then its reciprocal is \begin{align*}\frac{1}{x}\end{align*}.
Set up an equation:
The equation that describes the relationship between the numbers is \begin{align*}x+\frac{1}{x}=\frac{53}{14}\end{align*}
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \ \qquad \text{LCM} = 14x\!\\ \!\\ \text{Multiply all terms by} \ 14x: \qquad \qquad \qquad \quad \ 14x \cdot x + 14x \cdot \frac{1}{x}=14x \cdot \frac{53}{14}\end{align*}
(Notice that we’re multiplying the terms by \begin{align*}14x\end{align*} instead of by \begin{align*}\frac{14x}{14x}\end{align*}. We can do this because we’re multiplying both sides of the equation by the same thing, so we don’t have to keep the actual values of the terms the same. We could also multiply by \begin{align*}\frac{14x}{14x}\end{align*}, but then the denominators would just cancel out a couple of steps later.)
\begin{align*}\text{Cancel common factors in each term:} \qquad \qquad \ 14x \cdot x + 14x \cdot \frac{1}{x} = 14x \cdot \frac{53}{14}\!\\ \!\\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad 14x^2 + 14 = 53x\!\\ \!\\ \text{Write all terms on one side of the equation:} \qquad 14x^2 - 53x + 14 = 0\!\\ \!\\ \text{Factor:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (7x-2)(2x-7) = 0\!\\ \!\\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad x=\frac{2}{7} \ \text{and} \ x=\frac{7}{2}\end{align*}
Notice there are two answers for \begin{align*}x\end{align*}, but they are really parts of the same solution. One answer represents the number and the other answer represents its reciprocal.
Check:
\begin{align*}\frac{2}{7}+\frac{7}{2}=\frac{4+49}{14}=\frac{53}{14}\end{align*}. The answer checks out.
Work problems are problems where two people or two machines work together to complete a job. Work problems often contain rational expressions. Typically we set up such problems by looking at the part of the task completed by each person or machine. The completed task is the sum of the parts of the tasks completed by each individual or each machine.
To determine the part of the task completed by each person or machine we use the following fact:
\begin{align*}\text{Part of the task completed} = \text{rate of work} \times \text{time spent on the task}\end{align*}
It’s usually useful to set up a table where we can list all the known and unknown variables for each person or machine and then combine the parts of the task completed by each person or machine at the end.
Example C
Mary can paint a house by herself in 12 hours. John can paint a house by himself in 16 hours. How long would it take them to paint the house if they worked together?
Solution
Define variables:
Let \begin{align*}t =\end{align*} the time it takes Mary and John to paint the house together.
Construct a table:
Since Mary takes 12 hours to paint the house by herself, in one hour she paints \begin{align*}\frac{1}{12}\end{align*} of the house.
Since John takes 16 hours to pain the house by himself, in one hour he paints \begin{align*}\frac{1}{16}\end{align*} of the house.
Mary and John work together for \begin{align*}t\end{align*} hours to paint the house together. Using
\begin{align*}Part \ of \ the \ task \ completed = rate \ of \ work \cdot time \ spent \ on \ the \ task\end{align*}
we can write that Mary completed \begin{align*}\frac{t}{12}\end{align*} of the house and John completed \begin{align*}\frac{t}{16}\end{align*} of the house in this time.
This information is nicely summarized in the table below:
Painter | Rate of work (per hour) | Time worked | Part of task |
---|---|---|---|
Mary | \begin{align*}\frac{1}{12}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{12}\end{align*} |
John | \begin{align*}\frac{1}{16}\end{align*} | \begin{align*}t\end{align*} | \begin{align*}\frac{t}{16}\end{align*} |
Set up an equation:
In \begin{align*}t\end{align*} hours, Mary painted \begin{align*}\frac{t}{12}\end{align*} of the house and John painted \begin{align*}\frac{t}{16}\end{align*} of the house, and together they painted 1 whole house. So our equation is \begin{align*}\frac{t}{12}+\frac{t}{16}=1\end{align*}.
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \qquad \qquad \quad \qquad \text{LCM} = 48\!\\ \\ \text{Multiply all terms in the equation by the LCM:} \qquad \ \ 48 \cdot \frac{t}{12}+48 \cdot \frac{t}{16}=48 \cdot 1\!\\ \\ \text{Cancel common factors in each term:} \qquad \qquad \qquad \quad \ 4 \cdot \frac{t}{1}+3 \cdot \frac{t}{1}=48 \cdot 1\!\\ \\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4t+3t=48\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 7t=48 \Rightarrow t=\frac{48}{7}=6.86 \ hours\end{align*}
Check: The answer is reasonable. We’d expect the job to take more than half the time Mary would take by herself but less than half the time John would take, since Mary works faster than John.
Watch this video for help with the Examples above.
CK-12 Foundation: Solve Applications by Adding and Subtracting
Vocabulary
- The total resistance can be found using the expression: \begin{align*} R_{tot}=\frac{R_1R_2}{R_1+R_2} \end{align*}
Guided Practice
Suzie and Mike take two hours to mow a lawn when they work together. It takes Suzie 3.5 hours to mow the same lawn if she works by herself. How long would it take Mike to mow the same lawn if he worked alone?
Solution
Define variables:
Let \begin{align*}t =\end{align*} the time it takes Mike to mow the lawn by himself.
Construct a table:
Painter | Rate of work (per hour) | Time worked | Part of Task |
---|---|---|---|
Suzie | \begin{align*}\frac{1}{3.5}=\frac{2}{7}\end{align*} | 2 | \begin{align*}\frac{4}{7}\end{align*} |
Mike | \begin{align*}\frac{1}{t}\end{align*} | 2 | \begin{align*}\frac{2}{t}\end{align*} |
Set up an equation:
Since Suzie completed \begin{align*}\frac{4}{7}\end{align*} of the lawn and Mike completed \begin{align*}\frac{2}{t}\end{align*} of the lawn and together they mowed the lawn in 2 hours, we can write the equation: \begin{align*}\frac{4}{7}+\frac{2}{t}=1\end{align*}
Solve the equation:
\begin{align*}\text{Find the lowest common denominator:} \qquad \qquad \quad \qquad \text{LCM} = 7t\!\\ \\ \text{Multiply all terms in the equation by the LCM:} \qquad \ \ 7t \cdot \frac{4}{7}+7t \cdot \frac{2}{t}=7t \cdot 1\!\\ \\ \text{Cancel common factors in each term:} \qquad \qquad \qquad \quad \ t \cdot \frac{4}{1}+7 \cdot \frac{2}{1}=7t \cdot 1\!\\ \\ \text{Simplify:} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 4t+14=7t\!\\ \\ {\;} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 3t=14 \Rightarrow t=\frac{14}{3}=4 \frac{2}{3} \ hours\end{align*}
Check: The answer is reasonable. We’d expect Mike to work slower.
Practice
For 1-5, perform the indicated operation. Leave the denominator in factored form.
- \begin{align*}\frac{4x}{x+1}-\frac{2}{2(x+1)}\end{align*}
- \begin{align*}\frac{10}{21}+\frac{9}{35}\end{align*}
- \begin{align*}\frac{2x}{x-4}+\frac{x}{4-x}\end{align*}
- \begin{align*}\frac{5}{2x+3}-3\end{align*}
- \begin{align*}\frac{5x+1}{x+4}+2\end{align*}
For 6-8, find the missing resistance.
- \begin{align*}R_1=4, R_2=6, R_{tot}=?\end{align*}
- \begin{align*}R_1=1, R_2=?, R_{tot}=\frac{2}{3}\end{align*}
- \begin{align*}R_1=?, R_2=12, R_{tot}=\frac{36}{15}\end{align*}
Solve the following work problems.
- Andrea can wash the windows on their house in 30 minutes and Jorge can wash the windows on their house in 40 minutes. How long will it take for them to wash the windows together?
- A pool can be filled by one pipe in 5 hours and by a different pipe in 7 hours. How long will it take the pool to fill using both pipes?
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Here you'll learn how to solve circuitry and other real-world applications that involve adding and subtracting rational expressions.