module 6 solving rational equations assignment

General Mathematics Quarter 1 – Module 6: Solving Rational Equations and Inequalities

This module was designed and written for learners like you to determine a method and set of steps for solving rational equations and inequalities. Learners like you can also explore and develop new methods that you have synthesized and apply these techniques for performing operations with rational expressions.

In this module, you will able to explain the appropriate methods in solving rational equations and inequalities you used. You will also be able to check and explain extraneous solutions.

After going through this module, you are expected to:

1. Apply appropriate methods in solving rational equations and inequalities.

2. Solve rational equations and inequalities using algebraic techniques for simplifying and manipulating of expressions.

3. Determine whether the solutions found are acceptable for the problem by checking the solutions.

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7.6: Applications of Rational Equations

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Learning Objectives

• Solve applications involving relationships between real numbers.
• Solve applications involving uniform motion (distance problems).
• Solve work-rate applications.

Number Problems

Recall that the reciprocal of a nonzero number \(n\) is \(\frac{1}{n}\). For example, the reciprocal of \(5\) is \(\frac{1}{5}\) and \(5\cdot \frac{1}{5} = 1\). In this section, the applications will often involve the key word “reciprocal.” When this is the case, we will see that the algebraic setup results in a rational equation.

Example \(\PageIndex{1}\)

A positive integer is \(4\) less than another. The sum of the reciprocals of the two positive integers is \(\frac{10}{21}\). Find the two integers.

Begin by assigning variables to the unknowns.

Let \(n\) represent the larger positive integer.

Let \(n-4\) represent the smallest positive integer.

Next, use the reciprocals \(\frac{1}{n}\) and \(\frac{1}{n−4}\) to translate the sentences into an algebraic equation.

\(\begin{array}{ccc}{\color{Cerulean} { the\:sum\:of\:the\:reciprocals }} & {\color{Cerulean} { is }} \\ {\qquad \frac{1}{n}+\frac{1}{n-4}} & {=} & {\frac{10}{21}}\end{array}\)

We can solve this rational expression by multiplying both sides of the equation by the least common denominator (LCD). In this case, the LCD is \(21n(n−4)\).

\(\begin{array}{cl}{\frac{1}{n}+\frac{1}{n-4}=\frac{10}{21}}&{} \\ {\color{Cerulean}{21n(n-4)} \color{black}{\cdot \left(\frac{1}{n}+\frac{1}{n-4} \right)}=\color{Cerulean}{21n(n-4)}\color{black}{ \cdot\left(\frac{10}{21}\right)}}&{\color{Cerulean}{Multiply\:both\:sides}}\\{}&{\color{Cerulean}{by\:the\:LCD.}} \\ {\color{Cerulean}{21n (n-4)}\color{black}{ \cdot \frac{1}{n}+}\color{Cerulean}{21n (n-4)}\color{black}{ \cdot \frac{1}{n-4}}=\color{Cerulean}{21n (n-4)}\color{black}{ \cdot\left(\frac{10}{21}\right)}}&{\color{Cerulean}{Distribute\:and}}\\{}&{\color{Cerulean}{then\:cancel.}} \\ {21(n-4)+21n =10 n(n-4)}\end{array}\)

Solve the resulting quadratic equation.

\(\begin{array}{rlrl}{5 n-6} & {=0} & {\text { or }} & {n-7=0} \\ {5 n} & {=6} && {n=7} \\ {n} & {=\frac{6}{5}}\end{array}\)

The question calls for integers and the only integer solution is \(n=7\). Hence disregard \(\frac{6}{5}\). Use the expression \(n−4\) to find the smaller integer.

\(n-4=7-4=3\)

The two positive integers are \(3\) and \(7\). The check is left to the reader.

Example \(\PageIndex{2}\)

A positive integer is \(4\) less than another. If the reciprocal of the smaller integer is subtracted from twice the reciprocal of the larger, then the result is \(\frac{1}{30}\). Find the two integers.

Let \(n-4\) represent the smaller positive integer.

Set up an algebraic equation.

Solve this rational expression by multiplying both sides by the LCD. The LCD is \(30n(n−4)\).

\(\begin{aligned}\frac{2}{n}-\frac{1}{n-4}&=\frac{1}{30} \\ \color{Cerulean}{30 n(n-4)}\color{black}{ \cdot\left(\frac{2}{n}-\frac{1}{n-4}\right)}&=\color{Cerulean}{30 n(n-4)}\color{black}{ \cdot\left(\frac{1}{30}\right)} \\ \color{Cerulean}{30 n(n-4)}\color{black}{ \cdot \frac{2}{n}-}\color{Cerulean}{30 n(n-4)}\color{black}{ \cdot \frac{1}{n-4}}&=\color{Cerulean}{30 n(n-4)}\color{black}{ \cdot\left(\frac{1}{30}\right)}\end{aligned}\)

\(\begin{array}{rlrl}{n-10} & {=0 \quad \text { or }} & {n-24} & {=0} \\ {n} & {=10} & { n=24}\end{array}\)

Here we have two viable possibilities for the larger integer. For this reason, we will we have two solutions to this problem.

If \(n=10\), then \(n-4=10-4=6\).

If \(n=24\), then \(n-4=24-4=20\).

As a check, perform the operations indicated in the problem.

\(\begin{array}{r|r}{\text { Check } 6 \text { and } 10 .} &{\text{Check} \:20\:\text{and}\:24.}\\{2\left(\frac{1}{\color{OliveGreen}{10}}\right)-\frac{1}{\color{OliveGreen}{6}}=\frac{1}{5}-\frac{1}{6}}&{2\left(\frac{1}{\color{OliveGreen}{24}}\right)-\frac{1}{\color{OliveGreen}{20}}=\frac{1}{12}-\frac{1}{20}}\\{=\frac{6}{30}-\frac{5}{30}}&{=\frac{5}{60}-\frac{3}{60}}\\{=\frac{1}{30}\quad\color{Cerulean}{\checkmark}}&{=\frac{1}{30}\quad\color{Cerulean}{\checkmark}} \end{array}\)

Two sets of positive integers solve this problem: {\(6, 10\)} and {\(20, 24\)}.

Exercise \(\PageIndex{1}\)

The difference between the reciprocals of two consecutive positive odd integers is \(\frac{2}{15}\). Find the integers.

The integers are \(3\) and \(5\).

Uniform Motion Problems

Uniform motion problems, also referred to as distance problems, involve the formula

where the distance, \(D\), is given as the product of the average rate, \(r\), and the time, \(t\), traveled at that rate. If we divide both sides by the average rate, \(r\), then we obtain the formula

\[t=\frac{D}{r}\]

For this reason, when the unknown quantity is time, the algebraic setup for distance problems often results in a rational equation. Similarly, when the unknown quantity is the rate, the setup also may result in a rational equation.

We begin any uniform motion problem by first organizing our data with a chart. Use this information to set up an algebraic equation that models the application.

Example \(\PageIndex{5}\)

Mary spent the first 120 miles of her road trip in traffic. When the traffic cleared, she was able to drive twice as fast for the remaining 300 miles. If the total trip took 9 hours, then how fast was she moving in traffic?

First, identify the unknown quantity and organize the data.

Let \(x\) represent Mary's average speed (miles per hour) in traffic.

Let \(2x\) represent her average speed after the traffic cleared.

To avoid introducing two more variables for the time column, use the formula \(t=\frac{D}{r}\). Here the time for each leg of the trip is calculated as follows:

\(\begin{array}{c} {\color{Cerulean} { Time\:spent\:in\:traffic:}\:\:\color{black}{ t=\frac{D}{r}=\frac{120}{x}}} \\{\color{Cerulean}{Time\:clear\:of\:traffic:}\:\:\color{black}{t=\frac{D}{r}}=\frac{300}{2x}} \end{array}\)

Use these expressions to complete the chart.

The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 9 hours:

\(\begin{array}{ccccc}{\color{Cerulean}{time\:spent\:in\:traffic}}&{}&{\color{Cerulean}{time\:clear\:of\:traffic}}&{}&{\color{Cerulean}{total\:time\:of\:trip}} \\ {\overbrace{\frac{120}{x}}}&{+}&{\overbrace{\frac{300}{2x}}}&{=}&{\overbrace{9}} \end{array}\)

We begin solving this equation by first multiplying both sides by the LCD, \(2x\).

\(\begin{aligned} \frac{120}{x}+\frac{300}{2 x} &=9 \\ \color{Cerulean}{2x }\color{black}{ \cdot\left(\frac{120}{x}+\frac{300}{2 x}\right)} &=\color{Cerulean}{2 x}\color{black}{ \cdot 9} \\ \color{Cerulean}{2 x}\color{black}{ \cdot \frac{120}{x}+}\color{Cerulean}{2 x}\color{black}{ \cdot \frac{300}{2 x}} &=\color{Cerulean}{2 x}\color{black}{ \cdot 9} \\ 240+300 &=18 x \\ 540 &=18 x \\ 30 &=x \end{aligned}\)

Mary averaged 30 miles per hour in traffic.

Example \(\PageIndex{6}\)

A passenger train can travel, on average, \(20\) miles per hour faster than a freight train. If the passenger train covers \(390\) miles in the same time it takes the freight train to cover \(270\) miles, then how fast is each train?

First, identify the unknown quantities and organize the data.

Let \(x\) represent the average speed of the freight train.

Let \(x+20\) represent the speed of the passenger train.

Next, organize the given data in a chart.

Use the formula \(t=\frac{D}{r}\) to fill in the time column for each train.

\(\begin{array}{c} {\color{Cerulean} { Passenger\: train: }\:\:\color{black}{ t=\frac{D}{r}=\frac{390}{x+20}}} \\ {\color{Cerulean}{Freight\:train:}\:\:\color{black}{t=\frac{D}{r}=\frac{270}{x}}} \end{array}\)

Because the trains travel the same amount of time, finish the algebraic setup by equating the expressions that represent the times:

\(\frac{390}{x+20}=\frac{270}{x}\)

Solve this equation by first multiplying both sides by the LCD, \(x(x+20)\).

\(\begin{aligned} \color{Cerulean}{x(x+20)}\color{black}{ \cdot\left(\frac{390}{x+20}\right)} &=\color{Cerulean}{x(x+20)}\color{black}{ \cdot\left(\frac{270}{x}\right) }\\ 390 x &=270(x+20) \\ 390 x &=270 x+5400 \\ 120 x &=5400 \\ x &=45 \end{aligned}\)

Use \(x+20\) to find the speed of the passenger train.

\(x+20=\color{OliveGreen}{45}\color{black}{+20=65}\)

The speed of the passenger train is \(65\) miles per hour and the speed of the freight train is \(45\) miles per hour.

Example \(\PageIndex{7}\)

Brett lives on the river \(8\) miles upstream from town. When the current is \(2\) miles per hour, he can row his boat downstream to town for supplies and back in \(3\) hours. What is his average rowing speed in still water?

Let \(x\) represent Brett's average rowing speed in still water.

Rowing downstream, the current increases his speed, and his rate is \(x + 2\) miles per hour. Rowing upstream, the current decreases his speed, and his rate is \(x − 2\) miles per hour. Begin by organizing the data in the following chart:

Use the formula \(t=\frac{D}{r}\) to fill in the time column for each leg of the trip.

\(\begin{array}{c} {\color{Cerulean} { Trip\: downstream: } \:\:\color{black}{ t=\frac{D}{r}=\frac{8}{x+2}}}\\{\color{Cerulean}{Trip\:upstream:}\:\:\color{black}{t=\frac{D}{r}=\frac{8}{x-2}}} \end{array}\)

The algebraic setup is defined by the time column. Add the times for each leg of the trip to obtain a total of 3 hours:

\(\begin{array}{ccccc}{\color{Cerulean}{time\:rowing\:downstream}}&{}&{\color{Cerulean}{time\:rowing\:upstream}}&{}&{\color{Cerulean}{total\:time\:of\:trip}}\\{\overbrace{\frac{8}{x+2}}}&{+}&{\overbrace{\frac{8}{x-2}}}&{=}&{\overbrace{3}} \end{array}\)

Solve this equation by first multiplying both sides by the LCD, \((x+2)(x−2)\).

Next, solve the resulting quadratic equation.

\(\begin{array}{rlrl}{3 x+2} & {=0} & {\text { or }} & {x-6=0} \\ {3 x} & {=-2} & {} & {x=6} \\ {x} & {=\frac{-2}{3}}\end{array}\)

Use only the positive solution, \(x=6\) miles per hour.

His rowing speed is \(6\) miles per hour.

Exercise \(\PageIndex{2}\)

Dwayne drove \(18\) miles to the airport to pick up his father and then returned home. On the return trip he was able to drive an average of \(15\) miles per hour faster than he did on the trip there. If the total driving time was \(1\) hour, then what was his average speed driving to the airport?

His average speed driving to the airport was \(30\) miles per hour.

Work-Rate Problems

The rate at which a task can be performed is called a work rate.

Example \(\PageIndex{8}\)

For example, if a painter can paint a room in 8 hours, then the task is to paint the room, and we can write

\(\frac{1 \text { task }}{8 \text { hours }} \quad \color{Cerulean} { Work\: rate }\)

In other words, the painter can complete \(\frac{1}{8}\) of the task per hour.

If he works for less than 8 hours, then he will perform a fraction of the task. For example,

\(\begin{array}{cc} {\color{Cerulean}{work\:rate}\color{black}{\times}\color{Cerulean}{time}\color{black}{=}\color{Cerulean}{work\:completed}}&{}\\{\frac{1}{8}\times 2\text{hrs}=\frac{1}{4}}&{\color{Cerulean}{One-quarter\:of\:the\:room\:painted}}\\{\frac{1}{8}\times 4\text{hrs}=\frac{1}{2}}&{\color{Cerulean}{One-half\:of\:the\:room\:painted}}\\{\frac{1}{8}\times 8\text{hrs}=1}&{\color{Cerulean}{One\:whole\:room\:painted}} \end{array}\)

Obtain the amount of the task completed by multiplying the work rate by the amount of time the painter works. Typically, work-rate problems involve people working together to complete tasks. When this is the case, we can organize the data in a chart, just as we have done with distance problems.

Suppose an apprentice painter can paint the same room by himself in \(10\) hours. Then we say that he can complete \(\frac{1}{10}\) of the task per hour.

Let \(t\) represent the time it takes both of the painters, working together, to paint the room.

To complete the chart, multiply the work rate by the time for each person.

The portion of the room each can paint adds to a total of \(1\) task completed.

This is represented by the equation obtained from the first column of the chart:

This setup results in a rational equation that can be solved for \(t\) by multiplying both sides by the LCD, \(40\).

\(\begin{aligned} \frac{1}{8}t+\frac{1}{10}t &=1 \\ \color{Cerulean}{40}\color{black}{\cdot \left(\frac{1}{8}t+\frac{1}{10}t\right)}&=\color{Cerulean}{40}\color{black}{\cdot 1} \\ \color{Cerulean}{40}\color{black}{\cdot\frac{t}{8}+}\color{Cerulean}{40}\color{black}{\cdot\frac{t}{10}}&=\color{Cerulean}{40}\color{black}{\cdot 1}\\ 5t+4t &=40 \\ 9t & =40 \\ t & = \frac{40}{9} \\ t & = 4 \frac{4}{9} \end{aligned}\)

Therefore, the two painters, working together, complete the task in \(4 \frac{4}{9}\) hours.

In general, we have the following work-rate formula:

\[\frac{1}{t_{1}}t + \frac{1}{t_{2}}t = 1\]

Here \(\frac{1}{t_{1}}\) and \(\frac{1}{t_{2}}\) are the individual work rates and \(t\) is the time it takes to complete one task working together. If we factor out the time, \(t\), and then divide both sides by \(t\), we obtain an equivalent work-rate formula:

\(\begin{array}{c} {\frac{1}{t_{1}}t + \frac{1}{t_{2}}t = 1} \\ {t\left( \frac{1}{t_{1}}+\frac{1}{t_{2}} \right)=1} \\ {\frac{1}{t_{1}}+\frac{1}{t_{2}}=\frac{1}{t}} \end{array}\)

In summary, we have the following equivalent work-rate formulas:

\(\begin{array}{ccccc}{}&{}&{\color{Cerulean}{Work\:rate\:formulas:}}&{}&{} \\ {\frac{1}{t_{1}}t}&{\text{or}}&{\frac{t}{t_{1}}+\frac{t}{t_{2}}=1}&{\text{or}}&{\frac{1}{t_{1}}+\frac{1}{t_{2}}=\frac{1}{t}}\end{array}\)

Example \(\PageIndex{9}\)

Working alone, Billy’s dad can complete the yard work in 3 hours. If Billy helps his dad, then the yard work takes 2 hours. How long would it take Billy working alone to complete the yard work?

The given information tells us that Billy’s dad has an individual work rate of \(\frac{1}{3}\) task per hour. If we let \(x\) represent the time it takes Billy working alone to complete the yard work, then Billy’s individual work rate is \(\frac{1}{x}\), and we can write

Working together, they can complete the task in \(2\) hours. Multiply the individual work rates by \(2\) hours to fill in the chart.

The amount of the task each completes will total 1 completed task. To solve for \(x\), we first multiply both sides by the LCD, \(3x\).

\(\begin{aligned} \frac{1}{3}\cdot 2+\frac{1}{x}\cdot 2 &=1 \\ \color{Cerulean}{3x}\color{black}{\cdot \left(\frac{2}{3}+\frac{2}{x} \right)}&=\color{Cerulean}{3x}\color{black}{\cdot1} \\ \color{Cerulean}{3x}\color{black}{\cdot \frac{2}{3} +}\color{Cerulean}{3x}\color{black}{\cdot \frac{2}{x}}&=\color{Cerulean}{3x}\color{black}{\cdot 1} \\ 2x+6&=3x \\ 6&=x \end{aligned}\)

It takes Billy \(6\) hours to complete the yard work alone.

Of course, the unit of time for the work rate need not always be in hours.

Example \(\PageIndex{10}\)

Working together, two construction crews can build a shed in \(5\) days. Working separately, the less experienced crew takes twice as long to build a shed than the more experienced crew. Working separately, how long does it take each crew to build a shed?

Let \(x\) represent the time it takes the more experienced crew to build a shed.

Let \(2x\) represent the time it takes the less experienced crew to build a shed.

Working together, the job is completed in \(5\) days. This gives the following setup:

The first column in the chart gives us an algebraic equation that models the problem:

\(\begin{aligned} \frac{1}{x}\cdot 5+\frac{1}{2x}\cdot 5 &=1 \\ \frac{5}{x}+\frac{5}{2x}&=1 \end{aligned}\)

Solve the equation by multiplying both sides by \(2x\).

\(\begin{aligned} \color{Cerulean}{2x}\color{black}{\cdot \left(\frac{5}{x}+\frac{5}{2x} \right)}&=\color{Cerulean}{2x}\color{black}{\cdot 1}\\ \color{Cerulean}{2x}\color{black}{\cdot \frac{5}{x} +}\color{Cerulean}{2x}\color{black}{\cdot \frac{5}{2x}}&=\color{Cerulean}{2x}\color{black}{\cdot 1} \\ 10+5&=2x \\ 15&=2x \\ \frac{15}{2}&=x \quad\text{or}\quad x=7\frac{1}{2}\:\text{days} \end{aligned}\)

To determine the time it takes the less experienced crew, we use \(2x\):

\(\begin{aligned} 2x&= 2\color{black}{\left(\color{OliveGreen}{\frac{15}{2}} \right) } \\ &= 15\:\text{days} \end{aligned}\)

Working separately, the experienced crew takes \(7\frac{1}{2}\) days to build a shed, and the less experienced crew takes \(15\) days to build a shed.

Exercise \(\PageIndex{3}\)

Joe’s garden hose fills the pool in \(12\) hours. His neighbor has a thinner hose that fills the pool in \(15\) hours. How long will it take to fill the pool using both hoses?

It will take both hoses \(6\frac{2}{3}\) hours to fill the pool.

Key Takeaways

• In this section, all of the steps outlined for solving general word problems apply. Look for the new key word “reciprocal,” which indicates that you should write the quantity in the denominator of a fraction with numerator \(1\).
• When solving distance problems where the time element is unknown, use the equivalent form of the uniform motion formula, \(t=\frac{D}{r}\), to avoid introducing more variables.
• When solving work-rate problems, multiply the individual work rate by the time to obtain the portion of the task completed. The sum of the portions of the task results in the total amount of work completed.

Exercise \(\PageIndex{4}\) Number Problems

Use algebra to solve the following applications.

• A positive integer is twice another. The sum of the reciprocals of the two positive integers is \(\frac{3}{10}\). Find the two integers.
• A positive integer is twice another. The sum of the reciprocals of the two positive integers is \(\frac{3}{12}\). Find the two integers.
• A positive integer is twice another. The difference of the reciprocals of the two positive integers is \(\frac{1}{8}\). Find the two integers.
• A positive integer is twice another. The difference of the reciprocals of the two positive integers is \(\frac{1}{18}\). Find the two integers.
• A positive integer is \(2\) less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is \(\frac{5}{12}\), then find the two integers.
• A positive integer is \(2\) more than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is \(\frac{17}{35}\), then find the two integers.
• The sum of the reciprocals of two consecutive positive even integers is \(\frac{11}{60}\). Find the two even integers.
• The sum of the reciprocals of two consecutive positive odd integers is \(\frac{16}{63}\). Find the integers.
• The difference of the reciprocals of two consecutive positive even integers is \(\frac{1}{24}\). Find the two even integers.
• The difference of the reciprocals of two consecutive positive odd integers is \(\frac{2}{99}\). Find the integers.
• If \(3\) times the reciprocal of the larger of two consecutive integers is subtracted from \(2\) times the reciprocal of the smaller, then the result is \(\frac{1}{2}\). Find the two integers.
• If \(3\) times the reciprocal of the smaller of two consecutive integers is subtracted from \(7\) times the reciprocal of the larger, then the result is \(\frac{1}{2}\). Find the two integers.
• A positive integer is \(5\) less than another. If the reciprocal of the smaller integer is subtracted from \(3\) times the reciprocal of the larger, then the result is \(\frac{1}{12}\). Find the two integers.
• A positive integer is \(6\) less than another. If the reciprocal of the smaller integer is subtracted from \(10\) times the reciprocal of the larger, then the result is \(\frac{3}{7}\). Find the two integers.

1. {\(5, 10\)}

3. {\(4, 8\)}

5. {\(6, 8\)}

7. {\(10, 12\)}

9. {\(6, 8\)}

11. {\(1, 2\)} or {\(−4, −3\)}

13. {\(4, 9\)} or {\(15, 20\)}

Exercise \(\PageIndex{5}\) Uniform Motion Problems

• James can jog twice as fast as he can walk. He was able to jog the first \(9\) miles to his grandmother’s house, but then he tired and walked the remaining \(1.5\) miles. If the total trip took \(2\) hours, then what was his average jogging speed?
• On a business trip, an executive traveled \(720\) miles by jet aircraft and then another \(80\) miles by helicopter. If the jet averaged \(3\) times the speed of the helicopter and the total trip took \(4\) hours, then what was the average speed of the jet?
• Sally was able to drive an average of \(20\) miles per hour faster in her car after the traffic cleared. She drove \(23\) miles in traffic before it cleared and then drove another \(99\) miles. If the total trip took \(2\) hours, then what was her average speed in traffic?
• Harry traveled \(15\) miles on the bus and then another \(72\) miles on a train. If the train was \(18\) miles per hour faster than the bus and the total trip took \(2\) hours, then what was the average speed of the train?
• A bus averages \(6\) miles per hour faster than a trolley. If the bus travels \(90\) miles in the same time it takes the trolley to travel \(75\) miles, then what is the speed of each?
• A passenger car averages \(16\) miles per hour faster than the bus. If the bus travels \(56\) miles in the same time it takes the passenger car to travel \(84\) miles, then what is the speed of each?
• A light aircraft travels \(2\) miles per hour less than twice as fast as a passenger car. If the passenger car can travel \(231\) miles in the same time it takes the aircraft to travel \(455\) miles, then what is the average speed of each?
• Mary can run \(1\) mile per hour more than twice as fast as Bill can walk. If Bill can walk \(3\) miles in the same time it takes Mary to run \(7.2\) miles, then what is Bill’s average walking speed?
• An airplane traveling with a \(20\)-mile-per-hour tailwind covers \(270\) miles. On the return trip against the wind, it covers \(190\) miles in the same amount of time. What is the speed of the airplane in still air?
• A jet airliner traveling with a \(30\)-mile-per-hour tailwind covers \(525\) miles in the same amount of time it is able to travel \(495\) miles after the tailwind eases to \(10\) miles per hour. What is the speed of the airliner in still air?
• A boat averages \(16\) miles per hour in still water. With the current, the boat can travel \(95\) miles in the same time it travels \(65\) miles against it. What is the speed of the current?
• A river tour boat averages \(7\) miles per hour in still water. If the total \(24\)-mile tour down river and \(24\) miles back takes \(7\) hours, then how fast is the river current?
• If the river current flows at an average \(3\) miles per hour, then a tour boat makes the \(9\)-mile tour downstream with the current and back the \(9\) miles against the current in \(4\) hours. What is the average speed of the boat in still water?
• Jane rowed her canoe against a \(1\)-mile-per-hour current upstream \(12\) miles and then returned the \(12\) miles back downstream. If the total trip took \(5\) hours, then at what speed can Jane row in still water?
• Jose drove \(15\) miles to pick up his sister and then returned home. On the return trip, he was able to average \(15\) miles per hour faster than he did on the trip to pick her up. If the total trip took \(1\) hour, then what was Jose’s average speed on the return trip?
• Barry drove the \(24\) miles to town and then back in \(1\) hour. On the return trip, he was able to average \(14\) miles per hour faster than he averaged on the trip to town. What was his average speed on the trip to town?
• Jerry paddled his kayak upstream against a \(1\)-mile-per-hour current for \(12\) miles. The return trip downstream with the \(1\)-mile-per-hour current took \(1\) hour less time. How fast can Jerry paddle the kayak in still water?
• It takes a light aircraft \(1\) hour more time to fly \(360\) miles against a \(30\)- mile-per-hour head wind than it does to fly the same distance with it. What is the speed of the aircraft in calm air?

1. \(6\) miles per hour

3. \(46\) miles per hour

5. Trolley: \(30\) miles per hour; bus: \(36\) miles per hour

7. Passenger car: \(66\) miles per hour; aircraft: \(130\) miles per hour

9. \(115\) miles per hour

11. \(3\) miles per hour

13. \(6\) miles per hour

15. \(40\) miles per hour

17. \(5\) miles per hour

Exercise \(\PageIndex{6}\) Work-Rate Problems

• James can paint the office by himself in \(7\) hours. Manny paints the office in \(10\) hours. How long will it take them to paint the office working together?
• Barry can lay a brick driveway by himself in \(12\) hours. Robert does the same job in \(10\) hours. How long will it take them to lay the brick driveway working together?
• Jerry can detail a car by himself in \(50\) minutes. Sally does the same job in \(1\) hour. How long will it take them to detail a car working together?
• Jose can build a small shed by himself in \(26\) hours. Alex builds the same small shed in \(2\) days. How long would it take them to build the shed working together?
• Allison can complete a sales route by herself in \(6\) hours. Working with an associate, she completes the route in \(4\) hours. How long would it take her associate to complete the route by herself?
• James can prepare and paint a house by himself in \(5\) days. Working with his brother, Bryan, they can do it in \(3\) days. How long would it take Bryan to prepare and paint the house by himself?
• Joe can assemble a computer by himself in \(1\) hour. Working with an assistant, he can assemble a computer in \(40\) minutes. How long would it take his assistant to assemble a computer working alone?
• The teacher’s assistant can grade class homework assignments by herself in \(1\) hour. If the teacher helps, then the grading can be completed in \(20\) minutes. How long would it take the teacher to grade the papers working alone?
• A larger pipe fills a water tank twice as fast as a smaller pipe. When both pipes are used, they fill the tank in \(5\) hours. If the larger pipe is left off, then how long would it take the smaller pipe to fill the tank?
• A newer printer can print twice as fast as an older printer. If both printers working together can print a batch of flyers in \(45\) minutes, then how long would it take the newer printer to print the batch working alone?
• Working alone, Henry takes \(9\) hours longer than Mary to clean the carpets in the entire office. Working together, they clean the carpets in \(6\) hours. How long would it take Mary to clean the office carpets if Henry were not there to help?
• Working alone, Monique takes \(4\) hours longer than Audrey to record the inventory of the entire shop. Working together, they take inventory in \(1.5\) hours. How long would it take Audrey to record the inventory working alone?
• Jerry can lay a tile floor in \(3\) hours less time than Jake. If they work together, the floor takes \(2\) hours. How long would it take Jerry to lay the floor by himself?
• Jeremy can build a model airplane in \(5\) hours less time than his brother. Working together, they need \(6\) hours to build the plane. How long would it take Jeremy to build the model airplane working alone?
• Harry can paint a shed by himself in \(6\) hours. Jeremy can paint the same shed by himself in \(8\) hours. How long will it take them to paint two sheds working together?
• Joe assembles a computer by himself in \(1\) hour. Working with an assistant, he can assemble \(10\) computers in \(6\) hours. How long would it take his assistant to assemble \(1\) computer working alone?
• Jerry can lay a tile floor in \(3\) hours, and his assistant can do the same job in \(4\) hours. If Jerry starts the job and his assistant joins him 1 hour later, then how long will it take to lay the floor?
• Working alone, Monique takes \(6\) hours to record the inventory of the entire shop, while it takes Audrey only \(4\) hours to do the same job. How long will it take them working together if Monique leaves \(2\) hours early?

1. \(4 \frac{2}{17}\) hours

3. \(27 \frac{3}{11}\) minutes

5. \(12\) hours

7. \(2\) hours

9. \(15\) hours

11. \(9\) hours

13. \(3\) hours

15. \(6\frac{6}{7}\) hours

17. \(2 \frac{1}{7}\) hours

Module 10: Rational and Radical Functions

Applications with rational equations, learning outcomes.

• Solve a rational formula for a specified variable
• Solve an application using a formula that must be solved for a specified variable.
• Solve applications by defining and solving rational equations.

Rational Formulas

Rational formulas can be useful tools for representing real-life situations and for finding answers to real problems. You’ll see later in this module that certain equations representing relationships called direct , inverse , and joint variation are examples of rational formulas that can model many real-life situations.

When solving problems using rational formulas, after identifying the particular formula that represents the relationship between the known and unknown quantities, it is often helpful to then solve the formula for a specified variable. This is sometimes called  solving a literal equation .

The formula for finding the density of an object is [latex] D=\frac{m}{v}[/latex], where D is the density, m is the mass of the object, and v is the volume of the object. Rearrange the formula to solve for the mass ( m ) and then for the volume ( v ).

[latex] D=\frac{m}{v}[/latex]

Multiply both side of the equation by v to isolate m.

[latex] v\cdot D=\frac{m}{v}\cdot v[/latex]

Simplify and rewrite the equation, solving for m .

[latex]\begin{array}{l}v\cdot D=m\cdot \frac{v}{v}\\v\cdot D=m\cdot 1\\v\cdot D=m\end{array}[/latex]

To solve the equation [latex] D=\frac{m}{v}[/latex] in terms of v , you will need do the same steps to this point and then divide both sides by D .

[latex]\begin{array}{r}\frac{v\cdot D}{D}=\frac{m}{D}\\\\\frac{D}{D}\cdot v=\frac{m}{D}\\\\1\cdot v=\frac{m}{D}\\\\v=\frac{m}{D}\end{array}[/latex]

Therefore, [latex] m=D\cdot v[/latex] and [latex] v=\frac{m}{D}[/latex].

The formula for finding the volume of a cylinder is [latex]V=\pi{r^{2}}h[/latex], where V is the volume, r is the radius, and h is the height of the cylinder. Rearrange the formula to solve for the height ( h ).

[latex] V=\pi{{r}^{2}}h[/latex]

Divide both sides by [latex] \pi {{r}^{2}}[/latex] to isolate h.

[latex] \frac{V}{\pi {{r}^{2}}}=\frac{\pi {{r}^{2}}h}{\pi {{r}^{2}}}[/latex]

Simplify. You find the height, h , is equal to [latex] \frac{V}{\pi {{r}^{2}}}[/latex].

[latex] \frac{V}{\pi {{r}^{2}}}=h[/latex]

Therefore, [latex] h=\frac{V}{\pi {{r}^{2}}}[/latex].

Watch the following video for more examples of solving for a particular variable in a formula, a literal equation.

A work problem  is a useful real-world application involving literal equations. Let’s say you’d like to calculate how long it will take different people working at different speeds to finish a task.  You may recall the formula that relates distance, rate and time, [latex]d=rt[/latex]. A similar formula relates work performed to a work-rate and time spent working: [latex]W=rt[/latex]. The amount of work done [latex]W[/latex] is the product of the rate of work [latex]r[/latex] and the time spent working [latex]t[/latex]. Using algebra, you can write the work formula [latex]3[/latex] ways:

[latex]W=rt[/latex]

Solved for time [latex]t[/latex] the formula is [latex] t=\frac{W}{r}[/latex] (divide both sides by r).

Solved for rate [latex]r[/latex] the formula is   [latex] r=\frac{W}{t}[/latex] (divide both sides by t).

Rational equations can be used to solve a variety of problems that involve rates, times, and work. Using rational expressions and equations can help you answer questions about how to combine workers or machines to complete a job on schedule.

Some work problems include multiple machines or people working on a project together for the same amount of time but at different rates. In this case, you can add their individual work rates together to get a total work rate.

Myra takes [latex]2[/latex] hours to plant [latex]50[/latex] flower bulbs. Francis takes [latex]3[/latex] hours to plant [latex]45[/latex] flower bulbs. Working together, how long should it take them to plant [latex]150[/latex] bulbs?

Think about how many bulbs each person can plant in one hour. This is their planting rate.

Myra: [latex] \frac{50\,\,\text{bulbs}}{2\,\,\text{hours}}[/latex] or [latex] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Francis: [latex] \frac{45\,\,\text{bulbs}}{3\,\,\text{hours}}[/latex] or [latex] \frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Combine their hourly rates to determine the rate they work together.

Myra and Francis together:

[latex] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}+\frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}=\frac{40\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex]

Use one of the work formulas to write a rational equation, for example, [latex] r=\frac{W}{t}[/latex]. You know r , the combined work rate, and you know W , the amount of work that must be done. What you do not know is how much time it will take to do the required work at the designated rate.

[latex] \frac{40}{1}=\frac{150}{t}[/latex]

Solve the equation by multiplying both sides by the common denominator and then isolating t .

[latex]\begin{array}{c}1t\cdot\frac{40}{1} =\frac{150}{t}\cdot 1t\\\\40t=150\\\\t=\frac{150}{40}=\frac{15}{4}\\\\t=3\frac{3}{4}\text{hours}\end{array}[/latex]

It should take [latex]3[/latex] hours [latex]45[/latex] minutes for Myra and Francis to plant [latex]150[/latex] bulbs together.

Other work problems take a different perspective. You can calculate how long it will take one person to do a job alone when you know how long it takes people working together to complete the job.

Joe and John are planning to paint a house together. John thinks that if he worked alone, it would take him [latex]3[/latex] times as long as it would take Joe to paint the entire house. Working together, they can complete the job in [latex]24[/latex] hours. How long would it take each of them, working alone, to complete the job?

Choose variables to represent the unknowns. Since it takes John [latex]3[/latex] times as long as Joe to paint the house, his time is represented as [latex]3x[/latex].

Let [latex]x[/latex] = time it takes Joe to complete the job

[latex]3x[/latex] = time it takes John to complete the job

The work is painting [latex]1[/latex] house or [latex]1[/latex]. Write an expression to represent each person’s rate using the formula [latex] r=\frac{W}{t}[/latex] .

Joe’s rate: [latex] \frac{1}{x}[/latex]

John’s rate: [latex] \frac{1}{3x}[/latex]

Their combined rate is the sum of their individual rates. Use this rate to write a new equation using the formula [latex]W=rt[/latex].

combined rate: [latex] \frac{1}{x}+\frac{1}{3x}[/latex]

The problem states that it takes them [latex]24[/latex] hours together to paint a house, so if you multiply their combined hourly rate [latex] \left( \frac{1}{x}+\frac{1}{3x} \right)[/latex] by [latex]24[/latex], you will get [latex]1[/latex], which is the number of houses they can paint in [latex]24[/latex] hours.

[latex] \begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\frac{24}{x}+\frac{24}{3x}\end{array}[/latex]

Now solve the equation for x . (Remember that x represents the number of hours it will take Joe to finish the job.)

[latex]\begin{array}{l}\,\,\,1=\frac{3}{3}\cdot \frac{24}{x}+\frac{24}{3x}\\\\\,\,\,1=\frac{3\cdot 24}{3x}+\frac{24}{3x}\\\\\,\,\,1=\frac{72}{3x}+\frac{24}{3x}\\\\\,\,\,1=\frac{72+24}{3x}\\\\\,\,\,1=\frac{96}{3x}\\\\3x=96\\\\\,\,\,x=32\end{array}[/latex]

Check the solution in the original equation.

[latex]\begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\left[ \frac{\text{1}}{\text{32}}+\frac{1}{3\text{(32})} \right]24\\\\1=\frac{24}{\text{32}}+\frac{24}{3\text{(32})}\\\\1=\frac{24}{\text{32}}+\frac{24}{96}\\\\1=\frac{3}{3}\cdot \frac{24}{\text{32}}+\frac{24}{96}\\\\1=\frac{72}{96}+\frac{24}{96}\end{array}[/latex]

The solution checks. Since [latex]x=32[/latex], it takes Joe [latex]32[/latex] hours to paint the house by himself. John’s time is [latex]3x[/latex], so it would take him [latex]96[/latex] hours to do the same amount of work.

As shown above, many work problems can be represented by the equation [latex] \dfrac{t}{a}+\dfrac{t}{b}=1[/latex], where [latex]t[/latex] represents the quantity of time two people, [latex]A \text{ and } B[/latex], complete the job working together, [latex]a[/latex] is the amount of time it takes person [latex]A[/latex] to do the job, and [latex]b[/latex] is the amount of time it takes person [latex]B[/latex] to do the job. The [latex]1[/latex] on the right hand side represents [latex]1[/latex] job, the total work done—in this case, the work was to paint a house.

The key idea here is to figure out each worker’s individual rate of work. Then, once those rates are identified, add them together, multiply by the time t , set it equal to the amount of work done, and solve the rational equation.

If person [latex]A[/latex] works at a rate of [latex]1[/latex] job every [latex]a[/latex] hours, and person [latex]B[/latex] works at a rate of [latex]1[/latex] job every [latex]b[/latex] hours, and if [latex]t[/latex] represents the total amount of time it takes to paint [latex]1[/latex] house, we have

[latex]\begin{align}W&= \left(r_1 + r_2\right)t \\ 1 &= \left(\dfrac{1}{a}+\dfrac{1}{b}\right)t \\ 1 &= \dfrac{t}{a}+\dfrac{t}{b}\end{align}[/latex]

Watch the following video for an example of finding the total time given two people working together at known rates.

The following video shows an example of finding one person’s work rate given a known combined work rate.

Mixtures are made of ratios of different substances that may include chemicals, foods, water, or gases. There are many different situations where mixtures may occur both in nature and as a means to produce a desired product or outcome. For example, chemical spills, manufacturing, and even biochemical reactions involve mixtures. Mixture problems become mathematically interesting when components of the mixture are added at different rates and concentrations. The next example shows how to define an equation that models the concentration (a ratio) of sugar to water in a large mixing tank over time.

A large mixing tank currently contains [latex]100[/latex] gallons of water into which [latex]5[/latex] pounds of sugar have been mixed. A tap will open pouring [latex]10[/latex] gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of [latex]1[/latex] pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after [latex]12[/latex] minutes. Is that a greater concentration than at the beginning?

Let t  be the number of minutes since the tap opened. Since the water increases at [latex]10[/latex] gallons per minute, and the sugar increases at [latex]1[/latex] pound per minute, these are constant rates of change. This tells us the amount of water in the tank is a linear equation, as is the amount of sugar in the tank. We can write an equation independently for each:

The concentration, C , will be the ratio of pounds of sugar to gallons of water

[latex]C\left(t\right)=\frac{5+t}{100+10t}[/latex]

The concentration after [latex]12[/latex] minutes is given by evaluating [latex]C\left(t\right)[/latex] at [latex]t=\text{ }12[/latex].

[latex]C\left(12\right)=\frac{5+12}{100+10(12)}=\frac{17}{220}[/latex]

This means the concentration is [latex]17[/latex] pounds of sugar to [latex]220[/latex] gallons of water.

At the beginning, the concentration is

[latex]C\left(0\right)=\frac{5+0}{100+10(0)}=\frac{5}{100}=\frac{1}{20}[/latex]

Since [latex]\frac{17}{220}\approx 0.08>\frac{1}{20}=0.05[/latex], the concentration is greater after [latex]12[/latex] minutes than at the beginning.

The following video gives another example of how to use rational functions to model mixing.

• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Rational Function Application - Concentration of a Mixture. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/GD6H7BE_0EI . License : CC BY: Attribution
• Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology. Located at : http://nrocnetwork.org/dm-opentext . License : CC BY: Attribution
• Ex 1: Rational Equation Application - Painting Together. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/SzSasnDF7Ms . License : CC BY: Attribution
• Ex: Rational Equation App - Find Individual Working Time Given Time Working Together. Authored by : James Sousa (Mathispower4u.com) . Located at : https://www.youtube.com/watch?v=kbRSYb8UYqU&feature=youtu.be . License : CC BY: Attribution
• College Algebra: Mixture Problem. Authored by : Abramson, Jay et al.. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]:1/Preface
• Ex 2: Solve a Literal Equation for a Variable. Authored by : James Sousa (Mathispower4u.com). Located at : https://www.youtube.com/watch?v=ecEUUbRLDQs&feature=youtu.be . License : CC BY: Attribution

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