4.7 Solving Applications Involving Rational Equations

Learning Outcomes

Solve motion applications using rational equations.
Solve work applications using rational equations.

The reasoning behind both work and (uniform) motion applications is very similar. You should already be familiar with the formula [latex]d=r\cdot t[/latex] relating the quantities of distance, rate, and time. The rate, [latex]r[/latex], in the formula, is the rate of change of distance, [latex]d[/latex], with respect to time, [latex]t[/latex]. For example, if an object moves at a (constant) speed of [latex]30[/latex] miles per hour, then it travels 30 miles in one hour, [latex]30\cdot 2 = 60[/latex] miles in [latex]2[/latex] hours, [latex]30\cdot 3.5=105[/latex] miles in [latex]3.5[/latex] hours, etc. Note that to compute the distance traveled, we multiply the rate (expressed in a unit of distance by a unit of time) by the number of (compatible) units of time. A similar

“quantity” [latex]=[/latex] “the rate of change of the quantity with respect to a counting unit” [latex]\cdot[/latex] “the number of counting units”

principle is used in the following relationships:

[latex]R=x\cdot p[/latex] between the revenue, [latex]R[/latex] (the amount of money that comes into the business), the quantity, [latex]x[/latex], of a product sold, and the unit price, [latex]p[/latex] (i.e., dollars per unit sold). Note that the price per unit is the rate of change of revenue with respect to the quantity of a product sold. (Refer to this description.)
[latex]m=C\cdot V[/latex] between the mass, [latex]m[/latex], of a solvent, the volume, [latex]V[/latex], of a solution, and the concentration, [latex]C[/latex], the amount of solvent per unit volume of solution. Note that the concentration is the rate of change of mass with respect to volume. (Refer to this description.)

Motion Applications

Motion applications have already been introduced in Sec. 2.1. We continue working with uniform motion with a constant speed in this section. Equivalent forms of the [latex]d=r\cdot t[/latex] are [latex]r=\dfrac{d}{t}[/latex] and [latex]t=\dfrac{d}{r}[/latex] (the latter form has been discussed in Sec. 4.6).

Example

Avery can paddle her kayak [latex]4[/latex] miles per hour in still water. It takes her as long to paddle [latex]10[/latex] miles upstream as it takes her to travel [latex]22[/latex] miles downstream. Determine the speed of the river’s current.

Show Solution

Read and Understand:

Recall that when the kayak is going downstream, in the same direction as the river’s current, the current helps push the kayak, so the kayak’s actual speed is faster than its speed in still water. The actual speed at which the kayak is moving is the sum of the speeds in still water and the river’s current. When the kayak is going upstream, opposite to the river current though, the current is going against the kayak, so the kayak’s actual speed is slower than its speed in still water. The actual speed of the kayak is the difference between the speed in still water and the speed of the river’s current.

The speed of the kayak in still water is known while the speed of the river’s current is to be found. We do know the distances Avery paddles her kayak upstream and downstream. We do not know the times but we do know the two times are the same.

Define and Translate: Let [latex]c[/latex] be the unknown speed of the river’s current. Going upstream, the actual rate of the kayak is [latex]4-c[/latex]. Going downstream, the actual rate is [latex]4+c[/latex].

Write and Solve: Fill in the table with the information we know. The table is not necessary, but can help you organize.

	Distance	Rate (or speed)	Time
Upstream	[latex]10[/latex]	[latex]4-c[/latex]
Downstream	[latex]22[/latex]	[latex]4+c[/latex]

Since we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\dfrac{d}{r}[/latex]:

	Distance	Rate (or speed)	Time
Upstream	[latex]10[/latex]	[latex]4-c[/latex]	[latex]\dfrac{10}{4-c}[/latex]
Downstream	[latex]22[/latex]	[latex]4+c[/latex]	[latex]\dfrac{22}{4+c}[/latex]

Since the two times are the same (mathematically, equal), we obtain the following rational equation:

[latex]\dfrac{10}{4-c}=\dfrac{22}{4+c}[/latex]

Let’s acknowledge that the restricted values of [latex]c[/latex] are [latex]4[/latex] and [latex]-4[/latex]. (Can you explain why?) Now,

[latex]\require{color} \begin{align} \frac{10}{4-c}{\color{red}\: \cdot\: (4-c)\cdot (4+c)}&=\frac{22}{4+c}{\color{red}\: \cdot\: (4-c)\cdot (4+c)} && \color{blue}{\textsf{multiply by the LCD}}\\[5pt] \frac{10}{{\color{red}\cancel{\color{black}{(4-c)}}}}\cdot \frac{{\color{red}\cancel{\color{black}{(4-c)}}}\cdot (4+c)}{1} &= \frac{22}{{\color{red}\cancel{\color{black}{(4+c)}}}}\cdot \frac{(4-c)\cdot {\color{red}\cancel{\color{black}{(4+c)}}}}{1} && \color{blue}{\textsf{clear the fractions}}\\[5pt]10\cdot (4+c) &= 22\cdot (4-c)\\[5pt] \underset{\large{\color{red}{-40}}}{40}+\underset{\large{\color{red}{+22c}}}{10c} &= \underset{\large{\color{red}{-40}}}{88}-\underset{\large{\color{red}{+22c}}}{22c} && \color{blue}{\textsf{isolate $c$-terms}}\\[5pt] {\color{red}{\frac{\cancel{\color{black}{32}}{\color{black}{\: \cdot\: c}}}{\cancel{32} \cdot 1}}} &= {\color{red}{\frac{{\color{black}{48}}}{32}}} && \color{blue}{\textsf{divide to isolate $c$}}\\[5pt] c &= \frac{{\color{red}\cancel{\color{black}{16}}}\cdot 3}{{\color{red}\cancel{\color{black}{16}}}\cdot 2} && \color{blue}{\textsf{simplify}}\\[5pt] c &= \frac{3}{2} \end{align}[/latex]

Answer: The river’s current’s speed is [latex]\dfrac{3}{2}[/latex] miles per hour (or [latex]1.5[/latex] miles per hour).

Example

Hassan can bicycle [latex]72[/latex] kilometers in the same time as it takes him to walk [latex]30[/latex] kilometers. He can ride [latex]7[/latex] km/h faster than he can walk. How fast can he walk?

Show Solution

Example

The Vistula river current’s speed is [latex]4[/latex] miles per hour. A river tram in Kraków travels to a point [latex]6[/latex] miles upstream and back again in [latex]2[/latex] hours. What is the speed of the river tram in still water (without a current)?

Show Solution

Read and Understand:

The speed of the river’s current is known while the speed of the river tram in still water is to be found. We do know the distances the river tram goes upstream and downstream. We do not know the individual times but we do know the roundtrip total time.

Define and Translate: Let [latex]b[/latex] be the unknown speed of the river tram. Going upstream, the actual rate of the river tram is [latex]b-4[/latex]. Going downstream, the actual rate is [latex]b+4[/latex].

Write and Solve: Fill in the table with the information we know. The table is not necessary, but can help you organize. Note that sometimes the variable is first and other times it is last, like in our previous example. The boat or vehicle number/variable is always first.

	Distance	Rate (or speed)	Time
Upstream	[latex]6[/latex]	[latex]b-4[/latex]
Downstream	[latex]6[/latex]	[latex]b+4[/latex]

Since we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\dfrac{d}{r}[/latex]:

	Distance	Rate (or speed)	Time
Upstream	[latex]6[/latex]	[latex]b-4[/latex]	[latex]\dfrac{6}{b-4}[/latex]
Downstream	[latex]6[/latex]	[latex]b+4[/latex]	[latex]\dfrac{6}{b+4}[/latex]

Since the two times total (mathematically, their sum is) 2 hours, we obtain the following rational equation:

[latex]\dfrac{6}{b-4}+\dfrac{6}{b+4}=2[/latex]

Let’s acknowledge that the restricted values of [latex]b[/latex] are [latex]4[/latex] and [latex]-4[/latex]. Now,

[latex]\require{color} \begin{align} \frac{6}{b-4}+\frac{6}{b+4}&=2\\[5pt] {\color{red}{(b-4)\cdot(b+4)\cdot\: }}\left(\frac{6}{b-4}+\frac{6}{b+4}\right)&=2{\color{red}{\: \cdot\: (b-4)\cdot(b+4)}} && \color{blue}{\textsf{multiply by the LCD}}\\[5pt] \frac{{\color{red}\cancel{\color{black}{(b-4)}}}\cdot(b+4)}{1}\cdot \frac{6}{{\color{red}\cancel{\color{black}{(b-4)}}}\cdot 1} + \frac{(b-4)\cdot{\color{red}\cancel{\color{black}{(b+4)}}}}{1}\cdot \frac{6}{{\color{red}\cancel{\color{black}{(b+4)}}}\cdot 1}&= 2\cdot (b^2-16) && \color{blue}{\textsf{distribute and multiply}}\\[5pt] 6b+24+\underset{\large{\color{red}{-12b}}}{6b}-24 &= \underset{\large{\color{red}{-12b}}}{2b^2}-32 && \color{blue}{\textsf{simplify and isolate $0$}}\\[5pt] {\color{red}{\frac{{\color{black}{0}}}{2}}} &= {\color{red}{\frac{{\color{black}{2b^2-12b-32}}}{2}}} && \color{blue}{\textsf{divide out the GCF}}\\[5pt] 0 &= b^2-6b-16\\[5pt] 0 &= (b-8)(b+2) && \color{blue}{\textsf{factor}}\\[5pt] b-\underset{\large{\color{red}{+8}}}{8} &= \underset{\large{\color{red}{+8}}}{0}\ \ \textsf{or}\ \ b+\underset{\large{\color{red}{-2}}}{2} = \underset{\large{\color{red}{-2}}}{0} && \color{blue}{\textsf{Zero-Product Property}}\\[5pt] b &= 8 \ \ \textsf{or}\ \ b = -2 \end{align}[/latex]

Of course, the speed of the river tram in still water cannot be negative, so we discard [latex]b = -2[/latex].

Answer: The river tram’s speed in still water is [latex]8[/latex] miles per hour.

Example

Logan jogged to his grandma’s house [latex]5[/latex] miles away and then purchased a ride back home. It took him [latex]50[/latex] minutes longer to jog there than ride back. If his jogging rate was [latex]25[/latex] mph slower than the rate when he was riding, what was his jogging rate?

Show Solution

Read and Understand:

Neither the speed of jogging nor riding is known but we know Logan’s jogging rate was [latex]25[/latex] mph slower than the rate of the ride. Logan’s jogging rate is to be found. We do know the distances both ways. We do not know the times but we do know it took him [latex]50[/latex] minutes (five sixths of an hour) longer to jog there than ride back.

Define and Translate: Let [latex]j[/latex] be the unknown speed of Logan jogging. Since his jogging rate was [latex]25[/latex] mph slower than the rate of the ride, meaning that the ride speed was [latex]25[/latex] mph faster, the ride speed was [latex]j+25[/latex].

Write and Solve: Fill in the table with the information we know. The table is not necessary, but can help you organize.

	Distance	Rate (or speed)	Time
There	[latex]5[/latex]	[latex]j[/latex]
Back	[latex]5[/latex]	[latex]j+25[/latex]

Since we have the Distance and Rate in each row, we can determine the Time for each row accordingly using the formula [latex]t=\dfrac{d}{r}[/latex]:

	Distance	Rate (or speed)	Time
There	[latex]5[/latex]	[latex]j[/latex]	[latex]\dfrac{5}{j}[/latex]
Back	[latex]5[/latex]	[latex]j+25[/latex]	[latex]\dfrac{5}{j+25}[/latex]

Since it took him five sixths of an hour longer to jog there than ride back, meaning that the time There (the longer time) was five sixths of an hour more than the time Back (the shorter time), we obtain the following rational equation:

[latex]\dfrac{5}{j}=\dfrac{5}{j+25}+\dfrac{5}{6}[/latex]

Let’s acknowledge that the restricted values of [latex]j[/latex] are [latex]0[/latex] and [latex]-25[/latex]. Now,

[latex]\require{color} \begin{align} {\color{red}{6\cdot j\cdot (j+25)\: \cdot\: }}\frac{5}{j}&=\left(\frac{5}{j+25}+\frac{5}{6}\right){\color{red}{\: \cdot\: 6\cdot j\cdot (j+25)}}\\[5pt] \frac{6\cdot {\color{red}\cancel{\color{black}{j}}}\cdot (j+25)}{1}\cdot\frac{5}{{\color{red}\cancel{\color{black}{j}}}\cdot 1}&=\frac{5}{{\color{red}\cancel{\color{black}{(j+25)}}}\cdot 1}\cdot\frac{6\cdot j\cdot {\color{red}\cancel{\color{black}{(j+25)}}}}{1}+\frac{5}{{\color{red}\cancel{\color{black}{6}}}\cdot 1}\cdot\frac{{\color{red}\cancel{\color{black}{6}}}\cdot j\cdot (j+25)}{1}\\[5pt] \underset{\large{\color{red}{-30j}}}{30j}+\underset{\large{\color{red}{-750}}}{750} &= \underset{\large{\color{red}{-30j}}}{30j}+5j^2+\underset{\large{\color{red}{-750}}}{125j}\\[5pt] {\color{red}{\frac{{\color{black}{0}}}{5}}}&={\color{red}{\frac{{\color{black}{5j^2+125j-750}}}{5}}}\\[5pt] 0&=j^2+25j-150\\[5pt] 0&=(j+30)(j-5)\\[5pt] j+\underset{\large{\color{red}{-30}}}{30} &= \underset{\large{\color{red}{-30}}}{0}\ \ \textsf{or}\ \ j-\underset{\large{\color{red}{+5}}}{5} = \underset{\large{\color{red}{+5}}}{0}\\[5pt] j &= -30 \ \ \textsf{or}\ \ j = 5 \end{align}[/latex]

Of course, jogging rate cannot be negative, so we discard [latex]j = -30[/latex].

Answer: Logan’s jogging rate was [latex]5[/latex] miles per hour.

Note that in the example above, we could have presented the equation [latex]\dfrac{5}{j}=\dfrac{5}{j+25}+\dfrac{5}{6}[/latex] in equivalent form [latex]\dfrac{5}{j}-\dfrac{5}{j+25}=\dfrac{5}{6}[/latex]. Such setup can be perceived as one of the three common situations:

“set equal” for “the same time” problems,
“add together” for “the total time” problems, and
“subtract” for “the leaves earlier, or takes longer” problems.

Work Applications

Work problems often ask you to calculate how long it will take different people or machines working at different speeds to finish a task. Let [latex]W[/latex] represent a completed portion of a job (as a fraction or percentage) and [latex]t[/latex] the time an entity works on the job. For example, consider a math professor grading an exam. The job is to grade all exams. If it takes the professor [latex]5[/latex] hours to complete that job, we can say that [latex]100\%[/latex] of the job (or [latex]1[/latex] fully completed job) is completed in [latex]5[/latex] hours. We define the rate of work, [latex]r[/latex], expressing what portion of the job is completed in a unit of time, or mathematically, the rate of change of the completed portion of a job with respect to time, in a similar way as rates of change in the examples above. We divide the completed portion of the job by the time it took the entity to complete that portion. Assuming a constant rate of work, we can say that the rate of grading the exam by our math professor can be computed as [latex]1[/latex] (fully graded exam) divided by [latex]5[/latex] hours, or mathematically, [latex]\dfrac{1\ \text{job}}{5\ \text{hours}}=\dfrac{1}{5}\ \text{job per hour}[/latex]. That leads to a relationship based on the principle

“quantity” [latex]=[/latex] “the rate of change of the quantity with respect to a counting unit” [latex]\cdot[/latex] “the number of counting units”

of the form

“completed portion of a job” [latex]=[/latex] “the rate of change of the completed portion of a job with respect to time” [latex]\cdot[/latex] “the time an entity works on the job”,

or in simpler terms,

“completed portion of a job” [latex]=[/latex] “the rate of work” [latex]\cdot[/latex] “the time spent on the job”,

and mathematically,

[latex]W=r\cdot t[/latex].

Since we usually refer to completing [latex]100\%[/latex] of a task, [latex]W=1[/latex] in the equation above, resulting in a reciprocal relationship between the work rate and time to complete the task since we have [latex]1 = \textsf{rate}\cdot \textsf{time}[/latex] and hence [latex]\textsf{rate} = \dfrac{1}{\textsf{time}}[/latex] and [latex]\textsf{time} = \dfrac{1}{\textsf{rate}}[/latex].

Some work problems include multiple machines or people working on a task together for the same amount of time but at different rates. In that case, we can add their individual work rates together to get a total work rate. Let us look at an example.

Example

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?

Show Solution

If two entities with individual rates of work do a task together, we can add their individual work rates together to get a combined work rate.

In our example, Myra takes [latex]2[/latex] hours to plant [latex]50[/latex] flower bulbs. Hence, she can plant [latex]25[/latex] bulbs in [latex]1[/latex] hour. The full task is to plant [latex]150[/latex] bulbs. Thus, she can do [latex]\dfrac{25}{150}=\dfrac{1}{6}[/latex] of the task in [latex]1[/latex] hour, so her individual rate of work is [latex]\dfrac{1}{6}[/latex].

Similarly, Francis takes [latex]3[/latex] hours to plant [latex]45[/latex] bulbs and hence he can plant [latex]15[/latex] bulbs in [latex]1[/latex] hour, so his individual rate of work is [latex]\dfrac{15}{150}=\dfrac{1}{10}[/latex].

Their combined rate of work (when they work together) is then [latex]\dfrac{1}{6}+\dfrac{1}{10}[/latex]. Therefore, due to the reciprocal relationship between the work rate and time to complete the task, it will take them

[latex]\dfrac{1}{\; \dfrac{1}{6}+\dfrac{1}{10}\; }[/latex]

hours to complete the task working together. We recognize this expression as a complex fraction. Refer to Sec. 4.4 how to simplify complex fractions. Using the second method and noting that [latex]LCM(6,10)=LCM(2^1\cdot 3^1,2^1\cdot 5^1)=2^1\cdot 3^1\cdot 5^1=30[/latex], we obtain

[latex]\begin{align} \frac{1}{\; \dfrac{1}{6}+\dfrac{1}{10}\; }&=\frac{1{\color{red}{\: \cdot\: 30}}}{\; \left(\dfrac{1}{6}+\dfrac{1}{10}\right){\color{red}{\: \cdot\: 30}}\; }\\[5pt] &=\frac{30}{\; \dfrac{1}{{\color{red}\cancel{\color{black}{6}}}\cdot 1}\cdot\dfrac{{\color{red}\cancel{\color{black}{6}}}\cdot 5}{1}+\dfrac{1}{{\color{red}\cancel{\color{black}{10}}}\cdot 1}\cdot\dfrac{{\color{red}\cancel{\color{black}{10}}}\cdot 3}{1}\; }\\[5pt] &=\frac{30}{5+3}\\[5pt] &=\frac{30}{8}\\[5pt] &=\frac{{\color{red}\cancel{\color{black}{2}}}\cdot 15}{{\color{red}\cancel{\color{black}{2}}}\cdot 4}\\[5pt] &=\frac{15}{4} \end{align}[/latex]

Answer: Working together, it should take them [latex]\dfrac{15}{4}[/latex] hours (or [latex]3.75[/latex] hours, or [latex]3[/latex] hours and [latex]45[/latex] minutes).

Note that in the example above, if we let [latex]t[/latex] denote the time it should take Myra and Francis to plant [latex]150[/latex] bulbs together, there are two equivalent ways of expressing their combined rate of work:

as the sum of their individual rates, [latex]\dfrac{1}{6}+\dfrac{1}{10}[/latex], and
as the reciprocal of the time [latex]t[/latex], [latex]\dfrac{1}{t}[/latex].

That leads to the rational equation [latex]\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{1}{t}[/latex] that can also be used to solve the problem.

There are other types of work problems. 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 and their relative times.

Example

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?

Show Solution

In the video that follows, we show another example of finding one person’s work rate given a combined work rate.

As shown above, many work problems can be represented by the equation

[latex]\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{t}[/latex],

where [latex]t[/latex] is the time to do the job together, [latex]a[/latex] is the time it takes entity A to do the job alone, and [latex]b[/latex] is the time it takes entity B to do the job alone. The key idea here is to figure out each entity’s individual rate of work. Then, once those rates are identified, add them together, set it equal to the reciprocal of the time [latex]t[/latex], and solve the rational equation.

We conclude with another example of two people painting at different rates in the following video.

	Distance	Rate (or speed)	Time
Bicycling	[latex]72[/latex]	[latex]w+7[/latex]	[latex]\dfrac{72}{w+7}[/latex]
Walking	[latex]30[/latex]	[latex]w[/latex]	[latex]\dfrac{30}{w}[/latex]

Module 4: Rational Expressions, Functions, and Equations