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Study Guides > ALGEBRA / TRIG I

Applications with Rational Equations

Learning Outcomes

  • Solve work problems
  • Define and solve an equation that represents the concentration of a mixture

Rational Formulas

Rational formulas can be useful tools for representing real-life situations and for finding answers to real problems. Equations representing direct, inverse, and joint variation are examples of rational formulas that can model many real-life situations. As you will see, if you can find a formula, you can usually make sense of a situation. When solving problems using rational formulas, it is often helpful to first solve the formula for the specified variable. For example, work problems ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[/latex]. The amount of work done (W) is the product of the rate of work (r) and the time spent working (t). Using algebra, you can write the work formula [latex]3[/latex] ways: [latex-display]W=rt[/latex-display] Find the time (t): [latex] t=\frac{W}{r}[/latex] (divide both sides by r) Find the rate (r): [latex] r=\frac{W}{t}[/latex](divide both sides by t)


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).

Answer: Start with the formula for density. [latex-display] D=\frac{m}{v}[/latex-display] Multiply both side of the equation by v to isolate m. [latex-display] v\cdot D=\frac{m}{v}\cdot v[/latex-display] Simplify and rewrite the equation, solving for m. [latex-display]\begin{array}{l}v\cdot D=m\cdot \frac{v}{v}\\v\cdot D=m\cdot 1\\v\cdot D=m\end{array}[/latex-display] 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-display]\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-display] Therefore, [latex] m=D\cdot v[/latex] and [latex] v=\frac{m}{D}[/latex].

Now let us look at an example using the formula for the volume of a cylinder.


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).

Answer: Start with the formula for the volume of a cylinder. [latex-display] V=\pi{{r}^{2}}h[/latex-display] Divide both sides by [latex] \pi {{r}^{2}}[/latex] to isolate h. [latex-display] \frac{V}{\pi {{r}^{2}}}=\frac{\pi {{r}^{2}}h}{\pi {{r}^{2}}}[/latex-display] Simplify. You find the height, h, is equal to [latex] \frac{V}{\pi {{r}^{2}}}[/latex]. [latex-display] \frac{V}{\pi {{r}^{2}}}=h[/latex-display] Therefore, [latex] h=\frac{V}{\pi {{r}^{2}}}[/latex].

In the following video, we give another example of solving for a variable in a formula, or as they are also called, a literal equation. https://www.youtube.com/watch?v=ecEUUbRLDQs&feature=youtu.be


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.
Man with a lunch box walking. THere is a caption above him that says Finished after a good day's work.
A “work problem” is an example of a real life situation that can be modeled and solved using a rational equation. Work problems often ask you to calculate how long it will take different people working at different speeds to finish a task. The algebraic models of such situations often involve rational equations derived from the work formula, [latex]W=rt[/latex]. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or [latex]d=rt[/latex].) The amount of work done (W) is the product of the rate of work (r) and the time spent working (t). The work formula has [latex]3[/latex] versions.


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

Answer: 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-display] \frac{25\,\,\text{bulbs}}{1\,\,\text{hour}}+\frac{15\,\,\text{bulbs}}{1\,\,\text{hour}}=\frac{40\,\,\text{bulbs}}{1\,\,\text{hour}}[/latex-display] 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-display] \frac{40}{1}=\frac{150}{t}[/latex-display] Solve the equation by multiplying both sides by the common denominator and then isolating t. [latex-display]\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-display] It should take [latex]3[/latex] hours [latex]45[/latex] minutes for Myra and Francis to plant [latex]150[/latex] bulbs together.

https://www.youtube.com/watch?v=SzSasnDF7Ms&feature=youtu.be Other work problems go the other way. 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?

Answer: 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-display] \begin{array}{l}1=\left( \frac{1}{x}+\frac{1}{3x} \right)24\\\\1=\frac{24}{x}+\frac{24}{3x}\end{array}[/latex-display] Now solve the equation for x. (Remember that x represents the number of hours it will take Joe to finish the job.) [latex-display]\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-display] Check the solution in the original equation. [latex-display]\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-display] 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.


It takes [latex]32[/latex] hours for Joe to paint the house by himself and [latex]96[/latex] hours for John the paint the house himself.

In the video that follows, we show another example of finding one person's work rate given a combined work rate. https://www.youtube.com/watch?v=kbRSYb8UYqU&feature=youtu.be As shown above, many work problems can be represented by the equation [latex] \frac{t}{a}+\frac{t}{b}=1[/latex], where t is the time to do the job together, a is the time it takes person A to do the job, and b is the time it takes person B to do the job. The [latex]1[/latex] refers to the total work done—in this case, the work was to paint [latex]1[/latex] 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.

Try It

We present another example of two people painting at different rates in the following video. https://youtu.be/SzSasnDF7Ms


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. The thing that can make mixtures interesting mathematically is when components of the mixture are added at different rates and concentrations. In our final example, we will define an equation that models the concentration  - or ratio of sugar to water - in a large mixing tank over time. You are asked whether the final concentration of sugar is greater than the concentration at the beginning.


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:

[latex]\begin{cases}\text{water: }W\left(t\right)=100+10t\text{ in gallons}\\ \text{sugar: }S\left(t\right)=5+1t\text{ in pounds}\end{cases}[/latex]

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

[latex-display]C\left(t\right)=\frac{5+t}{100+10t}[/latex-display] The concentration after [latex]12[/latex] minutes is given by evaluating [latex]C\left(t\right)[/latex] at [latex]t=\text{ }12[/latex]. [latex-display]C\left(12\right)=\frac{5+12}{100+10(12)}=\frac{17}{220}[/latex-display] 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-display]C\left(0\right)=\frac{5+0}{100+10(0)}=\frac{5}{100}=\frac{1}{20}[/latex-display] 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.

In the following video, we show another example of how to use rational functions to model mixing. https://youtu.be/GD6H7BE_0EI


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CC licensed content, Original

  • Screenshot: A Good Day's Work. Provided by: Lumen Learning License: CC BY: Attribution.
  • 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. License: CC BY: Attribution.

CC licensed content, Shared previously

  • Unit 15: Rational Expressions, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology Located at: https://www.nroc.org/. License: CC BY: Attribution.
  • Ex 1: Rational Equation Application - Painting Together. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • Ex: Rational Equation App - Find Individual Working Time Given Time Working Together. Authored by: James Sousa (Mathispower4u.com) . License: CC BY: Attribution.
  • College Algebra: Mixture Problem. Authored by: Abramson, Jay et al.. Located at: https://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.