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Study Guides > College Algebra CoRequisite Course

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).

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].


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].

Watch the following video for more examples of solving for a particular variable in a formula, a literal equation. https://www.youtube.com/watch?v=ecEUUbRLDQs&feature=youtu.be


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:


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?

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.

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?

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.

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-display]\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-display] Watch the following video for an example of finding the total time given two people working together at known rates. https://youtu.be/SzSasnDF7Ms The following video shows an example of finding one person's work rate given a known combined work rate. https://www.youtube.com/watch?v=kbRSYb8UYqU&feature=youtu.be


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:

[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.

The following video gives another example of how to use rational functions to model mixing. https://youtu.be/GD6H7BE_0EI

Licenses & Attributions

CC licensed content, Original

  • 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.
  • Ex 2: Solve a Literal Equation for a Variable. Authored by: James Sousa (Mathispower4u.com). License: CC BY: Attribution.