Rational Functions, Equations, and Inequalities

Rational Functions are just a ratio of two polynomials (expression with constants and/or variables), and are typically thought of as having at least one variable in the denominator (which can never be 0 ).

Note that we talk about how to graph rationals using their asymptotes in the  Graphing Rational Functions, including Asymptotes  section. Also, since  limits  exist with Rational Functions and their asymptotes, limits are discussed here in the  Limits and Continuity  section . Since factoring is so important in algebra, you may want to revisit it first. Remember that we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section, and more Advanced Factoring can be found here .

Introducing Rational Expressions

Again, think of a  rational expression  as a  ratio of two polynomials . Here are some examples of expressions that are and are not rational expressions:

Multiplying, Dividing, and Simplifying  Rationals

Frequently, rational expressions can be simplified by factoring the numerator, denominator, or both, and crossing out factors. They can be multiplied and divided like regular fractions.

Here are some examples. Note that these look really difficult, but we’re just using a lot of steps of things we already know. That’s the fun of math!  Also, note in the last example, we are dividing rationals , so we flip the second and multiply .

Remember that when you cross out factors, you can cross out from the top and bottom of the same fraction, or top and bottom from different factors that you are multiplying. You can never cross out two things on top, or two things on bottom.

Also remember that at any point in the problem , when variables are in the denominator, we’ll have domain restrictions , since denominators can’t be $ 0$.

Multiplying, Dividing, and Simplifying Rationals

$ \displaystyle \require{cancel} \frac{{\left( {3{{x}^{2}}-5x-2} \right)}}{{{{x}^{2}}-4}}=\frac{{\left( {3x+1} \right)\cancel{{\left( {x-2} \right)}}}}{{\left( {x+2} \right)\cancel{{\left( {x-2} \right)}}}}=\frac{{3x+1}}{{x+2}};\,\,\,x\ne -2,\,\,2$

$ \displaystyle \require{cancel} \begin{align}\frac{{4{{x}^{2}}+12xy+2x+6y}}{{2{{x}^{2}}+6xy+6x+18y}}\,&=\,\frac{{2\left( {2{{x}^{2}}+6xy+x+3y} \right)}}{{2\left( {{{x}^{2}}+3xy+3x+9y} \right)}}\,=\,\frac{{2\left[ {2x\left( {x+3y} \right)+1\left( {x+3y} \right)} \right]}}{{2\left[ {x\left( {x+3y} \right)+3\left( {x+3y} \right)} \right]}}\,\,\,\\&=\,\frac{{\cancel{2}\left( {2x+1} \right)\cancel{{\left( {x+3y} \right)}}}}{{\cancel{2}\left( {x+3} \right)\cancel{{\left( {x+3y} \right)}}}}\,=\,\frac{{2x+1}}{{x+3}};\,\,\,\,x\ne -3,\,\,-3y\end{align}$

$ \displaystyle \require{cancel} \begin{align}\frac{{\frac{{7{{y}^{2}}}}{{{{y}^{2}}-16}}}}{{\frac{{14{{y}^{2}}+49y}}{{3{{y}^{2}}-10y-8}}}}\,&=\,\frac{{\frac{{7{{y}^{2}}}}{{\left( {y-4} \right)\left( {y+4} \right)}}}}{{\frac{{7y\left( {2y+7} \right)}}{{\left( {y-4} \right)\left( {3y+2} \right)}}}}\,=\,\frac{{\cancel{7}{{{\cancel{{{{y}^{2}}}}}}^{y}}}}{{\cancel{{\left( {y-4} \right)}}\left( {y+4} \right)}}\cdot \frac{{\cancel{{\left( {y-4} \right)}}\left( {3y+2} \right)}}{{\cancel{7}\cancel{y}\left( {2y+7} \right)}}\,\,\,\\&=\,\frac{{y\left( {3y+2} \right)}}{{\left( {y+4} \right)\left( {2y+7} \right)}};\,\,\,\,\,\,y\ne -4,\,-\frac{7}{2},\,\,-\frac{2}{3},\,\,0,\,\,4\end{align}$

Finding the Common Denominator

When we add or subtract two or more rationals, we need to find the least common denominator (LCD) , just like when we add or subtract regular fractions. If the denominators are the same, we can just add the numerators across, leaving the denominators as they are. We then must be sure we can’t do any further factoring:

$ \require{cancel} \displaystyle \frac{2}{{3x}}+\frac{4}{{3x}}=\frac{{(2+4)}}{{3x}}=\frac{{{{{\cancel{6}}}^{2}}}}{{{{{\cancel{3}}}^{1}}x}}=\frac{2}{x};\,\,\,\,x\ne 0$.

Just like with regular fractions, we want to use the factors in the denominators in every fraction, but not repeat them across denominators. When nothing is common, just multiply the factors.

Let’s find the least common denominators for the following denominators (ignore the numerators for now).

Adding and Subtracting Rationals

Now let’s add and subtract the following rational expressions. Note that one way to look at finding the LCD is to multiply the top by what’s missing in the bottom . For example, in the first example, the LCD is $ \left( {x+3} \right)\left( {x+4} \right)$, and we need to multiply the first fraction’s numerator by $ \left( {x+4} \right)$, since that’s missing in the denominator.

 Restricted Domains of Rational Functions

As we’ve noticed, since rational functions have variables in denominators, we must make sure that the denominators won’t end up as “ 0 ” at any point of solving the problem .

Thus, the domain of $ \displaystyle \frac{{x+1}}{{2x(x-2)(x+3)}}$ is $ \{x:x\ne -3,0,2\}$. This means if we ever get a solution to an equation that contains rational expressions and has variables in the denominator (which they probably will!), we must make sure that none of our answers would make any denominator in that equation “ 0 ” . These “answers” that we can’t use are called extraneous solutions . We’ll see this in the first example below.

Solving Rational Equations

When we solve rational equations, we can multiply both sides of the equations by the least common denominator , or LCD (which is $ \displaystyle \frac{{\text{least common denominator}}}{1}$ in fraction form), and not even worry about working with fractions! The denominators will cancel out and we just solve the equation using the numerators. Remember that with quadratics, we need to get everything to one side with 0 on the other side and either factor or use the Quadratic Formula .

Again, think of multiplying the top by what’s missing in the bottom from the LCD .

Notice that sometimes you’ll have to solve literal equations , which just means that you have to solve an equation for a variable, but you’ll have other variables in the answer. The last example shows this.

Rational Inequalities, including Absolute Values

Solving  rational inequalities  are a little more complicated since we are typically multiplying or dividing by variables, and we don’t know whether these are positive or negative. Remember that we have to  change the direction of the inequality when we multiply or divide by negative numbers . When we solve these rational inequalities, our answers will typically be a range of numbers .

Rational Inequalities from a Graph

It’s not too bad to see inequalities of rational functions from a graph . Look at this graph to see where $ y<0$ and $ y\ge 0$. Notice that we have ranges of $ x$  values in the two cases:

Solving Rational Inequalities Algebraically Using a Sign Chart

The easiest way to solve rational inequalities algebraically is using the sign chart method , which we saw here in the Quadratic Inequalities section . A sign chart or sign pattern is simply a number line that is separated into partitions (intervals or regions), with boundary points (called “ critical values “) that you get by setting the factors of the rational function, both in the numerator and denominator, to 0 and solving for $ x$.

With sign charts, you can pick any point in between the critical values , and see if the whole function is positive or negative . Then you just pick that interval (or intervals) by looking at the inequality. Generally, if the inequality includes the $ =$ sign, you have a closed bracket, and if it doesn’t, you have an open bracket. But any factor that’s in the denominator must have an open bracket for the values that make it 0 , since you can’t have 0 in the denominator.

The first thing you have to do is get everything on the left side (if it isn’t already there) and 0 on the right side , since we can see what intervals make the inequality true. We can only have one term on the left side , so sometimes we have to find a common denominator and combine terms.

Also, it’s a good idea to put open or closed circles  on the critical values to remind ourselves if we have inclusive points (inequalities with equal signs, such as $ \le $ and $ \ge $) or exclusive points (inequalities without equal signs, or factors in the denominators ).

Let’s do some examples; y ou can always use your  graphing calculator  to check your answers, too. Put in both sides of the inequalities and check the zeros, and make sure your ranges are correct!

And here’s another one where we have to do a lot of “organizing” first, including moving terms to one side, and then factoring:

Sometimes we get a funny interval with a single $ x$ value as part of the interval:

And here’s one where we have a removable discontinuity in the rational inequality, so we have to make sure we skip over that point:

Here are a couple that involves solving radical inequalities with absolute values . (You might want to review  Solving Absolute Value Equations and Inequalities   before continuing on to this topic.)

Here are more complicated ones, where the absolute value may need to be multiplied by other variables (think of if you had to cross multiply). Notice how it’s best to separate the inequality into two separate inequalities: one case when $ x$ is positive, and the other when $ x$ is negative. Notice also how we had to use the  Quadratic Formula  to get the critical points when $ x$ is negative:

The problem calls for $ >0$, so we look for the plus sign intervals. But we have to throw away any intervals where the sign doesn’t agree with our conditions of $ x$ (positive or negative), such as the interval $ \left( {.434,1} \right)$ ($ x$ is supposed to be negative). Tricky!

Look at the negative part of the sign chart when $ x$ is negative, and the positive part of the sign chart when $ x$ is positive.

When $ x$ is positive, the numerator yields no “real” critical points. When $ x$   is negative, use the Quadratic Formula to get the roots (critical points) of the numerator:

$ \displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}=\frac{{1\pm \sqrt{{{{{\left( {-1} \right)}}^{2}}-4\left( {-3} \right)\left( 1 \right)}}}}{{-6}}$

$ \displaystyle =\frac{{1\pm \sqrt{{13}}}}{{-6}}\approx -.768,\,\,\,.434$

The answer is $ \displaystyle \left( {.768,-\frac{1}{2}} \right)\cup \left( {1,\infty } \right)$.

Put the inequality in your graphing calculator to check your answer!

Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it. In this case, we have to separate in four cases , just to be sure we cover all the possibilities.

Use sign charts to determine what intervals the rational will be positive or negative. The problem calls for $ >0$, so look for the plus sign intervals, and make sure they work in the original inequality .

The answer is $ \displaystyle \left( {-1,-\frac{1}{3}} \right)\cup \left( {-\frac{1}{3},0} \right)$. We have to “skip over” (asymptote) $ \displaystyle -\frac{1}{3}$ (so we don’t divide by 0 ), and use soft brackets, since the inequality is $ >$ and not $ \ge $.

There’s one more that we did  here in the  Compositions of Functions, Even and Odd, and Increasing and Decreasing  section, when we worked on domains of composites.

Applications of Rationals

There are certain types of word problems that typically use rational expressions. These tend to deal with rates , since rates are typically fractions (such as distance over time). We also see problems dealing with plain fractions or percentages in fraction form.

Here is a rational “fraction” problem:

Here is a rational “percentage” problem:

Distance/Rate/Time Problems

With rational rate problems, we must always remember: $ \text{Distance}=\text{Rate }\times \,\,\text{Time}$ . It seems most of the problems deal with comparing times or adding times .

Shalini can run 3 miles per hour faster than her sister Meena can walk. If Shalini ran 12 miles in the same time it took Meena to walk 8 miles, what is the speed of each sister in this case ?

This is a “$ \text{Distance}=\text{Rate }\times \,\,\text{Time}$” problem, and let’s use a table to organize this information like we did in the Algebra Word Problems and  Systems of Linear Equations and Word Problems   sections. Let $ x=$ Meena’s speed, since Shalini runs faster and it’s easier to add than subtract:

Put it all together:

Work Problems:

Work problems typically have to do with different people or things working together and alone, at different rates. Instead of distance, we work with jobs (typically, 1 complete job).

I find that usually the easiest way to work these problems is to remember:

$ \displaystyle \frac{{\operatorname{time} \,\text{together}}}{{\text{time alone}}}\,\,+\,…\,\,+\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,=\,\,1$    (or whatever part of the job or jobs is done; if they do half the job, this equals $ \displaystyle \frac{1}{2}$; if they do a job twice, this equals 2 ).

(Use a “$ +$” between the terms if working towards the same goal, such as painting a room, and “$ -$” if working towards opposite goals, such as filling and emptying a pool.)

“Proof” : For work problems, $ \text{Rate }\times \,\text{Time = }1\text{ Job}$. Add up individuals’ portions of a job with this formula, using the time working with others (time together):

$ \displaystyle \begin{align}\left( {\text{individual rate }\times \text{ time }} \right)\text{ + }…\text{ +}\left( {\text{individual rate }\times \text{ time }} \right)&=1\\\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)\text{ + }…\text{ +}\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)&=1\\\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)\text{ + }…\text{ +}\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)&=1\end{align}$

Also, as explained after the first example below, often you see this formula as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.

Let’s do some problems:

Note : The formula above can also be derived by using the concept that you can figure out how much of the job the girls do per hour (or whatever the time unit is), both together and alone . Then you can add the individual “rates” to get the “rate” of their painting together.

We are actually adding the Work they complete (together and alone) using formula $ \text{Rate }\times \text{ }\text{Time }=\text{ Work}$, where the Time is 1 hour (or whatever the unit is).

In this example, Erica’s rate per hour is $ \displaystyle \frac{1}{5}$ (she can do $ \displaystyle \frac{1}{5}$ of the job in 1 hour); Rachel’s rate per hour is $ \displaystyle \frac{1}{R}$; we can add their rates to get the rate of their painting together: $ \displaystyle \frac{1}{5}+\frac{1}{R}=\frac{1}{3}$. If we multiply all the terms by 3 , we come up with the equation above! (Sometimes you see this as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.)

Here is a systems problem and also a work problem at the same time. We did this problem without using rationals here in the Systems of Linear Equations and Word Problems section (and be careful, since the variables we assigned were different).

Cost Problems

A third type of problem you might get while studying rationals has to do with average cost , or possibly costs per person (or unit cost) problems. Both of these types of costs can still be thought as rates , as they are an amount over time. Here are some examples:

Rational Inequality Word Problem

Here is a rational inequality word problem that we saw in the Algebra Word Problems section .

Note that there is a Rational Asymptote Application Problem here in the Graphing Rational Functions, including Asymptotes section .

Understand these problems, and practice, practice, practice!

For Practice : Use the Mathway  widget below to try a Rational Function  problem. Click on Submit (the blue arrow to the right of the problem) and click on Solve for x  to see the answer.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on Tap to view steps , or Click Here , you can register at Mathway for a free trial , and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to  Graphing Rational Functions, including Asymptotes     – you’re ready! 

Rational Equations Word Problems Lesson

  • Demonstrate an understanding of how to solve a word problem
  • Demonstrate an understanding of how to solve an equation with rational expressions
  • Learn how to solve word problems that involve rational equations

How to Solve a Word Problem with Rational Equations

Six-step method for solving word problems with rational expressions.

  • Read the problem carefully and determine what you are asked to find
  • Assign a variable to represent the unknown
  • Write out an equation that describes the given situation
  • Solve the equation
  • State the answer using a nice clear sentence
  • Check the result by reading back through the problem

Solving a Proportion Problem

Motion word problems with rational expressions, rate of work word problems, skills check:.

Solve each word problem.

Working alone, it takes Steve 11 hours to complete a restoration project on a truck. Jacob can perform the same task in 110 hours. How long would it take if they worked together?

Please choose the best answer.

Jamie’s hot tub has an outlet pipe that can empty the hot tub in 6 minutes. Additionally, her hot tub has an inlet pipe that can fill the hot tub in 3 minutes. If both pipes were turned on, how long would it take to fill a completely empty hot tub?

On her drive from Port Smith to Maryland, Stephanie averaged 51 miles per hour. If she had been able to average 60 miles per hour, she would have reached Maryland 3 hours earlier. What is the driving distance between Port Smith and Maryland?

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Ready for more? Watch the Step by Step Video Lesson   |   Take the Practice Test

  • 7.6 Solve Rational Inequalities
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.1 Solve Systems of Linear Equations with Two Variables
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.4 General Strategy for Factoring Polynomials
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Solve rational inequalities
  • Solve an inequality with rational functions

Be Prepared 7.16

Before you get started, take this readiness quiz.

Find the value of x − 5 x − 5 when ⓐ x = 6 x = 6 ⓑ x = −3 x = −3 ⓒ x = 5 . x = 5 . If you missed this problem, review Example 1.6 .

Be Prepared 7.17

Solve: 8 − 2 x < 12 . 8 − 2 x < 12 . If you missed this problem, review Example 2.52 .

Be Prepared 7.18

Write in interval notation: −3 ≤ x < 5 . −3 ≤ x < 5 . If you missed this problem, review Example 2.49 .

Solve Rational Inequalities

We learned to solve linear inequalities after learning to solve linear equations. The techniques were very much the same with one major exception. When we multiplied or divided by a negative number, the inequality sign reversed.

Having just learned to solve rational equations we are now ready to solve rational inequalities. A rational inequality is an inequality that contains a rational expression.

Rational Inequality

A rational inequality is an inequality that contains a rational expression.

Inequalities such as 3 2 x > 1 , 2 x x − 3 < 4 , 2 x − 3 x − 6 ≥ x , 3 2 x > 1 , 2 x x − 3 < 4 , 2 x − 3 x − 6 ≥ x , and 1 4 − 2 x 2 ≤ 3 x 1 4 − 2 x 2 ≤ 3 x are rational inequalities as they each contain a rational expression.

When we solve a rational inequality, we will use many of the techniques we used solving linear inequalities. We especially must remember that when we multiply or divide by a negative number, the inequality sign must reverse.

Another difference is that we must carefully consider what value might make the rational expression undefined and so must be excluded.

When we solve an equation and the result is x = 3 , x = 3 , we know there is one solution, which is 3.

When we solve an inequality and the result is x > 3 , x > 3 , we know there are many solutions. We graph the result to better help show all the solutions, and we start with 3. Three becomes a critical point and then we decide whether to shade to the left or right of it. The numbers to the right of 3 are larger than 3, so we shade to the right.

To solve a rational inequality, we first must write the inequality with only one quotient on the left and 0 on the right.

Next we determine the critical points to use to divide the number line into intervals. A critical point is a number which make the rational expression zero or undefined.

We then will evaluate the factors of the numerator and denominator, and find the quotient in each interval. This will identify the interval, or intervals, that contains all the solutions of the rational inequality.

We write the solution in interval notation being careful to determine whether the endpoints are included.

Example 7.54

Solve and write the solution in interval notation: x − 1 x + 3 ≥ 0 . x − 1 x + 3 ≥ 0 .

  • Step 1. Write the inequality as one quotient on the left and zero on the right.

Our inequality is in this form. x − 1 x + 3 ≥ 0 x − 1 x + 3 ≥ 0

Step 2. Determine the critical points—the points where the rational expression will be zero or undefined.

The rational expression will be zero when the numerator is zero. Since x − 1 = 0 x − 1 = 0 when x = 1 , x = 1 , then 1 1 is a critical point.

The rational expression will be undefined when the denominator is zero. Since x + 3 = 0 x + 3 = 0 when x = −3 , x = −3 , then −3 −3 is a critical point.

The critical points are 1 and −3 . −3 .

  • Step 3. Use the critical points to divide the number line into intervals.

The number line is divided into three intervals:

( − ∞ , −3 ) ( −3 , 1 ) ( 1 , ∞ ) ( − ∞ , −3 ) ( −3 , 1 ) ( 1 , ∞ )

Step 4. Test a value in each interval. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

To find the sign of each factor in an interval, we choose any point in that interval and use it as a test point. Any point in the interval will give the expression the same sign, so we can choose any point in the interval.

The number −4 −4 is in the interval ( − ∞ , −3 ) . ( − ∞ , −3 ) . Test x = −4 x = −4 in the expression in the numerator and the denominator.

Above the number line, mark the factor x − 1 x − 1 negative and mark the factor x + 3 x + 3 negative.

Since a negative divided by a negative is positive, mark the quotient positive in the interval ( − ∞ , −3 ) . ( − ∞ , −3 ) .

The number 0 is in the interval ( −3 , 1 ) . ( −3 , 1 ) . Test x = 0 . x = 0 .

Above the number line, mark the factor x − 1 x − 1 negative and mark x + 3 x + 3 positive.

Since a negative divided by a positive is negative, the quotient is marked negative in the interval ( −3 , 1 ) . ( −3 , 1 ) .

The number 2 is in the interval ( 1 , ∞ ) . ( 1 , ∞ ) . Test x = 2 . x = 2 .

Above the number line, mark the factor x − 1 x − 1 positive and mark x + 3 x + 3 positive.

Since a positive divided by a positive is positive, mark the quotient positive in the interval ( 1 , ∞ ) . ( 1 , ∞ ) .

  • Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

We want the quotient to be greater than or equal to zero, so the numbers in the intervals ( − ∞ , −3 ) ( − ∞ , −3 ) and ( 1 , ∞ ) ( 1 , ∞ ) are solutions.

But what about the critical points?

The critical point x = −3 x = −3 makes the denominator 0, so it must be excluded from the solution and we mark it with a parenthesis.

The critical point x = 1 x = 1 makes the whole rational expression 0. The inequality requires that the rational expression be greater than or equal to 0. So, 1 is part of the solution and we will mark it with a bracket.

Recall that when we have a solution made up of more than one interval we use the union symbol, ∪ , ∪ , to connect the two intervals. The solution in interval notation is ( − ∞ , −3 ) ∪ [ 1 , ∞ ) . ( − ∞ , −3 ) ∪ [ 1 , ∞ ) .

Try It 7.107

Solve and write the solution in interval notation: x − 2 x + 4 ≥ 0 . x − 2 x + 4 ≥ 0 .

Try It 7.108

Solve and write the solution in interval notation: x + 2 x − 4 ≥ 0 . x + 2 x − 4 ≥ 0 .

We summarize the steps for easy reference.

Solve a rational inequality.

  • Step 2. Determine the critical points–the points where the rational expression will be zero or undefined.
  • Step 4. Test a value in each interval. Above the number line show the sign of each factor of the numerator and denominator in each interval. Below the number line show the sign of the quotient.

The next example requires that we first get the rational inequality into the correct form.

Example 7.55

Solve and write the solution in interval notation: 4 x x − 6 < 1 . 4 x x − 6 < 1 .

Try It 7.109

Solve and write the solution in interval notation: 3 x x − 3 < 1 . 3 x x − 3 < 1 .

Try It 7.110

Solve and write the solution in interval notation: 3 x x − 4 < 2 . 3 x x − 4 < 2 .

In the next example, the numerator is always positive, so the sign of the rational expression depends on the sign of the denominator.

Example 7.56

Solve and write the solution in interval notation: 5 x 2 − 2 x − 15 > 0 . 5 x 2 − 2 x − 15 > 0 .

Try It 7.111

Solve and write the solution in interval notation: 1 x 2 + 2 x − 8 > 0 . 1 x 2 + 2 x − 8 > 0 .

Try It 7.112

Solve and write the solution in interval notation: 3 x 2 + x − 12 > 0 . 3 x 2 + x − 12 > 0 .

The next example requires some work to get it into the needed form.

Example 7.57

Solve and write the solution in interval notation: 1 3 − 2 x 2 < 5 3 x . 1 3 − 2 x 2 < 5 3 x .

Try It 7.113

Solve and write the solution in interval notation: 1 2 + 4 x 2 < 3 x . 1 2 + 4 x 2 < 3 x .

Try It 7.114

Solve and write the solution in interval notation: 1 3 + 6 x 2 < 3 x . 1 3 + 6 x 2 < 3 x .

Solve an Inequality with Rational Functions

When working with rational functions, it is sometimes useful to know when the function is greater than or less than a particular value. This leads to a rational inequality.

Example 7.58

Given the function R ( x ) = x + 3 x − 5 , R ( x ) = x + 3 x − 5 , find the values of x that make the function less than or equal to 0.

We want the function to be less than or equal to 0.

Try It 7.115

Given the function R ( x ) = x − 2 x + 4 , R ( x ) = x − 2 x + 4 , find the values of x that make the function less than or equal to 0.

Try It 7.116

Given the function R ( x ) = x + 1 x − 4 , R ( x ) = x + 1 x − 4 , find the values of x that make the function less than or equal to 0.

In economics, the function C ( x ) C ( x ) is used to represent the cost of producing x units of a commodity. The average cost per unit can be found by dividing C ( x ) C ( x ) by the number of items x . x . Then, the average cost per unit is c ( x ) = C ( x ) x . c ( x ) = C ( x ) x .

Example 7.59

The function C ( x ) = 10 x + 3000 C ( x ) = 10 x + 3000 represents the cost to produce x , x , number of items. Find ⓐ the average cost function, c ( x ) c ( x ) ⓑ how many items should be produced so that the average cost is less than $40.

More than 100 items must be produced to keep the average cost below $40 per item.

Try It 7.117

The function C ( x ) = 20 x + 6000 C ( x ) = 20 x + 6000 represents the cost to produce x , x , number of items. Find ⓐ the average cost function, c ( x ) c ( x ) ⓑ how many items should be produced so that the average cost is less than $60?

Try It 7.118

The function C ( x ) = 5 x + 900 C ( x ) = 5 x + 900 represents the cost to produce x , x , number of items. Find ⓐ the average cost function, c ( x ) c ( x ) ⓑ how many items should be produced so that the average cost is less than $20?

Practice Makes Perfect

In the following exercises, solve each rational inequality and write the solution in interval notation.

x − 3 x + 4 ≥ 0 x − 3 x + 4 ≥ 0

x + 6 x − 5 ≥ 0 x + 6 x − 5 ≥ 0

x + 1 x − 3 ≤ 0 x + 1 x − 3 ≤ 0

x − 4 x + 2 ≤ 0 x − 4 x + 2 ≤ 0

x − 7 x − 1 > 0 x − 7 x − 1 > 0

x + 8 x + 3 > 0 x + 8 x + 3 > 0

x − 6 x + 5 < 0 x − 6 x + 5 < 0

x + 5 x − 2 < 0 x + 5 x − 2 < 0

3 x x − 5 < 1 3 x x − 5 < 1

5 x x − 2 < 1 5 x x − 2 < 1

6 x x − 6 > 2 6 x x − 6 > 2

3 x x − 4 > 2 3 x x − 4 > 2

2 x + 3 x − 6 ≤ 1 2 x + 3 x − 6 ≤ 1

4 x − 1 x − 4 ≤ 1 4 x − 1 x − 4 ≤ 1

3 x − 2 x − 4 ≥ 2 3 x − 2 x − 4 ≥ 2

4 x − 3 x − 3 ≥ 2 4 x − 3 x − 3 ≥ 2

1 x 2 + 7 x + 12 > 0 1 x 2 + 7 x + 12 > 0

1 x 2 − 4 x − 12 > 0 1 x 2 − 4 x − 12 > 0

3 x 2 − 5 x + 4 < 0 3 x 2 − 5 x + 4 < 0

4 x 2 + 7 x + 12 < 0 4 x 2 + 7 x + 12 < 0

2 2 x 2 + x − 15 ≥ 0 2 2 x 2 + x − 15 ≥ 0

6 3 x 2 − 2 x − 5 ≥ 0 6 3 x 2 − 2 x − 5 ≥ 0

−2 6 x 2 − 13 x + 6 ≤ 0 −2 6 x 2 − 13 x + 6 ≤ 0

−1 10 x 2 + 11 x − 6 ≤ 0 −1 10 x 2 + 11 x − 6 ≤ 0

1 2 + 12 x 2 > 5 x 1 2 + 12 x 2 > 5 x

1 3 + 1 x 2 > 4 3 x 1 3 + 1 x 2 > 4 3 x

1 2 − 4 x 2 ≤ 1 x 1 2 − 4 x 2 ≤ 1 x

1 2 − 3 2 x 2 ≥ 1 x 1 2 − 3 2 x 2 ≥ 1 x

1 x 2 − 16 < 0 1 x 2 − 16 < 0

4 x 2 − 25 > 0 4 x 2 − 25 > 0

4 x − 2 ≥ 3 x + 1 4 x − 2 ≥ 3 x + 1

5 x − 1 ≤ 4 x + 2 5 x − 1 ≤ 4 x + 2

In the following exercises, solve each rational function inequality and write the solution in interval notation.

Given the function R ( x ) = x − 5 x − 2 , R ( x ) = x − 5 x − 2 , find the values of x x that make the function less than or equal to 0.

Given the function R ( x ) = x + 1 x + 3 , R ( x ) = x + 1 x + 3 , find the values of x x that make the function greater than or equal to 0.

Given the function R ( x ) = x − 6 x + 2 R ( x ) = x − 6 x + 2 , find the values of x that make the function less than or equal to 0.

Writing Exercises

Write the steps you would use to explain solving rational inequalities to your little brother.

Create a rational inequality whose solution is ( − ∞ , −2 ] ∪ [ 4 , ∞ ) . ( − ∞ , −2 ] ∪ [ 4 , ∞ ) .

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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High School Math : Solving Rational Equations and Inequalities

Study concepts, example questions & explanations for high school math, all high school math resources, example questions, example question #1 : solving rational equations and inequalities.

Solve the following rational expression:

solving rational equations and inequalities word problems

Solve the following rational equation:

solving rational equations and inequalities word problems

Example Question #3 : Solving Rational Equations And Inequalities

solving rational equations and inequalities word problems

Often, when solving equations involving rational expressions, it helps to elminate fractions by multiplying both sides of the equation by the denominators of each term intervolved. 

In the context of this problem, we can first multiply both sides of the equation by x+2 to eliminate the denominator of the first term.

solving rational equations and inequalities word problems

Be sure to distribute the x+2 to each term on the left side of the equation.

solving rational equations and inequalities word problems

Next, multiply both sides of the equation by x to elminate the term with an x in the denominator.

solving rational equations and inequalities word problems

Then, multiply both sides of the equation by 2.

solving rational equations and inequalities word problems

In order to solve this polynomial equation, we can use the Rational Root Theorem. According to this theorem, if there is a rational root to a polynomial equation, then that root must be in the form p/q, where p is a factor of the constant, and q is the factor of the coefficient of the highest term. 

In the context of this problem, p will be a factor of 4 (which is the constant), and q will be a factor of 2 (which is the coefficient of the highest term). 

solving rational equations and inequalities word problems

We will put -1/2 into the polynomial to see if it evaluates to zero.

solving rational equations and inequalities word problems

First, divide both sides of the equation by two. 

solving rational equations and inequalities word problems

Perhaps the most straightforward way to solve this is to use the quadratic formula:

solving rational equations and inequalities word problems

The question asks us to find the largest real value of x that solves the equation. Because -1/2 is the only real value that solves the equation, the answer must be -1/2.

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Solving Rational Inequalities

A Rational Expression looks like:

Inequalities

Sometimes we need to solve rational inequalities like these:

Solving inequalities is very like solving equations ... you do most of the same things.

These are the steps:

  • the "=0" points (roots), and
  • "vertical asymptotes" (where the function is undefined)
  • in between the "points of interest", the function is either greater than zero (>0) or less than zero (<0)
  • then pick a test value to find out which it is (>0 or <0)

Here is an example:

Example: 3x−10 x−4 > 2

First , let us simplify!

But You Cannot Multiply By (x−4)

Because "x−4" could be positive or negative ... we don't know if we should change the direction of the inequality or not. This is all explained on Solving Inequalities .

Instead, bring "2" to the left:

3x−10 x−4 − 2 > 0

Then multiply 2 by (x−4)/(x−4):

3x−10 x−4 − 2 x−4 x−4 > 0

Now we have a common denominator, let's bring it all together:

3x−10 − 2(x−4) x−4 > 0

x−2 x−4 > 0

Second , let us find "points of interest".

At x=2 we have: (0)/(x−4) > 0 , which is a "=0" point, or root

At x=4 we have: (x−2)/(0) > 0 , which is undefined

Third , do test points to see what it does in between:

  • x−2 = −2, which is negative
  • x−4 = −4, which is also negative
  • So (x−2)/(x−4) must be positive

We can do the same for x=3 and x=5 , and end up with these results:

That gives us a complete picture!

And where is it > 0 ?

  • Less than 2
  • More than 4

So our result is:

(−∞, 2) U (4, +∞)

We did all that without drawing a plot!

But here is the plot of (x−2)/(x−4) so you can see:

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Mathematics LibreTexts

4.3: Rational Inequalities and Applications

  • Last updated
  • Save as PDF
  • Page ID 119162

  • Carl Stitz & Jeff Zeager
  • Lakeland Community College & Lorain County Community College

Math 370 Learning Objectives

  • Solve rational inequalities using graphs and/or sign charts
  • Build models to solve applications involving rational functions

Solving Rational Inequalities

In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality.

Example \( \PageIndex{1} \)

  • Solve \(\frac{x^3-2x+1}{x-1} = \frac{1}{2}x-1\).
  • Solve \(\frac{x^3-2x+1}{x-1} \geq \frac{1}{2}x-1\).
  • Use graphing technology to check your answers to 1 and 2.
  • To solve the equation, we clear denominators\[\begin{array}{rclr} \dfrac{x^3-2x+1}{x-1} & = & \dfrac{1}{2}x-1 & \\  \left(\dfrac{x^3-2x+1}{x-1}\right) \cdot 2(x-1) & = & \left( \dfrac{1}{2}x-1 \right) \cdot 2(x-1) & \\ 2x^3 - 4x + 2 & = & x^2-3x+2 & \left( \text{expand} \right) \\ 2x^3 -x^2 - x & = & 0 & \\ x(2x+1)(x-1) & = & 0 & \left( \text{factor} \right) \\ x & = & -\dfrac{1}{2}, \, 0, \, 1 & \\ \end{array}\nonumber\]Since we cleared denominators, we need to check for extraneous solutions. Sure enough, we see that \(x=1\) does not satisfy the original equation and must be discarded. Our solutions are \(x=-\frac{1}{2}\) and \(x=0\).

Screen Shot 2022-04-01 at 1.16.25 AM.png

Subsection Footnotes

1  There is no asymptote at \(x = 1\) since the graph is well behaved near \(x = 1\). According to Theorem 4.1.1 , there must be a hole there.

Applications Involving Rational Equations

Next, we explore how rational equations can be used to solve some classic problems involving rates.

Example \( \PageIndex{2} \)

Carl decides to explore the Meander River, the location of several recent Sasquatch sightings. From camp, he canoes downstream five miles to check out a purported Sasquatch nest. Finding nothing, he immediately turns around, retraces his route (this time traveling upstream), and returns to camp 3 hours after he left. If Carl canoes at a rate of 6 miles per hour in still water, how fast was the Meander River flowing on that day?

We are given information about distances, rates (speeds) and times. The basic principle relating these quantities is: \[\text{distance} = \text{rate} \cdot \text{time}\nonumber\] The first observation to make, however, is that the distance, rate and time given to us aren’t "compatible": the distance given is the distance for only part of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the entire trip. Ultimately, we are after the speed of the river, so let’s call that \(R\) measured in miles per hour to be consistent with the other rate given to us. To get started, let’s divide the trip into its two parts: the initial trip downstream and the return trip upstream. For the downstream trip, all we know is that the distance traveled is \(5\) miles.

\[\begin{array}{rcl} \text{distance downstream} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ 5 \, \text{miles} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ \end{array}\nonumber\]

Since the return trip upstream followed the same route as the trip downstream, we know that the distance traveled upstream is also 5 miles.

\[\begin{array}{rcl} \text{distance upstream} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ 5 \, \text{miles} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ \end{array}\nonumber\]

We are told Carl can canoe at a rate of \(6\) miles per hour in still water. How does this figure into the rates traveling upstream and downstream? The speed the canoe travels in the river is a combination of the speed at which Carl can propel the canoe in still water, 6 miles per hour, and the speed of the river, which we’re calling \(R\). When traveling downstream, the river is helping Carl along, so we add these two speeds:

\[\begin{array}{rcl} \text{rate traveling downstream} & = & \text{rate Carl propels the canoe} + \text{speed of the river} \\ & = & 6 \dfrac{\text{miles}}{\text{hour}} + R \dfrac{\text{miles}}{\text{hour}} \\ \end{array}\nonumber\]

So our downstream speed is \((6+R) \frac{\text{miles}}{\text{hour}}\). Substituting this into our "distance-rate-time" equation for the downstream part of the trip, we get:

\[\begin{array}{rcl} 5 \, \text{miles} & = & \text{rate traveling downstream} \cdot \text{time traveling downstream} \\ 5 \, \text{miles} & = & (6+R) \dfrac{\text{miles}}{\text{hour}} \cdot \text{time traveling downstream} \\ \end{array}\nonumber\]

When traveling upstream, Carl works against the current. Since the canoe manages to travel upstream, the speed Carl can canoe in still water is greater than the river’s speed, so we subtract the river’s speed from Carl’s canoeing speed to get:

\[\begin{array}{rcl} \text{rate traveling upstream} & = & \text{rate Carl propels the canoe} - \text{river speed} \\ & = & 6 \dfrac{\text{miles}}{\text{hour}} - R \dfrac{\text{miles}}{\text{hour}} \\ \end{array}\nonumber\]

Proceeding as before, we get

\[\begin{array}{rcl} 5 \, \text{miles} & = & \text{rate traveling upstream} \cdot \text{time traveling upstream} \\ 5 \, \text{miles} & = & (6 - R) \dfrac{\text{miles}}{\text{hour}} \cdot \text{time traveling upstream} \\ \end{array}\nonumber\]

The last piece of information given to us is that the total trip lasted \(3\) hours. If we let \(t_{\text{down}}\) denote the time of the downstream trip and \(t_{\text{up}}\) the time of the upstream trip, we have: \(t_{\text{down}} + t_{\text{up}} = 3 \, \text{hours}\). Substituting \(t_{\text{down}}\) and \(t_{\text{up}}\) into the "distance-rate-time" equations, we get (suppressing the units) three equations in three unknowns: 2  \[\left\{\begin{array}{lrcl} E1 & (6+R) \, t_{\text{down}} & = & 5 \\ E2 & (6-R) \, t_{\text{up}} & = & 5 \\ E3 & t_{\text{down}} + t_{\text{up}} & = & 3 \end{array} \right.\nonumber\]

Since we are ultimately after \(R\), we need to use these three equations to get at least one equation involving only \(R\). To that end, we solve \(E1\) for \(t_{\text{down}}\) by dividing both sides 3  by the quantity \((6+R)\) to get \(t_{\text{down}} = \frac{5}{6+R}\). Similarly, we solve \(E2\) for \(t_{\text{up}}\) and get \(t_{\text{up}} = \frac{5}{6-R}\). Substituting these into \(E3\), we get: 4  \[\dfrac{5}{6+R} + \dfrac{5}{6 - R} = 3.\nonumber\] Clearing denominators, we get \(5(6-R) + 5(6+R) = 3(6+R)(6-R)\) which reduces to \(R^2 = 16\). We find \(R = \pm 4\), and since \(R\) represents the speed of the river, we choose \(R = 4\). On the day in question, the Meander River is flowing at a rate of \(4\) miles per hour.

One of the important lessons to learn from Example \( \PageIndex{2} \) is that speeds, and more generally, rates, are additive. As we see in our next example, the concept of rate and its associated principles can be applied to a wide variety of problems - not just "distance-rate-time" scenarios.

Example \( \PageIndex{3} \)

Working alone, Taylor can weed the garden in 4 hours. If Carl helps, they can weed the garden in 3 hours. How long would it take for Carl to weed the garden on his own?

The key relationship between work and time which we use in this problem is: \[\text{amount of work done} = \text{rate of work} \cdot \text{time spent working}\nonumber\]

We are told that, working alone, Taylor can weed the garden in 4 hours. In Taylor’s case then: \[\begin{array}{rcl} \text{amount of work Taylor does} & = & \text{rate of Taylor working} \cdot \text{time Taylor spent working} \\ 1 \, \text{garden} & = & (\text{rate of Taylor working}) \cdot (4 \, \text{hours}) \\ \end{array}\nonumber\]

So we have that the rate Taylor works is \(\frac{1 \, \text{garden}}{ 4 \, \text{hours}} = \frac{1}{4} \frac{\text{garden}}{\text{hour}}\). We are also told that when working together, Taylor and Carl can weed the garden in just 3 hours. We have:

\[\begin{array}{rcl} \text{amount of work done together} & = & \text{rate of working together} \cdot \text{time spent working together} \\ 1 \, \text{garden} & = & (\text{rate of working together}) \cdot (3 \, \text{hours}) \\ \end{array}\nonumber\]

From this, we find that the rate of Taylor and Carl working together is \(\frac{1 \, \text{garden}}{3 \, \text{hours}} = \frac{1}{3} \frac{\text{garden}}{\text{hour}}\). We are asked to find out how long it would take for Carl to weed the garden on his own. Let us call this unknown \(t\), measured in hours to be consistent with the other times given to us in the problem. Then:

\[\begin{array}{rcl} \text{amount of work Carl does} & = & \text{rate of Carl working} \cdot \text{time Carl spent working} \\ 1 \, \text{garden} & = & (\text{rate of Carl working}) \cdot (t \, \text{hours}) \\ \end{array}\nonumber\]

In order to find \(t\), we need to find the rate of Carl working, so let’s call this quantity \(R\), with units \(\frac{\text{garden}}{\text{hour}}\). Using the fact that rates are additive, we have:

\[\begin{array}{rcl} \text{rate working together} & = & \text{rate of Taylor working} + \text{rate of Carl working} \\[4pt] \dfrac{1}{3} \dfrac{\text{garden}}{\text{hour}} & = & \dfrac{1}{4} \dfrac{\text{garden}}{\text{hour}} + R \dfrac{\text{garden}}{\text{hour}} \\ \end{array}\nonumber\]

so that \(R = \frac{1}{12} \frac{\text{garden}}{\text{hour}}\). Substituting this into our ‘work-rate-time’ equation for Carl, we get:

\[\begin{array}{rcl} 1 \, \text{garden} & = & (\text{rate of Carl working}) \cdot (t \, \text{hours}) \\[4pt] 1 \, \text{garden} & = & \left(\dfrac{1}{12} \dfrac{\text{garden}}{\text{hour}} \right) \cdot (t \, \text{hours}) \\ \end{array}\nonumber\]

Solving \(1 = \frac{1}{12} t\), we get \(t = 12\), so it takes Carl 12 hours to weed the garden on his own.

As is common with "word problems" like Examples \( \PageIndex{2} \) and \( \PageIndex{3} \), there is no short-cut to the answer. We encourage the reader to carefully think through and apply the basic principles of rate to each (potentially different!) situation. It is time well spent. We also encourage the tracking of units, especially in the early stages of the problem. Not only does this promote uniformity in the units, it also serves as a quick means to check if an equation makes sense.

2  This is called a system of equations. No doubt, you’ve had experience with these things before, and we will study systems in greater detail in Chapter 9 .

3  While we usually discourage dividing both sides of an equation by a variable expression, we know \((6+R) \neq 0\) since otherwise we couldn’t possibly multiply it by \(t_{\text {down }}\) and get 5.

4  The reader is encouraged to verify that the units in this equation are the same on both sides. To get you started, the units on the "3" is "hours."

Applications Involving Rational Inequalities

Our next example deals with the average cost function, first introduced as applied to PortaBoy Game systems from Example 2.1.5 in Section 2.1 .

Example \( \PageIndex{4} \)

Given a cost function \(C(x)\), which returns the total cost of producing \(x\) items, recall that the average cost function, \(\overline{C}(x) = \frac{C(x)}{x}\) computes the cost per item when \(x\) items are produced. Suppose the cost \(C\), in dollars, to produce \(x\) PortaBoy game systems for a local retailer is \(C(x) = 80x + 150\), \(x \geq 0\).

  • Find an expression for the average cost function \(\overline{C}(x)\).
  • Solve \(\overline{C}(x) < 100\) and interpret.
  • Determine the behavior of \(\overline{C}(x)\) as \(x \to \infty\) and interpret.
  • From \(\overline{C}(x) = \frac{C(x)}{x}\), we obtain \(\overline{C}(x) = \frac{80x+150}{x}\). The domain of \(C\) is \(x \geq 0\), but since \(x=0\) causes problems for \(\overline{C}(x)\), we get our domain to be \(x>0\), or \((0, \infty)\).

Screen Shot 2022-04-01 at 1.37.15 AM.png

  • When we apply Theorem 4.1.2 to \(\overline{C}(x)\) we find that \(y=80\) is a horizontal asymptote to the graph of \(y=\overline{C}(x)\). To more precisely determine the behavior of \(\overline{C}(x)\) as \(x \to \infty\), we first use long division 5  and rewrite \(\overline{C}(x) = 80+\frac{150}{x}\). As \(x \to \infty\), \(\frac{150}{x} \to 0^{+}\), which means \(\overline{C}(x) \approx 80+\text { very small }(+)\). Thus the average cost per system is getting closer to \(\$ 80\) per system. If we set \(\overline{C}(x) = 80\), we get \(\frac{150}{x} = 0\), which is impossible, so we conclude that \(\overline{C}(x) > 80\) for all \(x > 0\). This means that the average cost per system is always greater than \(\$ 80\) per system, but the average cost is approaching this amount as more and more systems are produced. Looking back at Example 2.1.5 , we realize \(\$ 80\) is the variable cost per system \(-\) the cost per system above and beyond the fixed initial cost of \(\$150\). Another way to interpret our answer is that "infinitely" many systems would need to be produced to effectively "zero out" the fixed cost.

Our next example is another classic "box with no top" problem.

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  • High School Math Solutions – Inequalities Calculator, Radical Inequalities Last post, we went over how to solve absolute value inequalities. For today’s post, we will talk about how to solve... Read More

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  1. Rational Equations Word Problems Worksheet

    solving rational equations and inequalities word problems

  2. Solving Rational Inequalities Worksheet

    solving rational equations and inequalities word problems

  3. Rational Equations Word Problems Worksheet

    solving rational equations and inequalities word problems

  4. Rational Inequalities Worksheet

    solving rational equations and inequalities word problems

  5. 8.6B

    solving rational equations and inequalities word problems

  6. Solving rational inequalities with word problem

    solving rational equations and inequalities word problems

VIDEO

  1. 4.3 Solving Rational Equations & Inequalities 11-13-23

  2. Rational Equations Word Problems

  3. Solving Rational Inequalities Part I General Math by Teacher Jelyn

  4. Lesson 6 Solving Rational Equation and Inequalities

  5. Module 4.5 Solving Rational Equations and Inequalities

  6. حل المتباينات النسبية.mp4

COMMENTS

  1. Algebra

    Here is a set of practice problems to accompany the Rational Inequalities section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.

  2. PDF Notes, Examples, and practice (with solutions)

    Solving Rational Equations Notes, Examples, and practice (with solutions) Topics include cross multiplying, word problems, factoring, inequalities, extraneous answers, and more. Mathplane.com Practice Questions - Thanks for visiting. (Hope it helps!) If you have questions, suggestions, or requests, let us know. Cheers

  3. 9.7: Solve Rational Inequalities

    Write the inequality as one quotient on the left and zero on the right. Our inequality is in this form. Step 2. Determine the critical points—the points where the rational expression will be zero or undefined. The rational expression will be zero when the numerator is zero. Since when , then 1 is a critical point.

  4. Equations & inequalities word problems (practice)

    Algebra 2 > Modeling > Modeling with two variables Equations & inequalities word problems Google Classroom The Smiths and the Johnsons were competing in the final leg of the Amazing Race. In their race to the finish, the Smiths immediately took off on a 165 kilometer path traveling at an average speed of v kilometers per hour.

  5. Solving Rational Equations

    Solving Rational Equations | Inequalities | Word Problems Full Course GreeneMath.com 127K subscribers Join Subscribe Subscribed Share 2.3K views 1 year ago Math Full Courses...

  6. Rational equations word problem: combined rates (example 2)

    Substituting 1/x and 1/y for A and B in the equation, we get: 8 = 1/x + 1/y If we solve this equation for x and y, we will get different values than the correct answer. Therefore, it's important to use the correct units when defining the variables and writing the equation to ensure that the units cancel out correctly and we get the correct answer.

  7. 25 Fully Solved Rational Equations

    http://www.greenemath.com/Step-by-Step full solutions for our practice tests on solving rational equations, solving rational inequalities, and solving word p...

  8. Rational Functions, Equations, and Inequalities

    Restricted Domains of Rational Functions. Rational Functions are just a ratio of two polynomials (expression with constants and/or variables), and are typically thought of as having at least one variable in the denominator (which can never be 0 ). Since factoring is so important in algebra, you may want to revisit it first.

  9. Rational Equations Word Problems Lesson

    How to Solve a Word Problem with Rational Equations At this point in the course, we should be very comfortable with solving word problems. We will usually see some application problems that involve rational equations. Let's begin by reviewing our six-step method for solving word problems.

  10. 7.6 Solve Rational Inequalities

    To solve a rational inequality, we first must write the inequality with only one quotient on the left and 0 on the right. Next we determine the critical points to use to divide the number line into intervals. A critical point is a number which make the rational expression zero or undefined.

  11. Solving Inequality Word Questions

    Start with: S + A < 9 A = S + 3, so: S + (S + 3) < 9 Simplify: 2S + 3 < 9 Subtract 3 from both sides: 2S < 9 − 3 Simplify: 2S < 6 Divide both sides by 2: S < 3 Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals. Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals. Check:

  12. How to Solve a Word Problem Using a Rational Equation

    Step 1: Write the equation indicated by the word problem. Pay attention to words indicating mathematical operations or equalities. Step 2: Clear the equation of fractions. Step 3: Solve...

  13. Solving Rational Inequalities

    Solving Rational Inequalities The key approach in solving rational inequalities relies on finding the critical values of the rational expression which divide the number line into distinct open intervals. The critical values are simply the zeros of both the numerator and the denominator.

  14. High School Math : Solving Rational Equations and Inequalities

    Correct answer: y = 1 Explanation: Multiply the equation by (y − 3)(y + 3): y y − 3 + 6 y + 3 = 1 y(y + 3) + 6(y − 3) = 1(y − 3)(y + 3) y2 + 3y + 6y − 18 = y2 − 9 9y = 9 y = 1 Report an Error Example Question #1 : Solving Rational Equations And Inequalities Solve the following rational equation: 1 n − 2 = 2n + 1 n2 + 2n − 8 + 2 n + 4

  15. Solving Rational Inequalities

    Solving Rational Inequalities Rational. A Rational Expression looks like: Inequalities. Sometimes we need to solve rational ... less than or equal to (3−2x)/(x−1) ≤ 2. Solving. Solving inequalities is very like solving equations... you do most of the same things. When we solve inequalities we try to find interval(s), such as the ones ...

  16. Rational expressions, equations, & functions

    This topic covers: - Simplifying rational expressions - Multiplying, dividing, adding, & subtracting rational expressions - Rational equations - Graphing rational functions (including horizontal & vertical asymptotes) - Modeling with rational functions - Rational inequalities - Partial fraction expansion

  17. Word Problems Calculator

    Free Word Problems Calculator - solve word problems step by step

  18. 7.5: Solving Rational Equations

    Begin solving rational equations by multiplying both sides by the LCD. The resulting equivalent equation can be solved using the techniques learned up to this point. Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of ...

  19. 4.3: Rational Inequalities and Applications

    4.3: Rational Inequalities and Applications. In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality.

  20. Algebra

    Absolute Value Inequalities - In this final section of the Solving chapter we will solve inequalities that involve absolute value. As we will see the process for solving inequalities with a < < (i.e. a less than) is very different from solving an inequality with a > > (i.e. greater than). Here is a set of practice problems to accompany the ...

  21. Rational Equation Calculator

    Free rational equation calculator - solve rational equations step-by-step ... Get full access to all Solution Steps for any math problem By continuing, you agree to our Terms of ... Equations Inequalities Scientific Calculator Scientific Notation Arithmetics Complex Numbers Polar/Cartesian Simultaneous Equations System of Inequalities ...

  22. 4.3: Rational Inequalities and Applications

    In this section, we solve equations and inequalities involving rational functions and explore associated application problems. Our first example showcases the critical difference in procedure between solving a rational equation and a rational inequality. Example 4.3.1. Solve x3 − 2x + 1 x − 1 = 1 2x − 1. Solve x3 − 2x + 1 x − 1 ≥ 1 ...

  23. Rational Inequality Calculator

    Free rational inequality calculator - solve rational inequalities with all the steps. Type in any inequality to get the solution, steps and graph