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:

  • 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.

This figure shows the solution, the interval 3 to infinity, of the inequality x is greater than 3 on a number line. The values range from negative 5 to 5 on the number line. The inequality is modeled by an open parenthesis at the critical point 3 and shading 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.

This figure shows a number line divided into three intervals by its critical points marked at negative 3 and 0.

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.

This figure labels the expression, x minus 1, as the “numerator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 5. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when negative 4 is substituted into the expression for x, the result is negative 1. It labels the result “negative”.

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 ) .

This figure shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is negative, which is positive. It shows a number line divided into three intervals by its critical points marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3.

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

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 0 is substituted into the expression for x, the result is negative 1. It labels the result as “negative”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 0 is substituted into the expression for x, the result is 3. It labels the result “positive”.

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 ) .

This figure shows a shows the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line divided into three intervals by its critical points marked at negative 3 and 0. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1.

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

This figure labels the expression, x minus 1, as the “numerator”. It shows that when 2 is substituted into the expression for x, the result is 1. It labels the result as “positive”. It also labels the expression, x plus 3, as “the denominator”. It shows that when 2 is substituted into the expression for x, the result is 5. It labels the result “positive”.

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 , ∞ ) .

The figure shows that in the quotient of the quantity x minus 1 and the quantity x plus 3, the numerator is negative and the denominator is positive, which is negative. It shows a number line is divided into intervals by critical points at negative 3 and 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.

  • 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.

The number line is divided into intervals by critical points at negative 3 and 1. A closed parenthesis is used at 3 and an open bracket is used at 1. The number is shaded to the left of 3 and to the right of 1. The factors x minus 1 and x plus 3 are marked as negative above the number line for the interval negative infinity to negative 3. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative infinity to negative 3. The factor x minus 1 is marked as negative and the factor x plus 3 is marked as positive above the number line for the interval negative 3 to 1. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as negative below the number line for the interval negative 3 to 1. The factors x minus 1 and x plus 3 are marked as positive above the number line for the interval 1 to infinity. The quotient of the quantity x minus 1 and the quantity x plus 3 is marked as positive below the number line for the interval negative 1 to infinity.

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?

Section 7.6 Exercises

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.

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Solving Inequalities with Rational Numbers

More specific topics in solving inequalities with rational numbers.

  • Solving Inequalities with Fractions
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Related topics, other topics in rational numbers :.

  • Rational Numbers Definitions and Basics
  • Adding and Subtracting Fractions
  • Multiplying and Dividing Fractions
  • Decimal Operations
  • Solving Fraction Equations
  • Solving Decimal Equations
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IMAGES

  1. Solving Rational Inequalities Worksheet

    problem solving rational inequality

  2. 9. Solving Rational Inequalities Example 2

    problem solving rational inequality

  3. 8. Solving Rational Inequalities Example 1

    problem solving rational inequality

  4. Solving Rational Inequalities

    problem solving rational inequality

  5. Solving Rational Inequalities Worksheet

    problem solving rational inequality

  6. Rational Inequalities

    problem solving rational inequality

VIDEO

  1. Rational Inequality

  2. Week 8, Problem 16 from week 7

  3. Solving Rational Inequality Part 1

  4. Solve rational inequality by factoring and using test points

  5. Solving Rational Inequality Part 2

  6. Solving Rationals (Part 1)

COMMENTS

  1. 9.7: Solve Rational Inequalities

    9.7: Solve Rational Inequalities Expand/collapse global location

  2. 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.

  3. Rational inequalities: one side is zero (video)

    Alex Tamulis 11 years ago Your example looks like a rational equation, Sal has 3 videos on this topic (they are called solving rational equations). I solved your equation and you have to find a common denominator first, which is 21. Then you multiply 7 (x+1) + 3 (x+2), all that over 21, which equals 2.

  4. 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.

  5. Solving Rational Inequalities

    These are the steps: find "points of interest": 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

  6. 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.

  7. Rational inequalities: both sides are not zero

    11 years ago Hello Cyrus, I can see that your confusion is with the *"and"* and *"or"*. If there is a word *"and"* then you must see if they have common numbers: if they don't have, then the solution is invalid. If there is a word "or" then you don't need to determine if they have common numbers. ( 19 votes) Show more... adanbro7 11 years ago

  8. 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 ...

  9. 9.2: Rational Inequalities

    Step 1. Rewrite the inequality so that only zero is on the right side. Since x − 3 x + 1 > 0 x − 3 x + 1 > 0 already has zero on the right side, this step is done. Step 2. Determine where the rational expression is undefined or equals zero. To obtain where the rational expression equals zero, we set the numerator equal to zero:

  10. 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

  11. 6.6: Solving Polynomial and Rational Inequalities

    Solving Rational Inequalities. A rational inequality 19 is a mathematical statement that relates a rational expression as either less than or greater than another. Because rational functions have restrictions to the domain we must take care when solving rational inequalities. In addition to the zeros, we will include the restrictions to the ...

  12. 2.4: Solution of Rational Inequalities by Graphing

    In this section, we will use similar methods to solve rational inequalities. Rational inequalities involve ratios of polynomials or fractions. Because these types of problems involve fractions, the graphs of the functions that we work with will have what are known as asymptotes. This word comes from a Greek root having to do with two lines that ...

  13. Rational Inequalities

    This precalculus video tutorial provides a basic introduction into solving rational inequalitites using a sign chart on a number line and expressing the solu...

  14. Solving Rational Inequalities: Steps & Examples

    Steps to Solve a Rational Inequality. Step 1: Put the inequality in general form. General form means the inequality must be greater than, less than, greater than or equal to, or less than or equal ...

  15. PDF Solving Rational Inequalities

    Solve the simple rational inequality Algebraic Solution Recall that multiplying both sides of an inequality by a negative value reverses the inequality condition. Case 1: x —2<0 —x < 2 6>3x-6 —3x > —12 Since x < 2 and x < 4, therefore x < 2 Therefore, the solution is {x I x < 2 or x > 4, x e IR}. Case 2: x —2 > 0 2 6<3x-6 < -12

  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. Solving Rational Equations

    http://www.greenemath.com/In this course, we will learn how to solve rational equations, rational inequalities, and word problems that involve rational equat...

  18. Solving equations & inequalities

    Algebra 1 16 units · 184 skills. Unit 1 Algebra foundations. Unit 2 Solving equations & inequalities. Unit 3 Working with units. Unit 4 Linear equations & graphs. Unit 5 Forms of linear equations. Unit 6 Systems of equations. Unit 7 Inequalities (systems & graphs) Unit 8 Functions.

  19. Solving Inequalities with Rational Numbers

    This tutorial provides a great real world application of math. See how to turn a word problem into an inequality. Then solve the inequality by performing the order of operations in reverse. Don't forget that if you multiply or divide by a negative number, you MUST flip the sign of the inequality!

  20. Rational equations (practice)

    Course: Algebra 2 > Unit 10. Lesson 1: Rational equations. Rational equations intro. Rational equations intro. Equations with rational expressions. Equations with rational expressions (example 2) Rational equations. Finding inverses of rational functions. Find inverses of rational functions.

  21. 4.3: Rational Inequalities and Applications

    4: Rational Functions

  22. Algebra

    The process for solving rational inequalities is nearly identical to the process for solving polynomial inequalities with a few minor differences. Let's just jump straight into some examples. Example 1 Solve x +1 x −5 ≤ 0 x + 1 x − 5 ≤ 0 . Show Solution Example 2 Solve x2 +4x+3 x −1 > 0 x 2 + 4 x + 3 x − 1 > 0 . Show Solution

  23. Algebra

    Section 2.13 : Rational Inequalities. Back to Problem List. 1. Solve the following inequality. 4 −x x +3 >0 4 − x x + 3 > 0. Show All Steps Hide All Steps.