Solving Linear Inequalities in One Variable Digital Lesson

Solving Linear Inequalities in One Variable

Digital Lesson

Copyright © by Houghton Mifflin Company, Inc. All rights reserved. 2

Note that the “>” can be replaced by , <, or .

Examples: Linear inequalities in one variable.

2x – 2 < 6x – 5

A linear inequality in one variable is an inequality which can be put into the form

ax + b > c

can be written – 4x + (– 2) < – 5.

6x + 1 3(x – 5)

2x + 3 > 4

can be written 6x + 1 – 15.

where a, b, and c are real numbers.


The solution set for an inequality can be expressed in two ways.

1. Set-builder notation:

2. Graph on the real line:

{x | x < 3}

Example: Express the solution set of x < 3 in two ways.

1. Set-builder notation:

2. Graph on the real line:

{x | x 4}

Example: Express the solution set of x 4 in two ways.

0 1 2 3 4-4 -3 -2 -1

0 1 2 3 4-4 -3 -2 -1

Rounded parentheses indicate that the number is not included in the solution set.

]

)

Square brackets indicate that the number is included in the solution set.


A solution of an inequality in one variable is a number which, when substituted for the variable, results in a true inequality.

Examples: Are any of the values of x given below solutions of 2x > 5?

2 is not a solution.

2.6 is a solution.

x = 2 2(2) > 5 4 > 5

x = 2.6 2(2.6) > 5 5.2 > 5

The solution set of an inequality is the set of all solutions.

3 is a solution.x = 3 2(3) > 5 6 > 5

x = 1.5 2(1.5) > 5 3 > 5 1.5 is not a solution.False

False

True

True

? ?

? ?

? ?

? ?


• If a > b and c is a real number, then a > b, a + c > b + c, and a – c > b – c have the same solution set.

Addition and Subtraction Properties

Example: Solve x – 4 > 7.x – 4 > 7 Add 4 to each side of the inequality. + 4+ 4

x > 11Set-builder notation.{x | x > 11}

Example: Solve 3x 2x + 5.Subtract 2x from each side.3x 2x + 5

x 5

– 2x– 2x

Set-builder notation.

• If a < b and c is a real number, then a < b, a + c < b + c, and a – c < b – c have the same solution set.

{x | x 5}


Multiplication and Division Properties

• If c < 0 the inequalities a > b, ac < bc, and < have the same solution set.

c

a

c

b

• If c > 0 the inequalities a > b, ac > bc, and > have the same solution set.

c

a

c

b

Example: Solve 4x 12.

x 3

Divide by 4.

4 is greater than 0, so the inequality sign remains the same.

Example: Solve . 43

1 x

4 3

1 x (– 3) (– 3) Multiply by – 3.

– 3 is less than 0, so the inequality sign changes.

12x

4x 12 )4(

)4(


Example: Solve x + 5 < 9x + 1.

– 8x + 5 < 1 Subtract 9x from both sides.

– 8x < – 4 Subtract 5 from both sides.

x > 2

1 Divide both sides by – 8 and simplify.Inequality sign changes because of division by a negative number.

2

1| xx Solution set in set-builder notation.

Example: Solve .45

32

5

4 xx

Subtract from both sides.x5

3425

1x

Add 2 to both sides.65

1x

30x Multiply both sides by 5.

20 30 40 50 60-20 -10 0 10[ Solution set as a graph.


A compound inequality is formed by joining two inequalities with “and” or “or.”

Example: Solve x + 2 < 5 and 2x – 6 > – 8.

x + 2 < 5Solve the first inequality.

The solution set of the “and” compound inequality is the intersection of the two solution sets.

x < 3

{x | x > –1}{x | x < 3}

1|3| xxxx

Subtract 2.

Solution set

Solve the second inequality.2x – 6 > – 8

2x > – 2 Add 6.

Divide by 2. x > – 1Solution set

0 1 2 3 4-4 -3 -2 -1

31| xx

)(


Example: Solve 11 < 6x + 5 < 29.

6 < 6x < 24 Subtract 5 from each of the three parts.

1 < x < 4 Divide 6 into each of the three parts.

Solution set.

This inequality means 11 < 6x + 5 and 6x + 5 < 29.

When solving compound inequalities, it is possible to work with both inequalities at once.

Example: Solve .562

18 x

12

12 x

Multiply each part by – 2.24 x

Subtract 6 from each part.

Multiplication by a negative number changes the inequality sign for each part.

0 1 2 3 4-4 -3 -2 -1][ Solution set.

41| xx


Example: Solve x + 5 > 6 or 2x < – 4.

Solve the first inequality.

Since the inequalities are joined by “or” the solution set is the union of the solution sets.

Solution set

x + 5 > 6 2x < – 4

Solve the second inequality.

{ x | x > 1}

0 1 2 3 4-4 -3 -2 -1

2|1| xxxx

x > 1 x < – 2

Solution set{ x | x < – 2}

) (


Example: A cell phone company offers its customers a rate of

$89 per month for 350 minutes, or a rate of $40 per month

plus $0.50 for each minute used.

Solve the inequality 0.50x + 40 89 .

0.50x 49

x 24.5

Subtract 40.

Divide by 0.5.

Let x = the number of minutes used.

The customer can use up to 24.5 minutes per month before the

cost of the second plan exceeds the cost of the first plan.

How many minutes per month can a customer who chooses the

second plan use before the charges exceed those of the first plan?

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Solving Linear Inequalities in One Variable Digital Lesson