2.8 Solving Linear Inequalities

2.8 Solving Linear Inequalities

Definition.

An inequality is an algebraic expression related by< “is less than,” ≤ “is less than or equal to,”> “is greater than,” or ≥ “is greater than or equal to.”

We solve an inequality by finding all real number solutions of it. For example, the solution set {x | x ≤ 2} includes all real numbers that are less than or equal to 2, not just the integers less than or equal to 2.

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Linear Inequality in One VariableA linear inequality in one variable can be written in the form

where A, B, and C represent real numbers, and A ≠ 0.

,Ax B C ,Ax B C or,Ax B C ,Ax B C

Objective 1

Graph intervals on a number line.

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Graphing is a good way to show the solution set of an inequality. We graph all the real numbers belonging to the set {x | x ≤ 2} by placing a square bracket at 2 on a number line and drawing an arrow extending from the bracket to the left (to represent the fact that all numbers less than 2 are also part of the graph).

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Graph intervals on a number line.

The set of numbers less than or equal to 2 is an example of an interval on the number line. To write intervals, we use interval notation. For example, the interval of all numbers less than or equal to 2 is written (−∞, 2].

The negative infinity symbol −∞ does not indicate a number, but shows that the interval includes all real numbers less than 2.

As on the number line, the square bracket indicates that 2 is part of the solution.

A parentheses is always used next to the infinity symbol. The set of real numbers is written as (−∞, ∞).

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Graph intervals on a number line. (cont’d)

Write each inequality in interval notation, and graph the interval.

Solution:

Solution:

3x

4 x

[3, )

( , 4)

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Graphing Intervals on a Number LineCLASSROOM EXAMPLE 1

Keep the following important concepts regarding interval notation in mind:

1. A parenthesis indicates that an endpoint is not included in a solution set.

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Graph intervals on a number line. (cont’d)

Some texts use a solid circle ● rather than a square bracket to indicate the endpoint is included in a number line graph. An open circle is used to indicate noninclusion, rather than a parentheses.

2. A bracket indicates that an endpoint is included in a solution set.

3. A parenthesis is always used next to an infinity symbol, −∞ or ∞.

4. The set of all real numbers is written in interval notation as (−∞,∞).

Objective 2

Use the addition property of inequality.

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Addition Property of InequalityIf A, B, and C represent real numbers, then the inequalities

and

Have exactly the same solutions.That is, the same number may be added to each side of an inequality without changing the solutions.

Use the addition property of inequality.

A B A C B C

As with the addition property of equality, the same number may be subtracted from each side of an inequality.

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Use the addition property of inequality. (cont’d)

Because an inequality has many solutions, we cannot check all of them by substitutions as we did with the single solution of an equation. Thus, to check the solutions of an inequality, first substitute into the equation the boundary point of the interval and another number from within the interval to test that they both result in true statements. Next, substitute any number outside the interval to be sure it gives a false statement.

CLASSROOM EXAMPLE 2

Solve the inequality, and graph the solution set.

Solution:

1 8 7 2x x

71 8 7 2 7xx x x 11 12x x

( ,3)

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Using the Addition Property of Inequality

Objective 3

Use the multiplication property of inequality.

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Now multiply by each side of 3 < 7 by the negative number −5.

Multiply each side of the inequality 3 < 7 by the positive number 2.

To get a true statement when multiplying each side by −5, we must reverse the direction of the inequality symbol.

The addition property of inequality cannot be used to solve an inequality such as 4x ≥ 28. This inequality requires the multiplication.

6 2 3 72

True

15 75 3 5

False

15 75 3 5

True

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Use the multiplication property of inequality.

As with the multiplication property of inequality, the same nonzero number may be divided into each side of an inequality.

Multiplication Property of InequalityIf A, B, and C, with C ≠ 0,1. if C is positive, then the inequalities

and have exactly the same solutions;2. if C is negative, then the inequalities

and have exactly the same solutions.

That is, each of an inequality may be multiplied by the same positive number without changing the solutions. If the multiplier is negative, we must reverse the direction of the inequality symbol.

A B AC BC

A B AC BC

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Use the multiplication property of inequality. (cont’d)

Solution:

12r

2 2r

6r

( ,6)

Solve the inequality, and graph the solution set.

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Using the Multiplication Property of InequalityCLASSROOM EXAMPLE 3

Objective 4

Solve linear inequalities by using both properties of inequality.

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Solving a Linear InequalityStep 1: Simplify each side separately. Use the distributive property to clear parentheses and combine like terms on each side as needed.

Step 2: Isolate the variable terms on one side. Use the addition property of inequality to get all terms with variables on one side of the inequality and all numbers on the other side.

Step 3: Isolate the variable. Use the multiplication property of inequality to change the inequality to the form “variable < k” or “variable > k,” where k is a number.

Remember: Reverse the direction of the inequality symbol only when multiplying or dividing each side of an inequality by a negative number..

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Solve linear inequalities by using both properties of inequality.

Solution:

5 2 7 5x x x

74 72 7 5xx xx 23 5 22x

3 73 3x

73

x

7 ,3

Solve the inequality, and then graph the solution set.

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Solving a Linear InequalityCLASSROOM EXAMPLE 4

Solution:

4 1 3 15 2 1x x x

23 24 4x x xxx

3 13 3

2x

4x

4,

3 4 44x

Solve the inequality, and graph the solution set.

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Solving a Linear InequalityCLASSROOM EXAMPLE 5

Solve and graph the solution set.

1 3( 3) 2 ( 8)4 4x x

1 3( 3) 2 ( 8)4

44

4x x 1 3( 3) (24 4 4) ( 8)4 4x x

3 8 3( 8)x x 3 8 3 24x x

11 3 24x x 11 3 211 114x x

Multiply by 4.

Distributive property.

Multiply.

Subtract 11.

Distributive property.

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CLASSROOM EXAMPLE 6 Solving a Linear Inequality with Fractions

Solution:

11 3 211 114x x 3 13x x 3 33 13x xx x

2 13x

2 132 2x

132

xReverse the

inequality symbol when dividing by a negative number.

[

Subtract 11.

Subtract 3m.

Divide 2.

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Solving a Linear Inequality with Fractions (cont’d)

The solution set is the interval [−13/2, ).

CLASSROOM EXAMPLE 6

Objective 5

Solve applied problems by using inequalities.

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Inequalities can be used to solve applied problems involving phrases that suggest inequality. The table gives some of the more common such phrases, along with examples and translations.

In general, to find the average of n numbers, add the numbers and divide by n. We use the same six problem-solving steps from Section 2.4, changing Step 3 to “Write an inequality.”, instead of “Write an equation.”

Do not confuse statements such as “5 is more than a number” with phrases like “5 more than a number.” The first of these is expressed as 5 > x, while the second is expressed as x + 5 or 5 + x.

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Solve applied problems by using inequalities.

A rental company charges $5 to rent a leaf blower, plus $1.75 per hr. Marge Ruhberg can spend no more than $26 to blow leaves from her driveway and pool deck. What is the maximum amount of time she can use the rented leaf blower?

Step 1 Read the problem again. What is to be found?The maximum time Marge can afford to rent the blower.

What is given?The flat rate to rent the leaf blower, the additional

hourly charge to rent the leaf blower, and the maximum amount that Marge can spend.

Step 2 Assign a variable. Let h = the number of hours she can rent the blower.

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CLASSROOM EXAMPLE 7

Using a Linear Inequality to Solve a Rental Problem

Solution:

Step 3 Write an inequality. She must pay $5, plus $1.75 per hour for h hours

and no more than $26.Cost of is no

renting more than 26

5 + 1.75h ≤ 26

Step 4 Solve. 1.75h ≤ 21 h ≤ 12

Step 5 State the answer.She can use the leaf blower from a maximum of 12

hours.Step 6 Check.

If she uses the leaf blower for 12 hr, she will spend5 + 1.75(12) = 26 dollars, the maximum.

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CLASSROOM EXAMPLE 7

Using a Linear Inequality to Solve a Rental Problem (cont’d)

Subtract 5.Divide by 1.75.

Solution: Let x = Maggie’s fourth test score.

98 86 88 904

x

272 3 7260 2x

88x

2724

4 4x

Maggie must get greater than or equal to an 88.

Maggie has scores of 98, 86, and 88 on her first three tests in algebra. If she wants an average of at least 90 after her fourth test, what score must she make on that test?

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Finding an Average Test ScoreCLASSROOM EXAMPLE 8

Objective 6

Solve linear inequalities with three parts.

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Inequalities that say the one number is between two other numbers are three-part inequalities. For example,

says that 5 is between −3 and 7.

3

For some applications, it is necessary to work with a three-part inequality such as

where x +2 is between 3 and 8. To solve this inequality, we subtract 2 from each of the three parts of the inequality.

3 2x

3 22 2 2x 1 x

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Solve linear inequalities with three parts.

The idea is to get the inequality in the forma number < x < another number,

using “is less than.” The solution set can then easily be graphed.

When inequalities have three parts, the order of the parts is important. It would be wrong to write an inequality as 8 < x + 2 < 3, since this would imply 8 < 3, a false statement. In general, three-part inequalities are written so that the symbols point in the same direction and both point toward the lesser number.

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Solve linear inequalities with three parts. (cont’d)

Write the inequality in interval notation, and graph the interval.

Solution:

2 4x

2,4

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Solving Three-Part InequalitiesCLASSROOM EXAMPLE 9

Solve the inequality, and graph the solution set.

Solution:

2 3 1 8x

2 3 11 1 8 1x

1,3

93 33 3

3x

1 3x

Remember to work with all three parts of the inequality.

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Solving Three-Part InequalitiesCLASSROOM EXAMPLE 10