Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.3 Linear Inequalities.

Copyright © 2013, 2009, 2005 Pearson Education, Inc.

Section 3.3

Linear Inequalities


Objectives

• Basic Concepts

• Symbolic Solutions

• Numerical and Graphical Solutions

• An Application


Basic Concepts

An inequality results whenever the equals sign in an equation is replaced with any one of the symbols <, ≤, >, or ≥.

A solution to an inequality is a value of the variable that makes the statement true. The set of all solutions is called the solution set.

Slide 3


, 0 and 1,xf x a a a A linear inequality in one variable is an inequality that can be written in the form

ax + b > 0,

where a ≠ 0. (The symbol > may be replaced with ≥, <, or ≤.)

LINEAR INEQUALITY IN ONE VARIABLE


, 0 and 1,xf x a a a

The solution set for ax + b > 0 with a ≠ 0 is either {x│x < k} or {x│x > k}, where k is the solution to ax + b = 0 and corresponds to the x-intercept for the graph of y = ax + b. Similar statements can be made for the symbols <, ≤, and ≥.

SOLUTION SET FOR A LINEAR INEQUALITY


Let a, b, and c be real numbers.

1. a < b and a + c < b + c are equivalent. (The same number may be added to or subtracted from each side of an inequality.)

2. If c > 0, then a < b and ac < bc are equivalent. (Each side of an inequality may be multiplied or divided by the same positive number.)

3. If c < 0, then a < b and ac > bc are equivalent. (Each side of an equality may be multiplied or divided by the same negative number provided the inequality symbol is reversed.)

Similar properties exist for ≤ and ≥ symbols.

PROPERTIES OF INEQUALIES


Example

Solve the inequality. 5x – 2 > 9Solution

5x – 2 > 9

5x – 2 + 2 > 9 + 2

5x > 11 5 11

5 5

x

11

5x

The solution set is . 11

5x x


Example

Solve the inequality. Solution

12 5 1

3n n

12 5 1

3n n

2 3 5 3n n

2 15 3 3n n 4 17 3n

1717 3 74 1n

4 20n

5n

The solution set is {n│n ≤ 5}.


Example

Solve the inequality; 9 – 4x ≤ −x − 6.Solution

Next divide each side by −3. As when we are dividing by a negative number, Property 3 requires reversing the inequality by changing ≤ to ≥.

9 – 4x ≤ −x – 6 9 – 4x – 9 ≤ −x – 6 – 9

−4x ≤ −x – 15 −4x + x ≤ −x – 15 + x

−3x ≤ – 15 3 15

3 3

x

5x

The solution set is {x│x ≥ 5}.


Example

Use the table to find the solution set to each equation or inequality.a. b. c.

x −1 0 1 2 3

3 033

2x

9

23

2

3

2

33 0

2x

33 0

2x

33 0

2x

a. The expression equals 0 when x = 2. Thus the solution set is {x│x = 2}.

33

2x


Example (cont)


x −1 0 1 2 3

3 033

2x

9

23

2

3

2

33 0

2x

33 0

2x

33 0

2x

b. The expression is positive when x < 2. Thus the solution set is {x│x < 2}.

33

2x


Example (cont)


x −1 0 1 2 3

3 033

2x

9

23

2

3

2

33 0

2x

33 0

2x

33 0

2x

c. The expression is negative when x > 2. Thus the solution set is {x│x > 2}.

33

2x


Example

Use the graph to find the solution set to each equation or inequality. a.

The graph of crosses the x-axis at x = 2. Thus the solution set is

33 0

2x

33

2y x

2 .x x


Example (cont)

Use the graph to find the solution set to each equation or inequality. b.

The graph is above the x-axis when x < 2. Thus the solution set is {x│x < 2}.

33 0

2x


Example (cont)

Use the graph to find the solution set to each equation or inequality. c.

The graph is above the x-axis when x > 2. Thus the solution set is {x│x > 2}.

33 0

2x


Example

Use a graph to solve 7 – 4x ≤ x – 8. SolutionThe graph of y1 = 7 – 4x and y2 = x – 8 intersect at the point (3, −5). The graph of y1 is below the graph of y2 when x > 3. Thus 7 – 4x ≤ x – 8 is satisfied when x ≥ 3. Therefore the solution set is {x│x ≥ 3}.

Copyright © 2013, 2009, 2005 Pearson Education, Inc. Section 3.3 Linear Inequalities.

Documents

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