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Copyright © 2008 Pearson Education, Inc.
CHAPTER 2: Functions, Equations,
and Inequalities
2.1 Linear Equations, Functions, and Models
2.2 The Complex Numbers
2.3 Quadratic Equations, Functions, and Models
2.4 Analyzing Graphs of Quadratic Functions
2.5 More Equation Solving
2.6 Solving Linear Inequalities
Copyright © 2008 Pearson Education, Inc.
2.6 Solving Linear Inequalities
Solve linear inequalities. Solve compound inequalities. Solve inequalities with absolute value. Solve applied problems using inequalities.
Slide 2.6-4Copyright © 2008 Pearson Education, Inc.
Inequalities
An inequality is a sentence with <, >, , or as its verb.
Examples: 5x 7 < 3 + 4x
3(x + 6) 4(x 3)
Slide 2.6-5Copyright © 2008 Pearson Education, Inc.
Principles for Solving InequalitiesFor any real numbers a, b, and c:
The Addition Principle for Inequalities: If a < b is true,
then a + c < b + c is true.
The Multiplication Principle for Inequalities: If a < b and c > 0 are true, then ac < bc is true. If a < b and c < 0, then ac > bc is true.
Similar statements hold for a b.
When both sides of an inequality are multiplied or divided by a negative number, we must reverse the inequality sign.
Slide 2.6-6Copyright © 2008 Pearson Education, Inc.
Examples
Solve:
{x|x < 2} or (, 2)
Solve:
{x|x 4} or [4, )
4 6 2 10
4 2 4
2 4
2
x x
x x
x
x
6( 3) 7( 2)
6 18 7 14
4
4
x x
x x
x
x
0–5 5
)0–5 5
[
Slide 2.6-7Copyright © 2008 Pearson Education, Inc.
Compound Inequalities
When two inequalities are joined by the word and or the word or, a compound inequality is formed.
Conjunction contains the word and.
Example: 7 < 3x + 5 and 3x + 9 6
Disjunction contains the word or.
Example: 3x + 5 6 or 3x + 6 > 12
Slide 2.6-8Copyright © 2008 Pearson Education, Inc.
Examples
Solve: Solve:
4x 5 3 or 4x 5 > 3 4 3 8 11x
4 3 8 11
12 3 3
4 1
x
x
x
4 5 3 4 5 3
4 2 4 8
12
2
x or x
x x
x x
0–5 5
( ]0–5 5
(]
Slide 2.6-9Copyright © 2008 Pearson Education, Inc.
Inequalities with Absolute Value
Inequalities sometimes contain absolute-value notation. The following properties are used to solve them.
For a > 0 and an algebraic expression X:
|X| < a is equivalent to a < X < a.
|X| > a is equivalent to X < a or X > a.
Similar statements hold for |X| a and |X| a.
Slide 2.6-10Copyright © 2008 Pearson Education, Inc.
Example
Solve:
4 1 3
3 4 1 3
4 4 2
11
2
x
x
x
x
0–5 5
( )
Slide 2.6-11Copyright © 2008 Pearson Education, Inc.
Application
Johnson Catering charges $100 plus $30 per hour
to cater an event. Catherine’s Catering charges
$50 per hour. For what lengths of time does it cost
less to hire Catherine’s Catering?
1. Familiarize. Read the problem.
2. Translate. Catherine’s is less than Johnson
50x < 100 + 30x
Slide 2.6-12Copyright © 2008 Pearson Education, Inc.
Application continued
3. Carry out.
4. Check.
5. State. For values of x < 5 hr, Catherine’s Catering will cost less.
50 100 30
20 100
5
x x
x
x
50(5) ? 100 30(5)
250 ? 100 150
250 250
