Pat05 ppt 0106

1. Copyright © 2016, 2012 Pearson Education, Inc. 1 -1
Chapter 1
Graphs,
Functions,
and Models

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Section
1.1 Introduction to Graphing
1.2 Functions and Graphs
1.3 Linear Functions, Slope, and Applications
1.4 Equations of Lines and Modeling
1.5 Linear Equations, Functions, Zeros and Applications
1.6 Solving Linear Inequalities

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1.6
Solving Linear Inequalities
 Solve linear inequalities.
 Solve compound inequalities.
 Solve applied problems using inequalities.

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Inequalities
An inequality is a sentence with <, >, , or  as its
verb.
Example: 3x  5 < 6 – 2x
To solve an inequality is to find all values of the
variable that make the inequality true.
Each of these numbers is a solution of the inequality,
and the set of all such solutions is its solution set.
Inequalities that have the same solution set are called
equivalent inequalities.

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Principles for Solving Inequalities
For 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.
When both sides of an inequality are multiplied or divided by a
negative number, we must reverse the inequality sign.
Similar statements hold for a  b.

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Example
Solve each of the following. Then graph the solution set.
a. 3x – 5 < 6 – 2x b. 13 – 7x ≥ 10x – 4
3x  5  6  2x
5x  5  6
5x 11
x 
11
5
Solution:
11 11
, or ,
5 5
x x
   
    
   
13 7x 10x  4
1317x  4
17x  17
x 1
Solution:
x x £1{ }, or -¥,1( ùû

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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: –3 < 2x + 5 and 2x + 5  7
The sentence –3 < 2x + 5 ≤ 7 is an abbreviation.
Disjunction contains the word or.
Example: 2x – 5  –7 or 2x – 5 > 1

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Example
Solve each of the following. Then graph the solution set.
a. – 3 < 2x + 5 ≤ 7 b. 2x – 5 ≤ –7 or 2x – 5 > 1
3  2x  5  7
8  2x  2
4  x 1
Solution a:
x - 4 < x £1{ }, or -4,1( ]

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Example (continued)
Solve each of the following. Then graph the solution set.
a. – 3 < 2x + 5 ≤ 7 b. 2x – 5 ≤ –7 or 2x – 5 > 1
2x  5  7 or 2x  5 1
2x  2 or 2x  6
x  1 or x  3
Solution b:
 
   
1 3
or , 1 3,
x x or x  
   

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Application - Example
For her house painting job, Erica can be paid in one of
two ways:
Plan A: $250 plus $10 per hour
Plan B: $20 per hour
Suppose that a job takes n hours. For what values of n
is plan B better for Erica?

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Example (continued)
Solution:
1. Familiarize. Read the problem.
For a 30 hour job, n = 30:
Plan A: $250 + $10 • 30 = $550
Plan B: $20 • 30 = $600
Plan B is better for a 30 hour job.
For a 20 hour job, n = 20:
Plan A: $250 + $10 • 20 = $450
Plan B: $20 • 20 = $400
Plan A is better for a 20 hour job.

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Example (continued)
2. Translate. Translate to an inequality.
Income plan B is greater than Income plan A
20n > 250 + 10n
3. Carry out. Solve the inequality.
20n  250 10n
10n  250
n  25

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Application continued
4. Check. For n = 25, the income from plan A is $250 + $10 •
25, or $500, and the income from plan B is $20 • 25, or $500,
the same under either plan. We have seen that plan B pays
more for a 30-hr job. Since 30 > 25, this provides a partial
check. We cannot check all values of n.
5. State. For values of n greater than 25 hr, plan B is
better for Erica.