Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1

112 CHAPTER 2 Linear Equations, Inequalities, and Applications


OBJECTIVES Suppose that the government of a country decides that it will comply with a certain


restriction on greenhouse gas emissions within3 years of 2020. This means that the


differencebetween the year it will comply and 2020 is less than 3, without regard to


sign. We state this mathematically as


Absolute value inequality

where xrepresents the year in which it complies.


Reasoning tells us that the year must be between 2017 and 2023, and thus


makes this inequality true. But what general procedure is used to


solve such an inequality? We now investigate how to solve absolute value equations


and inequalities.


OBJECTIVE 1 Use the distance definition of absolute value.In Section 1.1,


we saw that the absolute value of a number x, written represents the distance


from xto 0 on the number line. For example, the solutions of are 4 and ,


as shown in FIGURE 29.


|x|= 4 - 4


|x|,


20176 x 62023


|x- 2020 | 6 3,


Absolute Value Equations and Inequalities


2.7


1 Use the distance
definition of
absolute value.
2 Solve equations
of the form

for
3 Solve inequalities
of the form
and
of the form

for
4 Solve absolute value
equations that
involve rewriting.
5 Solve equations
of the form

6 Solve special cases
of absolute value
equations and
inequalities.

|ax+b|=|cx+d|.

k 7 0.

|ax+b| 7 k,

|ax+b| 6 k

k 7 0.

|ax+b|=k,

–4 0
x = – 4 or x = 4

4 units from 0 4 units from 0

4

FIGURE 29

–4 04

More than
4 units from 0

More than
4 units from 0

x < – 4 or x > 4
FIGURE 30

Less than 4 units from 0

–4 04


  • 4 < x < 4
    FIGURE 31


Because absolute value represents distance from 0, we interpret the solutions


of to be all numbers that are morethan four units from 0. The set


fits this description. FIGURE 30shows the graph of the solution set


of Because the graph consists of two separate intervals, the solution set is


described using the word or:or.x6- 4 x 74


|x| 7 4.


1 - q, - 42 ́ 1 4, q 2


|x| 7 4


The solution set of consists of all numbers that are lessthan 4 units from


0 on the number line. This is represented by all numbers between and 4. This set


of numbers is given by as shown in FIGURE 31. Here, the graph shows that


- 46 x 6 4,which means x7- 4 and x 6 4.


1 - 4, 4 2 ,


- 4


|x| 64

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