Transcript Maath0312HCCS_0207

Chapter 2
Section 7
2.7 Absolute Value Equations and Inequalities
Objectives
1
Use the distance definition of absolute value.
2
Solve equations of the form |ax + b| = k, for k > 0.
3
Solve inequalities of the form |ax + b| < k and of the form |ax + b| > k,
for k > 0.
4
Solve absolute value equations that involve rewriting.
5
Solve equations of the form |ax + b| = |cx + d| .
6
Solve special cases of absolute value equations and inequalities.
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Objective 1
Use the distance definition of
absolute value.
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Slide 2.7- 3
Use the distance definition of absolute value.
The absolute value of a number x, written |x|, is the distance from x
to 0 on the number line.
For example, the solutions of |x| = 5 are 5 and 5, as shown below.
We need to understand the concept of absolute value in order to
solve equations or inequalities involving absolute values. We solve
them by solving the appropriate compound equation or inequality.
Distance is
5, so |5| = 5.
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Distance is
5, so |5| = 5.
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Use the distance definition of absolute value.
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Slide 2.7- 5
Objective 2
Solve equations of the form
|ax + b| = k, for k > 0.
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Slide 2.7- 6
Use the distance definition of absolute value.
Remember that because absolute value refers to distance from
the origin, an absolute value equation will have two parts.
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CLASSROOM
EXAMPLE 1
Solving an Absolute Value Equation
Solve |3x – 4| = 11.
Solution:
3x – 4 = 11
3x – 4 + 4 = 11 + 4
3x = 7
7
x
3
or
3x – 4 = 11
3x – 4 + 4 = 11 + 4
3x = 15
x=5
7 and 5 into the original absolute value
3
7 
equation to verify that the solution set is 

 ,5 .
 3 
Check by substituting 
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Slide 2.7- 8
Objective 3
Solve inequalities of the form
|ax + b| < k and of the form
|ax + b| > k, for k > 0.
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Slide 2.7- 9
CLASSROOM
EXAMPLE 2
Solving an Absolute Value Inequality with >
Solve |3x – 4|  11.
Solution:
3x – 4 ≤ 11
3x – 4 + 4 ≤ 11 + 4
3x ≤ 7
3x – 4  11
3x – 4 + 4  11 + 4
3x  15
x5
or
7
x
3
Check the solution. The solution set is
The graph consists of two intervals.
-5
-4
-3
]
-2
-1
0
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1
2
3
7

 ,  
3

4
[
5
6
7
5,   .
8
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CLASSROOM
EXAMPLE 3
Solving an Absolute Value Inequality with <
Solve |3x – 4| ≤ 11.
Solution:
11 ≤ 3x – 4 ≤ 11
11 + 4 ≤ 3x – 4 ≤ 11+ 4
7 ≤ 3x ≤ 15
7
  x5
3
Check the solution. The solution set is
 7
  3 ,

5 .

The graph consists of a single interval.
-5
-4
-3
[
-2
-1
0
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1
2
3
4
]
5
6
7
8
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Solve inequalities of the form |ax + b| < k and of the
form |ax + b| > k, for k > 0.
When solving absolute value equations and inequalities of the types in
Examples 1, 2, and 3, remember the following:
1. The methods describe apply when the constant is alone on one side of the
equation or inequality and is positive.
2. Absolute value equations and absolute value inequalities of the form |ax + b| > k
translate into “or” compound statements.
3. Absolute value inequalities of the form |ax + b| < k translate into “and” compound
statements, which may be written as three-part inequalities.
4. An “or” statement cannot be written in three parts.
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Slide 2.7- 12
Objective 4
Solve absolute value equations that
involve rewriting.
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Slide 2.7- 13
CLASSROOM
EXAMPLE 4
Solving an Absolute Value Equation That Requires Rewriting
Solve |3x + 2| + 4 = 15.
Solution:
First get the absolute value alone on one side of the equals sign.
|3x + 2| + 4 = 15
|3x + 2| + 4 – 4 = 15 – 4
|3a + 2| = 11
3x + 2 = 11
3x = 13
or
3x + 2 = 11
3x = 9
x=3
13
x
3
The solution set is
 13
  3 ,
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
3 .

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CLASSROOM
EXAMPLE 5
Solving Absolute Value Inequalities That Require Rewriting
Solve the inequality.
|x + 2| – 3 > 2
Solution:
|x + 2| – 3 > 2
|x + 2| > 5
x+2>5
or
x> 3
x+2<5
x < 7
Solution set: (, 7)  (3, )
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Slide 2.7- 15
CLASSROOM
EXAMPLE 5
Solving Absolute Value Inequalities That Require Rewriting (cont’d)
Solve the inequality.
|x + 2| – 3 < 2
Solution:
|x + 2| < 5
5 < x + 2 < 5
7 < x < 3
Solution set: (7, 3)
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Objective 5
Solve equations of the form
|ax + b| = |cx + d| .
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Slide 2.7- 17
Solve equations of the form |ax + b| = |cx + d|.
Solving |ax + b| = |cx + d|
To solve an absolute value equation of the form
|ax + b| = |cx + d| ,
solve the compound equation
ax + b = cx + d
or
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ax + b = (cx + d).
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CLASSROOM
EXAMPLE 6
Solving an Equation with Two Absolute Values
Solve |4x – 1| = |3x + 5|.
Solution:
4x – 1 = 3x + 5
or
4x – 1 = (3x + 5)
4x – 6 = 3x
or
4x – 1 = 3x – 5
or
7x =  4
6=x
x=6
or
Check that the solution set is
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4
x
7
 4 
 ,6 .
 7 
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Objective 6
Solve special cases of absolute
value equations and inequalities.
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Slide 2.7- 20
Solve special cases of absolute value equations and
inequalities.
Special Cases of Absolute Value
1. The absolute value of an expression can never be negative; that
is, |a|  0 for all real numbers a.
2. The absolute value of an expression equals 0 only when the
expression is equal to 0.
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CLASSROOM
EXAMPLE 7
Solving Special Cases of Absolute Value Equations
Solve each equation.
|6x + 7| = –5
Solution:
The absolute value of an expression can never be negative, so
there are no solutions for this equation. The solution set is .
1
x 3  0
4
1
The expression x  3 will equal 0 only if
4
1
x3
4
x  12.
The solution of the equation is 12.
The solution set is {12}, with just one element.
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CLASSROOM
EXAMPLE 8
Solving Special Cases of Absolute Value Inequalities
Solve each inequality.
|x | > – 1
Solution:
The absolute value of a number is always greater than or equal
to 0. The solution set is (, ).
|x – 10| – 2 ≤ –3
|x – 10| ≤ –1
Add 2 to each side.
There is no number whose absolute value is less than –1, so the
inequality has no solution. The solution set is .
|x + 2| ≤ 0
The value of |x + 2| will never be less than 0. |x + 2| will equal 0
when x = –2. The solution set is {–2}.
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