A. Solving Absolute Value Equations
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College Algebra G&M—
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Section 2.3 Absolute Value Functions, Equations, and Inequalities
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If an equation has more than one solution as in Example 1, they cannot be
simultaneously stored using the, X,T,,n key to perform a calculator check (in function or “Func” mode, this is the variable X). While there are other ways to “get
around” this (using Y 1 on the home screen, using a TABLE in ASK mode,
enclosing the solutions in braces as in {4, 10}, etc.), we can also store solutions
using the ALPHA keys. To illustrate, we’ll place the solution x ϭ 4 in storage location A,
using 4 STO ALPHA MATH (A). Using this “ STO ALPHA ” sequence we’ll next place the
solution x ϭ 10 in storage location B (Figure 2.40). We can then check both solutions
in turn. Note that after we check the first solution, we can recall the expression
using 2nd
and simply change the A to B (Figure 2.41).
ENTER
Figure 2.40
Figure 2.41
Absolute value equations come in many different forms. Always begin by isolating the absolute value expression, then apply the property of absolute value equations to solve.
EXAMPLE 2
ᮣ
Solving an Absolute Value Equation
Solve:
Solution
ᮣ
2
` 5 Ϫ x ` Ϫ 9 ϭ 8.
3
2
`5 Ϫ x ` Ϫ 9 ϭ 8
3
2
` 5 Ϫ x ` ϭ 17
3
2
5 Ϫ x ϭ Ϫ17
3
2
Ϫ x ϭ Ϫ22
3
x ϭ 33
Check
WORTHY OF NOTE
As illustrated in both Examples
1 and 2, the property we use to
solve absolute value equations can
only be applied after the absolute
value term has been isolated. As
you will see, the same is true for the
properties used to solve absolute
value inequalities.
ᮣ
2
For x ϭ 33: ` 5 Ϫ 1332 `
3
|5 Ϫ 21112 |
05 Ϫ 22 0
0 Ϫ17 0
17
original equation
add 9
or
or
or
Ϫ9ϭ8
Ϫ9ϭ8
Ϫ9ϭ8
Ϫ9ϭ8
Ϫ9ϭ8
8 ϭ 8✓
2
5 Ϫ x ϭ 17
3
2
Ϫ x ϭ 12
3
x ϭ Ϫ18
apply the property of absolute
value equations
subtract 5
multiply by Ϫ32
2
1Ϫ182 ` Ϫ 9 ϭ 8
3
| 5 Ϫ 21Ϫ62 | Ϫ 9 ϭ 8
0 5 ϩ 12 0 Ϫ 9 ϭ 8
0 17 0 Ϫ 9 ϭ 8
17 Ϫ 9 ϭ 8
8 ϭ 8✓
For x ϭ Ϫ18: ` 5 Ϫ
Both solutions check. The solution set is 5Ϫ18, 336.
Now try Exercises 19 through 22
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CHAPTER 2 More on Functions
For some equations, it’s helpful to apply the multiplicative property of absolute
value:
Multiplicative Property of Absolute Value
If A and B represent algebraic expressions,
then ͿABͿ ϭ ͿAͿͿBͿ.
Note that if A ϭ Ϫ1 the property says ͿϪ1 # BͿ ϭ ͿϪ1Ϳ ͿBͿ ϭ ͿBͿ. More generally
the property is applied where A is any constant.
EXAMPLE 3
ᮣ
Solution
ᮣ
Solving Equations Using the Multiplicative Property of Absolute Value
Solve: ͿϪ2xͿ ϩ 5 ϭ 13.
ͿϪ2xͿ ϩ 5 ϭ 13
ͿϪ2xͿ ϭ 8
ͿϪ2ͿͿxͿ ϭ 8
2ͿxͿ ϭ 8
ͿxͿ ϭ 4
x ϭ Ϫ4 or x ϭ 4
original equation
subtract 5
apply multiplicative property of absolute value
simplify
divide by 2
apply property of absolute value equations
Both solutions check. The solution set is 5Ϫ4, 46.
Now try Exercises 23 and 24
ᮣ
In some instances, we have one absolute value quantity equal to another, as in ͿAͿ ϭ ͿBͿ.
From this equation, four possible solutions are immediately apparent:
(1) A ϭ B
(2) A ϭ ϪB
(3) ϪA ϭ B
(4) ϪA ϭ ϪB
However, basic properties of equality show that equations (1) and (4) are equivalent, as
are equations (2) and (3), meaning all solutions can be found using only equations (1)
and (2).
EXAMPLE 4
ᮣ
Solving Absolute Value Equations with Two Absolute Value Expressions
Solve the equation Ϳ2x ϩ 7Ϳ ϭ Ϳx Ϫ 1Ϳ.
Solution
ᮣ
This equation has the form ͿAͿ ϭ ͿBͿ, where A ϭ 2x ϩ 7 and B ϭ x Ϫ 1. From our
previous discussion, all solutions can be found using A ϭ B and A ϭ ϪB.
AϭB
2x ϩ 7 ϭ x Ϫ 1
2x ϭ x Ϫ 8
x ϭ Ϫ8
solution template
substitute
subtract 7
subtract x
A ϭ ϪB
2x ϩ 7 ϭ Ϫ1x Ϫ 12
2x ϩ 7 ϭ Ϫx ϩ 1
3x ϭ Ϫ6
x ϭ Ϫ2
solution template
substitute
distribute
add x, subtract 7
divide by 3
The solutions are x ϭ Ϫ8 and x ϭ Ϫ2. Verify the solutions by substituting them
into the original equation.
A. You’ve just seen how
we can solve absolute value
equations
Now try Exercises 25 and 26
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Section 2.3 Absolute Value Functions, Equations, and Inequalities
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B. Solving “Less Than” Absolute Value Inequalities
Absolute value inequalities can be solved using the basic concept underlying the
property of absolute value equalities. Whereas the equation ͿxͿ ϭ 4 asks for all numbers
x whose distance from zero is equal to 4, the inequality ͿxͿ 6 4 asks for all numbers x
whose distance from zero is less than 4.
Distance from zero is less than 4
Figure 2.42
)
Ϫ5 Ϫ4
Ϫ3 Ϫ2 Ϫ1
)
0
1
2
3
4
5
As Figure 2.42 illustrates, the solutions are x 7 Ϫ4 and x 6 4, which can be written
as the joint inequality Ϫ4 6 x 6 4. This idea can likewise be extended to include
the absolute value of an algebraic expression X as follows.
Property I: Absolute Value Inequalities (Less Than)
If X represents an algebraic expression and k is a positive real number,
then ͿXͿ 6 k
implies Ϫk 6 X 6 k
Property I can also be applied when the “Յ” symbol is used. Also notice that if
k 6 0, the solution is the empty set since the absolute value of any quantity is always
positive or zero.
EXAMPLE 5
Solution
ᮣ
ᮣ
Solving “Less Than” Absolute Value Inequalities
Solve the inequalities:
Ϳ3x ϩ 2Ϳ
a.
Յ1
4
Ϳ3x ϩ 2Ϳ
a.
Յ1
4
Ϳ3x ϩ 2Ϳ Յ 4
Ϫ4 Յ 3x ϩ 2 Յ 4
Ϫ6 Յ 3x Յ 2
2
Ϫ2 Յ x Յ
3
b. Ϳ2x Ϫ 7Ϳ 6 Ϫ5
original inequality
multiply by 4
apply Property I
subtract 2 from all three parts
divide all three parts by 3
The solution interval is 3Ϫ2, 23 4.
b. Ϳ2x Ϫ 7Ϳ 6 Ϫ5
original inequality
Since the absolute value of any quantity is always positive or zero, the solution
for this inequality is the empty set: { }.
Now try Exercises 27 through 38
ᮣ
As with the inequalities from Section 1.5, solutions to absolute value inequalities
can be checked using a test value. For Example 5(a), substituting x ϭ 0 from the
solution interval yields:
1
Յ 1✓
2
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CHAPTER 2 More on Functions
In addition to checking absolute value inequalities using a test value, the TABLE
feature of a graphing calculator can be used, alone or in conjunction with a relational
test. Relational tests have the calculator return a “1” if a given statement is true, and a
“0” otherwise. To illustrate, consider the inequality 2Ϳx Ϫ 3Ϳ ϩ 1 Յ 5. Enter the
expression on the left as Y1, recalling the “abs(” notation is accessed in the MATH menu:
MATH
(NUM) “1:abs(”
(note this option gives only the left parenthesis, you must
supply the right). We can then simply inspect the Y1 column of the TABLE to find outputs that are less than or equal to 5. To use a relational test, we enter Y1 Յ 5 as Y2
(Figure 2.43), with the “less than or equal to” symbol accessed using 2nd MATH 6:Յ.
Now the calculator will automatically check the truth of the statement for any value of x
(but note we are only checking integer values), and display the result in the Y2 column
of the TABLE (Figure 2.44). After scrolling through the table, both approaches
show that 2Ϳx Ϫ 3Ϳ ϩ 1 Յ 5 for x ʦ [1, 5].
ENTER
Figure 2.43
Figure 2.44
B. You’ve just seen how
we can solve “less than”
absolute value inequalities
C. Solving “Greater Than” Absolute Value Inequalities
For “greater than” inequalities, consider ͿxͿ 7 4. Now we’re asked to find all numbers
x whose distance from zero is greater than 4. As Figure 2.45 shows, solutions are
found in the interval to the left of Ϫ4, or to the right of 4. The fact the intervals are
disjoint (disconnected) is reflected in this graph, in the inequalities x 6 Ϫ4 or x 7 4,
as well as the interval notation x ʦ 1Ϫq, Ϫ42 ´ 14, q 2.
Distance from zero
is greater than 4
Figure 2.45
)
Ϫ7 Ϫ6 Ϫ5 Ϫ4 Ϫ3 Ϫ2 Ϫ1
Distance from zero
is greater than 4
)
0
1
2
3
4
5
6
7
As before, we can extend this idea to include algebraic expressions, as follows:
Property II: Absolute Value Inequalities (Greater Than)
If X represents an algebraic expression and k is a positive real number,
then ͿXͿ 7 k
implies X 6 Ϫk or X 7 k
EXAMPLE 6
ᮣ
Solving “Greater Than” Absolute Value Inequalities
Solve the inequalities:
1
x
a. Ϫ ` 3 ϩ ` 6 Ϫ2
3
2
b. Ϳ5x ϩ 2Ϳ Ն Ϫ
3
2