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A. Solving Absolute Value Equations

# 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

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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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

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