1 Copyright © 2014, 2010, and 2006 Pearson Education, Inc. Absolute-Value Equations and Inequalities Equations with Absolute Value Inequalities with Absolute.

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Absolute-Value Equations and Inequalities

Equations with Absolute Value

Inequalities with Absolute Value

9.3

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Warm Up

– Solve and graph each compound inequality.

– 1. -3x – 4 < -16 and 2x - 5 < 15– 2. 4x + 2 > 10 or -2x + 6 > 16– 3. 5 < -3x+2 < 8– 4. 2x < -2 and x + 3 > 6– 5. 5x > 15 or 6x - 4 < 32

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Absolute Value

The absolute value of a number is its distance from 0. Since absolute value is distance it is never negative.

, if 0,

, if 0.

x xx

x x

Definition of absolute value.

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Find each absolute value.

|-3| |0| |4|

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Solution

Example

a) |x| = 6; b) |x| = 0; c) |x| = –2

a) |x| = 6 means that the distance from x to 0 is 6. Only two numbers meet that requirement. Thus the solution set is {–6, 6}.

b) |x| = 0 means that the distance from x to 0 is 0. The only number that satisfies this is zero itself. Thus the solution set is {0}.

c) Since distance is always nonnegative, |x| = –2 has no solution. Thus the solution set is .

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This brings us to…

The Absolute-Value Principlefor Equations If |x| = p, then x = p or x = -p.

Note: The equation |x| = 0 is equivalent to the equation x = 0 and the equation |x| = –p has no solution.

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Example: a) |2x +1| = 5; b) |3 – 4x| = –10

The solution set is {–3, 2}. The check is left for the student.

x = –3 or x = 2 2x = –6 or 2x = 4

Substituting

Solutiona) We use the absolute-value principle, knowing

that 2x + 1 must be either 5 or –5:

|2x +1| = 5

|x| = p

2x +1 = –5 or 2x +1 = 5

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Solution(continued)

The absolute-value principle reminds us that absolute value is always nonnegative. The equation |3 – 4x| = –10 has no solution. The solution set is

To apply the absolute value principle we must make sure the absolute value expression is ISOLATED.

.

b) |3 – 4x| = –10

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Solution Since we

Given that f (x) = 3|x + 5| – 4, find all x for which f (x) = 11

are looking for f(x) = 11, we substitute:

Replacing f (x) with 3|x + 5| − 4

x + 5 = –5 or x + 5 = 5 x = –10 or x = 0

The solution set is {–10, 0}. The check is left for the student.

3|x+5| – 4 = 11

f(x) = 11

3|x + 5| = 15

|x + 5| = 5

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Sometimes an equation has two absolute-value expressions like |x+1| = |2x|. This means that x+1 and 2x are the same distance from zero.

Since they are the same distance from zero, they are the same number or they are opposites.

So solve x+1 = 2x and x+1 = -2x separately to get your answers. As always check by substitution.

When more than one absolute value expression appears…

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Solve |3x – 5| = |8 + 4x|.

You are not done until you solve each equation.

3x – 5 = 8 + 4x

This equation sets both sides the same.

This equation sets them opposite. Notice parenthesis.

3x – 5 = –(8 + 4x) or

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3 5 8 4x x 3 5 (8 4 )x x

5 8 x 13 x

3 5 8 4x x 7 5 8x

7 3x 3

7x

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The Absolute-Value Inequalities Principle

For any positive number p and any expression x:

a) |x| < p is equivalent to –p < x < p. (conjunction)

b) |x| > p is equivalent to x < -p or x > p. (disjunction)

c) Similar rules apply for less than or equal, greater than or equal.

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So the solutions of |X| < p are those numbers that satisfy –p < X < p.

And the solutions of |X| > p are those numbers that satisfy X < –p or p < X.

–p p

–p p( )

()

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Solution

Solve |x| < 3. Then graph.

The solutions of |x| < 3 are all numbers whose distance from zero is less than 3. By substituting we see that numbers like –2, –1, –1/2, 0, 1/3, 1, and 2 are all solutions. The solution set is {x| –3 < x < 3}. In interval notation, the solution set is (–3, 3). The graph is as follows:

-3 0 3( )

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Solution

The solutions of are all numbers whose distance from zero is at least 3 units. The solution set is In interval notation, the solution set is

The graph is as follows:

Solve 3. Then graph.x

3x

{ | 3 3}. x x or x

( , 3] [3, ).

[-3 3]

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SolutionThe number 3x + 7 must be less than 8 units from 0.

|3x + 7| < 8

|X| < pReplacing X with 3x + 7 and p with 8

–8 < 3x + 7 < 8

The solution set is {x|–5 < x < 1/3}. The graph is as follows:

Solve |3x + 7| < 8. Then graph.

−5 < x < 1/3

–15 < 3x < 1

–5 1/3( )

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Solution

The number 3x + 7 must be greater than 8 units from 0.

|3x + 7| < 8

|X| < pReplacing X with 3x + 7 and p with 8

–8 < 3x + 7 < 8

The solution set is {x|–5 < x < 1/3}. The graph is as follows:

Solve |3x + 7| > 8. Then graph.

−5 < x < 1/3

–15 < 3x < 1

–5 1/3( )