3.7 Absolute Value Inequalities:

Inverse Operations: Operations that undo another operation.

Isolate: The use of inverse operations used to leave a variable by itself.

Absolute Value: The distance from zero to that place. It can be on the left (-) or on the right (+) of zero.

GOAL:

The absolute value is treated as any other inequality:

Absolute Value Inequalities:

1. First isolate the absolute value by using inverse operations | | <, >, ≤, ≥ something.

2. Set the inside of the absolute value as –( ) left side of zero

+( ) Right side of zero and isolate the variable. Remember to switch the sign if you divide OR

multiply by a negative number…!

Solving Absolute Value Inequalities:

EX:Solve :

| x+ 3| < 1

SOLUTION: Here the absolute value is already isolated.

| x+3 |< 1- (x + 3) < 1

Set Notation: {x| -4 < x < -2} Interval: (-4, -2)

+3 +3 -x < 4

-x - 3 < 1

x > - 4

x + 3 < 1 - 3 -3

x < - 2Divide by -1 an switch the sign

YOU TRY IT:

Solve:

| x – 5 | > – 4

SOLUTION: Here the absolute value is already isolated.

| x-5 |< -4- (x-5) < -4

Set Notation: {x| x < 1 or x > 9} Interval: (-∞, 1) U (9, ∞)

-5 -5 -x < -9

-x +5 < -4

x > 9

x - 5 < -4 + 5 +5

x < 1Divide by -1 an switch the sign

Solving Absolute Value Inequalities:

EX:Solve :

2| x+ 3| - 7 < 1

Remember to use inverse operations to isolate the absolute value

SOLUTION: Here we must isolate the absolute value first.

2| x+3 |-7 < 1

| x + 3| < 4

+ 7 +7

2 22| x+3 | < 8

SOLUTION: Here the absolute value is now isolated and we continue with the process. | x+3 |< 4

- (x + 3) < 4

Set Notation: {x| -7 < x < 1} Interval: (-7, 1)

+3 +3 -x < 7

-x - 3 < 4

x > - 7

x + 3 < 4 - 3 -3

x < 1Divide by -1 an switch the sign

SOLUTION:- 2 ≤ 2m – 4 < -1 Given (and)

4 - 2 ≤ 2m <-1 + 4 Inverse of subt.

2 ≤ 2m < 3 ≤ m <

Like terms

1 ≤ m < 1.5

Inverse of mult.

Interval: [1, 1.5)

Real-World:The official weight of a nickel is 5 g, but the actual weight can vary from this amount up to 0.194 g. Suppose a bank weights a roll of 40 nickels. The wrapper weights

1.5 g. What is the range of the possible

weights for the roll of nickels.

Real-World: (SOLUTION)Nickel = 5 g

Weight varies up to 0.194 |N – 5 | ≤ 0.194

| N -5 |< 0.194

- (N - 5) ≤ 0.194

- 5 - 5 -N ≤ - 4.806

-N + 5 ≤ 0.194

N ≥ 4.806

+ 5 + 5

Divide by -1 an switch the sign

N - 5 ≤ 0.194

N ≤ 5.194

Real-World: (SOLUTION CONTINUE)Roll 40 nickelsPaper 1.5 g

Total = 40 (nickel) + 1.5 Low Weight 4.806

High Weight 5.196

Low Total = 40 (4.806) + 1.5 193.74

High Total = 40 (5.196) + 1.5 209.26

Set Notation: {N| 193.74 < N < 209.26}

Interval: (193.74, 209.26)

VIDEOS: Absolute Value Inequalities

http://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/absolute-value-inequalities-example-1

http://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/absolute-value-inequalities

VIDEOS: Compoundinequalities

http://www.khanacademy.org/math/algebra/linear_inequalities/compound_absolute_value_inequali/v/absolute-value-inequalities-example-2

CLASSWORK:

Page 211-213

Problems: As many as needed to master the

concept.