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:
Absolute Value Inequalities:
The absolute value is treated as any other
inequality:
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
-x - 3 < 1
+3 +3
-x < 4
Divide by -1 an
switch the sign
x>-4
x+3<1
- 3 -3
x<-2
Set Notation: {x| -4 < x < -2} Interval: (-4, -2)
YOU TRY IT:
Solve:
|x–5|>–4
SOLUTION: Here the absolute value is
already isolated.
| x-5 |< -4
- (x-5) < -4
-x +5 < -4
-5 -5
-x < -9
Divide by -1 an
switch the sign
x>9
x - 5 < -4
+ 5 +5
x<1
Set Notation: {x| x < 1 or x > 9}
Interval: (-∞, 1) U (9, ∞)
Solving Absolute Value Inequalities:
Remember to use inverse operations to
isolate the absolute value
EX:
Solve :
2| x+ 3| - 7 < 1
SOLUTION: Here we must isolate the
absolute value first.
2| x+3 |-7 < 1
+ 7 +7
2| x+3 | < 8
2
2
| x + 3| < 4
SOLUTION: Here the absolute value is now
isolated and we continue with the
process.
| x+3 |< 4
- (x + 3) < 4
x+3<4
-x - 3 < 4
- 3 -3
+3 +3
x
<
1
-x < 7
Divide by -1 an
switch the sign
x>-7
Set Notation: {x| -7 < x < 1}
Interval: (-7, 1)
SOLUTION:
- 2 ≤ 2m – 4 < -1 Given (and)
4 - 2 ≤ 2m <-1 + 4 Inverse of subt.
2 ≤ 2m < 3
Like terms
𝟐
𝟐
Inverse of mult.
≤m<
𝟑
𝟐
1 ≤ m < 1.5 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
- (N - 5) ≤ 0.194
+
5
+
5
-N + 5 ≤ 0.194
-5 -5
N ≤ 5.194
-N ≤ - 4.806
Divide by -1 an
switch the sign
N ≥ 4.806
Real-World: (SOLUTION CONTINUE)
Roll  40 nickels
Paper  1.5 g
Low Weight  4.806
Total = 40 (nickel) + 1.5
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/linea
r_inequalities/compound_absolute_value_inequali
/v/absolute-value-inequalities
http://www.khanacademy.org/math/algebra/linea
r_inequalities/compound_absolute_value_inequali
/v/absolute-value-inequalities-example-1
VIDEOS:
Compound
inequalities
http://www.khanacademy.org/math/algebra/linea
r_inequalities/compound_absolute_value_inequali
/v/absolute-value-inequalities-example-2
CLASSWORK:
Page 211-213
Problems: As many as needed
to master the
concept.