Download Section 2.2 Inequalities Involving Absolute Value

Section 2.2
Solve Absolute Value Equations
and Inequalities
|x|
|x| = 3
|x| > 3
|x| < 3
Solving Absolute Value
Equations and Inequalities
Let k be a positive number and p and q be two
numbers.
To solve |ax + b| = k, solve the compound
equation
ax + b = k OR ax + b = -k
The solution set is usually of the form {p, q}.
Solving Absolute Value
Equations and Inequalities
Let k be a positive number and p and q be two
numbers.
To solve |ax + b| > k, solve the compound
inequality
ax + b > k OR ax + b < -k
The solution set is usually of the form
(-, p)  (q, )
Solving Absolute Value
Equations and Inequalities
Let k be a positive number and p and q be two
numbers.
To solve |ax + b| < k, solve the compound
inequality
-k < ax + b < k
The solution set is usually of the form
(p, q)
|x|
|x| = 3
|x| > 3
|x| < 3
Solve each equation and graph the solution set.
|x + 2| = 3
x + 2 = -3
x + 2 - 2 = -3 - 2
x=-5
x+2=3
x+2-2=3-2
x=1
x ={-5, 1}
Solve each equation and graph the solution set.
|x + 2| > 3
x + 2 < -3
x + 2 - 2 < -3 - 2
x<-5
x+2>3
x+2-2>3-2
x>1
(-, -5)  (1, )
)
(
Solve each equation and graph the solution set.
|x + 2| < 3
-3 < x + 2 < 3
-3 - 2 < x + 2 - 2 < 3 - 2
-5 < x < 1
(-5, 1)
(
)
Cautions
• These methods apply when the constant is
alone and positive.
• Absolute equations and |ax+b| > k
translate into OR compound statements.
• |ax+b| < k translate into AND compound
statements.
• An OR statement cannot be written into
three parts.
|ax + b| = |cx + d|
To solve an equation of this form solve the
compound equation
ax + b = cx + d OR ax + b = -(cx + d)
|4r - 1| = |3r + 5|
4r 1  3r  5
4r 1  1  3r  5  1
4r  3r  6
4r  1  (3r  5)
4r 1  3r  5
4r 1  1  3r  5  1
4r  3r  3r  3r  6
4r  3r  4
4r  3r  3r  3r  4
r 6
7r  4
  4
r  6,
4

r
 7 
7
|6x + 7| = - 5
Since the absolute value of an
expression can never be
negative, there are no
solutions to this equation.
Solution set = 
Special Cases
|x| > -1
(-, )
|y| < -5
Solution set = 
|k + 2| < 0
k = -2
Matching
E
|x| = 5
|x| < 5 C
|x| > 5 D
|x| < 5 B
|x| > 5 A
A
]
[
[
]
B
(
)
C
)
(
D
E
Solve
|2y + 3|= 19
2y + 3 = 19
2y = 16
y=8
2y + 3 = -19
2y = -22
y = -11
y = {8, -11}
Solve |3x - 1|> 8
3x - 1 > 8
3x > 9
x>3
]
[
3x - 1 < -8
3x < -7
7
x
3
7
x  (, ]  [3, )
3
Section 2.2 Homework
Page 30 #1-25 Odd
Page 31 #17 – 23 Odd
Review for Quiz
Page 44 #1 – 15 Odd
Page 30 #1-25 Odd