Download Chapter 6

Chapter 6
Solving Linear Inequalities
•
6.1 Solving Inequalities by Addition and
Subtraction
Solve by Addition
Set builder Notation:
– A way to write a solution set
•
Ex: if
t  58 the solution set is all #s less
than or equal to 58
{t | t  58}
ex: t – 45 < 13
+45 +45
t < 58
ex: 8 – 2y  -1y
+ 2y + 2y
8  y
{ t | t < 58}
{ y | y  8}
•If the variable is on the open side = shade right
•If the variable is on the closed side = shade left
<
Less than
Fewer
than
>
Greater
than
More
than


At most
At least
No more
than
Less than
or equal
to
No less
than
Greater
than or
equal to
48 50 52 54 56 58 60
5
6
7
8
9
Solve by Subtraction
ex: s + 19 > 56
ex: 5p + 7
-19 -19
-5p
s > 37
7
{s | s > 37}
{p|p
33 34 35 36 37 38 39
5
6
7
8
10 11
 6p
-5p
 p
 7}
9
10 11
6.2 Solving Inequalities by Multiplication and
Division
• If you multiply or divide by a
negative number, you must
change the inequality sign
Solve by Multilplication
2
4
Ex:
s  14

p  10
Ex:
5
3
3
4
10
3
5 2
14 5
  p   
 s 
4
3
1
4
2 5
1 2
30
70
p

s
4
2
s < 35
p  7.5
{ s | s < 35}
33 34 35 36 37 38 39
{p|p
5
6
7
8
9
 7.5}
10 11
Solve by Division
5
12x  60
/12
/12
x  5
-8q < 136
/-8 /-8
q > -17
{ x | x  5}
{ q | q > -17}
6
7
8
9
10 11
-19 -18 -17 -16 -15 -14 -13
6.3 Solving Multi-Step Inequalities
Multi-step Inequality
9
c  32  104
5
-32
3d – 2(8d – 9) > -2d - 4
3d – 16d + 18 > -2d - 4
-32
{c | c > 40}
9
c  72
5
5 9
72 5
 c

9 5
1 9
c > 40
38 39 40 41 42 43 44
Write and Solve an Inequality
Three times a number minus eighteen is at
least five times the number plus twenty-one
3x 18  5x  21
-3x
18  2x  21
 39  2x
/2
-21
-13d + 18 > -2d - 4
+13d
+13d
18 > 11d - 4
+4
+4
22 > 11d
/11 /11
2>d
{d | d < 2}
-1
0
1
2
3
4
5
Empty Sets and All Real Numbers
8(t +2)- 3(t – 4) < 5(t -7) + 8
8t +16 - 3t + 12 < 5t - 35 + 8
-3x
-21
Distributive Property
{ x | x  -19.5}
/2
5t + 28 < 5t - 27
-5t
- 5t
28 < -27
Empty set
•If it is true – all real numbers
{ x | x is a real number}
19.5  x
-24 -23 -22 -21 -20 -19 -18
•If it is false – it is an empty set
6.4 Solving Compound Inequalities
• Intersection = and
• Union = or
Ex: Solve and Graph a Compound
Inequality -5 < x – 4 < 2
x–4<2
+4 +4
-5 < x - 4
+4 +4
Ex: Graph a Compound Inequality
x  2
x3
-3
-2
-1
0
1
2
3
4
5
-3
-2
-1
0
1
2
3
4
5
-3
-2
-1
0
1
2
3
4
5
{x | -2  x < 3}
-1 < x
x<6
-3
-2
-1
0
1
2
3
4
5
-3
-2
-1
0
1
2
3
4
5
-3
-2
-1
0
1
2
3
4
{x | -1 < x < 6 }
6
5
6
6.5 Absolute Value Open
Make a frayerSentences
Foldable (diamond in
the center)
• Absolute Value: The distance from zero on
a number line
– The positive value of the number
Ex:
-7
|-5| = 5
|5| =
5
-6
-5
-4
-3
-2
-1
0
1
|-6| = 6
|2| = 2
2
3
4
5
6
7
•
Solving Absolute Value Equations:
a. The expression inside the absolute value bars is
positive
b. The expression inside the absolute value bars is
negative
Ex: | x + 7| = 4
a.
x+7=4
-7 -7
x = -3
b.
x + 7 = -4
-7
-7
x = -11
Solution set: (-3, -11)
• If the absolute value bars equal a negative
number- it is an empty set
• Write an absolute value equation:
|x-#|=#
3
Ex:
-3
-2
-1
0
2
3
1
2
3
2
Ex:
4
|x–2|=3
5
6
-19 -18 -17 -16 -15 -14 -13
| x + 16| = 2
# half way
between
#
spaces
from
each
point
• Absolute Value Function
f(x) = |x| and f(x) must be greater than or equal to zero
f(x) = |x| can be written as: f(x)= {
-x if x < 0
x if x
Graph of
f(x) = |x|
x
|x|
-1
1
1
1
0
0
-2
2
2
2
0
6.6 Solving Absolute Value
On the back of
Inequalities
the 6.5 frayer
• A. the expression in absolute value bars is
positive
• B. the expression in absolute value bars is
negative (also flip the inequality sign)
|x| = n
x = -n
|x| < n
x<n
|x| > n
x>n
or
x=n
and x > -n
or
x < -n
• Ex1:
|t+5|<9
t+5<9
-5 -5
t + 5 > -9
-5 -5
t<4
-16
t > -14
-14 -12 -10 -8
-6
-4
-2
-14 < t < 4
0
2
4
6
8
|2x + 8|  6
• Ex2:
2x + 8  6
-8 -8
2x
/2

2x + 8
-8
x  -7
-8
-7
-8
2x  -14
/2
/2
-2
/2
x  -1
-9
 -6
-6 -5
-4
-3
-2
-1
0
1
x  -7 or x  -1
• Ex3:
| 2y – 1 |  -4
the absolute value cannot be
less than zero so y is all
numbers
• Ex4: (your turn)
| 2k + 1| > 7

*If it is
a negative
number- the answer is
empty set
6.7 Graphing Inequalities with Two
Make a frayerVariables
Foldable (diamond in
the center)
•
•
The equation makes the line to define the boundary
The shaded region is the half-plane
1.
2.
Get the equation into slope-intercept form
List the intercept as an ordered-pair and the slope as a
ratio
Graph the intercept and use the slope to find at least 2
more points
Draw the line (dotted or solid)
Test an ordered-pair not on the line
3.
4.
5.
1.
2.
If it is true shade that side of the line
If it is false shade the other side of the line
Ex1:
< or >
 or 
Dotted Line
Solid Line
y 2x - 3
m=
2
1
b = -3 = (0, -3)
Use a solid line because it is 
Test: (0, 0)
0  2(0) – 3
0 0 – 3
0  -3
false (shade other side)
• Ex2: y – 2x < 4
y – 2x < 4
+ 2x +2x
y < 2x + 4
m=
2
Use a dotted line because it is <
1
b = 4 = (0, 4)
Test: (0, 0)
0 < 2(0) + 4
0 < 0+4
0 <4
true (shade this side)
• Ex3: 3y - 2 > -x + 7
3y – 2 > -x + 7
+2
+2
3y > -x + 9
/3
/3
/3
1
x+3
3
y>-
1
m=Use a dotted line because it is >
3
b = 3 = (0, 3)
Test: (0, 0)
1
3
0 > - (0) + 3
0 > 0+3
0 >3
false (shade other side)