Download Section 1.3 Operations on the Set of Real Numbers

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Linear Inequalities in One Variable
Basic ideas
Statements that express the inequality
of algebraic expressions are called
inequalities.
Inequality Symbols

is less than

is less than or equal to

is more than

is more than or equal to
Basic ideas
We know 4 < 5 but how do we compare -4 and -5
-5 -4 -3 -2 -1 0 1 2 3 4 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
 9  6
True
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
 9  6
False
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
 9  6
False
3  2
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
 9  6
False
3  2
True
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
 9  6
False
3  2
True
44
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
 9  6
False
3  2
True
44
True
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
 9  6
False
3  2
True
44
True
 4  5
-5 -4 -3 -2 -1 0 1 2 3 4 5
True or False
5  3
True
 9  6
False
3  2
True
44
True
 4  5 True
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
1.5<3
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
1.5<3
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
1.5<3
0<3
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
1.5<3
0<3
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
1.5<3
0<3
-2<3
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The statement x < 3 means that x is a
real number to the left of 3 on the
number line.
The numbers 1.5, 0 and -2 are all to
the left of 3 on the number line so the
following are true.
1.5<3
0<3
-2<3
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The set of all numbers that give a true
statement when used as a
replacement for x is called the
solution set to the inequality.
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The set of all numbers that give a true
statement when used as a
replacement for x is called the
solution set to the inequality.
The solution set to x< 3 is all the real
numbers to the left of 3 on a number
line.
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The set of all numbers that give a true
statement when used as a
replacement for x is called the
solution set to the inequality.
The solution set to x< 3 is all the real
numbers to the left of 3 on a number
line.
-5 -4 -3 -2 -1 0 1 2 3 4 5
Graphing Inequalities
The set of all numbers that give a true
statement when used as a
replacement for x is called the
solution set to the inequality.
The solution set to x< 3 is all the real
numbers to the left of 3 on a number
line.
-5 -4 -3 -2 -1 0 1 2 3 4 5
Open Interval
x x  a
Open Interval
x x  a
(
a
Open Interval
x x  a
(
a
Open Interval
x x  a
(
a
a, 
Open Interval
x a  x  b
Open Interval
x a  x  b
(
a
)
b
Open Interval
x a  x  b
(
a
)
b
Open Interval
x a  x  b
(
a
a, b
)
b
Open Interval
x x  b
Open Interval
x x  b
)
b
Open Interval
x x  b
)
b
Open Interval
x x  b
 , b
)
b
Half-open Interval
x x  a
Half-open Interval
x x  a
[
a
Half-open Interval
x x  a
[
a
Half-open Interval
x x  a
[
a
a, 
Half-opened Interval
x a  x  b
Half-opened Interval
x a  x  b
(
a
Half-opened Interval
x a  x  b
(
a
]
b
Half-opened Interval
x a  x  b
(
a
]
b
Half-opened Interval
x a  x  b
(
a
a, b
]
b
Half-opened Interval
x a  x  b
Half-opened Interval
x a  x  b
[
a
Half-opened Interval
x a  x  b
[
a
)
b
Half-opened Interval
x a  x  b
[
a
)
b
Half-opened Interval
x a  x  b
[
a
a, b
)
b
Half-open Interval
x x  b
Half-open Interval
x x  b
]
b
Half-open Interval
x x  b
]
b
Half-open Interval
x x  b
 , b
]
b
Closed Interval
x a  x  b
Closed Interval
x a  x  b
[
a
Closed Interval
x a  x  b
[
a
]
b
Closed Interval
x a  x  b
[
a
]
b
Closed Interval
x a  x  b
[
a
a, b
]
b
All real numbers
x x is real
All real numbers
x x is real
All real numbers
x x is real
 , 
Interval Notation
We write this set in set notation as
 x|x < 3
or in interval notation
(-,3)
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
We write this set in set notation as
 x|x < 3
or in interval notation
(-,3)
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
We write this set in set notation as
 x|x < 3
or in interval notation
(-,3)
)
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
We write this set in set notation as
 x|x < 3
or in interval notation
(-,3)
)
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
We write this set in set notation as
 x|x < 3
or in interval notation
(-,3)
)
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
We write this set in set notation as
 x|x < 3
or in interval notation
(-,3)
)
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
If an endpoint is to be included
we write this set in set notation as
 x|x  1
or in interval notation
1, 
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
If an endpoint is to be included
we write this set in set notation as
 x|x  1
or in interval notation
1, 
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
If an endpoint is to be included
we write this set in set notation as
 x|x  1
or in interval notation
1, 
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
If an endpoint is to be included
we write this set in set notation as
 x|x  1
or in interval notation
1, 
[
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
If an endpoint is to be included
we write this set in set notation as
 x|x  1
or in interval notation
1, 
[
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
If an endpoint is to be included
we write this set in set notation as
 x|x  1
or in interval notation
1, 
[
-5 -4 -3 -2 -1 0 1 2 3 4 5
Interval Notation
If an endpoint is to be included
we write this set in set notation as
 x|x  1
or in interval notation
1, 
[
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x > -5
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x > -5
(
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x > -5
(
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x > -5
or in interval notation
(
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x > -5
or in interval notation
 5, 
(
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x2
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x2
]
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x2
]
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x2
or in interval notation
]
-5 -4 -3 -2 -1 0 1 2 3 4 5
Write in interval notation and graph
x2
or in interval notation
 ,2
]
-5 -4 -3 -2 -1 0 1 2 3 4 5
Linear Inequality
A linear inequality in one variable x
is any inequality of the form
Ax + b < 0 where a and b are real
numbers, with a  0
Linear Inequality
A linear inequality in one variable x
is any inequality of the form
Ax + b < 0 where a and b are real
numbers, with a  0
In place of  we may also use
Linear Inequality
A linear inequality in one variable x
is any inequality of the form
Ax + b < 0 where a and b are real
numbers, with a  0
In place of  we may also use
,
Linear Inequality
A linear inequality in one variable x
is any inequality of the form
Ax + b < 0 where a and b are real
numbers, with a  0
In place of  we may also use
, ,
Linear Inequality
A linear inequality in one variable x
is any inequality of the form
Ax + b < 0 where a and b are real
numbers, with a  0
In place of  we may also use
, , or 
Properties of Inequality
Addition - Subtraction Propety
The same real number may be
added to or subtracted from
each side of an inequality
without changing the solution set.
Properties of Inequality
Addition - Subtraction Propety
The same real number may be
added to or subtracted from
each side of an inequality
without changing the solution set.
Properties of Inequality
Multiplication - Division Propety
The same real number may be
multiplied by or divided by
each side of an inequality
without changing the solution set.
Properties of Inequality
Multiplication - Division Propety
The same real number may be
multiplied by or divided by
each side of an inequality
without changing the solution set.
Properties of Inequality
Each side of an inequality may be
multiplied by or divided by the same
non-zero number without changing
the solution set.
If the inequality is multiplied by or
divided by a negative number, the
inequality symbol is reversed.
Properties of Inequality
Each side of an inequality may be
multiplied by or divided by the same
non-zero number without changing
the solution set.
If the inequality is multiplied by or
divided by a negative number, the
inequality symbol is reversed.
Equaivalent Inequalities
Equivalent inequalities are
inequalities with the same solution
set.
Solve x - 7 = -12
Solve x - 7 = -12
7
7
Solve x - 7 = -12
7
x
7

Solve x - 7 = -12
7
x
7
 -5
The solution set is
Solve x - 7 = -12
7
x
7
 -5
The solution set is- 5
Solve x - 7 = -12
7
x
7
 -5
The solution set is- 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
Solve x - 7 = -12
7
x
7
 -5
The solution set is- 5
-5 -4 -3 -2 -1 0 1 2 3 4 5
Solve 14  2m  3m
Solve 14  2m  3m
 2m  2m
Solve 14  2m  3m
 2m  2m
14
Solve 14  2m  3m
 2m  2m
14

Solve 14  2m  3m
 2m  2m
14
 1m
Solve 14  2m  3m
 2m  2m
14
 1m
23
Solve 14  2m  3m
 2m  2m
14
 1m
23
3 2
Solve 14  2m  3m
 2m  2m
 1m
14
m
23
3 2
Solve 14  2m  3m
 2m  2m
 1m
14
m
14
23
3 2
Solve 14  2m  3m
 2m  2m
14
 1m
m  14
23
3 2
Solve 14  2m  3m
 2m  2m
14
23
3 2
 1m
m  14
10
12
14
16
18
Solve 14  2m  3m
 2m  2m
14
23
3 2
 1m
m  14
10
12
[
14
16
18
Solve 14  2m  3m
 2m  2m
14
23
3 2
 1m
m  14
10
12
[
14
16
18
Solve 14  2m  3m
 2m  2m
14
 1m
m  14
10
12
23
3 2
[
14
14, 
16
18
Solve 5m  - 30
Solve 5m  - 30
5m
5
Solve 5m  - 30
5m
5

Solve 5m  - 30
5m
5

-30
5
Solve 5m  - 30
5m
5
m

-30
5
Solve 5m  - 30
5m
5

m 
-30
5
Solve 5m  - 30
5m
5

-30
5
m  6
Solve 5m  - 30
5m
5

-30
5
m  6
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
Solve 5m  - 30
5m
5

-30
5
m  6
]
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
Solve 5m  - 30
5m
5

-30
5
m  6
]
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
Solve 5m  - 30
5m
5

-30
5
m  6
 ,6
]
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
Solve - 3(x  4)  2  7 - x
Solve - 3(x  4)  2  7 - x
 3x
Solve - 3(x  4)  2  7 - x
 3x  12
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10 
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7  2x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7  2x
7
7
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7  2x
7
7
 17
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7  2x
7
7
 17 
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7  2x
7
7
 17   2x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7  2x
7
7
 17   2x
17
2
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
 7x
 3x  10
 3x
 3x
 10  7  2x
7
7
 17   2x
17
2  x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
 3x
 3x
 10  7  2x
7
7
 17   2x
17
2  x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
3 2
 3x
 3x
 10  7  2x
7
7
 17   2x
17
2  x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
3 2
 3x
 3x

2
x


10
7
17
x 2
7
7
 17   2x
17
2  x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
3 2
 3x
 3x

2
x


10
7
17
x 2
7
7
 17   2x
17
2  x
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
3 2
 3x
 3x

2
x


10
7
17
x 2
7
7
 17   2x
17
2  x
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
3 2
 3x
 3x

2
x


10
7
17
x 2
7
7
 17   2x
17
2  x
]
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
3 2
 3x
 3x

2
x


10
7
17
x 2
7
7
 17   2x
17
2  x
]
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
Solve - 3(x  4)  2  7 - x
 3x  12  2  7  x
23
 7x
 3x  10
3 2
 3x
 3x

2
x


10
7
17
x 2
7
7
 17   2x
17
 , 2 
17
2  x
]
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0
2
1 1
Solve  (r  3)   (5  r)
3
2 2
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2

2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
 4
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
 4
6
1

1
2
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
 4
6
1

1
2

6
2
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
 4
6
1

1
2

6
2
 3
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
 4
 4r  3
6
1

1
2

6
2
 3
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
 4
 4r  3  3 
6
1

1
2

6
2
 3
2
1 1
Solve  (r  3)   (5  r)
3
2 2
1
 2
1

6 (r  3)    6  (5  r) 
2
 3
2


6 2
1 3

12
3
 4
6
1

1
2

6
2
 4r  3  3  35  r 
 3
 4r  3  3  35  r 
 4r  3  3  35  r 
 4r
 4r  3  3  35  r 
 4r  12
 4r  3  3  35  r 
 4r  12  3
 4r  3  3  35  r 
 4r  12  3 
 4r  3  3  35  r 
 4r  12  3  15
 4r  3  3  35  r 
 4r  12  3  15  3r
 4r  3  3  35  r 
 4r  9
 15  3r
 4r  3  3  35  r 
 4r  9
 4r
 15  3r
 4r
 4r  3  3  35  r 
 4r  9
 4r
9
 15  3r
 4r
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
6
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
6 
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
23
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
23
3 2
 4r  3  3  35  r 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
23
3 2
 4r  3  3  35  r 
r  6
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
23
3 2
 4r  3  3  35  r 
r  6
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2
23
3 2
 4r  3  3  35  r 
r  6
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
(
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2
23
3 2
 4r  3  3  35  r 
r  6
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
(
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2
23
3 2
 4r  3  3  35  r 
r  6
 6, 
 15  3r
 4r  9
 4r
 4r
 9  15  r
 15  15
r
6 
(
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2
Solve - 2  - 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
-1
3k - 2
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k 
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k  1
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k  1
3k
3
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k  1
3k
3
1
3
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k 
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and
 3k  3k
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k 
1
3
- 3k - 1  5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1
5
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1
5

3k
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1
2 2
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
6
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
6
3k
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
3k  1
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k  1

3
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3
3
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3
3
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3
3
k
k 
1
3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3
3
2 k
1
k  3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3
3
2 k
1
k  3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3
3
2 k
1
k -2
k  3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3
3
2 k
1
k  -2
k  3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3 - 2, 1 
3

2

k
3
1
k  -2
k  3
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3 - 2, 1 
3

2

k
3
1
k  -2
k  3
-5 -4 -3 -2 -1 0 1 2 3 4 5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3 - 2, 1 
3

2

k
3
1
k  -2
k  3
[
-5 -4 -3 -2 -1 0 1 2 3 4 5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3 - 2, 1 
3

2

k
3
1
k  -2
k  3
[
]
-5 -4 -3 -2 -1 0 1 2 3 4 5
Solve - 2  - 3k - 1  5
- 2  - 3k - 1 and - 3k - 1  5
 3k
 3k
 3k  3k

1

5

3k
3k - 2  - 1

5

5
2 2
 6  3k
6
3k
3k  1

3 
3
1
3k
 3 - 2, 1 
3

2

k
3
1
k  -2
k  3
[
]
-5 -4 -3 -2 -1 0 1 2 3 4 5
A rental company charges $15 to
rent a chain saw, plus $2 per hour.
A rental company charges $15 to
rent a chain saw, plus $2 per hour.
Al Ghandi can spend no more than
$35 to clear some logs from his
yard.
A rental company charges $15 to
rent a chain saw, plus $2 per hour.
Al Ghandi can spend no more than
$35 to clear some logs from his
yard. What is the maximum amount
of time he can use the rented saw?
A rental company charges $15 to
rent a chain saw, plus $2 per hour.
Al Ghandi can spend no more than
$35 to clear some logs from his
yard. What is the maximum amount
of time he can use the rented saw?
$15  $2  x  $35
15  2x  35
15  2x  35
 15
 15
15  2x  35
 15
 15
2x
15  2x  35
 15
 15
2x 
15  2x  35
 15
 15
2x  20
15  2x  35
 15
 15
2x  20
2x
20
 2
2
15  2x  35
 15
 15
2x  20
2x
20
 2
2
x
15  2x  35
 15
 15
2x  20
2x
20
 2
2
x 
15  2x  35
 15
 15
2x  20
2x
20
 2
2
x  10
Martha has scores of 88, 86, and 90
on here three algebra scores.
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class.
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class. What possible
scores on her fourth test will earn
her an A average.
x
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class. What possible
scores on her fourth test will earn
her an A average.
x  88
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class. What possible
scores on her fourth test will earn
her an A average.
x  88  86
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class. What possible
scores on her fourth test will earn
her an A average.
x  88  86  90
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class. What possible
scores on her fourth test will earn
her an A average.
x  88  86  90
4
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class. What possible
scores on her fourth test will earn
her an A average.
x  88  86  90

4
Martha has scores of 88, 86, and 90
on here three algebra scores. An
average score of at least 90 will earn
an A in the class. What possible
scores on her fourth test will earn
her an A average.
x  88  86  90
 90
4
x  88  86  90
 90
4
x

88

86

90
4
 90
4
x

88

86

90
4
 90 4 
4
x

88

86

90
4
 90 4 
4
x  88  86  90
x

88

86

90
4
 90 4 
4
x  88  86  90 
x

88

86

90
4
 90 4 
4
x  88  86  90  360
x

88

86

90
4
 90 4 
4
x  88  86  90  360
x  264
 360
x

88

86

90
4
 90 4 
4
x  88  86  90  360
x  264
 264
 360
 264
x

88

86

90
4
 90 4 
4
x  88  86  90  360
x  264
 264
x
 360
 264
x

88

86

90
4
 90 4 
4
x  88  86  90  360
x  264
 264
x
 360
 264

x

88

86

90
4
 90 4 
4
x  88  86  90  360
x  264
 264
x
 360
 264
 96
L 1.5 # 1- 66