### Section 2.6

```2.6
Solving Linear Inequalities
1. Represent solutions to inequalities
graphically and using set notation.
2. Solve linear inequalities.
Inequalities
<
is less than
>
is greater than

is less than or equal to

is greater than or equal to
Inequality always points to the smaller number.
True or False?
44
4>4
True
False
Represent inequalities:
x > 4 is the same as {5, 6, 7…}
False
Graphically
Interval Notation
Set-builder Notation
Graphing Inequalities
Parentheses/bracket method :
If the variable is on the left,
the arrow points the same
direction as the inequality.
Parentheses: endpoint is not included <, >
Bracket: endpoint is included ≤, ≥
x<2
x≥2
Open Circle/closed circle method:
Open Circle: endpoint is not included <, >
Closed Circle: endpoint is included ≤, ≥
x<2
x≥2
Inequalities – Interval Notation
[( smallest, largest )]
Parentheses: endpoint is not included <, >
Bracket: endpoint is included ≤, ≥
Infinity: always uses a parenthesis
x<2
( –∞, 2)
x≥2
[2, ∞)
4<x<9
3-part inequality
(4, 9)
Inequalities – Set-builder Notation
{variable | condition }
pipe
{ x | x  5}
The set of all x such that x is greater than or equal to 5.
x<2
( –∞, 2)
{x|x<2}
[2, ∞)
{ x | x ≥ 2}
(4, 9)
{ x | 4 < x < 9}
x≥2
4<x<9
Inequalities
Graph, then write interval notation and set-builder notation.
x≥5
[
Interval Notation: [ 5, ∞)
Set-builder Notation: { x | x ≥ 5}
x < –3
)
Interval Notation: (– ∞, –3)
Set-builder Notation: { x | x < –3 }
Inequalities
Graph, then write interval notation and set-builder notation.
1<a<6
(
)
Interval Notation: ( 1, 6 )
Set-builder Notation: { a | 1 < a < 6 }
–7 < x ≤ 3
(
]
Interval Notation: (– 7, –3]
Set-builder Notation: { x | –7 < x ≤ 3 }
Inequalities
4<5
4<5
4+1<5+1
4–1<5–1
5<6
3<4
True
True
If a < b, then a + c < b + c
for all real numbers a, b, and c.
Also true for >, , or .
Inequalities
4<5
4 (2) < 5 (2)
8 < 10
True
4<5
4 (–2) < 5 (–2)
–8 < –10
–8 > –10
False
If we multiply (or divide) by a negative, reverse
the direction of the inequality!!!!!
The Multiplication Principle of Inequality
If a < b, then ac < bc if c is a positive real number.
If a < b, then ac > bc if c is a negative real number.
The principle also holds true for >, , and .
Solving Inequalities
If we multiply (or divide) by a negative, reverse
the direction of the inequality!!!!!
4 x  16
4
4
x  4
 4 x  16
 4
 4
x  4
4 x   16
4
4
x  4
Solving Inequalities
Solve then graph the solution and write it in interval notation
and set-builder notation.
3x  4  7
4 4
Don’t write = !
3x  3
3
(
3
x  1
Interval Notation: ( 1, ∞ )
Set-builder Notation: { x | x > 1 }
Solving Inequalities
Solve then graph the solution and write it in interval notation
and set-builder notation.
4  9 k   4 k  19
 4k
 4k
4  5 k  19
4
4
 5 k  15
 5
 5
]
k  3
Interval Notation: (– ∞, –3 ]
Set-builder Notation: { k | k ≤ –3 }
Solving Inequalities
Solve then graph the solution and write it in interval notation
and set-builder notation.

5
3
p   10
 5 
3
p   3   10 
 3

 5 p   30
 5
 5
)
p  6
Interval Notation: (– ∞, 6 )
Set-builder Notation: { p | p < 6 }
Solving Inequalities
Solve then graph the solution and write it in interval notation
Moving variable to the right.
and set-builder notation.
1
5
6 m  7  
1
2
12 m  14  15 m  5
3 m  1 
 12 m
1
1
10   6 m  7   10   3 m  1 
5
2
2 6 m  7   5 3 m  1 
12 m  14  15 m  5
 15 m
 14  3 m  5
 5
 5
 9  3m
3
3
3 m
 15 m
 3 m  14   5
 14
 3m  9
 3
 12 m
 3
m  3
[
 14
Interval Notation: [– 3, ∞ )
Set-builder Notation: { m | m ≥ – 3 }
```