Chapter 6

Solving Linear Inequalities

6.1 Solving Inequalities by Addition and Subtraction

• Set builder Notation:– A way to write a solution set

• Ex: if the solution set is all #s less

than or equal to 58

58t

}58|{ tt

< >Less than

Fewer than

Greater than

More than

At most

No more than

Less than or equal to

At least

No less than

Greater than or equal to

•If the variable is on the open side = shade right

•If the variable is on the closed side = shade left

Solve by Addition

Solve by Subtraction

ex: t – 45 < 13 ex: 8 – 2y -1y

ex: s + 19 > 56 ex: 5p + 7 6p

+45 +45 t < 58

{ t | t < 58}

60585654525048111098765

3938373635343310 115 6 7 8 9

+ 2y + 2y 8 y

{ y | y 8}

-19 -19 s > 37

{s | s > 37}

-5p -5p 7 p

{ p | p 7}

6.2 Solving Inequalities by Multiplication and Division

• If you multiply or divide by a negative number, you must change the inequality sign

145

2s

Solve by Multilplication

Ex:

2

5

1

14

5

2

2

5 s

2

70s

s < 35

39383736353433 111098765

Ex:

103

4 p

4

3

1

10

3

4

4

3 p

4

30p

p 7.5

Solve by Division

{ s | s < 35} { p | p 7.5}

12x 60/12 /12

x 5

{ x | x 5}

111098765

-8q < 136/-8 /-8

q > -17

{ q | q > -17}

-13-14-15-16-17-18-19

6.3 Solving Multi-Step InequalitiesMulti-step Inequality Distributive Property

Write and Solve an Inequality Empty Sets and All Real Numbers

104325

9c

9

5

1

72

5

9

9

5 c

-32 -32

725

9c

c > 40

{c | c > 40}

44434241403938

3d – 2(8d – 9) > -2d - 43d – 16d + 18 > -2d - 4

-13d + 18 > -2d - 4+13d +13d

18 > 11d - 4+4 + 4

22 > 11d/11 /11

2 > d

{d | d < 2}

543210-1

Three times a number minus eighteen is at least five times the number plus twenty-one

215183 xx-3x -3x

21218 x -21 -21

x239/2 /2

x 5.19

-18-19-20-21-22-23-24

{ x | x -19.5}

8(t +2)- 3(t – 4) < 5(t -7) + 8

8t +16 - 3t + 12 < 5t - 35 + 8

5t + 28 < 5t - 27-5t - 5t

28 < -27 Empty set

•If it is true – all real numbers

•If it is false – it is an empty set

{ x | x is a real number}

6.4 Solving Compound Inequalities

• Intersection = and• Union = or

3xEx: Graph a Compound Inequality

2x

543210-1-2-3

543210-1-2-3

543210-1-2-3

{x | -2 x < 3}

Ex: Solve and Graph a Compound Inequality -5 < x – 4 < 2

-5 < x - 4 x – 4 < 2+ 4 + 4

-1 < x

+ 4 + 4

x < 6

543210-1-2-3

543210-1-2-3

6

543210-1-2-3 6

{x | -1 < x < 6 }

6.5 Absolute Value Open Sentences

• Absolute Value: The distance from zero on a number line – The positive value of the number

Ex:

Make a frayer-Foldable (diamond in the center)

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7

|-5| =

|5| =

|-6| =

|2| =

5

5

6

2

• 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 b. x + 7 = -4

-7 -7 -7 -7x = -3 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:

543210-1-2-3 6

| x - # | = #

# half way between

# spaces from each point

3 3

| x – 2 | = 3

Ex: Ex:

-13-14-15-16-17-18-19

2 2

| x + 16| = 2

• 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 0

Graph of f(x) = |x|

x |x|

-1 1

1 1

0 0

-2 2

2 2

6.6 Solving Absolute Value Inequalities

• A. the expression in absolute value bars is positive

• B. the expression in absolute value bars is negative (also flip the inequality sign)

On the back of the 6.5 frayer

|x| = n x = -n or x = n

|x| < n x < n and x > -n

|x| > n x > n or x < -n

• Ex1: | t + 5 | < 9

t + 5 < 9 t + 5 > -9 -5 -5 -5 -5

t < 4 t > -14

64 20-2-4-6-8-10 8-12-14-16

-14 < t < 4

• Ex2: |2x + 8| 6

2x + 8 6 2x + 8 -6 -8 -8 -8 -8

2x -2 2x -14/2 /2 /2 /2

x -1 x -7

-8 -7 -6 -5 -4 -3 10-9 -1-2

x -7 or x -1

• Ex3: | 2y – 1 | -4

• Ex4: (your turn) | 2k + 1| > 7

the absolute value cannot be less than zero so y is all numbers *If it is a negative

number- the answer is empty set

6.7 Graphing Inequalities with Two Variables

• The equation makes the line to define the boundary• The shaded region is the half-plane

1. Get the equation into slope-intercept form2. List the intercept as an ordered-pair and the slope as a

ratio3. Graph the intercept and use the slope to find at least 2

more points4. Draw the line (dotted or solid)5. Test an ordered-pair not on the line

1. If it is true shade that side of the line2. If it is false shade the other side of the line

Make a frayer-Foldable (diamond in the center)

< or > or

Dotted Line Solid Line

Ex1: y 2x - 3

m =

b = -3 = (0, -3)1

2

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 =

b = 4 = (0, 4)1

2

Test: (0, 0)

0 < 2(0) + 4

0 < 0 + 4

0 < 4 true (shade this side)

Use a dotted line because it is <

• Ex3: 3y - 2 > -x + 7

3

1

3y – 2 > -x + 7

+2 +2

3y > -x + 9

/3 /3 /3

y > - x + 3

m = -

b = 3 = (0, 3)3

1

Test: (0, 0)

0 > - (0) + 3

0 > 0 + 3

0 > 3 false (shade other side)

3

1

Use a dotted line because it is >