can you help

Week Two

Discussion

One

-

Variable Compound Inequalities

compound inequality

and

or

intersection

union

This is my “and”

compound

inequality

:

-

7

≤

5

+

3x ≤ 20

What that means is the inequality must fulfill two conditions

at the same time. It means 5

+ 3x

must be equal to or less than 2

0

and

also at the same time greater than or equal to

-

7. I think of these as “between” inequalities because it turns out that the solution set for x

will be between two numbers. Now I will

find out what those two numbers are.

-

7

≤

5

+

3

x ≤ 20

Subtract 5 from all three parts of the inequality.

-

7

–

5 ≤ 5

–

5 + 3x ≤

20

–

5

-

12 ≤ 3x ≤ 15

Divide all three parts by 3

-

12

≤

3x

≤

15

3 3 3

-

4 ≤ x ≤ 5

So any value of x greater than or equal to

-

4 and less than

or e

qual to 5 will make this inequality true.

This

-

4 ≤ x ≤ 5

is how this

compound

inequality

is written algebraically.

As an

intersection

of sets it would look like [

-

4,

)

(

-

, 5] which equals [

-

4, 5] in

interval notation.

<

----------------

[

-----------

|

---------------

]

--------------

>

Here is a number line graph of the

-

4

0

5

solution set

.

The square brackets mean that the end points are included in the solution set; notice the

green highlighting extends through the square brackets as well.

T

his is my “

o

r”

compound

inequality

: 4

–

x ≥ 1

or

6x

–

3

> 27

What this means is that there are two conditions and one of them must be true with any

given x from the solution set but both cannot be true at the same time. Since the solution

will turn out to be two disjoint intervals, I am go

ing to solve each part of the inequality

separately.

4

–

x ≥ 1

Subtract 4 from both sides.

4

–

4

–

x ≥ 1

–

4

–

x ≥

–

3

We must pay close attention to that negative in front of x. To

remove it I must divide both sides of the inequality by

-

1 which

also means I must flip the

inequality symbol over so it points the other direction.

–

x

≤

–

3

Symbol is flipped.

-

1

-

1

3

This is one part of my

“

or

”

compound

inequality

.

6x

–

3

> 27

Add 3 to both sides.

6x

–

3

+ 3 > 27

+ 3

6x > 30

Divide bo

th sides by 6

, but it is positive, so no flipping involved.

6x

>

30

6

6

x > 5

This is the other part of my

“

or

”

compound inequality

.

The complete solution set written algebraically is

x ≤ 3 or x > 5

The solution set written in interval notation

is the

union

of two intervals

(

-

, 3]

(5,

)

Here is a number line graph of the solution set:

<

--------------

|

-------

]

------

(

------------------------------

>

0 3 5

Notice that the 3 is included in the solution set b

ut 5 is not.