x 12

9x

x 12

9x

x 12

9x

x 129x

912 x

9x

You already know inequalities.

Often they are used to place limits on variables.

That just means x can be any number equal to 9 or less than

9.

Sometimes we put more than one limit on the variable:

Now x is still less than or equal to 9, but it must also be greater than or

equal to –7.

7 and 9 xx

7 and 9 xx

Let’s look at the graph:

0 5 10 15-20 -15 -10 -5-25 20 25

The upper limit is 9. Because x can be equal to 9, we mark it with a filled-in

circle.

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The lower limit is -7. We also need to mark it with a filled-in

circle.

7 and 9 xx

Where are they found on the graph?

Yes!

It is less than or equal to 9?What about –15?

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There are other numbers that satisfy both conditions.

7 and 9 xx

15

It is also greater than or equal to -7?

What about –15?

0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

15 No!

Where are they found on the graph?

Because the word and is used, a number on the graph

needs to satisfy both parts of the inequality.

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7 and 9 xx

15

Yes!720

So let’s try 20. Does 20 satisfy both conditions?

0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

20

?

15

No!920

So let’s try 20. Does 20 satisfy both conditions?

0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

20

?

15

Since 20 does not satisfy both conditions, it can’t

belong to the solution set.

0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

2015

There is one region we have not checked.

0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

2015

We need to choose a number from that region.

0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

2015

You want to choose 0?Good choice! 0 is usually the easiest number to work

with.0

Does 0 satisfy both conditions?

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7 and 9 xx

2015 0

70 Yes!?

Does 0 satisfy both conditions?

0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

2015 0

90 Yes!?

If one number in a region completely satisfies an

inequality,

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7 and 9 xx

2015 0

you can know that every number in that region

satisfies the inequality.

And

Shade between the two numbers.

Make sure you think if it makes sense because no solutions can

exist.0 5 10 15-20 -15 -10 -5-25 20 25

7 and 9 xx

Let’s graph another inequality:

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1or 12 xx

tells us we want an open circle,The first sign

First we mark the boundary points:

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1or 12 xx

and the 12 tells us where the circle goes.

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1or 12 xx

and the 12 tells us where the circle goes.

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1or 12 xx

tells us we want a closed circle,The second sign

0 5 10 15-20 -15 -10 -5-25 20 25

1or 12 xx

and the -1 tells us where the circle goes.

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1or 12 xx

The boundary points divide the line into three regions:

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1or 12 xx

1 2 3

We need to test one point from each region.

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1or 12 xx

10 1210

No!?-110

Yes!?

Notice that the word used is or,

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1or 12 xx

10 1210

No! 1210

Yes!

instead of and.

Or means that a number

0 5 10 15-20 -15 -10 -5-25 20 25

1or 12 xx

10 1210

No! 1210

Yes!

only needs to meet one condition.

Because –10 meets one condition,

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1or 12 xx

1210 1210 Yes!

10

the region to which it belongs . . .. . . belongs to the graph.

Let’s check the next region:

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1or 12 xx

125 15 5

No!? No!?

Because –1 meets neither condition,

0 5 10 15-20 -15 -10 -5-25 20 25

1or 12 xx

125 15 5

No! No!

the numbers in that regionwill not satisfy the inequality.

Now the final region:

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1or 12 xx

1215 115 15

Yes!? No!?

Again, 15 meets one condition

0 5 10 15-20 -15 -10 -5-25 20 25

1or 12 xx

1215 115 15

Yes!

so we need to shade that region.

Or shade outside the boundaries

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1or 12 xx

Make sure you think because no solutions do exist

A quick review:

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1. Find and mark the boundary points.2. Test points from each region.3. Shade the regions that satisfy the inequality.or and or

To graph a compound inequality:

Given the graph below, write the inequality.

First, write the boundary points.

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Since x is between the boundary points on the

graph,

15 20

0 5 10 15-20 -15 -10 -5-25 20 25

it will be between the boundary points in the inequality.

xx

20 15 x

Since x is between the boundary points on the

graph,

x0 5 10 15-20 -15 -10 -5-25 20 25

it will be between the boundary points in the inequality.

0 5 10 15-20 -15 -10 -5-25 20 25

Try this one:

0 5 10 15-20 -15 -10 -5-25 20 25

And again, you need to choose the correct symbols:

10 5

0 5 10 15-20 -15 -10 -5-25 20 25

Because the x-values are not between the boundary points

on the graph,we won’t write x between the

boundary points in the equation.

x x 5 10

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x xx105x

We will use the word, or, instead:Remember that or means

a number has to satisfy only one of the conditions.

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Is there any one number that belongs to both shaded

sections in the graph?

xx 10or 5

Say NO!NO!

0 5 10 15-20 -15 -10 -5-25 20 25

xx 10or 5

So it would be incorrect to use and. And implies that a

number meets both conditions.

. . . you remember that a compound inequality is just two

inequalities put together.

9325 x

Solving compound inequalities is easy if . . .

932 x325 x 9325 x

9325 x

You can solve them both at the same time:

3 3628 x

2 2

3

234 x

Solve the inequality:

57415 xIs this what you did?

771248 x

4

7

4 42 x 3

You did remember to reverse the signs . . .

57415 x771248 x

4

7

4 42 x 3

. . . didn’t you?Good job!

HW: pg. 585 #2 to 40 even