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Compound Inequalities.

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Presentation on theme: "Compound Inequalities."— Presentation transcript:

1 Compound Inequalities

2 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.

3 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.

4 Let’s look at the graph:
The upper limit is 9. Because x can be equal to 9, we mark it with a filled-in circle. 5 10 15 -20 -15 -10 -5 -25 20 25

5 The lower limit is -7. We also need to mark it with a filled-in circle.
5 10 15 -20 -15 -10 -5 -25 20 25

6 There are other numbers that satisfy both conditions.
Where are they found on the graph? What about –15? It is less than or equal to 9? Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

7 Where are they found on the graph?
What about –15? It is also greater than or equal to -7? No! 5 10 15 -20 -15 -10 -5 -25 20 25

8 Because the word and is used, a number on the graph needs to satisfy both parts of the inequality.
5 10 15 -20 -15 -10 -5 -25 20 25

9 So let’s try 20. Does 20 satisfy both conditions?
Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

10 So let’s try 20. Does 20 satisfy both conditions?
No! 5 10 15 -20 -15 -10 -5 -25 20 25

11 Since 20 does not satisfy both conditions, it can’t belong to the solution set.
5 10 15 -20 -15 -10 -5 -25 20 25

12 There is one region we have not checked.
5 10 15 -20 -15 -10 -5 -25 20 25

13 We need to choose a number from that region.
You want to choose 0? Good choice! 0 is usually the easiest number to work with. 5 10 15 -20 -15 -10 -5 -25 20 25

14 Does 0 satisfy both conditions?
Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

15 Does 0 satisfy both conditions?
Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

16 If one number in a region completely satisfies an inequality,
you can know that every number in that region satisfies the inequality. 5 10 15 -20 -15 -10 -5 -25 20 25

17 Let’s graph another inequality:
5 10 15 -20 -15 -10 -5 -25 20 25

18 First we mark the boundary points: The first sign
tells us we want an open circle, 5 10 15 -20 -15 -10 -5 -25 20 25

19 and the 12 tells us where the circle goes.
5 10 15 -20 -15 -10 -5 -25 20 25

20 and the 12 tells us where the circle goes.
5 10 15 -20 -15 -10 -5 -25 20 25

21 tells us we want a closed circle,
The second sign tells us we want a closed circle, 5 10 15 -20 -15 -10 -5 -25 20 25

22 and the -1 tells us where the circle goes.
5 10 15 -20 -15 -10 -5 -25 20 25

23 The boundary points divide the line into three regions:
1 2 3 5 10 15 -20 -15 -10 -5 -25 20 25

24 We need to test one point from each region.
No! Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

25 Notice that the word used is or,
instead of and. No! Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

26 only needs to meet one condition.
Or means that a number only needs to meet one condition. No! Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

27 Because –10 meets one condition, the region to which it belongs . . .
belongs to the graph. Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

28 Let’s check the next region:
No! No! 5 10 15 -20 -15 -10 -5 -25 20 25

29 Because –1 meets neither condition, the numbers in that region
will not satisfy the inequality. No! 5 10 15 -20 -15 -10 -5 -25 20 25

30 Now the final region: Yes! No! 5 10 15 -20 -15 -10 -5 -25 20 25

31 Again, 15 meets one condition so we need to shade that region.
Yes! 5 10 15 -20 -15 -10 -5 -25 20 25

32 To graph a compound inequality:
A quick review: To graph a compound inequality: 1. Find and mark the boundary points. 2. Test points from each region. 3. Shade the regions that satisfy the inequality. ? ? ? 5 10 15 -20 -15 -10 -5 -25 20 25

33 1. Find and mark the boundary points.
A quick review: 1. Find and mark the boundary points. 2. Test points from each region. 3. Shade the regions that satisfy the inequality. or 5 10 15 -20 -15 -10 -5 -25 20 25

34 Given the graph below, write the inequality.
First, write the boundary points. 5 10 15 -20 -15 -10 -5 -25 20 25

35 Then look at the marks on the graph,
and write the correct symbol. 5 10 15 -20 -15 -10 -5 -25 20 25

36 Since x is between the boundary points on the graph,
it will be between the boundary points in the inequality. 5 10 15 -20 -15 -10 -5 -25 20 25

37 Since x is between the boundary points on the graph,
it will be between the boundary points in the inequality. 5 10 15 -20 -15 -10 -5 -25 20 25

38 Again, begin by writing the boundary points:
Try this one: Again, begin by writing the boundary points: 5 10 15 -20 -15 -10 -5 -25 20 25

39 And again, you need to choose the correct symbols:
5 10 15 -20 -15 -10 -5 -25 20 25

40 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. 5 10 15 -20 -15 -10 -5 -25 20 25

41 Because the x-values are not between the boundary points on the graph,
we won’t write them between the boundary points in the equation. 5 10 15 -20 -15 -10 -5 -25 20 25

42 We will use the word, or, instead:
Remember that or means a number has to satisfy only one of the conditions. 5 10 15 -20 -15 -10 -5 -25 20 25

43 We will use the word, or, instead:
Remember that or means a number has to satisfy only one of the conditions. 5 10 15 -20 -15 -10 -5 -25 20 25

44 Is there any one number that belongs to both shaded sections in the graph?
NO! Say NO! 5 10 15 -20 -15 -10 -5 -25 20 25

45 So it would be incorrect to use and
So it would be incorrect to use and. And implies that a number meets both conditions. 5 10 15 -20 -15 -10 -5 -25 20 25

46 Solving compound inequalities is easy if . . .
. . . you remember that a compound inequality is just two inequalities put together.

47 You can solve them both at the same time:

48 Write the inequality from the graph:
5 10 15 -20 -15 -10 -5 -25 20 25 3: Write variable: 1: Write boundaries: 2: Write signs:

49 Is this what you did? Solve the inequality:

50 You did remember to reverse the signs . . .
Good job! . . . didn’t you?


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