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# Compound Inequalities

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### Compound Inequalities

1. 1. Compound Inequalities
2. 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. 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. 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. 0 5 10 15 -20 -15 -10 -5 -25 20 25
5. 5. The lower limit is -7. We also need to mark it with a filled-in circle. 0 5 10 15 -20 -15 -10 -5 -25 20 25
6. 6. Where are they found on the graph? Yes! It is less than or equal to 9? What about –15? There are other numbers that satisfy both conditions. 0 5 10 15 -20 -15 -10 -5 -25 20 25
7. 7. It is also greater than or equal to -7? What about –15? No! Where are they found on the graph? 0 5 10 15 -20 -15 -10 -5 -25 20 25
8. 8. Because the word and is used, a number on the graph needs to satisfy both parts of the inequality. 0 5 10 15 -20 -15 -10 -5 -25 20 25
9. 9. Yes! So let’s try 20. Does 20 satisfy both conditions? 0 5 10 15 -20 -15 -10 -5 -25 20 25
10. 10. No! So let’s try 20. Does 20 satisfy both conditions? 0 5 10 15 -20 -15 -10 -5 -25 20 25
11. 11. 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
12. 12. There is one region we have not checked. 0 5 10 15 -20 -15 -10 -5 -25 20 25
13. 13. We need to choose a number from that region. Let’s go with 0? 0 is usually the easiest number to work with. 0 5 10 15 -20 -15 -10 -5 -25 20 25
14. 14. Does 0 satisfy both conditions? Yes! 0 5 10 15 -20 -15 -10 -5 -25 20 25
15. 15. Does 0 satisfy both conditions? Yes! 0 5 10 15 -20 -15 -10 -5 -25 20 25
16. 16. If one number in a region completely satisfies an inequality, you can say that every number in that region satisfies the inequality. 0 5 10 15 -20 -15 -10 -5 -25 20 25
17. 17. Let’s graph another inequality: 0 5 10 15 -20 -15 -10 -5 -25 20 25
18. 18. tells us we want an open circle, The first sign First we mark the boundary points: 0 5 10 15 -20 -15 -10 -5 -25 20 25
19. 19. and the 12 tells us where the circle goes. 0 5 10 15 -20 -15 -10 -5 -25 20 25
20. 20. and the 12 tells us where the circle goes. 0 5 10 15 -20 -15 -10 -5 -25 20 25
21. 21. tells us we want a closed circle, The second sign 0 5 10 15 -20 -15 -10 -5 -25 20 25
22. 22. and the -1 tells us where the circle goes. 0 5 10 15 -20 -15 -10 -5 -25 20 25
23. 23. The boundary points divide the line into three regions: 1 2 3 0 5 10 15 -20 -15 -10 -5 -25 20 25
24. 24. We need to test one point from each region. No! Yes! 0 5 10 15 -20 -15 -10 -5 -25 20 25
25. 25. Notice that the word used is or , No! Yes! instead of and . 0 5 10 15 -20 -15 -10 -5 -25 20 25
26. 26. Or means that a number No! Yes! only needs to meet one condition. 0 5 10 15 -20 -15 -10 -5 -25 20 25
27. 27. Because –10 meets one condition, Yes! the region to which it belongs . . . . . . belongs to the graph. 0 5 10 15 -20 -15 -10 -5 -25 20 25
28. 28. Let’s check the next region: No! No! 0 5 10 15 -20 -15 -10 -5 -25 20 25
29. 29. Because –1 meets neither condition, the numbers in that region will not satisfy the inequality. 0 5 10 15 -20 -15 -10 -5 -25 20 25 No! No!
30. 30. Now the final region: Yes! No! 0 5 10 15 -20 -15 -10 -5 -25 20 25
31. 31. Again, 15 meets one condition Yes! so we need to shade that region. 0 5 10 15 -20 -15 -10 -5 -25 20 25
32. 32. A quick review: 1. Find and mark the boundary points. 2. Test points from each region. 3. Shade the regions that satisfy the inequality. ? ? ? To graph a compound inequality: 0 5 10 15 -20 -15 -10 -5 -25 20 25
33. 33. 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 0 5 10 15 -20 -15 -10 -5 -25 20 25
34. 34. Given the graph below, write the inequality. First, write the boundary points. 0 5 10 15 -20 -15 -10 -5 -25 20 25
35. 35. Then look at the marks on the graph, and write the correct symbol. 0 5 10 15 -20 -15 -10 -5 -25 20 25
36. 36. Since x is between the boundary points on the graph, it will be between the boundary points in the inequality. 0 5 10 15 -20 -15 -10 -5 -25 20 25
37. 37. Since x is between the boundary points on the graph, it will be between the boundary points in the inequality. 0 5 10 15 -20 -15 -10 -5 -25 20 25
38. 38. Try this one: Again, begin by writing the boundary points: 0 5 10 15 -20 -15 -10 -5 -25 20 25
39. 39. And again, you need to choose the correct symbols: 0 5 10 15 -20 -15 -10 -5 -25 20 25
40. 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. 0 5 10 15 -20 -15 -10 -5 -25 20 25
41. 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. 0 5 10 15 -20 -15 -10 -5 -25 20 25
42. 42. We will use the word, or , instead: Remember that or means a number has to satisfy only one of the conditions. 0 5 10 15 -20 -15 -10 -5 -25 20 25
43. 43. We will use the word, or , instead: Remember that or means a number has to satisfy only one of the conditions. 0 5 10 15 -20 -15 -10 -5 -25 20 25
44. 44. Is there any one number that belongs to both shaded sections in the graph? Say NO! NO! 0 5 10 15 -20 -15 -10 -5 -25 20 25
45. 45. So it would be incorrect to use and . And implies that a number meets both conditions. 0 5 10 15 -20 -15 -10 -5 -25 20 25
46. 46. . . . you remember that a compound inequality is just two inequalities put together. Solving compound inequalities is easy if . . .
47. 47. You can solve them both at the same time:
48. 48. Write the inequality from the graph: 1: Write boundaries: 2: Write signs: 3: Write variable: 0 5 10 15 -20 -15 -10 -5 -25 20 25
49. 49. Solve the inequality:
50. 50. You did remember to reverse the signs . . . . . . didn’t you? Good job!