## A community for students. Sign up today

Here's the question you clicked on:

## anonymous one year ago I have some math problems involving that you have to solve an inequality in interval notations and I don't know how to do it.

• This Question is Open
1. anonymous

|dw:1438573641556:dw|

2. jim_thompson5910

Rule: $\Large |x| > k$ breaks down into $\Large x > k \ \ \text{or} \ \ x < -k$ where k is some positive number

3. jim_thompson5910

so in your case, |x-2| > 13 turns into x-2 > 13 or x-2 < -13

4. anonymous

I don't get it. What would my answer be?

5. anonymous

would it be 15??

6. jim_thompson5910

I'm assuming you added 2 to both sides of x-2 > 13, right?

7. anonymous

yes

8. jim_thompson5910

|dw:1438574103833:dw|

9. jim_thompson5910

you have the right idea |dw:1438574120370:dw|

10. jim_thompson5910

|dw:1438574134671:dw|

11. jim_thompson5910

so x-2 > 13 turns into x > 15

12. jim_thompson5910

what does x-2 < -13 turn into when you solve for x ?

13. anonymous

-11?

14. jim_thompson5910

not just -11, but something else is it x = -11 ? x < -11 ? x > -11 ?

15. anonymous

x<-11

16. jim_thompson5910

yes, x-2 < -13 turns into x < -11

17. jim_thompson5910

|x-2| > 13 turns into x-2 > 13 or x-2 < -13 and then that becomes x > 15 or x < -11

18. anonymous

so the answer is x is greater than 15 but less than -11??

19. anonymous

how would I write that?

20. jim_thompson5910

in interval notation you mean?

21. anonymous

yes

22. jim_thompson5910

let's just focus on x > 15 for now

23. jim_thompson5910

any ideas of what the interval notation of x > 15 is ?

24. anonymous

i really don't know.

25. jim_thompson5910

x > 15 means x is some number larger than 15 so if we describe an interval, we would write (15, infinity) the parenthesis say "don't include the enpoints"

26. jim_thompson5910

|dw:1438574679944:dw|

27. jim_thompson5910

|dw:1438574694799:dw|

28. anonymous

okay i get that part what about the -11?

29. jim_thompson5910

you tell me, what does x < -11 translate to?

30. anonymous

|dw:1438574720200:dw|

31. jim_thompson5910

close

32. anonymous

is that it?

33. jim_thompson5910

you need a negative before the infinity

34. anonymous

|dw:1438574761239:dw|

35. jim_thompson5910

$\Large x > 15 \rightarrow (15, \infty)$ $\Large x < -11 \rightarrow (-\infty, 11)$

36. jim_thompson5910

The "or" in the solution x > 15 or x < -11 means we will have a union symbol U Which is why this is the final answer $\Large (-\infty, 11) \cup (15, \infty)$

37. jim_thompson5910

oops I meant to write -11 instead of 11

38. jim_thompson5910

so it should be $\Large (-\infty, -11) \cup (15, \infty)$

39. anonymous

thank you very much for all of your help. Im probably going to struggle with the rest of these inequality problems. That seemed like an easy one compared to the others i have

40. jim_thompson5910

with more practice, the rest should come easier too

41. anonymous

can I message you for more help??

42. anonymous

the step by step process really helped!

43. jim_thompson5910

I'm glad it helped and yeah you can ask a few more questions

44. anonymous

|dw:1438575156359:dw|

45. anonymous

to get rid of the 3, do I multiply it by 8?

46. jim_thompson5910

first you need to use the rule |x| < k becomes -k < x < k

47. jim_thompson5910

where again, k is some positive number

48. jim_thompson5910

|dw:1438575356447:dw|

49. jim_thompson5910

notice |dw:1438575401382:dw|

50. jim_thompson5910

what would be the next step?

51. anonymous

8*3??

52. jim_thompson5910

no

53. jim_thompson5910

|dw:1438575499181:dw|

54. anonymous

do the opposite which is multiply

55. jim_thompson5910

yes, multiply ALL sides by 3 |dw:1438575563784:dw|

56. jim_thompson5910

|dw:1438575578883:dw|

57. jim_thompson5910

we are then left with this $\Large -36 \le x+8 \le 36$

58. jim_thompson5910

then what?

59. anonymous

okay that helped. and next we have to get x by itself

60. jim_thompson5910

yes so you have to somehow undo that +8

61. jim_thompson5910

btw, we are following PEMDAS in reverse when it comes to isolating variables

62. anonymous

so subtract 8 from each side?|dw:1438575726830:dw|

63. jim_thompson5910

good

64. anonymous

like that?

65. jim_thompson5910

correct, which ends you up with what?

66. anonymous

|dw:1438575822811:dw|

67. jim_thompson5910

you nailed it

68. jim_thompson5910

x is finally isolated so you're done

69. anonymous

|dw:1438576160100:dw|

70. anonymous

I tried putting in my answer all types of ways, but it still says it's incorrect.

71. jim_thompson5910

|dw:1438576344123:dw|

72. anonymous

what do you mean?

73. jim_thompson5910

|dw:1438576379769:dw|

74. jim_thompson5910

we include both endpoints so the interval notation would be [-44, 28]

75. jim_thompson5910

the square brackets say "include endpoints

76. jim_thompson5910

|dw:1438576439943:dw|