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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.

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  1. anonymous
    • one year ago
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    |dw:1438573641556:dw|

  2. jim_thompson5910
    • one year ago
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    Rule: \[\Large |x| > k\] breaks down into \[\Large x > k \ \ \text{or} \ \ x < -k\] where k is some positive number

  3. jim_thompson5910
    • one year ago
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    so in your case, |x-2| > 13 turns into x-2 > 13 or x-2 < -13

  4. anonymous
    • one year ago
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    I don't get it. What would my answer be?

  5. anonymous
    • one year ago
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    would it be 15??

  6. jim_thompson5910
    • one year ago
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    I'm assuming you added 2 to both sides of x-2 > 13, right?

  7. anonymous
    • one year ago
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    yes

  8. jim_thompson5910
    • one year ago
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    |dw:1438574103833:dw|

  9. jim_thompson5910
    • one year ago
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    you have the right idea |dw:1438574120370:dw|

  10. jim_thompson5910
    • one year ago
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    |dw:1438574134671:dw|

  11. jim_thompson5910
    • one year ago
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    so `x-2 > 13` turns into `x > 15`

  12. jim_thompson5910
    • one year ago
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    what does x-2 < -13 turn into when you solve for x ?

  13. anonymous
    • one year ago
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    -11?

  14. jim_thompson5910
    • one year ago
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    not just -11, but something else is it x = -11 ? x < -11 ? x > -11 ?

  15. anonymous
    • one year ago
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    x<-11

  16. jim_thompson5910
    • one year ago
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    yes, `x-2 < -13` turns into `x < -11`

  17. jim_thompson5910
    • one year ago
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    |x-2| > 13 turns into x-2 > 13 or x-2 < -13 and then that becomes x > 15 or x < -11

  18. anonymous
    • one year ago
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    so the answer is x is greater than 15 but less than -11??

  19. anonymous
    • one year ago
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    how would I write that?

  20. jim_thompson5910
    • one year ago
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    in interval notation you mean?

  21. anonymous
    • one year ago
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    yes

  22. jim_thompson5910
    • one year ago
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    let's just focus on x > 15 for now

  23. jim_thompson5910
    • one year ago
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    any ideas of what the interval notation of x > 15 is ?

  24. anonymous
    • one year ago
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    i really don't know.

  25. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    |dw:1438574679944:dw|

  27. jim_thompson5910
    • one year ago
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    |dw:1438574694799:dw|

  28. anonymous
    • one year ago
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    okay i get that part what about the -11?

  29. jim_thompson5910
    • one year ago
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    you tell me, what does x < -11 translate to?

  30. anonymous
    • one year ago
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    |dw:1438574720200:dw|

  31. jim_thompson5910
    • one year ago
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    close

  32. anonymous
    • one year ago
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    is that it?

  33. jim_thompson5910
    • one year ago
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    you need a negative before the infinity

  34. anonymous
    • one year ago
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    |dw:1438574761239:dw|

  35. jim_thompson5910
    • one year ago
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    \[\Large x > 15 \rightarrow (15, \infty)\] \[\Large x < -11 \rightarrow (-\infty, 11)\]

  36. jim_thompson5910
    • one year ago
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    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
    • one year ago
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    oops I meant to write -11 instead of 11

  38. jim_thompson5910
    • one year ago
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    so it should be \[\Large (-\infty, -11) \cup (15, \infty)\]

  39. anonymous
    • one year ago
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    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
    • one year ago
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    with more practice, the rest should come easier too

  41. anonymous
    • one year ago
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    can I message you for more help??

  42. anonymous
    • one year ago
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    the step by step process really helped!

  43. jim_thompson5910
    • one year ago
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    I'm glad it helped and yeah you can ask a few more questions

  44. anonymous
    • one year ago
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    |dw:1438575156359:dw|

  45. anonymous
    • one year ago
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    to get rid of the 3, do I multiply it by 8?

  46. jim_thompson5910
    • one year ago
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    first you need to use the rule |x| < k becomes -k < x < k

  47. jim_thompson5910
    • one year ago
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    where again, k is some positive number

  48. jim_thompson5910
    • one year ago
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    |dw:1438575356447:dw|

  49. jim_thompson5910
    • one year ago
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    notice |dw:1438575401382:dw|

  50. jim_thompson5910
    • one year ago
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    what would be the next step?

  51. anonymous
    • one year ago
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    8*3??

  52. jim_thompson5910
    • one year ago
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    no

  53. jim_thompson5910
    • one year ago
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    |dw:1438575499181:dw|

  54. anonymous
    • one year ago
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    do the opposite which is multiply

  55. jim_thompson5910
    • one year ago
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    yes, multiply ALL sides by 3 |dw:1438575563784:dw|

  56. jim_thompson5910
    • one year ago
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    |dw:1438575578883:dw|

  57. jim_thompson5910
    • one year ago
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    we are then left with this \[\Large -36 \le x+8 \le 36\]

  58. jim_thompson5910
    • one year ago
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    then what?

  59. anonymous
    • one year ago
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    okay that helped. and next we have to get x by itself

  60. jim_thompson5910
    • one year ago
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    yes so you have to somehow undo that +8

  61. jim_thompson5910
    • one year ago
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    btw, we are following PEMDAS in reverse when it comes to isolating variables

  62. anonymous
    • one year ago
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    so subtract 8 from each side?|dw:1438575726830:dw|

  63. jim_thompson5910
    • one year ago
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    good

  64. anonymous
    • one year ago
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    like that?

  65. jim_thompson5910
    • one year ago
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    correct, which ends you up with what?

  66. anonymous
    • one year ago
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    |dw:1438575822811:dw|

  67. jim_thompson5910
    • one year ago
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    you nailed it

  68. jim_thompson5910
    • one year ago
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    x is finally isolated so you're done

  69. anonymous
    • one year ago
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    |dw:1438576160100:dw|

  70. anonymous
    • one year ago
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    I tried putting in my answer all types of ways, but it still says it's incorrect.

  71. jim_thompson5910
    • one year ago
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    |dw:1438576344123:dw|

  72. anonymous
    • one year ago
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    what do you mean?

  73. jim_thompson5910
    • one year ago
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    |dw:1438576379769:dw|

  74. jim_thompson5910
    • one year ago
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    we include both endpoints so the interval notation would be [-44, 28]

  75. jim_thompson5910
    • one year ago
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    the square brackets say "include endpoints

  76. jim_thompson5910
    • one year ago
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    |dw:1438576439943:dw|