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SchoolFreak

  • 3 years ago

INEQUAILTY !!!!

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  1. SchoolFreak
    • 3 years ago
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    Please explain how to get it

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  2. Mimi_x3
    • 3 years ago
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    well..let's do the first first, alright?

  3. Mimi_x3
    • 3 years ago
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    first one*

  4. SchoolFreak
    • 3 years ago
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    okay please explain how to get answer i have no idea where to start on any of them ")

  5. Mimi_x3
    • 3 years ago
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    \[ x- 3 > -9\] do you know how to do it when its \((x-3) = -9\)

  6. SchoolFreak
    • 3 years ago
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    well i would make the x into -3

  7. Mimi_x3
    • 3 years ago
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    you are solving for \(x\)

  8. SchoolFreak
    • 3 years ago
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    is that what we have to ??

  9. Mimi_x3
    • 3 years ago
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    yes.. why not read this first to gain a better understanding.. http://www.sosmath.com/algebra/inequalities/ineq01/ineq01.html

  10. SchoolFreak
    • 3 years ago
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    okay i read it

  11. Mimi_x3
    • 3 years ago
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    if you have any problems then ask me..

  12. SchoolFreak
    • 3 years ago
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    okay for the first one what do we do??

  13. Mimi_x3
    • 3 years ago
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    \[x-3 >-9\] we are solving for \(x\) why not you try it first?

  14. SchoolFreak
    • 3 years ago
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    okay-3 is the answer because -3 times -3 is 9 and 9 is greater than -9 right

  15. SchoolFreak
    • 3 years ago
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    @Mimi_x3

  16. Mimi_x3
    • 3 years ago
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    what? \(x\) is variabe..not times..

  17. Mimi_x3
    • 3 years ago
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    a variable*

  18. SchoolFreak
    • 3 years ago
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    i thought were suppose to replace x with number??

  19. Mimi_x3
    • 3 years ago
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    looks like you have to solve for \(x\) not substituting..

  20. SchoolFreak
    • 3 years ago
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    ohhh k so we add 3 to both sides right?

  21. saifoo.khan
    • 3 years ago
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    @SchoolFreak , mini isn't already helping you!

  22. saifoo.khan
    • 3 years ago
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    is*

  23. SchoolFreak
    • 3 years ago
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    i know

  24. SchoolFreak
    • 3 years ago
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    @Mimi_x3 so then if we add 3 both sides we have x > -6

  25. Mimi_x3
    • 3 years ago
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    Yes!

  26. SchoolFreak
    • 3 years ago
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    thats the answer?

  27. Mimi_x3
    • 3 years ago
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    yeah lol

  28. Mimi_x3
    • 3 years ago
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    you are basically solving for \(x\) thats all

  29. SchoolFreak
    • 3 years ago
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    oh haha x is greater than -6 , the graph would have open circle on -6 shading to the right is that right to?

  30. Mimi_x3
    • 3 years ago
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    correct

  31. SchoolFreak
    • 3 years ago
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    okay lets try the next one

  32. SchoolFreak
    • 3 years ago
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    –2x + 1 ≤ –11

  33. Mimi_x3
    • 3 years ago
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    why dont you try and tell me what you get

  34. SchoolFreak
    • 3 years ago
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    okay i think the first step is to add 2 to both sides?

  35. Mimi_x3
    • 3 years ago
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    think first; why add to 2? since its +1?

  36. Mimi_x3
    • 3 years ago
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    just isolate the -2x in the first hand so you wont get confused

  37. SchoolFreak
    • 3 years ago
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    okay so then you add 1 to both sides?

  38. SchoolFreak
    • 3 years ago
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    @Mimi_x3 would substract 1 from both sides right?

  39. Mimi_x3
    • 3 years ago
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    yes you subtract because it was a positive in the rhs

  40. SchoolFreak
    • 3 years ago
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    –2x + 1 ≤ –11 Okay so now we have -2x ≤ 12

  41. Mimi_x3
    • 3 years ago
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    not quite.. whats -11 - 1?

  42. SchoolFreak
    • 3 years ago
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    ohhh -12

  43. Mimi_x3
    • 3 years ago
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    yes, now you know what to do next

  44. SchoolFreak
    • 3 years ago
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    -2x ≤ -12 now we divide both sides by -2 and we have x ≥ 6

  45. Mimi_x3
    • 3 years ago
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    yes!

  46. SchoolFreak
    • 3 years ago
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    woo hooo also x is greater than or equal to 6 the graph would have a closed circle on 6 shading to the right

  47. Mimi_x3
    • 3 years ago
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    yes

  48. SchoolFreak
    • 3 years ago
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    okay let me try this next one all by myself

  49. SchoolFreak
    • 3 years ago
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    10 < –3x + 1

  50. Mimi_x3
    • 3 years ago
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    alright, if you need help ask me :)

  51. SchoolFreak
    • 3 years ago
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    kk

  52. SchoolFreak
    • 3 years ago
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    -9 > x @Mimi_x3 is that the answer??

  53. Mimi_x3
    • 3 years ago
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    nope..show me what you did

  54. SchoolFreak
    • 3 years ago
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    10 < -3x + 1 9 < -3x -9 > x

  55. Mimi_x3
    • 3 years ago
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    you made a small mistake whats 9/-3?

  56. SchoolFreak
    • 3 years ago
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    3

  57. Mimi_x3
    • 3 years ago
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    nope 9/ - 3

  58. Mimi_x3
    • 3 years ago
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    like 9 divide by negative 3

  59. SchoolFreak
    • 3 years ago
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    -3 is the answer??

  60. Mimi_x3
    • 3 years ago
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    yes but dont forget the inequality signs

  61. SchoolFreak
    • 3 years ago
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    -3 > x Is that the answer @Mimi_x3

  62. Mimi_x3
    • 3 years ago
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    not quite..the sign is wrong

  63. SchoolFreak
    • 3 years ago
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    No because we divide by a negative so we flip the sign

  64. Mimi_x3
    • 3 years ago
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    10 < -3x + 1 9< -3x -3x > 9

  65. Mimi_x3
    • 3 years ago
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    understand so far?

  66. SchoolFreak
    • 3 years ago
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    why did u turn the equation on the last part

  67. Mimi_x3
    • 3 years ago
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    i dont know how to explain..but for inequalities to have to change the sign when you swap the equation.. you dont need to turn it; i just turned to prevent confusion

  68. Mimi_x3
    • 3 years ago
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    you have to change*

  69. SchoolFreak
    • 3 years ago
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    oh i didnt know that

  70. Mimi_x3
    • 3 years ago
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    lets not turn the equation then lets start from 9< -3x

  71. SchoolFreak
    • 3 years ago
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    okay then then we dived and we -3 > x

  72. Mimi_x3
    • 3 years ago
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    then you still have to change the x for it to be in the front so x> -3

  73. Mimi_x3
    • 3 years ago
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    i mean x<-3

  74. SchoolFreak
    • 3 years ago
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    ohh k

  75. SchoolFreak
    • 3 years ago
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    @Mimi_x3 2(x + 5) > 8x – 8 this is the next one it looks hard lol

  76. Mimi_x3
    • 3 years ago
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    not really..expand it

  77. Mimi_x3
    • 3 years ago
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    the first side first

  78. SchoolFreak
    • 3 years ago
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    if we do the first side we will have 2x + 10 right on the left side??

  79. Mimi_x3
    • 3 years ago
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    yeah. i gotta go now. sorry.

  80. SchoolFreak
    • 3 years ago
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    noo :(

  81. Mimi_x3
    • 3 years ago
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    hopefully someone else can help you..if not i'll be back later :)

  82. SchoolFreak
    • 3 years ago
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    okay

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