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Firejay5

  • 3 years ago

Simplify Each Expression. Show work and explain it. 32. m over m^2 - 4 + 2 over 3m + 6

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  1. Firejay5
    • 3 years ago
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    Well on #32, I am having trouble multiplying the denominators to make them the same

  2. wio
    • 3 years ago
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    "m over m^2 - 4 + 2 over 3m + 6" \[ \frac{m}{m^2-4}+\frac{2}{3m+6} \]

  3. Firejay5
    • 3 years ago
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    why

  4. wio
    • 3 years ago
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    Just use this: \[ \frac{a}{b}+\frac{c}{d}=\frac{ad+cb}{bd} \]

  5. wio
    • 3 years ago
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    I think the + is separating two fractions.

  6. Firejay5
    • 3 years ago
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    yes

  7. Firejay5
    • 3 years ago
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    yea

  8. Firejay5
    • 3 years ago
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    can you do it with me and then I'll do 33.) my self

  9. Firejay5
    • 3 years ago
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    what do you mean?

  10. Firejay5
    • 3 years ago
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    Equation editor

  11. Firejay5
    • 3 years ago
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    can we get back to the problem

  12. Firejay5
    • 3 years ago
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    or not @Hero

  13. Firejay5
    • 3 years ago
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    the example he gave isn't going nowhere for me

  14. wio
    • 3 years ago
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    let \(b=m^2-4\) and \(d=3m+6\)

  15. wio
    • 3 years ago
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    \[ \frac{m}{m^2-4}+\frac{2}{3m+6}=\frac{(m)(3m+6)+(2)(m^2-4)}{(m^2-4)(3m+6)} \]

  16. wio
    • 3 years ago
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    @Firejay5 Do you see the pattern?

  17. Firejay5
    • 3 years ago
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    Yes? Kind of how did you get that

  18. Firejay5
    • 3 years ago
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    I need full out steps, like getting the denominators equal, etc.

  19. Firejay5
    • 3 years ago
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    @Mertsj could you help with that

  20. mathslover
    • 3 years ago
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    hmn . \[\frac{m}{m^2-4}+\frac{2}{3m+6}\] Well I think that it would be Ok to factorize both denominators

  21. Mertsj
    • 3 years ago
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    Yes. mathslover will be able to show you.

  22. Firejay5
    • 3 years ago
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    @Mertsj you sure

  23. mathslover
    • 3 years ago
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    \(\color{blue}{m^2-4}\) is in the form of \(a^2-b^2\) \(a^2-b^2 = (a+b)(a-b)\) Use the above stated identity to factorize \(m^2-4 \) .

  24. Mertsj
    • 3 years ago
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    yes

  25. mathslover
    • 3 years ago
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    Can you factor \(m^2-4\) @Firejay5 ?

  26. Firejay5
    • 3 years ago
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    yes

  27. mathslover
    • 3 years ago
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    Can you tell me what would be the factorized form for \(x^2-4\) ?

  28. Firejay5
    • 3 years ago
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    (x - 2) & (x + 2)

  29. mathslover
    • 3 years ago
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    Great! So \(x^2-4 = \color{blue}{(x+2)(x-2)}\) Now. we have the complete question as : \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{3m+6}\)

  30. mathslover
    • 3 years ago
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    Now take 3 common from 3m + 6. What you get?

  31. Firejay5
    • 3 years ago
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    I did the other factorization as well

  32. Firejay5
    • 3 years ago
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    3(m + 2)

  33. mathslover
    • 3 years ago
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    Great! \(\cfrac{m}{\color{red}{(m+2)(m-2)}}+\cfrac{2}{\color{orange}{3(m+2)}}\). Is what we have now.

  34. mathslover
    • 3 years ago
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    Now take LCM .

  35. Firejay5
    • 3 years ago
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    m + 2

  36. mathslover
    • 3 years ago
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    Check it again

  37. Firejay5
    • 3 years ago
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    and m - 2

  38. mathslover
    • 3 years ago
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    LCM of \(\cfrac{1}{a+b} + \cfrac{1}{2(a-b)} \) (denominators) is (a+b)(a-b)2

  39. mathslover
    • 3 years ago
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    Similarly do that there. we have \((x+2)(x-2) \quad\textbf{and}\quad 3(m+2)\) as the denominators

  40. Firejay5
    • 3 years ago
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    it would be 3(m - 2) (m + 2)

  41. mathslover
    • 3 years ago
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    Yeah . and sorry there for using x on the place of m

  42. mathslover
    • 3 years ago
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    So I have : 3(m-2)(m+2) as LCM can you solve the fraction :

  43. Firejay5
    • 3 years ago
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    so what would it be entered in equation editor

  44. mathslover
    • 3 years ago
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    I didn't get you,, please explain!

  45. Firejay5
    • 3 years ago
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    type the problem in fraction form of what we have left

  46. mathslover
    • 3 years ago
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    hmn well it would be hard to say that I have : \(\cfrac{\textbf{yet to come}}{3(m+2)(m-2)}\)

  47. mathslover
    • 3 years ago
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    we have to solve numerator now

  48. mathslover
    • 3 years ago
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    can u do that?

  49. Firejay5
    • 3 years ago
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    wouldn't the numerator m + 2

  50. mathslover
    • 3 years ago
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    No! See : \(\cfrac{1}{(a+b)} + \cfrac{2}{(a-b)} = \cfrac{1(a-b) + 2(a+b)}{(a-b)(a+b)}\)

  51. Firejay5
    • 3 years ago
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    how does it work, you should + across

  52. mathslover
    • 3 years ago
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    |dw:1364264397027:dw|

  53. Firejay5
    • 3 years ago
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    doesn't make sense

  54. mathslover
    • 3 years ago
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    See i have LCM as 3(m+2)(m-2) I divided LCM by the first fraction's denominator that is (m+2)(m-2) to get 3 . Now I multiplied this 3 by the first fractions' numerator that is m

  55. mathslover
    • 3 years ago
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    getting it?

  56. wio
    • 3 years ago
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    Wow, finding the LCM is makes it complicated.

  57. Firejay5
    • 3 years ago
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    ????

  58. mathslover
    • 3 years ago
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    Similarly for the second one.. I have LCM as 3(m+2)(m-2) I divided this LCM by the denominator of the second fraction 3(m+2) \(\cfrac{\cancel{3(m+2)} (m-2)}{\cancel{3(m+2)} }\)

  59. mathslover
    • 3 years ago
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    Does that mean you don't get it @Firejay5 ?

  60. Firejay5
    • 3 years ago
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    kind of

  61. Firejay5
    • 3 years ago
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    are the denominators the same

  62. mathslover
    • 3 years ago
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    No.

  63. mathslover
    • 3 years ago
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    |dw:1364264678904:dw|

  64. Firejay5
    • 3 years ago
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    8/15

  65. mathslover
    • 3 years ago
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    But how?

  66. Firejay5
    • 3 years ago
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    whatever you multiply to the bottom, you must do to the top

  67. Firejay5
    • 3 years ago
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    work the problem a whole different way

  68. mathslover
    • 3 years ago
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    |dw:1364264834069:dw|

  69. Firejay5
    • 3 years ago
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    2(a - c) or 2(a + c)

  70. mathslover
    • 3 years ago
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    The LCM would be : 2(a+c)(a-c)

  71. Firejay5
    • 3 years ago
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    let's work the problem a different way and not factorize, and get the denominators the same

  72. mathslover
    • 3 years ago
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    Yes , that would be easy for u.

  73. mathslover
    • 3 years ago
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    Well which denominator you want to be changed

  74. Firejay5
    • 3 years ago
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    You're way is not what we were taught in class and is confusing me. 3m + 6

  75. mathslover
    • 3 years ago
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    Sorry, ok let us change 3(m+2)

  76. mathslover
    • 3 years ago
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    Would you try that for once?

  77. Firejay5
    • 3 years ago
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    what

  78. mathslover
    • 3 years ago
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    to change the denominator ?

  79. Firejay5
    • 3 years ago
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    I have been trying

  80. mathslover
    • 3 years ago
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    ok! I have first denominator as (m+2)(m-2) and other one as 3(m+2) I want to make 3(m+2) = (m+2)(m-2) For that I think I should multiply 3(m+2) by \(\cfrac{(m-2)}{3}\)

  81. Firejay5
    • 3 years ago
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    where did you get m - 2 over 3

  82. Firejay5
    • 3 years ago
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    don't jump ahead, it feels like you are just picking random numbers and variable from the problem

  83. Mertsj
    • 3 years ago
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    Are you all clear now, Firejay?

  84. Firejay5
    • 3 years ago
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    kind of would you explain it the same way

  85. Mertsj
    • 3 years ago
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    Do you want to talk about the original problem or what?

  86. Firejay5
    • 3 years ago
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    It feels like I am being taught a whole different then I learned it in class

  87. Firejay5
    • 3 years ago
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    different way

  88. Mertsj
    • 3 years ago
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    I just want to make sure we are talking about the right problem. Is this it:

  89. Mertsj
    • 3 years ago
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    |dw:1364265923023:dw|

  90. Firejay5
    • 3 years ago
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    yes! :D

  91. Mertsj
    • 3 years ago
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    |dw:1364265995425:dw|

  92. Mertsj
    • 3 years ago
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    When you add fractions you need to have the denominators the same. So we have to build the common denominator.

  93. Firejay5
    • 3 years ago
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    @mathslover you did good though! :D

  94. Mertsj
    • 3 years ago
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    For the first denominator we need the factors m-2 and m+2. For the second denominator we need the factors 3 and m+2. But we already have m+2. So the 3 factors we need are 3(m-2)(m+2)

  95. Mertsj
    • 3 years ago
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    |dw:1364266190884:dw|

  96. Mertsj
    • 3 years ago
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    Are you with me so far?

  97. Mertsj
    • 3 years ago
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    |dw:1364266301603:dw|

  98. Mertsj
    • 3 years ago
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    And finally: |dw:1364266416514:dw|

  99. Firejay5
    • 3 years ago
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    does it matter which order the m - 2 or m + 2 goes

  100. Firejay5
    • 3 years ago
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    uh???

  101. Firejay5
    • 3 years ago
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    I put 5m - 4 over 3(m + 2)(m - 2)

  102. Mertsj
    • 3 years ago
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    It doesn't matter.

  103. Firejay5
    • 3 years ago
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    How would you type 33.) y over y + 3 (-) 6y over y^2 - 9 in Wolframalpha

  104. Mertsj
    • 3 years ago
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    \[\frac{y}{y+3}-\frac{6y}{y^2-9}\]

  105. Firejay5
    • 3 years ago
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    yea

  106. Mertsj
    • 3 years ago
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    For wolf: y/(y+3)-(6y)/(y^2-9)

  107. Firejay5
    • 3 years ago
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    which of the alternative forms are the answer

  108. Mertsj
    • 3 years ago
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    |dw:1364267512777:dw|

  109. Mertsj
    • 3 years ago
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    That's the abbreviated version.

  110. Firejay5
    • 3 years ago
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    is it the first, second, or third one

  111. Mertsj
    • 3 years ago
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    |dw:1364267994242:dw|

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