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anonymous

  • one year ago

find the LCD of booth fractions and then simplify the expression completly assume the denominator equils 0 2x 3y ____ - ______ 6x^3y 4xy^2

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  1. anonymous
    • one year ago
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    this could help with finding the LCD http://www.helpwithfractions.com/math-homework-helper/least-common-denominator/

  2. anonymous
    • one year ago
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    ok let me check it out

  3. anonymous
    • one year ago
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    can you do live help please

  4. anonymous
    • one year ago
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    live help?

  5. anonymous
    • one year ago
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    yea like help me now walk through it with me it is easyer

  6. anonymous
    • one year ago
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    okay. well for this question, when you say to assume that the denominator equals zero, does that mean to just completely avoid the bottom of each fractions?

  7. anonymous
    • one year ago
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    idk

  8. Michele_Laino
    • one year ago
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    we can rewrite your expression as below: \[\Large \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{1}{{3{x^2}y}} - \frac{3}{{4xy}}\]

  9. anonymous
    • one year ago
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    what did u take out?

  10. anonymous
    • one year ago
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    The LCD is the smallest number or expression that is divisible by both denominators. You need to take each part of it at a time. Let's look at the numbers, then the x's then the y's. What is the smallest number that is divisible by both 3 and 6. In other words, what is the LCM of 3 and 6?

  11. anonymous
    • one year ago
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    3

  12. anonymous
    • one year ago
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    No, it is 6

  13. Michele_Laino
    • one year ago
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    we have to compute the LCM between: \[\Large 3{x^2}y,\quad and\quad 4xy\]

  14. anonymous
    • one year ago
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    This means that the number part we need for the LCD is 6. Now let's look at the x's. There is an x^2 and an x. The lowest expression that will divide both x^2 and x is x^2, so we need x^2 for the LCD. Now let's look at the y part. You have y and y^3. The simplest expression with y that will divide evenly by y and y^3 is y^3, so you need y^3 in the LCD. Once you put together the number part, the x part and rthe y part, you end up with: LCD = 6x^2y^3

  15. anonymous
    • one year ago
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    I am so confused I have 2 diffrent ppl telling me 2 different things

  16. anonymous
    • one year ago
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    I'm not sure where the other person is going with their answer but mine are clear and simple

  17. anonymous
    • one year ago
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    ok

  18. Michele_Laino
    • one year ago
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    I'm sure that my procedure is the right procedure

  19. anonymous
    • one year ago
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    .... aren't you a qualified helper lol

  20. anonymous
    • one year ago
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    ook

  21. anonymous
    • one year ago
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    but mine is correct I'm sure

  22. Michele_Laino
    • one year ago
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    ok! Then please continue! @aaldia678

  23. anonymous
    • one year ago
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    I want to know what each of ur answers r

  24. anonymous
    • one year ago
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    @Michele_Laino thank you

  25. anonymous
    • one year ago
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    reread what I wrote @sarahefal

  26. anonymous
    • one year ago
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    ok

  27. anonymous
    • one year ago
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    wait I'm sorry, I used the wrong number

  28. anonymous
    • one year ago
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    ok

  29. anonymous
    • one year ago
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    this is your equation right? \[\frac{ 2 }{ 6x ^{3y} } -\frac{ 3y }{ 4xy ^{2}}\]

  30. anonymous
    • one year ago
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    okay so what is the LCD of 6 and 4? figure that out first

  31. anonymous
    • one year ago
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    3*2^2

  32. anonymous
    • one year ago
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    ?

  33. anonymous
    • one year ago
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    the LCD

  34. anonymous
    • one year ago
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    so 12

  35. anonymous
    • one year ago
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    Correct. The smallest number that is divisible by both 4 and 6 is 12, That means the number part we need for the LCD is 12

  36. anonymous
    • one year ago
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    then what?

  37. anonymous
    • one year ago
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    Now let's look at the x's. There is an x^3 and an x. The lowest expression that will divide both x^3 and x is x^3, so we need x^3 for the LCD.

  38. anonymous
    • one year ago
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    Now let's look at the y part. You have y and y^2. The simplest expression with y that will divide evenly by y and y^2 is y^2, so you need y^2 in the LCD.

  39. anonymous
    • one year ago
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    ok so it is 12x^3y^2

  40. anonymous
    • one year ago
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    yes!!

  41. Michele_Laino
    • one year ago
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    that's right!

  42. anonymous
    • one year ago
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    :) thank you

  43. Michele_Laino
    • one year ago
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    good job! @aaldia678

  44. anonymous
    • one year ago
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    hb the numerators?

  45. anonymous
    • one year ago
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    12x^3y^2 is the final answer right. you just needed the LCD

  46. anonymous
    • one year ago
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    whatttt

  47. Michele_Laino
    • one year ago
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    we have to simplify the expression above

  48. anonymous
    • one year ago
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    btw it is assume no denomiator =0

  49. Michele_Laino
    • one year ago
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    yes!

  50. anonymous
    • one year ago
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    so what is the numerator if we found the denominator

  51. anonymous
    • one year ago
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    \[\left(\begin{matrix}2x \\ 0\end{matrix}\right) - \frac{ 3y }{ 0}\] so this is the equation that needs to be simplified?

  52. anonymous
    • one year ago
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    sorry, that first part came out weird

  53. anonymous
    • one year ago
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    no denominator =0 and thats ok

  54. Michele_Laino
    • one year ago
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    no one of the denominators can be equal to zero @aaldia678

  55. anonymous
    • one year ago
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    thx @Michele_Laino

  56. anonymous
    • one year ago
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    hmm okay so what would be the full answer?

  57. Michele_Laino
    • one year ago
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    step #1 we have to compute this division: \[\Large \frac{{12{x^3}{y^2}}}{{2x}} = ...?\]

  58. anonymous
    • one year ago
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    my question is why is the 12 term the numerator?

  59. anonymous
    • one year ago
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    that equals 6x^2y^2

  60. Michele_Laino
    • one year ago
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    ok! now we have to multiply the first numerator by 6x^2y^2, namely, what is: \[\Large 2x \cdot 6{x^2}{y^2} = ...?\]

  61. anonymous
    • one year ago
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    12x^3y^2

  62. Michele_Laino
    • one year ago
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    here is wha we are doing: |dw:1432936812802:dw|

  63. anonymous
    • one year ago
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    is that a 72?

  64. Michele_Laino
    • one year ago
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    please wait a moment the least common multiple is: \[\Large 24{x^3}{y^2}\]

  65. anonymous
    • one year ago
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    ok

  66. Michele_Laino
    • one year ago
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    not 12x^3y^2

  67. anonymous
    • one year ago
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    ohhh ok

  68. Michele_Laino
    • one year ago
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    ok! let's restart

  69. anonymous
    • one year ago
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    ok clean slate

  70. Michele_Laino
    • one year ago
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    we have to compute this: \[\Large \frac{{24{x^3}{y^2}}}{{2x}} = ...?\]

  71. anonymous
    • one year ago
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    can we start from the very beginging?

  72. Michele_Laino
    • one year ago
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    sorry I have made an error, we have to compute this quantity: \[\frac{{24{x^3}{y^2}}}{{6{x^3}y}} = ...?\]

  73. Michele_Laino
    • one year ago
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    yes! we are start from the beginning

  74. Michele_Laino
    • one year ago
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    starting*

  75. anonymous
    • one year ago
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    ok :) how did u get the 24?

  76. Michele_Laino
    • one year ago
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    since we have to factorize 6 and 4, so we get: 6 = 2*3 4=2^2*3 now we have to pick all common and uncommon factor with the highest exponent, so we pick 2^2 and 3, then we have: 2^2*3=12 so sorry @aaldia678 is right! the least common multiple is: 12 x^3 y^2

  77. Michele_Laino
    • one year ago
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    next step: \[\frac{{12{x^3}{y^2}}}{{6{x^3}y}} = 2y\]

  78. anonymous
    • one year ago
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    yeah lol I was right

  79. anonymous
    • one year ago
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    so do you guys know the final answer yet?

  80. Michele_Laino
    • one year ago
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    then we have to multiply the first numerator by 2y, so we can write: \[\Large \frac{{2x \cdot 2y}}{{12{x^3}{y^2}}}\] that is the first term

  81. Michele_Laino
    • one year ago
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    then, next we have to compute this: \[\Large \frac{{12{x^3}{y^2}}}{{4x{y^2}}} = 3{x^2}\]

  82. anonymous
    • one year ago
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    ok

  83. Michele_Laino
    • one year ago
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    then we have to multiply the second numerator by 3x^2, so we can write: \[\Large \frac{{3{x^2} \cdot 3y}}{{12{x^3}{y^2}}}\] that is the second term

  84. Michele_Laino
    • one year ago
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    so we got this expression: \[\Large \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}}\]

  85. anonymous
    • one year ago
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    ok

  86. Michele_Laino
    • one year ago
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    then we have to simplify, and after that simplification we get: \[\Large \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}}\]

  87. Michele_Laino
    • one year ago
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    now we can factor out, at numerator and at denominator the quantity xy, so we get: \[\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4 - 3x} \right)}}{{xy\left( {12{x^2}y} \right)}} \hfill \\ \end{gathered} \]

  88. Michele_Laino
    • one year ago
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    am I right?

  89. anonymous
    • one year ago
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    yea

  90. Michele_Laino
    • one year ago
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    ok! Now we can cancel out xy, then we get: \[\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4 - 3x} \right)}}{{xy\left( {12{x^2}y} \right)}} = \hfill \\ \hfill \\ = \frac{{4 - 3x}}{{12{x^2}y}} \hfill \\ \end{gathered} \]

  91. anonymous
    • one year ago
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    and is that it?

  92. Michele_Laino
    • one year ago
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    yes! we have finished!

  93. anonymous
    • one year ago
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    yey thank you!!!!!!

  94. Michele_Laino
    • one year ago
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    thank you!! :)

  95. Michele_Laino
    • one year ago
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    oops..there is a typo: \[\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right) - \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy - 9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4 - 9x} \right)}}{{xy\left( {12{x^2}y} \right)}} = \hfill \\ \hfill \\ = \frac{{4 - 9x}}{{12{x^2}y}} \hfill \\ \end{gathered} \]

  96. anonymous
    • one year ago
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    ok got u thanks !!!

  97. Michele_Laino
    • one year ago
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    thanks! again! :)

  98. anonymous
    • one year ago
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    why u thanking me i sould be thanking u

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