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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this could help with finding the LCD http://www.helpwithfractions.com/mathhomeworkhelper/leastcommondenominator/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok let me check it out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can you do live help please

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea like help me now walk through it with me it is easyer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay. 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we can rewrite your expression as below: \[\Large \frac{{2x}}{{6{x^3}y}}  \frac{{3y}}{{4x{y^2}}} = \frac{1}{{3{x^2}y}}  \frac{3}{{4xy}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what did u take out?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to compute the LCM between: \[\Large 3{x^2}y,\quad and\quad 4xy\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am so confused I have 2 diffrent ppl telling me 2 different things

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure where the other person is going with their answer but mine are clear and simple

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I'm sure that my procedure is the right procedure

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0.... aren't you a qualified helper lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but mine is correct I'm sure

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! Then please continue! @aaldia678

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I want to know what each of ur answers r

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Michele_Laino thank you

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0reread what I wrote @sarahefal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait I'm sorry, I used the wrong number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is your equation right? \[\frac{ 2 }{ 6x ^{3y} } \frac{ 3y }{ 4xy ^{2}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay so what is the LCD of 6 and 4? figure that out first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Correct. 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok so it is 12x^3y^2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1good job! @aaldia678

anonymous
 one year ago
Best ResponseYou've already chosen the best response.012x^3y^2 is the final answer right. you just needed the LCD

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to simplify the expression above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw it is assume no denomiator =0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so what is the numerator if we found the denominator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(\begin{matrix}2x \\ 0\end{matrix}\right)  \frac{ 3y }{ 0}\] so this is the equation that needs to be simplified?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry, that first part came out weird

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no denominator =0 and thats ok

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no one of the denominators can be equal to zero @aaldia678

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm okay so what would be the full answer?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1step #1 we have to compute this division: \[\Large \frac{{12{x^3}{y^2}}}{{2x}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my question is why is the 12 term the numerator?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! now we have to multiply the first numerator by 6x^2y^2, namely, what is: \[\Large 2x \cdot 6{x^2}{y^2} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here is wha we are doing: dw:1432936812802:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1please wait a moment the least common multiple is: \[\Large 24{x^3}{y^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! let's restart

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1we have to compute this: \[\Large \frac{{24{x^3}{y^2}}}{{2x}} = ...?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can we start from the very beginging?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1sorry I have made an error, we have to compute this quantity: \[\frac{{24{x^3}{y^2}}}{{6{x^3}y}} = ...?\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we are start from the beginning

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok :) how did u get the 24?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1since 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1next step: \[\frac{{12{x^3}{y^2}}}{{6{x^3}y}} = 2y\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah lol I was right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so do you guys know the final answer yet?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then, next we have to compute this: \[\Large \frac{{12{x^3}{y^2}}}{{4x{y^2}}} = 3{x^2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1so we got this expression: \[\Large \frac{{2x}}{{6{x^3}y}}  \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right)  \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1then we have to simplify, and after that simplification we get: \[\Large \frac{{2x}}{{6{x^3}y}}  \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right)  \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \frac{{4xy  9{x^2}y}}{{12{x^3}{y^2}}}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1now we can factor out, at numerator and at denominator the quantity xy, so we get: \[\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}}  \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right)  \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy  9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4  3x} \right)}}{{xy\left( {12{x^2}y} \right)}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! Now we can cancel out xy, then we get: \[\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}}  \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right)  \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy  9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4  3x} \right)}}{{xy\left( {12{x^2}y} \right)}} = \hfill \\ \hfill \\ = \frac{{4  3x}}{{12{x^2}y}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we have finished!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1oops..there is a typo: \[\Large \begin{gathered} \frac{{2x}}{{6{x^3}y}}  \frac{{3y}}{{4x{y^2}}} = \frac{{\left( {2x \cdot 2y} \right)  \left( {3{x^2} \cdot 3y} \right)}}{{12{x^3}{y^2}}} = \hfill \\ \hfill \\ = \frac{{4xy  9{x^2}y}}{{12{x^3}{y^2}}} = \frac{{xy\left( {4  9x} \right)}}{{xy\left( {12{x^2}y} \right)}} = \hfill \\ \hfill \\ = \frac{{4  9x}}{{12{x^2}y}} \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1thanks! again! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why u thanking me i sould be thanking u
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