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

- anonymous

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- anonymous

this could help with finding the LCD http://www.helpwithfractions.com/math-homework-helper/least-common-denominator/

- anonymous

ok let me check it out

- anonymous

can you do live help please

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## More answers

- anonymous

live help?

- anonymous

yea like help me now walk through it with me it is easyer

- anonymous

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?

- anonymous

idk

- Michele_Laino

we can rewrite your expression as below:
\[\Large \frac{{2x}}{{6{x^3}y}} - \frac{{3y}}{{4x{y^2}}} = \frac{1}{{3{x^2}y}} - \frac{3}{{4xy}}\]

- anonymous

what did u take out?

- anonymous

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?

- anonymous

3

- anonymous

No, it is 6

- Michele_Laino

we have to compute the LCM between:
\[\Large 3{x^2}y,\quad and\quad 4xy\]

- anonymous

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

- anonymous

I am so confused I have 2 diffrent ppl telling me 2 different things

- anonymous

I'm not sure where the other person is going with their answer but mine are clear and simple

- anonymous

ok

- Michele_Laino

I'm sure that my procedure is the right procedure

- anonymous

.... aren't you a qualified helper lol

- anonymous

ook

- anonymous

but mine is correct I'm sure

- Michele_Laino

ok! Then please continue! @aaldia678

- anonymous

I want to know what each of ur answers r

- anonymous

@Michele_Laino thank you

- anonymous

reread what I wrote @sarahefal

- anonymous

ok

- anonymous

wait I'm sorry, I used the wrong number

- anonymous

ok

- anonymous

this is your equation right?
\[\frac{ 2 }{ 6x ^{3y} } -\frac{ 3y }{ 4xy ^{2}}\]

- anonymous

okay so what is the LCD of 6 and 4? figure that out first

- anonymous

3*2^2

- anonymous

?

- anonymous

the LCD

- anonymous

so 12

- anonymous

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

- anonymous

then what?

- anonymous

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.

- anonymous

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.

- anonymous

ok so it is 12x^3y^2

- anonymous

yes!!

- Michele_Laino

that's right!

- anonymous

:) thank you

- Michele_Laino

good job! @aaldia678

- anonymous

hb the numerators?

- anonymous

12x^3y^2 is the final answer right. you just needed the LCD

- anonymous

whatttt

- Michele_Laino

we have to simplify the expression above

- anonymous

btw it is assume no denomiator =0

- Michele_Laino

yes!

- anonymous

so what is the numerator if we found the denominator

- anonymous

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

- anonymous

sorry, that first part came out weird

- anonymous

no denominator =0 and thats ok

- Michele_Laino

no one of the denominators can be equal to zero @aaldia678

- anonymous

thx @Michele_Laino

- anonymous

hmm okay so what would be the full answer?

- Michele_Laino

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

- anonymous

my question is why is the 12 term the numerator?

- anonymous

that equals 6x^2y^2

- Michele_Laino

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

- anonymous

12x^3y^2

- Michele_Laino

here is wha we are doing:
|dw:1432936812802:dw|

- anonymous

is that a 72?

- Michele_Laino

please wait a moment the least common multiple is:
\[\Large 24{x^3}{y^2}\]

- anonymous

ok

- Michele_Laino

not 12x^3y^2

- anonymous

ohhh ok

- Michele_Laino

ok! let's restart

- anonymous

ok clean slate

- Michele_Laino

we have to compute this:
\[\Large \frac{{24{x^3}{y^2}}}{{2x}} = ...?\]

- anonymous

can we start from the very beginging?

- Michele_Laino

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

- Michele_Laino

yes! we are start from the beginning

- Michele_Laino

starting*

- anonymous

ok :) how did u get the 24?

- Michele_Laino

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

- Michele_Laino

next step:
\[\frac{{12{x^3}{y^2}}}{{6{x^3}y}} = 2y\]

- anonymous

yeah lol I was right

- anonymous

so do you guys know the final answer yet?

- Michele_Laino

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

- Michele_Laino

then, next we have to compute this:
\[\Large \frac{{12{x^3}{y^2}}}{{4x{y^2}}} = 3{x^2}\]

- anonymous

ok

- Michele_Laino

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

- Michele_Laino

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

- anonymous

ok

- Michele_Laino

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

- Michele_Laino

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

- Michele_Laino

am I right?

- anonymous

yea

- Michele_Laino

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

- anonymous

and is that it?

- Michele_Laino

yes! we have finished!

- anonymous

yey thank you!!!!!!

- Michele_Laino

thank you!! :)

- Michele_Laino

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

- anonymous

ok got u thanks !!!

- Michele_Laino

thanks! again! :)

- anonymous

why u thanking me i sould be thanking u

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