anonymous
  • anonymous
find the LCD of booth fractions and then simplify the expression completly assume the denominator equils 0 2x 3y ____ - ______ 6x^3y 4xy^2
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
this could help with finding the LCD http://www.helpwithfractions.com/math-homework-helper/least-common-denominator/
anonymous
  • anonymous
ok let me check it out
anonymous
  • anonymous
can you do live help please

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
live help?
anonymous
  • anonymous
yea like help me now walk through it with me it is easyer
anonymous
  • 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
  • anonymous
idk
Michele_Laino
  • 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
  • anonymous
what did u take out?
anonymous
  • 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
  • anonymous
3
anonymous
  • anonymous
No, it is 6
Michele_Laino
  • Michele_Laino
we have to compute the LCM between: \[\Large 3{x^2}y,\quad and\quad 4xy\]
anonymous
  • 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
  • anonymous
I am so confused I have 2 diffrent ppl telling me 2 different things
anonymous
  • anonymous
I'm not sure where the other person is going with their answer but mine are clear and simple
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
I'm sure that my procedure is the right procedure
anonymous
  • anonymous
.... aren't you a qualified helper lol
anonymous
  • anonymous
ook
anonymous
  • anonymous
but mine is correct I'm sure
Michele_Laino
  • Michele_Laino
ok! Then please continue! @aaldia678
anonymous
  • anonymous
I want to know what each of ur answers r
anonymous
  • anonymous
@Michele_Laino thank you
anonymous
  • anonymous
reread what I wrote @sarahefal
anonymous
  • anonymous
ok
anonymous
  • anonymous
wait I'm sorry, I used the wrong number
anonymous
  • anonymous
ok
anonymous
  • anonymous
this is your equation right? \[\frac{ 2 }{ 6x ^{3y} } -\frac{ 3y }{ 4xy ^{2}}\]
anonymous
  • anonymous
okay so what is the LCD of 6 and 4? figure that out first
anonymous
  • anonymous
3*2^2
anonymous
  • anonymous
?
anonymous
  • anonymous
the LCD
anonymous
  • anonymous
so 12
anonymous
  • 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
  • anonymous
then what?
anonymous
  • 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
  • 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
  • anonymous
ok so it is 12x^3y^2
anonymous
  • anonymous
yes!!
Michele_Laino
  • Michele_Laino
that's right!
anonymous
  • anonymous
:) thank you
Michele_Laino
  • Michele_Laino
good job! @aaldia678
anonymous
  • anonymous
hb the numerators?
anonymous
  • anonymous
12x^3y^2 is the final answer right. you just needed the LCD
anonymous
  • anonymous
whatttt
Michele_Laino
  • Michele_Laino
we have to simplify the expression above
anonymous
  • anonymous
btw it is assume no denomiator =0
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
so what is the numerator if we found the denominator
anonymous
  • anonymous
\[\left(\begin{matrix}2x \\ 0\end{matrix}\right) - \frac{ 3y }{ 0}\] so this is the equation that needs to be simplified?
anonymous
  • anonymous
sorry, that first part came out weird
anonymous
  • anonymous
no denominator =0 and thats ok
Michele_Laino
  • Michele_Laino
no one of the denominators can be equal to zero @aaldia678
anonymous
  • anonymous
thx @Michele_Laino
anonymous
  • anonymous
hmm okay so what would be the full answer?
Michele_Laino
  • Michele_Laino
step #1 we have to compute this division: \[\Large \frac{{12{x^3}{y^2}}}{{2x}} = ...?\]
anonymous
  • anonymous
my question is why is the 12 term the numerator?
anonymous
  • anonymous
that equals 6x^2y^2
Michele_Laino
  • 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
  • anonymous
12x^3y^2
Michele_Laino
  • Michele_Laino
here is wha we are doing: |dw:1432936812802:dw|
anonymous
  • anonymous
is that a 72?
Michele_Laino
  • Michele_Laino
please wait a moment the least common multiple is: \[\Large 24{x^3}{y^2}\]
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
not 12x^3y^2
anonymous
  • anonymous
ohhh ok
Michele_Laino
  • Michele_Laino
ok! let's restart
anonymous
  • anonymous
ok clean slate
Michele_Laino
  • Michele_Laino
we have to compute this: \[\Large \frac{{24{x^3}{y^2}}}{{2x}} = ...?\]
anonymous
  • anonymous
can we start from the very beginging?
Michele_Laino
  • 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
  • Michele_Laino
yes! we are start from the beginning
Michele_Laino
  • Michele_Laino
starting*
anonymous
  • anonymous
ok :) how did u get the 24?
Michele_Laino
  • 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
  • Michele_Laino
next step: \[\frac{{12{x^3}{y^2}}}{{6{x^3}y}} = 2y\]
anonymous
  • anonymous
yeah lol I was right
anonymous
  • anonymous
so do you guys know the final answer yet?
Michele_Laino
  • 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
  • Michele_Laino
then, next we have to compute this: \[\Large \frac{{12{x^3}{y^2}}}{{4x{y^2}}} = 3{x^2}\]
anonymous
  • anonymous
ok
Michele_Laino
  • 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
  • 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
  • anonymous
ok
Michele_Laino
  • 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
  • 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
  • Michele_Laino
am I right?
anonymous
  • anonymous
yea
Michele_Laino
  • 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
  • anonymous
and is that it?
Michele_Laino
  • Michele_Laino
yes! we have finished!
anonymous
  • anonymous
yey thank you!!!!!!
Michele_Laino
  • Michele_Laino
thank you!! :)
Michele_Laino
  • 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
  • anonymous
ok got u thanks !!!
Michele_Laino
  • Michele_Laino
thanks! again! :)
anonymous
  • anonymous
why u thanking me i sould be thanking u

Looking for something else?

Not the answer you are looking for? Search for more explanations.