jay3390
Subtract.
[(u)/(6c^3*d)] - [(3c^3*w^3)/(8b^2*d^2)]
Simplify your answer as much as possible
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jay3390
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Help is greatly appreciated
saifoo.khan
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Find LCD.
jay3390
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dont know how to do this :/
saifoo.khan
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@Tomas.A . help.
jay3390
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that would be great if you could help!
saifoo.khan
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Ok.
Tomas.A
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\[[(u)/(6c^3*d)] - [(3c^3*w^3)/(8b^2*d^2)]\]
saifoo.khan
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|dw:1339902311825:dw|it's like that right?
jay3390
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yeah
saifoo.khan
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So what do u think of LCD?
jay3390
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im not really sure, i suck at fractions :/
saifoo.khan
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dont worry about fractions. it's easy.
you have two things:
6c^3*d and 8b^2*d^2
open up, the
numbers: 6,8
variables: c^3, d
variables: b^2 d^2
agree?
jay3390
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Right
saifoo.khan
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So the LCD will be,
6*8 = 48
for variables,
c^3 and b^2.
now what about d and d^2
we always take the bigger term,
so we will take d^2.
Therefore our LCD will be,
48*c^3*b^2*d^2
jay3390
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oh okay perfect, then what??
saifoo.khan
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now,
|dw:1339902810873:dw|Do you get this step?
jay3390
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for the most part, yeah
saifoo.khan
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What we have to do is, we have to make the denominators same as LCD.
so we had 6c^3d.. and we have to make it as 48c^3b^2d^2
we have to multiply 6c^3d with something to make it as 48c^3b^2d^2
so we multiplied by 8b^2d.
jay3390
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oh okay and for the numerator??
saifoo.khan
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When we multiply that for denominator, we multiply the same thing with numerator.
jay3390
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how do i multiply that out though? :/
saifoo.khan
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\[\Large \frac{48c^3b^2d^2}{6c^3d} \to8b^2d.\]
jay3390
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woah..thats the numerator?
saifoo.khan
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Yes sir.
saifoo.khan
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that's what u have to multiply with numerator.
jay3390
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i must be driving you crazy, lol, but i dont get how to multiply it and simplify afterwards..
:/
saifoo.khan
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C'mon. chill up!
saifoo.khan
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and tell me where are u stuck.
jay3390
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i got the denominator, now i need to get the numerator and then simplify. i know what to multiply but i dont know how :/ i get some super long answer with a bunch of variables
saifoo.khan
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whatever you multiply with denominator.. you have to multiply the same thing with the numerator.
jay3390
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yeah
saifoo.khan
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So now,
u(8b^2d) = 8b^2du
jay3390
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so, 8b^2du - 3c^3w^3?
saifoo.khan
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you forgot to multiply the right denominator and then the numerator.
saifoo.khan
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you did it with left one right? now do it with right one as well.
jay3390
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so it would be 6c^3*d(3c^3*w^3)?
saifoo.khan
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No. since the denominators are different. we will do something like:|dw:1339903883124:dw|
jay3390
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ohh
okay and once i get that, the top would be 8b^2du-18c^6w^3?
saifoo.khan
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Yes sir.
that answer over \[\Large \frac{8b^2du-18c^6w^3}{48c^3 b^2 d^2}\]
jay3390
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okay and does that get simplified further??
saifoo.khan
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Yes. you can take commons.
jay3390
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like the b^2?
saifoo.khan
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b^2 is not in the second term.
jay3390
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ahh :( idk how then
saifoo.khan
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HINT: 8 and 18
jay3390
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-10b^2*d*u*c^6*w^3/48c^3*b^2*d^2??
saifoo.khan
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umm... no
jay3390
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:/
saifoo.khan
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|dw:1339904363228:dw|
jay3390
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then theres more left to simplify or no?
jay3390
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lol okay, thank you SOOOO much!!
saifoo.khan
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No problem! (:
jay3390
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:) thanks again!
saifoo.khan
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Welcome.