compound fraction problem

[Guest]

[3x^4y^3/28n^4]/[12x^2y/7mn^3]

please help me...........I am sooooooooo lost. I need to express this in lowest terms and I don't even know where to start.

Thanks!!!

 

[skeeter]

multiply the fractions straight across ... the only "simplification" to be done is cancelling the greatest common factor of 12 (in the numerator) and 28 (in the denominator).

 

[Mrspi]

[3x^4y^3/28n^4]/[12x^2y/7mn^3]

please help me...........I am sooooooooo lost. I need to express this in lowest terms and I don't even know where to start.

Thanks!!!

Dividing by a fraction is the same thing as multipying by the reciprocal of that fraction. So,

[3x^4 y^3/28n^4 ] / [12x^2 y / 7mn^3] is the same thing as

[3x^4 y^3 / 28 n^4] * [7 m n^3 / 12 x^2 y]

Now, just multiply the fractions together (you can do some reducing first!)

 

[Denis]

[3x^4y^3 / 28n^4] / [12x^2y / 7mn^3]

Like Mrs Pie told you:
(a / b) / (c / d) = (a / b) * (d / c) = (a * d) / (b * c)

Doing that to yours:

3 x^4 y^3   7 m n^3
========= * =======
  28 n^4    12 x^2 y

You can rewrite that like this (less confusing!):
 3 x^4 y^3   7 m n^3
========== * =======
12 x^2 y      28 n^4 

You can now simplify a bit by cancelling 3 x^2 y on left, and 7 n^3 on right:
 x^2 y^2     m 
 ======= * ======
    4       4 n

Since that's a multiplication, then all you need is:
 m(x^2 y^2) 
 =========
   16 n

NOTE: I'm NEVER doing that again; took me 6 "edits" to line them fractions up :evil: