How do you simplify #(y^4/x^2)/((xy^2)/(2x^2))#?

2 Answers
Mar 12, 2017

Answer:

#(2y^2)/x#

Explanation:

A complex fraction such as #(a/b)/(c/d)# can be written in a more familiar way:

#(a/b)/(c/d)# means the same as #(a/b) div (c/d)#

To divide by a fraction is the same as multiplying by its reciprocal.

#a/b color(limegreen)(div c/d) = a/b color(limegreen)(xx d/c) = (ad)/(bc)#

However if you compare this result to the original fraction you will see that we can simplify a complex fraction immediately:

#(color(red)(a)/color(blue)(b))/(color(blue)(c)/color(red)(d)) = color(red)(ad)/(color(blue)(bc))#
Applying this to the complex fraction given gives us:

#(color(red)(y^4)/color(blue)(x^2))/(color(blue)(xy^2)/color(red)(2x^2)) = (color(red)(2x^2 xx y^4))/(color(blue)(x^2 xx xy^2))" "# which now simplifies to:

#=(2y^2)/x#

Mar 12, 2017

Answer:

#(2y^2)/x#

Explanation:

#(y^4/x^2)/((xy^2)/(2x^2))#

#:.y^4/x^2 xx (2x^2)/(xy^2#

#:.a^color(red)m*a^color(blue)n=a^(color(red)m+color(blue)n)#

#:.(y^color(red)(4-2) xx 2x^color(red)(2-2))/x^color(red)1#

#:.y^color(red)(4-2) xx 2x^color(red)(2-2-1)#

#:.y^color(red)2 xx 2x^color(red)(color(red)-1#

#:.y^color(red)2 xx 2 1/x^color(red)1#

#:.(2y^2)/x#