How do you divide #\frac { 6( 1- k ) } { k - 1} \div \frac { 4k ^ { 2} ( k + 2) } { 4k ^ { 2} )#?

2 Answers
Mar 14, 2018

Answer:

#= - 6/ (k + 2)#

Explanation:

#(6(1 - k))/(k - 1) -: (4k^2(k + 2))/(4k^2)#

change #(k-1) to -(-k + 1)# then rearrange

#= (6(1 - k))/-(-k + 1) -: (4k^2(k + 2))/(4k^2)#

#= (6(cancel(1 - k)))/- (cancel(1 - k )) -: (cancel(4 k^2)(k + 2))/(cancel (4k^2)#

#= 6/- 1 -: (k + 2)/1#

change sign #-: to *# then rearrange the operation of fraction.

#= 6/- 1 * 1/ (k + 2)#

#= - 6/ (k + 2)#

Mar 14, 2018

Answer:

The final expression is #(-6)/(k+2)#.

Explanation:

Factor the expressions, cancel like terms, then write the final division. Here's what it looks like:

#color(white)=(6(1-k))/(k-1)div(4k^2(k+2))/(4k^2)#

#=(6(-k+1))/(k-1)div(4k^2(k+2))/(4k^2)#

#=(-6(k-1))/(k-1)div(4k^2(k+2))/(4k^2)#

#=(-6color(red)cancelcolor(black)((k-1)))/color(red)cancelcolor(black)(k-1)div(color(red)cancelcolor(black)(4k^2)(k+2))/color(red)cancelcolor(black)(4k^2)#

#=-6div(k+2)#

#=(-6)/(k+2)#

This is as simplified as it gets. Hope this helped!