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

2 Answers
Mar 14, 2018

= - 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

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!