How do you find the quotient of #(k+3)/(k+2)divk/(5k+10)#?

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Answer 1 · 2017-06-03T17:10:23

Alternate solution:

Take the reciprocal of the right side and multiply:

#(k+3)/(k+2) xx (5k+10)/k#

Factor out a #5# from the right side:

#(k+3)/(k+2) xx(5(k+2))/k#

Cancel #k+2#:

#(k+3)/cancel(k+2) xx (5(cancel(k+2)))/k#

You're left with:

#(5(k+3))/k#

Distribute:

#(5k+15)/k#