How do you simplify #(8x-56)/(x^2-49)div(x-6)/(x^2+11x+28)#?

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Answer 1 · 2015-07-27T18:40:22

#(8x+32)/(x-6)#

First, do the division by inverting the second fraction and multiplying:

#(8x-56)/(x^2-49)\divide (x-6)/(x^2+11x+28)=(8x-56)/(x^2-49)*(x^2+11x+28)/(x-6)#

Next, factor each term as much as possible to see if anything cancels:

#(8x-56)/(x^2-49) * (x^2+11x+28)/(x-6)#

#=(8(x-7))/((x-7)(x+7))*((x+7)(x+4))/(x-6)#

#=(8cancel((x-7)))/(cancel((x-7))cancel((x+7)))*(cancel((x+7))(x+4))/(x-6)#

#=(8x+32)/(x-6)#

Answer 2 · 2015-07-27T18:43:22

Dividing by a fraction is multiplying by its inverse.

So we invert the right fraction, and then we factorise:

#=(8(x-7))/((x+7)(x-7))*((x+4)(x+7))/(x-6)#

And then we can cancel:

#=(8cancel((x-7)))/(cancel((x-7))cancel((x+7)))*(cancel((x+7))(x+4))/(x-6)#

#=8*(x+4)/(x-6)#