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

2 Answers
Jul 27, 2015

Answer:

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

Explanation:

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)#

Jul 27, 2015

Answer:

Dividing by a fraction is multiplying by its inverse.

Explanation:

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)#