How do you divide #\frac { x ^ { 2} + 5x - 14} { x y - 8x + 7y - 56} \div \frac { x y - 2y } { x y - 8x }#?

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Answer 1 · 2017-03-11T17:13:30

#(y(x-2)^2)/(x(y-8)^2)#

#(x^2+5x-14)/(xy-8x+7y-56)# ÷ #(xy-2y)/(xy-8x)#

when dividing two fractions, switch the numerator and denominator of the last one.

#(x^2+5x-14)/(xy-8x+7y-56)# ÷ #(xy-2y)/(xy-8x)# = #(x^2+5x-14)/(xy-8x+7y-56) * (xy-8x)/(xy-2y)#

before multiplying, a few things can be simplified:

#x^2+5x-14 = (x+7)(x-2)#

#xy-8x=x(y-8)#
#7y-56=7(y-8)#
#xy-8x+7y-56=x(y-8) + 7(y-8)# or #(x+7)(y-8)#

#xy-2y = y(x-2)#
#xy-8x = x(y-8)#

so now we have:

#((x+7)(x-2))/((x+7)(y-8)) * (y(x-2))/(x(y-8))#

cancel out #(x+7)#:

#(x-2)/(y-8) * (y(x-2))/(x(y-8))#

now multiply:

#(y(x-2)^2)/(x(y-8)^2)#

Answer 2 · 2017-03-11T19:16:16

#x/y#

#(x^2+5x-14)/(xy-8x+7y-56)-:(xy-2y)/(xy-8x)#

#:.=((x+7)(x-2))/(x(y-8)+7(y-8)) xx (xy-8x)/(xy-2y)#

#:.=((x+7)(x-2))/(x(y-8)+7(y-8)) xx (x(y-8))/(y(x-2))#

#:.=cancel((x+7)(x-2))/(cancel((y-8))cancel((x+7))) xx (color(red)x(cancel(y-8)))/(color(red)y(cancel(x-2)))#

#:.=x/y#