How do you simplify #((r+s)/(x^2-y^2))/((r+s)^2/(x-y))#?

2 Answers
May 29, 2018

Answer:

See Below.

Explanation:

We have,

#((r + s)/(x^2 - y^2))/((r + s)^2/(x - y)) = ((r + s)/((x + y)(x - y)))/(((r + s)(r + s))/((x - y))#

#=cancel((r +s))/((x + y)cancel((x - y))) xx cancel((x - y))/(cancel((r+ s))(r + s))# [As #a/b = a xx 1/b#]

#= 1/((r + s)(x + y))#

#= 1/(rx + sx + ry + sy)#

Hope this helps.

May 29, 2018

Answer:

#1/((x+y)(r+s))#

Explanation:

break it up like this:

#(r+s)/(x^2-y^2)div(r+s)^2/(x-y)#

when dividing remember to keep the dividend, then change the division sign to a multiplication sign, and then use the reciprocal of the divisor.

#(r+s)/(x^2-y^2)*(x-y)/(r+s)^2#

you can simplify the expression further. You can factor #x^2-y^2# to #(x+y)(x-y)#

#(r+s)/((x+y)(x-y))*(x-y)/((r+s)(r+s))#

now just multiply the two of these like you would do for normal fractions

#(cancel((r+s))cancel((x-y)))/((x+y)cancel((x-y))cancel((r+s))(r+s))#

some of the terms are going to cancel each other out leaving you with this answer
#1/((x+y)(r+s))#