How do you simplify #((10ab)/(x² - y²)) / ((5a(x - y))/(3ax(x + y))#?

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Answer 1 · 2018-05-08T09:46:03

#(6abx)/(x-y)^2#

Let us check which term can be simplified further:
#x^2-y^2# = #(x+y)(x-y)#

Then we know: #1/(a/b)# = #1xxb/a#
so now we simplify the original expression:

#((10ab)/(x^2-y^2))/((5a(x-y))/(3ax(x+y)))# = #(10ab)/((x+y)(x-y))xx(3ax(x+y))/(5a(x-y))#

#((10ab)/(x^2-y^2))/((5a(x-y))/(3ax(x+y)))# = #(10ab)/(cancel((x+y))(x-y))xx(3axcancel((x+y)))/(5a(x-y))#

#((10ab)/(x^2-y^2))/((5a(x-y))/(3ax(x+y)))# = #(10ab)/((x-y))xx(3ax)/(5a(x-y)#

#((10ab)/(x^2-y^2))/((5a(x-y))/(3ax(x+y)))# = #(30a^2bx)/((5a)(x-y)^2)#

#((10ab)/(x^2-y^2))/((5a(x-y))/(3ax(x+y)))# = #(30/5)xx((a^(2-1)xxbx))/(x-y)^2#

#((10ab)/(x^2-y^2))/((5a(x-y))/(3ax(x+y)))# = #(6abx)/(x-y)^2#