How do you simplify #(c/a+c/b) / (c/(ab))#?

1 Answer
Sep 28, 2015

Answer:

#a+b#

Explanation:

[1]#color(white)("XXX")color(blue)((c/a+c/b) = (bc+ac)/(ab) = (c(a+b))/(ab))#

[2]#color(white)("XXX")color(blue)((c/a+c/b))/color(red)((c/(ab))) = color(blue)((c/a+c/b))*color(red)(((ab)/c))#

Substituting result [1] into [2]
#color(white)("XXX")color(blue)((c/a+c/b))/color(red)((c/(ab))) = color(blue)((c(a+b))/(ab))*color(red)((ab)/c)#

#color(white)("XXX")(c/a+c/b)/(c/(ab)) = (cancel(c)(a+b))/(cancel(ab))*(cancel(ab))/cancel(c) = a+b#