What is #(1/4 + 11/8)/(2/13+1/26)#?

1 Answer
Jun 11, 2016

Nice to know: #(a/b)/(c/d) " simplifies "(a xx d)/(b xx c)#

Explanation:

There is a very nifty short cut when you need to divide layers of fractions such as in this example.

In #(a/b)/(c/d) " this simplifies to "(a xx d)/(b xx c)#

Just be sure that you know where the MAIN division is.

#(a/b)/c rArr (a/b)/(c/1)" while " a/(b/c) rArr (a/1)/(b/c) #

I have found this very useful!