How do you simplify #(6)/(7)+(5)/(2)#?

1 Answer
Aug 20, 2016

I found: #47/14#

Explanation:

Here you can add the two fractions BUT you must be careful becase you cannot add directly but you need to make their denominators equal first; if the denominators are equal you can add them directly adding the numerators.
Let us choose a common denominator:
A good one could be #14# that is small and common between #7# and #2#, and also is a multiple of both, making it easy to handle.
Let us do that:

#()/(14)+()/(14)=#

BUT what about the numerators?
We see that we "imposed" a new denominator but in order to mantain the fractions with the same original value we need to "manipulate" the numerator accordingly.

In the first fraction the old denominator #7# became #14#; this is the same as #7xx2=14#; so the old denominator was multiplyed by #2# and we do the same with the numerator:

In the second fraction the old denominator #2# became #14#; this is the same as #2xx7=14#; so the old denominator was multiplyed by #7# and we do the same with the numerator:

or:
#(6xxcolor(red)(2))/(14)+(5xxcolor(red)(7))/(14)=12/14+35/14=#

you can check that the two new fractions have exactly the same value of the original ones!
#6/7=12/14#
and
#5/2=35/14#

The good thing is that now the denominators are the same and we can add the two fractions as:
#(12+35)/14=47/14#