How would I add #{5n}/{4n^2-20n}# to #{5n}/{n-1}# (e.g., #{5n}/{4n^2-20n}+{5n}/{n-1}#)?

1 Answer
Alan P.
Feb 19, 2015

When adding two values with different denominators, convert your fractions so they have the same denominator and add the (converted) numerators.

#(5n)/(4n^2 - 20n) + (5n)/(n-1)#

#= 5/(4n - 20) + (5n)/(n-1)#

#= ( (5) * (n-1) ) / ( (4n -20) * (n-1) )+ ( (5n) * (4n - 20))/((n-1) * (4n - 20) ) #

#= ((5n -5) + (20n^2 -100n)) / (4n^2 -24n +20)#

#= (20n^2 - 95n -5) / (4n^2 -24n +20)#

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