How do you combine #n/(5-n)+(2n-5)/(n-5)#?

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Answer 1 · 2017-02-10T17:49:22

#1#

Using an example. Suppose we had -5

Another way of writing this is: #-(+5)#

So we can 'force' a change in sign so that both denominators are the same:

Write #n/(5-n)" as "n/(-(-5+n))#

Just changing the order we have: #n/(-(-5+n)) -> - n/(n-5)#

So write: #color(red)(n/(5-n))color(green)(+(2n-5)/(n-5))" as "color(green)((2n-5)/(n-5))color(red)( - n/(n-5))#

#=(2n-5-n)/(n-5)" " =" " (n-5)/(n-5)" "=" "1#