# How do you solve #(n^2-n-6)/(n^2-n)-(n-5)/(n-1)=(n-3)/(n^2-n)#?

##### 1 Answer

There are no solutions.

#### Explanation:

#(n^2-n-6)/(n^2-n)-(n-5)/(n-1)=(n-3)/(n^2-n)#

First, factor anything that can be factored:

#((n-3)(n+2))/(n(n-1))-(n-5)/(n-1)=(n-3)/(n(n-1))#

We should find a common denominator on the left-hand side so that we can combine the fractions. The least common denominator for everything here is

#((n-3)(n+2))/(n(n-1))-(n(n-5))/(n(n-1))=(n-3)/(n(n-1))#

We can now multiply everything through by

#(n-3)(n+2)-n(n-5)=n-3#

Expand the binomials:

#n^2-n-6-(n^2-5n)=n-3#

#n^2-n-6-n^2+5n-n+3=0#

#3n-3=0#

#n=1#

As we said before, *cannot* be included in our solution set. Trying to plug