Question

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

Answer 1

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(n-1)`.

`((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(n-1)` and clear the denominators. One thing that we have to be careful of, though, is that we don't forget that `n(n-1)` was originally in our denominator. That means that if `n=0` and/or `n=1` appear as an answer later, we must not include them in our solution.

`(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, `n=1` cannot be included in our solution set. Trying to plug `n=1` into the original problem would cause `0` to be in the denominators of the fractions, so `n=1` is not a valid solution. Thus, there are no valid solutions.