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

Aug 19, 2017

There are no solutions.

#### Explanation:

$\frac{{n}^{2} - n - 6}{{n}^{2} - n} - \frac{n - 5}{n - 1} = \frac{n - 3}{{n}^{2} - n}$

First, factor anything that can be factored:

$\frac{\left(n - 3\right) \left(n + 2\right)}{n \left(n - 1\right)} - \frac{n - 5}{n - 1} = \frac{n - 3}{n \left(n - 1\right)}$

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 $n \left(n - 1\right)$.

$\frac{\left(n - 3\right) \left(n + 2\right)}{n \left(n - 1\right)} - \frac{n \left(n - 5\right)}{n \left(n - 1\right)} = \frac{n - 3}{n \left(n - 1\right)}$

We can now multiply everything through by $n \left(n - 1\right)$ and clear the denominators. One thing that we have to be careful of, though, is that we don't forget that $n \left(n - 1\right)$ 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.

$\left(n - 3\right) \left(n + 2\right) - n \left(n - 5\right) = n - 3$

Expand the binomials:

${n}^{2} - n - 6 - \left({n}^{2} - 5 n\right) = n - 3$

${n}^{2} - n - 6 - {n}^{2} + 5 n - n + 3 = 0$

$3 n - 3 = 0$

$n = 1$

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.