# How do you solve 6/(n+3) + 20/(n^2+n-6) = 5/(n-2)?

Jun 5, 2015

The Least Common Denominator of the terms in the given equation is
$\textcolor{w h i t e}{\text{XXXX}}$$\left({n}^{2} + n - 6\right) = \left(n + 3\right) \left(n - 3\right)$

So the given equation could be written:
$\textcolor{w h i t e}{\text{XXXX}}$$\frac{6 \left(n - 2\right) + 20}{{n}^{2} + n - 6} = \frac{5 \left(n + 3\right)}{{n}^{2} + n - 6}$

and assuming $n \ne - 3$ and $n \ne 2$
(otherwise terms of the original equation would be undefined).

$\textcolor{w h i t e}{\text{XXXX}}$$6 \left(n - 2\right) + 20 = 5 \left(n + 3\right)$

$\textcolor{w h i t e}{\text{XXXX}}$n = 7#