## Algebra 1

$\dfrac{n^2+4n-5}{n+5} =n-1; n \ne -5$
Given: $\dfrac{n^2+4n-5}{n+5}$ Consider denominator $n+5=0$ to check the exclude value which makes the denominator $0$. $n+5=0$ Thus, $n=-5$ $\dfrac{n^2+4n-5}{n+5} = \dfrac{(n+5)(n-1)}{(n+5)}$ Hence, $\dfrac{n^2+4n-5}{n+5} =n-1; n \ne -5$