Question

Addula1" title=" \frac{k+5}{k+1} + \frac{8}{k^2-1} " alt=" \frac{k+5}{k+1} + \frac{8}{k^2-1} " align="absmiddle" class="latex-formula">

Answer 1

Get a common denominator which you can get by mutiplying by k-1
$\frac{(k + 5)(k - 1)}{(k + 1)(k - 1)} + \frac{8}{ {k}^{2} - 1}$
Now foil and distribute
$\frac{ {k}^{2} + 4k - 5}{ {k}^{2} - 1} + \frac{8}{ {k}^{2} - 1 }$
Now combine the fractions and like terms
$\frac{ {k}^{2} + 4k - 5 + 8}{ {k}^{2} - 1 } = \frac{ {k}^{2} + 4k + 3}{ {k}^{2} - 1 }$
Now factor out a k+1
$\frac{(k + 1)(k + 3)}{(k + 1)(k - 1)}$
you can cancel out the k+1 leaving
$\frac{k + 3}{k - 1}$