# How do you combine -10/(z^2-6z+5)+15/(z^2-4z-5)?

Jun 21, 2017

$\frac{5}{{z}^{2} - 1}$

#### Explanation:

$- \frac{10}{{z}^{2} - 6 z + 5} + \frac{15}{{z}^{2} - 4 z - 5}$

factorise denominator,

$= - \frac{10}{\left(z - 1\right) \left(z - 5\right)} + \frac{15}{\left(z + 1\right) \left(z - 5\right)}$

equalize the denominator,

$= - \frac{10 \left(z + 1\right)}{\left(z - 1\right) \left(z - 5\right) \left(z + 1\right)} + \frac{15 \left(z - 1\right)}{\left(z + 1\right) \left(z - 5\right) \left(z - 1\right)}$

$= \frac{- 10 z - 10}{\left(z - 1\right) \left(z - 5\right) \left(z + 1\right)} + \frac{15 z - 15}{\left(z + 1\right) \left(z - 5\right) \left(z - 1\right)}$

since it has a same denominator, combine all together and simplify,

$= \frac{- 10 z - 10 + 15 z - 15}{\left(z - 1\right) \left(z - 5\right) \left(z + 1\right)}$

$= \frac{5 z - 25}{\left(z - 1\right) \left(z - 5\right) \left(z + 1\right)}$

$= \frac{5 \cancel{\left(z - 5\right)}}{\left(z - 1\right) \cancel{\left(z - 5\right)} \left(z + 1\right)}$

$= \frac{5}{\left(z - 1\right) \left(z + 1\right)}$

$= \frac{5}{{z}^{2} - 1}$