# How do you combine 2/(4z^2-12z+9)-(z+1)/(2z^2-3z)?

Sep 20, 2016

$\frac{- 2 {z}^{2} + 3 z + 3}{z \left(2 z - 3\right) \left(2 z - 3\right)}$

#### Explanation:

In a question involving algebraic fractions, no matter what the operation is, factorize first.

$\frac{2}{4 {z}^{2} - 12 z + 9} - \frac{z + 1}{2 {z}^{2} - 3 z}$

$\frac{2}{\left(2 z - 3\right) \left(2 z - 3\right)} - \frac{z + 1}{z \left(2 z - 3\right)} \text{ } \leftarrow$find the LCD

=$\frac{2 z - \left(z + 1\right) \left(2 z - 3\right)}{z \left(2 z - 3\right) \left(2 z - 3\right)} \text{ } \leftarrow$ make equivalent fractions

$\frac{2 z - \left(2 {z}^{2} - 3 z + 2 z - 3\right)}{z \left(2 z - 3\right) \left(2 z - 3\right)} \text{ } \leftarrow$ simplify the numerator

$\frac{2 z - 2 {z}^{2} + 3 z - 2 z + 3}{z \left(2 z - 3\right) \left(2 z - 3\right)}$

$\frac{- 2 {z}^{2} + 3 z + 3}{z \left(2 z - 3\right) \left(2 z - 3\right)}$