Question

Subtract:$\frac{3}{z+3}-\frac{6z}{z^2-9}$.

Short answer

The simplification of the rational expression is$\frac{-3}{\left(z-3\right)}$

Step-by-step solution

Given information 

The given rational expression is$\frac{3}{z+3}-\frac{6z}{z^2-9}$.

The simplification of the rational expression

$\frac{3}{z+3}-\frac{6z}{z^2-9}$

Factor the denominator

$=\frac{3}{z+3}-\frac{6z}{\left(z+3\right)\left(z-3\right)}$

The LCD is $\left(z+3\right)\left(z-3\right)$.

localid="1648454254136" $$\begin{gathered} =\frac{3\left(z-3\right)}{\left(z+3\right)\left(z-3\right)}-\frac{6z}{\left(z+3\right)\left(z-3\right)} \\ =\frac{3z-9}{\left(z+3\right)\left(z-3\right)}-\frac{6z}{\left(z+3\right)\left(z-3\right)} \\ =\frac{3z-9-6z}{\left(z+3\right)\left(z-3\right)} \\ =\frac{-3z-9}{\left(z+3\right)\left(z-3\right)} \\ =\frac{-3\left(z+3\right)}{\left(z-3\right)\left(z+3\right)} \\ =\left(\frac{-3}{z-3}\right) \end{gathered}$$