Question

Divide Rational Expressions:$\frac{t^2\hspace{0.17em}-\hspace{0.17em}9}{2t}÷\left(t^2\hspace{0.17em}-\hspace{0.17em}6t\hspace{0.17em}+\hspace{0.17em}9\right)$

Short answer

The required answer is$\frac{t+3}{2t\left(t-3\right)}$.

Step-by-step solution

Step 1. Given information.

The given expression is$\frac{t^2\hspace{0.17em}-\hspace{0.17em}9}{2t}÷\left(t^2\hspace{0.17em}-\hspace{0.17em}6t\hspace{0.17em}+\hspace{0.17em}9\right)$.

Step 2.  Simplify the expression. 

$$\begin{gathered} \frac{t^2\hspace{0.17em}-\hspace{0.17em}9}{2t}÷\left(t^2\hspace{0.17em}-\hspace{0.17em}6t\hspace{0.17em}+\hspace{0.17em}9\right) \\ =\frac{\frac{t^2-9}{2t}}{t^2-6t+9} \end{gathered}$$

Apply the fraction rule.

$$\begin{gathered} \frac{t^2-9}{2t\left(t^2-6t+9\right)} \\ =\frac{\left(t+3\right)\left(t-3\right)}{2t\left(t^2-6t+9\right)} \\ =\frac{\left(t+3\right)\left(t-3\right)}{2t{\left(t-3\right)}^2} \end{gathered}$$

Apply exponent law.

$$\begin{gathered} \frac{\left(t+3\right)\left(t-3\right)}{2t{\left(t-3\right)}^2} \\ =\frac{\left(t+3\right)\left(t-3\right)}{2t\left(t-3\right)\left(t-3\right)} \\ =\frac{t+3}{2t\left(t-3\right)} \end{gathered}$$