Chapter 6.2, Problem 38WE

To determine

To calculate: The simplified form of the expression $\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}$.

Answer to Problem 38WE

The simplified form of the expression $\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}$ is $\frac{1}{2rt\left(3+t\right)}$.

Explanation of Solution

Given information:

The expression $\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}$.

Formula used:

Fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

The difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$.

Calculation:

Consider the expression $\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}$.

Recall that the difference of square of two numbers a and bis $a^2-b^2=\left(a-b\right)\left(a+b\right)$ .

Apply it in the denominator of second term,

  $\begin{array}{c}\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}=\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{\left(3-t\right)\left(3+t\right)}\\ =\frac{r\left(3-t\right)}{6r^{2}t}\cdot \frac{3}{\left(3-t\right)\left(3+t\right)}\end{array}$

Strike off the similar terms,

  $\begin{array}{c}\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}=\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{\left(3-t\right)\left(3+t\right)}\\ =\frac{r\left(3-t\right)}{6r^{2}t}\cdot \frac{3}{\left(3-t\right)\left(3+t\right)}\\ =\frac{r}{6r^{2}t}\cdot \frac{3}{\left(3+t\right)}\end{array}$

Recall that fractions are multiplied as $\frac{a}{b}\cdot \frac{c}{d}=\frac{ab}{cd}$.

Apply it,

  $\begin{array}{c}\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}=\frac{r}{6r^{2}t}\cdot \frac{3}{\left(3+t\right)}\\ =\frac{3r}{6r^{2}t\left(3+t\right)}\\ =\frac{1}{2rt\left(3+t\right)}\end{array}$

Thus, the simplified form of the expression $\frac{3r-rt}{6r^{2}t}\cdot \frac{3}{9-t^2}$ is $\frac{1}{2rt\left(3+t\right)}$.