Rational expressions: multiplying and dividing rational expressions

$$\frac{3}{4}·\frac{1}{2}=\frac{3·1}{4·2}=\frac{3}{8}$$

$$\frac{8}{9}·\frac{1}{6}=\frac{\stackrel{4}{\cancel{8}}}{9}·\frac{1}{\underset{3}{\cancel{6}}}=\frac{4·1}{9·3}=\frac{4}{27}$$

$$\frac{3x}{5y}·\frac{7}{12y}=\frac{\stackrel{1}{\cancel{3}}x}{5y}·\frac{7}{\underset{4}{\cancel{12}}y}=\frac{x·7}{5y·4y}=\frac{7x}{20y^2}$$

$$\begin{array}{lll}\frac{x+4}{x-2}·\frac{x+7}{x+4}\hfill & \hfill & \text{Divide out the common factor\hspace{0.17em}}x+4.\hfill \\ \frac{\cancel{x+4}}{x-2}·\frac{x+7}{\cancel{x+4}}\hfill & \hfill & \text{Multiply numerators and denominators together}\text{.}\hfill \\ \frac{x+7}{x-2}\hfill & \hfill & \hfill \end{array}$$

$$\begin{array}{l}\begin{array}{lll}\frac{x^2+x-6}{x^2-4x+3}·\frac{x^2-2x-3}{x^2+4x-12}.\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \frac{\left(x+3\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)}·\frac{\left(x-3\right)\left(x+1\right)}{\left(x+6\right)\left(x-2\right)}\hfill & \hfill & \text{Divide out the common factors\hspace{0.17em}}x-2\text{\hspace{0.17em}and\hspace{0.17em}}x-3.\hfill \\ \frac{\left(x+3\right)\cancel{\left(x-2\right)}}{\cancel{\left(x-3\right)}\left(x-1\right)}·\frac{\cancel{\left(x-3\right)}\left(x+1\right)}{\left(x+6\right)\cancel{\left(x-2\right)}}\hfill & \hfill & \text{Multiply}.\hfill \end{array}\\ \begin{array}{lllllllll}\frac{\left(x+3\right)\left(x+1\right)}{\left(x-1\right)\left(x+6\right)}\hfill & \hfill & \text{or}\hfill & \hfill & \frac{x^2+4x+3}{\left(x-1\right)\left(x+6\right)}\hfill & \hfill & \text{or}\hfill & \hfill & \frac{x^2+4x+3}{x^2+5x-6}\hfill \end{array}\\ \\ \text{Each of these three forms is an acceptable form of the same answer}.\end{array}$$

$$\begin{array}{l}\begin{array}{lll}\frac{2x+6}{8x-16}·\frac{x^2-4}{x^2-x-12}.\hfill & \hfill & \text{Factor}\text{.}\hfill \\ \frac{2\left(x+3\right)}{8\left(x-2\right)}·\frac{\left(x+2\right)\left(x-2\right)}{\left(x-4\right)\left(x+3\right)}\hfill & \hfill & \text{Divide out the common factors 2,\hspace{0.17em}}x+3\text{\hspace{0.17em}and\hspace{0.17em}}x-2.\hfill \\ \frac{\stackrel{1}{\cancel{2}}\cancel{\left(x+3\right)}}{\underset{4}{\cancel{8}}\cancel{\left(x-2\right)}}·\frac{\left(x+2\right)\cancel{\left(x-2\right)}}{\cancel{\left(x+3\right)}\left(x-4\right)}\hfill & \hfill & \text{Multiply}.\hfill \end{array}\\ \begin{array}{ccc}\frac{x+2}{4\left(x-4\right)}& \text{or}& \frac{x+2}{4x-16}\end{array}\\ \\ \text{Both these forms are acceptable forms of the same answer}.\end{array}$$

$$\begin{array}{lll}3x^2·\frac{x+7}{x-5}.\hfill & \hfill & \text{Rewrite\hspace{0.17em}}3x^2\text{\hspace{0.17em}as\hspace{0.17em}}\frac{3x^2}{1}.\hfill \\ \frac{3x^2}{1}·\frac{x+7}{x-5}\hfill & \hfill & \text{Multiply}.\hfill \\ \frac{3x^2\left(x+7\right)}{x-5}\hfill & \hfill & \hfill \end{array}$$