## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

The simplified form of the complex rational expression $\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}$ is $-\frac{1}{x}$.
$\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}$ The denominator of the complex rational expression is single. But, the numerator is not single, so perform the operation and get a single rational expression in the numerator. The LCD of $\frac{1}{2}-\frac{1}{x}$ is $2x$. Solve the expression by making all fractions equivalent to their LCD: \begin{align} & \frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{\frac{1}{2}\cdot \frac{x}{x}-\frac{1}{x}\cdot \frac{2}{2}}{\frac{2-x}{2}} \\ & =\frac{\frac{x}{2x}-\frac{2}{2x}}{\frac{2-x}{2}} \\ & =\frac{\frac{x-2}{2x}}{\frac{2-x}{2}} \end{align} Divide the numerator by the denominator, $\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{x-2}{2x}\div \frac{2-x}{2}$ \begin{align} & \frac{A}{B}\div \frac{C}{D}=\frac{A}{B}\cdot \frac{D}{C} \\ & =\frac{AD}{BC} \end{align} where $\frac{A}{B},\left( B\ne 0 \right)$, $\frac{C}{D},\left( D\ne 0 \right)$ are rational expressions with $\frac{C}{D}\ne 0$ The reciprocal of $\frac{2-x}{2}$ is $\frac{2}{2-x}$ So, multiply the reciprocal of the divisor, $\frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{x-2}{2x}\cdot \frac{2}{2-x}$ Simplify the terms, \begin{align} & \frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{\left( x-2 \right)\cdot 2}{\left( 2x \right)\left( 2-x \right)} \\ & =\frac{2\left( x-2 \right)}{2x\left( 2-x \right)} \\ & =\frac{x-2}{x\left( 2-x \right)} \end{align} Take out ‘–’ from the numerator, \begin{align} & \frac{\frac{1}{2}-\frac{1}{x}}{\frac{2-x}{2}}=\frac{-\left( -x+2 \right)}{x\left( 2-x \right)} \\ & =\frac{-\left( 2-x \right)}{x\left( 2-x \right)} \\ & =-\frac{1}{x} \end{align}