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

The simplified form of a complex rational expression $\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}$ is $\frac{{{\left( x+1 \right)}^{2}}}{\left( x-2 \right)\left( x+4 \right)}$.
$\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}$ The numerator and denominator of the complex rational expression are not single, so perform the operation and get a single expression. The numerator LCD of $x-\frac{3}{x-2}$ is $\left( x-2 \right)$, and the denominator LCD of $x-\frac{12}{x+1}$ is $\left( x+1 \right)$. Solve the expression by making all fractions equivalent to their LCD: \begin{align} & \frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\cdot \frac{\left( x-2 \right)}{\left( x-2 \right)}-\frac{3}{x-2}}{x\cdot \frac{\left( x+1 \right)}{\left( x+1 \right)}-\frac{12}{x+1}} \\ & =\frac{\frac{x\left( x-2 \right)}{x-2}-\frac{3}{\left( x-2 \right)}}{\frac{x\left( x+1 \right)}{\left( x+1 \right)}-\frac{12}{\left( x+1 \right)}} \\ & =\frac{\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}}{\frac{x\left( x+1 \right)-12}{\left( x+1 \right)}} \end{align} Divide the numerator by the denominator, $\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}\div \frac{x\left( x+1 \right)-12}{\left( x+1 \right)}$ \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 $\frac{x\left( x+1 \right)-12}{\left( x+1 \right)}$ is $\frac{\left( x+1 \right)}{x\left( x+1 \right)-12}$ So, multiply the reciprocal of the divisor, $\frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}\cdot \frac{\left( x+1 \right)}{x\left( x+1 \right)-12}$ Simplify the terms, \begin{align} & \frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{x\left( x-2 \right)-3}{\left( x-2 \right)}\cdot \frac{\left( x+1 \right)}{x\left( x+1 \right)-12} \\ & =\frac{{{x}^{2}}-2x-3}{\left( x-2 \right)}\cdot \frac{\left( x+1 \right)}{{{x}^{2}}+x-12} \\ & =\frac{\left( {{x}^{2}}-2x-3 \right)\left( x+1 \right)}{\left( x-2 \right)\left( {{x}^{2}}+x-12 \right)} \end{align} \begin{align} & \left( {{x}^{2}}-2x-3 \right)\left( x+1 \right)=\left( {{x}^{2}}+x-3x-3 \right)\left( x+1 \right) \\ & =\left( x\left( x+1 \right)-3\left( x+1 \right) \right)\left( x+1 \right) \\ & =\left( x+1 \right)\left( x-3 \right)\left( x+1 \right) \end{align} \begin{align} & \left( x-2 \right)\left( {{x}^{2}}+x-12 \right)=\left( x-2 \right)\left( {{x}^{2}}+4x-3x-12 \right) \\ & =\left( x-2 \right)\left( x\left( x+4 \right)-3\left( x+4 \right) \right) \\ & =\left( x-2 \right)\left( x+4 \right)\left( x-3 \right) \end{align} Thus, the expression becomes: \begin{align} & \frac{x-\frac{3}{x-2}}{x-\frac{12}{x+1}}=\frac{\left( x+1 \right)\left( x-3 \right)\left( x+1 \right)}{\left( x-2 \right)\left( x+4 \right)\left( x-3 \right)} \\ & =\frac{\left( x+1 \right)\left( x+1 \right)\left( x-3 \right)}{\left( x-2 \right)\left( x+4 \right)\left( x-3 \right)} \\ & =\frac{{{\left( x+1 \right)}^{2}}}{\left( x-2 \right)\left( x+4 \right)} \end{align}