## Precalculus (6th Edition) Blitzer

The required solution is $\frac{1}{y\left( y+2 \right)}$.
We have the given algebraic expression $\frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}$. For an algebraic expression, a rational expression is an expression that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator $q\ne 0$. And the given expression can also be written as $\frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}=\frac{\frac{1}{y}-\frac{1}{y+2}}{2}$. We know that a complex rational expression or a complex fraction is an algebraic rational expression in which either the numerator contains a rational expression or the denominator contains a rational expression or both the numerator and denominator contain a rational expression. Here the numerator is $\frac{1}{y}-\frac{1}{y+2}$. Solve the numerator of the given complex rational expression: \begin{align} & \frac{1}{y}-\frac{1}{y+2}=\frac{1}{y}\times \frac{\left( y+2 \right)}{\left( y+2 \right)}-\frac{1}{\left( y+2 \right)}\times \frac{y}{y} \\ & =\frac{y+2}{y\left( y+2 \right)}-\frac{y}{y\left( y+2 \right)} \\ & =\frac{y+2-y}{y\left( y+2 \right)} \\ & =\frac{2}{y\left( y+2 \right)} \end{align}. Therefore, the given complex rational expression becomes \begin{align} & \frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}=\frac{\frac{2}{y\left( y+2 \right)}}{2} \\ & =\frac{2}{y\left( y+2 \right)}\times \frac{1}{2} \\ & =\frac{1}{y\left( y+2 \right)} \end{align}. Hence, $\frac{{{y}^{-1}}-{{\left( y+2 \right)}^{-1}}}{2}$ $\frac{1}{y\left( y+2 \right)}$.