## Precalculus (6th Edition) Blitzer

The required solution is $\frac{1}{y\left( y+5 \right)}$.
We have the given algebraic expression: $\frac{{{y}^{-1}}-{{\left( y+5 \right)}^{-1}}}{5}$. 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 $q\ne 0$. So the given expression can also be written as $\frac{{{y}^{-1}}-{{\left( y+5 \right)}^{-1}}}{5}=\frac{\frac{1}{y}-\frac{1}{y+5}}{5}$. 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 $\frac{1}{y}-\frac{1}{y+5}$. Solve the numerator of the given complex rational expression: \begin{align} & \frac{1}{y}-\frac{1}{y+5}=\frac{1}{y}\times \frac{\left( y+5 \right)}{\left( y+5 \right)}-\frac{1}{\left( y+5 \right)}\times \frac{y}{y} \\ & =\frac{y+5}{y\left( y+5 \right)}-\frac{y}{y\left( y+5 \right)} \\ & =\frac{y+5-y}{y\left( y+5 \right)} \\ & =\frac{5}{y\left( y+5 \right)} \end{align}. Therefore, the given complex rational expression becomes \begin{align} & \frac{{{y}^{-1}}-{{\left( y+5 \right)}^{-1}}}{5}=\frac{\frac{5}{y\left( y+5 \right)}}{5} \\ & =\frac{5}{y\left( y+5 \right)}\times \frac{1}{5} \\ & =\frac{1}{y\left( y+5 \right)} \end{align}. Hence, $\frac{{{y}^{-1}}-{{\left( y+5 \right)}^{-1}}}{5}=$ $\frac{1}{y\left( y+5 \right)}$.