## College Algebra 7th Edition

$\frac{(x+y)^{2}}{xy}$
We multiply through by $xy(x+y)$ to get rid of excess fractions. Then we simplify: $\displaystyle \frac{x^{-1}+y^{-1}}{(x+y)^{-1}}=\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x+y}}=\frac{\frac{1}{x}+\frac{1}{y}}{\frac{1}{x+y}}*\displaystyle \frac{xy(x+y)}{xy(x+y)}=\frac{y(x+y)+x(x+y)}{xy}=\frac{xy+y^{2}+x^{2}+xy}{xy}=\frac{x^{2}+2xy+y^{2}}{xy}=\frac{(x+y)^{2}}{xy}$