Prealgebra (7th Edition)

$3\frac{3}{5}$
When $x=\frac{3}{4}$ and $y=-\frac{4}{7}$ $\frac{\frac{9}{14}}{x+y}$ $=\frac{9}{14}\div(x+y)$ $=\frac{9}{14}\div[(\frac{3}{4}+(-\frac{4}{7})]$ $=\frac{9}{14}\div(\frac{3}{4}-\frac{4}{7})$ $=\frac{9}{14}\div(\frac{21}{28}-\frac{16}{28})$ $=\frac{9}{14}\div\frac{5}{28}$ $=\frac{9}{14}\times\frac{28}{5}$ $=\frac{9}{1}\times\frac{2}{5}$ $=\frac{18}{5}$ $=3\frac{3}{5}$