Verified Answer :

\(\frac{15}{y}=\frac{40}{12} \\\\ y=\frac{15*12}{40} \\\\ y=\frac{15*3}{10} \\\\ y=\frac{3*3}{2} \\\\ \boxed{y=\frac{9}{2}}\)

\(\frac{\big{15}}{\big{y}} = \frac{\big{40}}{\big{12}}\ \ \ \Rightarrow\ \ \ the\ domain:\ \ \ D=R-\{0\} \\\\ \frac{\big{15}}{\big{y}} = \frac{\big{4\cdot10}}{\big{4\cdot4}} \\\\ \frac{\big{15}}{\big{y}} = \frac{\big{10}}{\big{3}} \ \ \ \Leftrightarrow\ \ \ 15\cdot3=10\cdot y\\\\45=10y\ /:10\\\\4.5=y\\\\4.5\in D\\\\Ans.\ \ y=4.5\)