\frac{7}{9 \frac{5}{11}} = \frac{ \frac{33}{104}}{y - 3.5} \\ 7(y - 3.5) = 9 \frac{5}{11} \times \frac{33}{104} \\ 7y - 24.5 = \frac{104}{11} \times \frac{33}{104} \\ 7y = 3 + 24.5 \\ 7y = 27.5 \\ y = \frac{275}{10 \times 7} = \frac{275}{70} = 3 \frac{65}{70} = 3 \frac{13}{14}