We have the following expression:
[tex]9-\frac{y}{\frac{5}{9}}[/tex]
If we substitute y=1/3, we have
[tex]9-\frac{\frac{1}{3}}{\frac{5}{9}}[/tex]
By applying the sandwhich rule
on the second term, we have
[tex]\begin{gathered} \frac{\frac{1}{3}}{\frac{5}{9}}=\frac{1\cdot9}{3\cdot5}=\frac{9}{15} \\ \text{which is equal to } \\ \frac{3}{5} \end{gathered}[/tex]
Then, our expression can be written as
[tex]9-\frac{3}{5}[/tex]
Now, 9 is equal to 45/5, then this last result is equal to
[tex]9-\frac{3}{5}=\frac{45}{5}-\frac{3}{5}=\frac{45-3}{5}=\frac{42}{5}[/tex]
That is, the answer is
[tex]\frac{42}{5}[/tex]
