The given expression is
[tex]\frac{1}{b}+\frac{1}{9}=\frac{1}{t}[/tex]First, we multiply the equation by t
[tex]\begin{gathered} (\frac{1}{b}+\frac{1}{9})\cdot t=\frac{1}{t}\cdot t \\ (\frac{1}{b}+\frac{1}{9})\cdot t=1 \end{gathered}[/tex]Now, we divide the equation by 1/b + 1/9
[tex]\begin{gathered} \frac{(\frac{1}{b}+\frac{1}{9})\cdot t}{(\frac{1}{b}+\frac{1}{9})}=\frac{1}{(\frac{1}{b}+\frac{1}{9})} \\ t=\frac{1}{(\frac{1}{b}+\frac{1}{9})} \end{gathered}[/tex]Now, we sum fractions
[tex]t=\frac{1}{\frac{9+b}{9b}}[/tex]Then, we solve this combined fraction
[tex]t=\frac{9b\cdot1}{9+b}=\frac{9b}{9+b}[/tex]Therefore, the final expression is
[tex]t=\frac{9b}{9+b}[/tex]