The given functions are
[tex]\begin{gathered} f(t)=\frac{11}{t-2} \\ g(t)=\frac{22t}{t^2-4} \end{gathered}[/tex]
The first part consists in adding the functions f(t) + g(t)
[tex]f(t)+g(t)=\frac{11}{t-2}+\frac{22t}{t^2-4}=\frac{11(t^2-4)+22t(t-2)}{(t-2)(t^2-4)}[/tex]
Let's factor the square difference
[tex]\frac{11(t+2)(t-2)+22t(t-2)}{(t-2)(t^2-4)}=\frac{11(t+2)+22t}{t^2-4}[/tex]
Now, we simplify
[tex]\frac{11t+22+22t}{t^2-4}=\frac{33t+22}{t^2-4}[/tex]
Hence, the addition of the given functions is
[tex]f(t)+g(t)=\frac{33t+22}{t^2-4}[/tex]
On other hand, let's solve the difference
[tex]\begin{gathered} f(t)-g(t)=\frac{11}{t-2}-\frac{22t}{t^2-4}=\frac{11(t^2-4)-22t(t-2)}{(t-2)(t^2-4)} \\ f(t)-g(t)=\frac{11(t+2)(t-2)-22t(t-2)}{(t-2)(t^2-4)} \\ f(t)-g(t)=\frac{11(t+2)-22t}{t^2-4}=\frac{11t+2-22t}{t^2-4} \end{gathered}[/tex]
Hence, the difference is
[tex]f(t)-g(t)=\frac{-11t+2}{t^2-4}[/tex]
