Let \(g(t) = \frac{2t+1}{3-t}\).
Evaluate the following:
a) \(g(2)\)
\[\begin{align*}
\frac{2t+1}{3-t} & \hspace{1.3cm}\mbox{Original Problem}\\[1ex]
\frac{2(2)+1}{3-(2)} & \hspace{1.3cm}\mbox{Substituted}\\[1ex]
\frac{5}{1} & \hspace{1.3cm}\mbox{Simplified}\\[1ex]
5 & \hspace{1.3cm}\mbox{Final Answer}\\
\end{align*}\]
b) \(g(a)\)
\[\begin{align*}
\frac{2t+1}{3-t} & \hspace{1.3cm}\mbox{Original Problem}\\[1ex]
\frac{2(a)+1}{3-(a)} & \hspace{1.3cm}\mbox{Substituted}\\[1ex]
\frac{2a+1}{3-a} & \hspace{1.3cm}\mbox{Simplified, Final Answer}\\
\end{align*}\]
c) \(g(a-4)\)
\[\begin{align*}
\frac{2t+1}{3-t} & \hspace{1.3cm}\mbox{Original Problem}\\[1ex]
\frac{2(a-4)+1}{3-(a-4)} & \hspace{1.3cm}\mbox{Substituted}\\[1.5ex]
\frac{2a-8+1}{3-a+4} & \hspace{1.3cm}\mbox{Distributed}\\[1.5ex]
\frac{2a-7}{7-a} & \hspace{1.3cm}\mbox{Simplified, Final Answer}\\
\end{align*}\]