Solution
Let y be the number of students involved.
let x be the years.
when
x = 1, y = 603
x = 3, y = 12545
Since the increase is linear,
[tex]\begin{gathered} \frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1} \\ \\ \Rightarrow\frac{y-603}{x-1}=\frac{12545-603}{3-1} \\ \\ \Rightarrow\frac{y-603}{x-1}=\frac{11942}{2} \\ \\ \Rightarrow\frac{y-603}{x-1}=5971 \\ \\ \Rightarrow y-603=(x-1)\times5971 \\ \\ \Rightarrow y-603=5971x-5971 \\ \\ \Rightarrow y=5971x-5971+603 \\ \\ \Rightarrow y=5971x-5368 \end{gathered}[/tex]
At the end of 5 years,
That is, when x = 5
[tex]y=5971(5)-5368=24487[/tex]