The given system is
[tex]\mleft\{\begin{aligned}y=\frac{1}{4}x+4 \\ 2x-8y=9\end{aligned}\mright.[/tex]To solve this system, we substitute the first equation in the second one.
[tex]2x-8(\frac{1}{4}x+4)=9[/tex]Now, we solve for x.
[tex]\begin{gathered} 2x-\frac{8}{4}x-32=9 \\ 2x-2x=9+32 \\ 0=41 \end{gathered}[/tex]In other words, the system represents parallel lines.