Given the system of equations
[tex]\begin{gathered} x=\frac{1}{4}y+2 \\ -4x+3y=4 \end{gathered}[/tex]Substitute x =1/4y+2 in the second equation
[tex]-4(\frac{1}{4}y+2)+3y=4[/tex]Simplify
[tex]\begin{gathered} -y-8+3y=4 \\ 2y-8=4 \end{gathered}[/tex]Solve for y
[tex]\begin{gathered} 2y-8+8=4+8 \\ 2y=12 \\ \frac{2y}{2}=\frac{12}{2} \\ y=6 \end{gathered}[/tex]Then substitute y = 6 in x
[tex]x=\frac{1}{4}(6)+2=\frac{6}{4}+2=\frac{7}{2}[/tex]Answer: (7/2, 6)