we are given
[tex] \frac{1}{4} x-3=\frac{1}{2} x+8 [/tex]
Let's assume it is equal to y
[tex] \frac{1}{4} x-3=\frac{1}{2} x+8=y [/tex]
so, we can write
[tex] y=\frac{1}{4} x-3 [/tex]
we can multiply both sides by 4
[tex] 4y=4*\frac{1}{4} x-4*3 [/tex]
[tex] 4y=x-12 [/tex]
[tex] 4y-x=-12 [/tex]
so, we got
first equation is
[tex] 4y-x=-12 [/tex]
now, we can find second equation
[tex] y=\frac{1}{2} x+8 [/tex]
we can multiply both sides by 2
[tex] 2y=2*\frac{1}{2} x+2*8 [/tex]
[tex] 2y= x+16 [/tex]
[tex] 2y-x=16 [/tex]
so, we got
system of equations as
[tex] 2y-x=16 [/tex]
[tex] 4y-x=-12 [/tex
so, option-B..............Answer
