Question

How can 1/4x-3=1/2x+8 be set up as a system of equations??

A) 4y+4x=-12
2y+2x=16

B) 4y-x=-12
2y-x=16

C) 4y+x=-12
2y+x=16

D) 4y-4x=-12
2y-2x=16

Answer 1

we are given

$\frac{1}{4} x-3=\frac{1}{2} x+8$

Let's assume it is equal to y

$\frac{1}{4} x-3=\frac{1}{2} x+8=y$

so, we can write

$y=\frac{1}{4} x-3$

we can multiply both sides by 4

$4y=4*\frac{1}{4} x-4*3$

$4y=x-12$

$4y-x=-12$

so, we got

first equation is

$4y-x=-12$

now, we can find second equation

$y=\frac{1}{2} x+8$

we can multiply both sides by 2

$2y=2*\frac{1}{2} x+2*8$

$2y= x+16$

$2y-x=16$

so, we got

system of equations as

$2y-x=16$

[tex] 4y-x=-12 [/tex

so, option-B..............Answer

Answer 2

B, is the final answer!!! :)