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

B, is the final answer!!! :)

Answer 2

we are given

`(1)/(4) x-3=(1)/(2) x+8`

Let's assume it is equal to y

`(1)/(4) x-3=(1)/(2) x+8=y`

so, we can write

`y=(1)/(4) x-3`

we can multiply both sides by 4

`4y=4*(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=(1)/(2) x+8`

we can multiply both sides by 2

`2y=2*(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