Can someone please answer this for me. I give medals.
How can one fourthx − 3 = one halfx + 8 be set up as a system of equations?

4y + 4x = −12
2y + 2x = 16

4y − x = −12
2y − x = 16

4y + x = −12
2y + x = 16

4y − 4x = −12
2y − 2x = 16

Question

Can someone please answer this for me. I give medals.
How can one fourthx − 3 = one halfx + 8 be set up as a system of equations? 

 4y + 4x = −12
2y + 2x = 16 

 4y − x = −12
2y − x = 16 

 4y + x = −12
2y + x = 16 

 4y − 4x = −12
2y − 2x = 16

gavin39 · Answer

really need an answer, if anyone could please tell me.

anonymous · Answer

Strange question.

You have only a single variable, and you have two functions that equal one another. The system required to solve this is exactly what you have.

$$\frac{1}{4}x - 3 = \frac{1}{2}x+8$$

I suppose you could set each equal to "y" and form them that way:

$$y = \frac{1}{4}x - 3$$
$$y= \frac{1}{2}x+8$$

Which you can then rearrange to:

$$y - \frac{1}{4}x = -3 \rightarrow 4y-x=-12$$
$$y-\frac{1}{2}x = 8 \rightarrow 2y-x = 16$$