How can (1/3)x – 2 = y and (1/4)x + 11 = y be set up as a system of equations?

A. 3y – x = –6
4y – x = 44

B. 3y + x = –6
4y + x = 44

C. 3y – 3x = –6
4y – 4x = 44

D. 3y + 3x = –6
4y + 4x = 44

Question

How can (1/3)x – 2 = y and (1/4)x + 11 = y be set up as a system of equations?

A. 3y – x = –6
4y – x = 44

B. 3y + x = –6
4y + x = 44

C. 3y – 3x = –6
4y – 4x = 44
	
D. 3y + 3x = –6
4y + 4x = 44

vera_ewing · Answer

I think it's C

unklerhaukus · Answer

hmm, are they the only options?

vera_ewing · Answer

yes those are the only options. i tried to get help yesterday, but nincompoop said all of the choices were wrong so I'm confused :/

unklerhaukus · Answer

\(\tfrac13x-2=y\]
multiplying the first equation by three
$$x-6=3y$$
rearranging
$$x-3y=6$$
 multiply by negative one, to get a match with one of the equations in the options

vera_ewing · Answer

So it must be A?

unklerhaukus · Answer

similarly, for the second equation
$$\tfrac14 x + 11 = y $$
multiply by the denominator (four in this case)

and rearrange to get the constants to the right-hand-side

vera_ewing · Answer

So the answer is A?

unklerhaukus · Answer

yep

vera_ewing · Answer

thanks! can you help me with a couple more?