7x - 3y = 4 2x - 4y = 1 which of the following systems is not the same as solving the system shown? 28x - 12y = 16 and -6x + 12y = -3 14x - 6y = 4 and -14x + 28y = 1 -28x + 12y = -16 and 28x - 56y = 14

Original equations : 7x - 3y = 4    2x - 4y = 1

Set of equations #1 : 28x - 12y = 16    -6x + 12y = -3

Set of equations #2 :  14x - 6y = 4    -14x + 28y = 1

Set of equations #3 : -28x + 12y = -16    28x - 56y = 14

OK, so first step is to check all of the answer choices, one by one, and see which one is incorrect.

Set #1 : For the first equation, divide the whole thin by 4, and you get the original equation. For the 2nd equation, divide the whole thing by 3, and get the original equation, so by doing this, we know that Set #1 is equal to the original equation, therefore Set #1 is not the right answer.

Set #2 : Do the same for the first equation, find a number that is divisible by all of the numbers in the original equation & with the set #2 equations. We see that if we divide those numbers, it does not equal the original equations. Therefore, Set #2 is not equal to the original equations, so Set #2 is the right answer. 

Because we see that Set #2 is correct, there is no need to check Set #3. However, if you would like to, just go through the steps again.

~Hope I helped!~