7x - 3y = 4 2x - 4y = 1 Which of the following system of equations is not equal to the system of equations shown above? 28x - 12y = 16 and -6x + 12y = -3 14x - 6y = 4 and -14x + 28y = 1 -28x + 12y = -16 and 28x - 56y = 14
\(\color{#000000 }{ \displaystyle 7x - 3y = 4 }\) \(\color{#000000 }{ \displaystyle 2x - 4y = 1 }\) I would multiply the first equation times 4, and second equation times -3.
How did you come up with that?
I want to make the y-coeffients equivalent, but one of them negative. I want to do this, so that the y goes away when I add the equations.
is this reasonable, or are you still confused?
Ohhh
This is very very reasonable... these are new equation..
ok, when system of equations do you get when you multiply as I directed?
28x -12y = 16 -6x + 12y = -3
\(\color{#000000 }{ \displaystyle 28x - 12y = 16 }\) \(\color{#000000 }{ \displaystyle -6x +12 y =-3 }\) good!
Oh look! That's the first option!
now, add the equations:)
I think I understand it now :D
ok, good:)
So...
now let's refer to the question again
7x - 3y = 4 2x - 4y = 1 Which of the following system of equations is not equal to the system of equations shown above? 28x - 12y = 16 and -6x + 12y = -3 14x - 6y = 4 and -14x + 28y = 1 -28x + 12y = -16 and 28x - 56y = 14
It's saying, which is NOT equal
Is it the middle one?
well, if you were to attempt to eliminate the x, what would you do with your system?
times what number would you multiply equation 1 and equation 2?
I would.... multiply the first equation by 2 and the second by ..
I'm not sure :(
and the second by 7?
yes, but, by -7, so that you have 14x in equation 1 and -14x in equation 2 (and addition will cancel them)
OHHH! lol you're a good teacher. What grade are u in?
I am in college
That makes sense. I am in grade 10 lol
Thank you !! :D
You are welcome!
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