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New post 21 Oct 2019, 18:43
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Given that both A,B≠0, how many solutions does this system of equations have: one, none, or infinitely many?

Ax+By=12

20x+5y=C

Statement 1: A=3B

Statement 2: C=12.
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New post 21 Oct 2019, 20:19
minustark wrote:
Given that both A,B≠0, how many solutions does this system of equations have: one, none, or infinitely many?

Ax+By=12

20x+5y=C

Statement 1: A=3B

Statement 2: C=12.



Analyzing the question:
It's an interesting question that can help students understand the concept of "linear independence". The equations above are linear and we have N equations N variables. Once we prove they are independent, we can guarantee a unique solution. To prove they are independent, we must have equations that contain different information. Hence as long as the second equation's left side isn't a multiple of the first equation's left side, the equations are independent.

Statement 1:
Confirmed the ratios are different and we have two independent, and of course linear, equations. This guarantees one unique solution, sufficient.

Statement 2:
With only C = 12 we know nothing on the first equation. We can have 20x + 5y = 12 for the first equation which results in the system simplifying to one equation of 20x + 5y = 12, which is infinitely many solutions. Insufficient.

Finally a note on how to get no solutions in a system, we basically need contradicting equations such as x + 2y = 10 and x + 2y = 0.

Ans: A
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User's self-made question   [#permalink] 21 Oct 2019, 20:19
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