Vectors, n-tuples, and headaches oh my

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SUMMARY

The discussion revolves around solving for the numerical values of x and y in the equations derived from the linear independence of n-tuples A and B. The equations presented are C = (x-3)A - yB and C = -yA - (x+2)B. Participants concluded that by equating coefficients, two equations can be formed: (x-3) = -y and -y = -(x+2), leading to a straightforward solution for x and y. The key takeaway is the importance of recognizing linear independence to simplify the problem-solving process.

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Homework Statement


An n-tuple C has two representation in terms of the linearly independent pair A,B
C = (x-3)A - YB
C = -yA - (x+2)B

Solve for x and y.

Homework Equations


aA+bB+cC = 0
where A,B,C are linearly dependent n-tuples and a,b,c are real numbers that don't all equal 0

The Attempt at a Solution


I tried this problem about a billion different ways; however, none yielded promising results. I know the answer, and it is numerical! I don't want x and y in terms of A and B; they have numerical values. It would be helpful to explain what you're doing as you do it, but I can probably figure it out if you just show the work; laziness is something that I understand well. :smile:
 
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I'm not sure I get why you are confused. If A and B are linearly independent and (x-3)A-yB=-yA-(x+2)B, doesn't that mean (x-3)=(-y) and (-y)=(-(x+2))? That's two equations in two unknowns. Just solve them.
 
I guess I was over-thinking it; thanks!
 

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