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EuRa · Feb 17, 2009, 02:34 PM

OK NM, I figured out the problem, ty.

ebaines · Feb 17, 2009, 02:57 PM

Hello Mike. The short way to solve: recognize that the only way A could have a known value is if the third equation is a linear combination of the first two. So, you show that it isn't, and hence conclude that the value for A can't be determined from the data you've been given.

The long way: try one of the proposed values for A and solve the 3 equations in 3 unknowns to determine X, Y and Z. Then do it again with a different value of A, and see that the values of X, Y and Z are different. Hence A cannot be determined.

EuRa · Feb 17, 2009, 03:18 PM

Nm

ebaines · Feb 17, 2009, 03:40 PM

Originally Posted by EuRa

Actually it can be solved. I will show you how. Feel free to take another crack at it.

Oops - I now see that the third equation is indeed a linear combination of the first two, if A = -6. Did this by calculating the determinant of the matrix of coefficients and setting it = 0 then solve for A.

EuRa · Feb 17, 2009, 03:40 PM

Actually... wait a minute. I thought I solved it but then I tried it again, and I was wrong. Maybe I didn't... ooh you might be right!

EuRa · Feb 17, 2009, 03:55 PM

Omg ebaines.. I think you were right the first time I believe! I thought I solved it, and I just tried it out, and I was entirely WRONG! Ty!!