Question 549766
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You have worded the question incorrectly because if *[tex \Large k\ =\ 0] or *[tex \Large k\ =\ 2] then the given system is either consistent/dependent (same line) or inconsistent (parallel lines).


Calculate the determinant:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left|\begin{array}k&k\\4&2k\end{array}\right|\ =\ 2k^2\ -\ 4k]


If the determinant is zero, then the system is either consistent/dependent (same line) or inconsistent (parallel lines).



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2k^2\ -\ 4k\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ 0]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ k\ =\ 2]


So *[tex \Large k\ =\ 0] or *[tex \Large k\ =\ 2] yields no unique solution.


In fact, the set of values of *[tex \Large k] for which your system yields a unique solution is all real numbers EXCEPT 0 and 2.  Note, however, that the unique solution is, in every case, the origin *[tex \Large (0,0)]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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