Question 767070
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In order for there to be no solution, the ratio of the coefficient on *[tex \LARGE x] to the coefficient on *[tex \LARGE y] in one of the equations has to be the same as the same ratio for the other equation AND the constant terms in the two equations must be different.


So, since the coefficients on *[tex \LARGE x] and *[tex \LARGE y] are 1 and 1 in the first equation, the coefficients on *[tex \LARGE x] and *[tex \LARGE y] must be in the ratio 1 to 1 in the second equation.  Therefore *[tex \LARGE k^2\ -\ 3\ =\ 1].


Solving the quadratic, *[tex \LARGE k\ =\ \pm 2]


For no solution, *[tex \LARGE k\ =\ -2]


For infinite solutions, *[tex \LARGE k\ =\ 2]


For a unique solution, *[tex \LARGE k\ \in\ \mathbb{R},\ k\ \not = \ \pm 2]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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