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Find all real constants k such that the system

x+3y=kx

3x+y=ky
has a solution other than (x,y)=(0,0)

 Sep 7, 2020
 #1
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oops this wasn't supposed to be off topic

 Sep 7, 2020
 #2
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If you take x = y = 1, then you get 4 = k, so k = 4 is the only solution.

 Sep 7, 2020
 #3
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Hmm!   What about k = -2?

 

Rearrange to get 

3y = (k-1)x  and

(k-1)y = 3x

 

Divide one by the other, the x and y cancel, and you will be left with a quadratic in k.

Alan  Sep 7, 2020
edited by Alan  Sep 7, 2020
 #4
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Exactly right! laugh

joliel3  Sep 7, 2020

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