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Find all real constants  such that the system

 

\(x+3y=kx,\)

\(3x+y=ky\)

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

 Aug 4, 2020
 #1
avatar+26735 
+2

From first equation   x = -3y/(1-k)    sub this in to the second equation

 

3 (-3y)/(1-k)  + y(1-k) = 0

-9y + y (1-k)^2 = 0

y ( k^2-2k-8) = 0                  already stated y cannot be 0

(k+2)(k-4) = 0       k = -2, 4   (if I did the math correctly !)  cheeky

 Aug 4, 2020
edited by ElectricPavlov  Aug 4, 2020
 #2
avatar+359 
+3

Thank you so much!!!

iamhappy  Aug 4, 2020

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