+96 Find all real constants k such that the system x+3y=kx and 3x+y=ky has a solution other than (x,y)=(0,0)
I'm really stuck and I tried 4 but it wasn't the right answer.
+37146 How did you get '4' Did you get another value for 'k' besides 4 ??? (Like when you factored a quadratic for k )
+569 First, I solved for y, \(y = \frac{kx - x}{3} = \frac{k - 1}{3} x.\)
Substituting is next, \(3x + \frac{k - 1}{3} x = k \cdot \frac{k - 1}{3} x.\)
Solve, and get values of k are 4, -2.
:)
+569 I think that should be right since I remember doing it a while ago in one of my courses, and here, geno also got the same answer...
https://web2.0calc.com/questions/halpp_3
Great!
+37146 Yah....that is why I was asking how kanisan got only '4'.....I wanted to see k's work so far....
I got 4 and -2 also.... ~EP
+118677 How do you know kanisan tried?
Maybe 4 was given to them by someone else (likely a dummy answer here).
Kanison did not state how the 4 was derived.
EP was right to ask.
And it is quite annoying when our 'teaching' answers are overridden.
Especially when the original person has not even responded to our response question yet.
