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.

kanisan Aug 16, 2020

#1**+1 **

How did you get '4' Did you get another value for 'k' besides 4 ??? (Like when you factored a quadratic for k )

ElectricPavlov Aug 16, 2020

#2**+1 **

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.

:)

iamhappy Aug 16, 2020

#3**+1 **

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!

iamhappy
Aug 16, 2020

#4**+1 **

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

ElectricPavlov Aug 17, 2020

#6**+2 **

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.

Melody
Aug 19, 2020