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# help asap algebra

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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.

Aug 16, 2020

#1
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How did you get '4'       Did you get another value for 'k' besides 4 ???   (Like when you factored a quadratic for k )

Aug 16, 2020
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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.

:)

Aug 16, 2020
edited by iamhappy  Aug 16, 2020
#3
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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
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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

Aug 17, 2020
#5
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yeah...

Maybe I shouldn't have given kenisen the answer, but I mean he/she did get half of the answer, so i guess, he/she already tried and did their work, not only seeking for the answer. :)

iamhappy  Aug 17, 2020
edited by iamhappy  Aug 17, 2020
#6
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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
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ok Melody, I do understand. :) I may have assumed something that wasn't true. :( I mean, kenisan could have done it her/himself, but the way of him/her's wording did make it seem like she got the answer randomly from another forum site or something like that. :(

iamhappy  Aug 19, 2020