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Mathematics 22 Online
OpenStudy (anonymous):

3x-2y=4; 6x+2y=k; Find k ( unique solution)

OpenStudy (tkhunny):

How would you solve this if instead if 'k', there were a number sitting there?

OpenStudy (anonymous):

Elimination. But the only problem here is that for unique solution as far as i know a1/a2 not equal to b1/b2 . with which i can conclude that k can take any value other than the answer i get. But here k is in the position of c2. so i cant procede!

OpenStudy (goldphenoix):

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OpenStudy (goldphenoix):

K = 3x +4. (Maybe) Not sure if this is how you do it.

OpenStudy (tkhunny):

Why not? Add them to get 9x = k+4. What's the problem with that?

OpenStudy (anonymous):

@GoldPhenoix - I am aware of this method...... but i am asked to use the a1/a2 formulaeto equate for K.

OpenStudy (jack1):

@tkhunny is it possible to get an answer given that there are 3 unknows and only 2 eqns...?

OpenStudy (goldphenoix):

Oh. I didn't see it saying unique solution. (Even though I don't know what that means)

OpenStudy (anonymous):

No probs Thanks a lot for your reply! :)

OpenStudy (tkhunny):

I'm not seeing the sense of it. If there were a 'k' on the left hand side, it would be something.

OpenStudy (anonymous):

@tkhunny - From yours and @GoldPhenoix's reply the value for k is also an equation but i need answer like the given system of equations will have a unique solution for all real values of k other than (whatever answer i get)

OpenStudy (tkhunny):

Well, it is the case that solutions are of this form \(\left(\dfrac{k+4}{9},\dfrac{2k-4}{6}\right)\). There is not a value for 'k' that does not provide a complete solution. I'm not sure where that leaves us, except wondering if there is more information in the problem statement.

OpenStudy (anonymous):

Ya, Probably! Thanks a lot!! :D :) Will continue on my assignments!! Cheers. :D.

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