It is given that 21x + 6y = k and (7x-20)^2 +(6x+10)^2 =200.

[maninthemirror]

Hi all,

Need your kind help for this question:

It is given that 21x + 6y = k and (7x-20)^2 +(6x+10)^2 =200.

a) If k =1, explain how you could tell that there are no solutions without actually solving for the equations.

b)Explain why there cannot be more than 2 solutions for all values of k.


Your kind help is deeply appreciated.

Thanks.

 

[Deleted member 4993]

Hi all,

Need your kind help for this question:

It is given that 21x + 6y = k and (7x-20)^2 +(6x+10)^2 =200.

a) If k =1, explain how you could tell that there are no solutions without actually solving for the equations.

b)Explain why there cannot be more than 2 solutions for all values of k.


Your kind help is deeply appreciated.

Thanks.

You do not have any y-component in this equation!! Please confirm or edit your post.

 

[Deleted member 4993]

Hi all,

Need your kind help for this question:

It is given that 21x + 6y = k and (7x-20)^2 +(6x+10)^2 =200.

a) If k =1, explain how you could tell that there are no solutions without actually solving for the equations.

b)Explain why there cannot be more than 2 solutions for all values of k.


Your kind help is deeply appreciated.

Thanks.

Asked and answered at:

http://math.stackexchange.com/quest...question-it-is-given-that-21x-6y-k-and-7x-202