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January 6th, 2019, 03:47 PM   #11
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What was the original problem that you were working on?
I solved it. The solution I found is the only solution TO THE ORIGINAL PROBLEM, but I had to use a different method to show it. I can show you through PM.

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January 6th, 2019, 04:00 PM   #12
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I solved it. The solution I found is the only solution, but I had to use a different method to show it. I can show you through PM.
So what romsek showed you must be wrong. Can you find the error in that post? I doubt it.
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January 6th, 2019, 04:03 PM   #13
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So what romsek showed you must be wrong. Can you find the error in that post? I doubt it.
Hi, I solved the original problem I was working on, but thanks for going out of your way to be rude!
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January 6th, 2019, 04:21 PM   #14
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If post #1 is the correct statement of the problem, then the solution is ...

$\displaystyle x=\dfrac{3a}{8}-\dfrac{1}{8a}$ and $\displaystyle y=\dfrac{3}{8a}-\dfrac{a}{8}$

for all values $\displaystyle -\sqrt{3}<a<\dfrac{-1}{\sqrt{3}}$ and $\displaystyle \dfrac{1}{\sqrt{3}}<a<\sqrt{3}$.

That is, there are an infinite number of solutions.
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January 6th, 2019, 04:25 PM   #15
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Hi everyone. What I posted was not the original problem statement, it was literally just a clarification of something I almost assumed, and I appreciate all of the help that I got.

I'm NOT SAYING that my solution is the only solution to that equation, but it turns out that it is the only solution to the original problem! THANKS!

Happy New Year to everyone!
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January 6th, 2019, 04:26 PM   #16
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Originally Posted by mrtwhs View Post
If post #1 is the correct statement of the problem, then the solution is ...

$\displaystyle x=\dfrac{3a}{8}-\dfrac{1}{8a}$ and $\displaystyle y=\dfrac{3}{8a}-\dfrac{a}{8}$

for all values $\displaystyle -\sqrt{3}<a<\dfrac{-1}{\sqrt{3}}$ and $\displaystyle \dfrac{1}{\sqrt{3}}<a<\sqrt{3}$.

That is, there are an infinite number of solutions.
Yes, great job! Thanks!
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