January 6th, 2019, 03:47 PM  #11 
Senior Member Joined: Oct 2016 From: Arizona Posts: 209 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  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.
Last edited by ProofOfALifetime; January 6th, 2019 at 04:05 PM. 
January 6th, 2019, 04:00 PM  #12 
Newbie Joined: Nov 2013 Posts: 28 Thanks: 8  
January 6th, 2019, 04:03 PM  #13 
Senior Member Joined: Oct 2016 From: Arizona Posts: 209 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  
January 6th, 2019, 04:21 PM  #14 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 
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. 
January 6th, 2019, 04:25 PM  #15 
Senior Member Joined: Oct 2016 From: Arizona Posts: 209 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. 
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! 
January 6th, 2019, 04:26 PM  #16  
Senior Member Joined: Oct 2016 From: Arizona Posts: 209 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  Quote:
 

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