My Math Forum Am I right to assume this?

 Algebra Pre-Algebra and Basic Algebra Math Forum

January 6th, 2019, 03:47 PM   #11
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From: Arizona

Posts: 209
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Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.
Quote:
 Originally Posted by skipjack 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.

Last edited by ProofOfALifetime; January 6th, 2019 at 04:05 PM.

January 6th, 2019, 04:00 PM   #12
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Joined: Nov 2013

Posts: 28
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Quote:
 Originally Posted by ProofOfALifetime 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.

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.
Quote:
 Originally Posted by Jomo 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!

 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 $\displaystyle -\sqrt{3}  January 6th, 2019, 04:25 PM #15 Senior Member Joined: Oct 2016 From: Arizona Posts: 209 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. Hi everyone. What I posted was not the original problem statement, it was literally just a clarification of something I almost assumed, and I appreciate all of the help that I got. I'm NOT SAYING that my solution is the only solution to that equation, but it turns out that it is the only solution to the original problem! THANKS! Happy New Year to everyone! January 6th, 2019, 04:26 PM #16 Senior Member Joined: Oct 2016 From: Arizona Posts: 209 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. Quote:  Originally Posted by mrtwhs If post #1 is the correct statement of the problem, then the solution is ...$\displaystyle x=\dfrac{3a}{8}-\dfrac{1}{8a}$and$\displaystyle y=\dfrac{3}{8a}-\dfrac{a}{8}$for all values$\displaystyle -\sqrt{3}
Yes, great job! Thanks!

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