January 6th, 2019, 11:30 AM  #1 
Senior Member Joined: Oct 2016 From: Arizona Posts: 207 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  Am I right to assume this?
Hi everyone. So, I've been working on a problem, and I ended up with the equation $$1=(3x+y)(x+3y)$$ where $x$ and $y$ are positive real numbers. So, I found that the solution is $(\frac{1}{4},\frac{1}{4})$ because the only way this can be true is if each factor is $1$. Am I correct that this should be the only solution? I'm basically asking, is it okay to just say, we get $(\frac{1}{4},\frac{1}{4})$ since this is only true if each factor is $1$. Last edited by ProofOfALifetime; January 6th, 2019 at 11:53 AM. 
January 6th, 2019, 11:38 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 534 Thanks: 81 
Equation has infinite pairs (x,y) $\displaystyle (\frac{1}{4} , \frac{1}{4} )$ is a solution , but not the only one 
January 6th, 2019, 11:50 AM  #3 
Senior Member Joined: Oct 2016 From: Arizona Posts: 207 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. 
Okay thank you! That makes sense.

January 6th, 2019, 12:00 PM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,452 Thanks: 1337 
$1 = 3x^2 + 10xy + 3y^2$ $3x^2 + 10 xy + 3y^2 1 = 0$ $x = \dfrac{10y \pm \sqrt{100y^2 12(3y^21)}}{6}$ $x = \dfrac{10y\pm\sqrt{64y^2+12}}{6}$ $x = \dfrac{5y\pm \sqrt{16y^2+3}}{3}$ 
January 6th, 2019, 12:08 PM  #5 
Senior Member Joined: Oct 2016 From: Arizona Posts: 207 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems. 
Thank you! I guess I will keep playing around with the original problem. You both are a big help!

January 6th, 2019, 12:13 PM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2642 Math Focus: Mainly analysis and algebra  Quote:
 
January 6th, 2019, 12:24 PM  #7 
Senior Member Joined: Oct 2016 From: Arizona Posts: 207 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  
January 6th, 2019, 12:26 PM  #8  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
$3x + y = a \ne 0 \text { and } x + 3y = \dfrac{1}{a} \implies (3x + y)(x + 3y) = 1.$ Last edited by JeffM1; January 6th, 2019 at 12:51 PM.  
January 6th, 2019, 12:37 PM  #9 
Senior Member Joined: Oct 2016 From: Arizona Posts: 207 Thanks: 37 Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.  Exactly why I asked the question! I was thinking the same thing, but since I work on these challenge problems alone, I like to get advice from people with more knowledge and experience.
Last edited by skipjack; January 6th, 2019 at 01:33 PM. 
January 6th, 2019, 01:54 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 20,746 Thanks: 2133 
What was the original problem that you were working on?


Tags 
assume 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Assume  the john  Calculus  2  March 26th, 2014 06:26 AM 
Would it be ok to assume this...  perfect_world  Calculus  4  August 17th, 2013 01:54 PM 
I assume someone already came up with this equation  Tallefred  Applied Math  2  August 13th, 2010 01:14 PM 
Would it be ok to assume this...  perfect_world  Algebra  0  December 31st, 1969 04:00 PM 