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 January 6th, 2019, 11:30 AM #1 Senior Member     Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 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: iPhone Posts: 436 Thanks: 68 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: 193 Thanks: 34 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: 2,370 Thanks: 1274 $1 = 3x^2 + 10xy + 3y^2$ $3x^2 + 10 xy + 3y^2 -1 = 0$ $x = \dfrac{-10y \pm \sqrt{100y^2 -12(3y^2-1)}}{6}$ $x = \dfrac{-10y\pm\sqrt{64y^2+12}}{6}$ $x = \dfrac{-5y\pm \sqrt{16y^2+3}}{3}$ Thanks from topsquark and ProofOfALifetime
 January 6th, 2019, 12:08 PM #5 Senior Member     Joined: Oct 2016 From: Arizona Posts: 193 Thanks: 34 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
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 Originally Posted by ProofOfALifetime 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?
$\frac12 \times 2 = 1$ too. And $\frac13 \times 3 = 1$. And...

January 6th, 2019, 12:24 PM   #7
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 Originally Posted by v8archie $\frac12 \times 2 = 1$ too. And $\frac13 \times 3 = 1$. And...
I'm thinking that those other options will result in nonpositive values for either $x$ or $y$, but I have more work to do. Thanks, you are right, there are many options for the factors!

January 6th, 2019, 12:26 PM   #8
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Quote:
 Originally Posted by ProofOfALifetime Hi everyone. So, I've been working on a problem, and I ended up with the equation $$1=(3x+y)(x+3y).$$ 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 $1$.
Frankly, I do not see why 3x + y = 1 = x + 3y.

$3x + y = a \ne 0 \text { and } x + 3y = \dfrac{1}{a} \implies (3x + y)(x + 3y) = 1.$

Last edited by JeffM1; January 6th, 2019 at 12:51 PM.

January 6th, 2019, 12:37 PM   #9
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Math Focus: I'm still deciding, but my recent focus has been olympiad problems and math journal problems.
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 Originally Posted by JeffM1 Frankly, I do not see why 3x + y = 1 = x + 3y. $3x + y = a \ne 0 \text { and } x + 3y = \dfrac{1}{a} \implies (3x + y)(x + 3y) = 1.$
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: 20,370 Thanks: 2007 What was the original problem that you were working on?

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