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$x$ and $y$ are both positive real numbers :

My attempt:

$$x^{2}+y^{2}=6xy \Leftrightarrow x^{2}+y^{2}-2xy=4xy \Leftrightarrow (x-y)^{2}=4xy \Leftrightarrow (x-y)=\frac{4xy}{x-y} $$ A little note here: $x$ doesn’t equal to $y$.

And now let’s plug this result to $\frac{x+y}{x-y} $: $$\frac{x+y}{x-y} = (x+y) \cdot \frac{x-y}{4xy}=\frac{x^{2}-y^{2}}{4xy}=\frac{x}{4y}-\frac{y}{4x}$$ And the problem here is to figure out the value of: $$\frac{x}{y}-\frac{y}{x} $$ Thank you for your help

PNT
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  • You could mention (somewhere in your proof) that $x=y$ in $x^2+y^2=6xy$ iff $x=y=0$, in which case $\frac{x+y}{x-y}$ is an indeterminate form. – PM 2Ring Jan 24 '21 at 19:53

5 Answers5

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Note that $$(x+y)^2-(x-y)^2=4xy$$

Use this in your equation $$(x-y)^2=4xy$$ $$(x-y)^2=(x+y)^2-(x-y)^2$$ $$\Rightarrow \dfrac{(x+y)^2}{(x-y)^2}=2$$ $$\Rightarrow \dfrac{x+y}{x-y}=\pm \sqrt{2}$$

Here both positive and negative values are possible depending on whether $x$ is larger or smaller than $y$.

cosmo5
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  • Note: You should make an argument for why that expression can take on both values. IE So far, you've only shown that it could take on either of those values, but not that it must take on either of those values. – Calvin Lin Jan 24 '21 at 19:27
  • @CalvinLin Yes. I did, thankyou :) – cosmo5 Jan 24 '21 at 19:31
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Hint: Treat the original equation as a quadratic in $x/y$, so you can find that value (2 possible values, corresponding to $\pm$).
Then, express the desired expression in term of $x/y$, and hence find the expression. (Again, there are 2 possible values).

Calvin Lin
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Another approach:

What is the numerator and denominator of $ \left( \frac{x+y}{x-y} \right) ^ 2$ in terms of $xy$?

Now, because $x,y$ can be swapped, explain why the expression can take on both positive and negative values.

Calvin Lin
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Proceed as you did: \begin{equation} x^2+y^2=6xy\Longrightarrow\left\{\begin{array}{l} x^2+y^2+2xy=8xy\Longrightarrow(x+y)^2=8xy,\\ x^2+y^2-2xy=4xy\Longrightarrow(x-y)^2=4xy. \end{array}\right. \end{equation} Then divide both right-side equations and you have it. $$\left(\frac{x+y}{x-y}\right)^2=\frac{8xy}{4xy}=2\Longrightarrow\frac{x+y}{x-y}=\pm\sqrt{2}.$$ Then, $$x>y\Longrightarrow\frac{x+y}{x-y}=\sqrt{2},\quad x<y\Longrightarrow\frac{x+y}{x-y}=-\sqrt{2}.$$

Juan Castaño
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  • Not quite. You're still missing the final step of verifying that these are indeed solutions (IE See my solution). Always be careful when manipulating non-linear system of equations, as you may have introduced extraneous solutions. – Calvin Lin Jan 24 '21 at 19:26
  • @Juan Castano you are wrong $(x-y)^{2}=4xy$ – PNT Jan 24 '21 at 19:33
  • @ Juan Castano Can you add the solutions to your answer – PNT Jan 24 '21 at 19:38
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hint $$x^2+y^2+2xy=6xy+2xy$$

$$\implies (x+y)^2=8xy$$ $$\implies (x+y)=\pm 2\sqrt{2xy}$$ and $$x^2+y^2-2xy=6xy-2xy$$ $$\implies (x-y)^2=4xy$$

hamam_Abdallah
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