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Solve the system of equations: $$\begin{cases}\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy} \\\left(\sqrt{3}+xy\right)^{\log_2x}+\dfrac{x}{\left(\sqrt{3}-xy\right)^{\log_2y}}= 1+\dfrac{x}{y}\end{cases}$$

My try: $\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy}\\\Leftrightarrow xy\left(x^2+y^2\right)+\left(xy\right)^2= x^2+y^2+1\\\Leftrightarrow\left(xy-1\right)\left(x^2+y^2+xy+1\right)=0\\\Leftrightarrow xy=1\,\,\,\mbox{(because:}\,\, x^2+y^2+xy+1>0)$

And here I'm stuck!

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if i answered your question then please mark it so! –  Atul Gangwar Nov 10 '13 at 6:33

1 Answer 1

up vote 1 down vote accepted

Subsitute $xy=1 , y=1/x,and \, log y=-logx$ :-

$log(3^{1/2}+1)^{logx}+ x*log(3^{1/2}-1)^{logx}=1+x^2$

then take Log at both sides of your equation then your 2nd equation becomes

$logx*log(3^{1/2}+1)+logx+ logx*log(3^{1/2}-1)=2logx$

$log(3^{1/2}+1)+ log(3^{1/2}-1)=1$


which is always true!

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of course i have leap frogged some steps. –  Atul Gangwar Nov 9 '13 at 17:45

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