Solve the following system of equations on $\mathbb R$

\begin{cases} \dfrac{x+y}{1+xy}= \dfrac{1-2y}{2-y}\\[6px] \dfrac{x-y}{1-xy}=\dfrac{ 1-3x}{3-x} \end{cases}


Setting $ \begin{cases} x=\dfrac{u-1}{u+1} \\[4px] y= \dfrac{v-1}{v+1}\end{cases}$

My question: How did they come up with that idea? How about your solution? :)

Thank you in advance !

  • 1
    $\begingroup$ Not sure if this is useful but if we substitute $x = \tan \alpha$ and $y = \tan \beta$, the LHS of the two equations become $\tan(\alpha + \beta)$ and $\tan(\alpha - \beta)$. We would also have that $\tan \alpha = \pi/4 + \tan^{-1} u$. Similarly for $\tan \beta$ $\endgroup$ – iamwhoiam Jul 23 '18 at 10:47
  • $\begingroup$ I guess $\tan(\alpha +\beta)$ would be $\frac{\tan(\alpha)+\tan(\beta)}{1\textbf{-}\tan(\alpha)\tan(\beta)}$ and not $\frac{\tan(\alpha)+\tan(\beta)}{1\textbf{+}\tan(\alpha)\tan(\beta)}$. $\endgroup$ – mrtaurho Jul 23 '18 at 11:16

The substitution reminds the fact that $$ \frac{e^{2t}-1}{e^{2t}+1}=\frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}=\tanh t $$ plus the fact that $$ \tanh(p\pm q)=\frac{\tanh p\pm\tanh q}{1\pm\tanh p\tanh q} $$ so this is essentially the same as setting $x=\tanh p$ and $y=\tanh q$.


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