Solutions for the given system with fractions I have to solve in $\Bbb{R}$ the following system :
$$
\ \left\{
    \begin{array}{ll}
        \frac{y}{x}+\frac{x}{y}=\frac{17}{4}  \\
        x^2-y^2=25
    \end{array}
\right.$$
For this one I am stuck, I tried to use the fact that $x^2-y^2=(x-y)(x+y)$ and multiply by $x$ (or $y$) in line $1$ but fractions 'bother' me. Any hint are welcome.

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Write $\ds{x\ \mbox{and}\ y}$ as
$\ds{x = 5\cosh\pars{\theta}\,,\  y =5\sinh\pars{\theta}}$ such that the second equation is satisfied. With the first equation, we'll get:
\begin{align}
{17 \over 4}&={\sinh\pars{\theta} \over \cosh\pars{\theta}}
+ {\cosh\pars{\theta} \over \sinh\pars{\theta}}
={2\cosh\pars{2\theta} \over \sinh\pars{2\theta}}\ \imp\ \tanh\pars{2\theta}
={8 \over 17}
\end{align}

$$
\theta = \half\,{\rm arctanh}\pars{8 \over 17}
={1 \over 4}\,\ln\pars{1 + 8/17 \over 1 - 8/17}
={1 \over 4}\,\ln\pars{25 \over 9}=
\ln\pars{\root{5 \over 3}}
$$

\begin{align}
&\color{#66f}{\Large x}=5\,{\expo{\theta} + \expo{-\theta} \over 2}
={5 \over 2}\,\pars{\root{5 \over 3} + \root{3 \over 5}}
=\color{#66f}{\large{4 \over 3}\,\root{15}} \approx {\tt 5.1640} 
\\[3mm]\mbox{Similarly,}\quad
&\color{#66f}{\Large y}=5\,{\expo{\theta} - \expo{-\theta} \over 2}
={5 \over 2}\,\pars{\root{5 \over 3} - \root{3 \over 5}}
=\color{#66f}{\large{1 \over 3}\,\root{15}} \approx {\tt 1.2910} 
\end{align}

$\ds{\large\tt\mbox{Note that}\ \pars{-x,-y}\ \mbox{is a solution too}}$.

