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.
 A: Let $s=x-y$ and $t=x+y$. Then the first equation becomes
$$\frac{t-s}{t+s}+\frac{t+s}{t-s}=\frac{17}{4}$$
which with some manipulation becomes $9t^2=25s^2$, or equivalently 
$t=\pm\frac{5}{3}s$.
The second equation is $st=25$. The rest should not be difficult. 
A: Hint:
$$x/y=t\Rightarrow y/x=1/t$$
from first equation
$$t+1/t=17/4\iff 4t^2-17t+4=0$$
$$t_{1,2}=\frac{17\pm15}{8}=4,1/4$$
$x=4y$ or $y=4x$
from second equation
$$(4y)^2-y^2=25,y^2=5/3$$ 
A: Multiplying by $4xy$ gives $4x^2+4y^2=17xy$ or $4x^2-17xy+4y^2=0$, so
$(4x-y)(x-4y)=0.$
Now substitute $y=4x$ and $x=4y$ into the second equation to find the possible solutions.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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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}

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

