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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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Try multiplying the first equation by $xy$. Then see if anything comes to you. – Paul Sundheim Aug 4 '14 at 21:21
Let $r = \dfrac{y}{x}$. Then the first equation becomes $r+\dfrac{1}{r} = \dfrac{17}{45}$, i.e. $r^2-\dfrac{17}{45}r+1 = 0$, which has no real solutions. Are you sure you have written the system of equations correctly? – JimmyK4542 Aug 4 '14 at 21:22
@JimmyK4542 Arfffff, you are absolutely right. I just watched. I will edit. Sorry! – user142836 Aug 4 '14 at 21:27
@downvoters Please share why the downvote. – user142836 Aug 4 '14 at 21:35
@JimmyK4542 Thanks you with your method I can solve my problem. – user142836 Aug 4 '14 at 21:38
up vote 5 down vote accepted

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.

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These seem like great (but contrived) substitutions. How did you come up with them? Is there a general method about what substition(s) one should use in this situation? – alexqwx Aug 4 '14 at 21:36
It has been a couple of hours since my last cup of coffee, so I do not have the imagination to make a great substitution. Comes up a lot, in various versions, most simply when we are expressing $n$ as a difference of squares. – André Nicolas Aug 4 '14 at 21:40
By the way, the approach of solving for $y/x$ is more natural. – André Nicolas Aug 4 '14 at 21:46
'with some manipulation' -> multiplying by $(s+t)(s-t)$ and factor. I got it. Cool substituion, thanks a lot! +1 and accept. – user142836 Aug 4 '14 at 21:47

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.

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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 $$(4y)^2-y^2=25,y^2=5/3$$

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Thanks, same method as JimmyK4542 but +1 anyway. – user142836 Aug 4 '14 at 21:50

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ 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}

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

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