Given $ \frac{1}{r}+\frac{1}{s}=a; \frac{1}{r}\times\frac{1}{s}=b; a+b=r; a\times b=s$, find $a$. (Brazilian Math Olympics, 2016) 
Given:  $\{a,b,r,s\}\subset \mathbb R$, $a>0$, 
$\frac{1}{r}$, $\frac{1}{s}$ are roots for $x^2-ax+b=0$, and 
$a$,$~b$ are roots for $x^2-rx+s$.
Find: the numeric value of $a$. 

This is question 3, level 2, phase 3, Brazilian Math Olympics (OBM, 2016). No answer provided. 
The first step is easy, using Girard relations, we can get the system
$$
\left\{ 
\begin{array}{c}
\frac{1}{r}+\frac{1}{s}=a \\ 
\frac{1}{r}\times\frac{1}{s}=b\\ 
a+b=r\\
a\times b=s
\end{array}
\right. 
$$
I'm having a hard time on solving this system of equations. All tricks I know seems to lead to nowhere. It was considered a hard question in the contest (level 2 in OBM is for students up to 9th grade).
Hints and solutions are appreciated. Sorry if this is a duplicate.
 A: $$\frac{1}{r}+\frac{1}{s}=a\tag1$$
$$\frac{1}{r}\times\frac{1}{s}=b\tag2$$
$$a+b=r\tag3$$
$$a\times b=s\tag4$$
Substituting $(3)(4)$ into $(1)(2)$ gives
$$(1)\implies\frac{1}{a+b}+\frac{1}{ab}=a\implies ab+a+b=a^2b(a+b)=a^3b+a^2b^2\tag5$$
$$(2)\implies \frac{1}{a+b}\times\frac{1}{ab}=b\implies 1=ab^2(a+b)\tag6$$
From $(5)(6)$, we have
$$(a^2b^2(a+b)=)\ \ b(ab+a+b)=a\implies (a+1)b^2=a-ab\tag7$$
Multiplying the both sides of $(5)$ by $a+1$ gives
$$(a+1)(ab+a+b)=(a+1)a^3b+a^2(a+1)b^2\tag8$$
Substituting $(7)$ into $(8)$ gives
$$(a+1)(ab+a+b)=(a+1)a^3b+a^2(a-ab),$$
i.e.
$$-(a^4-a^2-2a-1)b=a(a^2-a-1),$$
i.e.
$$-(a^2-(a+1)^2)b=a(a^2-a-1),$$
i.e.
$$-(a^2-a-1)(a^2+a+1)b=a(a^2-a-1)$$
If $a^2-a-1\not=0$, then $$(a+1)b^2=-ab-a^2b^2\tag9$$
From $(7)(9)$, we get
$$-ab-a^2b^2=a-ab\implies -ab^2=1$$
This is impossible since the LHS is negative while the RHS is positive.
So, we have $a^2-a-1=0\implies a=\frac{1+\sqrt 5}{2}$.
Therefore, the answer is $$\color{red}{a=\frac{1+\sqrt 5}{2}}$$
A: $
\left\{ 
\begin{array}{l}
\frac{1}{r}+\frac{1}{s}=a \\ 
\frac{1}{r}\cdot\frac{1}{s}=b\\ 
a+b=r\\
a b=s
\end{array}
\right.
$
$
\left\{ 
\begin{array}{l}
\frac{1}{a+b}+\frac{1}{a b}=a \\ 
\frac{1}{(a+b) (a b)}=b
\end{array}
\right.
$
$$a^5+a^4-2 a^3-2 a^2-2 a-1=0\to \left(a^2-a-1\right) \left(a^3+2 a^2+a+1\right)=0$$
The only positive solution is $\color{red}{a=\dfrac{1+\sqrt{5}}{2}}$
Because $a^3+2 a^2+a+1=0$ has only one real solution which is negative 
Indeed $P(a)=a^3+2 a^2+a+1$ 
has $P(-2)=-1$ and $P(-1)=1$ which means that there  is a zero in $(-2,-1)$
Furthermore $P'(a)=3 a^2+4 a+1=(a+1) (3 a+1)$ 
$P'(a)=0$ for $a=-1;\;a=-\frac13$
$P''(a)=6a+4$
$P''(-1)=-2$ means that $(-1,1)$ is a maximum
$P''(-1/3)=2$ means that $\left(-\frac13,\frac{23}{27}\right)$ is a minimum
Anyway the function doesn't have any other real zeros.
As you can see in the picture below.
Hope this can help. It required some work to factor correctly the equation in the unknown $a$ because at first I tried with $b$ which leads to more complicated algebraic equations. Nice problem anyway, thank you.
$$...$$

