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.

$$\frac{1}{r}+\frac{1}{s}=a\tag1$$
$$\frac{1}{r}\times\frac{1}{s}=b\tag2$$
$$a+b=r\tag3$$
$$a\times b=s\tag4$$
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 $$\color{red}{a=\frac{1+\sqrt 5}{2}}$$
