Solve the system of equations :
(EDIT : The problem does not say anything about the nature of $x$ and $y$ (integer, natural number ,..etc.) )
$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3} $
$2x + {\frac {1} {x+y}} = \frac {13} {3}$
I do not know how to approach these types of problems. I tried finding value of $\frac {1} {x+y}$ in terms of $x$ and $y$, but it complicates the problem even more.
Can anyone provide a pointer to what should be done ?
On
Less elegant than Anurag A's answer and using brute force.
Considering the equations $$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3}\tag 1$$ $$2x + {\frac {1} {x+y}} = \frac {13} {3}\tag 2$$ extract $y$ from $(2)$; this gives $$y=\frac{-6 x^2+13 x-3}{6 x-13}\tag 3$$ Replace $(3)$ in $(1)$ an simplify to get $$\frac{8 (x-2) (3 x-2) \left(24 x^2-118 x+149\right)}{(13-6 x)^2}=0\tag 4$$ and the quadratic term does not show real solution (so,two real roots and two complex conjugate roots for $x$).
The solutions of $(4)$ are simple; when you have them, go back to $(3)$ for the corresponding $y$'s.
Perhaps this can help: Let $x+y=v$ and $x-y=w$, then the equations can be written as \begin{align*} 3\left(v^2+\frac{1}{v^2}\right)+w^2 & = \frac{85}{3}\\ \left(v+\frac{1}{v}\right)+w & = \frac{13}{3}. \end{align*} Now let $v+\frac{1}{v}=t$, then the system can be rewritten as \begin{align*} 3t^2+w^2 & = \frac{85}{3}+6\\ t+w & = \frac{13}{3}. \end{align*} Now solve for $t$ and $w$...