How find the value of the $x+y$ Question:

let $x,y\in \Bbb R $, and such
  $$\begin{cases}
3x^3+4y^3=7\\
4x^4+3y^4=16
\end{cases}$$
Find the $x+y$

This problem is from china some BBS
My idea: since
$$(3x^3+4y^3)(4x^4+3y^4)=12(x^7+y^7)+x^3y^3(9y+16x)=112$$
$$(3x^3+4y^3)^2+(4x^4+3y^4)^2=9(x^6+y^8)+16(y^6+x^8)+24x^3y^3(1+xy)=305$$
then I can't  Continue
 A: Assuming the question is typed correctly as shown, there are two unique real solutions.  Let $$\begin{align*} u(z) &= -983749-111132z^3+786432z^4+71442z^6-196608z^8-20412z^9+18571z^{12}, \\ v(z) &= -178112-351232z^3+186624z^4+301056z^6-34992z^8-114688z^9+18571z^{12}. \end{align*}$$ These polynomials have exactly two distinct real roots; let $r(u,+)$, $r(u,-)$ be the positive and negative real roots of $u$, and $r(v,+)$, $r(v,-)$ be the positive and negative real roots of $v$, respectively.  Then $$(x,y) \in \{(r(u,-),r(v,+)), (r(u,+),r(v,-))\}$$ are the desired solutions.  The sum $x+y$ can then be expressed by the solution to a third polynomial  $$f(z) = 819447-537600z-8998752z^3+3291428z^3+22132992z^4-17875200z^5+3163146z^6+1042512z^8-437500z^9+18571z^{12},$$ for which there are again two real roots, both positive.  All of these polynomials are irreducible.  So I highly doubt that this is a problem that can be reasonably solved by hand.
A: Another idea for you to think about. We have:
$$\begin{cases}
3x^3+4y^3=7 \\
4x^4+3y^4=16
\end{cases}$$
Break it as: 
$$\begin{cases}
3x^3+3y^3 + y^3=7 \\
x^4 + 3x^4+3y^4=16
\end{cases}$$
and factor:
$$\begin{cases}
3(x+y)(x^2 - xy + y^2)=7 - y^3\\
3(x+y)(x^3 - x^2y + xy^2 - y^3)=16 - x^4
\end{cases}$$
And so: $$x+y = \frac{1}{3}\frac{7 - y^3}{x^2 - xy + y^2} = \frac{1}{3}\frac{(2-x)(8+4x+2x^2+x^3)}{x^3 - x^2y + xy^2 - y^3}$$
