Use implicit function theorem and show $u(x,y)^2+v(x,y)^2=\frac{16}{x+y}$

I am attempting exercise 11.6.4 on p. 368 of Wade's Introduction to Analysis. It asks:

"Find conditions on a point $(x_0, y_0, u_0, v_0)$ such that there exist functions $u(x,y)$ and v(x,y) that are C1 at $(x_0,y_0)$ and satisfy the simultaneous equations $$F(x,y,u,v)=xu^2 + yv^2 + xy =9 \\ G(x,y,u,v)=xv^2 + yu^2 - xy =7.$$Prove that the solutions satisfy $u^2 + v^2 = \frac{16}{x+y}.$"

The first part is straightforward Implicit function theorem, I get that $\frac{\partial(F,G)}{\partial(u,v)}=4uv(x+y)(x-y)$, so the conditions are $u,v \neq 0$ and $|x|\neq |y|$.

I am finding the relation on $x,y, u(x,y),v(x,y)$ more difficult to prove. If I differentiate implicitly, I get for instance

$$F_x=u^2 + 2ux\cdot u_x+2vy\cdot v_x + y = 0\\G_x = v^2 + 2xv\cdot v_x + 2uy \cdot u_x -y =0$$so by Cramer's rule

$$u_x = \frac{\left|\begin{matrix}-y-u^2 & 2vy \\ y-v^2 & 2xv\end{matrix}\right|}{\left|\begin{matrix}2ux & 2vy \\ 2uy & 2xv\end{matrix}\right|},$$

but I don't see how that is helping us.

Hunting around, I find that $(x,y,u,v)=(0,1,\sqrt{7},3)$ is a point where $F,G$ are satisfied and $\frac{\partial(F,G)}{\partial(u,v)}\neq 0$, and I note the relation holds there, although that doesn't help me to show it in general. Any ideas?

I apologize for the duplication with this recent question, which has not been answered (it was asked by someone else). I have tried to ask the question in a more clear, thorough way with my post.

• Look at $F(x,y,u,v) + G(x,y,u,v)$. – Daniel Fischer Jul 19 '13 at 16:03

As Daniel Fischer points out, all one needs to do is add together $F$ and $G$ and derive the relation by division, noting $x+y \neq 0$ at a point where $\frac{\partial{F,G}}{\partial(u,v)}\neq 0.$