I am asked to show that the set of equations $$x^2+4xy=1$$ $$x^2+3y^2=9$$ Has exactly one solution for $(x,y)\in [0,1]\times[1,2]$. Also, I shell give an iteration converging to this solution.
I know I will have to use Banach fixed-point theorem and I tried to define the obvious function $$\Phi(x,y)=(x^2+4xy+x-1, x^2+3y^2+y-9)^T$$ but this does not project into the given set... Did I miss something here?
You might try writing the equations as $y = \sqrt{3 - x^2/3}$, $x = (1 - x^2)/(4 y)$.
Let $z={x\over y}, w={1\over y}$. Divide both equations by $y^2$ to get $z^2+4z=w^2, z^2+3 = 9 w^2$, with $x \in [0,1], w\in [{1 \over 2} , 1]$. Taking 9 times the first equation minus the second equation yields a quadratic in $z$ with solutions $z = { -9 \pm \sqrt{87}\over 4}$, hence we must have $z={ -9 + \sqrt{87}\over 4}$ since this is the only solution in $[0,1]$.
Substituting this value into the first equation gives $w^2 = {12-\sqrt{87} \over 8}$, from which the only permissible solution is $w=\sqrt{{12-\sqrt{87} \over 8}}$.
Consequently the solution is $y= \sqrt{{8 \over 12-\sqrt{87} }}$, $x={ -9 + \sqrt{87}\over 4} \cdot \sqrt{{8 \over 12-\sqrt{87} }}$.