<s>[INST] square root system of equations 
I have a system of equations as follows that I need to solve for $x$:
  $$
\sqrt{1 - x^2} + \sqrt{4 - x^2} = z\\
\sqrt{4 - y^2} + \sqrt{9 - y^2} = z
$$

Originally, I was trying to put $x$ in terms of $z,$ but I’m at a loss how I would solve this system. Trying to solve it just makes it more complicated, no matter if I multiply by the reciprocal or square both sides.
 [/INST] Both the $x$ and $y$ equations have the form
$$\sqrt{a^2-h^2} + \sqrt{b^2 - h^2} = c$$
For everything to be real, we need $\sqrt{|a^2-b^2|} \le c \le a + b$. 
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side $a,b,c$ with respect to side $c$. We can use
Heron's formula to figure out the area $\Delta$ of the triangle and then
$$\begin{align}h = \frac{2\Delta}{c}  
&= \frac{1}{2c}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\\
&= \frac{1}{2c}\sqrt{4a^2b^2 - (c^2-a^2-b^2)^2}\end{align}$$
Apply this to the equations at hand. For positive $x$ and $y$ solutions, we obtain
$$\begin{align}
\sqrt{1 - x^2} + \sqrt{4-x^2} = z &\implies
x = \frac{1}{2z}\sqrt{16 - (z^2-5)^2}\\
\sqrt{4 - y^2} + \sqrt{9-y^2} = z &\implies
y = \frac{1}{2z}\sqrt{144 - (z^2 - 13)^2}
\end{align}$$
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