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.
On
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
$$\begin{align} \sqrt{1 - x^2} + \sqrt{4-x^2} = z &\implies x = \frac{1}{2z}\sqrt{16 - (z^2-5)^2}\\ \sqrt{4 - y^2} + \sqrt{9-y^2} = z &\implies y = \frac{1}{2z}\sqrt{144 - (z^2 - 13)^2} \end{align}$$
Without loss of generality, we can choose to deal only with positive $x,y,z$ values. Notice that $y\le2$ from the first radical in the second equation. Then $y$ varies between $0$ and $2$, so $z$ varies between $\sqrt 5$ and $5$. Similarly, in the first equation $x\in[0,1]$ so $z\in [\sqrt 3,3]$. Combining the two ranges for $z$ means a possible solution can be found if $z\in [\sqrt 5,3]$.
Let's focus now on the first equation, and do the substitution $u^2=1-x^2$, with $u\ge 0$. In fact $u\in [0,1]$. We can then rewite the first equation as $$u+\sqrt{3+u}=z$$ Leaving just the square root on the left, and squaring the equation, you get $$3+u=(z-u)^2$$ This is a simple quadratic in $u$, which you should be able to solve. Choose the solution that is in the $[0,1]$ interval.