Non-linear algebra with Cramer's Rule? I am asked to solve the following system of equations using Cramer's Rule
$$
A^2 + B = x\\
AB = 1-x^2
$$
where $A$ and $B$ are 2 unknowns that are to be solved for some given $x$. ie I'm expecting the solution to be $A = a(x)$  and $B = b(x)$ where $a(x)$ and $b(x)$ is some function of $x$.
Is it possible to do so? Alternatively, can I differentiate both equations explicitly w.r.t. $x$ and apply Cramer's rule? I have a feeling this is the right way since this involves linear algebra techniques.
$$
2AA' + B' = 1\\
A'B + B'A = -2x
$$
UPDATE
Alright, guys...the answer is actually to differentiate the equations w.r.t. $x$. I do not know if this is meaningful to any one. Basically, both $A$ and $B$ are implicitly defined in $x$. So after applying the differentiation, which is above, the next step is to construct the matrix:
$$
\begin{pmatrix}
2A & 1 \\
B & A
\end{pmatrix}\begin{pmatrix}
A'\\B'
\end{pmatrix} = \begin{pmatrix}
1\\-2x
\end{pmatrix}
$$
Applying Cramer's Rule,
$$
\begin{align*}
A' &= \frac{A + 2B}{2A^2 - B}\\
\\
B' &= \frac{4Ax + B}{B - 2A^2}
\end{align*}
$$
 A: Cramer's rule is for linear system of equations only. So, no, you can't apply it for the system above.
Update
Also, what's $x$ in your equations? Because if it's actually a variable, then your system is a functional equations. If it's just a number, never mind then.
A: You could write that system of equations as
$$ 
\left( \begin{matrix}A & 1 \\ 0 & A \end{matrix} \right)
\left( \begin{matrix}A \\ B \end{matrix} \right) =
\left( \begin{matrix}x \\ 1-x^2 \end{matrix} \right)
$$
and then solve, but then you get a solution for $A$ and $B$ in terms of $x$ and $A$; it's not clear that such a thing would be useful.
Or, you could write it as a system
$$ 
\left( \begin{matrix}1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right)
\left( \begin{matrix}A^2 \\ B \\ AB \end{matrix} \right) =
\left( \begin{matrix}x \\ 1-x^2 \end{matrix} \right)
$$
and get infinitely many putative solutions for $(A^2, B, AB)$, and you have the problem of figuring out which triples can be solved for $A$ and $B$. (e.g. the solution $(x,0,1-x^2)$ is a solution to the linear system, but you can't solve the system $A^2=x$, $B=0$, $AB=1-x^2$ for $A$ and $B$)
Your approach lets you write the equation as
$$ 
\left( \begin{matrix}2A & 1 \\ B & A \end{matrix} \right)
\left( \begin{matrix}A' \\ B' \end{matrix} \right) =
\left( \begin{matrix}1 \\ -2x \end{matrix} \right)
$$
So gain you could solve, but the answer is a solution for $(A',B')$ in terms of $A$ and $B$ -- i.e. a system of differential equation in two functions, so again it's not clear the result will be useful.
But then, it doesn't hurt to try any of these approaches to see if the result suggests what to do next.
A: Multiply both sides of the first equation to get $A^3+AB=Ax$. Plug in the second equation and rearrange the terms to get $A^3-Ax+(1-x^2)=0$. Solve this cubic either by hand or using WolfamAlpha. You can then solve $B$ by substituting the solution for $A$ in $B=x-A^2$.
