<s>[INST] Suppose $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$. What are $x,y,z,t$? Let $x^2+y^2=4$ and $z^2+t^2=9$ and $xt+yz=6$.
What are $x,y,z,t$?
 [/INST] Consider
$$
(x^2+y^2)(z^2+t^2)-(xt+yz)^2=0
$$
Expanding the left-hand side you get
$$
x^2z^2+\color{red}{y^2z^2}+\color{red}{x^2t^2}+y^2t^2
-\color{red}{x^2t^2}-2xyzt-\color{red}{y^2z^2}=0
$$
that simplifies to
$$
(xz-yt)^2=0
$$
so $xz=yt$.
Now consider the linear system in the unknowns $z$ and $t$:
\begin{cases}
yz+xt=6 \\
xz-yt=0
\end{cases}
With standard methods you get
$$
z=\frac{6y}{x^2+y^2}=\frac{3}{2}y,
\qquad
t=\frac{6x}{x^2+y^2}=\frac{3}{2}x
$$
Now any pair of numbers $x,y$ with $x^2+y^2=4$ will do.

Alternatively, consider the complex numbers
$$
A=x+yi,\qquad B=z+ti
$$
Then the hypotheses yield $|A|=2$, $|B|=3$ and that the imaginary part of $AB$ is $6$, that is
$$
AB=C+6i
$$
with $C$ real. Then $\bar{A}\bar{B}=C-6i$ and multiplying gives
$$
A\bar{A}B\bar{B}=C^2+36
$$
Thus $36=C^2+36$ and so $C=0$. Therefore
$$
AB=6i
$$
and
$$
z+ti=\frac{6i}{x+yi}=\frac{6i}{x-yi}{4}=\frac{3}{2}i(x-yi)=\frac{3}{2}(y+xi)
$$
whence again $z=\frac{3}{2}y$ and $t=\frac{3}{2}x$.
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