<s>[INST] Solve a system of equation 
Solve this system of equations.
  $$\begin{cases}
x^{2} + y^{2} = 4 \\ 
z^{2} + t^{2} = 9 \\
xt + yz = 6
\end{cases}$$

My attempts: 


*

*$(x^{2}+y^{2})(z^{2} + t^{2})=36 \xrightarrow{\text{by applying Lagrange formula}} (xt + zy)^{2} + (zx -ty)^{2}=36 \xrightarrow{xt + yz = 6}  36 + (zx -ty)^{2} =36\Rightarrow (zx -ty)^{2} = 0 \Rightarrow zx = ty \Rightarrow z = \frac{ty}{x}$

*$xt + yz = 6 \xrightarrow{z = \frac{ty}{x}}  xt + \frac{ty^{2}}{x} = 6 \Rightarrow x^{2}t + y^{2}t = 6x \Rightarrow t(x^{2} + y^{2})=6x \xrightarrow{x^{2} + y^{2} = 4} t=\frac{3}{2}x$

*$z= \frac{ty}{x} \xrightarrow{t=\frac{3}{2}x} z = \frac{3}{2}y$ 
Can we go further and determine the exact value of variables?
 [/INST] We have $4$ unknowns with three equations. So, we should not have deterministic solutions
$zx=ty\implies \dfrac xy=\dfrac tz=k$(say)
$$4=y^2(1+k^2)$$ $$9=z^2(k^2+1)$$
$$\text{and }6=yz(k^2+1)$$
$$yz\ne0,\dfrac96=\dfrac{z^2}{yz}=\dfrac zy$$
$\implies \dfrac z3=\dfrac y2=u$(say)
$$\implies\{x,y,z,t\}=\{k(2u),2u, 3u,k(3u)\}$$  where $u(\ne0),k$ are arbitrary numbers such that $$4=4u^2(1+k^2)$$
$$u^2(k^2+1)=1$$
</s>