<s>[INST] solving a system of equations dealing with Lorentz transformations Can anyone help me to find the solutions of this system of equations:
$$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
I know the answer:
$$x= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$y= \frac{1}{ \sqrt{1- \frac{ v^{2} }{ c^{2} } } } $$
$$z= -\frac{v}{c^2} \sqrt{1- \frac{ v^{2} }{ c^{2} } }  $$
But I can't follow the steps. If any one could say me how to solve this kind of problem that would be helpful for me. Thank you in advance.
 [/INST] $$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
mutliplying the second by $x^{2}$ we find
$$
y^2x^2 - c^2z^2x^2 = x^2
$$
using the third equation we find
$$
y^2x^2 - c^2z^2x^2 = y^2x^2 - c^2\left(\frac{vy^2}{c^2}\right)^2 = x^2
$$
or
$$
y^2x^2 - \frac{v^2}{c^2}y^4 = x^2
$$
now we can use the first equation to yield
$$
x^2y^2 = \frac{v^2y^4 + c^2y^2}{c^2}
$$
and subbing into the previous equation we obtain
$$
v^2y^4 + c^2y^2 - v^2y^4 = c^2 + v^2y^2
$$
or
$$
y^2 = \frac{c^2}{c^2-v^2} = \frac{1}{1-\left(\frac{v}{c}\right)^2}
$$
now you have y, you immediately obtain the remaining results.
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