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.

The second equality says that 
$$y^2=c^2z^2+1$$
Replace $y^2$ in the first and third equalities. You get:
$$c^2x^2-v^2(1+c^2z^2)=c^2,$$
$$v(1+c^2z^2)+c^2zx=0,$$
Therefore:
$$c^2x^2-v^2c^2z^2=c^2+v^2,\;\;(E_1)$$
$$vc^2z^2+c^2zx=-v,\;\;(E_2)$$
Now, you get from the first equality $(E_1)$:
$$x^2=\dfrac{c^2+v^2+c^2v^2z^2}{c^2},\;\;(S_1)$$
The previous second equality $(E_2)$ (squared) says that :
$$(c^2zx)^2=(-v-vc^2z^2)^2,\;\;(S_2)$$
Use $(S_1)$ and $(S_2)$ to get $z^2$:
$$z^2=\dfrac{v^2}{c^2(c^2-v^2)}$$
Now you get $$y^2=\dfrac{c^2}{c^2-v^2}$$
And $$x^2=y^2$$
P.S. I suppose that $c\neq v$
