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.
 A: 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 $c\neq v$
A: $$c^2x^2-v^2y^2=c^2$$
$$y^2-c^2z^2=1$$
$$vy^2+c^2zx=0$$
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.
