solving a system of equations dealing with Lorentz transformations

Question

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.

Answer 1

$$c^2x^2-v^2y^2=c^2$$ $$y^2-c^2z^2=1$$ $$vy^2+c^2zx=0$$

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.

Answer 2

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 $c\neq v$