How to solve system of equation, $\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3$, $\sqrt{y-4}+\sqrt{z-4}=4\sqrt3$ and $ \sqrt{x-9}+\sqrt{z-9}=4\sqrt3$ . $$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$ I tried somthing,like go to the power of two , and change of variables... but it became more complicated . Is there an idea to solve this system of equation ?
  Thanks in advance 

So here are the 3 equations:
$$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$
As suggested by transcenmental,
$\sqrt{x-1}-\sqrt{y-1}=4\sqrt 3 - 2 \sqrt{y-1}$ and multiplying with the first eq. gives 
$$
x - y = 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})
$$
For the second eq., use
$$-\sqrt{y-4}+\sqrt{z-4}=4\sqrt 3 - 2 \sqrt{y-4}$$ and multiplying with the second eq. 
$$
z - y = 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})
$$
Plugging into the last one gives an eq. in y:
$$
\sqrt{y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})-9}+\sqrt{ y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})-9}=4\sqrt3
$$
This is pretty akward, but $y = 28/3$ is a solution (by computer). From here the others follow, namely 
$$
x = 52/3$$ 
and 
$$
z = 76/3$$ 
EDIT:
with a little bit of hindsight and a little bit of psychology, you could argue 
as follows (with a twinkling of an eye):
suppose the person asking the question prefers a reasonably nicely looking solution (psychology 1). Then  all variables  should either be multiples of 3 or of $1/3$ to get rid of the $\sqrt 3$ on the RHS. Let's try $1/3$ (hindsight 1). So let $x = x' / 3$ etc. Now assume further that also the numerator of the variables should be nice, e.g. no roots etc. (psychology 2). Then we should have that $x'-3\cdot 1$ and  $x'-3\cdot 9$ are "nice" squares (likewise with the other variables). If we want it even nicer, they should be squares of integers (hindsight 2).
So $x' = 3 + n^2$ and $x' = 27 + m^2$. Now start playing. "Nice" integers n and m will be reasonably small (psychology 3). $x' = 52$ does it nicely, with $n=7$ and $m=5$. A small number of trials, to match all three variables, will then give the solution....
