ONE question: are you serious?
Well, I'd attack that (if attackable!) this way:
you have terms that contain :
x1x2; let a = that term
x1x3; let b...
x1x4; let c...
x2x3; let d...
x2x4; let e...
x3x4; let f...
So the equations can be rewritten:
sqrt(c) - sqrt(b) = sqrt(f) 
sqrt(c) - sqrt(a) = sqrt(e) 
sqrt(b) - sqrt(a) = sqrt(d) 
Put those 3 equations through the elimination process...and see what happens...
Don't give up! I'm playing with it now; getting interesting stuff!
I'll let you know later...
Have a look here; that's why I say "interesting":
solve sqrt(a^2-b^2)-sqrt(c^2-d^2)-sqrt(e^2-f^2)=0 for a - Wolfram|Alpha
Not from me!
I make that a "lost cause": 3 equations, 4 variables; you need minimum of 4 equations.
Also, almost impossible to manipulate even without the square roots.
And all of the "math heavyweights" at this site have (wisely!) stayed away from this mess...
Or one way is to set each term equal to a square, and get:
c^2 = a^2 + b^2
c^2 = d^2 + e^2
b^2 = d^2 + f^2
...in other words, that's 3 right triangles, possibly:
1073^2 = 448^2 + 975^2
1073^2 = 495^2 + 952^2
975^2 = 495^2 + 840^2
Then, using my c^2 (as example):
x1^2 x4^2 / 4 - (x1 - x4)^2 = 1073^4
But that also turns out as ugly as the 7 witches of MacBeth!
1) One eq and 2 vars: (x-2)^2+(y-3)^2=0. Answer: x=2, y=3
2) System of two eqns and three variables:
Answer: x=2, y=3, z=4.
Sometimes (a case when a number of eqns is less than a number of variables) we just have more than one solution. E.g.: system of two eqns and three variables:
Answer: (x,y,z) may take the form (x,4-x,1) so we get many roots.