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$$\begin{cases}\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3\\\sqrt{y-4}+\sqrt{z-4}=4\sqrt3\\\sqrt{x-9}+\sqrt{z-9}=4\sqrt3\end{cases}$$ 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

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  • $\begingroup$ Do you want an idea or an explanation, or both? I charge \$1 for an idea, \$2 for answer, $3 for an explanation. Kidding, but seriously, did you just want a hint? $\endgroup$ – transcenmental Sep 30 '17 at 15:27
  • $\begingroup$ If you just wanted an idea, try multiplying by the conjugate pairs of each equation. Like for the first one $\sqrt{x-1}+\sqrt{y-1}=4\sqrt 3$, it would be $\sqrt{x-1}-\sqrt{y-1}=4\sqrt 3 -2\sqrt{y-1}$. $\endgroup$ – transcenmental Sep 30 '17 at 15:33
  • $\begingroup$ No i want to train my brain ... $\endgroup$ – Khosrotash Sep 30 '17 at 15:52
  • $\begingroup$ There should be a clever substitution to solve this instead of a polynomial bash... $\endgroup$ – u8y7541 Sep 30 '17 at 17:07
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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....

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  • $\begingroup$ how to solve this $$\sqrt{y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})-9}+\sqrt{ y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})-9}=4\sqrt3$$ ? $\endgroup$ – Khosrotash Sep 30 '17 at 16:51
  • $\begingroup$ A formal way of solving this might turn pretty difficult: squaring twice will give a fourth order equaition, which you could try to factorize. I added the "EDIT" section to show a way to guess a reasonably looking solution. $\endgroup$ – Andreas Sep 30 '17 at 17:02
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Hint:

Eliminate $y$ and $z$,

$$y=(4\sqrt3-\sqrt{x-1})^2+1,\\z=(4\sqrt3-\sqrt{x-9})^2+9$$

and

$$\sqrt{44+x-8\sqrt3\sqrt{x-1}}+\sqrt{44+x-8\sqrt3\sqrt{x-9}}=4\sqrt3.$$

By plotting, you can see that this equation has two real solutions. It is possible, by successive squarings and regroupings, to turn it to a polynomial. But this will be tedious.

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  • $\begingroup$ Nothing new, I'm afraid. This type of double-rooted equation has been posted in a previous answer. $\endgroup$ – Andreas Sep 30 '17 at 17:04
  • $\begingroup$ @Andreas: not the suggestion to turn to a polynomial ;-) $\endgroup$ – Yves Daoust Sep 30 '17 at 17:08
  • $\begingroup$ ... which was given in a comment. Difficult, though. $\endgroup$ – Andreas Sep 30 '17 at 17:20
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Solving $1$:-

$\sqrt{x-1}+\sqrt{y-1}=4\sqrt3\\$

$\text{Squaring}$

$\begin{align}x-1+y-1+2\sqrt{(x-1)(y-1)}&=48\\to\\50-x-y&=2\sqrt{(x-1)(y-1)}\\\end{align}$

$\text{Squaring}$

$\begin{align}x^2 + y^2+ 2 x y- 100 x - 100 y + 2500&=4 x y - 4 x - 4 y + 4\\to\\x^2 + y^2 + 2496 &= 96 x+ 96 y +2 x y \end{align}$

Continue for the other equations, then solve.

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  • $\begingroup$ @:MalayTheDynamo : how would it help ? $\endgroup$ – Khosrotash Sep 30 '17 at 16:41
  • $\begingroup$ This is not answer. $\endgroup$ – Zaharyas Sep 30 '17 at 17:51
  • $\begingroup$ This idea leads to $\langle (3z-76)(3z^3-412z^2+13056z-119808),$ $60y-9z^3+1266z^2-42628z+413184,$ $160x+9z^3-1296z^2+46288z-489984\rangle$, but the three (actually real) roots in the cubic polynomial in $z$ don't give solutions to the original system. $\endgroup$ – Jan-Magnus Økland Sep 30 '17 at 18:04
  • $\begingroup$ @Khosrotash This removes the roots from the equation. After this you can simply solve with matrices, elemination, etc. $\endgroup$ – MalayTheDynamo Oct 1 '17 at 3:47
  • $\begingroup$ @Zaharyas Why not? $\endgroup$ – MalayTheDynamo Oct 1 '17 at 3:47

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