# 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

• Do you want an idea or an explanation, or both? I charge \$1 for an idea, \$2 for answer, $3 for an explanation. Kidding, but seriously, did you just want a hint? – transcenmental Sep 30 '17 at 15:27 • If you just wanted an idea, try multiplying by the conjugate pairs of each equation. Like for the first one$\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}$. – transcenmental Sep 30 '17 at 15:33 • No i want to train my brain ... – Khosrotash Sep 30 '17 at 15:52 • There should be a clever substitution to solve this instead of a polynomial bash... – u8y7541 Sep 30 '17 at 17:07 ## 3 Answers 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.... • how to solve this $$\sqrt{y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-1})-9}+\sqrt{ y + 4\sqrt 3 (4\sqrt 3 - 2 \sqrt{y-4})-9}=4\sqrt3$$ ? – Khosrotash Sep 30 '17 at 16:51 • 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. – Andreas Sep 30 '17 at 17:02 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. • Nothing new, I'm afraid. This type of double-rooted equation has been posted in a previous answer. – Andreas Sep 30 '17 at 17:04 • @Andreas: not the suggestion to turn to a polynomial ;-) – Yves Daoust Sep 30 '17 at 17:08 • ... which was given in a comment. Difficult, though. – Andreas Sep 30 '17 at 17:20 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. • @:MalayTheDynamo : how would it help ? – Khosrotash Sep 30 '17 at 16:41 • This is not answer. – Zaharyas Sep 30 '17 at 17:51 • 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. – Jan-Magnus Økland Sep 30 '17 at 18:04
• @Khosrotash This removes the roots from the equation. After this you can simply solve with matrices, elemination, etc. – MalayTheDynamo Oct 1 '17 at 3:47
• @Zaharyas Why not? – MalayTheDynamo Oct 1 '17 at 3:47