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Solve the system of equations: $\left\{\begin{array}{l}x^3-y^3+(3x^2+y-2)\sqrt{y+1}-(3y^2+x-2)\sqrt{x+1}=0\\x^2+y^2+xy-7x-6y+14=0\end{array}\right.$


I used wolframalpha.com and got the solution: $(x;y)\in\left\{(2;2);\left(\dfrac{7}{3};\dfrac{7}{3}\right)\right\}$

And combining with symmetry of first equation, I guess that we can get $x=y$ from first equation.

So who can help me?

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  • $\begingroup$ substitute for y in the second equation and solve a quadratic $\endgroup$ – David Quinn Jun 23 '15 at 10:31
  • $\begingroup$ Can you provide a detail solution for this problem? $\endgroup$ – idiots Jun 23 '15 at 10:40
  • $\begingroup$ Write $u = \sqrt{x+1}, v=\sqrt{y-1}$ so that your equations become polynomials in $(u, v)$. Since the first one is $f(u) - f(v) = 0$ with $f$ polynomial, you should be able to divide out $u-v$. $\endgroup$ – Circonflexe Jun 23 '15 at 11:13
  • $\begingroup$ @Circonflexe. $v=\sqrt{y-1}$ or $v=\sqrt{y+1}$ ? $\endgroup$ – Claude Leibovici Jun 23 '15 at 11:18
  • $\begingroup$ $+$ of course, thanks (I cannot edit my comment !?) $\endgroup$ – Circonflexe Jun 23 '15 at 11:28
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substitute for y in the second equation and solve a quadratic: $$3x^2-13x+14=0\Rightarrow(3x-7)(x-2)=0$$...etc...

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