# Solve the system of equations with one symmetrical equation

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?

• substitute for y in the second equation and solve a quadratic – David Quinn Jun 23 '15 at 10:31
• Can you provide a detail solution for this problem? – idiots Jun 23 '15 at 10:40
• 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 $u-v$. – Circonflexe Jun 23 '15 at 11:13
• @Circonflexe. $v=\sqrt{y-1}$ or $v=\sqrt{y+1}$ ? – Claude Leibovici Jun 23 '15 at 11:18
• $+$ of course, thanks (I cannot edit my comment !?) – Circonflexe Jun 23 '15 at 11:28

substitute for y in the second equation and solve a quadratic: $$3x^2-13x+14=0\Rightarrow(3x-7)(x-2)=0$$...etc...