Very Hard System of Equations

Question

Solve the system of equations:

\begin{cases} \sqrt{xy}(x + 3y)(3x + y) = 14 \\ (x + y)(x^2 + y^2 + 14xy) = 36. \end{cases}

Suppose $x + y = m$ and $xy = n$. So I get

\begin{cases} (3m^2+4n) \sqrt n = 14 \\ m^3+12mn=36 \end{cases}

So can express $m$ in terms of $n$ but I want only real solution. Is there a faster way?

Answer 1

Let $x+y=2k\sqrt{xy}.$ Thus, by AM-GM $k\geq1$ and from the first and the second equations we obtain: $$18\sqrt{xy}(3x^2+3y^2+10xy)=7(x+y)(x^2+y^2+14xy)$$ or $$18(12k^2+4)=7\cdot2k(4k^2+12)$$ or $$7k^3-27k^2+21k-9=0$$ or $$7k^3-21k^2-6k^2+18k+3k-9=0$$ or $$(k-3)(7k^2-6k+3)=0,$$ which gives $k=3$ and $$x+y=6\sqrt{xy}.$$ Can you end it now?

Answer 2

Solving the second equation for $n$ we get $$n=-1/12\,{\frac {{m}^{3}-36}{m}}$$ for $$m\neq 0$$ plugging this in your first equation we get $$1/6\, \left( 3\,{m}^{2}-1/3\,{\frac {{m}^{3}-36}{m}} \right) \sqrt {-3 \,{\frac {{m}^{3}-36}{m}}}=14 $$ so now we square and factorize this equation we obtain: $$-16\, \left( m-3 \right) \left( {m}^{2}+3 \right) \left( {m}^{2}+3\, m+3 \right) \left( {m}^{2}-3\,m+3 \right) \left( {m}^{2}+3\,m+9 \right) =0$$ I hope you will find all solutions.