Solving for $x$ and $y$ when $(3x + y)(x + 3y)\sqrt{xy} = 14$

Question

Solve for $x$ and $y$ when $$(3x + y)(x + 3y)\sqrt{xy} = 14$$ $$(x+y)(x^2 + 14xy + y^2) = 36.$$

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I was thinking of squaring the first equation and moving on from there, but I think it'll be a bit too messy. Is there a better way to start this problem?

Answer 1

Setting $p=x+y$ and $q=\sqrt{xy}$ (almost as suggested by Alexey in the comments, but $\sqrt{xy}$ looks like it will be more symmetric) and expanding gives \begin{align} 3p^2q + 4q^3 &= 14 \\ p^3 + 12pq^2 &= 36 \end{align} From here, a lucky coincidence is that $$ (p + 2q)^3 = p^3 + 6p^2q + 12pq^2 + 8q^3 = 36 + 2 \cdot 14 = 64 $$ and $$ (p - 2q)^3 = p^3 - 6p^2q + 12pq^2 - 8q^3 = 36 - 2 \cdot 14 = 8 $$ which gives us $p+2q = 4$ and $p-2q = 2$. Therefore $p=3$ and $q = \frac12$, giving us $x,y = \frac{3 \pm 2\sqrt2}{2}$.

Answer 2

From the first equation we have $xy>0$.

Thus, the second give $x+y>0,$ which says $x>0$ and $y>0$.

Let $y=t^2x,$ where $t>0$.

Thus, $$18(3t^2+1)(t^2+3)t=7(t^2+1)(t^4+14t^2+1)$$ or $$(t^2-6t+1)(7t^4-12t^3+26t^2-12t+7)=0$$ or $$t^2-6t+1=0,$$ which gives $y=(17-12\sqrt2)x$ or $y=(17+12\sqrt2)x.$

Let $y=(17-12\sqrt2)x.$

Thus, the first equation gives: $$(3+10(17-12\sqrt2)+3(17-12\sqrt2)^2)(3-2\sqrt2)x^3=14$$ or $$x^3=\frac{99+70\sqrt2}{8}$$ or $$x=\frac{1}{2}(3+2\sqrt2)$$ and $$y=(3-2\sqrt2)^2\cdot \frac{1}{2}(3+2\sqrt2),$$ which gives $$\left(\frac{3+2\sqrt2}{2},\frac{3-2\sqrt2}{2}\right).$$ The second case by the same way gives: $$\left(\frac{3-2\sqrt2}{2},\frac{3+2\sqrt2}{2}\right).$$