<s>[INST] How do we solve the system of equations?18-1-2013 Giải hệ phương trình:
(Editor's note: This translates to "Solve a system of equations:")
 $$ \begin{cases}\sqrt[4]{x}\left(\dfrac{1}{4}+\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right)=2 \\[8pt] \sqrt[4]{y}\left(\dfrac{1}{4}-\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right) =1\end{cases} $$
 [/INST] Divide the first equation by $\sqrt[4]{x}$ and the second by $\sqrt[4]{y}$. Then adding and subtracting the two resulting equations gives us the new pair of simultaneous equations:
$$
\frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt[4]{y}}=\frac{1}{2} \\
\frac{2}{\sqrt[4]{x}}-\frac{1}{\sqrt[4]{y}}=2\frac{2\sqrt{x}+\sqrt{y}}{x+y} \, .
$$
Multiplying these two equations together, we have:
$$
\frac{4}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{2\sqrt{x}+\sqrt{y}}{x+y} \, .
$$
Clearing denominators yields:
$$
(4\sqrt{y}-\sqrt{x})(x+y)=2x\sqrt{y}+y\sqrt{x}
$$
which after some algebra can be reduced to
$$
(x+2y)(2\sqrt{y}-\sqrt{x})=0 \, .
$$
So either $x=-2y$ or $\sqrt{x}=2\sqrt{y}$. If $x$ and $y$ are positive and real the first is clearly impossible and the second is equivalent to $x=4y$. (If they're not positive and real, we have to worry more about branch cuts than I really want to.) Now, if $x=4y$ we have
$$
\frac{\sqrt{2}}{\sqrt[4]{y}}+\frac{1}{\sqrt[4]{y}}=\frac{1}{2} \, ,
$$
from the top equation in this answer. So $\sqrt[4]{y}=2(1+\sqrt{2})$, yielding the solution
$$
x=64(1+\sqrt{2})^4 \\
y=16(1+\sqrt{2})^4 \, ,
$$
which upon expanding is precisely what Robert Israel got from Maple in the other answer.
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