G3.3 Let $x$ and $y$ be positive real numbers with $x<y$. If $\sqrt{x}+\sqrt{y}=1$ and $\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=\frac{10}{3}$ and $x<y$, find the value of $y-x$.
$$
\begin{array}{l}
(1)^{2}: x+y+2 \sqrt{x y}=1 \\
\sqrt{x y}=\frac{1-(x+y)}{2} \ldots \text { (3) }
\end{array}
$$
$$
\text { (2): } \frac{x+y}{\sqrt{x y}}=\frac{10}{3} \ldots
$$

Sub. (3) into (4): $\frac{x+y}{\frac{1-(x+y)}{2}}=\frac{10}{3}$
$6(x+y)=10(1-x-y)$
$16(x+y)=10$
$x+y=\frac{5}{8}$
$\sqrt{x y}=\frac{1-(x+y)}{2}=\frac{1}{2}\left(1-\frac{5}{8}\right)=\frac{3}{16}$
$x y=\frac{9}{256}$
$(y-x)^{2}=(x+y)^{2}-4 x y=\frac{25}{64}-\frac{9}{64}=\frac{1}{4}$
$$
y-x=\frac{1}{2}
$$

Method 2
Let $z=\sqrt{\frac{x}{y}}$, then $\frac{1}{z}=\sqrt{\frac{y}{x}}$
$$
\begin{array}{l}
\text { (2) becomes } z+\frac{1}{z}=\frac{10}{3} \\
3 z^{2}-10 z+3=0 \\
(3 z-1)(z-3)=0 \\
z=\frac{1}{3} \text { or } 3 \\
\because x<y \\
\therefore z=\sqrt{\frac{x}{y}}<1 \Rightarrow z=\frac{1}{3} \text { only } \\
\frac{\sqrt{y}-\sqrt{x}}{\sqrt{y}+\sqrt{x}}=\frac{1-\sqrt{\frac{x}{y}}}{1+\sqrt{\frac{x}{y}}}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{1}{2} \\
\because \sqrt{x}+\sqrt{y}=1 \quad \therefore \sqrt{y}-\sqrt{x}=\frac{1}{2} \\
y-x=(\sqrt{y}-\sqrt{x})(\sqrt{y}+\sqrt{x})=\frac{1}{2}
\end{array}
$$