SG. 3 Given that $x$ and $y$ are non-zero real numbers satisfying the equations $\frac{\sqrt{x}}{\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}}=\frac{7}{12}$ and $x-y=7$. If $w=x+y$, find the value of $w$.
The first equation is equivalent to $\frac{x-y}{\sqrt{x y}}=\frac{7}{12} \Rightarrow \sqrt{x y}=12 \Rightarrow x y=144$
Sub. $y=\frac{144}{x}$ into $x-y=7: \quad x-\frac{144}{x}=7 \Rightarrow x^{2}-7 x-144=0 \Rightarrow(x+9)(x-16)=0$ $x=-9$ or 16 ; when $x=-9, y=-16$ (rejected $\because \sqrt{x}$ is undefined); when $x=16 ; y=9$ $w=16+9=25$
Method 2 The first equation is equivalent to $\frac{x-y}{\sqrt{x y}}=\frac{7}{12} \Rightarrow \sqrt{x y}=12 \Rightarrow x y=144$
$\because x-y=7$ and $x+y=w$
$\therefore x=\frac{w+7}{2}, y=\frac{w-7}{2}$
Sub. these equations into (1): $\left(\frac{w+7}{2}\right)\left(\frac{w-7}{2}\right)=144$
$$
w^{2}-49=576 \Rightarrow w= \pm 25
$$
$\because$ From the given equation $\frac{\sqrt{x}}{\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}}=\frac{7}{12}$, we know that both $x>0$ and $y>0$
$$
\therefore w=x+y=25 \text { only }
$$