Solve the following system of equations:

$$
\begin{aligned}
& \lg (x-y)+\lg 2=\frac{1}{2}(\lg x-\lg y) \\
& \lg (x+y)-\lg 3=\frac{1}{2}(\lg y-\lg x)
\end{aligned}
$$
The logarithms involved are only defined for

$$
x>y>0
$$

satisfying this size relationship, and only solutions that meet this condition are acceptable. Based on the identities of logarithms, we can write both sides of (1) and (2) as a single logarithm:

$$
\begin{aligned}
\lg 2(x-y) & =\lg \sqrt{\frac{x}{y}} \\
\lg \frac{x+y}{3} & =\lg \sqrt{\frac{y}{x}}
\end{aligned}
$$

From these, using the inverse operation of the logarithm, we obtain simpler, equivalent equations. By raising 10 to the power of the exponents on both sides of (1a) and (2a), we get:

$$
10^{\lg z}=z \quad(\text { if } z>0)
$$

Based on this identity:

$$
\begin{aligned}
2(x-y) & =\sqrt{\frac{x}{y}} \\
\frac{x+y}{3} & =\sqrt{\frac{y}{x}}
\end{aligned}
$$

By writing the equality of the ratios of the left and right sides, we eliminate the square root:

$$
\frac{6(x-y)}{x+y}=\frac{x}{y}
$$

Dividing the numerator and denominator on the left side by $y$ and setting $\frac{x}{y}=u$, we obtain a quadratic equation:

$$
\frac{6(u-1)}{u+1}=u, \quad u^{2}-5 u+6=(u-2)(u-3)=0
$$

and both roots $u_{1}=2, u_{2}=3$ satisfy the condition $u>1$ derived from (3).

For $u_{1}=2$, we have $y_{1}=\frac{x_{1}}{2}$, substituting this into (1b) gives us the solution:

$$
\begin{aligned}
2\left(x_{1}-\frac{x_{1}}{2}\right) & =x_{1}=\sqrt{2} \text { (taking only the positive root) } \\
y_{1} & =\frac{1}{\sqrt{2}}
\end{aligned}
$$

For $u_{2}=3$, similarly,

$$
2\left(x_{2}-\frac{x_{2}}{3}\right)=\frac{4 x_{2}}{3}=\sqrt{3}, \quad x_{2}=\frac{3 \sqrt{3}}{4}, \quad y_{2}=\frac{\sqrt{3}}{4}
$$

According to all of this, the system of equations (1), (2) has two solutions.