The article in the fifth issue of our journal in 1983, titled "A Simple Method for Compiling Radical Equations," states that the equation
$$
\sqrt{5 \mathrm{x}-1}+\sqrt{2 \mathrm{x}}=3 \mathrm{x}-1
$$
"will produce a quartic equation after squaring twice, which may be quite troublesome to solve." In fact, this equation can be solved using a simpler method.
Solve the equation $3 x-1=(5 x-1)-2 x$
$=(\sqrt{5 x-1}+\sqrt{2 x})$
- $(\sqrt{5 x-1}-\sqrt{2 x})$

and $\sqrt{5 x-1}+\sqrt{2 x}>0$, from (1) we can get
$$
\sqrt{5 x-1}-\sqrt{2 x}=1 \text {. }
$$

This indicates $\sqrt{5 x-1} \geqslant 1$, i.e., $x \geqslant \frac{2}{5}$.
$$
\begin{array}{l}
\text { From (1)+(2) we get } \\
2 \sqrt{5 x-2}=3 x, \\
x \geqslant \frac{2}{5} \\
9 x^{2}-20 x+4=0, \\
x \geqslant \frac{2}{5}
\end{array}
$$

We get $x=2 . \quad\left(x=\frac{2}{9}0\right)$,
Similarly, we can get (2), and solve $x=2$.
From the above handling, we can summarize a unified solution method:
Let $f(x), g(x), \varphi(x)$ be polynomials:

The equation
$$
\sqrt{f(x)} \pm \sqrt{g(x)}=v(x)
$$
yields
$$
\sqrt{f(x)} \mp \sqrt{g(x)}=\psi(x) .
$$

Solving (A) and (B) together, we can find $\sqrt{f(x)}$ or $\sqrt{g(x)}$, and the equation can be simplified into a rational equation.