Example 15. Solve the equation

$$
\sqrt{x+7} \sqrt{3 x-2}=3 \sqrt{x-1} \sqrt{x+2}
$$
Solution. Solving the system of inequalities

$$
\left\{\begin{array}{r}
x+7 \geqslant 0 \\
3 x-2 \geqslant 0 \\
x-1 \geqslant 0 \\
x+2 \geqslant 0
\end{array}\right.
$$

we find the domain of the equation: $x \geqslant 1$. Squaring both sides of the equation and combining like terms, we obtain the equation

$$
3 x^{2}-5 x-2=0
$$

which is a consequence of the original equation. The roots of this equation are $x_{1}=-1 / 3$ and $x_{2}=2$.

The root $x_{1}=-1 / 3$ does not belong to the domain of the original equation.

The root $x_{2}=2$ belongs to the domain of the original equation and, when substituted, turns it into a true numerical equality.

Thus, $x_{1}=2$ is the only root of the original equation.

The equation $f(x)=g(x) \alpha(x)$ is a consequence of the equation

$$
\frac{f(x)}{\alpha(x)}=g(x)
$$

If the domain of the function $\varphi(x)$ is not narrower than the domain of the equation $f(x)=g(x)$, then the equation

$$
f(x) \varphi(x)=g(x) \varphi(x)
$$

is a consequence of the equation

$$
f(x)=g(x)
$$

Therefore, if in the process of solving an equation, transformations are performed that consist of eliminating the denominator from the equation or multiplying the equation by an expression whose domain is not narrower than the domain of the given equation, the domain of the original equation may be expanded, and, consequently, extraneous roots may appear. In this case, verification is also a necessary part of solving the equation.