Example 5. Solve the equation

$$
\left|\frac{x^{2}-10 x+21}{x^{2}-12 x+32}\right|=-\frac{x^{2}-10 x+21}{x^{2}-12 x+32}
$$
Solution. The given equation has the form $|f(x)|=-f(x)$, where $f(x)=\frac{x^{2}-10 x+21}{x^{2}-12 x+32}$. Such an equation is equivalent to the combination of systems

$$
\left\{\begin{array}{l}
f(x) = -f(x), \\
f(x) \geqslant 0,
\end{array} \quad \left\{\begin{array}{l}
f(x)=f(x) \\
f(x)<0
\end{array}\right.\right.
$$

The first system of this combination is equivalent to the equation $f(x) = 0$, and the second system is equivalent to the inequality $f(x)<0$; therefore, the combination of these systems is equivalent to the combination of the equation $f(x)=0$ and the inequality $f(x)<0$, i.e., the inequality $f(x) \leqslant 0$.

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-038.jpg?height=177&width=1056&top_left_y=1805&top_left_x=500)

Fig. 2.1

Thus, the original equation is equivalent to the inequality

$$
\frac{x^{2}-10 x+21}{x^{2}-12 x+32} \leqslant 0
$$

i.e., the inequality

$$
\frac{(x-3)(x-7)}{(x-4)(x-8)} \leqslant 0
$$

Solving it, for example, by the method of intervals (Fig. 2.1), we find the solution to the original equation—the union of intervals $[3 ; 4)$ and $[7 ; 8)$.

An equation of the form

$$
h(|f(x)|)=g(x)
$$

where $h, f, g$ are some functions, is equivalent to the combination of systems

![](https://cdn.mathpix.com/cropped/2024_05_21_aca6fef227914c05e0acg-039.jpg?height=137&width=857&top_left_y=388&top_left_x=611)