Let the function lac $(x)$ defined on the set of real numbers have the following assignment rule:

$$
\operatorname{lac}(x)= \begin{cases}x, & \text { if } x \in[2 n ; 2 n+1], \text { where } n \in \mathbb{Z} \\ -x+4 n+3, & \text { if } x \in] 2 n+1 ; 2 n+2[, \quad \text { where } n \in \mathbb{Z}\end{cases}
$$

Solve the equation lac $\left(2 x^{2}+x+4\right)=\operatorname{lac}\left(x^{2}+7 x-1\right)$ on the set of real numbers.
Solution. The lac $(x)$ function coincides with the function $f(x)=x$ in every $[2 n ; 2 n+1]$ interval (where $n \in \mathbb{Z})$, while in every other $] 2 n+1 ; 2 n+2[$ interval, its image is an open segment symmetric with respect to the point

$$
\left(2 n+\frac{3}{2} ; 2 n+\frac{3}{2}\right)
$$

parallel to the line $g(x)=-x$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_c083b90df942365e9064g-1.jpg?height=461&width=461&top_left_y=455&top_left_x=821)

Thus, the lac $(x)$ function is a bijective function (i.e., for every $x$ value, there is a unique lac $(x)$ function value, and vice versa).

From this, it follows that lac $\left(2 x^{2}+x+4\right)=$ lac $\left(x^{2}+7 x-1\right)$ holds if and only if $2 x^{2}+x+4=x^{2}+7 x-1$ also holds. Rearranging the equation:

$$
x^{2}-6 x+5=0
$$

From this: $x_{1}=1, x_{2}=5$.

Therefore, the original equation has two solutions.