Example 2. Solve the equation

$$
[x-1]=\left[\frac{x+2}{2}\right]
$$
Solution. 1st method. Let: $[x-1]=t$, $\left[\frac{x+2}{2}\right]=t$
Then

$$
t \leqslant x-1<t+1 \text { and } t \leqslant \frac{x+2}{2}<t+1
$$

or

$$
t+1 \leqslant x<t+2
$$

and

$$
2(t-1) \leqslant x<2 t
$$

The problem has been reduced to finding such integer values of $t$ for which there is a common part of the two specified intervals. Obviously, there is no solution if:

1) the interval (2) is to the "right" of interval (1), i.e., when $2(t-1) \geqslant t+2$, or $t \geqslant 4$
2) the interval (2) is to the "left" of interval (1), i.e., when $2 t \leqslant t+1$, or $t \leqslant 1$.

Therefore, a solution exists when $1<t<4$, and since $t \in \mathbf{Z}$, then $t_{1}=2, t_{2}=3$.

For $t=2$, from inequalities (1) and (2) we get:

$$
3 \leqslant x<4,2 \leqslant x<4
$$

from which $3 \leqslant x<4$.

For $t=3$, similarly we get: $4 \leqslant x<5$. Therefore, the solutions to the given equation are all values $x \in[3 ; 5[$.

2nd method. According to property $2^{0}$, we have:

$$
-1<x-1-\frac{x+2}{2}<1
$$

from which

$$
2<x<6
$$

This condition is necessary but insufficient. For example, when $x=2.5$, inequality (3) is satisfied, but the given equation is not.

Let's plot the graphs of the functions $y=x-1$ and $y=\frac{x+2}{2}$ for values of $x$ that satisfy inequalities (4).

We see (Fig. 2) that only in the second and third strips are there segments (shaded figures) where the integer values of $y$ coincide.

Substituting $y=2$ into the equation $y=x-1$, we get the lower boundary $x=3$, which satisfies the given equation. Substituting $y=4$ into the equation $y=x-1$, we find the upper boundary $x=5$. Checking confirms that the value $x=5$ does not satisfy the given equation. Therefore, the solutions to the equation are all real numbers $x$ that satisfy the inequality $3 \leqslant x<5$.

![](https://cdn.mathpix.com/cropped/2024_05_21_ee3d8c3d885a47ff5ccfg-25.jpg?height=551&width=668&top_left_y=1800&top_left_x=363)

Fig. 2 3rd method. Let's plot the graphs of the functions $y=[x-1]$ and $y=\left[\frac{x+2}{2}\right]$. From Fig. 3, we notice that the graphs have common points in the interval $[3 ; 5[$. Example 3. Solve the equation

$$
\left[-x^{2}+3 x\right]=x^{2}+0.5
$$

Solution. By the definition of the "integer part" function, the given equation is equivalent to the double inequality

24

$$
x^{2}+\frac{1}{2} \leqslant-x^{2}+3 x<x^{2}+\frac{3}{2}
$$

Solving this system of inequalities, we get the numerical interval

$$
\frac{3-\sqrt{5}}{4} \leqslant x \leqslant \frac{3+\sqrt{5}}{4}
$$

Consider the parabola $y=x^{2}+\frac{1}{2}$ in this interval (Fig. 4). The range of values of the function is determined by the ordinates of the points of intersection of the parabola with the lines

$$
\begin{gathered}
x=\frac{3-\sqrt{5}}{4} \text { and } x=\frac{3+\sqrt{5}}{4}: \\
\frac{11-3 \sqrt{5}}{8} \leqslant y \leqslant \frac{11+3 \sqrt{5}}{8} .
\end{gathered}
$$

We notice that within the range of $y$, there are two integers: $y_{1}=1$ and $y_{2}=2$. They correspond to the values: $x_{1}=$ $=\sqrt{0.5}$ and $x_{2}=\sqrt{1.5}$. These are the solutions to the given equation.