244. Solve the equation

$$
|x+1|-|x|+3|x-1|-2|x-2|=x+2
$$
244. To solve this equation, we will first look for its roots lying on the interval from 2 to $\infty$, then on the intervals from 1 to 2, from 0 to 1, from -1 to 0, and from $-\infty$ to -1.

$1^{\circ}$. Let $x \geqslant 2$. Then $x+1>0, x>0, x-1>0, x-2 \geqslant 0$; therefore, $|x+1|=x+1,|x|=x,|x-1|=x-1 ;|x-2|=$ $=x-2$, and we get the equation

$$
x+1-x+3(x-1)-2(x-2)=x+2
$$

which is satisfied identically.

Thus, any number greater than or equal to 2 is a root of our equation.

$2^{\circ}$. Let $1 \leqslant x<2$. Then $x+1>0, x>0, x-1 \geqslant 0$ and $x-2<0$ and, therefore, $|x+1|=x+1,|x|=x,|x-1|=x-1$, $|x-2|=-(x-2)$.

Thus, we get the equation

$$
x+1-x+3(x-1)+2(x-2)=x+2
$$

From this, $4 x=8, x=2$, i.e., it lies outside the interval $1 \leqslant x<2$. Therefore, our equation has no roots greater than or equal to 1 but less than 2.
$3^{\circ}$. Let, further, $0 \leqslant x<1$. Then we have:

$$
\begin{gathered}
|x+1|=x+1,|x|=x \\
|x-1|=-(x-1),|x-2|=-(x-2)
\end{gathered}
$$

Thus,

$$
x+1-x-3(x-1)+2(x-2)=x+2, x=-1
$$

i.e., it lies outside the interval $0 \leqslant x<1$.

We conclude that there are no roots greater than or equal to 0 but less than 1.

$4^{\circ}$. Let now $-1 \leqslant x<0$. Then $|x+1|=x+1$,

$$
\begin{gathered}
|x|=-x,|x-1|=-(x-1) \text { and }|x-2|=-(x-2) ; \\
x+1+x-3(x-1)+2(x-2)=x+2,
\end{gathered}
$$

which is impossible, i.e., there are no roots on this interval either.

5. Finally, let $x<-1$. In this case, $|x+1|=-(x+1)$,

$$
\begin{gathered}
|x|=-x,|x-1|=-(x-1) \text { and }|x-2|=-(x-2) \\
-(x+1)+x-3(x-1)+2(x-2)=x+2, x=-2
\end{gathered}
$$

We get another root $x=-2$.

Ultimately, the equation is satisfied by: the number -2 and any number greater than or equal to 2.

![](https://cdn.mathpix.com/cropped/2024_05_21_b937612abb2773f7abffg-273.jpg?height=660&width=550&top_left_y=876&top_left_x=436)

Fig. 28.

Note. The results of this problem become completely clear if we plot the graph of the function

$$
y=|x+1|-|x|+3|x-1|-2|x-2|-(x+2) .
$$

In Fig. 28, the graphs of the functions $y_{1}=|x+1|, y_{2}=-|x|, y_{3}=3|x-1|, y_{4}=-2|x-2|$ and $y_{5}=$
$=-(x+2)$ are shown with thin lines, and the graph of the function $y=y_{1}+y_{2}+$ $+y_{3}+y_{4}+y_{5}$ (the "sum of the graphs") is shown with a thick line. From this diagram, it is immediately clear that $y$ is zero on the ray $x \geqslant 2$ and at one other point $x=-2$.