Example 3. Find the solution set of the equation $|\mathrm{x}+\mathrm{y}|=|2 \mathrm{x}-\mathrm{y}+2|$ $+3 \mathbf{x}+2$.

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Solve: Let $\mathrm{x}+\mathrm{y}=0$ and $2 \mathrm{x}-\mathrm{y}+2=0$, draw these two lines, then the plane is divided into four angular regions (as shown in Figure 3).

(1) For the angular region I (i.e., the angular region $\angle A E C$ including the boundary rays), for any point $(x, y)$, we have
$$
\left\{\begin{array}{l}
x+y \geqslant 0, \\
2 x-y+2 \geqslant 0 .
\end{array}\right.
$$

Thus, in region I, the original equation is
$$
x+y=2 x-y+2+3 x+2,
$$

which simplifies to $2 x-y+2=0$. The points on the ray $\mathrm{EC}$ satisfy this equation;
(2) For the angular region $\mathbb{I}$, the inequalities hold:
$$
\left\{\begin{array}{l}
x+y \geqslant 0, \\
2 x-y+2<0 .
\end{array}\right.
$$

In this region, the original equation is $x+y=-2 x+y-2+3 x+2$, which simplifies to $0=0$. This indicates that the coordinates of every point in region $\mathbb{I}$ satisfy the equation;
(3) For the angular region II, the inequalities hold:
$$
\left\{\begin{array}{l}
x+y<0, \\
2 x-y+2<0
\end{array}\right.
$$

In this case, the original equation is $-x-y=-2 x+y-2+3 x+2$, which simplifies to $\mathrm{x}+\mathrm{y}=0$. This contradicts the conditions, so the coordinates of every point in this region do not satisfy the original equation.
(4) For the angular region $\mathbb{N}$, we have:
$$
\left\{\begin{array}{l}
x+y<0, \\
2 x-y+2 \geqslant 0 .
\end{array}\right.
$$

In this case, the original equation is
$$
\begin{array}{l}
-x-y=2 x-y+2+3 x+2, \\
x=-\frac{2}{3} \text{. Thus, the line } x=-\frac{2}{3}
\end{array}
$$

falls within region $\mathbb{N}$, and the segment of the ray $\boldsymbol{F} \boldsymbol{F}$ satisfies the original equation.

In summary, the solution set of the original equation is $\{(x, y) \mid \left\{\begin{array}{l}x+y \geqslant 0, \\ 2 x-y+2 \leqslant 0\end{array}\right.$ and $\left.x=-\frac{2}{3}\left(y \leqslant \frac{2}{3}\right)\right\}$, its graph is the angular region $\angle \mathrm{CEB}$ (including the boundaries) and the ray $\mathrm{EF}$.