$2 \cdot 87$ The number of distinct solutions to the equation $|x-| 2 x+1||=3$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(E) 4 .
(35th American High School Mathematics Examination, 1984)
[Solution 1] Let's consider the left side of the equation as the vertical distance between the graphs of the two functions $y=x$ and $y=12 x+1|$. When we plot their graphs in a Cartesian coordinate system, the latter has a V-shape with its vertex at $\left(-\frac{1}{2}, 0\right)$ and slopes of $\pm 2$.

From the graph, we can quickly see that there are exactly two values that make the vertical distance between them 3.
[Solution 2] According to the given equation, we have
$$
x-|2 x+1|=-3 \text { or } x-|2 x+1|=3 \text {. }
$$

Thus, we have $|2 x+1|=x+3$ or $|2 x+1|=x-3$.
In the first case, note that the left side of the equation is an absolute value, so the left side must also be $x+3 \geqslant 0$, i.e., $x \geqslant-3$, which is a constraint;
Similarly, in the second case, there must also be a constraint $x \geqslant 3$.
In the first case, $2 x+1$ must equal $x+3$ or $-x-3$, so we solve to get $x=2$ and $x=-\frac{4}{3}$, both of which satisfy the first constraint, so they are solutions to the original equation.

In the second case, $2 x+1$ must equal $x-3$ or $-x+3$, so we solve to get $x=-4$ and $x=\frac{2}{3}$, but neither satisfies the condition $x \geqslant 3$, so they are not solutions to the original equation. Therefore, the original equation has only two solutions.
Hence, the answer is (C).