$2 \cdot 85$ The number of real solutions to the equation $|2 x-1|+|x-2|=|x+1|$ is
(A) 1 .
(B) 2 .
(C) 3 .
(D) A conclusion different from the above.
(Shanghai Junior High School Mathematics Competition, China, 1984; Junior High School Mathematics Correspondence Competition in Some Provinces and Cities of China, 1988)
[Solution] First, find the $x$ values that make each algebraic expression inside the absolute values equal to zero. These values, arranged in ascending order, are $-1, \frac{1}{2}, 2$.

Then, divide the cases into $x<-1, \quad-1 \leqslant x<\frac{1}{2}, \quad \frac{1}{2} \leqslant x<2$ and $x \geqslant 2$, and remove the absolute value signs for each case to solve the equation.
When considering $\frac{1}{2} \leqslant x<2$, the original equation becomes
$(2 x-1)-(x-2)=x+1$. That is, $1=1$.
This is an identity, indicating that any $x$ value satisfying $\frac{1}{2} \leqslant x<2$ is a root of the equation, thus eliminating (A), (B), (C).
Therefore, the answer is $(D)$.