2004 AMC 10A Problem 4

On the number line, $|x-1|$ measures how far $x$ is from $1$, and $|x-2|$ measures how far $x$ is from $2$. The equation says these distances are equal, so $x$ sits exactly halfway between $1$ and $2$:
$$x = \frac{1+2}{2} = \frac32.$$

Check: $\left|\tfrac32 - 1\right| = \tfrac12$ and $\left|\tfrac32 - 2\right| = \tfrac12$.

The answer is $\boxed{\textbf{(D)}\ \frac32}$.

$|a - b|$ is the distance between $a$ and $b$. An equation of the form $|x - p| = |x - q|$ with $p \ne q$ therefore has exactly one solution, the midpoint $\frac{p+q}{2}$. Reading absolute values geometrically is faster and safer than casework whenever the problem is about distances.

Algebraically, $|u| = |v|$ means $u = v$ or $u = -v$. Here $x - 1 = x - 2$ is impossible, and $x - 1 = -(x - 2) = 2 - x$ gives $2x = 3$, so $x = \tfrac32$.

• Dropping the absolute values and solving x - 1 = x - 2, then concluding there is no solution.
• Squaring both sides but expanding $(x-2)^2$ incorrectly, which shifts the midpoint.