imc

$|x-1|$ is the distance between $x$ and $1$; $|x-2|$ is the distance between $x$ and $2$. Therefore, the given equation says $x$ is equidistant from $1$ and $2$, so $x=\frac{1+2}{2}=\frac{3}{2}\Rightarrow \boxed{\frac{3}{2}}$. Alternatively, we can solve by casework (a method which should work for any similar problem involving absolute values of real numbers). If $x\le 1$, then $|x-1|=1-x$ and $|x-2|=2-x$, so we must solve $1-x=2-x$, which has no solutions. Similarly, if $x\ge 2$, then $|x-1|=x-1$ and $|x-2|=x-2$, so we must solve $x-1=x-2$, which also has no solutions. Finally, if $1\le x\le 2$, then $|x-1|=x-1$ and $|x-2|=2-x$, so we must solve $x-1=2-x$, which has the unique solution $x=\frac{3}{2}$.