The real values of $x$ satisfying |$x-1|+|x-4$| = 3 make up:

For 1 ≤ $x$ ≤ 4 the sum is $(x-1)+(4-x)=3$ for every such $x$. For $x$ > 4 it is $2x-5$, which exceeds 3; for $x$ < 1 it is $5-2x$, which also exceeds 3. So the solution is the entire closed interval from 1 to 4 (for example $x=2$ gives 1 + 2 = 3).