Solve the equation |x| + |x - 1| = |x - 2| + |x - 3|.

Step-by-step solution

1. 1

   Locate the breakpoints and split the line. Each modulus changes formula where its inside hits zero, at $x=0,1,2,3$. These cut the real line into five intervals, and on each one every bar can be removed with a definite sign.

2. 2

   Interpret the equation before computing. The left side is the total distance from $x$ to $0$ and to $1$; the right side is the total distance to $2$ and to $3$. Since the pair $\{2,3\}$ sits entirely to the right of $\{0,1\}$, you should expect the solution somewhere in the middle — and this reading also predicts there will be exactly one.

3. 3

   Handle the three outer intervals. For $x\le 0$: $(-x)+(1-x)=(2-x)+(3-x)$ gives $1-2x=5-2x$, a contradiction. For $2\le x\le 3$: $x+(x-1)=(x-2)+(3-x)$ gives $2x-1=1$, so $x=1$, which is not in $[2,3]$ — rejected. For $x\ge 3$: $2x-1=2x-5$, another contradiction.

4. 4

   Handle the interval 0 <= x <= 1. Here $|x|=x$, $|x-1|=1-x$, $|x-2|=2-x$, $|x-3|=3-x$:

   $x+(1-x)=(2-x)+(3-x)\ \Longrightarrow\ 1=5-2x\ \Longrightarrow\ x=2,$

   which lies outside $[0,1]$ — rejected. Always test the candidate against the interval that produced it; this is where most attempts go wrong.

5. 5

   Handle the interval 1 <= x <= 2. Now $|x|=x$, $|x-1|=x-1$, $|x-2|=2-x$, $|x-3|=3-x$:

   $x+(x-1)=(2-x)+(3-x)\ \Longrightarrow\ 2x-1=5-2x\ \Longrightarrow\ 4x=6\ \Longrightarrow\ x=\tfrac32,$

   and $\tfrac32$ does lie in $[1,2]$, so it is accepted.

6. 6

   Check the solution in the original equation. At $x=\tfrac32$: the left side is $\tfrac32+\tfrac12=2$ and the right side is $\tfrac12+\tfrac32=2$ ✓. So the unique solution is $x=\tfrac32$ — the midpoint of the two pairs, exactly as the distance reading suggested.