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

Step-by-step solution

1. 1

   Locate the breakpoints. Each absolute value changes formula where its inside is zero: $x=0$ for $|x|$ and $x=1$ for $|x-1|$. These two points cut the real line into $x<0$, $0\le x<1$ and $x\ge 1$. Inside each piece both absolute values can be replaced by plain linear expressions, which is the whole point of the split.

2. 2

   Case 1: $x<0$. Both insides are negative, so $|x|=-x$ and $|x-1|=1-x$:

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

   Now test the case condition: $-1<0$ ✓, so $x=-1$ is a genuine solution.

3. 3

   Case 2: $0\le x<1$. Here $|x|=x$ but $|x-1|=1-x$:

   $x+(1-x)=3\ \Longrightarrow\ 1=3$

   The $x$ terms cancel and the statement is false, so this interval contributes nothing. Geometrically, $|x|+|x-1|$ is the total distance from $x$ to $0$ and to $1$, which equals exactly $1$ everywhere between them — it can never reach $3$.

4. 4

   Case 3: $x\ge 1$. Both insides are non-negative, so $|x|=x$ and $|x-1|=x-1$:

   $x+(x-1)=3\ \Longrightarrow\ 2x=4\ \Longrightarrow\ x=2$

   Check the condition: $2\ge 1$ ✓, so $x=2$ is valid.

5. 5

   Always test candidates against their own case. Discarding a value that solves the equation of one case but lies in another interval is the standard error here. Both surviving values pass, and substituting back confirms them: $|-1|+|-2|=1+2=3$ ✓ and $|2|+|1|=2+1=3$ ✓.

6. 6

   State the solution set. $x=-1$ or $x=2$. The distance interpretation explains why there are exactly two: you need total distance $3$ from the pair $\{0,1\}$, and that happens at one point on each side of the segment.