Solve the equation |x| + |x − 1| = 1.

Step-by-step solution

1. 1

   Read the left side as a total distance. $|x|$ is the distance from $x$ to $0$ and $|x-1|$ is the distance from $x$ to $1$. The equation asks for the points whose distances to $0$ and to $1$ add up to $1$ — and $1$ is exactly the gap between those two points.

2. 2

   Split into the three cases. The critical points $0$ and $1$ divide the line into three regions.

   Case $x<0$: $|x|=-x$ and $|x-1|=1-x$, so the equation reads $-x+1-x=1$, i.e. $-2x=0$, giving $x=0$ — which contradicts $x<0$. No solution here.

3. 3

   Case 0 ≤ x ≤ 1. Here $|x|=x$ and $|x-1|=1-x$, so

   $x+(1-x)=1$

   This is an identity — true for every $x$ in the interval. So the whole of $[0,1]$ solves the equation.

4. 4

   Case x > 1. Here $|x|=x$ and $|x-1|=x-1$, so $x+x-1=1$, i.e. $2x=2$ and $x=1$ — which contradicts $x>1$. No solution here either.

5. 5

   Collect the solution set.

   $\boxed{0\le x\le 1,\ \text{i.e. }x\in[0,1]}$

   The answer is an interval, not a finite list — a point worth stressing, since absolute-value equations are usually expected to have isolated solutions.

6. 6

   Check the endpoints and an interior point. At $x=0$: $0+1=1$ ✓. At $x=1$: $1+0=1$ ✓. At $x=0.3$: $0.3+0.7=1$ ✓. Outside: at $x=2$, $2+1=3\neq 1$ ✗; at $x=-1$, $1+2=3\neq 1$ ✗. This also matches the triangle inequality $|x|+|x-1|\ge|x-(x-1)|=1$, with equality exactly on $[0,1]$.