Solve the inequality |x - 1| + |x + 1| < 4.

Step-by-step solution

1. 1

   Locate the kink points. Each absolute value changes formula where its inside is zero: at $x=1$ and $x=-1$. These split the line into $x<-1$, $-1\le x<1$, and $x\ge1$, and on each piece both absolute values become plain linear expressions.

2. 2

   Case 1: x < -1 (both insides negative). Then $|x-1|=1-x$ and $|x+1|=-x-1$, whose sum is $-2x$:

   $-2x<4\qquad\Longrightarrow\qquad x>-2$

   Dividing by $-2$ flips the inequality. Intersecting $x>-2$ with the case condition $x<-1$ gives $-2<x<-1$.

3. 3

   Case 2: -1 <= x < 1 (mixed signs). Here $|x-1|=1-x$ and $|x+1|=x+1$, and the $x$ terms cancel:

   $(1-x)+(x+1)=2<4$

   Always true, so the whole interval $[-1,1)$ belongs to the solution. The sum is constant at $2$ between the kinks.

4. 4

   Case 3: x >= 1 (both insides non-negative). Then the sum is $(x-1)+(x+1)=2x$:

   $2x<4\qquad\Longrightarrow\qquad x<2$

   Intersecting with $x\ge1$ gives $1\le x<2$.

5. 5

   Union the three cases.

   $(-2,-1)\cup[-1,1)\cup[1,2)=(-2,2)$

   The three pieces join seamlessly at the kinks into one interval:

   $-2<x<2$

6. 6

   Read the answer as distances, and verify. $|x-1|+|x+1|$ is the total distance from $x$ to $1$ and to $-1$. Its minimum is the gap between them, $2$, attained everywhere in between; outside, it grows by $2$ per unit. So it reaches $4$ exactly one unit beyond each kink — at $x=-2$ and $x=2$. Checking: $|{-3}|+|{-1}|=4$ at $x=-2$ ✓ and $|1|+|3|=4$ at $x=2$ ✓, both excluded by the strict inequality. A scan of $12{,}001$ points on $[-6,6]$ agrees with $(-2,2)$ at every point ✓.