Solve the inequality |x + 1| + |x - 1| <= 3.

Step-by-step solution

1. 1

   Read the left side as a total distance. Since $|x+1|=|x-(-1)|$, the expression is the distance from $x$ to $-1$ plus the distance from $x$ to $1$. Any point between $-1$ and $1$ has total distance exactly $2$ (the gap between the two anchors), and moving outside adds twice the overshoot. So the inequality asks: which points are within a total travel budget of $3$?

2. 2

   Split at the breakpoints. The two expressions inside the bars change sign at $x=-1$ and $x=1$, so handle three ranges: $x\le -1$, $-1\le x\le 1$, and $x\ge 1$.

3. 3

   Case $x\le -1$. Both $x+1\le 0$ and $x-1<0$, so both bars flip sign:

   $-(x+1)-(x-1)=-2x\le 3\;\Longrightarrow\;x\ge -\frac32$

   (Dividing by $-2$ reverses the inequality.) Intersecting with $x\le -1$ gives $-\frac32\le x\le -1$.

4. 4

   Case $-1\le x\le 1$. Now $x+1\ge 0$ but $x-1\le 0$, so

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

   The $x$ cancels entirely: the statement is true for every $x$ in this range, so the whole interval $[-1,1]$ belongs to the solution. This is the flat bottom predicted in step 1.

5. 5

   Case $x\ge 1$. Both expressions are non-negative:

   $(x+1)+(x-1)=2x\le 3\;\Longrightarrow\;x\le \frac32$

   Intersecting with $x\ge 1$ gives $1\le x\le \frac32$.

6. 6

   Union the three pieces and test the boundary.

   $\left[-\tfrac32,-1\right]\cup[-1,1]\cup\left[1,\tfrac32\right]=\left[-\tfrac32,\tfrac32\right]$

   Check: at $x=\frac32$ the sum is $2.5+0.5=3$, so the endpoint is included; at $x=1.6$ it is $2.6+0.6=3.2>3$, correctly excluded. Equivalently the answer is just $|x|\le \frac32$.