Solve the inequality |2x - 1| <= 3.

Step-by-step solution

1. 1

   Translate $|A|\le b$ into a single chain. The absolute value measures distance from $0$, so requiring it to be at most $3$ confines $2x-1$ to the closed band between $-3$ and $3$:

   $-3\le2x-1\le3$

   Because the relation is at most (not at least), the result is one bounded interval rather than two rays — the direction of the inequality decides the shape of the answer.

2. 2

   Add 1 to all three parts. Adding a constant everywhere leaves both directions intact:

   $-3+1\le2x\le3+1\qquad\Longrightarrow\qquad-2\le2x\le4$

3. 3

   Divide all three parts by 2. The divisor is positive, so no sign flips:

   $-1\le x\le2$

4. 4

   Write the answer as a closed interval.

   $[-1,2]$

   Both endpoints are included because the original allowed equality — the square brackets record exactly that.

5. 5

   Verify the endpoints and the outside. At $x=-1$: $|2(-1)-1|=|-3|=3\le3$ ✓. At $x=2$: $|4-1|=3\le3$ ✓. At the centre $x=\tfrac12$: $|0|=0$ ✓. Just outside, $x=3$ gives $|5|=5$ ✗. Comparing the raw inequality with $[-1,2]$ at $90$ exact rational points agrees everywhere ✓. Note the interval is centred on $x=\tfrac12$, the point where $2x-1=0$, with radius $\tfrac32$.