## Precalculus (6th Edition) Blitzer

$(-\infty,-3]\cup[4,\infty)$
Step 1. Remove the absolute sign to get $2x-1\geq7$ and $2x-1\leq -7$ Step 2. For $2x-1\geq7$, we have $2x\geq 8$ or $x\geq4$ Step 3. For $2x-1\leq -7$, we have $2x \leq -6$ or $x\leq -3$ Step 4. The solutions are $(-\infty,-3]\cup[4,\infty)$