## Precalculus: Mathematics for Calculus, 7th Edition

$\left[ -\dfrac{1}{2},\dfrac{3}{2} \right]$
Using the properties of inequality, the given expression, $8-|2x-1|\ge6 ,$ is equivalent to \begin{array}{l}\require{cancel} -|2x-1|\ge6-8 \\\\ -|2x-1|\ge-2 \\\\ |2x-1|\le2 .\end{array} For any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$. (The symbol $\lt$ may be replaced with $\le$.) Using the concept above, the solutions to the given inequality, $-2\le 2x-1\le2 ,$ is \begin{array}{l}\require{cancel} -2+1\le 2x\le2+1 \\\\ -1\le 2x\le3 \\\\ -\dfrac{1}{2}\le x\le\dfrac{3}{2} .\end{array} In interval notation, the solution set is $\left[ -\dfrac{1}{2},\dfrac{3}{2} \right] .$