Question

Why is it incorrect

Answer 1

Keeping in mind that, an absolute value, is in effect a piece-wise function, so is a duplet, so |whatever here|   is really two siblings, the positive and negative siblings  +(whatever here) and -(whatever here), thus

$\bf \cfrac{|3v-1|}{5}\ge 4\implies |3v-1|\ge 20\implies \begin{cases} +(3v-1)\ge 20\\ -(3v-1)\ge 20 \end{cases}\\\\ -------------------------------\\\\ +(3v-1)\ge 20\implies 3v-1\ge 20\implies 3v\ge 21\implies v\ge \cfrac{21}{3} \\\\\\ \boxed{v\ge 7}\\\\ -------------------------------$

$\bf -(3v-1)\ge 20\implies -3v+1\ge 20\implies -3v\ge 19 \\\\\\ \stackrel{\textit{negative division, the sign changes}}{v\le \cfrac{19}{-3}}\implies \boxed{v\le -\cfrac{19}{3}}\\\\ -------------------------------\\\\ 7\le v\le -\cfrac{19}{3}$