## Thomas' Calculus 13th Edition

$1 \displaystyle \lt y\lt \frac{11}{3},\qquad$ or, $y\displaystyle \in(1,\frac{11}{3})$
Use properties from the table 'Absolute Values and Intervals". Applying Property 6: $|x|\lt a\quad\Leftrightarrow\quad -a\lt x\lt a, \quad$ for a = positive number $|3y-7|\lt 4\quad\Leftrightarrow\quad -4\lt 3y-7\lt 4$ Applying rule 1 for inequalities, add 7 to each part of the compound inequality $-4+7 \lt 3y\lt 4+7$ $3 \lt 3y\lt 11$ Apply rule 3: multiply with a positive number, $\displaystyle \frac{1}{3}$ $1 \displaystyle \lt y\lt \frac{11}{3}$ This is an open interval, $(1,\displaystyle \frac{11}{3})$ (See table: "Types of intervals".)