Answer
$1 \displaystyle \lt y\lt \frac{11}{3},\qquad$ or, $y\displaystyle \in(1,\frac{11}{3})$
Work Step by Step
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".)
