## Intermediate Algebra (6th Edition)

$(\frac{-3}{4},\frac{17}{4})$
$|\frac{4x-7}{5}| \lt 2$ By the property, $|X| \lt a$ is equivalent to $-a \lt X \lt a$, the absolute value expression becomes, $-2 \lt \frac{4x-7}{5} \lt 2$ $-10 \lt 4x-7 \lt 10$ $-10+7 \lt 4x-7+7 \lt 10+7$ $-3 \lt 4x\lt 17$ $\frac{-3}{4} \lt x \lt \frac{17}{4}$ Interval Notation : $(\frac{-3}{4},\frac{17}{4})$