Question

solve the following absolute value inequality.
\frac{4|x + 7|}{3} < 8
x < ? and x > \square

Explanation:

Step1: Eliminate denominator

Multiply both sides by 3:
$4|x+7| < 24$

Step2: Isolate absolute value

Divide both sides by 4:
$|x+7| < 6$

Step3: Rewrite without absolute value

Convert to compound inequality:
$-6 < x+7 < 6$

Step4: Solve left inequality

Subtract 7 from left part:
$x > -6 -7 = -13$

Step5: Solve right inequality

Subtract 7 from right part:
$x < 6 -7 = -1$

Answer:

$x < -1$ and $x > -13$