Question

Solve the absolute value inequality $2\times |3x+9|<36$

• A. - 9 > x > 3
• B. - 9 < x < 3
• C. x > - 9
• D. x < 3

Answer 1

The correct option is B - 9 < x < 3

$2\times |3x+9|<36$
Divide both side by 2
$\frac{2|3x+9|}{2}<\frac{36}{2}$

$|3x+9|<18$

Now, as we know that if $|x|<a$, then it means that $–a<x<a$
Therefore, $-18<3x+9<18$
Subtract 9 from each side
$-18-9<3x+9-9<18-9$
$-27<3x<9$
Divide by 3 on each side
$-9<x<3$