Question

solve the following inequality, then graph the solution set. |3x + 6| - 2 < 10. answer in interval notation.

Explanation:

Step1: Isolate the absolute - value expression

Add 2 to both sides of the inequality $|3x + 6|-2\lt10$.
$|3x + 6|\lt10 + 2$, so $|3x+6|\lt12$.

Step2: Rewrite the absolute - value inequality as a compound inequality

If $|u|\lt a$ ($a\gt0$), then $-a\lt u\lt a$. Here $u = 3x + 6$ and $a = 12$, so $-12\lt3x + 6\lt12$.

Step3: Solve the compound inequality for $x$

Subtract 6 from all parts of the compound inequality: $-12-6\lt3x+6 - 6\lt12 - 6$, which simplifies to $-18\lt3x\lt6$.
Then divide all parts by 3: $\frac{-18}{3}\lt\frac{3x}{3}\lt\frac{6}{3}$, so $-6\lt x\lt2$.

Answer:

$(-6,2)$

To graph the solution set on the number - line:

1. Place an open circle at $x=-6$ (because $x\gt - 6$ and $-6$ is not included in the solution set).
2. Place an open circle at $x = 2$ (because $x\lt2$ and 2 is not included in the solution set).
3. Draw a line segment connecting the two open circles.