The inequality given is:

$|3v + 12| > 3$

To solve this inequality, we need to break it down into two cases since it involves absolute value:

Case 1: $3v + 12 > 3$

Solve for $v$: $3v + 12 > 3$ Subtract 12 from both sides: $3v > 3 - 12$ $3v > -9$ Now, divide by 3: $v > -3$

Case 2: $-(3v + 12) > 3$

This is equivalent to: $-3v - 12 > 3$ Add 12 to both sides: $-3v > 3 + 12$ $-3v > 15$ Now, divide by -3 (remember to flip the inequality sign): $v < -5$

Combined solution:

The solution is: $v > -3 \quad \text{or} \quad v < -5$

This means that the inequality holds for $v$ in the intervals $(-\infty, -5) \cup (-3, \infty)$.

Graphing the solution:

• Draw a number line.
• There will be open circles at $v = -5$ and $v = -3$ since the inequality is strict (does not include equality).
• Shade the region to the left of $v = -5$ (for $v < -5$) and the region to the right of $v = -3$ (for $v > -3$).

Would you like further details on how to graph this, or do you have any questions?

Here are five related questions you might want to explore:

1. What is the difference between a strict and non-strict inequality in absolute value problems?
2. How do you solve absolute value inequalities with "less than" instead of "greater than"?
3. How do you represent solutions to inequalities on a number line?
4. Can you solve inequalities involving absolute values of quadratic expressions?
5. How does the solution change when the inequality is $\leq$ instead of $>$?

Tip: When solving absolute value inequalities, always remember to consider both the positive and negative cases of the expression inside the absolute value symbol!