Question

solve for p and graph the solution. |p + 3| ≥ 3 click two endpoints to graph a line segment, an endpoint and an arrowhead to graph a ray, or two arrowheads to graph a line. to change endpoints from filled - in circles to empty circles, click on them. (there is a number line with marks from - 9 to 3 and a submit button on the left, and learn with an example or watch a video options on the right.)

Explanation:

Step1: Solve the absolute - value inequality

The absolute - value inequality is \(|p + 3|\geq3\). By the definition of absolute - value inequalities, if \(|x|\geq a\) (\(a>0\)), then \(x\leq - a\) or \(x\geq a\).
Here, \(x = p + 3\) and \(a = 3\), so we have two cases:
Case 1: \(p+3\leq - 3\)
Subtract 3 from both sides of the inequality: \(p+3 - 3\leq - 3-3\), which simplifies to \(p\leq - 6\).
Case 2: \(p + 3\geq3\)
Subtract 3 from both sides of the inequality: \(p+3 - 3\geq3 - 3\), which simplifies to \(p\geq0\).

Step2: Graph the solution

• For the inequality \(p\leq - 6\), we draw a ray on the number line starting at \(p=-6\) (with a filled - in circle at \(p = - 6\) because the inequality is non - strict, \(\leq\)) and extending to the left (towards negative infinity).
• For the inequality \(p\geq0\), we draw a ray on the number line starting at \(p = 0\) (with a filled - in circle at \(p=0\) because the inequality is non - strict, \(\geq\)) and extending to the right (towards positive infinity).

Answer:

\( p \leq -6 \) or \( p \geq 0 \)