Question

solve for b and graph the solution.
2|3b + 300| + 2 ≥ 602
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.
number line with -1,000, -800, -600, -400, -200, 0, 200, 400, 600, 800, 1,000, 1,200, 1,400

Explanation:

Step1: Subtract 2 from both sides

Subtract 2 from each side of the inequality \(2|3b + 300| + 2 \geq 602\) to isolate the absolute - value term.
\(2|3b + 300|+2 - 2\geq602 - 2\)
\(2|3b + 300|\geq600\)

Step2: Divide both sides by 2

Divide each side of the inequality \(2|3b + 300|\geq600\) by 2 to get the absolute - value expression alone.
\(\frac{2|3b + 300|}{2}\geq\frac{600}{2}\)
\(|3b + 300|\geq300\)

Step3: Solve the compound inequality

The absolute - value inequality \(|x|\geq a\) (where \(a>0\)) is equivalent to the compound inequality \(x\geq a\) or \(x\leq - a\). So for \(|3b + 300|\geq300\), we have two cases:

Case 1: \(3b+300\geq300\)

Subtract 300 from both sides:
\(3b + 300-300\geq300 - 300\)
\(3b\geq0\)
Divide both sides by 3:
\(\frac{3b}{3}\geq\frac{0}{3}\)
\(b\geq0\)

Case 2: \(3b + 300\leq - 300\)

Subtract 300 from both sides:
\(3b+300 - 300\leq - 300-300\)
\(3b\leq - 600\)
Divide both sides by 3:
\(\frac{3b}{3}\leq\frac{-600}{3}\)
\(b\leq - 200\)

Answer:

The solution to the inequality is \(b\leq - 200\) or \(b\geq0\). To graph this, we draw a number line. For \(b\leq - 200\), we have a ray starting at \(-200\) (with a filled - in circle) and pointing to the left. For \(b\geq0\), we have a ray starting at \(0\) (with a filled - in circle) and pointing to the right.