Question

solve for r and graph the solution.
2r + 1 < 5 or 5r - 12 ≥ 3
plot the endpoints. select an endpoint to change it from closed to open. select the middle of a segment, ray, or line to delete it.

Explanation:

Step1: Solve \(2r + 1<5\)

Subtract \(1\) from both sides:
\(2r+1 - 1<5 - 1\)
\(2r<4\)
Divide both sides by \(2\):
\(r < 2\)

Step2: Solve \(5r-12\geq3\)

Add \(12\) to both sides:
\(5r-12 + 12\geq3 + 12\)
\(5r\geq15\)
Divide both sides by \(5\):
\(r\geq3\)

Answer:

The solution of the compound inequality \(2r + 1<5\) or \(5r - 12\geq3\) is \(r < 2\) or \(r\geq3\). On the number - line, for \(r < 2\), we have an open circle at \(r = 2\) and a ray pointing to the left. For \(r\geq3\), we have a closed circle at \(r = 3\) and a ray pointing to the right.