Question

solve for h and graph the solution.
2h + 87 ≤ 99 or \frac{-8h - 55}{5} ≥ 5
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 the first inequality \(2h + 87\leq99\)

Subtract 87 from both sides:
\(2h+87 - 87\leq99 - 87\)
\(2h\leq12\)
Divide both sides by 2:
\(h\leq6\)

Step2: Solve the second inequality \(\frac{-8h - 55}{5}\geq5\)

Multiply both sides by 5:
\(-8h-55\geq25\)
Add 55 to both sides:
\(-8h-55 + 55\geq25 + 55\)
\(-8h\geq80\)
Divide both sides by - 8 and reverse the inequality sign:
\(h\leq - 10\)

Answer:

The solution of the compound inequality \(2h + 87\leq99\) or \(\frac{-8h - 55}{5}\geq5\) is \(h\leq6\). On the number - line, we have a closed circle at \(h = 6\) (since the inequality is \(\leq\)) and the line extends to the left towards negative infinity. For the second inequality \(h\leq - 10\), it is already included in the solution \(h\leq6\) (because \(-10<6\)). So we just need to represent \(h\leq6\) on the number - line.