Question

solve the inequality.
$15(\frac{1}{3}b + 3) \leq 6(b + 9)$
solution:
graph the solution.
(there is a number line with -8, -6, -4, -2 marked and some graphing options above it)

Explanation:

Step1: Expand both sides

$15\times\frac{1}{3}b + 15\times3 \leq 6b + 6\times9$
$5b + 45 \leq 6b + 54$

Step2: Subtract $5b$ from both sides

$5b - 5b + 45 \leq 6b - 5b + 54$
$45 \leq b + 54$

Step3: Subtract 54 from both sides

$45 - 54 \leq b + 54 - 54$
$-9 \leq b$

Answer:

$b \geq -9$

For the graph: Use a closed dot at $-9$ on the number line, with an arrow pointing to the right (representing all values greater than or equal to $-9$; this corresponds to the "•→" option from the given graph tools).