To determine how many times Tom can play basketball while keeping his total spending at or below $200, we can set up the following equation.
- Membership Cost: $125 (one-time fee)
- Court Rental Fee for Members: $3 per use
- Total Budget: $200
Step 1: Calculate the remaining budget after the membership fee.
Remaining budget = Total budget - Membership cost
Remaining budget = $200 - $125 = $75
Step 2: Determine how many times Tom can rent the basketball court.
Let \( x \) be the number of times Tom plays basketball.
The cost for renting the court \( x \) times = \( 3x \).
We need to satisfy the following inequality:
\[ 3x \leq 75 \]
Step 3: Solve for \( x \).
Dividing both sides of the inequality by 3:
\[ x \leq \frac{75}{3} \]
\[ x \leq 25 \]
So, Tom can play basketball a maximum of 25 times while keeping his expenses at or below $200.
Summary
- Membership cost: $125
- Remaining budget for court rental: $75
- Cost per rental: $3
- Maximum number of rentals (times Tom can play basketball): 25
Number Line Illustrating Tom's Budget
You can visualize a number line to represent Tom's spending:
|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|----|
0 25 50 75 100 125 150 175 200
|----------------|
$200 (Total Budget)
|----------|
$125 (Membership Cost)
- The left part of the line represents the total money spent, reaching from $0 to $200.
- The segment from $125 to $200 indicates how much of his budget is left after paying for the membership.
- The maximum number of rentals \( x \) possible (up to $75) shows he can play basketball \( x \) times without exceeding the budget.
Thus, Tom can play basketball up to 25 times for the total cost to remain within his budget of $200.