Question

example
akira buys 50 tokens at the carnival. he gives his sister enough tokens for 5 kiddie rides. he wants to use some of his remaining tokens for thrill rides. how many thrill rides could he go on?
you can write and solve an inequality to solve the problem.
t = number of thrill rides
\\(5t + 3(5) \leq 50\\)
\\(5t + 15 \leq 50\\)
\\(5t + 15 - 15 \leq 50 - 15\\)
\\(5t \leq 35\\)
\\(\frac{5t}{5} \leq \frac{35}{5}\\)
\\(t \leq 7\\)
akira could go on 0, 1, 2, 3, 4, 5, 6, or 7 thrill rides.
1 akiras sister from the example is too young to ride alone. akira uses tokens to go on the 5 kiddie rides with her. now how many thrill rides can he go on? explain.
2 a truck holds at most 20,000 lb. there are 18 boxes on the truck. each box weighs 750 lb. how many additional boxes can the truck hold? graph the solution. show your work.

Explanation:

Problem 1:

Step1: Define variables and set up inequality

Let \( t \) be the number of thrill rides. Each kiddie ride (now Akira is on it too) uses 3 tokens, and there are 5 kiddie rides. Each thrill ride uses 5 tokens. Total tokens are 50. So the inequality is \( 5t + 3\times2\times5 \leq 50 \) (since Akira and his sister, so 2 people per kiddie ride, each needing 3 tokens? Wait, no, original example: sister's 5 kiddie rides, each needing 3 tokens. Now Akira is on the 5 kiddie rides with her, so each kiddie ride now uses \( 3\times2 = 6 \) tokens? Wait, no, maybe the token per ride per person. Wait, original example: sister's 5 kiddie rides, each needing 3 tokens (so 53=15 tokens for sister). Now Akira is on the 5 kiddie rides with her, so each kiddie ride needs 3 tokens for Akira and 3 for sister, so per kiddie ride: 6 tokens, 5 rides: 56=30 tokens. Then thrill rides: 5t tokens. Total tokens: 5t + 30 ≤ 50.

Wait, maybe I misread. Original example: sister's 5 kiddie rides, each 3 tokens (so 53=15). Now Akira is on the 5 kiddie rides with her, so each kiddie ride now has Akira and sister, so each kiddie ride uses 3 (sister) + 3 (Akira) = 6 tokens. So 5 kiddie rides: 56=30 tokens. Then thrill rides: 5t. So inequality: 5t + 30 ≤ 50.

Step2: Simplify the inequality

5t + 30 ≤ 50
Subtract 30 from both sides: 5t + 30 - 30 ≤ 50 - 30
5t ≤ 20

Step3: Solve for t

Divide both sides by 5: \( \frac{5t}{5} \leq \frac{20}{5} \)
t ≤ 4

Step1: Define variables and set up inequality

Let \( x \) be the number of additional boxes. Each box weighs 750 lb. Current boxes: 18, so current weight: 18*750 lb. Additional boxes: 750x lb. Total weight ≤ 20000 lb. So inequality: \( 750(18 + x) \leq 20000 \)

Step2: Simplify the inequality

First, calculate 18*750 = 13500. So:
750*18 + 750x ≤ 20000
13500 + 750x ≤ 20000

Subtract 13500 from both sides:
750x ≤ 20000 - 13500
750x ≤ 6500

Step3: Solve for x

Divide both sides by 750: \( x \leq \frac{6500}{750} \approx 8.666... \)

Since \( x \) must be an integer (number of boxes), \( x \leq 8 \).

Answer:

Akira could go on 0, 1, 2, 3, or 4 thrill rides.

Problem 2: