Question

6. A business rents bicycles and in-line skates Bicycle rentals cost $30 per day, and in-line skate rentals cost $20 per day. The business has 25 rentals and makes $650. c. Write a system of linear equations that represents this situation. d. How many bicycle rentals and in-line skate rentals did the business have today?​

Answer 1

• 15 bicycle rentals
• 10 in-line skate rentals

Explanation:

The system of equations will reflect the relations for the total number of rentals, and the total revenue. Let b and s represent the number of bicycle and skate rentals, respectively. We are told the relations are ...

b + s = 25 . . . . . . 25 rentals

30b +20s = 650 . . . . . . money from those rentals

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It is convenient to solve these equations by substitution. The first equation can be solved for s:

s = 25 -b

And that can be substituted into the second equation:

30b +20(25 -b) = 650 . . . . . substitute for s

10b +500 = 650 . . . . . . simplify

10b = 150 . . . . . . . subtract 500

b = 15 . . . . . . . divide by 10; the number of bike rentals

s = 25 -15 = 10 . . . . find the number of skate rentals

The business had 15 bicycle rentals and 10 in-line skate rentals today.

Answer 2

Answer: 15 bicycle rentals and 10 skate rentals

Explanation:

Let s represent the amount of skate rentals

Let b represent the amount of bicycle rentals

s + b = 25 --> Equation to represent amount of rentals per day

b = 25 - s --> Solved for b so it can be substituted

30b+20s=650 --> Equation to represent price per rental per day

Substitute equation 1 into equation 2:

`30b+20s=650`

`30\left(25-s\right)+20s=650`

`750-30s+20s=650`

`750-10s=650`

`750-10s-750=650-750`

`-10s=-100`

`(-10s)/(-10)=(-100)/(-10)`

`s=10`

Substitute s = 10 into equation 1:

`\left(10\right)\:+\:b\:=\:25`

`10+b-10=25-10`

`b=15`

Therefore, the business had 15 bicycle rentals and 10 skate rentals