Question
At the skating rink, David paid $
18.5
18.5 and skated for 3 hours while Jennifer skated for 2 hours and paid $
14.0
14.0. Both of the prices include the price of renting skates. Click Here for Help Video
The cost per hour is $
.
The initial skate rental cost is is $
.
18.5
18.5 and skated for 3 hours while Jennifer skated for 2 hours and paid $
14.0
14.0. Both of the prices include the price of renting skates. Click Here for Help Video
The cost per hour is $
.
The initial skate rental cost is is $
.
Answers
GPT-4o mini
Let's denote the skate rental cost as \( r \) and the cost per hour as \( c \).
From the problem, we can set up the following equations based on the information given:
For David:
\[
r + 3c = 18.5
\]
For Jennifer:
\[
r + 2c = 14.0
\]
Now, we can solve these two equations simultaneously.
First, let's subtract the second equation from the first:
\[
(r + 3c) - (r + 2c) = 18.5 - 14.0
\]
This simplifies to:
\[
c = 4.5
\]
Now that we have the cost per hour \( c = 4.5 \), we can substitute \( c \) back into one of the original equations to find the rental cost \( r \). Using Jennifer's equation:
\[
r + 2(4.5) = 14.0
\]
This simplifies to:
\[
r + 9 = 14.0
\]
\[
r = 14.0 - 9 = 5.0
\]
Thus, the answers are:
- The cost per hour is \( \boxed{4.5} \).
- The initial skate rental cost is \( \boxed{5.0} \).
From the problem, we can set up the following equations based on the information given:
For David:
\[
r + 3c = 18.5
\]
For Jennifer:
\[
r + 2c = 14.0
\]
Now, we can solve these two equations simultaneously.
First, let's subtract the second equation from the first:
\[
(r + 3c) - (r + 2c) = 18.5 - 14.0
\]
This simplifies to:
\[
c = 4.5
\]
Now that we have the cost per hour \( c = 4.5 \), we can substitute \( c \) back into one of the original equations to find the rental cost \( r \). Using Jennifer's equation:
\[
r + 2(4.5) = 14.0
\]
This simplifies to:
\[
r + 9 = 14.0
\]
\[
r = 14.0 - 9 = 5.0
\]
Thus, the answers are:
- The cost per hour is \( \boxed{4.5} \).
- The initial skate rental cost is \( \boxed{5.0} \).