Given data:
$39 for unlimited rides
$12 for admission plus $1 per ride
Let x be the number of rides
Let y be the cost in terms of the number of rides
First option is always $39 (it doesn't depend of the number of rides)
![y=39](https://img.qammunity.org/2023/formulas/mathematics/college/v9ty19j3uzmbuum5pbant69yj5asbts6ka.png)
Second option is $12 plus $1 multiplied by the number of rides:
![y=12+1x](https://img.qammunity.org/2023/formulas/mathematics/college/3jr99xxe1au8yfkrmutzf7o54ye6trdvgd.png)
System of equations:
![\begin{gathered} y=39 \\ y=12+x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/om69jtkhlqnx7lteigf1xkqxaehf4ahmjd.png)
The two options wind up costing the same amount when the equations are equal:
y=y
Subtitute the value of y corresponding in each equation:
![39=12+x](https://img.qammunity.org/2023/formulas/mathematics/college/4npk2v0edk5375sb7jthg5z6eblao0tnp0.png)
Solve x:
![\begin{gathered} 39-12=12-12+x \\ 27=x \\ \\ x=27 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/zykt0s3b3wyldlafgvv0vkbptyhxyp5rj3.png)
Then, the two options wind up costing the same amount at $39 and it correspond to 27 rides