The information provided in this question is incomplete. I'm adding some more data to make it solvable and help you with the procedure and results of your own problem
Matteo's schools selling tickets to a Christmas Musical. On the first day of ticket sales the schools sold 30 adult tickets and 90 children's tickets for a total of 330$. The school made 670 on the second day by selling 80 adult tickets and 135 children's tickets, What is the price of one adult ticket and one children's ticket?
Answer:
Each adult ticket cost $5 each and each children's ticket cost $2
Step-by-step explanation:
System of Linear Equations
When conditions are provided in the form
[tex]\left\{\begin{matrix}ax+by=c\\ dx+ey=f\end{matrix}\right.[/tex]
Where x and y are the unknown variables and a,b,c,d,e,f are known constants , we have a two-variable, two-equations system
Let x=price of one adult's ticket, y=price of one children's ticket
We know that when 30 adult and 90 children attended the Christmas Musical, $330 were made by the school, we also know that when 80 adult and 135 children attended the Christmas Musical, $670 were made by the school. We form this system of equations
[tex]\left\{\begin{matrix}30x+90y=330\\ 80x+135y=670\end{matrix}\right.[/tex]
To solve the system, we'll use the method of reduction. We first multiply the first equation by -8 and the second by 3
[tex]\left\{\begin{matrix}-240x-720y=-2640\\ 240x+405y=2010\end{matrix}\right.[/tex]
Adding both equations term by term we get
[tex]-720y+405y=-630[/tex]
[tex]-720y+405y=-630[/tex]
[tex]-315y=-630[/tex]
[tex]y=2[/tex]
Replacing this value in the first equation we have
[tex]30x+90(2)=330[/tex]
[tex]30x=330-180[/tex]
[tex]30x=150[/tex]
x=5
Answer: Each adult ticket cost $5 each and each children's ticket cost $2
