A school sells tickets for a show, (a) adult tickets and (c) child tickets.
1) each adult ticket costs $8 and each child ticket costs $5. the total takings are 2300.
Write an equation to represent this information.
2) solve algebraically the simultaneous equations to find the values of a and c.
i think the first one is 8a + 5c = 2300 but i'm not sure about the second one.
please help
The question as it is will allow you to formulate one equation which you did correctly.
It is normal that the question also tells you how many tickets have been sold in all. The total number of tickets sold should range between 289 to 460. This way, you can formulate a second equation of the form
a+c=N where 289≤N≤460
If this information is not given, there are many possible solutions.
forgot to write it, the school sells 370 tickets
So the second equation would be
a + c = 370 ....(2)
and the first equation is as you had it:
8a + 5c = 2300 ....(1)
You can solve it using substitution, namely, using (2),
a=370-c
and substitute a into equation (1) to get
8(370-c)+5c = 2300
Solve for c and subsequently for a.
Don't forget to put back your answers into equations (1) and (2) for a check.
You are correct with the first equation, which represents the total takings from selling adult and child tickets. The equation is:
8a + 5c = 2300
To solve the simultaneous equations algebraically, we'll need another equation that provides us with a relationship between the number of adult and child tickets.
Let's assume the school sold "a" adult tickets and "c" child tickets. Each adult ticket costs $8, so the total revenue from adult tickets would be 8 times the number of adult tickets, which gives us: 8a. Similarly, the total revenue from child tickets would be 5 times the number of child tickets, which gives us: 5c.
Since the total takings are $2300, we can create another equation that represents the total revenue from adult and child tickets:
Total Revenue = Revenue from Adult Tickets + Revenue from Child Tickets
2300 = 8a + 5c
Therefore, the second equation is:
8a + 5c = 2300
Now, we have a system of two equations:
1) 8a + 5c = 2300
2) Second equation to be solved.
To solve this system algebraically, we can use methods like substitution or elimination.