b. Alice’s school is selling tickets to the school carnival.
On the first day of ticket sales, the school sold 3 adult tickets and 9 student tickets for a total of $75.
The second day, the school sold $67 by selling 8 adult and 5 student tickets.
What is the price each of one adult and one student ticket?
so, what do you know?
3a+9s = 75
8a+5s = 67
Now just solve for a and s, then the answer is a+s
To find the price of one adult and one student ticket, we can set up a system of equations based on the given information.
Let's assign variables to unknowns:
- Let 'a' be the price of one adult ticket.
- Let 's' be the price of one student ticket.
From the first day's sales:
3 adult tickets were sold, so the revenue from adult tickets is 3a.
9 student tickets were sold, so the revenue from student tickets is 9s.
Therefore, the total revenue from the first day's sales is 3a + 9s, which equals $75.
From the second day's sales:
8 adult tickets were sold, so the revenue from adult tickets is 8a.
5 student tickets were sold, so the revenue from student tickets is 5s.
Therefore, the total revenue from the second day's sales is 8a + 5s, which equals $67.
We have now set up a system of equations:
3a + 9s = 75 ----(1)
8a + 5s = 67 ----(2)
We can solve this system of equations to find the values of 'a' and 's'. There are several methods we can use, such as substitution or elimination.
Let's solve it using the elimination method:
Multiply equation (1) by 8 and equation (2) by 3 to eliminate variable 'a':
24a + 72s = 600 ----(3)
24a + 15s = 201 ----(4)
Subtract equation (4) from equation (3):
(24a + 72s) - (24a + 15s) = 600 - 201
Simplifying, we get:
57s = 399
Divide both sides by 57:
s = 399 / 57
s ≈ 7
Now we can substitute the value of 's' into one of the original equations. Let's use equation (1):
3a + 9(7) = 75
3a + 63 = 75
3a = 75 - 63
3a = 12
Divide both sides by 3:
a = 12 / 3
a = 4
Therefore, the price of one adult ticket (a) is $4 and the price of one student ticket (s) is $7.