The combined cost of one advance ticket and one same-day ticket to a show was $50. It is known that 35 tickets were sold in advance and 25 the same day, for total receipts of $1600 . What was the price of each kind of ticket?
A+S=50
35A+25S=1600
I would solve if by substitution.
To find the price of each kind of ticket, let's assume that the price of an advance ticket is x dollars and the price of a same-day ticket is y dollars.
Given that 35 advance tickets and 25 same-day tickets were sold, we can create two equations:
1) x + y = 50 (since the combined cost of one advance ticket and one same-day ticket is $50)
2) 35x + 25y = 1600 (since the total receipts from selling tickets were $1600)
To solve this system of equations, we can use the method of substitution or elimination. Let's use the method of substitution.
From equation 1, we can rearrange it to get x = 50 - y.
Substitute this value of x into equation 2:
35(50 - y) + 25y = 1600
1750 - 35y + 25y = 1600
1750 - 10y = 1600
-10y = 1600 - 1750
-10y = -150
y = (-150)/(-10)
y = 15
Now substitute the value of y back into equation 1 to find the value of x:
x + 15 = 50
x = 50 - 15
x = 35
Therefore, the price of an advance ticket is $35 and the price of a same-day ticket is $15.