Devon purchased tickets to an air show for 9 adults and 2 children. The
total cost was $252. The cost of a childʹs ticket was $6 less than the cost of
an adultʹs ticket. Find the price of an adultʹs ticket and a childʹs ticket.
A=cost of adult ticket
C=cost of child ticket
9A+2C = 252
A=C+6
Use substitution
9(C+6)+2C = 252
9C+2C = 252-54
11C = 198
C=18
A=C+6=24
Check:
9A+2C=9*24+2*18=216+36=252 OK
To find the price of an adult's ticket and a child's ticket, let's break down the problem into equations.
Let's represent the cost of an adult's ticket as "a" and the cost of a child's ticket as "c".
According to the problem, Devon purchased tickets for 9 adults and 2 children, and the total cost was $252. We can write this information as follows:
9a + 2c = 252 ------(Equation 1)
The problem also states that the cost of a child's ticket was $6 less than the cost of an adult's ticket. Mathematically, this can be represented as:
c = a - 6 ------(Equation 2)
Now, we have a system of equations that we can solve to find the values of "a" and "c".
We can substitute Equation 2 into Equation 1 to eliminate "c":
9a + 2(a - 6) = 252
Simplifying the equation:
9a + 2a - 12 = 252
Combining like terms:
11a - 12 = 252
Adding 12 to both sides of the equation:
11a = 264
Dividing both sides of the equation by 11:
a = 24
Now that we have found the value of "a" (the cost of an adult's ticket), we can substitute it back into Equation 2 to find the value of "c" (the cost of a child's ticket):
c = a - 6
c = 24 - 6
c = 18
So, the price of an adult's ticket is $24 and the price of a child's ticket is $18.