Answer:
$8.00
Step-by-step explanation:
The problem statement gives two relations between the prices of two kinds of tickets. These can be used to write a system of equations for the ticket prices.
__
<h3>setup</h3>
Let 'a' and 'c' represent the prices of adult and children's tickets, respectively. The given relations can be expressed as ...
a - c = 1.50 . . . . . . . adult tickets are $1.50 more
175a +325c = 3512.5 . . . . . total revenue from ticket sales.
__
<h3>solution</h3>
We are only interested in the price of an adult ticket, so we can eliminate c to give one equation we can solve for 'a'. Using the first equation, an expression for c is ...
c = a -1.50
Substituting that into the second equation, we have ...
175a +325(a -1.50) = 3512.50
500a -487.50 = 3512.50 . . . . . . simplify
500a = 4000 . . . . . . add 487.50
a = 8 . . . . . . . . . divide by 500
An adult ticket costs $8.
Well, you lost the original $100 and then you gained back $70 but gave away $70 worth of merchandise, so it evens out and you lost $100 in total
A + s = 444
s = 2a
a + 2a = 444
3a = 444
a = 444/3
a = 148 <== adults
s = 2a
s = 2(148)
s = 296 ...students
X=5 because you add 2 to 13 and divide by 3
