Jennifer bought 28 tickets to the museum for students and adult chaperones. she paid a total of $350.80, which included an $8.00 donation to the museum. write a system of equations that can be used to determine the number of student tickets, s, and the number of adult tickets, a, jennifer purchased. algebraically solve the system of equations to determine the number of student tickets jennifer purchased. show or explain your work. the price for students is 14.50 and the adults are 19.50

We know there are total of 28 people:
s+a=28

Total price paid is:
total_student_price + total_adult_price + $8 = $350.80

We know:
student_price = $14.50
adult_price = $19.50
Also:
total_student_price = $14.50*s
total_adult_price = $19.50*a

Now we have:
$14.50*s + $19.50*a + $8 = $350.80

System of equations that we have is:
s+a=28
$14.50*s + $19.50*a + $8 = $350.80

Now we solve first equation and insert it into second one:
s=28-a
14.50*(28-a) + 19.50a + 8 = 350.80
406 - 14.50a + 19.50a + 8 = 350.80
414 + 5a = 350.80
5a = 350.80 - 414
5a = -63.2
a= -12.64

This problem does not have solution for three reasons:
1) we got negative number of persons which is impossible
2) we got decimal number of persons which is impossible
3) total price paid ends in .80 while price per ticket for both types of tickets ends in .50. There is no way to get that final price using whole number of tickets