Advance tickets for a school play went on sale. the price of each student ticket was $4 and everyone else paid $5. on the first day, no more than $80 in tickets were sold. describe and explain the possible values of s, the number of student tickets sold, and e, the number of tickets sold to nonstudents.

20 were sold because 80/4=20

1) Step 1
Esblish the equation and restrictions
4s + 5e ≤ 80
s ≥ 0
e ≥ 0
2) Stept 2
Establish the limits for the values os s and e.
If you draw the line 4s + 5e ≤ 80. it will be easier to visualize the following explanation.
The line 4s + 5e = 80 and the two axis bound the values of s and e. 
Find the vertices
-  minimum s = 0, => maximum e = [80 - 4s] / 5 = [80 - 0] / 5 = 16
- minimum e = 0 => maximum s = [80 - 5e] / 4 = 80 / 4 = 20  
Solution: 
- s, the number of student tickets sold, may be any integer value between 0 and 16, including the limits- e, the number of tickets sold to nonstudents, may be any integer value between 0 and 20, including the l

Answer for 2: his profit would be $2.88 explanation: you have to first multiply 72 and 0.12 together to get $8.64 once you do that you know how much money the shopkeeper made from selling the candy bars individually. so you have to do 8.64-5.76=2.88 hope that ; )answer for 3: yeah i’m sorry i have no idea what number 3 is. i’m sorry i couldn’t with that.