if they sold 310 tickets in one day equaling $2022.50 then set up this system of equations:

4.75x + 7.50y = 2022.50

x + y = 310

since you know that 310 tickets were sold (so that’s day tickets [x] plus evening tickets [y]) and you have the prices of the tickets and the amount that sold. so the next thing to do is to solve for either x or y, your choice. let’s say x…

i’d use substitution.

x = 310 – y

(see how you rearrange the 2nd equation to solve for x)

then plug in the x value in the other equation to solve for y.

4.75(310-y) + 7.50y = 2022.50

solve:

1472.5 – 4.75y + 7.50y = 2022.50

1472.5 + 2.75y = 2022.50

2.75y = 547

y = 547/2.75

y = 198.91

so approx. 198 were sold in the evening. so 310 – 198 = 112. so then 112 were sold in the day.

hope that helps!!

4.75x + 7.50y= 2022.5

and

x+y= 310

ok so now it could be kind of confusing. the 2nd equation could be changed to x= 310-y. So you plug the x into the first equation and solve for y. After you found the y value just use the second equation x+y=310 and plug in the y and solve.

x + y = 310

x=day tickets y=evening tickets 310=the total number of tickets

4.75(310-y) + 7.50y = 2022.50

1472.5 – 4.75y + 7.50y = 2022.50

1472.5 + 2.75y = 2022.50

2.75y = 550

y = 550/2.75

y = 200 tickets

x=310-200=110 tickets

http://www.math-videos-online.com/

4.75d + 7.50e = 2,022.50 thats all i got so far srry.