Question

I need help with this homework can anyone help me. I need work too, help is appreciated:)

Answer 1

Let the number sold advance be = x

Since those at the door was 206 more than advance, therefore number sold at the door = (x + 206)

Advance cost =$6, x number sold, total =6*x =6x

Door cost = $10, (x + 206) sold, total = 10*(x +206) =10x +2060

Total cost of tickets sold = $6828

Cost of Advance + Cost at Door = 6828

6x + 10x +2060 =6828

16x = 6828 - 2060

16x = 4768

x = 4768 / 16 = 298, recall x is the number sold advance.

At the door = x + 206 = 298 + 206 = 504

Therefore 298 tickets were sold advance and 504 were sold at the door.

Hope this explains it.

Answer 2

Let x = # of tickets sold in advance
Let y = # of tickets sold the day of


Cost of the tickets & total sales: 6x & 10y = 6828

You also know y = x + 206

Take the equation mentioned above y = x + 206 and sub it in anywhere the variable y is in the other equations so you'll have this:
6x + 10(x+206) = 6828

Now solve for x to get x = 298

To finish the problem, you must now find the number of y tickets sold.
Sub your x value that you found back into the equation y = x + 206 and you'll get y = 504.

So, 298 tickets were sold in advance and 504 tickets were sold the day of