1. ## Movie Theater

At a movie theater, a cashier sold 250 more adult admission tickets than children'admission tickets. The adult's tickets were $6.00 each and the children's tickets were$3.50 each. What is the least number of each type of ticket that the cashier had to sell for the total
receipts to be at least $2,750? 2. Hmmm. It took quite some time to solve this problem but finally i managed to figure it out. Hopefully it is correct let x be the number of adult ticket sold let y be the number of children ticket sold 6x + 3.5y =$2750 ............ Eqn 1
x - y = 250 ...................... Eqn2

We cannot solve the Eqn 1 and Eqn 2 directly as Eqn 1 involve cost and Eqn 2 involve quantity but it will help us solve the equation listed below

6(x-250) + 3.5y = 2750 - 6(250) where 250 are the number of tickets

Since x-250 = y ( derive from Eqn 2 )

6y + 3.5y = 1250
9.5y = 1250
y = 132
Resubstitute y = 132 into Eqn 1 or Eqn2 to get x, which is 382

I hope it helps

3. ## yes but....

Originally Posted by tester85
Hmmm. It took quite some time to solve this problem but finally i managed to figure it out. Hopefully it is correct

let x be the number of adult ticket sold
let y be the number of children ticket sold

6x + 3.5y = \$2750 ............ Eqn 1
x - y = 250 ...................... Eqn2

We cannot solve the Eqn 1 and Eqn 2 directly as Eqn 1 involve cost and Eqn 2 involve quantity but it will help us solve the equation listed below

6(x-250) + 3.5y = 2750 - 6(250) where 250 are the number of tickets

Since x-250 = y ( derive from Eqn 2 )

6y + 3.5y = 1250
9.5y = 1250
y = 132
Resubstitute y = 132 into Eqn 1 or Eqn2 to get x, which is 382

I hope it helps
Tell me, why is equation 2 x - y = 250 and NOT
x + y = 250?

4. It is due to this statement " a cashier sold 250 more adult admission tickets than children'admission tickets "

Since we let x be the number of adult ticket sold
we let y be the number of children ticket sold

x - y = 250