SOLUTION: Stuck on a ticket question: three kinds of tickets are sold for a school student parent dinner: a $1.00 ticket for one adult, a $1.50 ticket for one adult and one child and $2 tic

Let x be the number of $1.50 tickets for 1 adults and one child.

Let y be the number of $2    tickets for 2 adults and one child.

Then the number of $1 tickets for single adult is (62-x-2y).


Then you have these two equations:

   x +  y             = 32     (1)    (counting children)
1.5x + 2y + (62-x-2y) = 69     (2)    (counting money)


Simplify: 

   x +  y = 32,                (3)
0.5x      = 69-62 = 7          (4)


Simplify again:

   x +  y = 32,                (5)
0.5x      =  7.                (6)


It is clear now from (2) that  x = 2*7 = 14.

Then from (1)  y = 32 - x = 32 - 14 = 18.



Answer.  14 $1.50 tickets;  18 $2 tickets;  and the number of $1 tickets is 62-x-2y = 62-14-2*18 = 12.

     I wanted, most of all, to reduce the problem to 2 equations and avoid having 3 equations.

     And I succeeded in it.