Question 71898
Let the no. of adults and children who attended be 'x' and 'y' respectively.


Cost of each adult ticket = $20.
Cost of all adult tickets when 'x' adults were present = $(20x).


Cost of each children ticket = $8.
Cost of all children tickets when 'y' children were present = $(8y).


Hence the total cost of children's and adults' tickets = $(20x + 8y).
This value is given as $468.


So, 20x + 8y = 468 or 5x + 2y = 117. This is one equation.


Again, total no. of adults and children = (x + y).
This value is given to be 33.


So, x + y = 33. This is the other equation.


Now we have to solve 2 equations.
5x + 2y = 117 __________(1)
x + y = 33 ___________(2)


Multiplying equation (2) by 2 and then subtracting from equation (1)
{{{5x + 2y - 2(x + y)= 117 - 2*33}}}
{{{5x + 2y - 2x - 2y = 117 - 66}}}
{{{3x = 51}}}
{{{x = 51/3 = 17}}}


So, y 
= 33 - x [Recall eqn. (1) which says x + y =33]
= 33 - 17
= 16


So there were x = 17 adults and y = 16 children.