Question 74620
What you want to do here is to solve a system of equations by setting up two linear equations, with your x and y variables aligned on the same side.  Let x represent adult tickets, and y represent child tickets.  Then, 
x + y = 23  (means that the number of 'x' tickets plus the number of 'y' tickets total 23).  
Also, 9x+8y=193 (means that the number of x tickets, times the cost of $9 each, plus the number of y tickets, times the cost of $8 each, totalled $193).  

Now, line the two equations up with one above the other, and use one of the methods for solving systems of equations:
  x+y=23
9x+8y=193

Substitution would probably be your best bet here, since the first equation has variables with coefficients of 1.  So, solve for one of the variables.  If x+y=23, then subtract y from both sides to get x=23-y.  Plug this value into the second equation:  9(23-y)+8y=193.  Now, you only have one variable, which can be solved for.  Find the value of y.  First, distribute the 9.  Then, 
     9(23-y)+8y=193
(9*23)-(9*y)+8y=193
      207-9y+8y=193 
          207-y=193
              y=14 
This tells us that 14 'child' tickets were sold.  We can plug this value into the first equation to find that if 14 'child' tickets were sold, and a total of 23 tickets were sold, then 9 'adult' tickets were sold (23-14=9).  

Going back to the question, it asks for the number of children that attended.  We found that 14 'child' tickets were sold.  

Hope this helps!