Question 1204593
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Your question "Which is a viable solution for the system of equations" is very strange....<br>
The "viable solution" to a system of equations is THE solution to that system of equations.<br>
Presumably you were given some examples of systems of equations that might BE USED TO SOLVE the problem, and you were supposed to choose the one which was a valid setup.  But in that case, for us to answer your question, you would have to show us those examples.<br>
Without seeing those examples, the simplest set of equations is<br>
(1) 45x+37y=521  (total cost of x tickets at $45 each and y tickets at $37 each is $521)
(2) y=x+3  (the number of children tickets is 3 more than the number of adult tickets)<br>
As for the solution for the numbers of adult and children tickets, with the two equations in that form it seems substitution would be the most straightforward solution method:<br>
45x+37(x+3)=521<br>
and go from there.<br>
For a quick informal solution (which is good mental exercise), you can do this:<br>
The "extra" 3 children tickets cost 3($37) = $111, so the remaining tickets (equal numbers of adult and children tickets) cost $521-$111=$410.  Since one adult ticket and one children ticket cost $45+$37 = $82, the number of each kind of ticket is $410/$82 = 5.<br>
So she bought 5 adult tickets and 5+3 = 8 children tickets.<br>