Here's what I did:
I wrote 2 different equations using NC as number of children and NA as number of adults.
the first thing the problem tells us is that adults + children is 700. So NC + NA = 700.
Next, it tells us that the number of children at $7 per child is (7)NC. And the number of adults times $10 per adult is 10(NA). The total amount of money collected from both 7NC + 10NC is equal to 6,400.
So: 7NC + 10NA= 6,400
(Note: it's helpful to ignore the dollar signs for now.)
Next, set the two equations so that they are like an addition problem. Now, all you need to do is get rid of one Unknown (the 2 unknowns are NC and NA), but to do that, you need to have both of the same unknowns equal and one of them negative so they will cancel each other out. The easiest way to do this is to multiply both sides of the bottom equation by either -7 or -10. (You will get the right answer either way, you will just be solving for either NC or NA each time.)
And now, once you add, you're left with 3 NA= 1,500. divide the 1500 by the 3 and you're left with NA= 500
Finally, you can plug the 500 into your NC+NA=700 equation to get NC+500=700 which adds up to NC= 200.
So your two answers are:
200 Children and 500 adults :)
at 200($7)= $1,400 on children's tickets
at 500($10)= 5,000 on adult tickets
