In order to solve this, you have to set up a systems of linear equations.
Let's say that children = c and adults = a
30a + 12c = 19,080
a + c = 960
I'm going to show you how to solve this system of linear equations by substitution, the easiest way to solve in my opinion.
a + c = 960
- c - c
---------------------- ⇒ Step 1: Solve for either a or c in either equation.
a = 960 - c
20(960 - c)+ 12c = 19,080
19,200 - 20c + 12c = 19,080
19,200 - 8c = 19,080
- 19,200 - 19,200
---------------------------------- ⇒ Step 2: Substitute in the value you got for a or c
8c = -120 into the opposite equation.
------ ---------
8 8
c = -15
30a + 12(-15) = 19,080
30a - 180 = 19,080
+ 180 + 180
-------------------------------
30a = 19,260
------- -----------
30 30
a = 642
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I just realized that there can't be a negative amount of children, so I'm sorry if these results are all wrong.
