Question 1201132
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Answer: <font color=red>200 children</font>


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Work Shown:


c = number of children
s = number of students
a = number of adults
These are nonnegative integers  {0,1,2,3,...}


"<font color=blue>A theater has a seating capacity of 750 ... At a certain screening with full attendance</font>"
c+s+a = total number of people
c+s+a = total seating capacity = 750
c+s+a = 750
Let's refer to this as equation (1).


4c = revenue from the children only
6s = revenue from the students only
8a = revenue from the adults only
4c+6s+8a = total revenue = 4600 dollars
4c+6s+8a = 4600
2(2c+3s+4a) = 4600
2c+3s+4a = 4600/2
2c+3s+4a = 2300
Let's refer to this as equation (2).


"<font color=blue>there were half as many adults as children and students combined</font>"
c+s = number of children and students combined
0.5(c+s) = half the previous amount = number of adults
0.5(c+s) = a
Let's refer to this as equation (3).



System of equations:
c+s+a = 750
2c+3s+4a = 2300
0.5(c+s) = a


Let's multiply both sides of the third equation by 2, then get everything to one side.
0.5(c+s) = a
a = 0.5(c+s)
2*a = 2*0.5(c+s)
2a = c+s
0 = c+s-2a
c+s-2a = 0
This is the updated version of equation (3).


Updated system of equations:
c+s+a = 750
2c+3s+4a = 2300
c+s-2a = 0


Narrow your focus to equations (1) and (3)
c+s+a = 750
c+s-2a = 0
Both start with <font color=blue>c+s</font>
Let's subtract straight down to cancel out each copy of <font color=blue>c+s</font>
More specifically...<ul><li>The c variables cancel.</li><li>The s variables cancel.</li><li>For the 'a' variables we have a - (-2a) = a+2a = 3a</li><li>For the right hand sides we have 750-0 = 750</li></ul>So,
3a = 750 
3a/3 = 750/3
a = 250 is the number of adults


Plug that value of 'a' into the first equation of the system. 
Then isolate c.
c+s+a = 750
c+s+250 = 750
c+s = 750-250
c+s = 500
c = 500-s
We'll come back to this later.


Revisit the second equation.
Plug in a = 250 and do a bit of algebra.
2c+3s+4a = 2300
2c+3s+4*250 = 2300
2c+3s+1000 = 2300
2c+3s = 2300-1000
2c+3s = 1300


Then plug in c = 500-s
This will allow us to solve for s.
2c+3s = 1300
2(500-s)+3s = 1300 
1000-2s+3s = 1300
1000+s = 1300
s = 1300-1000
s = 300 is the number of students


We'll use that value of s to determine c.
c = 500-s
c = 500-300
<font color=red>c = 200 is the number of children</font>



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Summary


c = 200, s = 300, a = 250


<font color=red>200 children</font>
300 students
250 adults


Check:
c+s+a = 750
200+300+250 = 750
500+250 = 750
750 = 750 ....... first equation works
and
2c+3s+4a = 2300
2*200+3*300+4*250 = 2300
400+900+1000 = 2300
1300+1000 = 2300
2300 = 2300 ....... second equation works
and
0.5(c+s) = a
0.5(200+300) = 250
0.5(500) = 250
250 = 250  ....... third equation works
All three equations of the original system are true for the c, s and 'a' values mentioned.
Therefore, the answers are confirmed.


Other approaches are possible. Feel free to explore other options.
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