Asked by sup
                at a museum, adult admission is $15. and children admissions is $6.50. during one day, a total of 2300 adult and children admissions were collected, and $27,700 was collected in fees. how many adult admissions were collected? 
            
            
        Answers
                    Answered by
            Leo
            
    The algebra for this would be:
15a + 6.5c = 27700
where a is the number of adults and c is the number of children.
    
15a + 6.5c = 27700
where a is the number of adults and c is the number of children.
                    Answered by
            Farohw
            
    To expand on the previous response, 
you have to solve through a system of equations:
a + c = 2300
15a + 6.5c = 27700
In the first equation, solve for C:
c = 2300 - a
Then substitute into the second equation:
15a + 6.5(2300 - a) = 27700
15a + 14950 - 6.5a = 27700
15a - 6.5a = 27700 - 14950
8.5a = 12750
8.5a/8.5 = 12750/8.5
a = 1500
Therefore, 1500 adult admissions were collected.
You can also substitute a = 1500 into the first equation:
c = 2300 - 1500
c = 800 children admissions sold.
Verify,
a + c = 2300
1500 + 800 = 2300 total admissions.
Best,
Farohw
    
you have to solve through a system of equations:
a + c = 2300
15a + 6.5c = 27700
In the first equation, solve for C:
c = 2300 - a
Then substitute into the second equation:
15a + 6.5(2300 - a) = 27700
15a + 14950 - 6.5a = 27700
15a - 6.5a = 27700 - 14950
8.5a = 12750
8.5a/8.5 = 12750/8.5
a = 1500
Therefore, 1500 adult admissions were collected.
You can also substitute a = 1500 into the first equation:
c = 2300 - 1500
c = 800 children admissions sold.
Verify,
a + c = 2300
1500 + 800 = 2300 total admissions.
Best,
Farohw
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