SOLUTION: There were 166 paid admissions to a game. The price was $2.00 for aduls and $0.75 for children. The amount taken in was $293.25. How many adults and how many children attended?
Algebra ->
Systems-of-equations
-> SOLUTION: There were 166 paid admissions to a game. The price was $2.00 for aduls and $0.75 for children. The amount taken in was $293.25. How many adults and how many children attended?
Log On
Question 125089: There were 166 paid admissions to a game. The price was $2.00 for aduls and $0.75 for children. The amount taken in was $293.25. How many adults and how many children attended? Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! A+C=166 OR A=166-C
2A+.75C=293.25 NOW REPLACE A WITH (166-C) & SOLVE FOR C.
2(166-C)+.75C=293.25
332-2C+.75C=293.25
-1.25C=293.25-332
-1.25C=-38.75
C=-38.75/-1.25
C=31 THE NUMBER OF CHILDREN.
166-31=135 THE NUMBER OF ADULTS.
PROOF
2*135+.75*31=293.25
270+23.25=293.25
293.25=293.25
(