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Formulate a system of equations for the situation below and solve.
A theater has a seating capacity of 900 and charges $4 for children, $6 for students, and $8 for adults.
At a certain screening with full attendance, there were half as many adults as children and students combined.
The receipts totaled $5200. How many children attended the show?
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Since "there were half as many adults as children and students combined", you can momentarily conclude that there were 300 adults
(1/3 of the full capacity) and 600 children and students combined (2/3 of the full capacity).
Let C be the number of children.
Then the number of students was (600-C).
The "value" equation for the receipts totaled is THIS:
4C + 6*(600-C) + 8*300 = 5200 dollars.
*** One single equation for one single unknown. ***
4C + 3600 - 6C + 2400 = 5200
-2C = 5200 - 2400 - 3600 = -800 ====> C = 400.
Answer. 400 children.
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Notice. This problem is for one single equation, actually.
Therefore, my major wish was do not make it more complicated than it really should be.