Answer:
170 children
74 students
85 adults
Step-by-step explanation:
Given
Let:
[tex]C = Children; S = Students; A = Adults[/tex]
For the capacity, we have:
[tex]C + S + A = 329[/tex]
For the tickets sold, we have:
[tex]5C + 7S + 12A = 2388[/tex]
Half as many as adults are children implies that:
[tex]A = \frac{C}{2}[/tex]
Required
Solve for A, C and S
The equations to solve are:
[tex]C + S + A = 329[/tex] -- (1)
[tex]5C + 7S + 12A = 2388[/tex] -- (2)
[tex]A = \frac{C}{2}[/tex] -- (3)
Make C the subject in (3)
[tex]C = 2A[/tex]
Substitute [tex]C = 2A[/tex] in (1) and (2)
[tex]C + S + A = 329[/tex] -- (1)
[tex]2A + S + A = 329[/tex]
[tex]3A + S = 329[/tex]
Make S the subject
[tex]S = 329 - 3A[/tex]
[tex]5C + 7S + 12A = 2388[/tex] -- (2)
[tex]5*2A + 7S + 12A = 2388[/tex]
[tex]10A + 7S + 12A = 2388[/tex]
[tex]7S + 22A = 2388[/tex]
Substitute [tex]S = 329 - 3A[/tex]
[tex]7(329 - 3A) + 22A = 2388[/tex]
[tex]2303 - 21A + 22A = 2388[/tex]
[tex]2303 +A = 2388[/tex]
Solve for A
[tex]A = 2388 - 2303[/tex]
[tex]A = 85[/tex]
Recall that: [tex]C = 2A[/tex]
[tex]C = 2 * 85[/tex]
[tex]C = 170[/tex]
Recall that: [tex]S = 329 - 3A[/tex]
[tex]S = 329 - 3 * 85[/tex]
[tex]S = 329 - 255[/tex]
[tex]S = 74[/tex]
Hence, the result is:
[tex]C = 170[/tex]
[tex]S = 74[/tex]
[tex]A = 85[/tex]
