The given information is not sufficient to find a single answer.
x = # adults
y = # students
z = # children
(1) [number of people in the group is 14]
(2) [cost at Cinema Uno is $28]
(3) [cost at Cinema Doz is $42]
Subtract (1) from (2):
(4)
Subtract (2) from (3):
{5)
Equations (4) and (5) are the same; that tells us there will NOT be a single solution to the set of equations.
Since the problem requires solutions in non-negative integers, there might be a single solution; but there might be more than one.
We know that
and, since the total number in the group is 14, we know that
Comparing those two, we know that
So the number of adults is at least 7.
Now find all solutions, knowing that
(1) x >= 7
(2) y = 28-3x
(3) x+y+z = 14:
x y z
---------
7 7 0
8 4 2
9 1 4
Larger values of x will result in negative values for y, so these are all the solutions.
ANSWERS:
(1) 7 adults and 7 students; OR
(2) 8 adults, 4 students, and 2 children; OR
(3) 9 adults, 1 student, and 4 children