Question 1169857
<br>
The given information is not sufficient to find a single answer.<br>
x = # adults
y = # students
z = # children<br>
(1) {{{x+y+z = 14}}}  [number of people in the group is 14]
(2) {{{2.5x+1.5y+z = 28}}}  [cost at Cinema Uno is $28]
(3) {{{4x+2y+z = 42}}}  [cost at Cinema Doz is $42]<br>
Subtract (1) from (2):
(4) {{{1.5x+0.5y = 14}}}
{{{3x+y = 28}}}<br>
Subtract (2) from (3):
{5) {{{1.5x+0.5y = 14}}}
{{{3x+y = 28}}}<br><br>
Equations (4) and (5) are the same; that tells us there will NOT be a single solution to the set of equations.<br>
Since the problem requires solutions in non-negative integers, there might be a single solution; but there might be more than one.<br>
We know that
{{{3x+y = 28}}}<br>
and, since the total number in the group is 14, we know that
{{{x+y <= 14}}}<br>
Comparing those two, we know that
{{{2x >= 14}}}
{{{x >= 7}}}<br>
So the number of adults is at least 7.<br>
Now find all solutions, knowing that
(1) x >= 7
(2) y = 28-3x
(3) x+y+z = 14:<br><pre>
   x  y  z
  ---------
   7  7  0
   8  4  2
   9  1  4</pre>
Larger values of x will result in negative values for y, so these are all the solutions.<br>
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<br>