# This linear algebra question is killing me.

• January 12th 2010, 07:45 PM
complex123
This linear algebra question is killing me.
The question says:

A zoo charges $6 for adults,$3 for students, and $.50 for children. One morning 79 people enter and pay a total of$207. Determine the possible numbers of adults, students, and children.

So, I got the equations 6a+3s+0.5c = 207 and a+s+c=79. I then put them into a matrix and got it into REF that looks that

[ 1 0 0.83333 -10]
[ 0 1 1.83333 89]

I feel this should be really easy, but i cannot get a solution with guess and check.
• January 13th 2010, 03:44 AM
HallsofIvy
Quote:

Originally Posted by complex123
The question says:

A zoo charges $6 for adults,$3 for students, and $.50 for children. One morning 79 people enter and pay a total of$207. Determine the possible numbers of adults, students, and children.

So, I got the equations 6a+3s+0.5c = 207 and a+s+c=79. I then put them into a matrix and got it into REF that looks that

[ 1 0 0.83333 -10]
[ 0 1 1.83333 89]

I feel this should be really easy, but i cannot get a solution with guess and check.

You have two equations in three unknowns so, as a purely linear algebra problem, this has no single solutions. But here, of course, a, s, and c must be positive integers and the problem says "solutions", plural.

However, your row-reduce matrix is wrong- you dropped a sign. (Notice that the top row is equivalent to the equation a+ .83333 c= -10. a and c cannot both be non-negative!)

Your row reduced matrix should be $\begin{bmatrix}1 & 0 & -\frac{5}{6} & -10 \\ 0 & 1 & \frac{11}{6} & 89\end{bmatrix}$.

That is equivalent to $a- \frac{5}{6}c= -10$ and $s+ \frac{11}{6}= 89$ or $a= \frac{5}{6}c- 10$ and $s= 89- \frac{11}{6}c$.

Now, since a and s must be non-zero integers, it is clear that c must be a multiple of 6, at least 12, and no larger than 48 ( $89- \frac{11}{6}(48)= 1$). That is, c can be 12, 18, 24, 30, 36, 42, or 48.

If c= 12, a= 10-10= 0 and s= 89- 22= 67.
If c= 18, a= 15- 10= 5 and s= 89- 33= 56, etc.