Question 514480
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Like the man said, let *[tex \Large x] represent the number of children, let *[tex \Large y] represent the number of students, and let *[tex \Large z] represent the number of adults.


The first thing I did was represent the number of adults in terms of the combined number of children and students.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ \frac{x\ +\ y}{2}]


But since we know that there were 600 in attendance, we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ +\ z\ =\ 600]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ +\ \frac{x\ +\ y}{2}\ =\ 600]


A little algebra music:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 3y\ =\ 1200]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ =\ 400]


and therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ z\ =\ 200]


Furthermore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 400\ -\ x]


Now we write a value equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 5y\ +\ 7z\ =\ 3200]


And then substitute what we know about *[tex \Large y] and *[tex \Large z]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x\ +\ 5(400\ -\ x)\ +\ 7(200)\ =\ 3200]


All that is left is to solve for *[tex \Large x], the number of children.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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