Question 627773
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You aren't going to be able to solve this one using ordinary means.  You have three variables and only two equations.  However, all is not lost.  First realize that the solution must be three non-negative integers.  You can't watch part of a movie (or at least you won't be able to pay only part of the cost of a movie), so there will be no fractional parts of any of the three types of movie.  Also, you can't "unwatch" a movie, so no negative values in the solution either.


Now, notice that the cost of matinees is 4.50, so any odd number (1, 3, 5, 7, 9, or 11) matinees watched will have a cost component that ends in .50.  The other types of movies cost whole dollars, and since the number of those watched is an integer, neither of the other types can contribute the 50 cents needed in the total.  Hence, we have restricted the value of the variable *[tex \LARGE m] to a positive odd number less than or equal to 11.


From here it is trial and error.


Let's assume that he watched exactly 1 matinee, and that would have been 4.50 of the total cost.  Also, the sum of the other two types must then have been 10 movies and the total cost of the other two types had to be 45.50 minus 4.50 is 41.00.


Set up two equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ +\ e\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3r\ +\ 7e\ =\ 41]


-3 times the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3r\ -\ 3e\ =\ -30]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3r\ +\ 7e\ =\ 41]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0r\ +\ 4e\ =\ 11]


and *[tex \LARGE e\ =\ \frac{11}{4}].  Danger, Will Robinson!!!  Not an integer.  Keep looking.


Set up two new equations for the 3 matinee situation.  45.50 minus 3 times 4.50 = 32 and 11 minus 3 is 8, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ +\ e\ =\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3r\ +\ 7e\ =\ 32]


Oooo...even numbers! looks promising.


-3 times the first equation:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -3r\ -\ 3e\ =\ -24]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3r\ +\ 7e\ =\ 32]


So


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0r\ +\ 4e\ =\ 8]


and we have *[tex \LARGE e\ =\ 2] which means *[tex \LARGE r\ =\ 6]


Check.  6 times 3 is 18, 3 times 4.50 is 13.50, and 2 times 7 is 14.  18 plus 13.50 plus 14 is 45.50.  Checks.


I'll leave it to you to check the other 4 possibilities to make sure there aren't multiple answers, but it wouldn't make sense for other answers to exist.  We actually have two lines in 3 space that intersect in one point and we have already found it.


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