Question 1121103
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2 times 3.50 times *[tex \Large x] is the amount spent on salmon.  5 times *[tex \Large y] is the amount spent on trout and the total amount spent is 77.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7x\ +\ 5y\ =\ 77]


Which will be more convenient to represent as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5y\ =\ -7x\ +\ 77]


Since the weights in this problem must be positive integers, the smallest possible value for either *[tex \Large x] or *[tex \Large y] would be 1.


Clearly, if *[tex \Large x\ =\ 1], then *[tex \Large y\ =\ 14], and we have one possible answer.


We know that *[tex \Large y\ \not=\ 1] because *[tex \Large -7x\ +\ 77] is not evenly divisible by 5.


Note that for *[tex \Large 5y\ =\ -7x\ +\ 77] to have an integer solution, *[tex \Large y\ =\ 7k] where *[tex \Large k\ \in\ \mathbb{Z}].  But our previous work bounds the possible values for *[tex \Large y] between 1 and 14, so the only two possible values are 7 and 14.


Hence, 12 lbs salmon and 7 lbs trout, or 2 lbs salmon and 14 lbs trout.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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