Question 517902
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In order to do this problem you have to make an assumption that you really have no right to make, and that is you must assume that the problem actually has a solution.  Not every problem has a solution, and just because your math teacher gave you this one doesn't necessarily alter that fact.


Be that as it may, assuming that this is a solvable problem leads you to the conclusion that there were exactly three types of animals in this pet photo contest and no more.  If there were more, then you couldn't solve the problem.


Let *[tex \Large x] represent the total number of photos in the contest.  Then *[tex \Large \frac{x}{2}] are of cats, *[tex \Large \frac{x}{3}] are of dogs, and 7 are of guinea pigs.  So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 7\ +\ \frac{x}{2}\ +\ \frac{x}{3}\ =\ x]


Solve for *[tex \Large x].  Then calculate *[tex \Large \frac{x}{2}] and *[tex \Large \frac{x}{3}]


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