Question 782898
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Let *[tex \LARGE x] represent the number of pennies, *[tex \LARGE y] represent the number of nickels, and *[tex \LARGE z] represent the number of dimes.


First thing we know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ +\ y\ +\ z\ =\ 15]


Now since we know that at least one of each is used and that pennies are going to have to show up in groups of 5 so that you come out to an even $1, there has to be at least 5 pennies.  But if there were 10 pennies, that would only leave a total of 5 total dimes and nickels -- the most you could have would be 10 + 5 + 40 cents or 55 cents.  Therefore, the most pennies you could have is also 5, i.e. there are exactly 5 pennies and a total of 10 nickels and dimes.


So, knowing that we have a total of 10 nickels and dimes and that these 10 nickels and dimes have to add up to 95 cents (one dollar minus the 5 pennies),  we can now write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ +\ z\ =\ 10]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5y\ +\ 10z\ =\ 95]


Solve the 2X2 system for *[tex \LARGE y] and *[tex \LARGE z]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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