Question 975494
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Let *[tex \Large x] represent the number of 10 cent coins and *[tex \Large y] represent the number of 20 cent coins.  The total number of coins is 25, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  x\ +\ y\ =\ 25]


The 10 cent coins, being worth 10 cents each, must have a value of *[tex \Large 10x] cents if there are *[tex \Large x] of them.  Likewise, the value of *[tex \Large y] 20 cent coins is *[tex \Large 20y] cents.  The sum of these two values is 350 cents.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  10x\ +\ 20y\ =\ 350]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  0.10x\ +\ 0.20y\ =\ 3.50]


Would work just as well, but decimals are always messier than whole numbers, and neatness helps avoid errors.


So now all you need to do is solve the 2X2 linear system for *[tex \Large x] and *[tex \Large y].  This is a fairly straight-forward 2X2 system and either the substitution or the elimination method should work well enough.


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \