Question 802284
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You are getting yourself into trouble right off the bat.  Your claim is that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ +\ d\ +\ q\ =\ 5.00]


So your claim is that the number of nickels plus the number of dimes plus the number of quarters is the same as the value of the money he has.  Not even close.  The number of nickels plus the number of dimes plus the number of quarters is the NUMBER of coins he has, in this case, 30.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ n\ +\ d\ +\ q\ =\ 30]


Now you need a VALUE equation.  The VALUE of the nickels plus the VALUE of the dimes plus the VALUE of the quarters is the same as the VALUE of the money he has.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.05n\ +\ 0.10d\ +\ 0.25q\ =\ 5.00]


Then you are also given that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ q\ =\ 3n]


which you can rearrange thusly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3n\ -\ q\ =\ 0]


Now, you can use these three equations and set up a 3X4 augmented matrix, or you can substitute *[tex \Large 3n] for *[tex \Large q] in the first two equations, simplify, and then solve the much simpler 2X2 system.


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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