Question 1042100
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I'm assuming that the "There are 8 nickels and 8+9=17 dimes." is extraneous information that has nothing whatever to do with the problem you are asking about.


Next, I think you have left some words out of the problem or misstated it in some other way. If the total amount of money is $2.20 and there are three times as many quarters (as there are dimes one would assume from the way you wrote this) then you have an impossible situation.  The number of quarters would have to be a multiple of 3 since the number of dimes must be an integer.  That means that you either have 3 quarters (75 cents) or 6 quarters ($1.50).  You could not have 9 quarters because that would be worth $2.25 which is more than the total of the dimes and quarters.  If you have 3 quarters, then you would have one dime, 85 cents.  If you have 6 quarters, then you have 2 dimes, $1.70.  Neither adds up to $2.20.


What I think you meant to write was 'three times as many DIMES AS quarters".  Then you can say let *[tex \Large x] represent the number of quarters, then *[tex \Large 3x] is the number of dimes.  The value of the dimes is *[tex \Large 10(3x)] cents and the value of the quarters is *[tex \Large 25x] cents.  Since the total value of the money in the collection is 220 cents, you can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 30x\ +\ 25x\ =\ 220]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 55x\ =\ 220]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 4]


So 4 quarters is $1.00 and 3 times 4 equals 12 dimes is $1.20 and $1.00 plus $1.20 is $2.20.   Checks.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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