Question 910137
<pre>
Avoid decimals by thinking only in cents (pennies)
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3. Ted has $6.80 in quarters and dimes.  
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That's 680 pennies!

25q + 10d = 680
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The number of dimes is 3 times the number of quarters. 
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d = 3q
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Which system of equations can be used to find q, the number of quarters,
and d, the number of dimes, that Ted has?
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This system:

{{{system(25q + 10d = 680, d = 3q)}}}

Trouble is, there is no way there could be any solution.
12 quarters and 3 times as many dimes, which is 36 dimes, comes 
to only 660 ($6.60) <-- we're 20 cents short!
And then 13 quarters and 3 times as many dimes or 39 dimes comes 
to 715 ($7.15)
That's way over $6.80.  So we can't have 3 times as many dimes as
quarters and have $6.80.    

The closest you could come is 12 quarters and 38 dimes

[ 12*25 + 38*10 = 300 + 380 = 680 ($6.80)

Your problem is botched.

Maybe it was supposed to be:

The number of quarters is 3 times the number of dimes.

Then you could have 24 quarters and 8 dimes because 
8*3=24 and 24*25+8*10 = 600+80 = 680 ($6.80)

Edwin</pre>