Avoid decimals by thinking only in cents (pennies)
3. Ted has $6.80 in quarters and dimes.
That's 680 pennies!
25q + 10d = 680
The number of dimes is 3 times the number of quarters.
d = 3q
Which system of equations can be used to find q, the number of quarters,
and d, the number of dimes, that Ted has?
This system:
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