Question 41648: I am having trouble setting up this problem:
*John found a box of coins. The box contained nickels,dimes, and quarters. He added up the change and there were 85 coins totaling $6.25 in the box. If there were 3 times as many nickels as dimes, how many quarters were there?
-I came up with the equation 3x+y=85 (x=nickels, y=dimes),but this equation alone doesn't help so I think that I may need to use substitution, but I can't come up with another equation.
Please Help! Thank YOu
Answer by 303795(602)   (Show Source):
You can put this solution on YOUR website! There are 3 nickels for each dime so (each dime + its associated 3 nickels) is worth 25 cents.
Use x for the number of dimes and y for the number of quarters
Using the value of the coins form equation 1
.25x +.25y=6.25
Multiply this equation by 4 to remove the decimals
x + y = 25
Using the number of coins form equation 2
4x+y=85
Subtract the first equation from the second to get
3x=60
so x = 20
so y (the number of quarters must be 5)
(There were 60 nickels, 20 dimes and 5 quarters (85 coins) with a value of $3.00+$2.00+$1.25=$6.25)
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