SOLUTION: Joe has a collection of nickels and dimes that is worth $6.80. If the number of dimes were doubled and the number of nickels increased by 6, the value of the coins would be $11.10
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Question 198645: Joe has a collection of nickels and dimes that is worth $6.80. If the number of dimes were doubled and the number of nickels increased by 6, the value of the coins would be $11.10 How many dimes does he have?
Nickels = X
Dimes = Y
X+Y=6.80
X+6+2Y=$11.10
-(x+y=6.80)
-x-y=-6.80
x+6+y=11.10
6+y=4.30
y=-1.7
This is not the answer, but its what I keep going back to.
Found 2 solutions by scott8148, nerdybill:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
try this ___ X(.05) + Y(.10) = 6.80
and this ___ (X+6)(.05) + (2Y)(.10) = 11.10
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Joe has a collection of nickels and dimes that is worth $6.80. If the number of dimes were doubled and the number of nickels increased by 6, the value of the coins would be $11.10 How many dimes does he have?
.
Let n = nickels
and d = dimes
.
From:"If the number of dimes were doubled and the number of nickels increased by 6, the value of the coins would be $11.10"
.05(n+6) + .10(2d) = 11.10
.05n+ .30 + .20d = 11.10
.05n + .20d = 10.80
.
From: "Joe has a collection of nickels and dimes that is worth $6.80"
.05n + .10d = 6.80
.
Our two equations:
.05n + .20d = 10.80
.05n + .10d = 6.80
.
Multiplying the bottom equation by -1 and adding to the first:
.05n + .20d = 10.80
-.05n - .10d = -6.80
---------------------
.10d = 4.00
d = 40 dimes
.
Plug the above into the following equation and solve for n:
.05n + .10d = 6.80
.05n + .10(40) = 6.80
.05n + 4.00 = 6.80
.05n = 2.80
n = 56 nickels
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