Question 187585: Please Help me solve this word problem...
Camille has 19 coins consisting of nickels, dimes, and quarters. The value of the coins is $1.60. If she has six times as many nickels as quarters, then how many of each does she have?
I tried:
x= nickles
Y = quarters
so 6x + y = 1.60? I thought there was a way to try and isolate each variable. Like doing y= 1.60 -6x , y= 1.60 - 6(19) I don't think that's right...
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Camille has 19 coins consisting of nickels, dimes, and quarters. The value of the coins is $1.60. If she has six times as many nickels as quarters, then how many of each does she have?
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Equations:
Quantity Equation: n + d + q = 19
Value Equation...:5n + 10d +25q = 160 cents
Quantity Eq......:n = 6q
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Rearrange the 3 equation:
n + d + q = 19
5n + 10d +25q = 160
n + 0 - 6q = 0
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Use any method you know to solve this system of three equations to get:
n = 12
d = 5
q = 2
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Cheers,
Stan H.
Answer by Alan3354(69443)  (Show Source):
You can put this solution on YOUR website! Camille has 19 coins consisting of nickels, dimes, and quarters. The value of the coins is $1.60. If she has six times as many nickels as quarters, then how many of each does she have?
I tried:
x= nickles
Y = quarters
so 6x + y = 1.60? I thought there was a way to try and isolate each variable. Like doing y= 1.60 -6x , y= 1.60 - 6(19) I don't think that's right...
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x and y are ok, I would have used n, d and q
Add d for dimes
x + y + d = 19 (number of coins)
x = 6y (6 times as many nickels)
5x + 10d + 25y = 160 ($1.60 total)
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Sub 6y for x in the 1st and 3rd eqn
6y + y + d = 19
30y + 10d + 25y = 160
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7y + d = 19 --> d = 19-7y
55y+10d = 160
Sub for d in eqn above
55y + 10(19-7y) = 160
55y + 190 - 70y = 160
-15y = -30
y = 2 ******* 2 quarters
x = 12 ***** 12 nickels
d = 5 ******* 5 dimes
You can check it.
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