SOLUTION: Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $1

Question 34459: Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
Found 2 solutions by mukhopadhyay, stanbon:
Answer by mukhopadhyay(490) (Show Source):
You can put this solution on YOUR website!
Nickels = x
Dimes = y
.05x+ .10y = 5.65 (based on total collection)
=> 5x + 10y = 565.....(1)
(.10)(2y)+(x+8)(.05) = 10.45 (Based on adjusted collection)
=> .2y+.05x+.40 = 10.45
=> .2y+.05x = 10.05
=> 5x+20y = 1005.......(2)
(2) minus (1) gives,
10y = 440
=> y = 44;
From (1) 5x = 565-10y = 565-440 = 125
=> x = 125/5 = 25;
Answer: Number of dimes = 44;

Answer by stanbon(48545) (Show Source):
You can put this solution on YOUR website!
Now Data:
Let number of dimes be "x" and number of nickels be "y".
Value of dimes is 10x cents
Value of nickels is 5y cents
EQUATION: 10x+5y=565
Later Data:
Number of dimes is "2x" and number of nickels is "y+8".
Value of dimes is 10(2x)=20x cents
Value of nickels is 5(y+8)=5y+40 cents
EQUATION: 20x+5y+40=1045
System of equations is:
#1 10x+5y=565
#2 20x+5y=1005
Subtract #1 from #2 to get 10x=440
Then x=44 (original number of dimes)
Cheers,
Stan H.