SOLUTION: This question is for Applications of Linear Equations. 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

Question 164621: This question is for Applications of Linear Equations.
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?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
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?
:
n = no. of dimes
d = no. of nickels
:
Write an equation for each statement
:
"Joe has a collection of nickels and dimes that is worth $5.65."
.05n + .10d = 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."
.05(n+8) + .10(2d) = 10.45
.05n + .40 + .20d = 10.45
.05n + .20d = 10.45 - .40
.05n + .20d = 10.05
:
Multiply the 1st equation by 2, subtract the above equation from it:
.10n + .20d = 11.30
.05n + .20d = 10.05
---------------------subtraction eliminates d
.05n = 1.25
n =
n = 25 nickels
:
How many dimes does he have?
:
Use the 1st equation to find d, substitute 25 for n
.05(25) + .10d = 5.65
1.25 + .10d = 5.65
.10d = 5.65 - 1.25
.10d = 4.40
d =
d = 44 dimes
;
;
Confirm our solutions, using the 2nd statement: 33 nickels and 88 dimes
.05(33) + .10(88) = 10.45
1.65 + 8.80 = 10.45
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