Question 387709: Joe has a collection of nickels and dimes that is worth $2.10. If the number of dimes was tripled and the number of nickels was increased by 24, the value of the coins would be $6.30. How many nickels and dimes does he have?
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! Let N= number of Nickels, each worth 5 cents
Let D= number of Dimes, each worth 10 cents
5N + 10D = 210
5(N+24)+ 10(3D) = 630
5N + 10D = 210
5N+120 + 30D=630
In the second equation, subtract 120 from each side:
5N + 10D = 210
5N + 30D = 510
Multiply both sides of first equation by -1 to eliminate the N variables:
-5N - 10D = -210
5N + 30D = 510
Add the two equations together:
20D = 300
Divide both sides by 20
D=300/20=15 Dimes
5N + 10D = 210
5N + 10*15= 210
5N + 150 = 210
5N = 60
N= 12 Nickels
Check:
5(N+24) + 10(3D) = 630
5(36) + 10(45) = 630
180 + 450 = 630
IT CHECKS!!
Dr. Robert J. Rapalje, Retired
|
|
|
