Joe has a collection of nickels and dimes that is worth $6.10. If the number of dimes was tripled and the number of nickels was increased by 49, the value of the coins would be $18.95. How many nickels and dimes does he have?

Question

hero · Answer

.05n + .10d = 6.10
.05(n + 49) + .10(3d) = 18.95

Solve for n and d

anonymous · Answer

.05N + .10D = 6.10
 .05(N+49) + .10*3*D = 18.95
 
.05(N+49) + .30D = 18.95
 -3[.05N + .10D = 6.10]
 
.05N + 2.45 + .30D = 18.95
 -.15N - .30D = -18.30
 
-.10N + 2.45 = .65
 2.45 - .65 = .10N
 1.80 = .10N
 N = 1.8/.1 = 18
 0.90 in nickels 
6.1-.9 = 5.20 
52 dimes and 18 nickels

anonymous · Answer

thats what i got