Question 567635
Let D = the number of dimes, N = the number of nickels, and P = the number of pennies.
From the problem description, you can write:
1) D+N+P = 8 "Victor has 8 coins."
2) D = N-2 "He has 2 fewer dimes than nickels."
3) P = N+1 "...and one more penny than nickels."
Replace the D and the P in equation 1) with their equivalents from equations 2) and 3).
(N-2)+N+(N+1) = 8 Simplify this.
3N-1 = 8 Add 1 to both sides.
3N = 9 Now divide both sides by 3.
N = 3 and...
D = N-2
D = 3-2
D = 1 and...
P = N+1
P = 3+1
P = 4
So victor has 1 dime ($0.10), 3 nickels ($0.15), and 4 pennies ($0.04).
Add them together to get:
$0.29 or 29 cents.