A collection of coins has value of 64 cents. There are two more nickels than dimes and three times as many pennies as dimes in this collection. How many of each kind of coin are in the collection?
d = dimes p = pennies = 3d n = nickles = d+2
10(d) + 3d(1) + (d+2)(5) = 64 solve for d
17d = 54
d =3 p=3d=9 n= d+2 = 5
+129849 Let the number of dimes = x
Then the number of nickels = x + 2
And the number of pennies = 3x
And we have
10x + 5(x + 2) + 3x = 64 simplify
10x + 5x + 10 + 3x = 64
18x + 10 = 64 subtract 10 from each side
18x = 54 divide both sides by 18
x = 3 this is the number of dimes
x + 2 = 3 + 2 = 5 = the number of nickels
3x = 3(3) = 9 = the number of pennies
