A box contains $6.70 in nickels, dimes, and quarters. There are 39 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?

n = nickels

d = dimes

q = quarters

1) n + d + q = 39

n+d = q - 3

Translate to: 2) n + d - q = -3

.05n + .1d + .25q = 6.7 amount of each coin times the number of coins must add to 6.7

Multiply both sides by 20...

3) 1n + 2d + 5q = 134

Combine 1) & 2)

n + d + q = 39

n + d - q = -3

Add together

2n + 2d = 36

Divide by 2:

4) n + d = 18

Combine 1) and 3)

n + d + q = 39 Multiply by -5 => -5n -5d -5q = -195

n + 2d + 5q = 134 n + 2d + 5q = 134

Add together-> 5) -4n - 3d = -61

Combine 4) and 5)

n + d = 18 Multiply by 4 => 4n + 4d = 72

-4n - 3d = 61 -4n - 3d = -61

Add: d = 11

11 dimes...

Plug that into 4)

n + 11 = 18

n = 18-11 = 7 nickels.

Plug that into 1.

n + d + q = 39

7 + 11 + q = 39

18 + q = 39

q = 39-18 = 21 quarters

21 quarters, 7 nickels, and 11 dimes.

21(.25) + 7(.05) + 11(.1) = 5.25 + .35 + 1.1 = 5.6 + 1.1 = 6.7 Check.

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