Question 928118
Here's a similar problem:
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A man has 52 coins, all of which are dimes, quarters, and nickels. The total value of his change is $8.20. If the number of quarters is twice the number of nickels, how many of each does he have?
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Use d, q & n, makes it clearer
d + q + n = 52
10d + 25q + 5n = 820
q = 2n
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Sub for q in the 1st 2 eqns
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d + 3n = 52 times 2 --> 2d + 6n = 104
10d + 50n + 5n = 820
10d + 55n = 820 /5 --> 2d + 11n = 164
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2d + 6n = 104
2d + 11n = 164
------------------- Subtract
-5n = -60
n = 12 nickels
etc